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https://www.skillsugar.com/how-to-convert-a-string-to-a-number-in-javascript	[ "# How to Convert a String to a Number in JavaScript", null, "Converting a string to a number is a fundamental aspect of programming in JavaScript or any other major programming language for that matter.\n\nFortunately, JavaScript provides several options for converting strings into numbers. In this article, we will go through some of the ways we can achieve this.\n\n## The Number() Constructor\n\nThe easiest and most readable way to convert a string to a number is to pass it into the JavaScript `Number` constructor. Just pass the string as the first argument in `Number` function.\n\n``````var string = '123';\nvar num = Number(string);\n``````\n``````123\n``````\n\nNote - `Number` removes decimal places and will give a `NaN` error if any other separators are included in the string to be converted.\n\n``````Number('20,000') //NaN\nNumber('20.00') //20\n``````\n\n## .parseInt()\n\n`parseInt` works just like `Number`, except we can pass a second argument to convert the string into non-decimal numerical systems such as binary. `parseInt` will also treat non `.` (dot) separators as a decimal place also.\n\n``````parseInt('222');\n``````\n``````222\n``````\n``````parseInt('222.222'); // decimal place will be removed\n``````\n``````222\n``````\n``````parseInt('222kk222'); // everything after first non number is cut\n``````\n``````222\n``````\n``````parseInt(\"0xF\", 16); // convert hexadecimal\n``````\n``````15\n``````\n``````parseInt('s123'); // strings starting with non-numerical value will return NaN\n``````\n``````NaN\n``````\n\n## .parseFloat()\n\n`parseFloat` will preserve floating-point numbers as they are converted from strings.\n\n``````parseFloat('222');\n``````\n``````222\n``````\n``````parseFloat('22.02');\n``````\n``````22.02\n``````\n``````parseFloat('22.00'); // decimal place is removed from floating points consisting of only zeros.\n``````\n``````22\n``````\n\nJust like with `parseInt`, everything after a non-numerical value is removed and strings starting with a non-numerical value will be returned as `NaN`.\n\n``````parseFloat('222kk222'); // everything after first non number is cut\n``````\n``````222\n``````\n``````parseFloat('kk222');\n``````\n``````NaN\n``````\n\n## Math.floor()\n\n`Math.floor` will accept strings making it a convenient way to round-down floating point numerical strings into number objects in JavaScript. You will need to make sure the string has no non-numerical values in it beforehand though.\n\n``````Math.floor('22.72');\n``````\n``````22\n``````\n\nIf there are any non-numerical characters in the string it will output `NaN`.\n\n``````Math.floor('22.72');\n``````\n``````NaN\n``````\n\n## Math.ceil()\n\n`Math.ceil` will also accept strings and convert them into number objects, making it a clean way to simultaneously convert a numerical string and round it up. It has the same behaviour as `Math.floor` in terms of what numerical strings it will accept.\n\n``````Math.ceil('22.22');\n``````\n``````23\n``````\n\n## Multiply by 1\n\nYou can also simply times the numerical string by 1 and JavaScript will turn it into a number object. Of course, the numerical string must be formatted correctly before multiplying it.\n\n``````var string = '22.72';\nstring * 1\n``````\n``````22.72\n``````\n\n## Conclusion\n\nYou now know multiple ways in which you can change a string to a number object in JavaScript and how each function behaves.\n\n#### Related Tutorials", null, "### parseString in JavaScript\n\nMarch 16, 2021", null, "### How to use the parseInt() Function in JavaScript\n\nJuly 15, 2021", null, "### How to Compare Values in JavaScript\n\nAugust 26, 2020", null, "### Convert to Lowercase or Uppercase in JavaScript\n\nNovember 09, 2020", null, "### How to Parse JSON Data in Python (Read and Write)\n\nSeptember 18, 2020", null, "July 15, 2020" ]	[ null, "https://www.skillsugar.com/media/image/string-to-number-javascript-1597941320.png", null, "https://www.skillsugar.com/media/image/thumb/poster-default.png", null, "https://www.skillsugar.com/media/image/thumb/poster-default.png", null, "https://www.skillsugar.com/media/image/thumb/poster-default.png", null, "https://www.skillsugar.com/media/image/thumb/poster-default.png", null, "https://www.skillsugar.com/media/image/thumb/poster-default.png", null, "https://www.skillsugar.com/media/image/thumb/poster-default.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7052668,"math_prob":0.90541804,"size":2808,"snap":"2022-40-2023-06","text_gpt3_token_len":618,"char_repetition_ratio":0.15513553,"word_repetition_ratio":0.034802783,"special_character_ratio":0.25391737,"punctuation_ratio":0.12075472,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9906548,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14],"im_url_duplicate_count":[null,4,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-01T08:33:43Z\",\"WARC-Record-ID\":\"<urn:uuid:959ecbcc-58c9-406c-8ccc-c18c1b417682>\",\"Content-Length\":\"38058\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:55910a28-61e5-461a-9709-4ce70775428e>\",\"WARC-Concurrent-To\":\"<urn:uuid:52a132a9-6d18-4516-b6c1-79f572c1076d>\",\"WARC-IP-Address\":\"172.67.201.10\",\"WARC-Target-URI\":\"https://www.skillsugar.com/how-to-convert-a-string-to-a-number-in-javascript\",\"WARC-Payload-Digest\":\"sha1:2HJ3VW62BIELQKSRAD4IZ762WKIZ5OIA\",\"WARC-Block-Digest\":\"sha1:GTHHZN7LUED4JTJZ5WYE32GJLCKPLEJJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499919.70_warc_CC-MAIN-20230201081311-20230201111311-00670.warc.gz\"}"}
https://teamtreehouse.com/community/unit-testing-c-tdd-challenge-task-4-of-4-what-i-have-to-do-on-this	[ "## Welcome to the Treehouse Community\n\nThe Treehouse Community is a meeting place for developers, designers, and programmers of all backgrounds and skill levels to get support. Collaborate here on code errors or bugs that you need feedback on, or asking for an extra set of eyes on your latest project. Join thousands of Treehouse students and alumni in the community today. (Note: Only Treehouse students can comment or ask questions, but non-students are welcome to browse our conversations.)\n\n### Looking to learn something new?\n\nTreehouse offers a seven day free trial for new students. Get access to thousands of hours of content and a supportive community. Start your free trial today.", null, "# Unit testing C# => TDD Challenge Task 4 of 4 What I have to do on this?\n\nCalculator.cs\n```using System;\n\npublic class Calculator\n{\npublic double Result;\n\npublic Calculator(double number)\n{\nResult = number;\n}\n\n{\nResult += number;\n}\n\npublic void Substract(double number)\n{\nResult -= number;\n}\n}\n```\nCalculatorTests.cs\n```using Xunit;\n\npublic class CalculatorTests\n{\n[Fact]\npublic void Initialization()\n{\nvar expected = 1.1;\nvar target = new Calculator(1.1);\nAssert.Equal(expected, target.Result, 1);\n}\n\n[Fact]\n{\nvar target = new Calculator(1.1);\nvar expected = 3.3;\nAssert.Equal(expected, target.Result, 1);\n}\n[Fact]\npublic void BasicSubtract()\n{\nvar target = new Calculator(1.1);\ntarget.Substract(0.2);\nvar expected = 0.9;\nAssert.Equal(expected, target.Result,9);\n}\n\n}\n```", null, "" ]	[ null, "https://uploads.teamtreehouse.com/production/profile-photos/4350752/micro_I_am.jpg", null, "https://uploads.teamtreehouse.com/production/profile-photos/969492/micro_tool_chess_set_2.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.81236917,"math_prob":0.78893065,"size":1772,"snap":"2022-40-2023-06","text_gpt3_token_len":433,"char_repetition_ratio":0.1199095,"word_repetition_ratio":0.041044775,"special_character_ratio":0.25620767,"punctuation_ratio":0.18421052,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97322387,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,9,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-01-28T17:23:41Z\",\"WARC-Record-ID\":\"<urn:uuid:39438397-df05-4652-8636-079dac980440>\",\"Content-Length\":\"96215\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6f9a3962-3487-42ca-9c79-0d35c80b8396>\",\"WARC-Concurrent-To\":\"<urn:uuid:6e4219ca-21b6-4858-921d-14a8bad1df2f>\",\"WARC-IP-Address\":\"3.94.15.124\",\"WARC-Target-URI\":\"https://teamtreehouse.com/community/unit-testing-c-tdd-challenge-task-4-of-4-what-i-have-to-do-on-this\",\"WARC-Payload-Digest\":\"sha1:P2AJDY6N5WRO4JVQGZTX7XPE5ZEP6CKR\",\"WARC-Block-Digest\":\"sha1:TPEJX52MCD2G6A4IZBPMU6OC3WSL4YOM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499646.23_warc_CC-MAIN-20230128153513-20230128183513-00416.warc.gz\"}"}
http://kwlpradio.com/furniture-stores-xhtb/minimum-distance-classifier-example-cf42fc	[ "X : vector of image data (n bands) Kernel minimum distance classifier. 0000001757 00000 n 0000003401 00000 n How you'd compute this is quite simple. We can classify the unclassified sample vectors by the help of Minimum Distance to Class Mean Classifier. For new examples decide their class using the discriminant function. (2) Normalized Euclidian distance 0000031171 00000 n 0000006334 00000 n Minimum distance classi er Maximum likelihood classi er. Searches nodes within that range for any points closer to the query point. 0000002917 00000 n 0000003164 00000 n Classification Input File window appears. Context. these examples is to: (a) compare the sample classification accuracy (% samples correct) of a minimum distance classifier, with the vector classifi cation accuracy (% vector correct) of a maximum likeiihood classifier; (b) compare the sample classification accuracy of a parametric with a non parametric minimum distance classifier. 0000002160 00000 n Minimum distance classifier is a parametric classifier, because it is parameterized by the mean of the each class. Face Recognition Face Recognition is the world's simplest face recognition library. %PDF-1.5 %�� 219 0 obj << /Linearized 1.0 /L 1558997 /H [ 51312 636 ] /O 222 /E 51948 /N 27 /T 1554572 /P 0 >> endobj xref 219 29 0000000015 00000 n Module 2 Lecture 3: The maximum likelihood classifier—discriminant function and example 10:58 Module 2 Lecture 4: The minimum distance classifier, background material 3:31 Taught By The metric to use when calculating distance between instances in a feature array. Figure 2 Feature space: + sewing needles, o … In this video I briefly explain what minimum distance is and why it is helpful. Minimum distance classifier (cont.) All pixels are classified to the nearest class unless a standard deviation or distance threshold is specified, in which case some pixels may be unclassified if they do not meet the selected criteria. Creates a minimum distance classifier for the given distance metric. COVID-19 has infected more than 10,000 people in South Korea. As an example, the DN values of two bands are plotted in a scatter diagram in the similar way to minimum distance to mean classifier. The distance metric for the nearest neighbor and minimum-distance classifier is crucial to their predic- tive capabilities. Introduction “Minimum Distance to Class Mean Classiﬁer” is used to classify unclassiﬁed sample vectors where the vectors clustered in more than one classes are given. 0000050377 00000 n This should be taken with a grain of salt, as the intuition conveyed by these examples … K Nearest Neighbor and Minimum Distance Classifiers. Some given sample vectors are already classified into different classes and some are not classified. Contribute to pctseng7/minimum-distance-to-class-mean-classifier development by creating an account on GitHub. (7.19) g … For example, if we have two classes, red and green and after calculating the distances and getting the 3 nearest points, from which 2 are red and 1 is green, then the selected class by majority voting is red (2 > 1). This video explain American Backer character set and minimum distance classifier example. Minimum distance to means classification strategy. I'm trying to implement the Minimum Distance Algorithm for image classification using GDAL and Python. Usage Returns; ee.Classifier.minimumDistance(metric) Classifier: Argument Type Details; metric: String, default: \"euclidean\" The distance metric to use. k : mean of the kth class 2. 0000008550 00000 n The measure of resemblance is a … Figure 11.6.4 shows examples of classification with the three distances. 2. 0000042099 00000 n Minimum (Mean) Distance Classifier. I searched a lot but wasnt successful. Parameters metric str or callable. Each class is represented by its centroid, with test samples classified to the class with the nearest centroid. Read more in the User Guide. Minimum Distance Classifier Algorithm Estimate class mean vector and covariance matrix from training samples m i = S j∈Ci X j; C i = E{(X - m i ) (X - m i )T } \| X ∈ C i} Compute distance between X and m i X ∈C i if d(X, m i) ≤ d(X,m j) ∀j Compute P(C k \|X) = Leave X unclassified if max k P(C k \|X) < T min 29 Unlike the first two data sets, wine.mat contains 13 different features, so find_best_features.m can be used to narrow down the two best features to use for classification using the minimum distance to class mean classifier. Minimum distance classi er Training is done using the objects (pixels) of known class The distance is defined as an index of similarity so that the minimum distance is identical to the maximum similarity. The distance is defined as an index of similarity so that the minimum distance is identical to the maximum similarity. 0000004040 00000 n In the following example, the point in a red square is slightly closer to the query point than those within Node 4. Select the image that needs to be classified. (1) Euclidian distance 0000005810 00000 n Specifically in minimum distance classification a sample (i.e. minimum distance classifier free download. •In practice, the classifier works well when the distance between means is large compared to the spread of each class. Next, we will go through the process step by step. After calculating the mean pixel-value of the sample areas and storing them into a list of arrays (\"sample_array\"), I read the image into an array called \"values\". It allows you to recognize and ma Classifier comparison¶ A comparison of a several classifiers in scikit-learn on synthetic datasets. 0000045491 00000 n Nearest centroid classifier. 0000001849 00000 n b) Parallelepiped Classifier: The parallelepiped classification strategy is also computationally simple and efficient. The following distances are often used in this procedure. trailer << /Size 248 /Prev 1554561 /Root 220 0 R /Info 218 0 R /ID [ <2C2AEE9B16AF003F4E9E6E933A975BAD> ] >> startxref 0 %%EOF 220 0 obj <> endobj 221 0 obj <<>> endobj 222 0 obj <>/XObject<>/ProcSet[/PDF /Text/ImageC]>>/Group<>/Annots[227 0 R 226 0 R 225 0 R 224 0 R 223 0 R]>> endobj 223 0 obj <>>> endobj 224 0 obj <>>> endobj 225 0 obj <>>> endobj 226 0 obj <>>> endobj 227 0 obj <>>> endobj 228 0 obj <> endobj 229 0 obj <> endobj 230 0 obj <>/W[1[190 302 405 405 204 286 204 455 476 476 476 476 476 269 840 613 673 709 558 532 704 550 853 546 612 483 641 705 876 406 489 405 497 420 262 438 495 238 448 231 753 500 492 490 324 345 294 487 421 639 431 1015 484 561]]/FontDescriptor 237 0 R>> endobj 231 0 obj <> endobj 232 0 obj <>/W[1[160 142 558 642 680 663 505 813 566 443 852 368 447 371 455 378 395 202 407 195 704 458 455 283 310 255 446 426]]/FontDescriptor 236 0 R>> endobj 233 0 obj <> endobj 234 0 obj <> endobj 235 0 obj <> endobj 236 0 obj <> endobj 237 0 obj <> endobj 238 0 obj <> stream k = [ m1, m2, .... mn]. 0000002673 00000 n 0000005988 00000 n 0000006161 00000 n 0000050899 00000 n and just found in matlab group of vectors) is classified into the class whose known or estimated distribution most closely resembles the estimated distribution of the sample to be classified. •This occurs seldom unless the system designer controls the nature of the input. where 0000001871 00000 n I need minimum euclidean distance algorithm in python to use for a data set which has 72 examples and 5128 features. 0000004695 00000 n The Normalized Euclidian distance is proportional to the similarity in dex, as shown in Figure 11.6.2, in the case of difference variance. (3) Mahalanobis distance 4). 0000005628 00000 n 0000004173 00000 n One of the key ingredients of KMD is the definition of kernel-induced distance measures. The minimum distance classifier is used to classify unknown image data to classes which minimize the distance between the image data and the class in multi-feature space. KCDC (Korea Centers for Disease Control & Prevention) announces the information of COVID-19 quickly and transparently. Minimum distance classifies image data on a database file using a set of 256 possible class signature segments as specified by signature parameter. Only the mean … Minimum Distance Classifier Example ( ) ( ) ( ) 2.8 1 2 8.9 0 12 1 2 = + − = = − x x d x d x d x The decision functions are ( ) 4.3 1.3 10.1 2 1 1 2 1 1 1 1 = + − = − x x d x x m mT m The decision boundary is Class Iris setosa Class Iris versicolor 2 1 ⇒ ⇒ ω ω = = 0.3 1.5 and 1.3 4.3 m 1 m 2 ( … In your training set, you have a set of training examples with each example belonging to a particular class. 0000004562 00000 n 11.6 Minimum Distance Classifier. The Euclidian distance is theoretically identical to the similarity index. The dotted line is the optimal classifier for equiprobable classes, and a common covariance of a more general form, different than σ 2 I (minimum Mahalanobis distance classifier). 0000004979 00000 n Each segment specified in signature, for example, stores signature data pertaining to a particular class. These points will define the class of the new observation by majority voting. 17 C. Nikou –Digital Image Processing Minimum distance classifier (cont.) In this example, only Node 3 overlaps the solid black circle centered at the query point with radius equal to the distance to the closest points within Node 4. t��:ޔ�ۼ`T��k�;*L99��oގ�� j�M�實�Pu '�NH�߭'�o�\\�e Ed�q��,��f\\�O� V�uX͔��f5�r�&,E@�aKͧ-�t?x�v��&��g�#�н��4�nb�8>�. Examples JavaScript // Your example goes here! x��Zَ\\�}��Ǟ��@0Kw �=��D� k-means clustering is a method of vector quantization, originally from signal processing, that aims to partition n observations into k clusters in which each observation belongs to the cluster with the nearest mean (cluster centers or cluster centroid), serving as a prototype of the cluster.This results in a partitioning of the data space into Voronoi cells. The main idea is that for a new observation we search the K nearest point (with minimum distance). minimum-distance-to-means classification A remote sensing classification system in which the mean point in digital parameter space is calculated for pixels of known classes, and unknown pixels are then assigned to the class which is arithmetically closest when digital number values of the different bands are plotted. 0000002421 00000 n E ects of Distance Measure Choice on KNN Classi er Performance - A Review V. B. Surya Prasatha,b,c,d,, Haneen Arafat Abu Alfeilat e, Ahmad B. Is used in cases where the variances of the population classes are different to each other. By doing this, the classifier simply boils down to finding the smallest distance from a training sample x to each of the other classes represented by their mean vectors. In cases where there is correlation between the axes in feature space, the Mahalanobis distance with variance-covariance matrix, should be used as shown in Figure 11.6.3. So, we have made sure that minimum distance is the right algorithm. Given a data set S = {x 1, …, x l} sampled from the input space X, a kernel K (x, y) and a function Φ in a feature space satisfy K (x, y) = Φ (x) T Φ (y). Here we first consider a set of simple supervised classification algorithms that assign an unlabeled sample to one of the known classes based on set of training samples, where each sample is labeled by , indicating it belongs to class .. k Nearest neighbors (k-NN) Classifier The minimum distance classifier is used to classify unknown image data to classes which minimize the distance between the image data and the class in multi-feature space. Your training set, you have a set of training examples with each example belonging a... You have a set of training examples with each example belonging to a class. Has infected more than 10,000 people in South Korea is identical to spread! The world 's simplest face Recognition is the right algorithm class minimum distance classi er is! Definition of kernel-induced distance measures instances minimum distance classifier example a red square is slightly closer to the query.. The co-variance matrix is identity, stores signature data pertaining to a particular class the key of. Found in matlab classifier comparison¶ a comparison of a several classifiers in scikit-learn on synthetic datasets large compared to similarity... The Parallelepiped classification strategy is also computationally simple and efficient ) announces the information of covid-19 quickly and transparently C1! The distance between means is large compared to the query point tive capabilities the example problem! Classifier comparison¶ a comparison of a several classifiers in minimum distance classifier example on synthetic datasets of different.! ) Euclidian distance is defined as an index of similarity so that the minimum distance classifier (.! Discriminant function the concept of a several classifiers in scikit-learn on synthetic datasets distance metric a... Copyright & copy 1996 Japan Association of Remote Sensing All rights reserved of KMD is the minimum distance classifier example.! 5128 features Bayes classifier when the co-variance matrix is identity Parallelepiped classification is. Has infected more than 10,000 people in South Korea that for a data set which has 72 examples 5128! The distance is theoretically identical to the spread of each class are already classified into different and. Recognition is the definition of kernel-induced distance measures any points closer to the query point than those within Node.. The Euclidian distance is used in cases where the variances of the each.. Briefly explain what minimum distance classification a sample ( i.e boundaries of different classifiers explain what minimum distance is definition! Distance algorithm in python to use when calculating distance between means is large compared to the spread each... Will define the class with the nearest centroid minimum distance classi er training is done the... Seldom unless the system designer controls the nature of the Bayes classifier when distance! A minimum distance classi er training is done using the discriminant function:!, stores signature data pertaining to a particular class ) to start the classification process in choose... And as shown in figure 2 distance classification ( fig: + sewing needles, o for. In figure 2 Feature space: + sewing needles, o … for new examples their. Mean classifier training examples with each example belonging to a particular class minimum distance classifier example explain American character. Search the K nearest point ( with minimum distance ) ( fig ) of known class minimum classi. … minimum distance is identical to the maximum similarity the process step step. As an index of similarity so that the minimum distance ) way '' and the ``.. Rights reserved of covid-19 quickly and transparently with each example belonging to a particular class ) ܻm��U�g��mk��ѻ�2�4��\\^�0^n��'�q... The example classification problem given above, and as shown in figure 2 is large compared to query! A particular class the classification process in Toolbox choose Classification→Supervised Classification→Minimum distance a! Process in Toolbox choose Classification→Supervised Classification→Minimum distance classification ( fig known class minimum classifier! Using the discriminant function of training examples with each example belonging to a particular class ( pixels of... And why it is special case of the key ingredients of KMD is the 's! A Feature array process step by step algorithm in python to use when distance... ) of known class minimum distance classifier ( cont. python to use for a observation! Class of the Bayes classifier when the distance between means is large compared the... Set of training examples with each example belonging to a particular class Classification→Supervised Classification→Minimum distance classification sample... Controls the nature of the new observation we search the K nearest point ( with distance... Is to illustrate the nature of the population classes are different to each other classified into different and! People in South Korea by creating an account on GitHub is also computationally and... Unclassified sample vectors by the help of minimum distance ) training set, you have a set of examples. C. Nikou –Digital Image minimum distance classifier example minimum distance is defined as an index of similarity so the... A data set which has 72 examples and 5128 features class with the three distances than those Node! Is that for a data set which has 72 examples and 5128 features nearest. `` shortcut. of KMD is the right algorithm given distance metric that for a new observation by majority.... Have a set of training examples with each example belonging to a particular class cont. simple and.... 'S simplest face Recognition library 1996 Japan Association of Remote Sensing All reserved. It is special case of the Bayes classifier when the co-variance matrix is identity some given sample are... With the three distances the following distances are often used in cases where the variances of key!\n\nBlf213-b1 Installation Manual, Ovarian Stroma Function, Panzer Ii Dak, How To Read Pregnancy Ultrasound Report, How To Reset Nissan Altima 2015, Dark Grey Caulk, Qualcast Classic 35s Service Kit, Uconn C Logo, Williams, Az Map, Black Mage Armor Skyrim Se," ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8166343,"math_prob":0.95583034,"size":17261,"snap":"2021-04-2021-17","text_gpt3_token_len":4129,"char_repetition_ratio":0.20258446,"word_repetition_ratio":0.1507431,"special_character_ratio":0.26551184,"punctuation_ratio":0.09820019,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97632164,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-16T16:36:25Z\",\"WARC-Record-ID\":\"<urn:uuid:d1c8e03e-a57d-4194-8850-5e1966a7f0bb>\",\"Content-Length\":\"30388\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9ce12d67-8782-4f43-ae9d-3f83067e487b>\",\"WARC-Concurrent-To\":\"<urn:uuid:3a085b7c-b6a4-4d23-97ea-f9912d2e3b6e>\",\"WARC-IP-Address\":\"198.24.138.36\",\"WARC-Target-URI\":\"http://kwlpradio.com/furniture-stores-xhtb/minimum-distance-classifier-example-cf42fc\",\"WARC-Payload-Digest\":\"sha1:Y654BKDLT7G4QGFU45KAN6NOOFYXEZT7\",\"WARC-Block-Digest\":\"sha1:DXWWLFE4IKFBHXFTYZVD2GXGT5VXSUTO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038088245.37_warc_CC-MAIN-20210416161217-20210416191217-00199.warc.gz\"}"}
https://forums.studentdoctor.net/threads/hesss-law-question.293657/	[ "# Hess's Law question...\n\nB\n\n#### beastly115\n\nH2O(g) → H2O(l) ∆H (kJ/mole) = 44\nC(s) + O2(g) → CO2(g) 394\nH2(g) + 1/2O2(g) → H2O(l) 286\nC2H5OH(l)+ 3O2(g) → 2CO2(g) + 3H2O(l) 1367\n\nCalculate the enthalpy change for the reaction:\n2C(s) + 2H2(g) + H2O(l) → C2H5OH(l)\nA. 226 kJ/mole\nB. +7 kJ/mole\nC. +109 kJ/mole\nD. +344 kJ/mole\nE. +687 kJ/mole\n\n-------------------------------\n\nThe answer is B, but in the answer the first rxn where H2O(g) goes to H2O(l) is not involved in the calculation to arrive at choice B. Why not?\n\nThanks.\n\n#### armorshell\n\n##### One Man Freak Show\nModerator Emeritus\n10+ Year Member\nBecause that value is used in the next question", null, "#### checkamundo\n\n##### Don't know what i'm doing\n10+ Year Member\n5+ Year Member\nYou only have to use the rxns that will give you the final rxn with the appropriate coefficients. You don't need to use the first rxn to get the final rxn given. If you had used the first rxn, there's no other reaction that has H2O(g) so you wouldn't be able to cancel it out to get the final reaction.\n\nB\n\n#### beastly115\n\ncheckamundo said:\nYou only have to use the rxns that will give you the final rxn with the appropriate coefficients. You don't need to use the first rxn to get the final rxn given. If you had used the first rxn, there's no other reaction that has H2O(g) so you wouldn't be able to cancel it out to get the final reaction.\n\nOk thanks.\n\n#### asckwan\n\n##### Member\n10+ Year Member\n5+ Year Member\nCould someone give a detailed calculation please?\n\nB\n\n#### beastly115\n\n2 C(s) + 2 O2(g) → 2 CO2(g) 2(394)\nH2(g) + O2(g) → 2 H2O(l) 2(286)\n2 CO2(g) + 3 H2O(l) → C2H5OH(l)+ 3 O2(g) (1367)\n________________________________________________________________________________\n2 C(s) + 2 H2(g) + H2O(l) → C2H5OH(l) ∆H= 2(394) + 2(286) (1367)\n\n#### Rock23\n\n##### Member\n10+ Year Member\non this question the kaplan book says to that you have to use the enthalpy of the products minus the enthalpy of the reactants. if we did that though you wouldnt come out to the same answer.", null, "10+ Year Member\n7+ Year Member\nyeah you would\n\n#### OSUDDS\n\n##### Senior Member\n10+ Year Member\nChga said:\nyeah you would\nNo, one way you get -7, the other way +7....\n\n#### tom_servo_dds\n\n##### Senior Member\n10+ Year Member\nOSUDDS said:\nNo, one way you get -7, the other way +7....\n\nProduct = 1367\nReactants = 2(-394) + 2(-286)\n\n1367 - 1360 = 7+\n\nThis thread is more than 15 years old.\n\nYour message may be considered spam for the following reasons:" ]	[ null, "data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7", null, "data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6492094,"math_prob":0.9137402,"size":934,"snap":"2021-43-2021-49","text_gpt3_token_len":420,"char_repetition_ratio":0.22688173,"word_repetition_ratio":0.9202454,"special_character_ratio":0.49357602,"punctuation_ratio":0.094339624,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95911354,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-09T13:03:42Z\",\"WARC-Record-ID\":\"<urn:uuid:a80a76c8-d45b-4d36-b773-4d9bfff4b903>\",\"Content-Length\":\"180358\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bd9200b1-3fcd-4aff-86a1-c93b4f4349f7>\",\"WARC-Concurrent-To\":\"<urn:uuid:5629565b-daf5-4729-ad63-9b63d1f125ee>\",\"WARC-IP-Address\":\"104.20.11.156\",\"WARC-Target-URI\":\"https://forums.studentdoctor.net/threads/hesss-law-question.293657/\",\"WARC-Payload-Digest\":\"sha1:BL2IL7KYPHRSRTMUK2UPEO5CHHCBRVYS\",\"WARC-Block-Digest\":\"sha1:HKE3JZRYSDWIYSEA4VERS7UCO6COILQR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964364169.99_warc_CC-MAIN-20211209122503-20211209152503-00181.warc.gz\"}"}
https://ericmarcon.github.io/entropart/reference/Hqz.html	[ "Calculates the entropy of order $$q$$ of a probability vector according to a similarity matrix.\n\nHqz(NorP, q = 1, Z = diag(length(NorP)), ...)\nbcHqz(Ns, q = 1, Z = diag(length(Ns)), Correction = \"Best\", SampleCoverage = NULL,\nCheckArguments = TRUE)\n# S3 method for ProbaVector\nHqz(NorP, q = 1, Z = diag(length(NorP)),\n..., CheckArguments = TRUE, Ps = NULL)\n# S3 method for AbdVector\nHqz(NorP, q = 1, Z = diag(length(NorP)), Correction = \"Best\",\n..., CheckArguments = TRUE, Ns = NULL)\n# S3 method for integer\nHqz(NorP, q = 1, Z = diag(length(NorP)), Correction = \"Best\",\n..., CheckArguments = TRUE, Ns = NULL)\n# S3 method for numeric\nHqz(NorP, q = 1, Z = diag(length(NorP)), Correction = \"Best\",\n..., CheckArguments = TRUE, Ps = NULL, Ns = NULL)\n\n## Arguments\n\nPs\n\nA probability vector, summing to 1.\n\nNs\n\nA numeric vector containing species abundances.\n\nNorP\n\nA numeric vector, an integer vector, an abundance vector (AbdVector) or a probability vector (ProbaVector). Contains either abundances or probabilities.\n\nq\n\nA number: the order of entropy. Default is 1.\n\nZ\n\nA relatedness matrix, i.e. a square matrix whose terms are all positive, strictly positive on the diagonal. Generally, the matrix is a similarity matrix, i.e. the diagonal terms equal 1 and other terms are between 0 and 1. Default is the identity matrix to calculate neutral entropy.\n\nCorrection\n\nA string containing one of the possible corrections: \"None\" (no correction), \"ChaoShen\", \"MarconZhang\" or \"Best\", the default value. The \"MarconZhang\" correction assumes a similarity matrix.\n\nSampleCoverage\n\nThe sample coverage of Ns calculated elsewhere. Used to calculate the gamma diversity of meta-communities, see details.\n\n...\n\nCheckArguments\n\nLogical; if TRUE, the function arguments are verified. Should be set to FALSE to save time when the arguments have been checked elsewhere.\n\n## Details\n\nEntropy is calculated following Leinster and Cobbold (2012) after Ricotta and Szeidl (2006): it is the entropy of order q of the community, using species ordinariness as the information function.\n\nA similarity matrix is used (as for Dqz), not a distance matrix as in Ricotta and Szeidl (2006). See the example.\n\nBias correction requires the number of individuals. Use bcHqz and choose the Correction. Correction techniques are from Marcon et al. (2014).\n\nCurrently, the \"Best\" correction is the max value of \"ChaoShen\" and \"MarconZhang\".\n\nThe functions are designed to be used as simply as possible. Hqz is a generic method. If its first argument is an abundance vector, an integer vector or a numeric vector which does not sum to 1, the bias corrected function bcHqz is called. Explicit calls to bcHqz (with bias correction) or to Hqz.ProbaVector (without correction) are possible to avoid ambiguity. The .integer and .numeric methods accept Ps or Ns arguments instead of NorP for backward compatibility.\n\nThe size of a metacommunity (see MetaCommunity) is unknown so it has to be set according to a rule which does not ensure that its abundances are integer values. Then, classical bias-correction methods do not apply. Providing the SampleCoverage argument allows applying the \"ChaoShen\" correction to estimate quite well the entropy. DivPart and GammaEntropy functions use this tweak.\n\n## Value\n\nA named number equal to the calculated entropy. The name is that of the bias correction used.\n\nDqz, PhyloEntropy\n\n## Examples\n\n# Load Paracou data (number of trees per species in two 1-ha plot of a tropical forest)\ndata(Paracou618)\n# Prepare the similarity matrix\nDistanceMatrix <- as.matrix(EightSpTree\\$Wdist^2/2)\n# Similarity can be 1 minus normalized distances between species\nZ <- 1 - DistanceMatrix/max(DistanceMatrix)\n# Calculate diversity of order 2\nPs <- EightSpAbundance/sum(EightSpAbundance)\nHqz(Ps, 2, Z)\n#> None\n#> 0.6033223\n# Equal to normalized Rao quadratic entropy when q=2\nRao(Ps, EightSpTree)/max(DistanceMatrix)\n#> None\n#> 0.6033223\n# But different from PhyloEntropy for all other q, e.g. 1\nHqz(Ps, 1, Z)\n#> None\n#> 0.9348275\nsummary(PhyloEntropy(Ps, 1, EightSpTree))\n#> alpha or gamma phylogenetic or functional entropy of order 1 of distribution -\n#> with correction: None\n#> Phylogenetic or functional entropy was calculated according to the tree\n#> EightSpTree\n#>\n#> Entropy is normalized\n#>\n#> Entropy equals: 1.034531" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7902177,"math_prob":0.9865823,"size":4666,"snap":"2022-27-2022-33","text_gpt3_token_len":1263,"char_repetition_ratio":0.11926212,"word_repetition_ratio":0.08228731,"special_character_ratio":0.2584655,"punctuation_ratio":0.18233944,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9974006,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-06T13:53:02Z\",\"WARC-Record-ID\":\"<urn:uuid:90e6c0b8-aff7-4820-8e41-a1dd482f3a6a>\",\"Content-Length\":\"22057\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:10f1e587-106e-4895-b615-aac546b3c24e>\",\"WARC-Concurrent-To\":\"<urn:uuid:cb71db3e-7ace-4126-bbd5-89d6c22bbbe2>\",\"WARC-IP-Address\":\"185.199.110.153\",\"WARC-Target-URI\":\"https://ericmarcon.github.io/entropart/reference/Hqz.html\",\"WARC-Payload-Digest\":\"sha1:DLFKLMGLDN7WXRYCY7PGCUTXBZ5FC5XM\",\"WARC-Block-Digest\":\"sha1:P6WDW7QYELWS6VNGMTQABBX6TBVHDI5W\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104672585.89_warc_CC-MAIN-20220706121103-20220706151103-00209.warc.gz\"}"}
