Stepper motor for ebike

Question: In these days I'm going to design a e bike project. I'm going to use BLDC motor for that. but normally (according to my knowledge) there are two types of BLDC motors. there are, normal BLDC motor and stepper motor. my problem in can I use a Large stepper motor for the e bike?
Comment: Seems like a better question for the [Electrical Engineering Stack Exchange](http://electronics.stackexchange.com/).
Comment: I'm voting to close this question as off-topic because it has been cross-posted to http://electronics.stackexchange.com/
Comment: While it wouldn't be impossible to use a stepper motor and there may even be some advantages if you're going to use the motor from a standing start it's not likely to be ideal. Whatever motor you use will also need integrating with the drive chain. Kits are available with hub motors for the front or rear wheel. These are DC motors in every kit I've looked at, and from what I've seen the control circuit is usually some form of PWM. The hub motors are also available for you to build your own wheel but this would be an expensive option.
Comment: @Chris thanks. please let me know if i go for a stepper motor(integrated as side motor not a hub motor) is it possible?
Comment: I've got the hilltopper retrofit kit, and its got a three phase AC motor in the front wheel.  So there's heaps of options, do some research online.  http://electric-bike-kit-forum.com/ may be a good start.
Answer: Stepper motors have very poor maximum RPMs and dreadful torque at high rpm. Also, controlling stepper motors requires a special controller and it's hard to find high amperage controllers. High-amperage stepper motors are also very expensive.
There's really no advantage to using a stepper motor. You don't need its precision (you can tell a stepper to rotate CW 145 degrees then back CCW 32 degrees, but why would you need that in an e-bike).
Go with an appropriate brushless motor.
Answer: Stepper motors are generally used for servo (positioning) applications.  They work great in Shimano's Di2 electronic derailleurs (obviously a positioner).  But your application is straight velocity mode, stick with BLDC.

Stepper motor for ebike


Latest updates froze - reboot to emergency mode

Question: After running the updates today I think appcentre said there were 8 updates, it seemed to hang the process while running 2 "operating system" updates. I waited a few hours for it to finish (5 I think) but it never did. I tried to reboot the system but it didn't respond so I (against my better wisdom) forced the appcentre closed and force rebooted. Now I can only boot into emergency mode. I'm not sure what to do now. I'm reading that I should try a live distro and try to fschk the root partition.  I'm about to try that now.
Anyone know what to try here, or need more info?
I can't seem to get network access in emergency mode; at least I don't know how to...
Comment: Upon reboot I get a message, /dev/sda1: clean. A start job is running for dev-disk-by\x2duuid-9608803308801481.device with a 1minute, 30 second count down. Then it continues and boots to emergency mode. Fsck is clean. I've tried "systemctl reboot" with the same thing and journalctl -xb shows the same error as above with dev-disk-by\x2duuid-9608803308801481.device/start timed out.  That disk is a network SMB drive that I mount at startup.
Comment: OK, so I edited my fstab and commented out that entry. Issued "systemctl reboot", and after the grub menu it booted fine! So thankful...
Comment: You can add your procedures as an answer if it solved your problem :)
Answer: Commenting out the mount point listed in the error in my fstab solved my problem. I don't have a clue why this became an issue today as I've had that for many many months.  Probably over a year with that setting.
Answer: I had the same behaviour yesterday and thanks to Your comments I realized that the last chance I did to my machine was precisely in the fstab file.
I removed the problematic line and voyla! it worked.
here it is the procedure:
1.in the terminal showed while booting Press  CTRL-D to enter emergency mode
2. enter the command vi /etc/fstab
3. move to the problematic entry with arrow keys,  in vi you could remove the entire line by pressing  letter D twice
4. Exit vi saving the file by pressing ESC : WQ in secuence
5. Reboot the machine with systemctl reboot
This saves Me from reinstalling everything


Why can humans feel electricity?

Question: What evolutionary process has provided humans with the ability of feeling electric current? Besides lightning and electric eel, what natural hazards include electricity that poses a threat to humans?
Answer: Electrical currents stimulate neurons aspecifically. For example, the BrainPort artificial vision device conveys visual information through electrical stimulation of mechanoreceptors in the tongue. Similarly, cochlear implants and retinal implants convey acoustic and visual information through electrical stimulation of the inner ear and retina, respectively. The list goes on. Bottom line is, none of these organs (tongue, ear or eye) evolved to process electrical stimuli. And in fact, there are no specific current or voltage sensors in man. Electric currents aspecifically stimulate neurons, causing them to fire. When sensory afferents are activated in this way, sensory perceptions are generated.
In case of lightning (electrical shock to the skin) it is mainly pain and heat receptors that mediate the sensation. In the case of the BrainPort, it is the tactile receptors on the tongue that are mainly stimulated (taste perceptions are rare). In case of cochlear implants it is auditory nerve fibers and in retinal implants it is optic nerve fibers that are stimulated. Hence, various, and in fact all, neurons can be stimulated by electric current.
Note that sharks feature Ampullae of Lorenzini. These sensory organs are electroreceptors, which allow sharks to sense the bio-electrical fields of prey under water.
Answer: Electricity directly stimulate neurons. You have to understand that neurons basically work via electric currents , which occur due to a difference in membrane charge. So when you give electricity to a neuron, you basically give electrons to the extracellular membrane, which makes it much more negative compared to the intracellular or the stuff that is inside the membrane. This offsets the concentration gradient that kept the careful balance until now, and then neuron fires. Anyway, i hope you got that. Note that you can't directly "feel" electricity, or electrostatic force. That is impossible, nature has not given us the mechanism to do so. What you feel is a direct consequence of electricity interfering with the functioning of your muscles and nerves.
Answer: Electricity is essential in order to control and coordinate neurons and skeletal muscles activities. You should imagine a neuron as a very complicated logic port of an electronic circuit, that receives information through its dendrits from other neuronal cells and then integrate this information in its body. If the spatial and temporal summation of inhibiting and stimulating signals is overthreshold, then at level of the first part of the axon, there is the creation of the all-or-null response of the neuronal cell: the action potential. In general, there is the simultaneous activation of the voltage-gated sodium channels that elicit spikes spreading them along the whole axon that is connected to other neuronal dendrits or skeletal muscle cells. These ones, receive and integrate in the same manner the stimuli sent to them. As you can easily imagine, external electric stimuli, like electrostimulators or instrument used by neurologist to analyze neuronal and muscle activities in diagnostic tasks are able to alterate the natural electric potential present across the cell membranes of these excitable cell types and response to this  electric field. Indeed, cells possess and actively keep an electric potential among the two sides of the phospholipidic bilayer of the cell membrane of about -70mV. This one is due to the asimmetric distribution of ionic species (sodium, potassium and chloride mainly) sustained using chemical energy obtained from exogenous sources. Thus, in general, our cells are not able to feel electricity directly but every phenomena, either phisical or chemical, able to alterate the electrical equilibium (homeostasis) of the cell membranes induce a response at inner cellular level leading, for example, to neuronal cell response or skeletal muscle contraction. You can think to the use of defibrilator to treat cardiac arithmias that send to the hearth strong current pulse-wave, that in case of fibrillation or ventricular tachycardia is frequently capable to restore normal cardiac activity and save life. Also pacemaker acts in a similar manner. Some pathologies like ALS, Alzheimer's, Parkinson, Myotonia involve also alteration of this electric mechanism leading to modified capacity of the cells to handle and interpret these electric message. In electrophysiology research, external stimulation of cells by electrodes (see patch clamp technique) is used to test the effect of drugs to treat these pathologic conditions.
Finally, I can say that evolution didn't give us the ability to feel electricity, instead it has been adapted as a cellular-level instrument handled by cells to communicate and integrate (or store in brain) stimuli and information.

Why can humans feel electricity?