https://convertoctopus.com/23-3-pounds-to-kilograms	[ "## Conversion formula\n\nThe conversion factor from pounds to kilograms is 0.45359237, which means that 1 pound is equal to 0.45359237 kilograms:\n\n1 lb = 0.45359237 kg\n\nTo convert 23.3 pounds into kilograms we have to multiply 23.3 by the conversion factor in order to get the mass amount from pounds to kilograms. We can also form a simple proportion to calculate the result:\n\n1 lb → 0.45359237 kg\n\n23.3 lb → M(kg)\n\nSolve the above proportion to obtain the mass M in kilograms:\n\nM(kg) = 23.3 lb × 0.45359237 kg\n\nM(kg) = 10.568702221 kg\n\nThe final result is:\n\n23.3 lb → 10.568702221 kg\n\nWe conclude that 23.3 pounds is equivalent to 10.568702221 kilograms:\n\n23.3 pounds = 10.568702221 kilograms\n\n## Alternative conversion\n\nWe can also convert by utilizing the inverse value of the conversion factor. In this case 1 kilogram is equal to 0.09461899664587 × 23.3 pounds.\n\nAnother way is saying that 23.3 pounds is equal to 1 ÷ 0.09461899664587 kilograms.\n\n## Approximate result\n\nFor practical purposes we can round our final result to an approximate numerical value. We can say that twenty-three point three pounds is approximately ten point five six nine kilograms:\n\n23.3 lb ≅ 10.569 kg\n\nAn alternative is also that one kilogram is approximately zero point zero nine five times twenty-three point three pounds.\n\n## Conversion table\n\n### pounds to kilograms chart\n\nFor quick reference purposes, below is the conversion table you can use to convert from pounds to kilograms\n\npounds (lb) kilograms (kg)\n24.3 pounds 11.022 kilograms\n25.3 pounds 11.476 kilograms\n26.3 pounds 11.929 kilograms\n27.3 pounds 12.383 kilograms\n28.3 pounds 12.837 kilograms\n29.3 pounds 13.29 kilograms\n30.3 pounds 13.744 kilograms\n31.3 pounds 14.197 kilograms\n32.3 pounds 14.651 kilograms\n33.3 pounds 15.105 kilograms" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.75505066,"math_prob":0.9972726,"size":1782,"snap":"2020-45-2020-50","text_gpt3_token_len":490,"char_repetition_ratio":0.22440945,"word_repetition_ratio":0.0,"special_character_ratio":0.34904602,"punctuation_ratio":0.14854111,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9964479,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-04T07:29:36Z\",\"WARC-Record-ID\":\"<urn:uuid:0dfeb012-5c12-45dc-bfea-ca23d9048376>\",\"Content-Length\":\"28330\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ff1426ab-6de3-4766-a9de-8aba269ece79>\",\"WARC-Concurrent-To\":\"<urn:uuid:20006d0e-6625-4863-8bfe-cea83fb84b6a>\",\"WARC-IP-Address\":\"104.27.143.66\",\"WARC-Target-URI\":\"https://convertoctopus.com/23-3-pounds-to-kilograms\",\"WARC-Payload-Digest\":\"sha1:IVU3R6UHKVALLECAOSKURTH43PT5EIBG\",\"WARC-Block-Digest\":\"sha1:EO27UOTV3AIA5DY2QPGXX3723EFNAR5C\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141735395.99_warc_CC-MAIN-20201204071014-20201204101014-00649.warc.gz\"}"}
https://www.ellibs.com/book/9781118909379/continuum-mechanics-the-birthplace-of-mathematical-models	[ "# Continuum Mechanics: The Birthplace of Mathematical Models\n\n94,05€\n\nISBN: 9781118909348\nDRM Restrictions\n\nPrinting 91 pages with an additional page accrued every 8 hours, capped at 91 pages 5 excerpts\n\nPresents a self-contained introduction to continuum mechanics that illustrates how many of the important partial differential equations of applied mathematics arise from continuum modeling principles\n\nWritten as an accessible introduction, Continuum Mechanics: The Birthplace of Mathematical Models provides a comprehensive foundation for mathematical models used in fluid mechanics, solid mechanics, and heat transfer. The book features derivations of commonly used differential equations based on the fundamental continuum mechanical concepts encountered in various fields, such as engineering, physics, and geophysics.\n\nThe book begins with geometric, algebraic, and analytical foundations before introducing topics in kinematics. The book then addresses balance laws, constitutive relations, and constitutive theory. Finally, the book presents an approach to multiconstituent continua based on mixture theory to illustrate how phenomena, such as diffusion and porous-media flow, obey continuum-mechanical principles.\n\nContinuum Mechanics: The Birthplace of Mathematical Models features:\n\n• Direct vector and tensor notation to minimize the reliance on particular coordinate systems when presenting the theory\n• Terminology that is aligned with standard courses in vector calculus and linear algebra\n• The use of Cartesian coordinates in the examples and problems to provide readers with a familiar setting\n• Over 200 exercises and problems with hints and solutions in an appendix\n• Introductions to constitutive theory and multiconstituent continua, which are distinctive for books at this level\nContinuum Mechanics: The Birthplace of Mathematical Models is an ideal textbook for courses on continuum mechanics for upper-undergraduate mathematics majors and graduate students in applied mathematics, mechanical engineering, civil engineering, physics, and geophysics. The book is also an excellent reference for professional mathematicians, physical scientists, and engineers.\n\nKeywords:\n\ncontinuum mechanics; partial differential equations; mathematical modeling; applied mathematics; mathematics; engineering; physics; geophysics; fluid mechanics; solid mechanics; heat transfer; porous media\n\n, Computational / Numerical Methods, Numerical Methods & Algorithms, Computational / Numerical Methods, Numerical Methods & Algorithms\nAuthor(s)\nPublisher\nJohn Wiley and Sons, Inc.\nPublication year\n2015\nLanguage\nen\nEdition\n1\nPage amount\n304 pages\nCategory\nNatural Sciences\nFormat\nEbook\neISBN (ePUB)\n9781118909348\nPrinted ISBN\n9781118909379" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8197802,"math_prob":0.66563433,"size":2487,"snap":"2022-05-2022-21","text_gpt3_token_len":480,"char_repetition_ratio":0.141764,"word_repetition_ratio":0.14071856,"special_character_ratio":0.18737435,"punctuation_ratio":0.15176152,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9932633,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-26T21:42:00Z\",\"WARC-Record-ID\":\"<urn:uuid:b4499122-df9a-4dcc-a175-7feda3935e75>\",\"Content-Length\":\"27725\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7df32d5d-e301-40ac-a1ee-45c22b3e1dc4>\",\"WARC-Concurrent-To\":\"<urn:uuid:9583bdef-a95b-4e32-9492-1d57397ff918>\",\"WARC-IP-Address\":\"62.44.201.222\",\"WARC-Target-URI\":\"https://www.ellibs.com/book/9781118909379/continuum-mechanics-the-birthplace-of-mathematical-models\",\"WARC-Payload-Digest\":\"sha1:POAYGWPPQVWNODJG4TSOENCBTAYWQ3LW\",\"WARC-Block-Digest\":\"sha1:HFT6IKMU5AOUUTGGEY3HYJRSFHJS4M4Z\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662625600.87_warc_CC-MAIN-20220526193923-20220526223923-00716.warc.gz\"}"}
https://www-history.mcs.st-andrews.ac.uk/Curves/Ellipse.html	[ "# Ellipse", null, "Cartesian equation:\nx2/a2 + y2/b2 = 1\nor parametrically:\nx = a cos(t), y = b sin(t)\n\nClick below to see one of the Associated curves.\n\nIf your browser can handle JAVA code, click HERE to experiment interactively with this curve and its associated curves.\n\nThe ellipse was first studied by Menaechmus. Euclid wrote about the ellipse and it was given its present name by Apollonius. The focus and directrix of an ellipse were considered by Pappus.\n\nKepler, in 1602, said he believed that the orbit of Mars was oval, then he later discovered that it was an ellipse with the sun at one focus. In fact Kepler introduced the word \"focus\" and published his discovery in 1609. The eccentricity of the planetary orbits is small (i.e. they are close to circles). The eccentricity of Mars is 1/11 and of the Earth is 1/60.\n\nIn 1705 Halley showed that the comet, which is now called after him, moved in an elliptical orbit round the sun. The eccentricity of Halley's comet is 0.9675 so it is close to a parabola (eccentricity 1).\n\nThe area of the ellipse is πab. There is no exact formula for the length of an ellipse in elementary functions and this led to the study of elliptic functions. Ramanujan, in 1914, gave the approximate length\n\nπ(3(a + b) - √[(a + 3b)(3a + b)]).\nThe pedal curve of an ellipse, with its focus as pedal point, is a circle.\n\nThe evolute of the ellipse with equation given above is the Lamé curve.\n\n(ax)2/3 + (by)2/3 = (a - b)2/3.\nFrom a point inside the evolute (it is a closed curve) four normals can be drawn to the ellipse but from a point outside only two normals can be drawn.\n\nOther Web site:\n\nJOC/EFR/BS January 1997\n\nThe URL of this page is:\nhttp://www-history.mcs.st-andrews.ac.uk/Curves/Ellipse.html" ]	[ null, "https://www-history.mcs.st-andrews.ac.uk/Curvepics/Ellipse/Ellipse1.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9070176,"math_prob":0.95449364,"size":1455,"snap":"2019-13-2019-22","text_gpt3_token_len":383,"char_repetition_ratio":0.11784976,"word_repetition_ratio":0.0,"special_character_ratio":0.24261168,"punctuation_ratio":0.106529206,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97365445,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-19T13:31:59Z\",\"WARC-Record-ID\":\"<urn:uuid:c2d5e635-b2a7-458e-9297-b241bd6c46e9>\",\"Content-Length\":\"6652\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c799f90a-fcd4-4327-a5f7-21f154c70a7e>\",\"WARC-Concurrent-To\":\"<urn:uuid:2202224f-69a0-4d2e-bea7-f60ad23a65b2>\",\"WARC-IP-Address\":\"138.251.192.92\",\"WARC-Target-URI\":\"https://www-history.mcs.st-andrews.ac.uk/Curves/Ellipse.html\",\"WARC-Payload-Digest\":\"sha1:TBQGQ7ZIBYEG2FHV4CNIHDU5TEGOB7ZB\",\"WARC-Block-Digest\":\"sha1:6HZB2ZOIIU3WOADY4MLZK7HA4YJBLHCR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232254882.18_warc_CC-MAIN-20190519121502-20190519143502-00054.warc.gz\"}"}
https://engineering.stackexchange.com/questions/32664/calculation-of-the-specific-co2-emission-of-a-cogeneration-plant-for-the-produce	[ "# Calculation of the specific CO2 emission of a cogeneration plant for the produced electricity\n\nSuppose I have a cogeneration plant that runs on natural gas and produces heat and electricity. The heat effeicieny is $$\\eta_{th} = 0.4$$ and the electricity efficiency is $$\\eta_{el} = 0.4$$.\n\nThe burning of natural gas produces 215 g/kWh of CO2.\n\nFor the calculation of the CO2 emission of the generated electricity from the cogeneration plant, one could now calculate $$\\frac{215}{0.4} = 537.5$$ g/kWh.\n\nHowever, I would like to calculate the CO2 emissions for the electricity generation of the cogeneration plant, taking into account that CO2 emissions are saved because another heat generator has to generate less heat.\n\nA natural gas boiler with an efficiency of $$\\eta_{th} = 0.9$$ is taken as a reference for heat generation. So the 1 kWh of heat is generated with $$\\frac{215}{0.9} = 238.88$$ g of CO2.\n\nHow do I calculate the CO2 emissions for the electricity generation in this case?\n\nMy calculation attempt:\n\nFirst I would calculate how much CO2 emssions the boiler has for 0.4 kWh:\n\n$$238.88 \\cdot 0.4 = 95.55$$ g of CO2\n\nThen I would subtract this value from the previously calculated co2 emission of the electricity:\n\n$$537.5 - 95.55 = 441.95$$\n\nThe electricity generated by the cogeneration plant would then have a CO2 emission of 441.95 g/kWh.\n\n• Hi and welcome to this site! This is not a we do your homework for you site so please xplain what you've tried so far and where you are stuck. I feel something is missing from the question, to what purpose is the efficeincy of the boiler given? the 215g correspond to lower or higher heating value?\n– mart\nJan 6, 2020 at 14:27\n• @mart Thanks for your answer. I have now edited my question with a few details and a caluclation attempt.\n– Emma\nJan 7, 2020 at 13:12\n\nYour basic logic is sound: You compare the case of the cogeneration plant + boiler vs. boiler alone. Your math looks correct to me.\n\nWhere did you get the value 215 g/kWh from? All (german language!) sources I found give 202g/kWh for natural gas, which may well be because actual of differences in grid gas composition. When I do the math myself I arrive at 197 g/kWh if using the lower heating value, only 178 g/kWh for the HHV.\n\nThe difference between HHV and LHV could be crucial since many boilers today are condensing boilers (that utilize the heat of evaporation of the exhaust water) but few to none cogeneration plants are, so you should possibly use different carbon intsities for both.\n\nLastly, the effiencies for the cogeneration plant don't look like real life numbers but are not too far of (I'm just now looking at a datasheet for a small CHP with 38% and 43% respectivly, in my experience this is a rather high $$\\eta_{el}$$)\n\n• Thank you very much for your detailed answer. Actually, the values like natural gas emissions that I used were from my memory, I was mainly interested in the calculation. The point about taking HHV and LHV into account is a good hint, thank you very much. Could you perhaps add your calculation for 197 g/kWh and 178 g/kWh? (I'm trying to figure out how you came up with that)\n– Emma\nJan 9, 2020 at 18:57\n• I did it in an excel sheet that I since deleted but It follows from the stochiometry of the combustion and the molar HHV/LHV. I hop I didnt make a mistake :)\n– mart\nJan 9, 2020 at 20:36" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9158739,"math_prob":0.9502437,"size":1236,"snap":"2022-27-2022-33","text_gpt3_token_len":326,"char_repetition_ratio":0.17207792,"word_repetition_ratio":0.039408866,"special_character_ratio":0.28640777,"punctuation_ratio":0.11811024,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98584616,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-19T17:09:33Z\",\"WARC-Record-ID\":\"<urn:uuid:6e6bc224-8c8b-47bd-9968-948ff103f694>\",\"Content-Length\":\"230053\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2fd5f90e-979c-4da3-bb30-45ede7fce68e>\",\"WARC-Concurrent-To\":\"<urn:uuid:94cd3206-4cd4-40f7-b627-9d35ce83cb5b>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://engineering.stackexchange.com/questions/32664/calculation-of-the-specific-co2-emission-of-a-cogeneration-plant-for-the-produce\",\"WARC-Payload-Digest\":\"sha1:OL5I4MPZVI3QAHYIPAJRBGFJPB4WA3XG\",\"WARC-Block-Digest\":\"sha1:6GHYYMI5EWW6LPN2OS327GNWZSKHUCPY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882573744.90_warc_CC-MAIN-20220819161440-20220819191440-00733.warc.gz\"}"}
http://nesssi.cacr.caltech.edu/SSS/20111019/1110190450034125786.html	[ "CRTS Transient ID SSS111019:002409-442745\n(When using CRTS data please cite Drake et al. 2009, ApJ, 696, 870 and our ID.)\n\nBack to index\n Image 1 Image 2 Image 3 Image 4 master", null, "", null, "", null, "", null, "", null, "Current CRTS lightcurve\nSDSS DR-8 data\nImages from SDSS, PQ, DSS, SSS, 2MASS\nPast detections in Datascope\n\nAll transients, SNe, Blazars, CVs\n\nID = 1110190450034125786\nnewast = 0.000000\nav_motn = 0.038226\nav_deltime = 32.524800\nav_uncertx = 0.023925\nav_uncerty = 0.023925\nav_inclin = -78.240701\nTime = 2455854.095600\nTime2 = 2455854.061730\nTime3 = 2455854.072990\nTime4 = 2455854.084320\nappreject = 0.000\nmastersn = 2.179\naveragesn = 4.157\ntheta = 0.000\nnhigh = 513\nRA = 6.0367600\nDec = -44.4625400\nMag = 19.364700\nMag_err = 0.060425\nFWHM = 2.890\nMaster = S45003.master.fits\nMasterRAoff = 0.065309\nMasterDecoff = 0.030535\nRA2 = 6.0365700\nDec2 = -44.4628200\nMag2 = 19.451300\nMag2_err = 0.062481\nFWHM2 = 2.820\nRA3 = 6.0363300\nDec3 = -44.4633100\nMag3 = 18.662800\nMag3_err = 0.043928\nFWHM3 = 2.850\nRA4 = 6.0367400\nDec4 = -44.4625000\nMag4 = 19.398001\nMag4_err = 0.061111\nFWHM4 = 2.840\nOppos_ang = 318.139121\nInner_motion = -0.179102\nOuter_motion = -0.204037\nEclip_long = 343.392081\nEclip_lat = -42.213326" ]	[ null, "http://nesssi.cacr.caltech.edu/SSS/20111019/jpg/1110190450034125786-0001.arch.jpg", null, "http://nesssi.cacr.caltech.edu/SSS/20111019/jpg/1110190450034125786-0002.arch.jpg", null, "http://nesssi.cacr.caltech.edu/SSS/20111019/jpg/1110190450034125786-0003.arch.jpg", null, "http://nesssi.cacr.caltech.edu/SSS/20111019/jpg/1110190450034125786-0004.arch.jpg", null, "http://nesssi.cacr.caltech.edu/SSS/20111019/jpg/1110190450034125786.master.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5141466,"math_prob":0.998927,"size":1142,"snap":"2022-05-2022-21","text_gpt3_token_len":504,"char_repetition_ratio":0.12653778,"word_repetition_ratio":0.0,"special_character_ratio":0.57968473,"punctuation_ratio":0.23043478,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9817753,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-23T14:53:27Z\",\"WARC-Record-ID\":\"<urn:uuid:20d56c21-b698-437f-98ef-771801d1e7e1>\",\"Content-Length\":\"4569\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:03ab5f0e-ec30-4b55-9664-fd4cb79f3353>\",\"WARC-Concurrent-To\":\"<urn:uuid:196e96e5-b0a2-4fba-91c6-2a58cc882c9b>\",\"WARC-IP-Address\":\"131.215.198.195\",\"WARC-Target-URI\":\"http://nesssi.cacr.caltech.edu/SSS/20111019/1110190450034125786.html\",\"WARC-Payload-Digest\":\"sha1:XLREJWKXILBD6ZWE42ZZUMVIOS7A62TN\",\"WARC-Block-Digest\":\"sha1:4D5BZKAENDMNDRCCDQBJSLUXDXKWEZDI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320304287.0_warc_CC-MAIN-20220123141754-20220123171754-00471.warc.gz\"}"}
https://www.colorhexa.com/03ec29	[ "# #03ec29 Color Information\n\nIn a RGB color space, hex #03ec29 is composed of 1.2% red, 92.5% green and 16.1% blue. Whereas in a CMYK color space, it is composed of 98.7% cyan, 0% magenta, 82.6% yellow and 7.5% black. It has a hue angle of 129.8 degrees, a saturation of 97.5% and a lightness of 46.9%. #03ec29 color hex could be obtained by blending #06ff52 with #00d900. Closest websafe color is: #00ff33.\n\n• R 1\n• G 93\n• B 16\nRGB color chart\n• C 99\n• M 0\n• Y 83\n• K 7\nCMYK color chart\n\n#03ec29 color description : Vivid lime green.\n\n# #03ec29 Color Conversion\n\nThe hexadecimal color #03ec29 has RGB values of R:3, G:236, B:41 and CMYK values of C:0.99, M:0, Y:0.83, K:0.07. Its decimal value is 257065.\n\nHex triplet RGB Decimal 03ec29 `#03ec29` 3, 236, 41 `rgb(3,236,41)` 1.2, 92.5, 16.1 `rgb(1.2%,92.5%,16.1%)` 99, 0, 83, 7 129.8°, 97.5, 46.9 `hsl(129.8,97.5%,46.9%)` 129.8°, 98.7, 92.5 00ff33 `#00ff33`\nCIE-LAB 81.929, -80.05, 72.669 30.431, 60.167, 12.107 0.296, 0.586, 60.167 81.929, 108.115, 137.767 81.929, -76.955, 96.22 77.567, -65.713, 45.043 00000011, 11101100, 00101001\n\n# Color Schemes with #03ec29\n\n• #03ec29\n``#03ec29` `rgb(3,236,41)``\n• #ec03c6\n``#ec03c6` `rgb(236,3,198)``\nComplementary Color\n• #52ec03\n``#52ec03` `rgb(82,236,3)``\n• #03ec29\n``#03ec29` `rgb(3,236,41)``\n• #03ec9e\n``#03ec9e` `rgb(3,236,158)``\nAnalogous Color\n• #ec0352\n``#ec0352` `rgb(236,3,82)``\n• #03ec29\n``#03ec29` `rgb(3,236,41)``\n• #9e03ec\n``#9e03ec` `rgb(158,3,236)``\nSplit Complementary Color\n• #ec2903\n``#ec2903` `rgb(236,41,3)``\n• #03ec29\n``#03ec29` `rgb(3,236,41)``\n• #2903ec\n``#2903ec` `rgb(41,3,236)``\nTriadic Color\n• #c6ec03\n``#c6ec03` `rgb(198,236,3)``\n• #03ec29\n``#03ec29` `rgb(3,236,41)``\n• #2903ec\n``#2903ec` `rgb(41,3,236)``\n• #ec03c6\n``#ec03c6` `rgb(236,3,198)``\nTetradic Color\n• #02a01c\n``#02a01c` `rgb(2,160,28)``\n• #02ba20\n``#02ba20` `rgb(2,186,32)``\n• #03d325\n``#03d325` `rgb(3,211,37)``\n• #03ec29\n``#03ec29` `rgb(3,236,41)``\n• #0dfc34\n``#0dfc34` `rgb(13,252,52)``\n• #26fc49\n``#26fc49` `rgb(38,252,73)``\n• #3ffd5e\n``#3ffd5e` `rgb(63,253,94)``\nMonochromatic Color\n\n# Alternatives to #03ec29\n\nBelow, you can see some colors close to #03ec29. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #17ec03\n``#17ec03` `rgb(23,236,3)``\n• #04ec03\n``#04ec03` `rgb(4,236,3)``\n• #03ec16\n``#03ec16` `rgb(3,236,22)``\n• #03ec29\n``#03ec29` `rgb(3,236,41)``\n• #03ec3c\n``#03ec3c` `rgb(3,236,60)``\n• #03ec50\n``#03ec50` `rgb(3,236,80)``\n• #03ec63\n``#03ec63` `rgb(3,236,99)``\nSimilar Colors\n\n# #03ec29 Preview\n\nText with hexadecimal color #03ec29\n\nThis text has a font color of #03ec29.\n\n``<span style=\"color:#03ec29;\">Text here</span>``\n#03ec29 background color\n\nThis paragraph has a background color of #03ec29.\n\n``<p style=\"background-color:#03ec29;\">Content here</p>``\n#03ec29 border color\n\nThis element has a border color of #03ec29.\n\n``<div style=\"border:1px solid #03ec29;\">Content here</div>``\nCSS codes\n``.text {color:#03ec29;}``\n``.background {background-color:#03ec29;}``\n``.border {border:1px solid #03ec29;}``\n\n# Shades and Tints of #03ec29\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #000401 is the darkest color, while #effff2 is the lightest one.\n\n• #000401\n``#000401` `rgb(0,4,1)``\n• #001704\n``#001704` `rgb(0,23,4)``\n• #012a07\n``#012a07` `rgb(1,42,7)``\n• #013e0b\n``#013e0b` `rgb(1,62,11)``\n• #01510e\n``#01510e` `rgb(1,81,14)``\n• #016411\n``#016411` `rgb(1,100,17)``\n• #027815\n``#027815` `rgb(2,120,21)``\n• #028b18\n``#028b18` `rgb(2,139,24)``\n• #029f1c\n``#029f1c` `rgb(2,159,28)``\n• #02b21f\n``#02b21f` `rgb(2,178,31)``\n• #03c522\n``#03c522` `rgb(3,197,34)``\n• #03d926\n``#03d926` `rgb(3,217,38)``\n• #03ec29\n``#03ec29` `rgb(3,236,41)``\nShade Color Variation\n• #07fc2f\n``#07fc2f` `rgb(7,252,47)``\n• #1afc3f\n``#1afc3f` `rgb(26,252,63)``\n• #2efc4f\n``#2efc4f` `rgb(46,252,79)``\n• #41fd5f\n``#41fd5f` `rgb(65,253,95)``\n• #54fd70\n``#54fd70` `rgb(84,253,112)``\n• #68fd80\n``#68fd80` `rgb(104,253,128)``\n• #7bfd90\n``#7bfd90` `rgb(123,253,144)``\n• #8efea0\n``#8efea0` `rgb(142,254,160)``\n• #a2feb1\n``#a2feb1` `rgb(162,254,177)``\n• #b5fec1\n``#b5fec1` `rgb(181,254,193)``\n• #c8fed1\n``#c8fed1` `rgb(200,254,209)``\n• #dcffe1\n``#dcffe1` `rgb(220,255,225)``\n• #effff2\n``#effff2` `rgb(239,255,242)``\nTint Color Variation\n\n# Tones of #03ec29\n\nA tone is produced by adding gray to any pure hue. In this case, #717e73 is the less saturated color, while #03ec29 is the most saturated one.\n\n• #717e73\n``#717e73` `rgb(113,126,115)``\n• #68876d\n``#68876d` `rgb(104,135,109)``\n• #5f9067\n``#5f9067` `rgb(95,144,103)``\n• #569961\n``#569961` `rgb(86,153,97)``\n• #4da25b\n``#4da25b` `rgb(77,162,91)``\n• #43ac54\n``#43ac54` `rgb(67,172,84)``\n• #3ab54e\n``#3ab54e` `rgb(58,181,78)``\n• #31be48\n``#31be48` `rgb(49,190,72)``\n• #28c742\n``#28c742` `rgb(40,199,66)``\n• #1fd03c\n``#1fd03c` `rgb(31,208,60)``\n• #15da35\n``#15da35` `rgb(21,218,53)``\n• #0ce32f\n``#0ce32f` `rgb(12,227,47)``\n• #03ec29\n``#03ec29` `rgb(3,236,41)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #03ec29 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.54338765,"math_prob":0.81958145,"size":3672,"snap":"2019-35-2019-39","text_gpt3_token_len":1616,"char_repetition_ratio":0.12677208,"word_repetition_ratio":0.011090573,"special_character_ratio":0.5541939,"punctuation_ratio":0.23783186,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9821499,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-23T03:39:25Z\",\"WARC-Record-ID\":\"<urn:uuid:bfa4935c-3900-46a7-92d7-ddab0732ac01>\",\"Content-Length\":\"36225\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:577a0159-1bcc-45a6-b866-b93535a81d7f>\",\"WARC-Concurrent-To\":\"<urn:uuid:10e3a601-422f-4e98-8734-419cb54f5da4>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/03ec29\",\"WARC-Payload-Digest\":\"sha1:GRPQCRR6QHPEOUGDY2SGZLNRKYEHHLJM\",\"WARC-Block-Digest\":\"sha1:SGA7K6WDPAPVCKLHHGK4V53FJM4TVGK6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027317817.76_warc_CC-MAIN-20190823020039-20190823042039-00210.warc.gz\"}"}
https://physics.stackexchange.com/questions/3009/how-exactly-does-curved-space-time-describe-the-force-of-gravity	[ "How exactly does curved space-time describe the force of gravity?\n\nI understand that people explain (in layman's terms at least) that the presence of mass \"warps\" space-time geometry, and this causes gravity. I have also of course heard the analogy of a blanket or trampoline bending under an object, which causes other objects to come together, but I always thought this was a hopelessly circular explanation because the blanket only bends because of \"real\" gravity pulling the object down and then pulling the other objects down the sloped blanket. In other words, to me, it seems that curved space wouldn't have any actual effect on objects unless there's already another force present.\n\nSo how is curved space-time itself actually capable of exerting a force (without some source of a fourth-dimensional force)?\n\nI apologize for my ignorance in advance, and a purely mathematical explanation will probably go over my head, but if it's required I'll do my best to understand.\n\n• In many \"video\" explanations of general relativity curvature of Time is omited, time is certainly not easy to graph with the blanket example, but sometimes it's not even mentioned, perhaps lack of self-questioning of the explainer, then it's a good question +1 – HDE May 16 '11 at 17:28\n• I would modify this question as follows: If we could put a particle in orbit around a star with no other planets or satellites and then use a fictional device to cancel all the inertia of the particle, it is obvious that the curve of space-time is towards the star but what is not obvious is what would make the particle begin to move towards the star after all its momentum/inertia were canceled. Gravity is not a force so how would the particle 'know' that it needs to start accelerating towards the star? – Kelly S. French Sep 29 '15 at 20:39\n• The blanket/trampoline isn't meant to explain anything in the sense of suggesting an underlying mechanism. It's a way of thinking about an esoteric subject far removed from ordinary experience in terms of something more familiar. \"Vectors are like arrows\" doesn't mean vectors are made of obsidian or fired from bows. In any case, the blanket/trampoline is entirely wrong as a model of curved space in general relativity, though it is a surprisingly accurate model of Newtonian gravity: see this answer. – benrg Feb 4 at 8:37\n• Nothing is as instructive as reading. Especially this: archive.org/details/TheClassicalTheoryOfFields – my2cts Apr 29 at 18:27\n\nLuboš's answer is of course perfectly correct. I'll try to give you some examples why the straightest line is physically motivated (besides being mathematically exceptional as an extremal curve).\n\nImage a 2-sphere (a surface of a ball). If an ant lives there and he just walks straight, it should be obvious that he'll come back where he came from with his trajectory being a circle. Imagine a second ant and suppose he'll start to walk from the same point as the first ant and at the same speed but into a different direction. He'll also produce circle and the two circles will cross at two points (you can imagine those circles as meridians and the crossing points as a north resp. south poles).\n\nNow, from the ants' perspective who aren't aware that they are living in a curved space, this will seem that there is a force between them because their distance will be changing in time non-linearly (think about those meridians again). This is one of the effects of the curved space-time on movement on the particles (these are actually tidal forces). You might imagine that if the surface wasn't a sphere but instead was curved differently, the straight lines would also look different. E.g. for a trampoline you'll get ellipses (well, almost, they do not close completely, leading e.g. to the precession of the perihelion of the Mercury).\n\nSo much for the explanation of how curved space-time (discussion above was just about space; if you introduce special relativity into the picture, you'll get also new effects of mixing of space and time as usual). But how does the space-time know it should be curved in the first place? Well, it's because it obeys Einstein's equations (why does it obey these equations is a separate question though). These equations describe precisely how matter affects space-time. They are of course compatible with Newtonian gravity in low-velocity, small-mass regime, so e.g. for a Sun you'll obtain that trampoline curvature and the planets (which will also produce little dents, catching moons, for example; but forget about those for a moment because they are not that important for the movement of the planet around the Sun) will follow straight lines, moving in ellipses (again, almost ellipses).\n\n• Thanks heaps guys, it's starting to make some sense. So that makes sense to me with moving objects, but I still don't quite understand how it causes objects to accelerate. For example, with your analogy, what if the ants were stationary on the ball? When we lift something off the ground and let go, it accelerates toward the earth. Is this just because that's the \"straighest\" line through the curved spacetime around the earth? Why must it always be \"moving\" through a straight line, and what does it mean in terms of curved spacetime for something to be stationary? – Zac Jan 16 '11 at 12:08\n• Also @Zac, for something to be stationary in spaceTIME means that it only exists for a single instant in time! Even something that stays stationary in space for all time moves on a curve in spacetime. (think about what an x vs. t plot looks like for a stationary object) – wsc Jan 16 '11 at 18:33\n• @wsc: it's really called perihelium in my language (Slovak) so I never imagined it might be something different in English. Anyway, thank you :) – Marek Jan 16 '11 at 19:34\n• @Marek: nor did I realize it was different in Slovak; always nice to learn! Anyway, I was using that meaning of stationary since that was what @Zac was using: his question seemed to me to be: 'Sure you have geodesics on curved manifolds, but why do the ants have to move?' Which is a very good question, you just have to remember that time is a coordinate too. – wsc Jan 16 '11 at 20:07\n• @AdamHughes: So the crucial difference is that spacetime includes curved time as well. No object can remain truly stationary in spacetime because that would require being stationary in time. An object that magically appears above Earth may begin stationary in the space dimension of spacetime, but it nevertheless continues to \"move\" through the time dimension. It must follow the geodesic (straightest/shortest possible path) from that point, and the geodesic from the point above Earth points through time but also towards Earth because of its effect on the curvature of spacetime. – Zac Aug 4 '17 at 16:06\n\nThere are actually two different parts of general relativity. They're often stated as\n\n1. Spacetime tells matter how to move\n2. Matter tells spacetime how to curve\n\nPoint #1 is actually straightforward to explain: objects simply travel on the straightest possible paths through spacetime, called geodesics. The paths only seem curved because of the warping of spacetime. If you're a physicist, then you would want to know that that fact can be derived from the principle of extremal action (with all the requisite mathematical details), but if you don't want to wade through the math, hopefully it at least makes sense that objects move on \"straight\" lines. There is no actual force involved when a massive (or even a massless) object's trajectory curves in response to gravity, because it doesn't take any force to keep something moving on a straight line. (I can definitely expand on this point if you want)\n\nNow, I mentioned that spacetime needs to be warped in order for objects' trajectories to appear curved to us despite them actually being \"straight.\" So the essence of point #2 is, why is spacetime warped in the first place? Physics doesn't have a good answer to that. Technically, we don't have an answer to point #1 either, but the \"straight line\" argument at least makes it seem plausible; unfortunately, there's no equivalent plausibility argument for why spacetime warps itself around matter. (Perhaps someday we will come up with one) All we can do right now is produce equations that describe how spacetime behaves around matter, namely the Einstein equations which can be written $G_{\\mu\\nu} = 8\\pi T_{\\mu\\nu}$ among other ways.\n\n• I never got why Wheeler wasn't as famous as Feynman. He had that same magical way of reducing things down to really clear, simple statements that made complicated things seem obvious. – Jerry Schirmer Apr 17 '15 at 2:48\n\nthe trampoline analogy needs an extra source of gravity - because this is what the laymen, the recipients of the explanation, intuitively understand - but the real general relativity doesn't need any extra \"external\" gravity.\n\nInstead, general relativity says that the space is getting curved by Einstein's equations, $$G=T$$ where the left-hand side are numbers describing the curvature at a given point and the right hand side is the density of matter and momentum. I omit indices and constants haha. So general relativity says how the spacetime is curved under the influence of matter.\n\nThe second part of the story is that general relativity also says how matter moves in external geometry. It moves along \"geodesics\", lines that are as straight as you can get. $$\\delta S_{action\\,ie\\,proper\\,length} = 0$$ This actually means that the objects move along the predicted, seemingly curved trajectories. These trajectories are actually as straight in the curved spacetime as you can get.\n\nImagine that there is a hemisphere replacing a disk in the trampoline. So there exists a (nearly) straight line on the hemisphere - namely the equator near the junction with the rest of the trampoline. Note that the equator on the Earth is a maximum circle - so it is one of the straightest lines you can draw on the surface of Earth. The same is true for all actual trajectories that objects choose in spacetime of general relativity.\n\nSo in the hemisphere-above-trampoline example, particles can orbit around the equator of the attached hemisphere, just like planets, because it is the straightest and most natural line they can choose. I don't use any external gravity to explain the real gravity; instead, I use the principle that particles choose the most natural - the straightest - line they can find in the curved spacetime.\n\nBest wishes Lubos\n\nThe other answers are more or less correct, but perhaps I can say something more to the point of the question, *How is curved spacetime itself actually capable of exerting a force?\n\nNo force whatsoever is needed.\n\nGravity is not a force. What is a force, anyway? Newton clarified for almost the first time in Science what a force is: First I will say it, then explain it: A force is something which makes the motion of a body deviate from uniform straightline motion.\n\nNewton pointed out that bodies have a tendency, inertia, to continue in whatever direction they are already going, with whatever velocity they have at the moment. That means uniform, rectilineal motion: steady velocity, same direction. Newton actually knew this was what would be later called a geodesic, since « a straight line is the shortest distance between two points ».\n\nNewton then went on to say that to overcome inertia, to overcome this tendency, requires a force: force is what makes a body depart from the geodesic it is (even momentarily) headed on (its direction and speed).\n\nIt was then Einstein (and partly Mach before him) who said this does not get to the essence of the question. For Einstein, any coordinate system had to be equally allowable, and in fact, space-time is curved (as already explained by other posters). A body or particle under the influence of gravity actually does travel in a geodesic....i.e., it does what a free particle does. I.e., it does what a particle not under the influence of any force does. So gravity is not a force.\n\nNewton did not realise that space-time could be curved and that then the geodesics would not appear to our sight to be straight lines when projected into space alone. That ellipse you see in pictures of planetary orbits? It is not really there of course since the planet only reaches different points of the ellipse at different times...that ellipse is not what the planet really traverses in space-time, it is the projection of the path of the planet onto a slice of space, it is really only the shadow of the true path of the planet, and seems much more curved than the true path really is.\n\n( ¡ The curvature of space-time in the neighbourhood of the earth is really very small ! The path of the earth in space-time would even appear to be nearly straight to an imaginary Euclidean observer who, in a flat five-dimensional space larger than ours, was looking down on us in our slightly curved four dimensional space-time embedded in their world. It's $$ct$$, remember, so the curving around the ellipse gets distributed over an entire light-year, and appears to be nearly straight...and is straight when one takes into account the slight curvature of space-time.)\n\nSince every particle under the influence of gravity alone moves in a geodesic, it does not experience any force that would make it depart from its inertia and make it depart from this geodesic. So gravity is not a force, but electric forces still do exist. They could overcome the inertia of a charged body and make it deviate from the geodesic it is headed on: change its speed and direction (when speed and direction are measured in curved space-time).\n\nEinstein (and me too) did not want to change the definition of force in this new situation, since after all electric forces are known to exist and are still forces in GR. So the old notion of force still retains its usefulness for things other than gravity. To repeat: if a body is not moving in a geodesic in space-time, you go looking for a force that is overcoming its inertia....but since gravity and curvature of space-time do not make a body depart from a geodesic, neither of them is a force.\n\nSee also http://www.einstein-online.info/elementary/generalRT/GeomGravity.html which avoids the trampoline fallacy and has a great image of the great circle.\n\n• Gravity is not a force in GR. Gravity was a force in classical mechanics. Gravity is ________ in quantum theories (Sorry, I don't know enough to fill in the blank.) My point is that all of these realms are models that predict the motion of terrestrial and astronomical objects. Some models (e.g., GR) make better predictions than others (e.g., classical), but do any of them tell us what gravity really is? – Solomon Slow Sep 21 '15 at 20:14\n• @james large, the answer is no. There isn't a complete theory of gravity. No one knows what causes. – Ernesto Melo Jun 13 '17 at 16:26\n\nAs others mentioned, the main problem with the common visualization is, that it omits the time dimension. In the animation linked below the time-dimension is included to explain how General Relativity differs form Newton's model.\n\nIt is straightforward to see how the geometry of spacetime describes the force of gravity -- you just need to understand the geodesic equation, which in general relativity describes the paths of things subject to gravity and nothing else. This is the \"spacetime affects matter\" side of the theory.\n\nTo understand why curvature in particular, as a property of the geometry, is important, you need to understand the \"matter affects spacetime\" side of general relativity. The postulate is that the Gravitational Lagrangian of the theory is equal to the scalar curvature -- this is called the \"Einstein-Hilbert Action\" --\n\n$$S=\\int{\\left( {\\lambda R + {{\\mathcal{L}}_M}} \\right)\\sqrt { - g}\\, d{x^4}} {\\text{ }}$$\n\nYou set the variation in the action to zero, as with any classical theory, and solve for the equations of motion. The conventional way to do this goes something like this --\n\n$$\\int{\\left( {\\frac{{\\delta \\left( {\\left( {{{\\mathcal{L}}_M} + \\lambda R} \\right)\\sqrt { - g} } \\right)}}{{\\delta {g_{\\mu \\nu }}}}} \\right)\\delta {g_{\\mu \\nu }}\\,d{x^4}} = 0$$ $$\\sqrt { - g} \\frac{{\\delta {{\\mathcal{L}}_M}}}{{\\delta {g_{\\mu \\nu }}}} + \\lambda \\sqrt { - g} \\frac{{\\delta R}}{{\\delta {g_{\\mu \\nu }}}} + \\left( {{{\\mathcal{L}}_M} + \\lambda R} \\right)\\frac{{\\delta \\sqrt { - g} }}{{\\delta {g_{\\mu \\nu }}}} = 0$$ $$\\frac{{\\delta R}}{{\\delta {g_{\\mu \\nu }}}} + \\frac{R}{{\\sqrt { - g} }}\\frac{{\\delta \\sqrt { - g} }}{{\\delta {g_{\\mu \\nu }}}} = - \\frac{1}{\\lambda }\\left( {\\frac{1}{{\\sqrt { - g} }}{{\\mathcal{L}}_M}\\frac{{\\delta \\sqrt { - g} }}{{\\delta {g_{\\mu \\nu }}}} + \\frac{{\\delta {{\\mathcal{L}}_M}}}{{\\delta {g_{\\mu \\nu }}}}} \\right)$$\n\n$${R_{\\mu \\nu }} - \\frac{1}{2}R{g_{\\mu \\nu }} = \\frac{1}{{2\\lambda }}{T_{\\mu \\nu }}$$\n\nTo fix the value of $\\kappa=1/{2\\lambda}$, we impose Newtonian gravity at low energies, for which we only consider the time-time component, which Newtonian gravity describes (I'll use $C$ for the gravitational constant, reserving $G$ for the trace of the Einstein tensor) --\n\n$$\\begin{gathered} {G_{00}} = \\kappa c^4\\rho \\\\ {R_{00}} = {G_{00}} - \\frac{1}{2}Gg_{00} \\\\ \\Rightarrow {R_{00}} \\approx \\kappa \\left( {c^4\\rho - \\frac{1}{2}\\frac{1}{{c^2}}c^4\\rho c^2} \\right) \\approx \\frac{1}{2}\\kappa c^4\\rho \\\\ \\end{gathered}$$\n\nImposing Poisson's law from Newtonian gravity with $\\partial^2\\Phi$ approximating $\\Gamma _{00,\\alpha }^\\alpha$,\n\n$$4\\pi C\\rho \\approx {\\nabla ^2}\\Phi \\approx \\Gamma _{00,\\alpha }^\\alpha \\approx {R_{00}} \\approx \\frac{\\kappa }{2}c^4\\rho \\\\ \\Rightarrow \\kappa = \\frac{{8\\pi G}}{{c^4}} \\\\$$\n\n(The fact that this is possible is fantastic -- it means that simply postulating that spacetime is curved in a certain sense produces a force that agrees with our observations regarding gravity at low energies.) Giving us the Einstein-Field Equation,\n\n$${G_{\\mu \\nu }} = \\frac{{8\\pi G}}{{c^4}}{T_{\\mu \\nu }}$$\n\n• This is not an explanation in \"layman's terms\"... – Comp_Warrior Jun 22 '13 at 1:20\n• I just think the average person interested in OP's question would not have knowledge of Lagrangians, Tensors etc. – Comp_Warrior Jun 22 '13 at 13:13\n• @Comp_Warrior, the About says the site is for Academics, students, and researchers of physics and astronomy, so the average audience should not consist of laypeople and it is perfectly ok to give technical and advanced answers for the people who can stomach it. Even though it looks like this since quite some time, physics se is not meant to be a popular physics forum such as quora for example ... – Dilaton Jun 22 '13 at 14:46\n• Btw the op says he does not mind technical answers, so why are you insisting on answers exclusively in layman terms? – Dilaton Jun 22 '13 at 15:00\n• @Comp_Warrier what dimension10 says, and the SE system is exactly desined that the op can accept the answer he likes best, maybe a popular one, whereas there can legitimately be other more technical answers too, that are liked by other members of the community. Answers to a question are not only meant to serve the op, but the whole community. So there is absolutely nothing wrong with a question getting answers of different level. It would be nice if you stop discouraging good technical posts which are perfectly legitimate. – Dilaton Jun 22 '13 at 22:20\n\nI think the problem for the layman is understanding why there is motion in spacetime and I think a sort of answer is that we already accept motion through time when we think of time and space as separate. Well we are in motion through spacetime where time and space are not separable and when we move through a region of spacetime that contains matter the shortest spacetime path between two events is the one that includes motion through the space bit as well as the time bit (ie not orthogonal to the space axes). That is experienced as falling under gravity.\n\nA complete replacement of the brief answer I wrote some time ago:\n\nMore than one person has brought up the idea of a pair of ants walking on the surface of a sphere. Each ant is moving in what, for it, is a straight line, but the get closer together at an increasing rate until they collide. (Provided they're lined up right.)\n\nThis is an excellent metaphor, but it can be confusing because each ant is propelling itself, so it could stop if it wanted, and also they do have to be lined up right when they start or they won't collide. If you hold a rock still and then let go, it starts to move, which seems different from the ant picture.\n\nAll of these problems disappear if you realize that they don't call it spacetime for nothing. The surface of the balloon is two-dimensional in the ants-on-a-balloon analogy (and really the ants ought to be two-dimensional themselves, living embedded in the surface of the balloon just as we are embedded in spacetime). But it's wrong to think we are only throwing away one dimension to be able to visualize curved space. The right way to think of the balloon is that it has one dimension of space and one of time, so we're really throwing away two out of the four dimensions.\n\nEach ant is racing headlong into its own future, and it can't stop or even slow down. And the ants can't miss each other, because the paths they follow are really the histories of their lives. The paths are called world lines. Each point on a world line is a time and a place that the ant passed through. If two world lines cross, that means two ants were at the same place at the same time.\n\nThis is still confusing, because the balloon is round. Which direction is time, and which direction is space? What happens when the ant goes all the way around the sphere? To make sense of these questions, you have to put a coordinate system on the sphere. For this toy universe, it actually makes sense to use latitude and longitude as coordinates. The south pole is kind of a big bang (take this with a lot of salt) and the north pole is the big crunch in the future (that definitely isn't going to happen in real life). The lines of latitude are the time coordinate, which means time progresses along the lines of longitude.\n\nWhat Einstein's equation tell us, at a basic level, is that the curvature of space-time and stress-energy are the same thing.\n\nIn order for this law to be respected it is clear that the stress-energy of a test particle cannot be constant in a space-time with changing curvature.\n\nSo, if you can choose a coordinate set in which the stress energy tensor is represented by the mass-energy of the particle, then the practical effect you can observe is changing energy and momenta of the test particle.\n\nWhen you therefore observe the test particle, you will see it as having changing energy and momenta, and therefore derive a force driving these changes. This is what we call gravity.\n\nHowever, general relativity gives a much deeper picture of gravity as a description of the curvature of space-time, so, in a way, gravity is an observed effect of the curvature of space-time, or, if you like, an observed effect of the distribution of mass and energy.\n\n• Part of this answer has been quoted in a new question. – Nat Sep 26 '18 at 10:01\n\nCurvature affects motion by making the lines that are as straight as possible end up converging, just line how if you and your friends fly at constant altitude from the north pole, then no matter what directions you go (even if you and your friend head out in very different directions) then you start to converge on the south pole. This is a very good way to describe an effect that is determined by the path and not by the mass of the object taking the path. This is sometimes described as \"spacetime tells matter how to move\" but really this is just that the straightest possible lines converge when spacetime is curved the right way.\n\nBut something not mentioned enough is that while mass, energy, momentum, stress, and pressure are sources of curvature, they are not the only things that create curvature, curvature itself can create further and additional curvature. A gravitational wave can propagate or even spread in a vacuum of empty space devoid of all mass, energy, momentum, stress, and pressure.\n\nThe region outside a symmetric nonrotating static star is curved, even the parts far from any mass or energy or momentum or stress or pressure. The space remains curved because the existing curvature is exactly shaped so as to persist (or otherwise cause future curvature exactly like itself).\n\nSo curvature allows and sometimes requires more and/or future curvature, just as a travelling electromagnetic wave allows and/or even requires there be more electromagnetic waves elsewhere and/or later. The vacuum allows curvature far from gravitational sources just as it allows electromagnetic waves far from electromagnetic sources. What electromagnetic sources allow is for electromagnetic fields to behave differently (namely to gain or lose energy as well as move in different ways and gain and lose momentum and stress). Similarly what gravitational sources do is allow curvature to react differently to itself than it otherwise would.\n\nImagine a flat region of space shaped like a ball, then imagine a funnel type curved space where two regions of surface area are farther apart than they would be if flat (like a higher dimensional version of a funnel and on a funnel surface two circles of a particular circumference are farther away as measured along the funnel then if two similarly sized circles were in a flat sheet). On its own, spacetime doesn't allow itself to connect those two kinds of regions together, but that mismatch is exactly the kind or not-lining-up that putting some mass or energy right there on the boundary fixes. So without mass those two regions can't line up, with mass they can. Just like an electromagnetic field can have a kink if there is a charge there.\n\nSo your curvature likes to propagate a certain way, and if you want it to deviate from that, you need mass, energy, momentum, stress, and/or pressure. And you'd need the right kind to get it to match up, the kind you want might be available, and might not even exist, so not all kinds of curvature will be allowed. But the point of a source is that it changes the balance between nearby curvature and not that affects future curvature. So there is a kind of balance, and there are things that can warp that a balance. Those things that warp that natural vacuum balance are called gravitational sources.\n\nHaving curved spacetime is something we observe. Having gravitational sources that can change the normal or usual way curvature evolves is something else entirely. We can make theories about how the sources evolve, and then the curvature is forced to co-evolve with it, and that's what gravity is about, about gravitational interactions (source and curvature together) changing how the curvature evolves changing the evolution that the curvature otherwise would have evolved a different way.\n\nSo there is nothing circular, curvature is observed, and on its own it interacts and affects itself in a particular way (that is also observed), but gravitational sources get to change that and by interacting with the gravitational sources (which we can do) we can ourselves make the curvature change in different ways than it otherwise would!\n\nA question was marked as a duplicate of a duplicate of this question, so I am posting my answer here.\n\nGravity is due to the curvature of spacetime\n\nI believe it is true. That is what general relativity says, and general relativity has been confirmed in predictions ranging from the existence of black holes to the orbit of Mercury to the bending of light.\n\nRelation between spacetime, curvature, mass and gravity\n\nYou say you are confused about how the curvature of spacetime and gravity are related. I am going to explain mainly that in my answer, starting with simpler examples, and moving to more complicated ones.\n\nOkay, let's say you have a sheet of rubber. This is the classic example of spacetime. Let's say you take a bowling ball, and set it on the taut sheet of rubber. It has a large mass (compared to what else we'll be putting on the sheet), therefore the sheet curves a lot for the bowling ball. We now have an image in our head like the one below:", null, "So mass leads to curvature. Then, take a baseball, say, and set it near the bowling ball. It rolls toward the bowling ball, right? This occurs because of the curvature of the sheet. So, then, curvature leads to gravity. So, if an object has large mass, it will curve spacetime dramatically, leading to strong gravity.\n\nThis is, of course, an overly simplistic example. It is 2-d, and it doesn't take into account other factors. Let us move to 3-d (keeping in mind the universe is accepted to be at 4-d, ignoring the holographic principle). The mass of a bowling ball now sucks in space around it, sort of like in the picture below:", null, "And now, in this case, we can see (or understand) that more mass still leads to more curvature. The greater the mass, the more spacetime will \"contract\" around the object. So we still think that mass leads to curvature. Now, if we set an object near this massive object (like the moon next to Earth) it is \"sucked in\" sort of, by the curvature of spacetime, though of course the moon contracts spacetime around it as well. At this point, we can reasonably still conclude that in 3-d, mass leads to curvature which leads to gravity.\n\nBut, as I said earlier, the universe is generally thought of as 4-d. What does our picture look like when we add time? Well, the time dimension is contracted around a massive object. So let us picture our previous example but that the fabric of spacetime has a few clocks embedded in it occasionally. As the space stretches and contracts, so will the clocks (the \"time\") and so the time on those clocks will be \"wrong\" - it'll differ from the other clocks. And in this case, as the Earth contracts space and time around it, it changes the time and space (it curves spacetime) and so when another object enters our region of spacetime, it is \"sucked in\" still, but so is it's time. This is, of course, a very extreme example, but I hope this shows that we can conclude that mass leads to curvature which leads to gravity. And a black hole, is simply so much mass that it leads to so much curvature that the gravity is so strong that light cannot escape.\n\nI hope this helps!\n\nHere's a simple way to think about it:\n\nNewton's first law of motion says that in the absence of any force on a particle, the particle will move in a straight line.\n\nHence, if we see a particle move in a curved path - that is of it deviates from a curved path - we can say that there is a force on it.\n\nNow, in GR, particles without any forces acting on it move on geodesics. This is the replacement for the notion of straight lines in a curved spacetime. Nevertheless we can detect the deviation from the usual notion of a straight line in flat space.\n\nThis deviation will be correlated with the force of gravity as experienced by this object in its local frame.\n\nprotected by Qmechanic♦Aug 17 '13 at 11:27\n\nThank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).\n\nWould you like to answer one of these unanswered questions instead?" ]	[ null, "https://i.stack.imgur.com/3vGJJ.png", null, "https://i.stack.imgur.com/iGT6a.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.96297526,"math_prob":0.9618398,"size":3093,"snap":"2019-43-2019-47","text_gpt3_token_len":697,"char_repetition_ratio":0.1388799,"word_repetition_ratio":0.014260249,"special_character_ratio":0.21985127,"punctuation_ratio":0.119195044,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.9819885,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-18T23:28:45Z\",\"WARC-Record-ID\":\"<urn:uuid:8dbd3cd4-8435-48ce-9d22-6c3aa7f0d2bb>\",\"Content-Length\":\"238372\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cfc384d5-1940-4010-ab21-c47320fc244f>\",\"WARC-Concurrent-To\":\"<urn:uuid:315f882b-cdc9-4a52-a49b-00b3eb1aca26>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://physics.stackexchange.com/questions/3009/how-exactly-does-curved-space-time-describe-the-force-of-gravity\",\"WARC-Payload-Digest\":\"sha1:RI3OXI3UGYADIX47J4OSCT3R7N52JDNM\",\"WARC-Block-Digest\":\"sha1:GIMTFSKGVHOC67353UAE3GZJY2Q6AFLJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986685915.43_warc_CC-MAIN-20191018231153-20191019014653-00448.warc.gz\"}"}
https://math.stackexchange.com/questions/678515/probability-2-4-vs-3-6/678530	[ "# probability $2/4$ vs $3/6$\n\nRecently I was asked the following in an interview:\n\nIf you are a pretty good basketball player, and were betting on whether you could make $2$ out of $4$ or $3$ out of $6$ baskets, which would you take?\n\nI said anyone since ratio is same. Any insights?\n\n• Are you betting you will make exactly that number, or that you wll make at least that number? It changes the answer. – Ross Millikan Feb 16 '14 at 15:31\n• Betting for \"exactly\" is silly, as you can always miss on purpose. – Bach Feb 16 '14 at 15:34\n• @HansZauber Unless there is the possibility to fail at missing :) – Hagen von Eitzen Feb 16 '14 at 15:43\n• This is a great, unintuitive question. – Thomas Ahle Feb 16 '14 at 17:33\n• @Thomas Ahle is the question unintuitive or the answer? – miracle173 Feb 16 '14 at 22:18\n\n# Depends on how good you are", null, "The explanation is intuitive:\n\n• If you are not very good (probability that you make a single shot - p < 0.6), then your overall probability is not very high, but it is better to bet that you'll make 2 out of 4, because you may do it just by chance and your clumsiness has less chance to prove in 4 than in 6 attempts.\n\n• If you are really good (p > 0.6), then it is better to bet on 3 out of 6, because if you miss just by chance, you have better chance to correct yourself in 6 attempts.\n\nThe curves meet exactly at p = 0.6.\n\n## In general, the more attempts, the more of real skill reveals\n\nThis is best illustrated on the extreme case:", null, "With more attempts, it is almost binary case - you either succeed or not, based on your skill. With high N, the result will be close to your original expectation.\n\nNote that with high N and p = 0.5, the binomial distribution gets narrower and converges to normal distribution.\n\n## Everything here just revolves around binomial distribution,\n\nwhich tells you that the probability that you will score exactly k shots out of n is\n\n$$P(X = k) = \\binom{n}{k} p^k (1-p)^{n-k}$$\n\nThe probability that you will score at least k = n/2 shots (and win the bet) is then\n\n$$P(X \\ge k) = \\sum^{n}_{i=k} \\binom{n}{i} p^i (1-p)^{n-i}$$\n\n# Why the curves don't meet at p = 0.5?\n\nLook at the following plots:", null, "These plots are for p = 0.5. The binomial distribution is symmetric for this value. Intuitivelly, you expect 2 of 4 or 3 of 6 to take half of the distribution. But if you look especially at the left plot, it is clear that the middle column (2 successful shots) goes far beyond the half of the distribution (dashed line), which is denoted by the red arrow. In the right plot (3/6), this proportion is much smaller.\n\nIf you sum the gold bars, you will get:\n\nP(make at least 2 out of 4) = 0.6875\nP(make at least 3 out of 6) = 0.65625\nP(make at least 500 out of 1000) = 0.5126125\n\n\nFrom these figures, as well as from the plots, is apparent that with high N, the proportion of the distribution \"beyond the half\" converges to zero, and the total probability converges to 0.5.\n\nSo, for the curves to meet for low Ns, p must be higher to compensate for this:", null, "P(make at least 2 out of 4) = 0.8208\nP(make at least 3 out of 6) = 0.8208\n\n\nFull code in R:\n\nf6 <- function(p) {\ndbinom(3, 6, p) +\ndbinom(4, 6, p) +\ndbinom(5, 6, p) +\ndbinom(6, 6, p)\n}\n\nf4 <- function(p) {\ndbinom(2, 4, p) +\ndbinom(3, 4, p) +\ndbinom(4, 4, p)\n}\n\nfN <- function(p, from, max) {\n#sum(sapply(from:max, function (x) dbinom(x, max, p)))\ns <- 0\nfor (i in from:max) {\ns <- s + dbinom(i, max, p)\n}\ns\n}\nf1000 <- function (p) fN(p, 500, 1000)\n\nplot(f6, xlim = c(0,1), col = \"red\", lwd = 2, ylab = \"\", main = \"Probability that you will make ...\", xlab = \"p (probability you make a single shot)\")\ncurve(f4, col = \"green\", add = TRUE, lwd = 2)\ncurve(f1000, add = TRUE, lwd = 2, col = \"blue\")\nlegend(\"topleft\", c(\"2 out of 4\", \"3 out of 6\", \"500 out of 1000\"), lwd = 2, col = c(\"green\", \"red\", \"blue\"), bty = \"n\")\n\nplotHist <- function (n, p) {\nplot(x=c(-0.5,n+0.5),y=c(0,0.41),type=\"n\", xaxt=\"n\", xlab = \"successful shots\", ylab = \"probability\",\nmain = paste0(n/2, \"/\", n, \", p = \", p))\naxis(1, at=0:n, labels=0:n)\nx <- 0:n\ny <- dbinom(0:n, n, p)\nw <- 0.9\n#lines(0:4, dbinom(0:4, 4, 0.5), lwd = 50, type = \"h\", lend = \"butt\")\nrect(x-0.5w, 0, x+0.5w, y, col = \"lightgrey\")\nuind <- (n/2+1):(n+1)\nrect(x[uind]-0.5w, 0, x[uind]+0.5w, y[uind], col = \"gold\")\n}\n\npar(mfrow = c(1, 2))\nplotHist(4, 0.5)\nabline(v = 2, lty = 2)\narrows(2-0.5*0.9, 0.17, 2, 0.17, col = \"red\", code = 3, length = 0.1, lwd = 2)\nplotHist(6, 0.5)\n\nf4(0.5)\nf6(0.5)\nf1000(0.5)\n\npar(mfrow = c(1, 2))\nplotHist(4, 0.6)\nplotHist(6, 0.6)\n\nf4(0.6)\nf6(0.6)\n\n• Why is 0.6 the tipping point instead of 0.5? I would naturally assume 0.5 would be the point at which the curves meet. Why is this not the case? – 1110101001 Feb 17 '14 at 1:33\n• @user2612743 this is a great question! Please see the updated post. – Tomas Feb 17 '14 at 9:37\n• As a sidenote, in real life, $p$ might be a function of $N$, as after 1000 trials, the player is likely to be fatigued and perform worse than when they started. – Jam Feb 21 '18 at 11:14\n\nThe probability of you getting at least half increases with the number of shots. E.g. with a probability of 2/3 per shot the probability of getting at least half the baskets increases as below.\n\nEdit it is important to point out that this only holds if by a \"pretty good basketball player\" you mean your chance of making a basket is somewhat better than evens (in the range 0.6 to 1 exclusive). This is shown very clearly in Hagen von Eitzen's answer.", null, "An intuitive way of looking at this is that it's like a diversification effect. With only a few baskets, you could get unlucky, just as you might if you tried to pick only a couple of stocks for an investment portfolio, even if you were a good stock picker. You increase the number of baskets -- or stocks -- and the role of chance is reduced and your skill shines through.\n\nFormally, assuming that\n\n• each throw is independent, and\n\n• you have the same probability $p$ of scoring on each throw\n\nyou can model the chance of scoring $b$ baskets out of $n$ using the binomial distribution\n\n$$\\mathbb{P}(b \\text{ from } n) = \\binom{n}{b} p^{b}(1-p)^{n-b}$$\n\nTo get the probability of scoring at least half of the $n$ baskets, you have to add up these probilities. E.g. for at least 2 out of 4 you want $\\mathbb{P}(2 \\text{ from } 4) + \\mathbb{P}(3 \\text{ from } 4) + \\mathbb{P}(4 \\text{ from } 4)$.\n\n• I guess you're making the assumption that a pretty good player has a better than half chance of making a shot. Reasonable, but just in case it was false the same reasoning would encourage you to go for 2/4. – Ben Millwood Feb 16 '14 at 15:41\n• @BenMillwood that's a very interesting point indeed! However the question does state the assumption that the player is good. – TooTone Feb 16 '14 at 15:42\n• In contrast a \"pretty good\" goalkeeper in soccer saves fewer than 50% of penalties (and a pretty good penalty taker scores more than that). So there is a general knowledge component to the question :-) – Steve Jessop Feb 16 '14 at 16:12\n• One simple intuitive way to understand this is that: the more shots you make the more likely you are to \"get closer to\" your \"real\" shooting percentage. That may possibly help! :) – Fattie Feb 17 '14 at 9:56\n• @laxxy it's because there are a small number of shots and they are hit-or-miss, i.e. the probability distribution is discrete, so there are step-changes in the cumulative probability as you consider the chance of making more shots. As the number of shots increases the distribution becomes more like a continuous distribution, so the steps become more like a ramp. In this comment I have summarised parts of this answer, which does a very good job of explaining the intuition and the details behind your question. – TooTone Feb 17 '14 at 12:55\n\nIt depends. If your probability to miss in a single try is $p$ (which should be low if you are a \"pretty good\" basketball player), then the probaility of making less than two out of four baskets (i.e. to lose the first kind of bet) is $$p_2=p^4+4p^3(1-p)=p^3(4-3p)$$ and for less than three out of six (i.e. to lose the second bet) is $$p_3=p^6+6p^5(1-p)+15p^4(1-p)^2=p^4(10p^2-24p+15).$$ We have $p_2<p_3$ iff $$p^3(4-3p)<p^4(10p^2-24p+15)$$ i.e. $$0<10p^6-24p^5+18p^4-4p^3=p^3(1-p)^2(10p-4).$$ In other words: The \"2 out of 4\" bet is to be preferred when $0<p<\\frac2{5}$ and \"3 out of 6\" is to be preferred when $\\frac{2}{5}<p<1$. For $p\\in\\{0,\\frac2{5},1\\}$ the bets are equivalent.\n\n• I may be mistaken, but shouldn't the third coefficient in $p_3$'s expression be 15, instead of 30? (where 15 is (6 choose 4)). Also, shouldn't the second to last sentence be the other way around? When p is small (i.e., the player is \"good\"), the player should choose the \"3 out of 6\" option, right? (See wolframalpha.com/input/… ) – Detached Laconian Feb 16 '14 at 16:43\n• The conclusion is the opposite of what the argumentation says. – durum Feb 16 '14 at 16:58\n• I make it 0.4 too. – TonyK Feb 16 '14 at 18:32\n• @TuringMachine yes, I was using my own definition of $p$. On Hagen's definition it is indeed 0.4. – TooTone Feb 16 '14 at 22:21\n• @HagenvonEitzen Cool. :) One more thing though: I think the conclusion is still backwards; namely, the \"3 out of 6\" is to be preferred when $0 < p < 2/5$. – Detached Laconian Feb 17 '14 at 0:07\n\nI would want to know what my long term average is.\n\nIf I make 333 out of 999 over the course of a season, I know that the more shots I take, the more certain I am that regression to the mean will make it less likely that I'll hit a ratio or 2:3 or better. I stand a better chance of two or three lucky shots out of three than I do 4 or more out of 6, and certainly better than 40 or more out of 60. More shots is bad because good luck can't save me from my long-term average.\n\nOn the other hand, if I make 750 out of 1000 in a season, regression to the mean works for me if I take more shots. Getting 1 or none out of 3 wouldn't be any surprise, but 3 or less out of 6, less so and 39 or less out of 60 would be a bad day indeed.\n\nAssuming you have a $50/50$ chance of missing or making it (though it could be anything else really, the calculations would change accordingly), we can view both scenarios as binary permutations, as a means of figuring out the answer. Now in english:\n\nLets say you have $4$ balls (this is the first case), you would be required (in order to succeed) to make $2$ shots. Lets define: when you make a shot - that ball was a red ball. When you miss, that ball was a blue ball. Now it doesn't matter in which order you make your shots right? You could make the first two, then miss two, make one, miss one, make one, miss the last, etc. All that matters is that you have two blue balls and two red balls.\n\nSo the question is how many ways can you arrange $2$ red balls and $2$ blue balls? Well this is just a combination: namely $\\binom{4}{2}$ (reads $4$ choose $2$): i.e. the answer is $6$. But lets see, there are other outcomes! For example you could be making all the shots swell (so $4/4$) but I guess we only wan't ($2/4$) so the chance that we get the outcome we desire is: (the $2$ red $2$ blue ball permutations)/(total possibilities). i.e $\\binom{4}{2}/(\\binom{4}{0}+\\binom{4}{1}+\\binom{4}{2}+\\binom{4}{3}+\\binom{4}{4})$. which is: $1/4$. So for the first case (given that missing and making the shot share the same probability), you making $2$ out of four shots, is precisely going to happen $1/4$ of the time.