How to add Reporting Services to an existing SQL Server 2019 Clustered Instance

Question: I have already set up my clustered environment. But now, there is a requirement came for reporting service for only one instance.
When I try to add it using the "Add feature to the existing environment" option in setup. However, I can't see Reporting Service on there.
So how I install and set up reporting service to the required instance.
Comment: Note the "Looking for Reporting Services" note above the features list. RS is no longer included with the SQL Server installation.
Answer: As you may know, SSRS is not a cluster aware feature, so the install requires some extra steps. Also, if you want high availability for SQL Server Reporting Services you should deploy it in a farm which provides both availability and load balancing.
Adding SSRS on an installed SQL Server Cluster Instance is not so straight forward. When we try to add SSRS on an existing SQL Server cluster instance, the install does not allow us to proceed after the Installation rule check page because one of the rules fails during the rule check process. One solution is to install SSRS as a separate instance, but since this configuration was already a multi-instance cluster I was not in favor of increasing the number of instances by installing SSRS as a separate instance.  Below I will explain the step by step method to configure SSRS on an existing SQL Server cluster instance without deploying a farm
Step 1
To add/install SSRS on an installed SQL Server Instance, we need to run setup again on each cluster node. First run SQL Server setup on your active node (Node 1). Follow all necessary steps in the setup windows. Make sure to run this setup to add SSRS on your existing instance rather than creating a new instance. Choose Reporting Services on the feature selection page of the setup window. After we start the install, we will get an error because one of the rules will fail as shown below.
Step 2
When you get the above error, you cannot proceed with the install because the "Next" button will be disabled due to this failed rule.
To get a detailed report about this failed rule, click on "view detailed report" shown in the above screenshot and you will get details about this rule as shown below.
Step 3
As we already saw that we cannot add SSRS to an existing SQL Server cluster, the solution is to run setup and skip the installation rules to install SQL Server Reporting Services in an existing clustered instance.
Run the below command at the Windows command prompt to start SQL Server setup on the active node.  Make sure to run this command after changing the root directory of the command prompt to the location where you have placed the SQL Server setup files.
Setup.exe /SkipRules=StandaloneInstall_HasClusteredOrPreparedInstanceCheck /Action=Install
Once you press enter to run the command, the SQL Server product version will display on the command prompt, as shown above, and an installation window named "Program Compatibility Assistant" will appear. Now click on "Run program" to proceed with this installation.
Step 4
Now follow the same process which you normally do in an installation. Again choose the existing instance to add SSRS and select Reporting Services in the feature selection page which we need to install.
Now this time you can see the installation rule check passes without an error, because we skipped the installation rule process to make this installation possible.
Step 5
Here we can see the "Next" button is enabled, so click on Next to install SQL Server Reporting services on the active node.
Step 6
Once you are done with the installation on the active node (Node 1), follow the same process on each of the other nodes in the cluster.
Step 7
Now that each of the nodes has Reporting Services installed, it is time to configure Reporting Services on each node using "Reporting Services Configuration Manager" which can be found under configuration tools.
Here the configurations would be the same as you would have for any Reporting Services configuration. One thing will change though, we will use the SQL Server failover cluster network/virtual name while making a connection to the report database server. If we use node names in place of the failover cluster network name, the report server will be unable to connect to the report server database in case failover occurs.
Configure Reporting Services on each node in the same way with SQL Server failover cluster network name otherwise you will get the below Report Manager error after failover when an instance is online from another node.
Step 8
Now you are done with your SSRS configuration. Go ahead and launch Report Manager to check whether SSRS is configured properly. You can also test your Report Manager accessibility after failover and failback to verify whether it's providing the necessary failover functionality.


Is there a quicker algorithm for calculating 'drifted' portfolio weights? (R, Pandas/NumPy, MATLAB)

Question: 'Sup, QuantSX.
BLOT (Bottom Line On Top): Is there a nice clean algorithm for rapidly calculating portfolio weight drift? In R, Python or MATLAB - I'm not fussed which.
Details
I'm in the final stages of extending a stress testing script (in MATLAB) that I've used for a while, and I've hit a performance wall on what should be a pretty easy updating function.
The guts of the script is a medium-sized Monte Carlo - 1000 iterations of 240 x n[i] for ~70 portfolios, where n[i] is the number of portfolio components in the ith portfolio (n[i] ranges from ~5 to ~25). Gaussian copula, so generates the pseudo-data super fast.
In the past the output has just been presented under a monthly-rebalance assumption, so the pseudo-observations for the portfolio itself are just
$w\cdot R = \sum_i\bar{w}_i \times R_{i,t}$
where $\bar{w}_i$ are the original-and-target portfolio weights. And of course that's also super-quick - O(msec). In fact it's the quickest module of the whole stress-testing framework, which completes in ~35 seconds per portfolio.
Part of the extension is to calculate the portfolio path under a range of alternative rebalancing assumptions - full-drift (i.e., no rebalancing, buy-and-hold); quarterly; annual; and 'threshold' by asset class.
In testing the full-drift version, it's taking 2 seconds per iteration to calculate the drifted weights; that seems a very long time for 240 rows of data.
And 70 lots of 1000 × ~2 seconds is a long time.
The weight transition equation is really straightforward -
$w_{i,t+1} = w_{i,t}\cdot\left(\frac{1+R_{i,t}}{1+R_{p,t}}\right)$
$w_{i,0} = \bar{w}_i$
The drifted weight depends on the portfolio return, which is the weighted sum of the component returns in period t -
$R_{p,t} = \sum_i(w_{i,t}\cdot R_{i,t}) = w'̌\cdot R$
So far, no rocket surgery.
To do this in MATLAB, the only way I could think of was to initialise an aggregator, and to accumulate the recalculated weights (and as a byproduct, the 'drifted' portfolio returns).
So my function is presented below (although my function's properly indented)
function[outPVals, outWts, wtDiff] = DriftWts(inTable, inWts, outName)
%% Calculates Portfolio Value and Component Weights over time if weights drift.
xT = inTable;
xP = inTable;
xT.Properties.DimensionNames{2} = 'Weights';
xT.Weights = zeros(size(xT)); %% pre-allocate weights table
xD = xT;                      %% pre-allocate weight-delta table
tBeg = min(xT.MthYr);
xT(tBeg,:).Weights = transpose(inWts); %% initialises weights at current AA
xR = inTable(tBeg,:).Returns/100;
xP.(outName)(tBeg,:) = 100*(xR * inWts);
xT = sortrows(xT,'MthYr','ascend');
for mmm = 2:length(xT.MthYr)
    m = xT.MthYr(mmm);
    m_l = eomdate(m - calmonths(1));
    w_l = xT(m_l,:).Weights;
    R_l = inTable(m_l,:).Returns/100;
    xR = inTable(m,:).Returns/100;
    Rp_l = w_l * transpose(R_l);
    w_0 = w_l .* (1 + R_l)/( 1 + Rp_l);
    xT(m,:).Weights = w_0;
    xD(m,:).Weights = w_0 - w_l;
    xP.(outName)(m,:) = 100*(w_0 * transpose(xR));
 end
 outWts = xT;
 outPVals = xP(:,outName);
 wtDiff = xD;
 end

I considered an alternative - i.e., generating cumprod(1 +$R_{i,t}$ /100) of the component returns, weighting them by $\bar{w}_i$ at t=0, summing them, then dividing everything through by the row sum.
It's not obvious to me that this alternative would be quicker (but I have no idea what goes on under the hood in cumprod(), so I'll try it today)
Answer: OK, so the 'alternative' mechanism (cumprod, weight, divide by row sum) is O(msec) and produces identical results.

Old stupid method: 1.705sec for a 200×9 table.
New sexy method:   0.004sec for the same table.

And it's only 4 lines of code.
T01 = cumprod(1+inTable/100,'reverse');
T02 = inWts' .* T01;
R01 = sum(T02, 2);
W01 = T02 ./ R01;

Comment: Makes sense, I was about to say "find a way to vectorise", but that's what cumprod is doing. It's often the answer to slow matlab code.


Step by step guide on how to connect to peers

Question: I am using the command line wallet, and mining on my PC. The daemon says I have '8+0 connections', which I believe is the default. I believe this means I am connected to 8 nodes, but no one is connected to me. Am I correct in assuming that every block I mine is empty because I am not connected to anyone? Is there a guide that someone could link to to help me open port 18080 to allow others to connect to me? Also, what are the pros and cons for connecting to others? Would it strengthen the network more, increase my orphan rate, give me transaction fees? Is it worth it?
Answer: It's tough to write a step by step guide for port forwarding, because port forwarding is something your router does, and people have different routers.
If your router has UPnP, that would be the easiest.  Many people will argue that it's safer for UPnP to be off, which is correct; but you can enable it long enough for it to allow the daemon to punch a hole through the firewall (a non-technical term for port forwarding), and then turn it back off.  The port forwarding would have been performed automatically in that case.
But anyway, just look at your router's instructions for port forwarding.  Nearly all modern routers allow port forwarding, and it is not complicated.

Your other questions can either be asked in separate questions, or they may actually have already been asked and answered, so please search a bit.
Answer: When you open up port 18080, you strengthen the network by allowing others to connect to you to download the block chain.
Look at your modem/router and get the exact name, model, as well as your internet provider name.  Then "google" with that information on how to do port forwarding.  Takes a little bit of effort but it's fairly simple, after you do it one time.  The "know how" you need is on the internet.