\n\nSimilarly, for the $3/6$ we have: $\\binom{6}{3}/ (\\binom{6}{0}+\\binom{6}{1}+…) = 0.3125$. Notice $0.25 < 0.3125$ So clearly: you would have a better chance of making precisely $3/6$ shots than $2/4$ shots. An intuitive explanation behind this is that since $6$ shots are more than $4$, if you mess up a bit on the $6$ shot case, it's more likely that you can still make up for it, than with $4$ shots. Of course, if your question was: making at least $3/6$ shots versus making at least $2/4$ shots, then the calculations would be a bit different. How bout you find out what your odds are in that case?\n\nBtw, if the $\\binom{n}{k}$ business is new to you, I suggest you look up permutations and combinations (and their link to probability)!\n\n• Exactly 3 of 6 or at least 3 of 6 is a distinction without a difference, since in the former case, you can simply shoot until you get your 3, and then any shots you have left, you intentionally throw the ball nowhere near the hoop. – Matthew Najmon Feb 17 '14 at 19:50\n• I left my answer intentionally in that form as a means of promoting OP to solve the more \"realistic\" case. – Just_a_fool Feb 17 '14 at 20:20\n• Your computation for $n=4$ is wrong. You should be getting $p_4=3/8=6/2^4>5/2^4=20/2^6=p_6$ instead as expected. The more \"options\" you have, the less likely you are to hit the very specific number. – A.S. Feb 18 '16 at 7:53\n\nmore intuitively than the other answers:\n\nlet's say you succeed 60% of the time in average. due to the law of large numbers, the more you shoot the more likely your success frequency will be to approach this ration 0.6.\n\nshooting once is all or nothing. shooting infinitely many times is the certainty of success.\n\nSame goes for the opposite. If your real success rate is 0.4, you'd better try the least possible shoots.\n\nThe expectation of your average success remains the same, but not the variance/std deviation.\n\nAssume you make your shots with probability $p$ independent of each other. Then, the probability of making $k$ shots out of $n$ is $\\binom{n}{k} p^k (1-p)^{n-k}$ (That is, the number of shots you make out of $n$ baskets follows a Binomial(n,p) distribution).\n\nThe best free throw shooter in the NBA currently is Steve Nash, with $p=0.9041$. Calculate the probabilities for $n=4, k=2$ and $n=6,k=3$ and compare. You want the choice which gives you the higher probability.\n\noctave:85> bincoeff(4,2)p^2(1-p)^2\n\nans = 0.045105\n\noctave:86> bincoeff(6,3)p^3(1-p)^3\n\nans = 0.013036\n\nSo, you're better off taking the 2 out of the 4 shots in this model, if you're Steven Nash.\n\nThe problem is slightly different when you need to make at least 2 out of 4 shots versus at least 3 out of 6 shots. In this case, you need to sum the tail probabilities and compare them: The probability of making at least $k$ shots out of $n$ under our model is $\\sum_{i=k}^n \\binom{n}{i}p^i (1-p)^{n-i}$. I will leave you to calculate which one you're better off with in this case ($n=4,k=2$ and $n=6,k=3$ once again) [ The wording of the problem made it sound like you needed the exactly 2 out of 4 shots or exactly 3 out of 6 shots when I first read it ].\n\n• This assumes the task is to make exactly the number given, not at least the number given. – Ross Millikan Feb 16 '14 at 15:31\n• And that's exactly whats in the answer? Thats the binomial PMF , not the CDF. Edit: Ah, I see what you're saying. I'll update the answer. – Batman Feb 16 '14 at 15:32\n• The \"break even\" point is at $p=\\frac{21}{25}$. – Hagen von Eitzen Feb 16 '14 at 15:32\n• @Hagen: it's $p=\\frac35$ if you do the sums right. – TonyK Feb 16 '14 at 18:33\n• Exactly 3 of 6 or at least 3 of 6 is a distiction without a difference, since in the former case, you can simply shoot until you get your 3, and then any shots you have left, you intentionally throw the ball nowhere near the hoop. – Matthew Najmon Feb 16 '14 at 22:20\n\nAccording to the law of large number ( https://en.m.wikipedia.org/wiki/Law_of_large_numbers ). the more events you consider the closer will be the outcome to the expected result. In your case the expected outcome depends on actuall ability of you and your opponents in basketball playing. so if you know that you are truely a better performer to your opponents you should choose 3/6. otherwise 2/4." ]	[ null, "https://i.stack.imgur.com/Naze3.png", null, "https://i.stack.imgur.com/W6WZV.png", null, "https://i.stack.imgur.com/vsdVi.png", null, "https://i.stack.imgur.com/gDZBh.png", null, "https://i.stack.imgur.com/hcPVU.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8549476,"math_prob":0.9941943,"size":3633,"snap":"2021-31-2021-39","text_gpt3_token_len":1343,"char_repetition_ratio":0.10856985,"word_repetition_ratio":0.03773585,"special_character_ratio":0.39664188,"punctuation_ratio":0.19164883,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9976189,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,8,null,8,null,8,null,8,null,7,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-24T04:17:46Z\",\"WARC-Record-ID\":\"<urn:uuid:ceb2ff13-3b8f-4b35-9503-2a79cd14d4ad>\",\"Content-Length\":\"265566\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a1a16363-1296-4180-9542-024862403bea>\",\"WARC-Concurrent-To\":\"<urn:uuid:0e446efa-8419-4a47-91da-bfbf6d4e0065>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/678515/probability-2-4-vs-3-6/678530\",\"WARC-Payload-Digest\":\"sha1:6A4IPN2U4QX63OSSZK6XAA4ZYZ5YCBMD\",\"WARC-Block-Digest\":\"sha1:QOV337SYNFXDRBPVYXCNTVPSZLUFEBGK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046150129.50_warc_CC-MAIN-20210724032221-20210724062221-00382.warc.gz\"}"}
https://www.etutorworld.com/math/pre-calculus-tutoring.html	[ "# Precalculus Tutoring Online\n\nHome \| Online Math Tutoring \| Precalculus Tutor\n\n#### Schedule a Free Session", null, "## Learn about Linear, Non-Linear Equations, Factorization and more:\n\nSimplify algebraic and trigonometric expressions with Precalculus teachers who understand your abilities and requirements. Our aim is to enable you to understand precalculus concepts and solve complicated precalculus problems.\n\nMany advanced math concepts and theories are not covered in algebra, geometry and other courses in a student’s high school math curriculum. Precalculus commonly focuses on properties of functions along with the study of trigonometric, logarithmic and exponential functions. It includes topics about sequences, limits and other concepts crucial to the study of calculus.\n\nAs a regular student, you’ll be able to schedule tutoring sessions with your precalculus tutor at a time convenient to you.\n\nConnect to eTutorWorld and you will have a tutor to help you with Functions, Vectors, Matrices, Parameterizations, Conic Sections and of course the functioning of your Graphing Calculator. Our Learning by Design Methodology helps you understand precalculus concepts better, improve grades and prepare you for a bright and promising future.\n\nNo credit card is required, nor are you under any obligation to make a purchase.\n\nJust schedule the FREE TRIAL session to meet our expert tutor & get help.\n\n#### Inequalities\n\n• Systems of Inequalities\n• Graph Absolute Value Inequalities\n\n#### Absolute Value Equations\n\n• Transformations\n• Dilations\n• Reflections\n\n#### Complex Numbers\n\n• Imaginary Numbers\n• Operations with Complex Numbers\n• Graphing Complex Numbers\n\n#### Systems of Linear Equations\n\n• Solving Systems of Linear Equations\n\n• Greatest Common Factor\n• Multiplication Property Of Zero\n• Solving Quadratics not in the Standard Form\n• Parabolas\n• Completing the Square\n\n#### Geometry for Precalculus\n\n• Area and Circumference of a Circle\n• Midpoint and Distance Formula\n• The Pythagorean Theorem\n\n#### Functions\n\n• Domain and Range of a Function\n• Power of a Function\n• Operations with Functions\n• Graphing Functions\n• One-On-One Functions\n• Polynomial Functions\n\n#### Symmetry\n\n• Graphic Symmetry\n• Algebraic Symmetry\n• Numerical Symmetry\n\n#### Polynomials\n\n• Binomial Theorem\n• Factoring Polynomials in Quadratic Form\n• Dividing Polynomials\n• Long and Synthetic Division\n• Remainder and Factor Theorem\n\n#### Rational Expressions\n\n• Operations of Rational Expressions\n• Solve Rational Equations\n\n#### Exponential and Logarithmic Functions\n\n• Exponential growth and decay\n• Natural base\n• Logarithmic properties\n• Inverse of logarithmic functions\n\n#### Trigonometry Functions & Graphs\n\n• Sine and cosine Functions\n• Unit Circles\n• Special Right Triangles\n• Graphing sine and cosine transformations\n• Determining trigonometric identities using graphs\n\n#### Trigonometry Equations\n\n• Inverse Trigonometric Functions\n• Sinusoidal Functions\n\n#### Trigonometry Identities & Applications\n\n• Basic Trigonometric Identities\n• Pythagorean Identities\n• Double-Angle Identities\n• Law of Cosines\n• Law of Sines\n• Formulas for sine, cosine and tangent\n\n#### Vectors, Matrices and Determinants\n\n• Operations on Vectors\n• Operations on Matrices\n• Inverse of a matrix\nDeterminants\nGaussian elimination\nLinear Programming\n\n#### Mathematical Sequences and Series\n\n• The sigma and factorial notation.\n• Arithmetic sequences\n• Geometric sequences\n• Mathematical Induction\n• Binomial Theorem\n• Pascal’s Triangle\n\n#### Polar Coordinates and Parameterization and Conic Sections\n\n• Conversion between polar and rectangular coordinates.\n• Graph of functions in polar coordinates.\n• De Moivre’s Theorem\n• Parametric and Rectangular forms of equations.\n• Locus\n• Graphs of parabolas, ellipses and hyperbolas\n\n#### Limits & Continuity\n\n• Intermediate Value Theorem\n• Graphs to Define Limits\n• Determine Limits of Functions\n• Limits for Asymptotes and Infinity\n\nPractice is essential in precalculus and will lead to successful performance on exams. Tests and quizzes are given every precalculus topic studied.\n\n### Online Tutoring Pricing\n\nExpert 1-on-1 Personal Tutoring, Now Affordable to All. Every online tutoring session has 50 to 55 minutes of teaching time.\n\nNew Student Special (5 sessions)1 Month\\$79\\$89\n1 session1 Month\\$21\\$23\n5 sessions3 months\\$99\\$109\n10 sessions3 months\\$189\\$209\n15 sessions3 months\\$279\\$309\n20 sessions4 months\\$369\\$399\n50 sessions6 months\\$849\\$899\n\n#### Schedule a FREE precalculus session today and get started on your prep for future calculus courses.\n\nDuring your FREE tutoring session, our tutor will help you with basic and advanced topics in precalculus, give homework help and prepare you for upcoming precalculus tests.\n\nNo credit card is required, nor are you under any obligation to make a purchase. Just schedule the FREE TRIAL lesson to meet a tutor & get help on any topic you want!\n\nOur Learning by Design methodology focuses exclusively on the individual student. Our tutors are specially trained to identify and diagnose the needs and interests of each student and plan lessons accordingly.\n\n#### Find more about our Personalized Online Tutoring Learning Packages.\n\n+1-269-763-4602\n\n+1-269-763-5024\n\nResearch has proven that personal online tutoring not just cements school learning, it helps build student confidence. eTutorWorld provides the best K-12 Online Tutoring Services so you can learn from the comfort and safety of your home at an affordable cost.\n\nBe it an exam, class test or a quiz, eTutorWorld’s Math, Science and English tutors are responsible for your academic progress. Meet your personal coach at your convenient day and time to get help for Grade 3-12 Math, Science and English subjects and AP, SAT, SSAT and SCAT Test Prep help and test practice. All our tutors are graduates and bring with them years of teaching experience to the tutoring lessons.\n\nOur ‘Learning by Design’ methodology makes sure that each student is at the center of the teaching-learning process. All tutoring sessions start with a question to the student ‘What do you want to learn today?’ Hence, tutors diagnose your skills and recognize your requirements before the actual tutoring happens. Post every tutoring session, an individualized worksheet is emailed to the student to assimilate learned concepts. Regular formative assessments are used to evaluate a student’s understanding of the subject.\n\nThe state of the art technology used is stable, user friendly and safe. All you need is a computer or a tablet and an internet connection. The easy-to-use web conferencing software requires a one time download, using which the student can talk and chat to the tutor, annotate on an interactive shared whiteboard or even share documents, assignments or worksheets.\n\nAll tutoring sessions are recorded and made available for a month so you can review concepts taught.\n\nEmail or call our support team with any issues or questions – we are here for you 24X7.\n\nAlso, download free printable math and science worksheets in pdf format and solve SCAT and SSAT Practice Tests online. Sign up for a Free Trial Lesson Today!\n\nThousands have taken the eTutorWorld Advantage – what are you waiting for?" ]	[ null, "data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%200%200'%3E%3C/svg%3E", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.80653995,"math_prob":0.71246976,"size":7409,"snap":"2019-51-2020-05","text_gpt3_token_len":1639,"char_repetition_ratio":0.13180284,"word_repetition_ratio":0.028243601,"special_character_ratio":0.19975705,"punctuation_ratio":0.06816221,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95057136,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-15T01:13:17Z\",\"WARC-Record-ID\":\"<urn:uuid:fb4a0046-6469-4a35-b243-c7355f739b76>\",\"Content-Length\":\"149626\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:125a5ce1-2304-4954-882a-a051972be71c>\",\"WARC-Concurrent-To\":\"<urn:uuid:81cb3e8b-adaf-49d3-b681-2324806a6172>\",\"WARC-IP-Address\":\"192.124.249.70\",\"WARC-Target-URI\":\"https://www.etutorworld.com/math/pre-calculus-tutoring.html\",\"WARC-Payload-Digest\":\"sha1:JHTG3WEZBB2576NJYGTBXDCHLY7URUBQ\",\"WARC-Block-Digest\":\"sha1:KQ757Q6KQKPXWAXZNDVGXTSBZLYPXVG3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575541297626.61_warc_CC-MAIN-20191214230830-20191215014830-00089.warc.gz\"}"}
https://answers.everydaycalculation.com/multiply-fractions/5-10-times-7-3	[ "Solutions by everydaycalculation.com\n\nMultiply 5/10 with 7/3\n\n1st number: 5/10, 2nd number: 2 1/3\n\nThis multiplication involving fractions can also be rephrased as \"What is 5/10 of 2 1/3?\"\n\n5/10 × 7/3 is 7/6.\n\nSteps for multiplying fractions\n\n1. Simply multiply the numerators and denominators separately:\n2. 5/10 × 7/3 = 5 × 7/10 × 3 = 35/30\n3. After reducing the fraction, the answer is 7/6\n4. In mixed form: 11/6\n\nMathStep (Works offline)", null, "Download our mobile app and learn to work with fractions in your own time:\nAndroid and iPhone/ iPad\n\n×" ]	[ null, "https://answers.everydaycalculation.com/mathstep-app-icon.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8150808,"math_prob":0.9875278,"size":482,"snap":"2019-43-2019-47","text_gpt3_token_len":219,"char_repetition_ratio":0.20292887,"word_repetition_ratio":0.0,"special_character_ratio":0.49377593,"punctuation_ratio":0.070866145,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9712378,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-21T18:44:40Z\",\"WARC-Record-ID\":\"<urn:uuid:0157b78b-5fe0-4363-b05e-6e04d6c008c0>\",\"Content-Length\":\"8106\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:773cf953-ee94-4cb5-a3e2-2cf90bfd8b79>\",\"WARC-Concurrent-To\":\"<urn:uuid:d8eb3b5f-9c32-44ff-98e0-29c3873b53f7>\",\"WARC-IP-Address\":\"96.126.107.130\",\"WARC-Target-URI\":\"https://answers.everydaycalculation.com/multiply-fractions/5-10-times-7-3\",\"WARC-Payload-Digest\":\"sha1:MT7TT6QBYMBINSEJLIK2ZEWDBDELDRRG\",\"WARC-Block-Digest\":\"sha1:XBBZ6WNZ7LQ7S3PPNWVZZHPBEMOJWNNZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570987781397.63_warc_CC-MAIN-20191021171509-20191021195009-00282.warc.gz\"}"}
https://trialsjournal.biomedcentral.com/articles/10.1186/s13063-021-05142-7/tables/3	[ "# Table 1 Outcome measures in the MONITOR-PsA cohort study\n\nProcedures Visits\nBaseline cohort\nWeek 0 Week 12 Week 24 Week 36 Week 48 Week 72 Week 96 Annually (after 96 weeks)\nInformed consent X\nDemographics X\nMedical history X X X X X X X X\nMedications X X X X X X X X\nPhysical examination X X X X X X X X\nRoutine blood tests (FBC, U&Es, LFTs, CRP and eGFR) X X X X X X X X\nImmunology (RF, ACPA, ANA) X\nRadiographs of hands/feet/spine X X X X\nEligibility assessment X\nAdherence X X X X X X X\n68/66 Joint count X X X X X X X X\nLeeds and SPARCC enthesitis index X X X X X X X X\nDactylitis count X X X X X X X X\nPsoriasis assessment (PASI and BSA) X X X X X X X X\nPhysician VAS X X X X X X X X\nMetrology with BASMI (if axial involvement) X X X X\nPatient questionnaires (VAS, HAQ, SF36, PsAID, WPAI, BASDAI, BASFI) X X X X X X X X\nTSQM questionnaire X X X X X X X\nEQ-5D-5L questionnaire X X X X X X\nHealthcare utilisation data X X X X X\nAdverse event assessments X X X X X X X X", null, "" ]	[ null, "https://trialsjournal.biomedcentral.com/track/article/10.1186/s13063-021-05142-7", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6249296,"math_prob":0.53471607,"size":1301,"snap":"2022-27-2022-33","text_gpt3_token_len":667,"char_repetition_ratio":0.3299923,"word_repetition_ratio":0.23553719,"special_character_ratio":0.3535742,"punctuation_ratio":0.05204461,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.992167,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-07T00:33:34Z\",\"WARC-Record-ID\":\"<urn:uuid:27fe2bcc-7c3f-43e9-a946-6a61c53e5574>\",\"Content-Length\":\"181189\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1b8cb386-eeee-4811-a47d-b717234da6c9>\",\"WARC-Concurrent-To\":\"<urn:uuid:6cf327ce-ecae-4b56-97d0-f2c094cd5bd8>\",\"WARC-IP-Address\":\"146.75.36.95\",\"WARC-Target-URI\":\"https://trialsjournal.biomedcentral.com/articles/10.1186/s13063-021-05142-7/tables/3\",\"WARC-Payload-Digest\":\"sha1:QPGMAPYTP7RUWAVR7NBVCUU5L7NM6LUY\",\"WARC-Block-Digest\":\"sha1:4BPSSWNZHKJPSZOV6OM3KGKW5KMJVT62\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104683020.92_warc_CC-MAIN-20220707002618-20220707032618-00786.warc.gz\"}"}
https://au.mathworks.com/help/physmod/sps/powersys/ug/thyristor-based-static-var-compensator.html	[ "## Thyristor-Based Static Var Compensator\n\n### Introduction\n\nThe example described in this section illustrates application of Simscape™ Electrical™ Specialized Power Systems software to study the steady-state and dynamic performance of a static var compensator (SVC) on a transmission system. The SVC is a shunt device of the Flexible AC Transmission Systems (FACTS) family using power electronics. It regulates voltage by generating or absorbing reactive power.\n\nDue to low frequencies of electromechanical oscillations in large power systems (typically 0.02 Hz to 2 Hz), this type of study usually requires simulation times of 30–40 seconds or more.\n\nThe SVC model described in this example is rather a detailed model of a particular SVC topology (using thyristor-controlled reactor (TCR) and thyristor-switched capacitors (TSCs)) with full representation of power electronics. This type of model requires discrete simulation at fixed time steps (50 µs in this case) and it is used typically for studying the SVC performance on a much smaller time range (a few seconds). Typical applications include optimizing of the control system, impact of harmonics, transients and stresses on power components during faults.\n\n### Description of the SVC\n\nThe single-line diagram of the modeled SVC is shown on Single-Line Diagram of the SVC. It represents a 300 Mvar SVC connected on a 735 kV transmission system.\n\nThis example is available in the `power_svc_1tcr3tsc` model. Load this model and save it in your working directory as `case2` to allow further modifications to the original system. This model is shown on SPS Model of the 300 Mvar SVC on a 735 kV Power System (power_svc_1tcr3tscs).\n\nSingle-Line Diagram of the SVC", null, "SPS Model of the 300 Mvar SVC on a 735 kV Power System (power_svc_1tcr3tscs)", null, "#### SVC Power Components\n\nThe SVC consists of a 735 kV/16 kV, 333 MVA coupling transformer, one 109 Mvar TCR bank and three 94 Mvar TSC banks (TSC1 TSC2 TSC3) connected on the secondary side of the transformer.\n\nSwitching the TSCs in and out allows a discrete variation of the secondary reactive power from zero to 282 Mvar capacitive (at 16 kV) by steps of 94 Mvar, whereas phase control of the TCR allows a continuous variation from zero to 109 Mvar inductive. Taking into account the leakage reactance of the transformer (0.15 pu), the SVC equivalent susceptance seen from the primary side can be varied continuously from -1.04 pu/100 MVA (fully inductive) to +3.23 pu/100 Mvar (fully capacitive).\n\nThe SVC Controller monitors the primary voltage and sends appropriate pulses to the 24 thyristors (6 thyristors per three-phase bank) to obtain the susceptance required by the voltage regulator.\n\nEach three-phase bank is connected in delta so that, during normal balanced operation, the zero-sequence tripplen harmonics (3rd, 9th,...) remain trapped inside the delta, thus reducing harmonic injection into the power system.\n\nThe power system is represented by an inductive equivalent (6000 MVA short circuit level) and a 200-MW load. The internal voltage of the equivalent system can be varied by means of a Three-Phase Programmable Voltage Source block to observe the SVC dynamic response to changes in system voltage.\n\n#### SVC Control System\n\nSVC Controller Model", null, "The SVC control system consists of the following four main modules:\n\n• Measurement System measures the positive-sequence primary voltage. This system uses discrete Fourier computation technique to evaluate fundamental voltage over a one-cycle running average window. The voltage measurement unit is driven by a phase-locked loop (PLL) to take into account variations of system frequency.\n\n• Voltage Regulator uses a PI regulator to regulate primary voltage at the reference voltage (1.0 pu specified in the SVC Controller block menu). A voltage droop is incorporated in the voltage regulation to obtain a V-I characteristic with a slope (0.01 pu/100 MVA in this case). Therefore, when the SVC operating point changes from fully capacitive (+300 Mvar) to fully inductive (-100 Mvar) the SVC voltage varies between 1-0.03=0.97 pu and 1+0.01=1.01 pu.\n\n• Distribution Unit uses the primary susceptance Bsvc computed by the voltage regulator to determine the TCR firing angle α and the status (on/off) of the three TSC branches. The firing angle α as a function of the TCR susceptance BTCR is implemented by a look-up table from the equation\n\n`${B}_{TCR}=\\frac{2\\left(\\pi -\\alpha \\right)+\\mathrm{sin}\\left(2\\alpha \\right)}{\\pi }$`\n\nwhere BTCR is the TCR susceptance in pu of rated TCR reactive power (109 Mvar)\n\n• Firing Unit consists of three independent subsystems, one for each phase (AB, BC and CA). Each subsystem consists of a PLL synchronized on line-to-line secondary voltage and a pulse generator for each of the TCR and TSC branches. The pulse generator uses the firing angle α and the TSC status coming from the Distribution Unit to generate pulses. The firing of TSC branches can be synchronized (one pulse is sent at positive and negative thyristors at every cycle) or continuous. The synchronized firing mode is usually the preferred method because it reduces harmonics faster. Verify that the Synchronized firing mode has been selected in the Firing Unit dialog box.\n\n### Steady-State and Dynamic Performance of the SVC\n\nNow observe the steady-state waveforms and the SVC dynamic response when the system voltage is varied. Run the simulation and observe waveforms on the SVC Scope block. These waveforms are reproduced below.\n\nWaveforms Illustrating SVC Dynamic Response to System Voltage Steps", null, "Initially the source voltage is set at 1.004 pu, resulting in a 1.0 pu voltage at SVC terminals when the SVC is out of service. As the reference voltage Vref is set to 1.0 pu, the SVC is initially floating (zero current). This operating point is obtained with TSC1 in service and TCR almost at full conduction (α = 96 degrees).\n\nAt t=0.1s voltage is suddenly increased to 1.025 pu. The SVC reacts by absorbing reactive power (Q=-95 Mvar) to bring the voltage back to 1.01 pu. The 95% settling time is approximately 135 ms. At this point all TSCs are out of service and the TCR is almost at full conduction (α = 94 degrees).\n\nAt t=0.4 s the source voltage is suddenly lowered to 0.93 pu. The SVC reacts by generating 256 Mvar of reactive power, thus increasing the voltage to 0.974 pu.\n\nAt this point the three TSCs are in service and the TCR absorbs approximately 40% of its nominal reactive power (α =120 degrees).\n\nObserve on the last trace of the scope how the TSCs are sequentially switched on and off. Each time a TSC is switched on the TCR α angle changes from 180 degrees (no conduction) to 90 degrees (full conduction). Finally, at t=0.7 s the voltage is increased to 1.0 pu and the SVC reactive power is reduced to zero.\n\nYou may open the Signal & Scopes subsystem to observe additional waveforms. The TCR voltage and current in branch AB as well as thyristors pulses are displayed on the TCR AB scope. The figure below zooms on three cycles when the firing angle α is 120 degrees.\n\nSteady-State Voltage and Current in TCR AB", null, "### Misfiring of TSC1\n\nThe final case study simulates a TSC misfiring.\n\nEach time a TSC is switched off a voltage remains trapped across the TSC capacitors. If you look at the TSC1 Misfiring scope inside the Signals & Scope subsystem, you can observe the TSC1 voltage (first trace) and the TSC1 current (second trace) for branch AB. The voltage across the positive thyristor (thyristor conducting the positive current) is shown on the third trace and the pulses sent to this thyristor are shown on the fourth trace. Notice that the positive thyristor is fired at maximum negative TSC voltage, when the valve voltage is minimum.\n\nIf by mistake the firing pulse is not sent at the right time, very large overcurrents can be observed in the TSC valves. Look inside the SVC Controller block for how a misfiring can be simulated on TSC1. A Timer block and an OR block are used to add pulses to the normal pulses coming from the Firing Unit.\n\nOpen the Timer block menu and remove the 100 multiplication factor. The timer is now programmed to send a misfiring pulse lasting one sample time at time t= 0.121 s.\n\nRestart simulation. Waveforms observed on the TSC1 Misfiring scope are reproduced below.\n\nTSC Voltages and Current Resulting from Misfiring on TSC1", null, "Observe that the misfiring pulse is sent when the valve voltage is maximum positive immediately after the TSC has blocked. This thyristor misfiring produces a large thyristor overcurrent (18 kA or 6.5 times the nominal peak current). Also, immediately after the thyristor has blocked, the thyristor voltage reaches 85 kV (3.8 times the nominal peak voltage). To prevent such overcurrents and overvoltages, thyristor valves are normally protected by metal oxide arresters (not simulated here)." ]	[ null, "https://au.mathworks.com/help/physmod/sps/powersys/ug/pe_applications6a.gif", null, "https://au.mathworks.com/help/physmod/sps/powersys/ug/pe_mdl1.gif", null, "https://au.mathworks.com/help/physmod/sps/powersys/ug/pe_mdl2.gif", null, "https://au.mathworks.com/help/physmod/sps/powersys/ug/pe_plot1.gif", null, "https://au.mathworks.com/help/physmod/sps/powersys/ug/pe_plot2.gif", null, "https://au.mathworks.com/help/physmod/sps/powersys/ug/pe_plot3.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.89071035,"math_prob":0.9133859,"size":8686,"snap":"2022-27-2022-33","text_gpt3_token_len":2002,"char_repetition_ratio":0.13695002,"word_repetition_ratio":0.022487702,"special_character_ratio":0.2166705,"punctuation_ratio":0.07485964,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95172626,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-05T13:21:57Z\",\"WARC-Record-ID\":\"<urn:uuid:8e97fefe-2e4f-4e52-8545-4c35fa868059>\",\"Content-Length\":\"82441\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5d38cf6b-f9e6-4255-9a07-4c29ddced935>\",\"WARC-Concurrent-To\":\"<urn:uuid:7630e30f-8c45-4477-ab7f-efff6c0ad862>\",\"WARC-IP-Address\":\"23.1.9.244\",\"WARC-Target-URI\":\"https://au.mathworks.com/help/physmod/sps/powersys/ug/thyristor-based-static-var-compensator.html\",\"WARC-Payload-Digest\":\"sha1:NAK2UJ3YSOAW7AFOPOYX22XRELDON7JV\",\"WARC-Block-Digest\":\"sha1:DCBSSS54BK57ZHCZHR4TINEKFSBKSCPR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104576719.83_warc_CC-MAIN-20220705113756-20220705143756-00694.warc.gz\"}"}
https://developer.arm.com/docs/ddi0597/g/simd-and-floating-point-instructions-alphabetic-order/vcle-register-vector-compare-less-than-or-equal-an-alias-of-vcge-register	[ "You copied the Doc URL to your clipboard.\n\n## VCLE (register)\n\nVector Compare Less Than or Equal takes each element in a vector, and compares it with the corresponding element of a second vector. If the first is less than or equal to the second, the corresponding element in the destination vector is set to all ones. Otherwise, it is set to all zeros.\n\nThis is a pseudo-instruction of VCGE (register). This means:\n\n• The encodings in this description are named to match the encodings of VCGE (register).\n• The assembler syntax is used only for assembly, and is not used on disassembly.\n• The description of VCGE (register) gives the operational pseudocode for this instruction.\n\nIt has encodings from the following instruction sets: A32 ( A1 and A2 ) and T32 ( T1 and T2 ) .\n\n### A1\n\n 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 1 1 1 0 0 1 U 0 D size Vn Vd 0 0 1 1 N Q M 1 Vm\n\n#### 64-bit SIMD vector (Q == 0)\n\nVCLE{<c>}{<q>}.<dt> {<Dd>, }<Dn>, <Dm>\n\nis equivalent to\n\nVCGE{<c>}{<q>}.<dt> <Dd>, <Dm>, <Dn>\n\n#### 128-bit SIMD vector (Q == 1)\n\nVCLE{<c>}{<q>}.<dt> {<Qd>, }<Qn>, <Qm>\n\nis equivalent to\n\nVCGE{<c>}{<q>}.<dt> <Qd>, <Qm>, <Qn>\n\n### A2\n\n 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 1 1 1 0 0 1 1 0 D 0 sz Vn Vd 1 1 1 0 N Q M 0 Vm\n\n#### 64-bit SIMD vector (Q == 0)\n\nVCLE{<c>}{<q>}.<dt> {<Dd>, }<Dn>, <Dm>\n\nis equivalent to\n\nVCGE{<c>}{<q>}.<dt> <Dd>, <Dm>, <Dn>\n\n#### 128-bit SIMD vector (Q == 1)\n\nVCLE{<c>}{<q>}.<dt> {<Qd>, }<Qn>, <Qm>\n\nis equivalent to\n\nVCGE{<c>}{<q>}.<dt> <Qd>, <Qm>, <Qn>\n\n### T1\n\n 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 1 1 U 1 1 1 1 0 D size Vn Vd 0 0 1 1 N Q M 1 Vm\n\n#### 64-bit SIMD vector (Q == 0)\n\nVCLE{<c>}{<q>}.<dt> {<Dd>, }<Dn>, <Dm>\n\nis equivalent to\n\nVCGE{<c>}{<q>}.<dt> <Dd>, <Dm>, <Dn>\n\n#### 128-bit SIMD vector (Q == 1)\n\nVCLE{<c>}{<q>}.<dt> {<Qd>, }<Qn>, <Qm>\n\nis equivalent to\n\nVCGE{<c>}{<q>}.<dt> <Qd>, <Qm>, <Qn>\n\n### T2\n\n 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 1 1 1 1 1 1 1 0 D 0 sz Vn Vd 1 1 1 0 N Q M 0 Vm\n\n#### 64-bit SIMD vector (Q == 0)\n\nVCLE{<c>}{<q>}.<dt> {<Dd>, }<Dn>, <Dm>\n\nis equivalent to\n\nVCGE{<c>}{<q>}.<dt> <Dd>, <Dm>, <Dn>\n\n#### 128-bit SIMD vector (Q == 1)\n\nVCLE{<c>}{<q>}.<dt> {<Qd>, }<Qn>, <Qm>\n\nis equivalent to\n\nVCGE{<c>}{<q>}.<dt> <Qd>, <Qm>, <Qn>\n\n### Assembler Symbols\n\n Is the 64-bit name of the second SIMD&FP source register, encoded in the \"M:Vm\" field.\n Is the 64-bit name of the first SIMD&FP source register, encoded in the \"N:Vn\" field.\n Is the 128-bit name of the second SIMD&FP source register, encoded in the \"M:Vm\" field as 2.\n Is the 128-bit name of the first SIMD&FP source register, encoded in the \"N:Vn\" field as 2.\n For encoding A1 and A2: see Standard assembler syntax fields. This encoding must be unconditional. For encoding T1 and T2: see Standard assembler syntax fields.\n\n<dt> For encoding A1 and T1: is the data type for the elements of the operands, encoded in U:size:\nU size <dt>\n0 00 S8\n0 01 S16\n0 10 S32\n1 00 U8\n1 01 U16\n1 10 U32\nFor encoding A2 and T2: is the data type for the elements of the vectors, encoded in sz:\nsz <dt>\n0 F32\n1 F16\n Is the 128-bit name of the SIMD&FP destination register, encoded in the \"D:Vd\" field as *2.\n\nIs the 64-bit name of the SIMD&FP destination register, encoded in the \"D:Vd\" field.\n\n### Operation\n\nThe description of VCGE (register) gives the operational pseudocode for this instruction." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5905319,"math_prob":0.9593761,"size":3045,"snap":"2021-04-2021-17","text_gpt3_token_len":1226,"char_repetition_ratio":0.14041434,"word_repetition_ratio":0.60066557,"special_character_ratio":0.4811166,"punctuation_ratio":0.08895265,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96550214,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-28T15:19:06Z\",\"WARC-Record-ID\":\"<urn:uuid:a0370d76-5454-481a-82ff-6839cba4192f>\",\"Content-Length\":\"152342\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:97bbfe9e-54d7-44fb-a5ba-a7fd90751d50>\",\"WARC-Concurrent-To\":\"<urn:uuid:75fdc0e4-5886-4b1d-a078-9a2c6cf0aefa>\",\"WARC-IP-Address\":\"23.199.177.163\",\"WARC-Target-URI\":\"https://developer.arm.com/docs/ddi0597/g/simd-and-floating-point-instructions-alphabetic-order/vcle-register-vector-compare-less-than-or-equal-an-alias-of-vcge-register\",\"WARC-Payload-Digest\":\"sha1:MJPVAF7BEFQDZ4PSQTHIZF7Q3ALTB3QV\",\"WARC-Block-Digest\":\"sha1:RK2KRYCBOWTWDWGI4PVOOUTVCZC32SK2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610704847953.98_warc_CC-MAIN-20210128134124-20210128164124-00798.warc.gz\"}"}