Temp and time for using half the recipe of cake

Question: I have a recipe which makes cake for 6-8 ppl. I want to bake the same recipe using just the half recipe. It's original temp. Is 180 C using 40 minutes. Now after using just the half recipe what will my temp. And time be.. Please reply
Comment: Are you using the same size pan or a smaller one? Generally you can keep the temp the same and halve the time... and then watch until it's done.
Comment: Start checking it after 1/2 the time ... but it'll likely take closer to 3/4 of the time, as it doesn't scale linearly when you move things to a smaller pan.  (it might be 1/2 the time if you kept it in the same size pan, so it's 1/2 as thick ... but that also tends to lead to overcooking)
Comment: To continue, if you use a pan that is half the size and comes up to the same depth, the cooking time should be the same according to the box mix I found.  Remember that shiny metal pans (or non-stick pans) have a higher temperature and glass pans are 25 F degrees lower.  Again, sorry that you will have to figure that one out for yourself.  Also, do you have any box cake mixes that you can check for yourself at home or call someone who uses them.  That could be helpful to you.  All this info is based on a pretty simple cake.  Good Luck
Comment: I found a box cake mix that adds water, oil and eggs. If you use a 13 inch by 9 inch pan you bake 28-33 minutes at 350 F degrees in  shiny metal pan or 325 F degrees in glass pan. Two 8 inch round pans 28-33 minutes 350 F or 325 F, two 9 inch pans 24-29 minutes at 350 F degrees or 325 F degrees. Sorry, pls do the math to convert your 180 C to my 325 F and 350 F. I checked my pans and the 13 by 9 was the same depth as the 8 inch but the 9 inch was larger so half would be a flatter dough hence, less cooking time. I would put toothpick or knife to check if dry at the earlier time to see if baked
Answer: As @joe said, it does not scale linearly.
There are various factors that effects cooking time:
Thickness: the heat transfers mainly from the top and bottom for a round on a not too thick to thin dish, so for half a cake with the same thickness the time will stay about the same. Think of a pizza for example, whether it's 5 or 15 inches in diameter the cooking time is nearly exactly the same. For a very thick cake, though (sponge cake for a wedding cake for example), the lateral heat transfer will matter far more.
Chemical reactions: Some chemical reactions (egg coagulation, Maillard, unfolding of proteins etc.) happens at a very specific temp under which they don't happen at all or takes forever. For example  the  coagulation an egg at ambient temperature will take several years, where it would happen in a few minutes at 65 centigrades. Try browning a toast at 90 c (half of the temp needed), it'll never happens. It takes 180c to start burning bread, anything under will stop the expected result from happening .
Most chemical or physical reactions (water boiling) that gives the expected taste or texture to something you bake happens between 60 to 180 c so do not change the temp when changing the amount in a recipe if the temp indicated is within that range.
If you cook a bigger item a higher temperature it won't have a chance to cook internally before being burnt on the outside, on the opposite, a smaller item with the same shape will be crisp and nice at the same temp.
As for the time needed, only on very compact items, nearly spherical, such as scones, choux pastry, or poultry (think of the time needed to roast a breast of chicken vs a whole chicken vs a turkey) the amount of time will be more closely proportional to the size or weight. You cook the smaller item less longer.
In brief: don't change the temp if it's equal or less than 180c. If your cake is compact (like a banana bread cooked in a tin) drop maybe  around a quarter  of the time for a halving of the recipe, and maybe a sixth of the time for a thinner cake.


Bitcoin Difficulty granularity

Question: Difficulty adjusts by very granular percentages to target for 10 minutes blocks. But adding another zero at the end of the chain of zeroes from the header hash requirement for a valid block would increase the difficulty exponential. So how does the Alogrithm target so precisely the right difficuly only through the use of "zeroes"?
Comment: The number of leading zeroes is never, at any point, ever, taken into account.
Answer: The difficulty is actually represented by the target threshold encoded in the nBits value in the block header. Where difficulty represents the human readable representation ("how often do we need to try to find a solution"), the target threshold defines the prefix a block must undershoot in order to be valid. This means that the 256-bit block hash interpreted as a number must be lower than the target threshold.
Although, nBits is only a 4-byte value, it is a compressed representation of a 256-bit (32-byte) number. The first byte defines the exponent, the remaining three bytes give a 24-bit mantissa for the target.
While this leaves most of the 32 byte in the target threshold to be just composed of zeroes, the difficulty can adjust in a much more granular fashion than just adding leading zeros.
David Harding has elaborated the full details in How is the target section of a block header calculated?.
Comment: @zndtoshi I think your confusion stems from the common (but wrong) saying that PoW is about "the number of leading zeroes". It is not, at all. The PoW rule is that the block hash, when interpreted as a number, must be below the target. `nBits` is a compact encoding of the target.
Comment: Replacing a leading one with a leading zero doubles the difficulty, while lowering a 24-bit prefix by one count only increases the difficulty by a factor of 1.00000006.
Comment: Ok I understand that nBits is actually the difficulty target, which is part of the header. I still don't understand what this has to do with the header hash leading zeroes: I assume a full node takes the header, double sha256s it and gets a header hash that it compares to the nBits to see if it's right. Correct? My question is how are those leading zeroes more granular for finding the right hash of the header? I think I'm missing something here..
Comment: Ok. So the target is given by nBits (which is very granular and that's perfect). But if this is true, why do you need zeroes in front of the header hash? Wouldn't you just take the nBits target and verify the header hash is correct?
Or are the hash header leading zeros exponentially increasing difficulty?
Comment: I'm not sure I understand the question. The target is so low that for a header hash to undershoot it, there are a number of leading zeroes. If you always represent your numbers with 9 digits, but have to pick one below two thousand, you'll always have five leading zeroes. But it's still a difference whether you have hit below 2000 or below 1999.
Answer: As a complement to @Murch's answer, I'd like to cite an example from Grokking Bitcoin:

The target is written in the block header as 4 bytes, ABCD; the
  32-byte target is calculated as BCD × 2^(8*(A-3)). That's BCD with
  A-3 zero bytes after it. It's this awkward because we must be able
  to express a wide range of targets, 1–2^256, with only 32 bits. The
  target in Qi's [a character in the book] block is written as
  1c926eb9, meaning 926eb9 with 25 zero bytes after (1c–3 = 19,
  hex code for 25).

Bitcoin Difficulty granularity


Equal angles formed by the tangent lines to an ellipse and the lines through the foci.

Question: Given an ellipse with foci $F_1, F_2$ and a point $P$. Let $T_1, T_2$ the points of tangency on the ellipse determined by the tangent lines through $P$. Show that $\widehat {T_1 P F_1} = \widehat {T_2 P F_2}$.
Comment: Picture:
http://sgernesto.files.wordpress.com/2013/02/equal-angles-tangents1.png
Comment: See very simple proof [here](https://www.cut-the-knot.org/Curriculum/Geometry/AngleBisectorsInEllipse2.shtml#proof) using the concept of confocal ellipses.
Answer: Let $F_1'$ be the image of $F_1$ under reflection in the tangent at $T_1$.  Since the tangent at $T_1$ is the external bisector of $\angle F_1T_1F_2$, the points $F_2T_1F_1'$ are collinear.  Likewise, let $F_1''$ be the image of $F_1$ under reflection in the tangent at $T_2$; then $F_2T_2F_1''$ are collinear.

Now,

$F_1'P = F_1P = F_1''P$ by reflection,
$F_2F_1' = F_2T_1+T_1F_1 = F_2T_2+T_2F_1 = F_2F_1''$ by reflection and the defining property of an ellipse,
$F_2P = F_2P$.

By SSS, $\triangle F_2PF_1' \cong \triangle F_2PF_1''$.  In particular,
$$ \angle F_2PF_1' = \angle F_2PF_1'' = \tfrac12\angle F_1'PF_1'' \tag{1} $$
Next, by construction, reflecting in the tangent at $T_1$ and then in the tangent at $T_2$ sends $F_1'$ to $F_1$ and then to $F_1''$.  But the composition of reflections in two lines is a rotation around their point of intersection, by twice the (directed) angle between the lines.  Thus $F_1''$ is the image of $F_1'$ under a rotation around $P$ by $2\angle T_1PT_2$; in particular,
$$ \angle F_1'PF_1'' = 2\angle T_1PT_2 \tag{2} $$
Combining (1) and (2) yields
$$ \angle T_1PT_2 = \angle F_1'PF_2 $$
Subtracting $\angle T_1PF_2$ from both sides yields
$$ \angle F_2PT_2 = \angle F_1'PT_1 $$
Finally, by reflection, $\angle F_1'PT_1 = \angle F_1PT_1$, so
$$ \angle F_2PT_2 = \angle F_1PT_1 $$
as desired.

Incidentally, from $\triangle F_2PF_1' \cong \triangle F_2PF_1''$ it also follows that, letting $Q$ be the intersection of $F_2T_1$ and $F_1T_2$, the triangle $F_1T_1Q$ and $F_2T_2Q$ have a common excircle, whose centre is $P$.
Answer: Let the ellipse be:
$$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$
Set $P(a,t)$, the tangent line $PT_1$ is $y=k(x-a)+t$, $k=\tan A_1$, since it is tangent line, so put in ellipse equation, we have:
$$ b^2x^2+a^2(kx-ka+t)^2=a^2b^2$$
i.e.,
$$(b^2+a^2k^2)x^2+2a^2k(t-ka)x+a^2(t-ka)^2-a^2b^2=0$$
Solving for the point of tangency, the discriminant $\Delta$ must be zero:
$$\Delta$=$4a^4k^2(t-ka)^2-4(b^2+a^2k^2)[a^2(t-ka)^2-a^2b^2] $$
At $\Delta=0$, we get $ 2kta-t^2+b^2=0$, that is, $k=\dfrac{t^2-b^2}{2ta}$.
Let $k_1$ be $PF_1$'s slope, $k_1=\dfrac{t}{a+c}=\tan A_2$, the $ \angle T_1PF_1=|A_1-A_2|$, now we calculate $\tan(A_1-A_2)$:
$$\tan(A_1-A_2)=\dfrac{\tan A_1-\tan A_2}{1+\tan A_1 \tan A_2}=\dfrac{k-k_1}{1+k k_1}=\dfrac{\dfrac{t^2-b^2}{2ta}-\dfrac{t}{a+c}}{1+\dfrac{t^2-b^2}{2ta}\dfrac{t}{a+c}}=\dfrac{(t^2-b^2)(a+c)-2at^2}{2at^2+2act+t^3-tb^2}$$
Since $b^2=a^2-c^2$, we get:
$$ RHS=\dfrac{(t^2-a^2+c^2)(a+c)-2at^2}{t(2a^2+2ac+t^2-a^2+c^2)}=\dfrac{c-a}{t}$$
Let $k_2$ be line $PF_2$'s slope, then $k_2=\dfrac{t}{a-c}=\tan\angle PF_2T_2$. Since $\angle T_2PF_2=\dfrac{\pi}{2}-\angle PF_2T_2$, so $\tan \angle T_2PF_2=\dfrac{1}{\tan\angle PF_2T_2}=\dfrac{a-c}{t}=\tan(A_2-A_1)$. Since the angles are acute, we have:
$\angle T_2PF_2=A_2-A_1=\angle T_1PF_1$
Answer: From Geometry Expressions
Equal by inspection.
Answer: HINT: It is enough to prove that

Where n is normal unit vector.
NOTE: parametric equation is γ(t)= (p cos t, q sin t)


Customs and baggage when transferring in Schengen and going outside

Question: I am flying from Prague through Frankfurt to Toronto. Do I need to go through customs in Frankfurt and will my baggage go directly onto the next plane?
Comment: You will very likely check in for both flights in Prague, and your luggage should make it on its own. You of course don't go through passport check in Prague since it's an in-Schengen flight. The passport control will be on the Schengen boundary, which is in Frankfurt for you. I'm sure that this has been asked before, I'll try to find the original question.
Comment: Do you have one ticket or two?
Comment: @Rish Yes, you are wrong. E.g. when going to Israel, you may be asked to present all your luggage for inspection by Mossad prior boarding. When you're transporting something very expensive into a country with the intention to transport it back, it's the place where you leave the country where you have to claim it. So for instance if you go from Prague to NY and then to Miami, you do all the customs on the US side in NY, and not in Miami.
Comment: I am booking the trip as a whole, so should have all boarding passes printed at check in, just did not know if I needed extra time between flights to deal with customs. I assumed baggage will go right through on next flight to Toronto, so would need less time if I do not have to recheck it
Comment: I presume if you have all the boarding passes right to your destination printed at the origin, you'd never have to deal with customs or luggage in between (not only in this case, but in general). Correct me if I'm wrong.
Answer: (I didn't manage to find the duplicate, so I'll try to make a summary answer.)
It all depends on two big aspects:

Do you have one ticket (with transfer) or two separate tickets?
To which country do you fly?

If you have one ticket with transfer, then you should get checked in for both flights in Prague. I recommend to arrive the advised 2 hours before departure so that you manage to sort out things nicely. If you get checked in for both flights, then your luggage will make it on its own, and you just move yourself to the correct terminal/gate in Frankfurt. (Check your luggage receipt to ensure it did though! It has to show the code YYZ for Toronto rather than FRA for Frankfurt.) You will pass the passport control (Schengen exit) in Frankfurt (in Prague T2, it's just a brief control and you can even use your ID for that one). It can happen that they don't issue the boarding pass for Frankfurt. Ask the clerk then if the luggage gets checked-in to Toronto or not, and read the next section carefully.
If you have two separate tickets you can still try to get checked in for the flight Frankfurt--Toronto in Prague, but the chances of success are low. (May work if both airlines are in the same alliance, and none of them are low cost carriers, but you'll need to ask nicely at check-in.) Then, you'll have to get your luggage in Frankfurt like if you were terminating there, move (with your luggage) and go the departure terminal at the airport and check-in yourself and your luggage like someone who starts there. Remember that this may take quite some time. This paragraph applies if your luggage wasn't checked in to Toronto in Prague, too.
If your luggage was checked to Toronto in Prague, but your boarding pass wasn't issued (this can happen in very extra-ordinary cases), you just proceed to a transfer desk in Frankfurt, which will issue your boarding pass to Toronto.
If you travel to a country which does a detailed check (USA knows to issue these sometimes, only to change their mind couple minds later, Israel is another such one, and there are surely more, but it all changes every couple months or so), then it can happen that they want to inspect your luggage before loading it on the overseas flight. I would recommend asking for the procedures on the airlines' hotline in advance, they ought to know the procedures.
For the customs: If you have any goods to declare on leaving (like an apparatus worth €1M that you plan to take back home), you better do that in Prague. You should be able to do it in Frankfurt too (since most stuff can travel between CZ and DE without any problem), but it's better to solve this stuff as soon as possible. I better don't give more advice since it's impossible to answer that in general.
Comment: (+1) It might useful to stress that Schengen and customs are unrelated (basically Schengen is about people, not things).
Comment: (I don't have much experience with these transits, so comments are welcome.)
Comment: @Relaxed I think that this question is just general "how is my luggage treated" question, and for that, it's already too complicated. I'll just add a short notice about it.
Comment: @Donna You're welcome. If you needed more specific information, just ask for this. Since you didn't provide any, I tried to give a general advice on the topic.
Comment: Tohecz, I think you are in Prague, do you know a company called ICTAXI, they have a web site, I am looking for transportation from, Passau to Prague, are they reputable, thanks
Comment: @Donna I've never heard of them, I generally avoid such companies. I think you're much safer to travel by train or by bus. The shortest train journey takes a bit over 5 hours.
Comment: http://ictaxi.cz/contact/.   This is the web contact info, I would be so grateful if you could check them out for me, they seem to be ok, I have seen reviews from a few people on Facebook. Just want to be sure, would hate to be stranded in Passau, there r 6 of us travelling
Comment: @Donna Well, as I said, I don't know them and the webpage wouldn't really reveal much. They seem fine, just don't seem to have Passau (Pasov in Czech) as one of their destinations, so you better call there and make sure they transport you and make sure what the price is gonna be.


Lumia 640 screen brightness

Question: Can the screen brightness be increased somehow on the Lumia 640 to be more visible in sunlight. The screen setting is already set to bright.
Answer: Try this:

Go to Setting >> System >> Display
Disable automatically adjust display brightness
Comment: Thank you for your suggestions however they have already been initiated without sucess. Has anybody with a Lumia 640 actually been able to read the screen in bright sunlight/daylight and if so how have you achieved this result?

Lumia 640 screen brightness


How are green OA publications cited?

Question: Suppose one of my papers is published by journal A and I upload it in addition to my university repository B. When people cite this paper, would some cite the paper in A and some the one in B? Or is it ensured somehow that people will only cite the one in the journal?
Comment: I think it is best to provide the peer-reviewed journal reference (whether it is OA or not).  Then, if you wish, mention additional sources for the manuscript.  The exception is when the journal publication has not yet occurred.
Comment: When you're writing your bibliography manually, you can just say "A preprint is available at ..." (or "Also available at", if you know that the preprint *is* the published version). With BibTeX, I've got no idea (I don't even know how to make it print DOIs...).
Comment: Sorry if I miss-phrased it, but I wanted to ask whether this actually (always) works in practice. After publication, many people will only access the paper via B, as they would need to pay for A. My question is whether they are in practice always citing A or if it can happen that they cite B. E.g. someone might refer to B in their paper, without mentioning A at all. This would be a disadvantage of publishing via green OA.
Comment: A citation should always include a DOI (if available), which identifies the paper uniquely. Personally, I cite the published version with the doi linking to the journal source via doi.org and also add the e.g. arxiv identifier (and link).
Answer: If it's a direct copy of the published version (same PDF, with formatting, headers etc.) I'd be very surprised if anyone would fail to cite the journal version. I mean, what's next? Providing the address to their university library instead of using a standard citation format for a printed publication?
If it isn't a direct copy, but the copy in university repository B lists the journal reference, readers will generally cite the journal version. This also tends to happen with papers on preprint repositories such as arXiv, except when people find the preprint version before it's published, and don't remember to check their bibliography for updates before they submit/publish their own work. If no journal reference is listed, you can't really blame the readers for not noticing the version in A, so make it easy for them to get it right.
Now, this isn't really enforced. The closest thing I've encountered is that many journals will ask authors to double check references that don't look like peer reviewed and published versions during the copy-edit phase. As you say, this could be considered a disadvantage to green OA, but it's likely outweighed by the visibility advantage of having a non-paywalled version.
Comment: Ok, I sort of suspected this, but just wanted to ask some people with more experience. Thanks for your detailed answer.
Answer: In addition to Anyon's answer: Some journals might have specific rules about copies of papers on university repositories and preprint servers. E.g. Molecular Ecology Resources state in their editorial policies:

...
Given that the measurable impact of the article is diminished when citations are split between the preprint and the published article, authors are required to:

update the entry on the preprint server so that it links to and cites the DOI for the published version
cite only the published article themselves.