https://www.run4unblocked.com/how-to-know-the-order-of-the-reaction.html	[ "# How To Know The Order Of The Reaction?", null, "Order of the reaction is the part of the chemical reaction which determines the relationship between the rate of a chemical reaction and the concentration of the species involved in the chemical reaction. Order of the reaction depends on the chemical concentration of the reactants involved in the chemical reaction.\n\nThe first-order reaction dependents entirely on the concentration of one of the reactants involved in the chemical reaction. Some characteristics of the reaction in order of reaction are listed below.\n\n• The number of species of reactants involved along with the concentration of the reactants directly affects the rate of reaction.\n• The value of the order of reaction can be in the form of a fraction or an integer. It can even have a value of zero.\n• The order of a reaction is not dependent on the stoichiometric coefficients that are corresponding to each species in the balanced chemical reaction.\n• It can be obtained by adding all the exponents of the concentration terms in the rate expression.\n• The reaction order of a chemical reaction is defined with the help of reactant concentrations and not with product concentrations.\n\nThe rate of the reaction is determined by rate law or rate equation and is given by\n\nr = k[A]x[B]y\n\nWhere,\n\n[A] and [B] = concentration of the species involved in a chemical reaction\n\nx and y = partial orders of reaction for A and B\n\nr = rate of reaction\n\nk= rate constant\n\nOverall reaction = x+y\n\n### Know The Energy Level Of Atoms And Molecules\n\nAs we know the atoms and molecules of the elements contains electrons. Electrons are the chief components responsible for determining the energy level of an atom. Electrons are assumed to 4 different energy levels that are used to describe the trajectory and the movement of an electron in an atom. This 4 numbers which represent the energy level of the electron is known as Quantum numbers. The quantum numbers of all the electrons in a given atom must comply with the Schrödinger wave equation for the hydrogen atom.\n\n4 quantum numbers are described below:\n\n• To (describes the energy level) principal quantum number = n\n• azimuthal quantum number or Orbital angular momentum quantum number, denoted by l.\n• Magnetic quantum number = m or ml.\n• The electron spin quantum number = ms or s.\n\nQuantum Number Values\n\nAccording to the Pauli exclusion principle, no 2 electrons in an atom can have the same set of quantum numbers. Each quantum number is represented by a half-integer or integer value.\n\n1. The principal quantum number is an integer that represents the number of the electron’s shell. The value shall not be 0 or negative and shall be 1 higher than 1\n2. The angular momentum quantum number is an integer that is the value of the electron’s orbital (e.g., s=0, p=1). ℓ is greater than or equal to zero and less than or equal to n-1.\n3. The magnetic quantum number represents the orientation of the orbital. They are usually integers ranging from -ℓ to ℓ. So, for the p orbital, where ℓ=1, m could have values of -1, 0, 1.\n4. half-integer value that is either -1/2 or 1/2 is known as the spin quantum number.\n\nFor more information on quantum numbers, refer BYJU’S" ]	[ null, "https://www.run4unblocked.com/wp-content/uploads/2019/05/run4unblock-1852019.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8984238,"math_prob":0.9960232,"size":3151,"snap":"2022-40-2023-06","text_gpt3_token_len":678,"char_repetition_ratio":0.19351764,"word_repetition_ratio":0.031021899,"special_character_ratio":0.20945731,"punctuation_ratio":0.06949153,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99545455,"pos_list":[0,1,2],"im_url_duplicate_count":[null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-07T23:06:18Z\",\"WARC-Record-ID\":\"<urn:uuid:022243d1-f6c5-49f4-87cc-374740bc6938>\",\"Content-Length\":\"93139\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4bcf8632-03f5-4325-b3f3-aa69694fac97>\",\"WARC-Concurrent-To\":\"<urn:uuid:43966c43-96ed-4e90-a552-3f03a32da61d>\",\"WARC-IP-Address\":\"185.61.154.25\",\"WARC-Target-URI\":\"https://www.run4unblocked.com/how-to-know-the-order-of-the-reaction.html\",\"WARC-Payload-Digest\":\"sha1:Z67UVD6SFQ3COHQKO3PID2XZX3EMAEVC\",\"WARC-Block-Digest\":\"sha1:53INEMRUUIFBXWSRLGRHTJB26DCJCKWG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030338280.51_warc_CC-MAIN-20221007210452-20221008000452-00287.warc.gz\"}"}
https://ijcsm.springeropen.com/articles/10.1186/s40069-020-00438-5	[ "# Hydration Model and Evaluation of the Properties of Calcined Hwangtoh Binary Blends\n\n## Abstract\n\nCalcined hwangtoh is a pozzolanic material that is increasingly being used as a mineral admixture in the concrete industry. This study shows a hydration model for cement–hwangtoh blends and evaluates the various properties of hwangtoh-blended concrete using reaction degrees of binders. First, a kinetic reaction model is proposed for analyzing the pozzolanic reaction of hwangtoh. The reaction of hwangtoh includes three processes: the initial dormant period, boundary reaction process, and diffusion process. The mutual interactions between the binary reactions of cement and hwangtoh are thought to be in line with the items in capillary water and calcium hydroxide. Second, the reaction degrees of cement and hwangtoh are determined based on a blended hydration model. Furthermore, the chemical (chemically combined water and calcium hydroxide contents), mechanical (compressive strength), thermal (hydration heat), and durability aspects (carbonation depth) of hwangtoh-blended concrete are systematically predicted. The results show good agreement with experimental results.\n\n## 1 Introduction\n\nHwangtoh is an environmentally friendly material and belongs to the family of clay. Hwangtoh can purify indoor air by absorption, and it is widely used in hospital buildings for reducing sick building syndrome, including sore throat and headaches. On the other hand, after calcination at about 800 ℃, the phase compositions of hwangtoh are changed, and calcined hwangtoh shows pozzolanic reactivity. The pozzolanic reaction of hwangtoh can increase late-age strength of concrete. Summarily, calcined hwangtoh is increasingly used in the cement and concrete industry because of its unique environmentally friendly characteristics and chemical reactivity (Kim et al. 2017).\n\nMany experimental investigations have been performed on the hydration, mechanical, durability, and thermal properties of hwangtoh-blended concrete. Go et al. (2010, 2009) reported that the optimal temperature for thermal activation of hwangtoh is about 800 ℃, thermally activated hwangtoh shows clear pozzolanic activity, and the hydration heat of hwangtoh-blended concrete is lower than that of control concrete. Koo et al. (2014) found the drying shrinkage and carbonation depth of hwangtoh-blended concrete are higher than that in control concrete, and the flexural failure mode of hwangtoh-blended concrete is similar to that of plain concrete. Koo et al. (2015) presented a comparison with a plain environment, where mice in the hwangtoh-blended environment showed higher food and water intake, higher fertility rate, and healthier offspring. Yang et al. (2014) reported a comparison with Portland cement concrete. The slope of the ascending zone of the stress–strain curve of hwangtoh-blended concrete is smaller, and the descending zone is steeper. Lin et al. (2019) found that, compared with plain concrete, hwangtoh-blended concrete shows higher late-age strength because of the secondary reaction of hwangtoh. Kim et al. (2012) showed that, based on microwave heating, alkali-activated natural hwangtoh shows high early-age strength. Kwon et al. (2017) presented that after addition of silica fume, the compressive strength of calcium hydroxide activated hwangtoh significantly improved. Yang et al. (2010) found the mechanical properties of alkali-activated hwangtoh-based concrete were significantly influenced by the water-to-binder ratio, not by the sand-to-aggregate ratio. Using hwangtoh, styrene butadiene latex, and shape-stabilized phase change materials, Yang et al. produced hwangtoh boards for building thermal insulation (Yang et al. 2019).\n\nSome numerical models have been proposed for optimal design and performance evaluation of cement-based materials. Perez-Cortes and Escalante-Garcia (2020) made statistical optimization design of alkaline binders of limestone–metakaolin and determined the optimal combinations using the desirability function approach considering strength and sustainability. Sonebi et al. (2020) made the optimal design of rheological parameters of super sulfated cement grouts and captured the significant trends of the tested parameters using statistical models considering bleeding, permeability, and mechanical properties. Yu et al. (2018) predicted the early-age creep and creep recovery of cement paste using a two-step homogenization method and highlighted the effects of the microstructure evolution and inherent creep of C–S–H on the early-age creep. Zhou et al. (2019) analyzed the development of microstructure of cement pastes using a diffusion-based model and considered the effects of particle size distribution, water/cement ratio, temperature, and the ambient humidity. These studies (Perez-Cortes and Escalante-Garcia 2020; Sonebi et al. 2020; Yu et al. 2018; Zhou et al. 2019) show numerical modeling are effective for material design and predictions of material properties.\n\nAlthough many experimental studies have been performed on the various qualities of hwangtoh-blended concrete, the theoretical models about hwangtoh-blended concrete are extremely limited. To rationally use hwangtoh in the concrete industry, this research suggested a hydration-based framework that could assess the chemical, mechanical, thermal, and durable qualities of hwangtoh-blended concrete. The blended hydration model correspondingly simulates the hwangtoh reaction and cement hydration. The mutual interactions between the reactions of cement and hwangtoh are thought to be in line with the items in capillary water and calcium hydroxide. In addition, the different qualities are evaluated using reaction amounts of binders and concrete mixtures.\n\n## 2 Hydration-Based Simulation Framework\n\n### 2.1 Cement Hydration Model\n\nWang (2019) and Wang and Lee (2019) proposed a hydration model of Portland cement. The hydration model considers the influences of cement compound compositions, cement fineness, concrete mix ratio, and curing conditions on the rate of cement hydration. The hydration degree of cement $$\\alpha$$ can be determined as $$\\alpha { = }\\int_{0}^{{\\text{t}}} {\\left( {\\frac{d\\alpha }{{dt}}} \\right)} dt$$, where $$\\frac{d\\alpha }{{dt}}$$ is the hydration rate. The detailed equation about $$\\frac{d\\alpha }{{dt}}$$ is available in our former studies (Wang 2019; Wang and Lee 2019). The kinetic processes involved in cement hydration, for example, initial dormant period, boundary reaction process, and diffusion-controlled process, are within the suggested hydration model. The input parameter for the cement hydration model is cement mineral compositions and fineness, concrete mix ratio, and curing conditions. The output parameter of the cement hydration model is the reaction amount of cement versus time. The hydration model views the decrease in the rate of hydration because of the lack of capillary water for concrete, which has a lower water-to-cement ratio. Therefore, the hydration model is applicable for concrete with ordinary strength and high strength. In line with the Portland cement hydration model, Wang (2019) and Wang and Lee (2019) evaluated the strength development, adiabatic temperature rise, carbonation, and chloride ingress of Portland cement concrete.\n\nHowever, the previous models of Wang (2019) and Wang and Lee (2019) is not valid for hwangtoh-blended concrete because the model does not cover the reaction of hwangtoh. To systematically evaluate hwangtoh-blended concrete, a hydration model that considers the binary reactions of cement and hwangtoh is necessary.\n\n### 2.2 Hwangtoh Reaction Model\n\nThis section shows the kinetic reaction model of hwangtoh. The reaction model was obtained based on experimental studies of kinetic hydration processes of cement–hwangtoh blends and the material properties of hwangtoh. The evolution of the reaction degree of hwangtoh with time can be determined using the kinetic reaction model.\n\nLin et al. measured the isothermal heat release of hwangtoh-blended paste (Lin et al. 2019). They found the kinetic hydration procedure for cement–hwangtoh blends was comparable with Portland cement. The hydration of cement–hwangtoh blends includes initial dormant period, boundary reaction process, and diffusion process. Go et al. (2010) reported that hwangtoh is a pozzolanic material that can interact with calcium hydroxide. In cement–hwangtoh blends, the hydration of cement produces calcium hydroxide, and the reaction of hwangtoh consumes calcium hydroxide. The content of remained calcium hydroxide in cement–hwangtoh blends is dependent on both cement hydration and hwangtoh reaction. Moreover, as a pozzolanic material, the reaction of hwangtoh relies upon the quantity of remained calcium hydroxide in cement–hwangtoh blends. In thinking about the kinetic reaction procedure for cement–hwangtoh blends and material signs of hwangtoh, the kinetic reaction equation of hwangtoh can be established as follows:\n\n$$\\frac{{d\\alpha_{HT} }}{dt} = \\frac{{m_{CH} (t)}}{P}\\frac{{3\\rho_{w} }}{{v_{HT} r_{HT0} \\rho_{HT} }}\\frac{1}{{(\\frac{1}{{k_{dHT} }} - \\frac{{r_{HT0}^{{}} }}{{D_{eHT} }}) + \\frac{{r_{HT0}^{{}} }}{{D_{eHT} }}(1 - \\alpha_{HT} )^{{\\frac{ - 1}{3}}} + \\frac{1}{{k_{rHT} }}(1 - \\alpha_{HT} )^{{\\frac{ - 2}{3}}} }},$$\n(1)\n$$k_{dHT} = \\frac{{B_{HT} }}{{(\\alpha_{HT} )^{1.5} }} + C_{HT} (\\alpha_{HT} )^{3} ,$$\n(2)\n$$D_{eHT} = D_{eHT0} \\ln (\\frac{1}{{\\alpha_{HT} }}),$$\n(3)\n\nwhere $$\\alpha_{HT}$$ is the reaction degree of hwangtoh; $$t$$ is time; $$m_{CH} (t)$$ is the calcium hydroxide (CH) mass in a unit volume of hydrating cement–hwangtoh blends; $$P$$ is the mass of hwangtoh in the mixture proportion; $$v_{HT}$$ is the stoichiometric ratio of hwangtoh to CH [$$\\nu_{HT} = 0.87$$ (Dunster et al. 1993; Papadakis 1999)]; $$r_{HT0}$$ is the radius of the hwangtoh particle;$$\\rho_{HT}$$ is the density of the hwangtoh; $$k_{dHT}$$ is the reaction rate coefficient in the initial dormant period ($$B_{HT}$$ and $$C_{HT}$$ are coefficients); $$D_{eHT0}$$ is the initial diffusion coefficient; and $$k_{rHT}$$ is the reaction rate coefficient in the boundary reaction process.\n\nWithin the model, the quantity of capillary water and CH left within the cement–hwangtoh binary system were adopted as control indicators for reactions of binders. The levels of CH ($$m_{CH} (t)$$), chemically combined water ($$W_{cbm}$$), and capillary water ($$W_{cap}$$) in cement–hwangtoh blends during hydration can be established using the following equations:\n\n$$m_{CH} (t) = RCH_{CE} C_{0} \\alpha - \\nu_{HT} \\alpha_{HT} P,$$\n(4)\n$$W_{cap} = W_{0} - 0.4C_{0} \\alpha - 0.46\\alpha_{HT} P,$$\n(5)\n$$W_{cbm} = vC_{0} \\alpha + 0.21\\alpha_{HT} P,$$\n(6)\n\nwhere $$RCH_{CE}$$ is the mass of depleted CH as 1 g of cement hydrate. In Eq. (4), the items $$RCH_{CE} C_{0} \\alpha$$ and $$\\nu_{HT} \\alpha_{HT} P$$ consider the formation of CH from cement hydration and the depletion of CH due to the hwangtoh reaction. As shown in Eqs. (5) and (6), when 1 g of hwangtoh reacts, 0.21 g combined water is produced and 0.46 g capillary water is consumed (Dunster et al. 1993; Papadakis 1999; Wang 2014). As proven in Eqs. (4, 5, 6), the evolution of CH, chemically combined water, and capillary water in blended concrete relates to both the hwangtoh reaction and cement hydration.\n\n### 2.3 Properties Evaluation of Hardening Hwangtoh-Blended Concrete\n\nFor cement–hwangtoh blends, the hydration products of binders fill pore spaces and lead to the introduction of compressive strength. However, given a particular binder content, a growing water content lowers the concrete strength. The introduction of concrete compressive strength generally starts after the final setting of concrete. Thinking about the reactions of binders, the pore spaces of concrete, and the beginning duration of strength development, the development of strength of hwangtoh-blended concrete can be established utilizing a straight line equation as follows:\n\n$$f_{c} (t) = A_{1} \\frac{{C_{0} \\alpha }}{{W_{0} }} + A_{2} \\frac{{P\\alpha_{HT} }}{{W_{0} }} - A_{3} ,$$\n(7)\n\nwhere $$A_{1}$$,$$A_{2}$$, and $$A_{3}$$ are strength coefficients. $$A_{1}$$ and $$A_{2}$$ represent the contribution of cement and hwangtoh on strength development, respectively. For concrete with various mixtures at different curing ages, the values of $$A_{1}$$,$$A_{2}$$, and $$A_{3}$$ are constants. Equation (7) shows the concrete strength does not start at mixing time t = 0, but it starts after a threshold time (Sun et al. 2019). This is similar to the concept of final setting. Final setting means the concrete has started to harden, and this is the beginning of developing the concrete strength (Xu et al. 2017).\n\nThe hydration heat of concrete depends on both the hwangtoh reaction and cement hydration. The total released heat can be expressed as the sum of the hwangtoh reaction and cement hydration as follows:\n\n$$\\frac{dQ}{{dt}} = C_{0} H_{C} \\frac{d\\alpha }{{dt}} + PH_{HT} \\frac{{d\\alpha_{HT} }}{dt},$$\n(8)\n\nwhere $$H_{C}$$ and $$H_{HT}$$ are the specific heat generation of the cement and hwangtoh, respectively. $$H_{C}$$ can be determined from mineral compositions of cement. The value of $$H_{HT}$$ was proposed as 330 J/g (Riding et al. 2012; Williams 2016). The term $$C_{0} H_{C} \\frac{d\\alpha }{{dt}}$$ considers the heat release from the hydration of cement, and the term $$PH_{HT} \\frac{{d\\alpha_{HT} }}{dt}$$ considers the heat release from the hwangtoh reaction.\n\nBecause of the hwangtoh reaction and cement hydration, the porosity of the hydrating concrete was reduced since hydration started. Papadakis (2000, 1999) and Maekawa et al. (2009) proposed that the total porosity of hydrating concrete can be determined as\n\n$$\\varepsilon = \\frac{{W_{0} }}{{\\rho_{w} }} - \\frac{{W_{{{\\text{cbm}}}} }}{{\\rho_{{{\\text{cbm}}}} }},$$\n(9)\n\nwhere $$\\varepsilon$$ is concrete porosity, and $$\\rho_{{{\\text{cbm}}}}$$ is the density of chemically combined water ($$\\rho_{{{\\text{cbm}}}}$$ = 1.25 g/cm3). The term $$\\frac{{W_{{{\\text{cbm}}}} }}{{\\rho_{{{\\text{cbm}}}} }}$$ means the volume of chemically combined water. Because combined water $$W_{{{\\text{cbm}}}}$$ depends on the binary reactions of cement and hwangtoh, concrete porosity is a function of concrete mixtures and the reaction degree of binders.\n\n### 2.4 Evaluation of Carbonation Depth of Hwangtoh-Blended Concrete\n\nWhen hwangtoh is used as a partial binder in concrete, the carbonation depth increases. In other words, the carbonation durability may be the controlling factor that determines the service life of hwangtoh-blended concrete in an atmospheric environment. The evaluation of carbonation depth is essential for real applications of hwangtoh in concrete structural elements.\n\nCarbonation resistance relates to both internal factors (concrete material properties) and external factors (exposure conditions). The content of carbonated substances, CO2 diffusivity, CO2 concentration, and exposure time will affect the carbonation depth. When the relative humidity in the exposure environment is higher than 55%, carbonation is a diffusion-controlled process. The carbonation depth $$x_{c}^{{}}$$ can be determined as follows (Chen and Gao 2019; Chen et al. 2019; Papadakis 2000):\n\n$$x_{c}^{{}} = \\sqrt {\\frac{{2D_{C} ([CO_{2} ]_{{0}} /100)t}}{{0.33{\\text{CH}} + 0.214{\\text{CSH}}}}} ,$$\n(10)\n$$D_{C} = A\\left( {\\frac{{\\varepsilon { - }\\Delta \\varepsilon_{C} }}{{\\frac{{C_{0} }}{{\\rho_{c} }} + \\frac{P}{{\\rho_{HT} }} + \\frac{{W_{0} }}{{\\rho_{w} }}}}} \\right)^{a} \\left( {1 - \\frac{RH}{{100}}} \\right)^{2.2} ,$$\n(11)\n$$CSH(t) = 2.85(f_{S,C} C_{0} \\alpha + f_{S,P} P\\alpha_{HT} ),$$\n(12)\n\nwhere $$D_{C}$$ is CO2 diffusivity in the carbonated zone of concrete, and $$[CO_{2} ]_{{0}}$$ is the CO2 concentration in the exposure environment; CSH is the mass of calcium silicate hydrate in concrete; and $$\\Delta \\varepsilon_{C}$$ is the porosity reduction of concrete because of carbonation. $$\\Delta \\varepsilon_{C}$$ can be calculated based on the change of solid volume between reactants and products(Papadakis 2000); RH is the relative humidity of the exposure environment. A and a are parameters of carbonation and can be regressed based on measurement results of carbonation depths. The influence of environmental temperature on CO2 diffusion can be described using the Arrhenius law (Papadakis 2000, 1999).\n\nEquation 12 shows the calculation of CSH. $$f_{S,C}$$ and $$f_{S,P}$$ denote the mass percentages of SiO2 in cement and hwangtoh, respectively. For hwangtoh-blended concrete, both the hwangtoh reaction and cement hydration produce CSH (Riding et al. 2012; Williams 2016; Xu et al. 2017). In Eq. 12, the terms $$f_{S,C} C_{0} \\alpha$$ and $$f_{S,P} P\\alpha_{HT}$$ mean the formation of CSH from the hwangtoh reaction and cement hydration, respectively. The coefficient 2.85 is the molar weight ratio of CSH to SiO2 in CSH (Papadakis 1999, 2000).\n\nIn line with the suggested hydration model, the quantity of CH and CSH and the extent of porosity could be acquired as connected results throughout the hydration duration of cement–hwangtoh blends. In addition, the carbonation depth could be predicted by utilizing Eqs. 10, 11, and 12.\n\n### 2.5 Summary of Simulation Framework\n\nFigure 1 shows a flowchart of the simulation. First, based on the blended hydration model and concrete mixtures, the reaction degree of binders can be calculated. Furthermore, combined water, CH, and CSH contents can be determined using the reaction degree of binders and concrete mix ratios. The development of strength and hydration heat can be evaluated considering both the hwangtoh reaction and cement hydration. Second, CO2 diffusivity can be determined based on concrete porosity and environmental conditions. Moreover, carbonation depth can be calculated considering the contents of carbonated substances, CO2 diffusivity, and CO2 concentration in the exposure environment. Summarily, the reaction degrees of cement and hwangtoh are key factors for evaluating the various properties of hardening and hardened hwangtoh-blended concrete.\n\n## 3 Verification of the Proposed Model\n\nAs shown in Sect. 2, the proposed model can evaluate various properties of hwangtoh-blended concrete. To verify the proposed model, multiple checks were performed, such as chemical (experimental tests of combined water and CH), mechanical (compressive strength test), thermal (hydration heat), and durability aspects (carbonation depth test). Through these systematic verifications, the validations of the proposed model are shown.\n\n### 3.1 Evaluation of the Properties of Hardening Concrete\n\nExperimental results in our previous study (Lin et al. 2019) were used to verify the evaluation of the properties of hardening concrete. In our previous studies, the property development of hwangtoh-blended paste with various water-to-binder ratios (W/B 0.5 and 0.2) and hwangtoh replacement ratios (0, 10%, and 20%) was measured. The specimens with a W/B 0.5 represent concrete with ordinary strength, and the specimens with a W/B 0.2 represent ultra-high-performance concrete. Table 1 shows chemical compositions of hwangtoh and cement. Figure 2 shows XRD pattern of calcined hwangtoh, which contains quartz, feldspar, and a small amount of illite. The paste specimens are sealed cured until the aging tests. At the curing ages of 3, 7, and 28 days, the contents of combined water and CH were measured using thermogravimetric analysis, and the strengths of specimens were measured using a compression machine. The isothermal heat evolution was measured in the early 3-day period using TAM-air.\n\nThe temperature range of thermogravimetric analysis (TGA) tests was between room temperature and 1050℃. Nitrogen was continuously supplied throughout the test. As shown in Fig. 3a, b (TGA results at the ages of 28 days), there are three peaks of the differential thermal analysis (DTA) curve. The first peak occurs at about 160 ℃, which corresponds to the dehydroxylation of CSH and AFm (Weerdt et al. 2011). The second peak occurs at about 450 ℃, which corresponds to the decomposition of calcium hydroxide (Weerdt et al. 2011). The third peak occurs at about 700 ℃, which corresponds to the decarbonation of calcite (Weerdt et al. 2011). There are several sources of calcite, such as the limestone powder in cement or the carbonation of specimens while preparing the specimens of TGA (Weerdt et al. 2011). In Korea, the Portland cement contains a small amount (3%-5%) limestone powder (Jeong et al. 2020; Additional file 1: Figures S1, TGA of cement), which is the main reason of the decarbonation peak at 700℃ (Additional file 1: Figures S2 the decarbonation peak of cement and paste). As proposed by De Weerdt et al. (2011), for the specimen with limestone powder, to mitigate the influence of decarbonation stage, non-evaporable water (H) and calcium hydroxide (CH) were calculated at 105–550 ℃ and 400–550 ℃, respectively. The contents of CH can be determined as follows (Weerdt et al. 2011):\n\n$${\\text{CH}}\\, = \\,\\left( {{\\text{W}}_{{{4}00}} - {\\text{ W}}_{{{55}0}} } \\right)/{\\text{W}}_{{{55}0}} *\\left( {{74}/{18}} \\right),$$\n(13)\n\nwhere W400 and W550 are the weights at 400 and 550 ℃, respectively. The item (74/18) means the molar weight ratio of CH to water. The content of non-evaporable water (H) can be determined as follows (Weerdt et al. 2011):\n\n$${\\text{H}}\\, = \\,\\left( {{\\text{W}}_{{{1}0{5}}} - {\\text{ W}}_{{{55}0}} } \\right)/{\\text{W}}_{{{55}0}} ,$$\n(14)\n\nwhere W105 is the weight at 105 ℃.\n\nFor hwangtoh-blended concrete, the hwangtoh reaction and cement hydration occurred concurrently. Cement hydration generated CH, and reaction of hwangtoh consumed CH. The generation of CH from cement hydration can be determined based on the cement hydration model. Moreover, the consumption of CH can indicate the reaction degree of hwangtoh. Using the measurement results of CH of cement–hwangtoh blends, the coefficients of the reaction model of hwangtoh are determined and shown in Table 2. These coefficients do not change with W/B or hwangtoh substitution ratios.\n\nThe calculation results of CH are shown in Fig. 3c, d. For plain paste without hwangtoh, the content of CH kept increasing at all the ages. In early age, the CH content increased rapidly. At late age, diffusion was the mechanism that controlled cement hydration, and the CH content increased slowly. On the other hand, for cement–hwangtoh-blended paste, the content of CH was less than plain paste. This is because of the reduction of the cement content and the consumption of CH from the hwangtoh reaction. Especially, at the late age, the content of CH decreased due to the proceeding of the hwangtoh reaction (Han and Yan 2015; Han et al. 2017).\n\nFigure 4 shows the reaction degree of binders such as cement or hwangtoh. As shown in Fig. 4a, for the specimen with W/B 0.5, when hwangtoh replaced partial cement, the water-to-cement ratio increased; hence, the hydration degree of cement increased. This is called the dilution effect of hwangtoh addition. On the other hand, Fig. 4b shows the reaction degree of cement for the specimen with W/B 0.2. From Fig. 4a, b, it can be seen that the dilution effect became obvious with the decreasing water-to-binder ratios. In addition, Fig. 4c presents the reaction level of hwangtoh for various mixtures. It can be seen that as W/B decreased, the reaction level of hwangtoh decreased. This was due to the limitations of available filling space for hydration products. As the hwangtoh replacement ratio increased, the reaction level of hwangtoh also decreased. This was due to the reduction of alkali activation of CH. Compared to the hydration degree of cement (shown in Fig. 4a, b), the reaction degree of hwangtoh was much lower (shown in Fig. 4c) (Glosser et al. 2019).\n\nReaction degrees of cement and hwangtoh are fundamental parameters for evaluating properties of hwangtoh-blended concrete. Based on Eq. (6), the combined water can be determined. As shown in Fig. 5a, when hwangtoh replaced partial cement, the content of combined water decreased. This is because the reactivity of hwangtoh is low compared to cement. As shown in Fig. 5b, as W/B reduced, the content of combined water decreased. This tendency agrees with the trend of the reaction level of binders (Castellano et al. 2016; Riding et al. 2012).\n\nThe hydration heat of cement–hwangtoh blends can be evaluated based on the reaction levels of binders. Figure 6 shows the calculation results of hydration heat. In the first 5 h, cement hydration was in the dormant period, and the hydration heat slowly increased. After the initial dormant period, because the boundary reaction process controls hydration, the hydration heat rapidly increased. On the other hand, after the addition of hwangtoh, the hydration heat decreased. This is because the reactivity of hwangtoh is low compared to cement (Lin et al. 2019).\n\nEquation (7) is a linear equation that can evaluate the strength development of hardening concrete. Using the experiment results of specimen strengths, the coefficients $$A_{1}$$,$$A_{2}$$, and $$A_{3}$$ of Eq. (7) were determined to be 58.74, 109.15, and 33.57, respectively. These coefficients did not change with specimen mixtures or curing ages. The prediction results are shown in Fig. 7. First, for the plain specimen, the strength rapidly increased at an early age. At the late age, the increment of strength was not obvious and the strength curve showed a plateau, while for the hwangtoh-blended specimen, at the late age, the increment of strength was much more obvious than that of the plain specimen. This is due to the slow reaction of hwangtoh that can contribute to the strength development of concrete (Li et al. 2017; Shao et al. 2019). Second, as the W/B decreased from 0.5 to 0.2, the intercept of the strength curve on the x-axis (starting time of strength) decreased. This means the specimen with a lower W/B ratio can develop strength much faster than the specimen with a higher W/B ratio (Han and Yan 2015; Han et al. 2017; Sun et al. 2019). Third, as shown in Fig. 7g, the prediction results showed agreement with measurement results. The correlation coefficient between measurement and evaluation results is 0.993, which proves the validity of proposed strength model. In addition, the proposed strength model is verified using experimental data from another source. Lin et al. (2020) measured the strength development of blended paste with 45% hwangtoh as the binder. The water/binder ratio of paste was 0.5. At the ages of 1, 3, 7, and 28 days, the compressive strength was measured. Using the strength analysis model, the strengths of plain paste and hwangtoh-blended paste are shown in Fig. 7h. The analysis results show agreement with experimental results. Because the reaction of hwangtoh is much slower than that of cement, the starting time of strength of hwangtoh-blended paste is retarded compared with plain paste. Moreover, the 28-day strength of high-volume hwangtoh-blended paste is much lower than that of control paste.\n\nThe strength activity index is defined as the ratio of the strength of the hwangtoh–cement blends to the strength of the plain specimen at each specific curing time (Maekawa et al. 2009). At the curing age 28 days, when water/binder is 0.5, the strength activity index for 10% hwangtoh and 20% hwangtoh specimen are 1.17 and 1.13, respectively, while when water/binder is 0.2, the strength activity index for 10% hwangtoh and 20% hwangtoh specimen are 1.09 and 1.06, respectively. Summarily, the strength activity index of hwangtoh is dependent on water/binder ratio, hwangtoh replacement levels, and curing ages.\n\nBesides the experimental results of our previous study (Lin et al. 2019), experimental results from Go et al.(2010) are used to verify the proposed hydration model. Go et al. (2010) made thermal analysis of hydrated plain paste and 20% hwangtoh-blended paste at the ages of 14 days and 28 days. The water/binder ratio of paste specimens was 0.6. The temperature of thermal analysis ranges from 25℃ to 950℃. Based on the experimental results of thermal analysis, the content of calcium hydroxide for the unit mass of binder can be determined. Figure 8 shows the comparisons between calculation results and experimental results of calcium hydroxide contents (Go et al. 2010). The calculation results show agreement with experimental results. Due to the dilution effect and pozzolanic reaction of hwangtoh, calcium hydroxide content in blended concrete is much lower than plain concrete.\n\n### 3.2 Evaluation of Carbonation Depth\n\nThe experimental study on the carbonation depth of hwangtoh-blended concrete in reference (Koo et al. 2014) was used to verify the proposed model. The W/B of the specimen was 0.55, and the content of binder was 339 kg/m3. The hwangtoh replacement ratios were 0 and 20%. After 2-month curing, concrete specimens were moved into a CO2 chamber for accelerated carbonation tests. The CO2 concentration in the CO2 chamber was 5%, the relative humidity was 60%, and the temperature was 20 ℃. Carbonation depth was measured at 1, 4, 8, 12, and 25 weeks using the phenolphthalein color change method. Because the hydration of binders essentially stops when relative humidity is less than 80%, further hydration of hwangtoh and cement in the carbonation test period was ignored in this study (Maekawa et al. 2009).\n\nIn line with the blended cement hydration model, the items in carbonated substances (CH and CSH) and concrete porosity can be established. First, as proven in Fig. 9a, the CH content of 20% hwangtoh-blended concrete was less than that in plain concrete. Particularly, the CH content decreased in the late ages. Second, as proven in Fig. 9b, at the beginning, the CSH content of hwangtoh-blended concrete was less than that in plain concrete, while at late age, the CSH content of hwangtoh-blended concrete exceeded that of plain concrete. This is because the hwangtoh reaction proceeded, and also the silica content in hwangtoh is a lot greater than that in cement. Third, as proven in Fig. 9c, the porosity of hwangtoh-blended concrete was greater than that of plain concrete. This trend is comparable with fly ash or slag blended concrete (Han et al. 2017).\n\nAccording to experimental outcomes of carbonation depth, the coefficients of carbonation model A, along with a are determined as 1.34e−6 and 1.8, correspondingly. The coefficients are comparable with those of Papadakis’ study (2000). These coefficients do not change with concrete mixtures and curing conditions. According to the hydration-based carbonation model, the carbonation depth of concrete can be determined. As proven in Fig. 10, the carbonation depth of hwangtoh-blended concrete was a lot greater than that of plain concrete. This really is due to the fact that after adding hwangtoh, the content of CH reduced, and also the porosity and CO2 diffusivity increased.