How are green OA publications cited?


How can we identify, sometimes, the angel of the Lord as God if the prophets also spoke as God?

Question: I am not saying the prophets are God. What I am saying is that the prophets have a tendency to speak like God in the first person. Which can make it difficult to use the OT when discussing the premises of the Trinity.
This was an argument used by Muslims.
Comment: Can you provide some examples of prophets speaking as God? In the ones I can think of, the prophets begin with "Thus says the Lord ..."
Comment: After speaking with a friend about this (it's a fascinating question), he couldn't think of such an instance in the OT and pointed out the Jews were zealous when it came to making sure who was speaking when.  However, he pointed out it's possible that (especially in Isaiah), a prophet began with "Verily thus saith..." and was still speaking in Jehovah's voice a chapter and a half later, by which time the reader has forgotten Isaiah properly announced what he was doing; leaving the reader to believe Isaiah was speaking "as God."  This might be a case of mistaken identity.
Comment: I also thought they always used introductions like that, I first heard this argument from French Muslims a week ago. The guy quoted Isaiah I think. But I can't remember the chapter and verse, sorry. He spoke super fast and then a Christian began doing a rebuttal immmediatly. However the rebuttal put an emphasis on how extraordinary an event is as the standard. So he noted that the angel of the Lord is always equated with God by those who meet him. Whereas the prophets are never spoken to as if they were God, but merely as messangers.
Comment: Welcome to Christianity.SE. For a quick overview, please take the [Site Tour](http://christianity.stackexchange.com/tour). For more on what this site is all about, see: [How we are different than other sites](http://meta.christianity.stackexchange.com/q/1808/20394). Meanwhile, I hope you'll browse some of the other questions and answers on this site.
Answer: The key to identifying the difference between the two lies in the appearance of the one making the statement. Prophets were ordinary men who were known members of the community. Their presence was not a one time or sudden event. Whereas with the Angel of the Lord appeared at the moment of the announcement, and most often only for a few minutes.
Since you mentioned the Book of Isaiah the Angel of the Lord only appears once in the book of Isaiah. That is in Chapter 37 verse 36, and that particular chapter is most indicative of the point I am making.
When King Hezekiah was about to acquiesce to the king of Assyria Isaiah sent a message to Hezekiah which you will notice begins with Thus saith the LORD God of Israel. While in verse 36 we find:

Isaiah 37:36 KJV  Then the angel of the LORD went forth, and smote in the camp of the Assyrians a hundred and fourscore and five thousand: and when they arose early in the morning, behold, they were all dead corpses.

What we find in this verse is a quick and decisive action by the Angel of the Lord.
In all 63 instances of The Angel of the Lord appearing in the Old Testament you will find that the Angel appeared and only remained as long as necessary to accomplish the will of God.
Hope this helps.
Answer: The prophets were always sent, and the books declare each prophet's name and to whom they were sent.  When we read the book of Isaiah we know that the prophet was Isaiah, and he was speaking on behalf of God, relaying God's words to the people.
Isa. 1:1, Isaiah sent to Judah and Jerusalem:

"The vision of Isaiah the son of Amoz, which he saw concerning Judah and Jerusalem..."  (KJV)

Isa. 13:1,  Isaiah speaking the prophesy to befall Babylon:

"The burden of Babylon, which Isaiah the son of Amoz did see."  (KJV)

By studying through both the OT and the NT we can determine that the angel of the Lord in the OT was the prefigured Christ.
Ex. 23:20-23,

"20 Behold, I send an Angel before thee, to keep thee in the way, and to bring thee into the place which I have prepared.
21 Beware of him, and obey his voice, provoke him not; for he will not pardon your transgressions: for my name is in him.
22 But if thou shalt indeed obey his voice, and do all that I speak; then I will be an enemy unto thine enemies, and an adversary unto thine adversaries."  (KJV)

The angel had the authority and name of God. He also was the warrior for the people.
In Exodus 3:2 the angel of the Lord appears to Moses in a flame of fire, in a burning bush, and verse 6 He identifies himself as

"...I am the God of thy father, the God of Abraham, the God of Isaac, and the God of Jacob. And Moses hid his face; for he was afraid to look upon God."

In contrast to other instances when angels forbade a man to worship them (Gabriel in Dan. c. 9 -12), this angel accepted Moses worship.
This is also the same "captain of the Lord's hosts" of Josh 5:14-15 whom Joshua worshiped. Just as the angel told Moses in Ex. 3:5, the captain of the host told Joshua to remove his shoes for the place he stood was holy. It is the same angel of the Lord whom David worshiped in 1 Chron. 21: 15-19.
In Judges 13:16-21 the angel of the Lord appeared to Manoah and his wife to tell them of their son Samson, and Manoah asks for the angel's name. The angel said in verse 18 that His name was secret.
Rev. 19:12, speaking of Christ, the warrior of His people (Psa. 34:7; Dan. 12:1),

" His eyes were as a flame of fire, and on his head were many crowns; and he had a name written, that no man knew, but he himself."  (KJV)

In Judges 13:20 the angel of the Lord ascends in the flame of the altar, just as he had appeared to Moses in the flame of the burning bush.  The flames, the fire are symbols of God, and especially of His judgment.  (See Rev. 1:14; 2:18; Psa. 83:14; Isa. 10:17; Isa. 30:30, etc.)
Hosea 12:3-5 makes it clear that the angel Jacob had wrestled with in Gen. 32:24-25 was the "the Lord God of Hosts."
The word angel means "messenger".  The angel of the Lord was the very special messenger of God better than all the angels (Heb. 1:4), the very special begotten Son of God (Heb. 1:5), the one who created all things (Col. 1:16) which would include the angels (Psa. 148: 1-6), the one sent to be the savior for all who will call on His name through the gospel of Christ (Rom. 1:6; 1 Cor. 1:2; 2 Tim. 1:9).
Benson's commentary on Exodus 3:2,

"Exodus 3:2. The Angel of the Lord appeared to him — Not a created angel, but the Angel of the covenant, Christ, who then and ever was God, and was to be man, and a messenger from God to man. He, termed the Angel of God's presence, (Isaiah 63:9,) had wrestled with Jacob, (Genesis 32:24;) and had redeemed him from all evil, (Genesis 48:16;) and afterward conducted his posterity through the wilderness, 1 Corinthians 10:4. These his temporary appearances were presages of his more solemn mission and coming, on account of which he is fitly called the Angel or Messenger. That this angel was no creature, appears from his saying, I am the Lord, a language which angels never speak; but, I am sent from God — I am thy fellow-servant. In a flame of fire — Representing God's majesty, purity, and power, and showing that he was about to bring terror and destruction to his enemies, and light and comfort to his people, and to display his glory before all. The bush burned and was not consumed — An emblem of the church now in bondage in Egypt, burning in the brick-kilns, yet not consumed; cast down, but not destroyed; for God was in the burning bush, was and always will be present with his people in their sufferings; Isaiah 43:2; Daniel 3:25."  Source: here
Answer: Deuteronomy 18:22 "If the prophet speaks in the name of the Lord, and the thing does not occur and does not come about, that is the thing the Lord did not speak. The prophet has spoken it wantonly; you shall not be afraid of him."
If they speak of something that never happens they're not a prophet. With this it's not always completely clear because you have to look at the details. Like if the person gives a timeline for when they say it'll happen. Like before that generation dies out, or something that can be used as a proof it happened like they said in the time period they said, or it didn't. Something that can be verified.
What ever the person says would have to agree 100% with what god already established, because it's everlasting and unchangeable. (Deuteronomy 13, Ezekiel 37:26)
So if they speak any objection to what god taught they do not speak for God.
Miracles are not evidence that they speak for god. Deuteronomy Ch 13 describes how even false prophets can do miracles. God allows it because he says he's testing to see if you love him with all your heart and soul.
Comment: This answer doesn't seem to address the question about the Angel of the Lord.


What software can I use to encrypt all files that are copied out of a Windows system?

Question: What software can I use to encrypt all files that are copied out of the system (USB, network, email, etc.) such that only systems with the same encryption will be able to read the files? OS is Windows.
Comment: Welcome to Software Recommendations! We will need much more information to give good recommendations here. Please take a look at [What is required for a question to contain "enough information"?](//softwarerecs.meta.stackexchange.com/q/336/185) Then please [edit] your question and see if you can incorporate some of these improvements. For example: what OSes are involved? How much are you willing to spend when it comes to paid solutions? Are you expecting encryption to kick in automatically and only on "export" (I doubt that such a thing exists)?
Comment: I think that you want this to happen automatically. If so, please update the question to make it clear (I believe that the answer is that it cannot be easily done)
Answer: If you want to use a password, there are many options. VeraCrypt is one of them and perhaps the most reliable. You can create a key file in it and prevent the data from being decrypted on any system that has no access to that file.


Arranging "DEFINITION"

Question:
How many arrangements can be made by the letters of the word
  "DEFINITION"

a) if the letter 'i' do not occupy the first or last place,
b) if the 'i'-s are together?

How to approach this problem?
Comment: See here for a few ideas: http://math.stackexchange.com/q/58849/
Answer: a) We are making a $10$-letter word. We are told that all the occurrences of the letter I must be on the "inside."  There are therefore $\binom{8}{3}$ ways to choose where the I's will go.  For every decision about where the I's go, there are $\binom{7}{2}$ ways to decide where the N's go. For every decision about the I's and N's, there are $5!$ ways to arrange the remaining letters in the $5$ remaining vacant places, for a total of
$$\binom{8}{3}\binom{7}{2}5! \quad\text{words.}$$
b) The I's must be together, so let's tie them together into a single "megaletter" like this: $\overline{\text{III}}$.  We are forming $8$-letter words from the "letters" D, E, F, $\overline{\text{III}}$, N, N, O, and T.
The location of the two N's can be chosen in $\binom{8}{2}$ ways.  For each choice of location for the N's, there are $6!$ ways to arrange the remaining "letters" in the $6$ remaining open slots, giving a total of
$$\binom{8}{2}6!\quad\text{words.}$$
Comment: As a general heuristic, one should try to deal with the most restrictive conditions first. In part a), it is important to place the $I$'s first. If we place too many other letters first, for some placements it may no longer be true that there is "room" for the I's on the inside.
Another strategy (which we did not use) is to overcount in a systematic way, and then take care of the overcounting. So in part b), we could put a red sticker on one of the N's, to make the N's temporarily into different letters. Then the resulting $8$ distinct "letters" (including red N, and $\overline{\text{III}}$), can be arranged in $8!$ ways. For every word $\mathcal{W}$, there is exactly one other word that differs from $\mathcal{W}$ only in the placement of the $N's$.  When we remove the red stickers, any such word pair collapses into one, giving a total of $8!/2$ words.
Answer: DEFINITION has 10 letters. However, there are repetitions. In particular, there are 2 N's and 3 I's. Thus we can't just use binomial coefficients and the 'choose' notation.
Instead, we use "multinomial coefficients."
Let's count the number of ways directly, using only binomial coefficients. Without restrictions on individual letter placement, suppose we just had 10 symbols. To account for the letter I, we want to know how many ways we can choose 3 of the 10 symbols and call them I. This can be done in $10 \choose 3$ ways. From the remaining, to account for the N's, we want to know how many ways we can choose 2 of the 7 remaining symbols and call them N. This can be done in $7 \choose 2$ ways. Then we have $5 \choose 1$ ways to place D, $4 \choose 1$ ways to place E, and so on. Then we have considered the number of ways to scramble the letters from DEFINITION.
Putting this together, we get ${10 \choose 3} { 7 \choose 2}{5 \choose 1}{4 \choose 1}{3\choose 1}{ 2 \choose 1}{1 \choose 1} = \dfrac{10! \cdot 7! \cdot 5! \cdot 4! \cdot 3! \cdot 2! \cdot 1!}{3!7! \cdot2!5! \cdot1!4! \cdot1!3! \cdot 1!2! \cdot 1!1! \cdot 1!0!}$
$= \dfrac{10!}{3!2!}$
And this has the special notation ${ 10 \choose 3,2,1,1,1,1,1}$
For part a, rework this idea. For part b, consider the single 'letter' "iii" instead of three individual letters.
Answer: At the basic level, you need to apply the so-called

 Multiplication Principle:
Suppose two experiments are to be performed in succession.  Suppose
  that the first experiment has exactly $n_1$ possible outcomes.
  Suppose that the second  experiment always has exactly $n_2$ outcomes
  (regardless of what happened in experiment~1).  Then the total number
  of outcomes from performing both experiments is $n_1\cdot n_2$.

Or rather, the  generalization of the above to the case where $m$ experiments are performed in succession:

Generalized Multiplication Principle: Suppose that $m$
  experiments are to be performed in succession. If, for each admissable
  $i$, the $i^{\rm th}$ experiment has $n_i$ outcomes regardless of what
  occurred in the experiments before, then the total number of outcomes
  from performing all the experiments is $n_1\cdot n_2\cdot
  n_3\cdot\,\cdots\,\cdot n_m$.

To apply the above to a particular problem, you must envision the given process, in your case winding up with a rearangement of the letters in "definition", as having been obtained from a series of actions, or choices (the "experiments" above). You then figure out in how many ways each choice can be made and then multiply these numbers.

For your problem where the letters of "definition" are rearanged with the restriction that
the $i$'s are not at the ends, you could do the following:
1) choose where the three $i$'s go (they can't be at the ends)
2) choose where the two $n$'s go
3) choose where the next unused letter (the $d$, say) goes
$\ \vdots$
7) choose where the last letter (the $o$, say) goes.
Note the first  step. There are three indistinguishable $i$'s, so we arrange those  as a group so to speak. Also note that the above series of choices will indeed produce all the desired rearangements of the letters in "definition" (in particular, we arrange the $i$'s first to ensure this).

Now towards figuring out the number of ways to do each step, you may think of filling in ten slots.  Each slot gets a letter.
For experiment 1), we select three slots and place the $i$'s in them.
Pause here: with the restriction that the $i$'s are not on the end, this means we select three slots from the  8 "non-end"-slots.
The number of ways of doing this, as one can show, is ${8\cdot7\cdot6\over 6}$.
So there are ${8\cdot7\cdot6\over 6}$ outcomes for experiment 1).
For experiment 2), we select 2 of the remaining 7 slots in which to put the two $n$'s
There are $7\cdot6\over 2$ outcomes for experiment 2).
Ok, the hard part's over. In each of the remaining experiments, we are choosing one of the remaining slots in which to put the letter.
There are $5$ outcomes for experiment 3).
There are $4$ outcomes for experiment 4).
There are $3$ outcomes for experiment 5).
There are $2$ outcomes for experiment 6).
There are $1$ outcomes for experiment 7).
So the total number of desired rearangements is
$$
{8\cdot7\cdot6\over 6}\cdot{7\cdot6\over 2}\cdot5\cdot4\cdot3\cdot2\cdot1.
$$

For your second problem, take the same sequence of experiment.  The number of outcomes for experiment 1 would be 8 (if all the $i$'s are together, there are 8 choices of slots for the left-most $i$).

Arranging "DEFINITION"


wireshark capture the traffic of other devices in LAN

Question: I am using wireshark on Windows to capture my traffic.
Is there a way to capture the traffic of other computers which are connected to the same LAN. If it is not possible with wireshark, is there other tool capable of doing this.
Answer: If your switch(es) support it you can configure port mirroring/monitoring which will allow the switch to mirror traffic on the monitored ports to the monitor port.
http://en.wikipedia.org/wiki/Port_mirroring
Comment: +1. This is a pretty standard way of going about this without interrupting the end-user or putting the target NIC into the somewhat worrisome promiscuous mode.
Answer: One way to achieve what you want is to use an arp poisoning tool, such as Ettercap. You should be able to Wireshark/tcpdump the information you require.
The problem you have is that an Ethernet switch is designed so that it learns the MAC addresses on each port and uses this to "route" Ethernet frames to the correct port based on their MAC address. This is to reduce collisions associated with Ethernet hubs (something you rarely see these days). Therefore, you will only see Ethernet frames destined to or originating from your NIC including broadcast Ethernet frames, such as ARP, but not foreign traffic. This is a good thing :)
Most managed switches (not a dumb desktop one) allow you to designate a port mirror so that all Ethernet frames are replicated on a specific port where you can attach a machine in promiscuous mode and capture "foreign" Ethernet frames using tcpdump/Wireshark.
This still won't let them be captured by Wireshark/tcpdump, however. So you need a way to act as an Ethernet bridge between the interesting hosts and their gateway but without being physically in the path. Enter Ettercap which is an arp poisoning tool. It tricks your interested hosts (and the switch) that your machine MAC address now owns the IP of the old IP gateway by sending out a "gratuitous arp". The interesting machines will unwittingly send all gateway/default route destined traffic to your machine. Your machine will now forward packets through its IP stack as if it was the gateway.
Where this won't work is when "port security" has been enabled on the switch, a not uncommon practice. This is to stop arp poisoning by blocking gratuitous arps where an IP is moving from one Ethernet port to another.
Answer: It's not about the tool, it's about capturing from a device which is in the traffic path you're interested in. Since switches only transmit packets on the ports they are destined for, and edge device is not going to see traffic between two other devices. The usual approach is to capture from a device which is either a bridge or router for the target device.