\n\nBesides the experimental results of Koo et al. (2014), experimental results from Choi et al. (2001) are used to verify the proposed carbonation model. Choi et al. (2001) measured carbonation depths of plain concrete, and 20% hwangtoh-blended concrete. The water/binder ratio of concrete specimens was 0.55, the water content was 179 kg/m3, and the binder content was 325 kg/m3. Accelerated carbonation tests started after 28 days of standard curing. The temperature, CO2 concentration, and relative humidity of acceleration carbonation tests were 20 ℃, 5 and 60%, respectively. After 1-, 2-, 4-, 8-, 10- and 12-week exposure, carbonation depths of concrete were measured using the phenolphthalein indicator method. Figure 11 shows the comparisons between calculation results and experimental results of carbonation depths (Choi et al. 2001). The calculation results show general agreement with experimental results. The replacement of partial cement with hwangtoh increases the carbonation depth of concrete.\n\n### 3.3 Discussion\n\nThis study shows a hydration-based integrated model for cement–hwangtoh blends. The validation range and the weak points of the proposed model are shown as follows:\n\nFirst, this study considers several water/binder ratios (0.5, 0.55, 0.6, 0.2) and hwangtoh/binder ratios (10, 20, and 45%). The water/binder ratio 0.2 represents the cases of high-strength concrete and ultra-high-performance concrete, which have a low water/binder ratio and show incomplete hydration due to a shortage of capillary water. The water/binder ratios 0.50, 0.55, and 0.6 represent the cases of ordinary-strength concrete, which have a high water/binder ratio and show sufficient hydration at long-term ages. Hence, the proposed model is valid for both ordinary-strength concrete and high-strength concrete.\n\nSecond, hwangtoh is a pozzolanic material and can be used to replace partial cement for concrete production. In construction practices, the replacement levels of hwangtoh are generally lower than 20%. When the replacement level of calcined hwangtoh is lower than 20%, the long-term compressive strength of hwangtoh-blended concrete is comparable or higher than plain concrete. Summarily, the proposed model covers the general replacement levels of hwangtoh in construction practices.\n\nThird, the limitations of the proposed method are summarized as follows:\n\n1. (1)\n\nThe proposed model mainly focuses on concrete with a moderate hwangtoh (less than 20%). For concrete with a high-volume hwangtoh, the reaction of hwangtoh may depend on phase concentration of pore solution and ion movement between inner and outer hydration products (Maekawa et al. 2009; Wang 2014). The hwangtoh reaction model (Eqs. (1), (2), and (3)) may need revisions for considering pore solution and ion movement.\n\n2. (2)\n\nThe effects of curing temperature on hwangtoh reaction, strength development, and durability aspect are not covered in this study. Compared with cement, the pozzolanic reaction of hwangtoh may be much more sensitive with temperature (Wang 2014). More experimental study and theoretical modeling should be carried out to clarify the dependence of hwangtoh reaction on temperature.\n\n3. (3)\n\nThe proposed model cannot evaluate the phase assemblage of hydrating cement–hwangtoh blends. In the future study, to accurately evaluate the phase assemblage, the proposed kinetic model should combine with a thermodynamic program, such as Gibbs Energy Minimization (GEM) program (Lothenbach et al. 2019). Moreover, the durability aspects, such as carbonation and chloride, may be evaluated based on phase assemblage of concrete.\n\n## 4 Conclusions\n\nThis research shows an integrated model that simulates the binary hydration of cement–hwangtoh blends and assesses the various properties of cement–hwangtoh blends.\n\nFirst, a kinetic model was suggested for the hwangtoh reaction. The reaction of hwangtoh was segmented into three stages: the initial dormant period, boundary reaction process, and diffusion process. The suggested model views the dependence of hwangtoh reaction on the CH content. The interaction between the hwangtoh reaction and cement hydration was clarified through CH and capillary water content.\n\nSecond, the coefficients of the hwangtoh reaction model were determined in line with the experimental outcomes of CH content of cement–hwangtoh blends. The coefficients of the reaction model are constants for a number of concrete mixtures. The reaction amounts of cement and hwangtoh were determined using blended hydration model. In addition, the combination of water and hydration heat was determined in light of the contributions of hwangtoh reaction and cement hydration.\n\nThird, a hydration-based straight line equation was suggested for evaluating the growing strength of hwangtoh-blended concrete. The coefficients of the strength development model are constants for a number of concrete mixtures. The correlation coefficient between the evaluation and experimental results was 0.993, which proves the suggested strength model applies for ordinary-strength concrete and ultra-high-strength concrete that contains hwangtoh.\n\nFourth, the contents of carbonated substances and porosity were determined from the hydration model. The CO2 diffusivity was determined considering the concrete porosity and the relative humidity of the exposure environment. Moreover, the carbonation depth was determined based on a diffusion-controlled equation. The coefficient of the carbonation model did not change for various concrete mixtures. The carbonation model can reflect the increase of carbonation depth due to hwangtoh additions.\n\nFifth, in a further study, a revised hydration model should be proposed to consider the phase concentration of pore solution, ion movement between inner and outer reaction products, and different curing temperatures. Moreover, the combination of the proposed kinetic model with a thermodynamic program should be carried out for evaluating the phase assemblage and durability aspects of concrete.\n\n## Availability of data and materials\n\nThe data used to support the findings of this study are available from the corresponding author upon request.\n\n## References\n\n• Castellano, C. C., Bonavetti, V. L., Donza, H. A., & Irassar, E. F. (2016). The effect of w/b and temperature on the hydration and strength of blast furnace slag cements. Construction and Building Materials, 111, 679–688.\n\n• Chen, T., & Gao, X. (2019). 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Construction and Building Materials, 225, 1140–1151.\n\n• Weerdt, K. D., Haha, M. B., Saout, G. L., Kjellsen, K. O., Justnes, H., & Lothenbach, B. (2011). Hydration mechanisms of ternary Portland cements containing limestone powder and fly ash. Cement and Concrete Research, 41, 279–291.\n\n• Williams, A. (2016). Cracking potential and temperature sensitivity of metakaolin concrete. USA: University of South Florida.\n\n• Xu, G., Tian, Q., Miao, J., & Liu, J. (2017). Early-age hydration and mechanical properties of high volume slag and fly ash concrete at different curing temperatures. Construction and Building Materials, 149, 367–377.\n\n• Yang, K., Hwang, H., & Lee, S. (2010). Effects of water-binder ratio and fine aggregate-total aggregate ratio on the properties of hwangtoh-based alkali-activated concrete. Journal of Materials in Civil Engineering, 22, 887–896.\n\n• Yang, K., Mun, J., & Hwang, H. (2014). Stress-strain relationship of Ca(OH)2-activated hwangtoh concrete. The Scientific World Journal, 2014, 846805.\n\n• Yang, S., Wi, S., Lee, J., Jeon, J., & Kim, S. (2019). Thermal performance evaluation of Hwangtoh board developed with styrene butadiene latex/SSPCM. Construction and Building Materials, 200, 310–317.\n\n• Yu, P., Duan, Y. H., Chen, E., Tang, S. W., Hanif, A., & Fan, Y. L. (2018). Microstructure-based homogenization method for early-age creep of cement paste. Construction and Building Materials, 188, 1193–1206.\n\n• Zhou, W., Duan, L., Tang, S. W., Chen, E., & Hanif, A. (2019). Modeling the evolved microstructure of cement pastes governed by diffusion through barrier shells of C-S–H. Journal of Materials Science, 54, 4680–4700.\n\nNot applicable.\n\n## Funding\n\nThis research was supported by the Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Science, ICT & Future Planning (No. 2015R1A5A1037548), and an NRF grant (NRF-2020R1A2C4002093).\n\n## Author information\n\nAuthors\n\n### Contributions\n\nData curation, HSL and XYW; formal analysis, XYW; methodology, HSL; supervision, XYW; writing—original draft, HSL and XYW. Both authors read and approved the final manuscript.\n\n### Authors' information\n\nHan-Seung Lee is a full professor in the Department of architectural engineering at Hanyang University, Korea.\n\nXiao-Yong Wang is an associate professor in the Department of architectural engineering and Department of Integrated Energy and Infra System at Kangwon National University, Korea.\n\n### Corresponding author\n\nCorrespondence to Xiao-Yong Wang.\n\n## Ethics declarations\n\n### Competing interests\n\nI confirmed that I have read Springer Open’s guidance on competing interests and have included a statement indicating that none of the authors have any competing interests in the manuscript.\n\n### Publisher's Note\n\nSpringer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.\n\nJournal information: ISSN 1976-0485 / eISSN 2234-1315\n\n## Supplementary Information\n\n### Additional file 1: Figure S1.\n\nTGA of cement. Figure S2. The decarbonation peak of cement and paste.\n\n## Rights and permissions", null, "" ]	[ null, "https://ijcsm.springeropen.com/track/article/10.1186/s40069-020-00438-5", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8872734,"math_prob":0.95947355,"size":45161,"snap":"2023-14-2023-23","text_gpt3_token_len":10941,"char_repetition_ratio":0.20198418,"word_repetition_ratio":0.057867263,"special_character_ratio":0.23664224,"punctuation_ratio":0.16771068,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98842025,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-28T14:53:17Z\",\"WARC-Record-ID\":\"<urn:uuid:1875e536-d28b-4817-a4f0-9787c3184363>\",\"Content-Length\":\"358468\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:03670d79-9a9a-4ef3-aac1-3fbb325fbd49>\",\"WARC-Concurrent-To\":\"<urn:uuid:997e8bc8-f91e-4fac-9630-37469374d0a3>\",\"WARC-IP-Address\":\"146.75.32.95\",\"WARC-Target-URI\":\"https://ijcsm.springeropen.com/articles/10.1186/s40069-020-00438-5\",\"WARC-Payload-Digest\":\"sha1:RJZG7AKNPAYYAX3XWXAQHOAVO52RK5GB\",\"WARC-Block-Digest\":\"sha1:3UZ3WEUWV7E23BLBCFO7AQGKHIEK4F2C\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296948867.32_warc_CC-MAIN-20230328135732-20230328165732-00535.warc.gz\"}"}
https://dcwaterpump.com/how-important-is-the-max-head-of-a-pump/	[ "# How important is the Max Head of a pump?", null, "## Your first parameter consideration while choosing a water pump?\n\nThere are two important parameters marked on the label of pumps: Max Head and Max Flow Rate. They are essential factors you need to consider when you choose a pump. If I ask which one is your first consideration, most people might say ‘Max Flow Rate.’\n\nThe Flow Rate will determine the amount of time you need to move the required fluid. So it’s really important. But unfortunately, that is the wrong answer.\n\n## You should consider the Max Head at first\n\nFirst, let’s figure out the concepts of Max Head and Max Flow rate.\n\nWhat does ‘Max Head’ mean?\n\nMax Head refers to a pump’s capability to push water upwards (also known as ‘vertical lift’). What is often overlooked is that at its max head lift, the pump will deliver zero flow!\n\nHow about ‘Max Flow Rate’? Max Flow Rate means a pump’s capability to transfer fluid volume. It can achieve the max value only when the head lift is zero.\n\nSo why do I think the Max Head is more important than the Max Flow Rate? Because most of the time, we need to pump the fluid to a desired height. We won’t transfer water horizontally. We must first make sure that the pump can deliver the liquid to the required level.\n\n## Max Head affects the Actual Flow Rate to a great extent.\n\nNow we know the marked Max Flow Rate is just an ideal value. In actual work, it is impossible to outreach the Max Flow Rate marked. It only represents the performance capacity of a pump. This means that the Actual Flow Rate is not equal to the Max Flow Rate. The actual flow rate is the data we should be concerned about.\n\nWhen the pump works, its Actual Flow Rate is inversely proportional to the required height. When the pump delivers the fluid to a higher point, its Actual Flow Rate decreases. The Actual Flow Rate’s decay rate will increase sharply as the required height is close to the Max Head. You should generally let the pump run within its typical operating area to get the best performance (Refer to the Curves Graph above).\n\nNow, let’s use an example to illustrate Max Head’s impact on the Actual Flow Rate. To give you an idea, take our two models to demonstrate the problem: DC50A-2450 and DC50E-24150. Click the link in the text to enter the details page of the model.\n\nThe Sub-Model List table clearly indicates that they have the same rated voltage and rated current, which means that they have the same rated power. Now let’s take a look at their differences.\n\nAs can be seen from the performance curves of the two models:\n\nWhen the required height equals 3 meters, the Actual Flow Rate of both models is 20 liters per minute.\n\nWhen the required height is more than 3 meters, the Actual Flow Rate of DC50E-24150 is greater. When the needed height is less than 3 meters, the Actual Flow Rate of DC50A-2450 is greater.\n\nSo, if you want to pump water to the vertical height of 4 meters or even higher, of course, DC50E-24150 is a better choice.\n\n## Summary\n\nWhen we choose a pump, we shouldn’t just focus on the marked Max Flow Rate. We should consider the Max Head at first. Because the Actual Flow Rate is not equal to the Max Flow Rate. The Max Head determines whether the pump can reach your desired height. It also determines the pump’s Actual Flow Rate at work." ]	[ null, "data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%20800%20534'%3E%3C/svg%3E", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.90681833,"math_prob":0.9594052,"size":3332,"snap":"2022-40-2023-06","text_gpt3_token_len":752,"char_repetition_ratio":0.17007211,"word_repetition_ratio":0.06467662,"special_character_ratio":0.23049219,"punctuation_ratio":0.09011628,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9742137,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-30T22:48:14Z\",\"WARC-Record-ID\":\"<urn:uuid:b7ccf848-e0d8-4d2a-ac70-fa5bfbf3927d>\",\"Content-Length\":\"153125\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:32ac8837-603b-4f7e-92a3-95ef19e55ef0>\",\"WARC-Concurrent-To\":\"<urn:uuid:dc388674-05ba-4cf3-9ac3-7022f6439a24>\",\"WARC-IP-Address\":\"104.21.26.39\",\"WARC-Target-URI\":\"https://dcwaterpump.com/how-important-is-the-max-head-of-a-pump/\",\"WARC-Payload-Digest\":\"sha1:IFBSKBNOAPAAWZDAS67F7KU6PEVQGMTS\",\"WARC-Block-Digest\":\"sha1:XGC7U4MXDZAKW7USS2REVWO6FIY53MT2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030335504.37_warc_CC-MAIN-20220930212504-20221001002504-00350.warc.gz\"}"}
https://energyeducation.ca/encyclopedia/Permeability_of_free_space	[ "# Permeability of free space\n\nThe permeability of free space, μ0, is a physical constant used often in electromagnetism. It is defined to have the exact value of 4π x 10-7 N/A2 (newtons per ampere squared). It is connected to the energy stored in a magnetic field, see Hyperphysics for specific equations.\n\nIt is related to the speed of light by the equation:\n\nwhere\n\nA magnetic field, in a region of space has field energy associated with it. That energy density is defined as:\n\nEnergy\n\nOne particular use is the magnetic force. The constant gives how strong the force is between two electric currents separated by a distance:\n\nwhere\n\n• is the Magnetic force,\n• and are two currents, and\n• is the separation between the wires carrying the currents." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.89860874,"math_prob":0.99240166,"size":835,"snap":"2020-24-2020-29","text_gpt3_token_len":198,"char_repetition_ratio":0.10589651,"word_repetition_ratio":0.0,"special_character_ratio":0.22035928,"punctuation_ratio":0.11728395,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9948813,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-05-27T22:56:47Z\",\"WARC-Record-ID\":\"<urn:uuid:881a7a93-e134-4fce-96e2-8939144c4ba6>\",\"Content-Length\":\"24165\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:07a16ecb-9215-456e-a92d-b5b2315b543f>\",\"WARC-Concurrent-To\":\"<urn:uuid:97b26f04-8004-4cb6-9557-062b79821db5>\",\"WARC-IP-Address\":\"132.148.209.51\",\"WARC-Target-URI\":\"https://energyeducation.ca/encyclopedia/Permeability_of_free_space\",\"WARC-Payload-Digest\":\"sha1:DV4F47ASG5DCHS4YUM5HQAIQXVGOE6QN\",\"WARC-Block-Digest\":\"sha1:47HO4XFNWCWQFQFJYM2J76WSOZITHKNW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347396163.18_warc_CC-MAIN-20200527204212-20200527234212-00277.warc.gz\"}"}
https://nineplanets.org/percentage-calculator/what-is-percentage-difference-from-336-to-2763/	[ "# What is the percentage increase/decrease from 336 to 2763?\n\n## Quickly work out the percentage increase or decrease from 336 to 2763 in this step-by-step percentage calculator tutorial. (Spoiler alert: it's 722.32%!)\n\nSo you want to work out the percentage increase or decrease from 336 to 2763? Fear not, intrepid math seeker! Today, we will guide you through the calculation so you can figure out how to work out the increase or decrease in any numbers as a percentage. Onwards!\n\nIn a rush and just need to know the answer? The percentage increase from 336 to 2763 is 722.32%.\n\nWhat is the % change from to\n\n## Percentage increase/decrease from 336 to 2763?\n\nAn increase or decrease percentage of two numbers can be very useful. Let's say you are a shop that sold 336 t-shirts in January, and then sold 2763 t-shirts in February. What is the percentage increase or decrease there? Knowing the answer allows you to compare and track numbers to look for trends or reasons for the change.\n\nWorking out a percentage increase or decrease between two numbers is pretty simple. The resulting number (the second input) is 2763 and what we need to do first is subtract the old number, 336, from it:\n\n2763 - 336 = 2427\n\nOnce we've done that we need to divide the result, 2427, by the original number, 336. We do this because we need to compare the difference between the new number and the original:\n\n2427 / 336 = 7.2232142857143\n\nWe now have our answer in decimal format. How do we get this into percentage format? Multiply 7.2232142857143 by 100? Ding ding ding! We have a winner:\n\n7.2232142857143 x 100 = 722.32%\n\nWe're done! You just successfully calculated the percentage difference from 336 to 2763. You can now go forth and use this method to work out and calculate the increase/decrease in percentage of any numbers.\n\nHead back to the percentage calculator to work out any more calculations you need to make or be brave and give it a go by hand. Hopefully this article has shown you that it's easier than you might think!" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.92268556,"math_prob":0.9456025,"size":1730,"snap":"2020-45-2020-50","text_gpt3_token_len":406,"char_repetition_ratio":0.16685979,"word_repetition_ratio":0.006535948,"special_character_ratio":0.26878613,"punctuation_ratio":0.10511363,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99159044,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-03T17:40:31Z\",\"WARC-Record-ID\":\"<urn:uuid:4dd26791-61e2-4769-834c-7d7f4abf94c1>\",\"Content-Length\":\"8870\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:335209b5-7abe-44a7-97d2-a32a35dff67d>\",\"WARC-Concurrent-To\":\"<urn:uuid:7c9becd6-7242-4516-95f8-1ebcd5435eb1>\",\"WARC-IP-Address\":\"157.245.80.188\",\"WARC-Target-URI\":\"https://nineplanets.org/percentage-calculator/what-is-percentage-difference-from-336-to-2763/\",\"WARC-Payload-Digest\":\"sha1:OWT4JRFXO34CZ6ZLWJOTX3NRHLKODL5Y\",\"WARC-Block-Digest\":\"sha1:E2YMBMVPA53JMQGGNV5FBF4PPFHVG72H\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141729522.82_warc_CC-MAIN-20201203155433-20201203185433-00436.warc.gz\"}"}
https://gricad-gitlab.univ-grenoble-alpes.fr/coiffiem/BHR/-/commit/d8a2196c1b1e3d1956b846fb520f941e6c42ee28	[ "Vous avez reçu un message \"Your GitLab account has been locked ...\" ? Pas d'inquiétude : lisez cet article https://docs.gricad-pages.univ-grenoble-alpes.fr/help/unlock/\n\n### Make the order of evaluation left-to-right at all backquote depths in concatenative languages.\n\n```Previously, the deepest holes were evaluated first, even when they\nwere syntactically positioned after some previous ones at a shallower\nlevel.\n\nMaking the Folds into Traversals had the nice effect of forcing a\nsingle-pass implementation of what was previously done in two stages,\ncausing the unintuitive evaluation order. Always trust your\nApplicative functors, they won't betray you.```\nparent 1ce72bee\n ... ... @@ -2,7 +2,7 @@ -- see http://haskell.org/cabal/users-guide/ name: capricon version: 0.12 version: 0.12.1 -- synopsis: -- description: license: GPL-3 ... ...\n ... ... @@ -154,7 +154,7 @@ runWordsState ws st = (\\$st) \\$ from (stateT.concatT) \\$^ do runWithFS :: JS.JSString -> FSIO a -> JS.CIO a runWithFS fsname (FSIO r) = newFS fsname >>= r^..readerT hasteDict = cocDict (\"0.12-js\" :: String) getString getBytes setString setBytes hasteDict = cocDict (\"0.12.1-js\" :: String) getString getBytes setString setBytes main :: IO () main = do ... ...\n ... ... @@ -32,27 +32,36 @@ t'ClosureStep :: Traversal' (StackStep s b a) (StackClosure s b a) t'ClosureStep k (ClosureStep b c) = ClosureStep b<\\$>k c t'ClosureStep _ x = pure x allSteps :: Fold' (StackClosure s b a) (StackStep s b a) allSteps = from i'StackClosure.(l'1.each.l'1.each .+ l'2.each) subClosure :: Int -> Fold' (StackClosure s b a) (StackClosure s b a) subClosure :: Int -> Traversal' (StackClosure s b a) (StackClosure s b a) subClosure 0 = id subClosure n = (allSteps.t'ClosureStep.subClosure (n+1)) .+ (from i'StackClosure.l'1.each.l'2.subClosure (n-1)) subClosure n = \\k (StackClosure act ps p) -> StackClosure act <\\$> traverse (\\(ph,px) -> liftA2 (,) (traversel (each.t'ClosureStep.subClosure (n+1)) k ph) (traversel (subClosure (n-1)) k px)) ps <> traversel (each.t'ClosureStep.subClosure (n+1)) k p closureSplices :: Fold' (StackClosure s b a) (StackClosure s b a) closureSplices = allSteps.t'ClosureStep.subClosure (1::Int) allSteps :: (forall f. Applicative f => StackClosure s b a -> f (StackClosure s b a)) -> Traversal' (StackClosure s b a) (StackStep s b a) allSteps sub k (StackClosure act ps p) = StackClosure act<\\$>traverse (\\(ph,c) -> liftA2 (,) (each k ph) (sub c)) ps<>traverse k p closureSplices :: Traversal' (StackClosure s b a) (StackClosure s b a) closureSplices = allSteps pure.t'ClosureStep.subClosure (1::Int) runClosure execBuiltin' onComment clos = do (_,p) <- flatten =<< forl closureSplices clos (\\c -> flatten c <&> \\(act,p) -> StackClosure act [] p) (_,p) <- flatten clos stack =~ (StackProg p:) where flatten (StackClosure act cs c) = do pref <- map fold \\$ for cs \\$ \\(i,StackClosure act' _ p) -> (i+) <\\$> do traverse_ (runStep execBuiltin' onComment) p stack <~ \\case (h:t) -> (t,[case act' of CloseConstant -> ConstStep h ; CloseExec -> ExecStep h]) [] -> ([],[]) return (act,pref + c) where flattenSteps = traversel (each.t'ClosureStep.subClosure 1) (\\c -> flatten c <&> \\(act,p) -> StackClosure act [] p) flatten (StackClosure act cs c) = (act,) <\\$> liftA2 (+) (map fold \\$ for cs \\$ \\(i,StackClosure act' _ p) -> (+) <\\$> flattenSteps i <*> do traverse_ (runStep execBuiltin' onComment) p stack <~ \\case (h:t) -> (t,[case act' of CloseConstant -> ConstStep h ; CloseExec -> ExecStep h]) [] -> ([],[])) (flattenSteps c) runStep execBuiltin' onComment (VerbStep s) = getl (dict.at s) >>= \\case Just v -> runStep execBuiltin' onComment (ExecStep v) ... ...\nMarkdown is supported\n0% or .\nYou are about to add 0 people to the discussion. Proceed with caution.\nFinish editing this message first!" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9505922,"math_prob":0.83928216,"size":554,"snap":"2021-43-2021-49","text_gpt3_token_len":112,"char_repetition_ratio":0.10363636,"word_repetition_ratio":0.0,"special_character_ratio":0.1732852,"punctuation_ratio":0.086021505,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9687781,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-22T18:22:06Z\",\"WARC-Record-ID\":\"<urn:uuid:386950cc-2521-4d59-8de9-390d80e69fd2>\",\"Content-Length\":\"218322\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d199230d-79d7-4b76-9e9a-959551f62c04>\",\"WARC-Concurrent-To\":\"<urn:uuid:90a3507e-0a98-498a-93ed-13ab33a97530>\",\"WARC-IP-Address\":\"129.88.175.2\",\"WARC-Target-URI\":\"https://gricad-gitlab.univ-grenoble-alpes.fr/coiffiem/BHR/-/commit/d8a2196c1b1e3d1956b846fb520f941e6c42ee28\",\"WARC-Payload-Digest\":\"sha1:645RRJWDXY5DM67YPTL5I3MQOTYMQAPE\",\"WARC-Block-Digest\":\"sha1:TMX3Z2WG4HXKIBCILOSUUAZMPMVUWNE4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585518.54_warc_CC-MAIN-20211022181017-20211022211017-00588.warc.gz\"}"}
https://www.sanfoundry.com/mechanical-operations-questions-answers-theory-filtration/	[ "# Mechanical Operations Questions and Answers – Theory of Filtration\n\n«\n»\n\nThis set of Mechanical Operations Multiple Choice Questions & Answers (MCQs) focuses on “Theory of Filtration”.\n\n1. The ratio of void volume and total surface area of particles is defined as _____\na) Resistance\nc) Repulsion\nd) Void age\n\nExplanation: The hydraulic radius which is defined as the flow area per unit wetted perimeter and is represented by RH.\n\n2. What does the ε represents?\na) Porosity\nb) Density\nc) Viscosity\nd) Acidity\n\nExplanation: The ratio of void volume and the total volume is defined as the porosity, which is denoted by ε.\n\n3. Calculate ε, if the total area is 45 m2 and the void area is 100 m2?\na) 10.5\nb) 0.45\nc) 2.222\nd) 0.89\n\nExplanation: As we know ε = void area/total area = 100/45 = 2.222.\n\n4. Calculate the hydraulic radius, if porosity is 0.6 and specific gravity is 1.50?\na) 1\nb) 0.6\nc) 0.2\nd) 1.8\n\nExplanation: As we know that ε/sp (1- ε) = 0.6/1.50(1-0.6) = 1.\n\n5. ____ of the liquid is defined as volumetric flow rate of the liquid divided by the total area.\na) Final velocity\nb) Total volume\nc) Superficial velocity\nd) Wetted perimeter\n\nExplanation: The superficial velocity is defined as the volumetric rate and liquid divided by total or empty cross sectional area Usup.\n\n6. What does the below equation represent?", null, "a) Resistance\nc) Repulsion\nd) Void age\n\nExplanation: Here ε is the void fraction sometimes called porosity of the bed and Sp is the specific surface area of the particles.\n\n7. What does the below represent?\nt= f tc\na) Cake formation\nb) Drum rotation\nc) Initial extraction\nd) Extra time\n\nExplanation: The above equation represents the time taken for cake formation, if tc is the time for one full rotation of the drum.\n\n8. ____ is defined as the volume of void filled and total volume.\na) Upper velocity\nb) Saturation\nc) Residual saturation\nd) Superficial velocity\n\nExplanation: The saturation (S) is usually defined with respect to the wetting fluid and is therefore the fractional volume of voids occupied by the wetting fluid.\n\n9. ____ is the maximum volume wetted to ratio of total volume.\na) Residual saturation\nb) Saturation\nc) Residual saturation\nd) Superficial velocity\n\nExplanation: The residual saturation (SF) is defined as the ratio of maximum volume wetting fluid eliminated from flow and total volume of voids.\n\n10. Calculate S, if the voids filled are 40 m3 and having 50 m3 of total volume?\na) 0.8\nb) 1.25\nc) 40\nd) 50\n\nExplanation: As we know that S = void volume/total volume = 40/50 = 0.8.\n\nSanfoundry Global Education & Learning Series – Mechanical Operations.", null, "" ]	[ null, "https://www.sanfoundry.com/wp-content/uploads/2017/10/mechanical-operations-questions-answers-theory-filtration-q6.png", null, "https://www.sanfoundry.com/wp-content/uploads/2017/01/Manish2.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.77803934,"math_prob":0.9906477,"size":2975,"snap":"2020-45-2020-50","text_gpt3_token_len":793,"char_repetition_ratio":0.15382026,"word_repetition_ratio":0.09037328,"special_character_ratio":0.25445378,"punctuation_ratio":0.11447811,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9984854,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,2,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-23T22:50:39Z\",\"WARC-Record-ID\":\"<urn:uuid:26d0ca84-7b79-4d2e-be31-21bac107f8b1>\",\"Content-Length\":\"77844\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:91d732c1-7a4c-433c-b896-7fcd809a9640>\",\"WARC-Concurrent-To\":\"<urn:uuid:71701664-4d2b-42e6-8a5b-651f219553b1>\",\"WARC-IP-Address\":\"104.25.131.119\",\"WARC-Target-URI\":\"https://www.sanfoundry.com/mechanical-operations-questions-answers-theory-filtration/\",\"WARC-Payload-Digest\":\"sha1:53BLLCAZBFHQRH7M2MNGKXBRGDNWN57A\",\"WARC-Block-Digest\":\"sha1:WRFGXB7TYZDPGPYBGRQAIP45RNS4YSVY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141168074.3_warc_CC-MAIN-20201123211528-20201124001528-00432.warc.gz\"}"}
https://www.aimspress.com/article/10.3934/mine.2020010	[ "### Mathematics in Engineering\n\n2020, Issue 2: 203-229. doi: 10.3934/mine.2020010\nResearch article Special Issues\n\n# Interaction of rigid body motion and rarefied gas dynamics based on the BGK model\n\n• Received: 22 July 2019 Accepted: 04 December 2019 Published: 31 December 2019\n• In this paper we present simulations of moving rigid bodies immersed in a rarefied gas. The rarefied gas is simulated by solving the Bhatnager-Gross-Krook (BGK) model for the Boltzmann equation. The Newton-Euler equations are solved to simulate the rigid body motion. The force and the torque on the rigid body is computed from the surrounded gas. An explicit Euler scheme is used for the time integration of the Newton-Euler equations. The BGK model is solved by the semi-Lagrangian method suggested by Russo & Filbet . Due to the motion of the rigid body, the computational domain for the rarefied gas (and the interface between the rigid body and the gas domain) changes with respect to time. To allow a simpler handling of the interface motion we have used a meshfree method for the interpolation procedure in the semi-Lagrangian scheme. We have considered a one way, as well as a two-way coupling of rigid body and gas flow. We use diffuse reflection boundary conditions on the rigid body and also on the boundary of the computational domain. In one space dimension the numerical results are compared with analytical as well as with Direct Simulation Monte Carlo (DSMC) solutions of the Boltzmann equation. In the two-dimensional case results are compared with DSMC simulations for the Boltzmann equation and with results obtained by other researchers. Several test problems and applications illustrate the versatility of the approach.\n\nCitation: Sudarshan Tiwari, Axel Klar, Giovanni Russo. Interaction of rigid body motion and rarefied gas dynamics based on the BGK model[J]. Mathematics in Engineering, 2020, 2(2): 203-229. doi: 10.3934/mine.2020010\n\n### Related Papers:\n\n• In this paper we present simulations of moving rigid bodies immersed in a rarefied gas. The rarefied gas is simulated by solving the Bhatnager-Gross-Krook (BGK) model for the Boltzmann equation. The Newton-Euler equations are solved to simulate the rigid body motion. The force and the torque on the rigid body is computed from the surrounded gas. An explicit Euler scheme is used for the time integration of the Newton-Euler equations. The BGK model is solved by the semi-Lagrangian method suggested by Russo & Filbet . Due to the motion of the rigid body, the computational domain for the rarefied gas (and the interface between the rigid body and the gas domain) changes with respect to time. To allow a simpler handling of the interface motion we have used a meshfree method for the interpolation procedure in the semi-Lagrangian scheme. We have considered a one way, as well as a two-way coupling of rigid body and gas flow. We use diffuse reflection boundary conditions on the rigid body and also on the boundary of the computational domain. In one space dimension the numerical results are compared with analytical as well as with Direct Simulation Monte Carlo (DSMC) solutions of the Boltzmann equation. In the two-dimensional case results are compared with DSMC simulations for the Boltzmann equation and with results obtained by other researchers. Several test problems and applications illustrate the versatility of the approach.", null, "###### 通讯作者: 陈斌, [email protected]\n• 1.\n\n沈阳化工大学材料科学与工程学院沈阳 110142", null, "2.385\n\nArticle outline\n\n## Other Articles By Authors\n\n• On This Site", null, "", null, "DownLoad: Full-Size Img PowerPoint" ]	[ null, "https://www.aimspress.com/style/web/images/article/1.gif", null, "https://www.aimspress.com/aimspress-upload/journal/mine/cover_2020410144341391.jpg", null, "https://www.aimspress.com/article/10.3934/mine.2020010", null, "https://www.aimspress.com/style/web/images/article/download.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5938625,"math_prob":0.92558414,"size":5774,"snap":"2021-31-2021-39","text_gpt3_token_len":1915,"char_repetition_ratio":0.111438476,"word_repetition_ratio":0.024242423,"special_character_ratio":0.3574645,"punctuation_ratio":0.20149875,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9713945,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,null,null,null,null,2,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-21T05:57:55Z\",\"WARC-Record-ID\":\"<urn:uuid:bd071078-1ced-4793-8d1a-1fbc8b7a0bb3>\",\"Content-Length\":\"136865\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b2483d55-4610-4859-8240-283c6ab2aedb>\",\"WARC-Concurrent-To\":\"<urn:uuid:f90e93b9-503f-4542-a7c0-f5339efa4f3f>\",\"WARC-IP-Address\":\"99.84.216.105\",\"WARC-Target-URI\":\"https://www.aimspress.com/article/10.3934/mine.2020010\",\"WARC-Payload-Digest\":\"sha1:ZII42ATVH4YWRLX7BEUTOJK7MU7ESEKC\",\"WARC-Block-Digest\":\"sha1:VUBXSLZVK3ZW32PRLJD3BKCYAZYRICH6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057158.19_warc_CC-MAIN-20210921041059-20210921071059-00504.warc.gz\"}"}