How did the Witch-king break Gandalf's staff in the movie?

Question: In the movie, the Witch-king breaks Gandalf's staff. How did this happen and why (since it's not in the book)?
Comment: Just standard Jacksonian blustery injury to the tale.  It never happened.  Maketh no sense.
Answer: While Gandalf faces the Witch-king in Minas Tirith, there is no record of Gandalf's staff being broken. The flaming sword is seen in the scene in the book, from Return of the King, "The Siege of Gondor":

'You cannot enter here,' said Gandalf, and the huge shadow halted. 'Go back to the abyss prepared for you! Go back! Fall into the nothingness that awaits you and your Master. Go!' The Black Rider flung back his hood, and behold! he had a kingly crown; and yet upon no head visible was it set. The red fires shone between it and the mantled shoulders vast and dark. From a mouth unseen there came a deadly laughter. 'Old fool!' he said. 'Old fool! This is my hour. Do you not know Death when you see it? Die now and curse in vain!' And with that he lifted high his sword and flames ran down the blade. Gandalf did not move. And in that very moment, away behind in some courtyard of the City, a cock crowed. Shrill and clear he crowed, recking nothing of wizardry or war, welcoming only the morning that in the sky far above the shadows of death was coming with the dawn.

After that scene, Rohan arrives, and the Witch-king flies off to handle them.
The Witch-king's ability to destroy others' weapons is found in the books, however. From The Fellowship of the Ring, "Flight to the Ford":

Then the leader, who was now half across the Ford, stood up menacing in his stirrups, and raised up his hand. Frodo was stricken dumb. He felt his tongue cleave to his mouth, and his heart labouring. His sword broke and fell out of his shaking hand. The elf-horse reared and snorted. The foremost of the black horses had almost set foot upon the shore.

From the same chapter, after the wraiths are driven off, Aragorn is examining the cloak left by the Witch-king:

"Look!" he cried; and stooping he lifted from the ground a black cloak that had lain there hidden by the darkness. A foot above the lower hem there was a slash. "This was the stroke of Frodo's sword," he said. "The only hurt that it did to his enemy, I fear; for it is unharmed, but all blades perish that pierce that dreadful King."

It's likely that Jackson interpreted Gandalf's raised staff as a "blade to pierce" the Witch-king. It doesn't say in the books whether the Witch-king could destroy an Istar's staff, but I think that scene is Jackson's interpretation of these scenes in the book.
Comment: +1. I really like the nugget about "power stare" shattering weapons being canon in the books even if not in the same scene. I think it's the most reasonable explanation. Frankly I'd change the accepted answer to this from mine if I was the OP
Comment: Also worth noting that after the Witch King has fallen (and, I think, after Eomer come and left despairing), Merry finds the blade he used to stab the Witch King.  It blackens and crumbles to ash or dust.
Comment: The singular of _Istari_ is _Istar_.
Answer: As far as I know, there was no Jackson commentary on that specific scene, so one can only speculate.

As for "why", the most cogent speculation I have seen on the forums is that the scene served to:

Highlight the Witch-king's power
Therefore make Merry's and Éowyn's takedown of him all the more heroic.

A second theory was that this was to highlight the power and contribution of the World of Men - Gandalf is almost beaten and can be killed by such a powerful adversary, yet saved by the fact that Witch-king needs to go away and deal with Rohirrim - which, if you know the backstory of Istari, actually jibes with Tolkien's overall vision of the order of things (Olórin's mission is to inspire Middle-earthers to fight Sauron, not to fight him himself with his Maiar powers).
This has further symbolic significance in that this mirrors Gandalf the White's breaking of Saruman's staff.

As for "how", there just isn't any info. The Witch-king holds a flaming sword, stares at Gandalf, and the staff explodes.
Comment: Gandalf was a Maiar.  The Witch-King was a shade of a Man.  Very different levels of power, so I don't think there's a internally consistent way of how the Witch-King shattered Gandalf's staff unless you ignore the books completely and treat the movies as a separate world.  In which case, the Force, a Care Bear Stare or the power of Heart all work. ;)
Comment: @Eelvex - The Force. That's my story and I'm sticking to it. Seriously, there's zero info from Jackson and co, and the image in the movie is basically WK staring at Gandalf till the staff explodes.
Comment: @Oldcat - Gandalf was not allowed to use Maia powers against non-Maia (e.g. anyone outside Balrog)
Comment: the same way that gandalf shattered saruman's staff?
Comment: Disagree - if that were so, Gandalf would have just used Maiar powers to wipe out the army around Minas Tirith, and all the need to bring Rohan or Aragorn to help were just wastes of time.  It is a good way to show that the WK and his army were supreme in that moment, until the forces Gandalf had set into motion arrived and changed things.
Comment: Very interesting. Any ideas on *how* did he do it?
Answer: The short answer is, Peter Jackson thought it would be cool.
Gandalf is far more powerful than the Witch-king for a number of reasons.

Gandalf was created at the literal beginning of time, while the Witch-king has been around for about an age.
Gandalf is a Maiar who has been sent to Middle-earth, with reduced powers, the Witch-king is a human wraith with some of Sauron's power.
Gandalf bears Narya, the elvish ring of fire, the Witch-king used to carry one of the nine human rings (Sauron took them back)
Gandalf fought an unknown number of Nazgûl at Weathertop back when he was Gandalf the Grey (it is entirely possible that he fought all nine) and has only become more powerful.
He defeated a Balrog (another Maiar) and then became even more powerful.

As cool as the scene may look, it has no basis in the world of the books. If the Witch-king was able to overpower Gandalf as easily as he did, there is literally no way that Éowyn and Merry would be able to touch him, let alone kill him.
Comment: It is worth noting that in reference to the Witch King Gandalf notes that he is vested with Sauron's power in that hour, and that Gandalf has not been tested against WK. It's likely he is assuming he could prevail against the WK alone, but as an Istari he's not technically allowed to display his full power. But he may be thinking of the prophecy, and wondering if it technically applies to him. Besides, in Tolkien's magic, as in psychedelics, a lot depends on the setting and frame of mind in which you attempt to do things. That was WK's power hour.
Comment: The Prophecy  ? Which one  ?
Comment: "No man can kill me""?
Answer: I think it is important to keep in mind that Gandalf the Grey did not operate in a mode of "If it kills me, I will come back." His sacrifice at Khazad-dûm was absolutely done at risk and his arrival in Fangorn was very much a shock to everyone, himself included. You see in his manner of speech, not "I came back, I can do that you know." but "I have been sent back" and "I am Saruman as he was meant to be." It clearly came across to me that he was baffled by the turn of events, but pleased with the second chance.
I don't have a reference handy but I was told by a Scholar of Arda much greater than me that once the Nine had been fully subsumed by the power of Sauron that he took their rings back. This is why when he falls they don't have a ring of power in the pile of Nazgûl detritus.
Back to the original topic, there is no literary reason in the narrative for the Witch-king to shatter Gandalf's staff. It is in the movie for two intertwined reasons:

Dramatic effect
Peter Jackson likes to repeat himself (the Breaking of Saruman is cool, let's use that again. See also Hobbit2-Kili-Tauriel-OrcPoison vs. Fellowship-ArwennotGlorfindel-Frodo-MorgulBlade)
Answer: I think that it is not too hard to explain how the Witch-king did it: using his already known power of destroying enemies' weapons. The point is to know if he could do what he has done or not. I think not.
In the chapter "The White Rider", Gandalf says:

'Dangerous!' cried Gandalf. 'And so am I, very dangerous: more dangerous than anything you will ever meet, unless you are brought alive before the seat of the Dark Lord'. (p. 499)

So, one could easily say that the Witch-king is less powerful than Gandalf because only Sauron himself could defeat the White Rider. From that, the answer is clear: there is no situation with could get the Witch-king the power to destroy Gandalf's staff. It is a mistake Peter Jackson made in order to give much more glory to Merry and Éowyn.
Answer: In my opinion this is a quite clear indication of how Gandalf feels about the Witch-king's power.