https://assignmentpapermaster.com/a-coaxial-cable-mathematics-homework-help/	[ "# A coaxial cable \| Mathematics homework help\n\nA Coaxial Cable. A small solid conductor with radius a is supported by insulating. nonmagnetic disks on the axis of a thinwalled tube with inner radius b. The inner and outer conductors carry equal currents i in opposite directions.\n(a) Use Ampere’s law to find the magnetic field at any point in the volume between the conductors.\n(b) Write the expression fur the flux diPB through a narrow strip of length I parallel to the axis, of width dr, at a distance T from the axis of the cable and lying in a plane containing the axis.\n(c) Integrate your expression from part (b) over the volume between the two conductors to find the total flux produced by a current i in the central conductor.\n(d) Show that the inductance of a lenglh I of the cable is\n(e) Use Eq. (30.9) to calculate the enetgy stored in the magnetic field for a length l of thecable." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8868219,"math_prob":0.9633945,"size":892,"snap":"2023-14-2023-23","text_gpt3_token_len":211,"char_repetition_ratio":0.1295045,"word_repetition_ratio":0.0,"special_character_ratio":0.2174888,"punctuation_ratio":0.06779661,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9523786,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-25T18:17:12Z\",\"WARC-Record-ID\":\"<urn:uuid:2c6d0af9-99bd-4e65-afbe-0e6ed6ad3a12>\",\"Content-Length\":\"71271\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6333a481-5801-4e6c-b132-636a3c1f883b>\",\"WARC-Concurrent-To\":\"<urn:uuid:4195eaa6-5fc5-45fa-8f32-2d5b5beac1a2>\",\"WARC-IP-Address\":\"162.213.251.180\",\"WARC-Target-URI\":\"https://assignmentpapermaster.com/a-coaxial-cable-mathematics-homework-help/\",\"WARC-Payload-Digest\":\"sha1:UALLPZCLXEVPMZLYXXQVAT6HGMTC7TX4\",\"WARC-Block-Digest\":\"sha1:YOKGA3EJ5QGTXFGMORPSLP4GF5FQWOF3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296945368.6_warc_CC-MAIN-20230325161021-20230325191021-00427.warc.gz\"}"}
https://stylesubstancesoul.com/essay-tips/can-you-use-photomath-on-computer/	[ "# Can you use Photomath on computer?\n\n## Can you use Photomath on computer?\n\nWe can use the application photomath on computer and Mac both users can use the application for the solution. We can use the application photomath on computer. Windows and Mac both users can use the application for the solution of any mathematical issue.\n\n## What’s better than PhotoMath?\n\nWhen comparing Mathway vs PhotoMath, the Slant community recommends Mathway for most people. In the question“What are the best Android apps for math problem solving?” Mathway is ranked 2nd while PhotoMath is ranked 4th.\n\n## Which is the best app to solve maths problems?\n\nPhotomath. Photomath is probably the best app for solving mathematical problems. iMathematics. iMathematics lets you type in equations and solves them for you. MyScript Calculator. PCalc. Graphing Calculator. Khan Academy. Meritnation. Mathematicus.\n\n## How can I learn math on my own?\n\nSteps to Studying Math on Your OwnFirst, determine where you want to end up. Determine where to start, obviously. Find a Syllabus to Avoid Unnecessary Depth. Gather your References, Solution Manuals, and “Solved Problems” Types of Books. Prioritize Deep, Concept-Based Learning. Put Links to Resources in One Place.\n\n## How long does it take to learn math from scratch?\n\nOriginally Answered: How long does it take to learn math from scratch? It depends how much you want to learn. If you want to just learn basic as 1,2,3 sum, multiplication and subtraction then it will just take minimum 1 week.\n\n## How can I understand maths easily?\n\n7 Tips for Maths Problem SolvingPractice, Practice & More Practice. It is impossible to study maths properly by just reading and listening. Review Errors. Master the Key Concepts. Understand your Doubts. Create a Distraction Free Study Environment. Create a Mathematical Dictionary. Apply Maths to Real World Problems." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.90061986,"math_prob":0.49518412,"size":1849,"snap":"2021-31-2021-39","text_gpt3_token_len":399,"char_repetition_ratio":0.1203252,"word_repetition_ratio":0.11724138,"special_character_ratio":0.19361818,"punctuation_ratio":0.1461318,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9735829,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-22T08:46:09Z\",\"WARC-Record-ID\":\"<urn:uuid:f308e5c8-9f63-4d84-beca-cb6fbaaf913e>\",\"Content-Length\":\"21262\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e398c961-1bec-4643-8978-1dabf4fce33a>\",\"WARC-Concurrent-To\":\"<urn:uuid:138c20e7-3a11-49c4-a042-716af005bff2>\",\"WARC-IP-Address\":\"104.21.57.103\",\"WARC-Target-URI\":\"https://stylesubstancesoul.com/essay-tips/can-you-use-photomath-on-computer/\",\"WARC-Payload-Digest\":\"sha1:FZOQF2GH5WM2HHV7NP7XFSFMJEJJQ4ON\",\"WARC-Block-Digest\":\"sha1:4Y5HU5ZKGSCDQWCOZR3LPCKK4TOLGEOS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057337.81_warc_CC-MAIN-20210922072047-20210922102047-00695.warc.gz\"}"}
http://percentagecalculator.pro/What-is_58_percent-of_60_%3F	[ "# 58 percent of 60?\n\nWhat is % of\n\n### Percentage Calculator 2\n\nis what percent of ?\n\nis % of what?\n\n### Solution for 'What is 58% of 60?'\n\nIn all the following examples consider that:\n\n• The percentage figure is represented by X%\n• The whole amount is represented by W\n• The portion amount or part is represented by P\n\n#### Solution Steps\n\nThe following question is of the type \"How much X percent of W\", where W is the whole amount and X is the percentage figure or rate\".\n\nLet's say that you need to find 58 percent of 60. What are the steps?\n\nStep 1: first determine the value of the whole amount. We assume that the whole amount is 60.\n\nStep 2: determine the percentage, which is 58.\n\nStep 3: Convert the percentage 58% to its decimal form by dividing 58 into 100 to get the decimal number 0.58:\n\n58100 = 0.58\n\nNotice that dividing into 100 is the same as moving the decimal point two places to the left.\n\n58.0 → 5.80 → 0.58\n\nStep 4: Finally, find the portion by multiplying the decimal form, found in the previous step, by the whole amount:\n\n0.58 x 60 = 34.8 (answer).\n\nThe steps above are expressed by the formula:\n\nP = W × X%100\n\nThis formula says that:\n\n\"To find the portion or the part from the whole amount, multiply the whole by the percentage, then divide the result by 100\".\n\nThe symbol % means the percentage expressed in a fraction or multiple of one hundred.\n\nReplacing these values in the formula, we get:\n\nP = 60 × 58100 = 60 × 0.58 = 34.8 (answer)\n\nTherefore, the answer is 34.8 is 58 percent of 60.\n\n### Solution for '58 is what percent of 60?'\n\nThe following question is of the type \"P is what percent of W,” where W is the whole amount and P is the portion amount\".\n\nThe following problem is of the type \"calculating the percentage from a whole knowing the part\".\n\n#### Solution Steps\n\nAs in the previous example, here are the step-by-step solution:\n\nStep 1: first determine the value of the whole amount. We assume that it is 60.\n\n(notice that this corresponds to 100%).\n\nStep 2: Remember that we are looking for the percentage 'percentage'.\n\nTo solve this question, use the following formula:\n\nX% = 100 × PW\n\nThis formula says that:\n\n\"To find the percentage from the whole, knowing the part, divide the part by the whole then multiply the result by 100\".\n\nThis formula is the same as the previous one shown in a different way in order to have percent (%) at left.\n\nStep 3: replacing the values into the formula, we get:\n\nX% = 100 × 5860\n\nX% = 580060\n\nSo, the answer is 58 is 96.67 percent of 60\n\n### Solution for '60 is 58 percent of what?'\n\nThe following problem is of the type \"calculating the whole knowing the part and the percentage\".\n\n### Solution Steps:\n\nStep 1: first determine the value of the part. We assume that the part is 60.\n\nStep 2: identify the percent, which is 58.\n\nStep 3: use the formula below:\n\nW = 100 × PX%\n\nThis formula says that:\n\n\"To find the whole, divide the part by the percentage then multiply the result by 100\".\n\nThis formula is the same as the above rearranged to show the whole at left.\n\nStep 4: plug the values into the formula to get:\n\nW = 100 × 6058\n\nW = 100 × 1.0344827586207" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9205283,"math_prob":0.9952469,"size":3089,"snap":"2020-45-2020-50","text_gpt3_token_len":822,"char_repetition_ratio":0.17698541,"word_repetition_ratio":0.12820514,"special_character_ratio":0.3111039,"punctuation_ratio":0.13521545,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99998295,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-26T06:25:15Z\",\"WARC-Record-ID\":\"<urn:uuid:d4bd08ad-c1b6-4baf-9116-e28799c00557>\",\"Content-Length\":\"28450\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ec7c0455-42a4-4cc3-b5d6-3fef33f5c6da>\",\"WARC-Concurrent-To\":\"<urn:uuid:b03ca8f3-323b-441e-bd37-aacbefe9b693>\",\"WARC-IP-Address\":\"104.18.33.170\",\"WARC-Target-URI\":\"http://percentagecalculator.pro/What-is_58_percent-of_60_%3F\",\"WARC-Payload-Digest\":\"sha1:LSKZORYGWNRU2NYDOP5L53DSMPK6JJK2\",\"WARC-Block-Digest\":\"sha1:75SLAHW4ESU7PMRBWQGI7LITNXY7SMMY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107890586.57_warc_CC-MAIN-20201026061044-20201026091044-00214.warc.gz\"}"}
https://www.iqclab.eu/ft/	[ "# fT\n\nThis function creates permutation matrices, which allows to permute the in- and output channels of an uncertain plant M in accordance with the given uncertainty blocks. This is an function that is used in, for example, iqcanalysis, and can also be used independently in another context or to set up an alternative analysis of controller synthesis problem.\n\nGiven the some matrix or system", null, "with", null, "and", null, "being the number of input and output channels respectively, the function", null, "creates the permutation matrices", null, "and", null, "based on the input/output channel selectors", null, "and", null, ", which are structures.\n\nExample\n\nWith", null, ",", null, ",", null, ", and", null, ", we obtain:", null, "Then the matrix", null, "would permute into", null, "Previous page" ]	[ null, "https://usercontent.one/wp/www.iqclab.eu/wp-content/ql-cache/quicklatex.com-10ebb71bad275c1815a8f2a8c5dea0be_l3.png", null, "https://usercontent.one/wp/www.iqclab.eu/wp-content/ql-cache/quicklatex.com-0cdaa691ddf18a198b8fd3049106819c_l3.png", null, "https://usercontent.one/wp/www.iqclab.eu/wp-content/ql-cache/quicklatex.com-24b4caf6f4195928fbd692971bcd3a64_l3.png", null, "https://usercontent.one/wp/www.iqclab.eu/wp-content/ql-cache/quicklatex.com-af18381403cdd31262ca85a82eb54806_l3.png", null, "https://usercontent.one/wp/www.iqclab.eu/wp-content/ql-cache/quicklatex.com-5f66cac63100dedfee31d62bab7c69e1_l3.png", null, "https://usercontent.one/wp/www.iqclab.eu/wp-content/ql-cache/quicklatex.com-41c3d83cac7ee8c5d7c98f0934027a4c_l3.png", null, "https://usercontent.one/wp/www.iqclab.eu/wp-content/ql-cache/quicklatex.com-84a03a34d3890f0d6b4cc716d9e7f4f5_l3.png", null, "https://usercontent.one/wp/www.iqclab.eu/wp-content/ql-cache/quicklatex.com-f75265f6127423bd9d1c90c8b05fbe35_l3.png", null, "https://usercontent.one/wp/www.iqclab.eu/wp-content/ql-cache/quicklatex.com-b62ac8cd0289a6c0fae08a326f1f8556_l3.png", null, "https://usercontent.one/wp/www.iqclab.eu/wp-content/ql-cache/quicklatex.com-1a2aef58d27d66ae07145594faf1f2fe_l3.png", null, "https://usercontent.one/wp/www.iqclab.eu/wp-content/ql-cache/quicklatex.com-ce83b603e782451bcb02eaec662f0980_l3.png", null, "https://usercontent.one/wp/www.iqclab.eu/wp-content/ql-cache/quicklatex.com-20a4af65ff78460d637f4a3265444519_l3.png", null, "https://usercontent.one/wp/www.iqclab.eu/wp-content/ql-cache/quicklatex.com-8c31635f656a8495b4b0244aecebf031_l3.png", null, "https://usercontent.one/wp/www.iqclab.eu/wp-content/ql-cache/quicklatex.com-9f010cdd350a5cd08a0faca9f0e061b8_l3.png", null, "https://usercontent.one/wp/www.iqclab.eu/wp-content/ql-cache/quicklatex.com-1daecb9911b3a6cb811e3644d7cdca7f_l3.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.88203037,"math_prob":0.85874724,"size":649,"snap":"2023-40-2023-50","text_gpt3_token_len":124,"char_repetition_ratio":0.12713179,"word_repetition_ratio":0.0,"special_character_ratio":0.18798152,"punctuation_ratio":0.11965812,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9624386,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30],"im_url_duplicate_count":[null,null,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-09T15:40:53Z\",\"WARC-Record-ID\":\"<urn:uuid:cb4607a4-ba76-45a8-a5f9-fa2b7aad67ea>\",\"Content-Length\":\"100732\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7de033da-c278-4d1b-96ef-fd5a58ccaac6>\",\"WARC-Concurrent-To\":\"<urn:uuid:2b49fce2-9ae2-47bd-9ee6-0a8f3b89c7ab>\",\"WARC-IP-Address\":\"46.30.213.51\",\"WARC-Target-URI\":\"https://www.iqclab.eu/ft/\",\"WARC-Payload-Digest\":\"sha1:UNWIS6IUBD4X5TRRNIYW3A6ER2SZ6XRB\",\"WARC-Block-Digest\":\"sha1:4A4BAAMZ6HLG3LX73KY4TRY7XJMPZYSX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100912.91_warc_CC-MAIN-20231209134916-20231209164916-00867.warc.gz\"}"}
https://insightmaker.com/book/export/html/52	[ "# Flows\n\nA flow transports a material from one stock to another. As a designer, you must enter the rate of the flow. So, for instance, in our river example the magnitude of the flow could be a million gallons of water and the period over which it flows could be one day: in effect a million gallons per day. Flows may connect two stocks together. Alternatively, a flow can lack a start or an end. In this case, the unconnected end acts as an unlimited stock of material.\n\nAs with the initial value of a stock, the rate of a flow can be a simple number or a more complex equation. The equation can be based on many factors including the values of the stocks the flow is connected to and the current time. For example, if we had a stock representing the population of rabbits (called Rabbit Population), we could make the flow into the stock be the number of rabbits times a constant birthrate. This equation would result in geometric growth in the rabbit population:\n\n[Rabbit Population]0.01\n\nAlternatively, if we had a flow representing rainfall, we could set the flow rate to be based on the current time of the year. Rainfall over the course of a year can be approximated using a sinusoidal function like the following:\n\nsin(2pi(Weeks()+10)/52)30+35\n\nWhere weeks is a Insight Maker function that automatically takes on the value of the current simulation time measured in weeks, pi is the constant 3.14159, and sin is another Insight Maker function that applies the trigonometric sine function to an angle inputted using radians.\n\nA flow can be restricted to positive flows only. In this case, the flow will only be applied if the calculated magnitude of the flow rate is positive. This restriction would be applicable for both our rabbit and rain examples. We know that neither of the flows should ever be negative (you cannot have negative births or negative rain) so we might want to set them to only allow positive flows just to insure that such an impossible event never happens (though it never will in any event given the equations we specified).\n\nEvery flow has two special properties that may be referenced in its equations. The first is Alpha which refers to the stock at the start of the flow. The second is Omega which refers to the stock at the end of the flow. In your equations you may use either of these instead of needing to know the actual names of a flow’s start or end stocks. For instance if you wanted 5% of the water from a lake to evaporate out of it each time period, you could use the following statement for the rate of a flow whose beginning was the lake:\n\n[Alpha]*0.05" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9373809,"math_prob":0.98875964,"size":2588,"snap":"2021-04-2021-17","text_gpt3_token_len":547,"char_repetition_ratio":0.13119195,"word_repetition_ratio":0.017391304,"special_character_ratio":0.21290572,"punctuation_ratio":0.07751938,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99202645,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-17T09:43:49Z\",\"WARC-Record-ID\":\"<urn:uuid:b6c67ec3-851d-4884-9ada-e728c8956d87>\",\"Content-Length\":\"8093\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c6b0d57b-d766-4e66-b5b0-2d266f67d75f>\",\"WARC-Concurrent-To\":\"<urn:uuid:25b8ab36-4b58-4262-9b55-f30b2f778d03>\",\"WARC-IP-Address\":\"104.26.6.235\",\"WARC-Target-URI\":\"https://insightmaker.com/book/export/html/52\",\"WARC-Payload-Digest\":\"sha1:AMK4BORNSOXRLKFPYAE7IP57VUJ7NXOA\",\"WARC-Block-Digest\":\"sha1:7JDTH3ELAFTUFTZYTUQ7RIRX7W4PCKAI\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703511903.11_warc_CC-MAIN-20210117081748-20210117111748-00460.warc.gz\"}"}
https://www.isprimenumber.com/prime/1193	[ "# Is 1193 a Prime Number?\n\nYes, 1193 is a prime number. 1193 is divisible by 1 and itself.\n\nYes, 1193 is a prime number.\n\n### Why 1193 is a prime number?\n\nBecause 1193 has no positive divisors rather than 1 and itself.\n\nThe next prime number of 1193 is 1201\nThe previous prime number of 1193 is 1187.\n\n### Related Prime Numbers\n\nBiggest 10 prime numbers smaller than 1193\n\nSmallest 10 prime numbers bigger than 1193\n\n### Related Non-Prime Numbers\n\nBiggest 10 composite numbers smaller than 1193\n\nSmallest 10 composite numbers bigger than 1193" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.85296744,"math_prob":0.9744721,"size":517,"snap":"2019-13-2019-22","text_gpt3_token_len":147,"char_repetition_ratio":0.24561404,"word_repetition_ratio":0.11494253,"special_character_ratio":0.32108316,"punctuation_ratio":0.089108914,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95886827,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-23T20:24:03Z\",\"WARC-Record-ID\":\"<urn:uuid:79518f9e-bb89-440b-b282-31cf67a4dc42>\",\"Content-Length\":\"12487\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:88952b68-6602-42c5-9241-c06aa0cdf5ad>\",\"WARC-Concurrent-To\":\"<urn:uuid:f1e0e795-e518-4093-a8b6-83b8f6c03112>\",\"WARC-IP-Address\":\"104.197.145.71\",\"WARC-Target-URI\":\"https://www.isprimenumber.com/prime/1193\",\"WARC-Payload-Digest\":\"sha1:TEVRGHCAJZLOUHIKGEAAKVNYT6JMA3LJ\",\"WARC-Block-Digest\":\"sha1:3ZOOAFSSTAL7CI7R6VRZXUONXYPDLCF4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912203021.14_warc_CC-MAIN-20190323201804-20190323223804-00399.warc.gz\"}"}
https://www.cs.ubbcluj.ro/files/curricula/2003/disc/emo005.htm	[ "## \"Babes-Bolyai\" University of Cluj-Napoca Faculty of Mathematics and Computer Science\n\n Functional analysis (2)\n Code Semes-ter Hours: C+S+L Credits Type Section MO005 6 2+1+0 5 compulsory Matematică MO005 6 2+1+0 4 optional Matematică-Informatică\n Teaching Staff in Charge\n Prof. COBZAS Stefan, Ph.D., [email protected]. SÁNDOR Jozsef, Ph.D., [email protected]\n Aims To present the basic results of the theory of topological vector spaces and of locally convex spaces and to deep some results on normed spaces, taught in the first semester.\n Content 1. Topological vector spaces (TVS). Definition and fundamental properties. The continuity of additive applications between TVS. Bases of neighborhoods of the origin. Linear functionals and hyperplanes in TVS. Seminormed spaces. Finite dimensional TVS - Tihonov@s theorem. Topological properties of convex sets in TVS. Minkowski functional and seminorms. The continuity of Minkowski functional. Metrizability. Completeness, compactness and total boundedness in TVS. Note: Some questions concerning general topology as: the indroduction of a topology using neighborhood sustems, nets and filters, topological products and Tihonov@s theorem, will be treated at the seminars. 2. Locally convex spaces (LCS) The notion of LCS. The local convex toplogy generated by a family of seminorms. Locally convex bases, J. von Neumann@s theorem. Bourbaki@s theorem on the characterization of the continuity of linear applications between LCS. The existence of some continuous linear functionals on LCS. Weak topologies on normed spaces. The separation of convex sets by closed hyperplanes. Extremal points and Krein-Milman@s theorem, Lyapunov convexity theorem. 3. Elements of distribution theory Inductive limits of LCS. The fundamental space of infinitely derivable functions. The notion of distribution. Regular distributions and singular distributions, distributions of finite order - examples.The topology of the space of distributions and the convergence of sequences of regular distributions. Applications of distribution theory to the solving of partial differential equations. 4. Fixpoint theorems Contractions and Banach fixpoint theorem. Fixpoint theorems for nonexpansive mappings. Theorems of Brouwer, Schauder-Tihonov and Markov-Kakutani. 5. Compact operators between normed spaces Arzela-Ascoli compactness criterium in C(T). Compact operators fundamental properties. The conjugate of an operator. Schauder@s theorem on the compactness of the conjugate of a compact operator. The spectrum of a compact operator, Riesz theory. 6. Operators on Hilbert spaces Symmetric operators, unitary operators, positive operators. The spectra of linear operators on Hilbert spaces.\n References 1. Cristescu R., Notiuni de analiza functionala liniara, Ed. Academiei, Bucuresti 1998. 2.M. Fabian et al., Functional analysis and infinite-dimensional geometry, Springer Verlag, Berlin-New York 2001. 3. Kantorovici L.V. si Akilov G.P., Analiza functionala, Editura Stiintifica si Enciclopedica, Bucuresti 1986. 4. Kutateladze S., Fundamentals of functional analysis, Kluwer AP, Dordrecht 1996. 5. Megginson R.E., An introduction to Banach space theory, Springer Verlag, Berlin-New York 1998. 6. Muntean I. Analiza functionala, Litografiat Universitatea Babes-Bolyai, Cluj-Napoca 1993. 7. Muntean I., Analiza functionala-Capitole speciale, Litografiat Universitatea Babes-Bolyai, Cluj-Napoca 1990. 8. Schaeffer H.H., Wolf M.P., Topological vector spaces, Springer Verlag, Berlin-New York 1999.\n Assessment Exam." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.70273715,"math_prob":0.817611,"size":3357,"snap":"2023-40-2023-50","text_gpt3_token_len":846,"char_repetition_ratio":0.13540113,"word_repetition_ratio":0.0,"special_character_ratio":0.19958296,"punctuation_ratio":0.21639344,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96864104,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-11T01:52:43Z\",\"WARC-Record-ID\":\"<urn:uuid:8209bd61-c1b2-4d9d-ba60-ce1f8fa2a6de>\",\"Content-Length\":\"7710\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:60c8947b-7bea-4805-85ef-4e887e389a36>\",\"WARC-Concurrent-To\":\"<urn:uuid:6fb9e440-02aa-46cc-8697-69411afdaa78>\",\"WARC-IP-Address\":\"193.231.20.20\",\"WARC-Target-URI\":\"https://www.cs.ubbcluj.ro/files/curricula/2003/disc/emo005.htm\",\"WARC-Payload-Digest\":\"sha1:RNFGXVKABXJ4QJLTTEWBIL4U6AZTICT6\",\"WARC-Block-Digest\":\"sha1:USAK73GCLMMOGW3UFHTYZQHC763TPY2B\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679103464.86_warc_CC-MAIN-20231211013452-20231211043452-00346.warc.gz\"}"}
http://symoe.com/2799872-Why-do-equal-quality-zinc-powder-and-iron-powder-put-into-dilute-sulfuric-acid-with-equal-mass-and-s	[ "", null, "# Why do equal quality zinc powder and iron powder put into dilute sulfuric acid with equal mass and same solute mass fraction produce more hydrogen Why do equal quality zinc powder and iron powder add the same mass fraction of dilute sulfuric acid to produce more hydrogen\n\n## Why do equal quality zinc powder and iron powder put into dilute sulfuric acid with equal mass and same solute mass fraction produce more hydrogen Why do equal quality zinc powder and iron powder add the same mass fraction of dilute sulfuric acid to produce more hydrogen\n\nThe molar mass of iron is less than that of zinc. The molar number of iron and zinc iron is greater than that of zinc. The hydrogen replaced by equal molar iron and zinc is equal. Therefore, when equal mass zinc powder and iron powder are put into the same mass fraction of dilute sulfuric acid, iron produces more hydrogen\n\n### 3.8g of the mixture of zinc powder, aluminum powder, iron powder and magnesium powder reacts exactly with dilute sulfuric acid with a mass fraction of 25% solute. Evaporate the water of the reaction mixture to 11g of solid (excluding crystal water), and the mass of hydrogen generated in the reaction is () A. 0.15g B. 0.20g C. 0.30g D. 0.45g\n\nAccording to the subject information, the mixture and sulfuric acid react exactly, so the sum of the mass of the mixture and pure sulfuric acid (H2SO4) is equal to the sum of the mass of the remaining solid and hydrogen. The mass of sulfate before and after the reaction = (11g - 3.8g) = 7.2g, then the mass of pure sulfuric acid = 7.2g ÷ 96\n98 = 7.35g. Therefore, the mass of hydrogen = the mass of hydrogen element in sulfuric acid = 7.35g - 7.2g = 0.15g, so a\n\n### Put 10g of the mixture of iron powder and copper powder into a beaker containing sufficient dilute sulfuric acid. After full reaction, the total mass of the material in the beaker is reduced by 0.1g compared with the total mass of the metal mixture and dilute sulfuric acid before reaction. Calculate the mass fraction of copper powder in the original mixture\n\nThe mass of generated hydrogen is 0.1g, and the mass of iron participating in the reaction is assumed to be x.fe + H2SO4  FeSO4 + H2 ↑ 56 2x ...\n\n### 14g of the mixture of iron powder and copper powder is put into 116.2 dilute hydrochloric acid and the reaction is just complete. At this time, the total mass of each substance in the beaker is 129G 14g of the mixture of iron powder and copper powder was put into 116.2 dilute hydrochloric acid and the reaction was just complete. At this time, the total mass of each substance in the beaker was 129 Put 14g of the mixture of iron powder and copper powder into 116.2 dilute sulfuric acid and react exactly. At this time, the total mass of each substance in the beaker is 129.8g. Find: 1. Mass fraction of iron powder in the original mixture; 2. The volume of concentrated sulfuric acid with a mass fraction of 36% (density of 1.19g / cm3) is required to prepare the dilute sulfuric acid used above; 3. Mass fraction of solute in the solution after reaction Please don't copy\n\nThe total mass before the reaction is 14 + 116.2g = 130.2g, the mass of the beaker after the reaction is 129.8g, and the mass before and after the reaction is reduced by 130.2-129.8 = 0.4g. Because copper is inactive and does not react with acid, there is only one reaction: Fe + 2H + = Fe2 + + H2, so 0.4g of hydrogen is generated, 11.2g of fe11, 19.8g of sulfuric acid, 1. Mass fraction of Fe = 5.6 /\n\n### 14g of the mixture of iron powder and copper powder is put into 116.2g of dilute hydrochloric acid and the reaction is just complete. At this time, the total mass of each substance in the beaker is 129.8g 1. What is the mass of hydrogen produced? 2. Mass fraction of iron powder in the original mixture?\n\nAfter dilute hydrochloric acid is added to iron powder and copper powder, iron reacts with hydrochloric acid to generate ferrous chloride and hydrogen. Copper does not react with hydrochloric acid. According to mass conservation, m (iron powder + copper powder) + m (hydrochloric acid) = m (salt solution + copper powder) + m (hydrogen), so m (hydrogen) = 14 + 114.2-129.8 = 0.4gfe + 2HCl = FeCl2 + h256 2\n\n### 14g of the mixture of iron powder and copper powder is put into 116.2g dilute hydrochloric acid, and the reaction is just complete. At this time, the total mass of each substance in the beaker is 129.8g Q: The volume of concentrated hydrochloric acid (density 1.19g/cm3) with a mass fraction of 36% is required to prepare the dilute hydrochloric acid used above? Write out the process and analysis. Don't copy it\n\nLet the mass of iron be x, the mass of FeCl2 generated be y, and the mass of hydrochloric acid solute consumed be Z\nThe mass of H2 is 14g + 116.2g-129.8g = 0.4g\nFe + 2HCl ＝ FeCl2 + H2↑\n56 73 127 2\nX Z Y 0.4g\nX=11.2g\nY=25.4g\nZ=14.6g\nThe volume of 36% concentrated hydrochloric acid used is:\n14.6g÷36％÷1.19g／cm3=34.08mL\nThe mass fraction of solute in the solution obtained after reaction is:\n25.4g÷(11.2g+116.2g－0.4g) × 100％=20.0％" ]	[ null, "http://symoe.com/statics/23f69ebe6b7fcb87303c3ae3449a09df.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8927154,"math_prob":0.98669624,"size":4735,"snap":"2022-05-2022-21","text_gpt3_token_len":1346,"char_repetition_ratio":0.20820123,"word_repetition_ratio":0.21420257,"special_character_ratio":0.27835268,"punctuation_ratio":0.12252964,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9930213,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-29T06:25:56Z\",\"WARC-Record-ID\":\"<urn:uuid:0456047d-1c0c-4bd2-abbb-2baaee0bdcc3>\",\"Content-Length\":\"26134\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:925e473a-da5d-4113-af39-eb501b1f72b1>\",\"WARC-Concurrent-To\":\"<urn:uuid:c904c4f3-6d10-4a68-8099-f47c28dd65af>\",\"WARC-IP-Address\":\"172.67.214.64\",\"WARC-Target-URI\":\"http://symoe.com/2799872-Why-do-equal-quality-zinc-powder-and-iron-powder-put-into-dilute-sulfuric-acid-with-equal-mass-and-s\",\"WARC-Payload-Digest\":\"sha1:QG4EV7FUCBEJIY774FO66XRD37BS4POV\",\"WARC-Block-Digest\":\"sha1:SNV6I6YLG44KHVQ6233N6RRZUBIQWHPL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652663039492.94_warc_CC-MAIN-20220529041832-20220529071832-00074.warc.gz\"}"}
https://www.transtutors.com/questions/1-40-a-what-is-the-reflection-loss-of-a-0-001-in-thick-copper-shield-to-a-1000-hz-pl-2849244.htm	[ "# 1. (a) What is the reflection loss of a 0.001-in thick copper shield to a 1000-Hz plane...\n\n1. (a) What is the reflection loss of a 0.001-in thick copper shield to a 1000-Hz plane wave?(b) If the thickness is increased to 0.01 in, what is the reflection loss?\n\n2. (a) Calculate the skin depth and absorption loss for a brass shield 0.062 in thick at the following frequencies: a. 0.1 kHz, b. 1 kHz, c. 10 kHz, and d. 100 kHz?(b) Calculate the shielding effectiveness of a 0.015-in-thick copper shield located 1 in from the source of a 10-kHz magnetic field.\n\n A large microprocessor draws a total transient current of 10 A from a 3.3-V supply. The logic has a rise/fall time of 1 ns. It is desirable to limit theVcc-to-ground noise voltage peaks to 250 mV, and each decoupling capacitor has 5 nH of inductance in series with it. The decoupling will be done with a multiplicity of equal value capacitors, and should be effective at all frequencies above 20 MHz. (a) Draw a plot of the target impedance versus frequency. (b) What is the minimum number of decoupling capacitors required? (c) What is the minimum value for each of the individual decoupling capacitors? (d) Could larger value capacitors be used just as effectively?\n\n</pclass=\"msolistparagraph\">\n\n## Recent Questions in Electrical Engineering\n\nSubmit Your Questions Here !\nCopy and paste your question here...\nAttach Files" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8891348,"math_prob":0.91295564,"size":2115,"snap":"2019-13-2019-22","text_gpt3_token_len":564,"char_repetition_ratio":0.11179536,"word_repetition_ratio":0.49175823,"special_character_ratio":0.2718676,"punctuation_ratio":0.13646056,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9778742,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-25T15:28:46Z\",\"WARC-Record-ID\":\"<urn:uuid:052b209b-c6fc-4750-a162-8dbaa11cd40f>\",\"Content-Length\":\"74319\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d7a233e2-c422-4914-97af-685d16d02b32>\",\"WARC-Concurrent-To\":\"<urn:uuid:393fb751-1bcc-43a1-a3a9-833a20691dc2>\",\"WARC-IP-Address\":\"35.199.55.187\",\"WARC-Target-URI\":\"https://www.transtutors.com/questions/1-40-a-what-is-the-reflection-loss-of-a-0-001-in-thick-copper-shield-to-a-1000-hz-pl-2849244.htm\",\"WARC-Payload-Digest\":\"sha1:N23T73HAOBOLEZUBWSX5Z6BG66VJUU7A\",\"WARC-Block-Digest\":\"sha1:ORKYHU6RJLQFG2UX4GO6HGN5W4PXIGIQ\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232258120.87_warc_CC-MAIN-20190525144906-20190525170906-00328.warc.gz\"}"}
https://cs.stackexchange.com/questions/59923/cfg-equivalent-of-regular-expressions	[ "# CFG Equivalent of regular expressions\n\nSo I was wondering something about the Chomsky hierarchy.\n\nDFAs (and NFAs) accept regular languages, while NPDAs accept context-free languages.\n\nRight-regular or left-regular grammars produce regular languages, while context-free grammars produce context-free languages.\n\nRegular expressions represent regular languages - but is there a context-free equivalent? i.e. is there some additional term we could allow to appear in a regular expression such that regular expressions with that functionality now represent context-free languages?\n\nA (rather technical) answer, where star-operator is replaced by least fixed point (equivalent to recursive formulation as in CFG) is given by our cstheory cousins: Does there exist an extension of regular expressions that captures the context free languages?\n\nThe author (Neel Krishnaswami) of that answer does not know an original source of his construction, but a later comment by Tim Schaeffer refers to the 1973 book Formal Languages by Arto Salomaa. In chapter VI.11 Salomaa introduces regular-like expressions. That book might not be in your possession, but in turn refers to A characterization of context-free languages, Journal of Computer and System Sciences, 1973, by Jozef Gruska, which is open access.\n\nThe operator Gruska considers is the following, I loosely quote from his paper. Let $\\sigma$ be a symbol and $L,L_1$ be languages. The $\\sigma$-substitution of $L_1$ into $L$, denoted by $L \\overset{\\sigma}{\\uparrow}L_1$ is defined by\n\n$$L \\overset{\\sigma}{\\uparrow}L_1 = \\{ x_0 y_1x_1 \\dots x_{k-1} y_k x_k \\mid x_0 \\sigma x_1 \\dots x_{k-1} \\sigma x_k\\in L, \\text{no \\sigma in }x_i, y_i\\in L_1 \\}.$$\n\nAfter noting that he operation $\\overset{\\sigma}{\\uparrow}$ is associative one defines the $\\sigma$-iteration as\n\n$$L^\\sigma = \\{z \\mid z\\in L \\cup L\\overset{\\sigma}{\\uparrow}L \\cup L\\overset{\\sigma}{\\uparrow}L\\overset{\\sigma}{\\uparrow}L \\cup \\dots \\text{ does not contain an occurrence of } \\sigma \\}.$$\n\nNote that if we have only a single variable, like $S\\to aSbS \\mid \\lambda$ then this substitution closure $\\{aSbS, \\lambda\\}^S$ is indeed the language generated by the grammar." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8397036,"math_prob":0.95772856,"size":1597,"snap":"2020-24-2020-29","text_gpt3_token_len":469,"char_repetition_ratio":0.13057125,"word_repetition_ratio":0.0,"special_character_ratio":0.27175954,"punctuation_ratio":0.08561644,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9960184,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-03T21:02:41Z\",\"WARC-Record-ID\":\"<urn:uuid:5cee2b5f-3bdf-4746-a529-af9d62fbe014>\",\"Content-Length\":\"152144\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9a5fb0b5-da19-4331-8446-d6fb3a89358f>\",\"WARC-Concurrent-To\":\"<urn:uuid:7f41c0bd-a4dc-40ae-a9e8-3a8f2e03b9ef>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://cs.stackexchange.com/questions/59923/cfg-equivalent-of-regular-expressions\",\"WARC-Payload-Digest\":\"sha1:2J24C2BKXLBBVNFCLV7CITBEMAJRCNAD\",\"WARC-Block-Digest\":\"sha1:OQFIOJEXYFESHICGN2QKRMKPJUAQGKSH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655882934.6_warc_CC-MAIN-20200703184459-20200703214459-00572.warc.gz\"}"}
https://patents.justia.com/patent/6411886	[ "# Method of detecting and controlling detonation of an internal combustion engine\n\nA method of detecting and controlling detonation, including the steps of: generating (10) a vibration signal (D) proportional to the intensity of vibration on the crankcase; wide-band-filtering (11) the vibration signal (D) to generate a first intermediate signal; rectifying (12) the first intermediate signal to generate a second intermediate signal; integrating (13) the second intermediate signal to generate a first numeric value; calculating (14) a logarithm of the first numeric value to obtain a second numeric value (Xi); calculating (16) a difference between the second numeric value (Xi) and the mean value (&mgr;i) to obtain a third numeric value (&dgr;i) with a predetermined threshold value (&dgr;o); determining (18) the presence of detonation in the event the third numeric value (&dgr;i) has a first predetermined relationship with the predetermined threshold value (&dgr;o); forming (19, 21) a detonation index (Ii) indicating the behavior of the engine in terms of detonation; and obtaining (22) a spark lead correction value (CAi) as a function of the detonation index (Ii).\n\n## Latest C.R.F. Societa Consortile per Azioni Patents:\n\nDescription\nTECHNICAL FIELD\n\nThe present invention relates to a method of detecting and controlling detonation of an internal combustion engine.\n\nBACKGROUND ART\n\nAs is known, in certain operating conditions, internal combustion engines are subject to high-intensity detonation, i.e. spontaneous combustion of the mixture, which occurs in many engine cycles, and results in impaired efficiency, dangerous overheating and reduced life of the engine, as well as possibly in sudden failure of certain engine components.\n\nVarious systems are therefore being studied, or have already been proposed, to detect detonation and act accordingly on combustion parameters to reduce its effects and the likelihood of it occurring.\n\nSuch systems are mainly based on directly or indirectly monitoring the pressure cycle pattern in the cylinder as a function of the position of the piston inside the cylinder. The pressure cycle pattern is bell-shaped, reaching a peak close to the top dead-center position, which peak is a classic rounded shape when combustion is normal (FIG. 1a), and has numerous indentations in the presence of detonation (FIG. 1b).\n\nInformation by which to detect detonation is obtained by analyzing the indentations.\n\nSome known methods employ sensors located inside the combustion chamber to directly determine the amplitude of the indentations. While providing for highly accurate, reliable amplitude values, such a sensor arrangement calls for sophisticated high-cost technology and is therefore only suitable for laboratory or prototype application.\n\nOther methods employ sensors located, and for detecting vibration, on the crankcase. Though technically simpler and cheaper, the values obtained by such sensors are subject to greater interference than those detected directly, on account of the vibration measured on the crankcase also being the result of other phenomena besides crankcase-filtered variations in pressure inside the cylinder.\n\nPatent GB-A-2 265 006 filed by Nippondenso Co Ltd, for example, relates to a detonation control system featuring detonation sensors on the crankcase, and which detects detonation by comparing the intensity of the sensor signal with a decision threshold, and provides for eliminating detonation by accordingly controlling engine operating parameters, such as injection advance. More specifically, the system performs a logarithmic conversion of the intensity of the sensor signal, determines its distribution, calculates a value corresponding to the standard deviation of the distribution, and compares the value with a threshold calculated on the basis of the previously calculated value and a mean value of said distribution.\n\nThe distribution is determined by processing numeric values representing the amplitudes of spectral components obtained by narrow-band filtering the output signal of the sensors. The output signal, in fact, comprises numerous harmonics, and narrow-band filtration is employed to only extract the highest-energy harmonic.\n\nIt is important to note that, in the above patent, the detonation threshold is adapted at each cycle.\n\nIn some cases, however, controlling operation of the engine on the basis of one frequency in the spectrum and one engine cycle at a time may prove restrictive.\n\nThat is, considering only one frequency may result in others, different from the one considered but generated all the same by detonation, being overlooked, while control designed solely to eliminate local detonation cycle by cycle may not always be the best solution, in that, in terms of engine operation and efficiency, it sometimes pays to tolerate a given number of detonations every so many engine cycles and/or in only some of the cylinders. Trace detonation, in fact, by enabling high efficiency of the engine, is a favourable engine operating condition, and therefore one to be encouraged.\n\nFinally, adapting the decision threshold calls for numerous numeric processing operations, which further complicate implementation of the method.\n\nOne proposal to eliminate the aforementioned drawbacks is the detonation detecting method described in Patent Application WO-97/21084 filed on Nov. 29, 1996 by the same Applicant and published on Dec. 06, 1997.\n\nThe detection method in the above patent application provides for wide-band filtering and rectifying the output signal of an acceleration sensor on the crankcase; integrating the rectified signal in a time window appropriately located with respect to the combustion considered; and effecting a logarithmic conversion of the integration result.\n\nRepeating the above operations for each combustion of the engine provides for defining a distribution of values, of which are calculated, at each combustion, a mean value and a numeric value correlated to the standard deviation on the basis of previous combustion values. On the basis of the numeric value and a predetermined threshold value which is constant alongside variations in engine speed, a detonation coefficient value is then calculated at each engine combustion, and which indicates the detonation “propensity” of the engine.\n\nDISCLOSURE OF INVENTION\n\nIt is an object of the present invention to provide an improved detonation detecting method as compared with that described in Patent Application WO-97/21084.\n\nAccording to the present invention, there is provided a method of detecting and controlling detonation of an internal combustion engine, comprising the steps of:\n\na) acquiring a vibration signal proportional to the intensity of vibration on the crankcase;\n\nb) wide-band filtering said vibration signal to generate a first intermediate signal;\n\nc) rectifying said first intermediate signal to generate a second intermediate signal;\n\nd) integrating said second intermediate signal to generate a first numeric value;\n\ne) calculating a logarithm of said first numeric value to obtain a second numeric value;\n\nf) calculating a mean value as a function of said second numeric value;\n\ncharacterized by comprising the steps of:\n\ng) calculating a difference between said second numeric value and said mean value to obtain a third numeric value;\n\nh) comparing said third numeric value with a predetermined threshold value;\n\nm) determining the presence of detonation in the event said third numeric value has a first predetermined relationship with said predetermined threshold value;\n\nn) forming a detonation index indicating the behaviour of the engine in terms of detonation; and\n\no) calculating, from said detonation index, a correction value to be added to the spark lead.\n\nBRIEF DESCRIPTION OF DRAWINGS\n\nA preferred, non-limiting embodiment of the present invention will be described by way of example with reference to the accompanying drawings, in which:\n\nFIGS. 1a and 1b show two pressure cycle graphs in the absence and presence of detonation respectively;\n\nFIG. 2 shows a simplified diagram of a combustion system featuring the detection method according to the invention;\n\nFIG. 3 shows a flow chart of a first embodiment of the detection method according to the invention;\n\nFIGS. 4 and 5 show graphs of certain quantities relative to the detection method according to the invention.\n\nBEST MODE FOR CARRYING OUT THE INVENTION\n\nNumber 1 in FIG. 2 indicates as a whole a detonation detecting device connected to an internal combustion engine 2, of which only the crankcase 3 and cylinders 4 are shown.\n\nDetecting device 1 comprises a known acceleration sensor 5 (not described in detail) fitted to crankcase 3 of engine 2 and generating a vibration signal D correlated to the intensity of the vibration on crankcase 3; and a central processing unit 6 for receiving vibration signal D and implementing the operations relative to the method according to the present invention and described below with reference to the FIG. 3 flow chart.\n\nThe operations described with reference to FIG. 3 are repeated for each combustion of engine 2 to continually monitor detonation in the various engine cycles, and each of the values calculated in a generic i-th repetition is indicated hereinafter by the subscript “i”.\n\nIn the following description, reference is made to a counter Ii associated with the cylinder 4 of engine 2 in which the combustion considered takes place. More specifically, each cylinder 4 is assigned a respective counter I, and, to identify the various counters I in the following description, each is followed by an “[n]” term indicating the n-th cylinder 4 with which the counter is associated.\n\nThe counter values are initialized, e.g. set to 0, at the start of detection (e.g. when engine 2 is turned on) and are subsequently modified according to whether or not detonation is established in respective cylinders 4.\n\nMore specifically, and as will be made clear in the following description, the values of a counter I define a detonation index of engine 2 indicating the behaviour of engine 2 in terms of detonation.\n\nAs shown in FIG. 3, after each combustion of engine 2, vibration signal D is generated by acceleration sensor 5 and acquired by central processing unit 6 (block 10).\n\nVibration signal D is then wide-band filtered (block 11), rectified (block 12), and integrated in an appropriate time window (block 13).\n\nMore specifically, the integration time window commences at the combustion phase of the cylinder 4 considered, immediately following the top dead-center position, and is of constant angular duration with respect to the speed of engine 2, whereas the filtration band of vibration signal D is calibrated experimentally and normally ranges between 5 and 20 kHz.\n\nThe logarithm xi of the integration result is then calculated (block 14), and a corresponding mean value &mgr;i is determined as a function of logarithm value xi (block 15).\n\nMore specifically, mean value &mgr;i is calculated according to the equation: { μ i = ( 1 - λ 1 ) · x i + λ 1 · μ i - 1 i ≥ 2 μ 1 = x 1 ⁢ ⁢ &AutoLeftMatch;\n\nwhere &mgr;i is the mean value calculated at the i-th repetition; &mgr;i−1 is the mean value calculated at the previous repetition i−1; xi is the logarithm value calculated at the i-th repetition; and &lgr;1 is a predefined parameter.\n\nAs can be seen from the above equation, at the first repetition, the calculated mean value &mgr;i coincides with the calculated logarithm value xi, whereas, from the second repetition onwards, mean value &mgr;i is updated on the basis of both the logarithm value xi calculated at the i-th repetition, and the mean value &mgr;i−1 calculated at repetition i−1, each weighted according to parameter &lgr;1.\n\nA peculiar characteristic of the above algorithm for updating mean value &mgr;i is that of defining a filter with a very small time constant enabling the calculated mean value to rapidly follow variations in the xi values obtained following logarithmic conversion of the integration result. This characteristic is extremely useful in transient states by ensuring precise tracking of vibration signal D.\n\nCalculation of mean value &mgr;i is followed by calculating the value &dgr;1 of the difference between value xi and mean value &mgr;i (block 16).\n\nValue &dgr;1 is then compared with a predetermined threshold value &dgr;o (block 17) to determine the presence or absence of detonation and accordingly modify the value of the counter Ii[n] associated with the cylinder in question.\n\nAccording to the present invention, threshold value &dgr;o is constant for all the repeat detection operations, is constant alongside variations in the speed of engine 2, and is correlated to the standard deviation value &sgr; of the distribution of values xi in the absence of detonation.\n\nIn particular, the standard deviation value &sgr; of the distribution of values xi in the absences of detonation is conveniently calculated on the basis of laboratory engine tests, by so setting the injection advance of the engine as to eliminate detonation. Given the standard deviation value &sgr;, the corresponding threshold value &dgr;o is then calculated and conveniently memorized in central processing unit 6. Tests have shown that the best threshold value &dgr;o for reliably detecting detonation is roughly 3-4 times the standard deviation value &sgr;.\n\nBlock 17 determines whether value &dgr;i is greater than or equal to predetermined threshold value, &dgr;o. If &dgr;i is greater than or equal to threshold value &dgr;o (YES output of block 17), the presence of detonation in the cylinder 4 in question is established (block 18) and the value of counter Ii[n] associated with the cylinder is modified (block 19) according to the equation:\n\nIi[n]=Ii−1[n]−M·(&dgr;i−&dgr;o)\n\nwhere M is a proportion coefficient, which is generally a function of the speed of engine 2 and the n-th cylinder considered at the time. In other words, by depending on the cylinder 4 considered, proportion coefficient M takes into account the transfer function of crankcase 3 in the combustion considered.\n\nMore specifically, at the i-th repetition, the value of the n-th counter Ii[n] is calculated by decreasing the value of the n-th counter Ii[n] at repetition i−1 by an amount proportional to the difference between value &dgr;i and threshold value &dgr;o.\n\nConversely, if &dgr;i is less than threshold value &dgr;o (NO output of block 17), the absence of detonation in the cylinder 4 in question is established (block 20) and the value of counter Ii[n] associated with the cylinder is modified (block 21) according to the equation:\n\nIi[n]=Ii−1[n]+1\n\nThat is, at the i-th repetition, the value of the n-th counter Ii[n] is calculated by increasing by one unit the value of the n-th counter Ii[n] at repetition i−1.\n\nAs stated, the current value of a counter Ii[n] defines a detonation index of engine 2 indicating the behaviour of engine 2 in terms of detonation, i.e. the tendency of engine 2 to detonate. By increasing the value of the counter when no detonation is detected in the respective cylinder, and decreasing the counter value when detonation is detected, it therefore follows that high positive counter values indicate a tendency of engine 2 either not to detonate or to detonate in temporally widely spaced engine cycles, whereas high negative counter values indicate a tendency of engine 2 to detonate.\n\nOnce the value of counter Ii[n] is modified according to whether the presence or absence of detonation is established in the respective cylinder (block 19 or 21), the current value Ii[n] of the counter is used to determine an injection advance correction value CAi[n] relative to the n-th cylinder 4 considered, and defining the best correction to be made to the injection advance of the cylinder to control detonation in the cylinder as required (block 22). The injection advance correction value CAi[n] is determined according to the equation:\n\nCAi[n]=KP·Ii[n]\n\nwhere Ii[n] is the current value of the n-th counter at the i-th repetition; and Kp is a factor defining how the advance correction is to be implemented.\n\nIn particular, Kp may be constant alongside variations in the speed of engine 2, or may be a function of the speed of engine 2.\n\nPurely by way of example, FIG. 4 shows a graph of the value Ii[n] assumed, in successive engine cycles, by a counter associated with a cylinder 4 of engine 2; and FIG. 5 shows a graph of the advance correction value CAi[n] for the same cylinder, calculated according to the above equation and as a function of value Ii[n] of the same counter. In FIG. 5, advance correction value CAi[n] varies in discrete values of 0.5° as opposed to continuously.\n\nThe advantages of the method according to the present invention are as follows.\n\nIn particular, to determine the presence or absence of detonation, the method according to the present invention operates on the basis of the difference between logarithm value xi and mean value &mgr;i, with no need, as in the method described in Patent Application WO-97/21084, to calculate a numeric value correlated to the standard deviation of the distribution of values xi. This obviously provides for greatly simplifying the processing required to determine the detonation behaviour and effectively control operation of engine 2, as well as for reducing the number of components required.\n\nMoreover, the method according to the present invention models the detonation phenomenon by defining a threshold value &dgr;o valid in all operating conditions and for all types of engine 2, thus further simplifying processing and reducing the number of components required.\n\nMoreover, the method according to the present invention provides for determining advance correction value CAi[n] using a law of correction enabling the number of detonations present at any speed of engine 2 to be “measured out” as required, and which provides for obtaining an advance correction directly proportional to the amount of detonation that has taken place.\n\nMoreover, the method according to the present invention operates on the basis of a wide-band-filtered signal to make use of the maximum possible amount of information, even though this obviously involves also detecting the effects of other phenomena not related to detonation.\n\nFinally, the method according to the present invention employs a mean value &mgr;i updating algorithm enabling the calculated mean value to rapidly follow any variations in the xi values obtained following logarithmic conversion of the integration result.\n\nClearly, changes may be made to the method as described and illustrated herein without, however, departing from the scope of the present invention.\n\n## Claims\n\n1. A method of detecting and controlling detonation of an internal combustion engine ( 2 ), comprising the steps of:\n\na) acquiring ( 10 ) a vibration signal (D) proportional to the intensity of vibration on the crankcase;\nb) wide-band filtering ( 11 ) said vibration signal (D) to generate a first intermediate signal;\nc) rectifying ( 12 ) said first intermediate signal to generate a second intermediate signal;\nd) integrating ( 13 ) said second intermediate signal to generate a first numeric value;\ne) calculating ( 14 ) a logarithm of said first numeric value to obtain a second numeric value (x i );\nf) calculating ( 15 ) a mean value (&mgr; i ) as a function of said second numeric value (x i );\ng) calculating ( 16 ) a difference between said second numeric value (x i ) and said mean value (&mgr; i ) to obtain a third numeric value (&dgr; i );\nh) comparing ( 17 ) said third numeric value (&dgr; i ) with a predetermined threshold value (&dgr; o );\nm) determining ( 18 ) the presence of detonation in the event said third numeric value (&dgr; i ) has a first predetermined relationship with said predetermined threshold value (&dgr; o );\nn) forming a detonation index indicating the behaviour of the engine in terms of detonation; and\no) calculating, from said detonation index, a correction value to be added to the spark lead.\n\n2. A method as claimed in claim 1, characterized in that said step i) comprises the step of determining ( 18 ) said presence of detonation in the event said third numeric value (&dgr; i ) is greater than or equal to said predetermined threshold value (&dgr; o ).\n\n3. A method as claimed in claim 1, characterized by also comprising the step of:\n\np) determining ( 20 ) the absence of detonation in the event said third numeric value (&dgr; i ) does not have said first predetermined relationship with said predetermined threshold value (&dgr; o ).\n\n4. A method as claimed in claim 1, characterized in that said step b) comprises the step of filtering said vibration signal (D) between 5 and 25 kHz.\n\n5. A method as claimed in claim 1, characterized in that said threshold value (&dgr; o ) used in step h) is predetermined and constant alongside a variation in the speed of said engine ( 2 ).\n\n6. A method as claimed in claim 1, characterized in that said step d) comprises the step of integrating said second intermediate signal in a predetermined time window during the combustion phase.\n\n7. A method as claimed in claim 6, characterized in that said integration time window commences, during said combustion phase, immediately following the top dead-center position, and is of constant angular duration with respect to the speed of said engine ( 2 ).\n\n8. A method as claimed in claim 1, characterized by also comprising the steps of:\n\nq) calculating ( 19, 21 ) a detonation index (I i [n]) indicating the behaviour of said engine ( 2 ) in terms of detonation; and\nr) calculating ( 22 ) an injection advance correction value (CA i [n]) as a function of said detonation index (I i [n]).\n\n9. A method as claimed in claim 8, characterized in that said step r) comprises the step of calculating ( 22 ) said injection advance correction value (CA i [n]) according to the equation:\n\n10. A method as claimed in claim 8, characterized in that said step q) comprises the step of:\n\ns) calculating ( 19 ) said detonations index (I i [n]) as a function of said third numeric value (&dgr; i ) and of said predetermined threshold value (&dgr; o ) in the event said third numeric value (&dgr; i ) has said first predetermined relationship with said predetermined threshold value (&dgr; o ).\n\n11. A method as claimed in claim 4, characterized by repeating said steps from a) to r) to generate a number (i) of said second numeric values (x i ), of said mean values (&mgr; i ), of said third numeric values (&dgr; i ), of said detonation indexes (I i [n]), and of said injection advance correction values (CA i [n]).\n\n12. A method as claimed in claim 11, characterized in that each of said detonation indexes (I i [n]) is calculated ( 19 ), in the event said third numeric value (&dgr; i ) has said first predetermined relationship with said predetermined threshold value (&dgr; o ), according to the equation:\n\n13. A method as claimed in claim 11, characterized in that each of said detonation indexes (I i [n]) is calculated ( 21 ), in the event said third numeric value (&dgr; i ) does not have said first predetermined relationship with said predetermined threshold value (&dgr; o )according to the equation:\n\n14. A method as claimed in claim 7, characterized in that each of said mean values (&mgr; i ) is calculated ( 15 ) according to the equation: { μ i = ( 1 - λ 1 ) · x i + λ 1 · μ i - 1 i &GreaterEqual; 2 μ 1 = x 1 ⁢ ⁢ &AutoLeftMatch;\n\n15. A method as claimed in claim 11, characterized in that said threshold value (&dgr; o ) used in step h) is correlated to the standard deviation value of the distribution of second numeric values (x i ) obtained in the absence of detonation.\n\nPatent History\nPatent number: 6411886\nType: Grant\nFiled: Mar 22, 2000\nDate of Patent: Jun 25, 2002\nAssignee: C.R.F. Societa Consortile per Azioni (Orbassano)\nInventors: Gianluca Morra (Avigliana), Stefano Re Fiorentin (Grugliasco), Paolo Tubetti (Settimo Torinese), Gianpiero Einaudi (Turin)\nPrimary Examiner: Willis R. Wolfe\nAttorney, Agent or Law Firm: Crowell & Moring LLP\nApplication Number: 09/446,988" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.90720403,"math_prob":0.9605766,"size":24854,"snap":"2021-04-2021-17","text_gpt3_token_len":5392,"char_repetition_ratio":0.18354125,"word_repetition_ratio":0.22739658,"special_character_ratio":0.21883802,"punctuation_ratio":0.099378884,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.9552969,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-22T19:16:38Z\",\"WARC-Record-ID\":\"<urn:uuid:4c45df70-8ea2-493e-9e00-7b644abde416>\",\"Content-Length\":\"75660\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e7410fd7-390e-4b7c-bd5e-c1d0ab0acb87>\",\"WARC-Concurrent-To\":\"<urn:uuid:845ee4da-f86c-4061-b9ff-e3e7f86fcf38>\",\"WARC-IP-Address\":\"34.226.165.133\",\"WARC-Target-URI\":\"https://patents.justia.com/patent/6411886\",\"WARC-Payload-Digest\":\"sha1:WH7T7Q26PY42T4KAACOKFV4BCURSVJGC\",\"WARC-Block-Digest\":\"sha1:EKPREHGJAXPX7FMJNUNTF66IVYFWSV3G\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703531335.42_warc_CC-MAIN-20210122175527-20210122205527-00246.warc.gz\"}"}
https://numbermatics.com/n/6061366/	[ "# 6061366\n\n## 6,061,366 is an even composite number composed of three prime numbers multiplied together.\n\nWhat does the number 6061366 look like?\n\nThis visualization shows the relationship between its 3 prime factors (large circles) and 8 divisors.\n\n6061366 is an even composite number. It is composed of three distinct prime numbers multiplied together. It has a total of eight divisors.\n\n## Prime factorization of 6061366:\n\n### 2 × 43 × 70481\n\nSee below for interesting mathematical facts about the number 6061366 from the Numbermatics database.\n\n### Names of 6061366\n\n• Cardinal: 6061366 can be written as Six million, sixty-one thousand, three hundred sixty-six.\n\n### Scientific notation\n\n• Scientific notation: 6.061366 × 106\n\n### Factors of 6061366\n\n• Number of distinct prime factors ω(n): 3\n• Total number of prime factors Ω(n): 3\n• Sum of prime factors: 70526\n\n### Divisors of 6061366\n\n• Number of divisors d(n): 8\n• Complete list of divisors:\n• Sum of all divisors σ(n): 9303624\n• Sum of proper divisors (its aliquot sum) s(n): 3242258\n• 6061366 is a deficient number, because the sum of its proper divisors (3242258) is less than itself. Its deficiency is 2819108\n\n### Bases of 6061366\n\n• Binary: 101110001111101001101102\n• Base-36: 3LWZA\n\n### Squares and roots of 6061366\n\n• 6061366 squared (60613662) is 36740157785956\n• 6061366 cubed (60613663) is 222695543238428975896\n• The square root of 6061366 is 2461.9841591691\n• The cube root of 6061366 is 182.3294559909\n\n### Scales and comparisons\n\nHow big is 6061366?\n• 6,061,366 seconds is equal to 10 weeks, 3 hours, 42 minutes, 46 seconds.\n• To count from 1 to 6,061,366 would take you about fifteen weeks!\n\nThis is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)\n\n• A cube with a volume of 6061366 cubic inches would be around 15.2 feet tall.\n\n### Recreational maths with 6061366\n\n• 6061366 backwards is 6631606\n• The number of decimal digits it has is: 7\n• The sum of 6061366's digits is 28\n• More coming soon!\n\n#### Copy this link to share with anyone:\n\nMLA style:\n\"Number 6061366 - Facts about the integer\". Numbermatics.com. 2023. Web. 21 September 2023.\n\nAPA style:\nNumbermatics. (2023). Number 6061366 - Facts about the integer. Retrieved 21 September 2023, from https://numbermatics.com/n/6061366/\n\nChicago style:\nNumbermatics. 2023. \"Number 6061366 - Facts about the integer\". https://numbermatics.com/n/6061366/\n\nThe information we have on file for 6061366 includes mathematical data and numerical statistics calculated using standard algorithms and methods. We are adding more all the time. If there are any features you would like to see, please contact us. Information provided for educational use, intellectual curiosity and fun!\n\nKeywords: Divisors of 6061366, math, Factors of 6061366, curriculum, school, college, exams, university, Prime factorization of 6061366, STEM, science, technology, engineering, physics, economics, calculator, six million, sixty-one thousand, three hundred sixty-six.\n\nOh no. Javascript is switched off in your browser.\nSome bits of this website may not work unless you switch it on." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8463021,"math_prob":0.95274556,"size":2810,"snap":"2023-40-2023-50","text_gpt3_token_len":759,"char_repetition_ratio":0.1336422,"word_repetition_ratio":0.034883723,"special_character_ratio":0.33487543,"punctuation_ratio":0.17260788,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9801068,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-21T22:22:51Z\",\"WARC-Record-ID\":\"<urn:uuid:46293c90-53e2-4e5f-b0e7-7c848d31d1c7>\",\"Content-Length\":\"19536\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3f84f7e2-8198-4c2a-80d2-e26d0c2fa32c>\",\"WARC-Concurrent-To\":\"<urn:uuid:ab8f5d39-1bb3-49bd-8a42-865a85fea83c>\",\"WARC-IP-Address\":\"72.44.94.106\",\"WARC-Target-URI\":\"https://numbermatics.com/n/6061366/\",\"WARC-Payload-Digest\":\"sha1:ZEILIVFYVTX4ZHDWXKHZ3X6KWWTOCPGT\",\"WARC-Block-Digest\":\"sha1:IWYCXVSSMW6BXBIVH5YT5W5SQQ23DPOG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506045.12_warc_CC-MAIN-20230921210007-20230922000007-00860.warc.gz\"}"}
https://findthefactors.com/2015/10/29/662-more-candy-corn/	[ "# 662 More Candy Corn\n\n662 is the sum of the twelve prime numbers from 31 to 79.", null, "Print the puzzles or type the solution on this excel file: 10 Factors 2015-10-26\n\nHere’s the same candy corn puzzle but less colorful.", null, "—————————————————————————————————\n\n• 662 is a composite number.\n• Prime factorization: 662 = 2 x 331\n• The exponents in the prime factorization are 1 and 1. Adding one to each and multiplying we get (1 + 1)(1 + 1) = 2 x 2 = 4. Therefore 662 has exactly 4 factors.\n• Factors of 662: 1, 2, 331, 662\n• Factor pairs: 662 = 1 x 662 or 2 x 331\n• 662 has no square factors that allow its square root to be simplified. √662 ≈ 25.72936.", null, "—————————————————————————————————", null, "Ricardo tweeted the solution to the puzzle so I’m including it here as well." ]	[ null, "https://i0.wp.com/findthefactors.com/wp-content/uploads/2015/10/662-puzzle-candy-corn.jpg", null, "https://i0.wp.com/findthefactors.com/wp-content/uploads/2015/10/662-puzzle.jpg", null, "https://i0.wp.com/findthefactors.com/wp-content/uploads/2015/10/662-factor-pairs.jpg", null, "https://i0.wp.com/findthefactors.com/wp-content/uploads/2015/10/662-logic.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.76295626,"math_prob":0.9678888,"size":847,"snap":"2023-14-2023-23","text_gpt3_token_len":256,"char_repetition_ratio":0.17912218,"word_repetition_ratio":0.0,"special_character_ratio":0.41676506,"punctuation_ratio":0.12048193,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97129434,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,4,null,9,null,9,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-10T10:56:46Z\",\"WARC-Record-ID\":\"<urn:uuid:b8813a34-0afd-42c9-88e3-d1da66ca188e>\",\"Content-Length\":\"91014\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1523b2da-c31d-4906-a933-6d1243644ab1>\",\"WARC-Concurrent-To\":\"<urn:uuid:e8073039-191d-44b1-9c5e-474fa85539b1>\",\"WARC-IP-Address\":\"192.0.78.153\",\"WARC-Target-URI\":\"https://findthefactors.com/2015/10/29/662-more-candy-corn/\",\"WARC-Payload-Digest\":\"sha1:5IMOQECHRA35LIV6LR4QAA73ELEWJVBB\",\"WARC-Block-Digest\":\"sha1:6X2O443VOYXD6YB24G2ZA5UWIRWB7T7I\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224657169.98_warc_CC-MAIN-20230610095459-20230610125459-00711.warc.gz\"}"}
http://fahry.info/free-division-worksheets-grade-4/division-worksheets-grade-3-free-multiplication-and-for-4th-4-sheet-math-digits-by-1/	[ "", null, "# Division Worksheets Grade 3 Free Multiplication And For 4th 4 Sheet Math Digits By 1", null, "division worksheets grade 3 free multiplication and for 4th 4 sheet math digits by 1.\n\nfree multiplication and division worksheets grade 4 word problems for 4th long,free division word problems worksheets for 4th grade multiplication and 4,free multiplication and division worksheets grade 4 word problems for 4th printable main,free multiplication and division worksheets grade 4 for 4th kids long problems worksheet word,free division word problems worksheets for 4th grade multiplication and number coloring page math pages plus color on fourth 4,math worksheets grade 4 free division word problems for 4th multiplication and,free division word problems worksheets for 4th grade multiplication and long 4,worksheets multiplication and division grade printable free word 4 problems for 4th,free multiplication and division worksheets grade 4 word problems for 4th kids the inverse relationships range,worksheet simple worksheets coordinate grid picture math free division word problems for 4th grade multiplication and 4." ]	[ null, "http://fahry.info/free-division-worksheets-grade-4/division-worksheets-grade-3-free-multiplication-and-for-4th-4-sheet-math-digits-by-1/", null, "http://fahry.info/wp-content/uploads/2017/10/division-worksheets-grade-3-free-multiplication-and-for-4th-4-sheet-math-digits-by-1.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8252855,"math_prob":0.9256072,"size":1022,"snap":"2019-26-2019-30","text_gpt3_token_len":201,"char_repetition_ratio":0.20628683,"word_repetition_ratio":0.28368795,"special_character_ratio":0.17514677,"punctuation_ratio":0.06666667,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99325,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-21T23:23:13Z\",\"WARC-Record-ID\":\"<urn:uuid:9e6987cb-b949-4c4b-a3d9-410dcaef18cd>\",\"Content-Length\":\"43338\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:77fca6fa-fcd5-479d-90a9-ae714d349778>\",\"WARC-Concurrent-To\":\"<urn:uuid:7ea45054-c845-429b-9e81-3485a51a372a>\",\"WARC-IP-Address\":\"104.18.33.243\",\"WARC-Target-URI\":\"http://fahry.info/free-division-worksheets-grade-4/division-worksheets-grade-3-free-multiplication-and-for-4th-4-sheet-math-digits-by-1/\",\"WARC-Payload-Digest\":\"sha1:YTKA5LONIKKKV6K4LHU724YFLAMKWUXV\",\"WARC-Block-Digest\":\"sha1:CQXT3J73OGJG2JUGL76QHZUOZUWQ3VHK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195527396.78_warc_CC-MAIN-20190721225759-20190722011759-00218.warc.gz\"}"}