"My friends,' he said, 'and all you people of this city and of the Western lands! Things of great sorrow and renown have come to pass. Shall we weep or be glad? Beyond hope the Captain of our foes has been destroyed, and you have heard the echo of his last despair. But he has not gone without woe and bitter loss. And that I might have averted but for the madness of Denethor."
Answer: Plot reasons is what stands out to me. I'm pretty sure Gandalf could have bodied the Witch-king. He wasn't "allowed" because of the plan to have Merry and Éowyn destroy the Witch-king. Peter Jackson also used Gandalf's staff being destroyed to emphasize the Witch-king's power. This way, it shows the great courage Merry and Éowyn had to face him.
Before the battle Théoden had said both of them had no place in battle and Éomer says something similar adding that they don't have courage like men. Further showing Peter's message that power is not what makes a person "strong" but courage, determination and compassion.
Comment: This is a nice answer from an out of universe perspective and is a good theory for _why_ it happened. It could be better if you found evidence that Jackson actually did it for this reason and [edit]ed that in. The question also seems to be asking how the Witch King could do it in universe which would strengthen your answer if you could add an explanation about that.
Comment: Eomer said "I do not doubt his heart, only the reach of his arm" in the movies. Merry's courage was never doubted, it was his ability to apply his courage to combat that was doubted.
Comment: @TheLethalCarrot *The question also seems to be asking how the Witch King could do it in universe* This appears to be you projecting something on the question.  If you think that needs to be clarified, suggest you comment under the question and see if the focus of the question is 'in universe" or "out of universe" "either" "Neither" or "Both" ... let's not ask answerers to mind read the OP.
Comment: @KorvinStarmast I'm not projecting anything, the question is asking both in and out of universe. I was just a bit cautious in my wording because I didn't want the answered to feel forced to edit because it is a nice answer as it is and I also didn't want my comment to make people think this isn't an answer which it certainly is.
Answer: Gandalf's power was greatest when it was needed to guide and protect, the narrative always indicated that although his role is vital to the survival of the free peoples, his fate was never to overcome the dark adversary(s) himself. Not only that but, it's stated very early on that the fate (and power) of the Nine is bound to Sauron, so it would follow that the Witch-king's power was at its apex at the siege of Minas Tirith, when Sauron's might (albeit greatly diminished without the Ring) was at its greatest since his fall at the hands of Isildur.
My interpretation is that the destruction of Gandalf's staff by the Witch-king fits entirely to the narrative. The most common theme to the storytelling in both The Lord of the Rings and The Hobbit is that 'small is not necessarily weaker than big', so Peter Jackson's dramatic attempt to emphasise the strength of will of Merry, Éowyn and Sam and Frodo is not misplaced.
Answer: The scene of the Witch-king breaking Gandalf's staff seemed always odd and inconsistent with the text of Tolkien, as clearly pointed out by a citation before. It makes even less sense when one considers that Gandalf is now the highest member of the Wizard order and carrying the third Elven ring of power as it is revealed later on (at least in the books as I remember it). His power must be on the same level as Elrond's and Galadriel's, thus, superior to the power of the nine rings of men.
Peter Jackson probably wanted to increase the impact of the scene, where the Witch-king on his Nazgûl attacks Theoden and claims that no living man can kill him, but then Éowyn reveals herself as woman and destroys him. This resembles a crucial story of the Mahabharata, in which the invincible warrior Bhisma could only be defeated by someone who was not born as a man, Shikhandi/ni.
Answer: I think the reason the staff breaks is because

Sauron has infused a good amount of his power into the ring that the Witch King bears, and since Sauron is much much much more powerful than Gandalf, he could have given the Witch-king enough power to be more powerful than a normal Miar.
The staff is only a part of Gandalf's power, and Gandalf is not allowed to use his full power while on Middle-earth, only a small fraction of it. If he could use his full power, he could have destroyed Smaug, instead of getting a bunch of Dwarves to do it. The Balrog that Gandalf killed was probably much weaker than a Sauron-aided/power-induced Witch-king.

Therefore, for these reasons, mainly because the Witch-king was heavily power-fuelled by Sauron, it is very likely that the Witch-king can hurt Gandalf. But who becomes the victor in a one-on-one battle remains to be seen.
Comment: Welcome to SF&F:SE. These are pretty sweeping statements so you might want to back them up with a quote or two (to prove your points).
Comment: It's stated in the book that Gandalf is a Maiar (Olórin), therefore he cannot be fully destroyed, only his physical form, just like Sauron's was. So, even though the Witch King had more power at that specific scene, he could not vanquish Gandalf, who would come back once more, and maybe even more powerful, as he did before.


What would be the configuration and performance for Saturn V with all stages RP-1/LOX?

Question: Saturn V had RP-1/LOX 1st stage with 5 F-1 rocket engines. The 2nd and 3rd stage used LH2/LOX J-2 rocket engines, 5 and 1 respectively. At that time NASA had an advantage to launch heavy payloads because it was successful in the production of LH2/LOX rocket engines. What situation would it be if even the 2nd and 3rd stages of Saturn V would have been with RP-1/LOX engines? I am speaking for that situation where the 1st stage remains the same. What configuration would the Saturn V have had for the 2nd and 3rd stages (with what rocket engines, how many of them, what amount of fuel in their tanks, or even a possible 4th stage) and what performance? Probably it would have lower payload values and would be a shorter rocket since RP-1 is denser than LH2, but what would be the values and how would it look like?
Comment: @PaulJordan as Russell Borogove has explained it would have a low performance, not enough for a manned landing. A performance that would make the rocket not too practical with 4 launches at least and also many docking procedures.If NASA wouldn't be successful with LH2/LOX than they would not use the F-1 or H-1, they would have tried to build new more efficient RP-1 engines.
Comment: Completing the trilogy: the all-hydrogen Saturn (http://space.stackexchange.com/questions/17629/how-would-the-saturn-v-have-differed-if-the-first-stage-was-also-lh2-lox) and the methalox Saturn (http://space.stackexchange.com/questions/17684/how-would-the-saturn-v-have-differed-if-the-first-stage-was-using-liquid-methane).
Comment: I am interested to know for all stages RP-1 not all stages LH2? What would have done NASA, how would it be the project for this rocket in this case
Answer: If we hold the loaded masses of the Saturn V's three stages constant, but switch to kerosene tankage, engines, and propellants for the upper stages, we get a rocket that can send perhaps 55 tons to LEO and 9 tons translunar -- not enough for a crewed landing mission, but certainly enough for a flyby and possibly a crewed lunar orbit mission. Two launches and an Earth orbit rendezvous could possibly get a bare bones, Soyuz/LK lander mission, and three launches and EOR could probably pull off an Apollo CSM/LM mission.
First stage remains a 5-engine S-IC, 2148 tons propellant, 132 tons dry.
Second stage would be 447t propellant, 34t dry, 1x F-1 engine.
Third stage would be 106t propellant, 9t dry, 1x H-1 engine.
For translunar flight, the third-stage H-1 would have to be modified to provide restart capability; it would burn once to get into LEO and then again for TLI, like the J-2 on the Saturn V third stage.
All-up launchpad mass would be 2885 tons for the translunar mission, 2931 tons for 60t to LEO.
The upper stages would be much more compact than on the Saturn V, as you note. The second stage would be about the size of the Saturn V third stage, in fact, so the stack could look rather like the proposed Saturn INT-20 configuration.
(Amended tonnages to reflect more conservative fuel tank mass fractions.)
Comment: Weren't the upper stages size-limited, rather than mass-limited?  If so, using kerosene/LOX would allow the use of more propellant mass, improving your performance somewhat.
Comment: For an unchanged Saturn V first stage, the upper stages are mass limited. The full stack liftoff TWR of the Saturn V was only about 1.16:1, which is about as low as you want to go -- any lower and you worry about gusting winds blowing you into the launch tower.
Answer: It would probably not have been practical to do a three-stage, single launch LOR Saturn/Apollo all on kerosene; the specific impulse advantage of the J-2 engines is just too great.
According to my spreadsheet estimates, a four-stage kerosene rocket 3 times the size of Saturn V could do it.
The translunar stage, S-IV-K, is 138 tons propellant, 12 tons dry, 47 tons payload (Apollo CSM and LM). 1x H-1 engine (as used on the Saturn 1B).
Third stage, S-III-K, 465 propellant, 35t dry, 1x F-1 engine.
Second stage, S-II-K, 1674t propellant, 126t dry, 3x F-1 engines.
First stage, S-I-K, 6392t propellant, 408t dry, 16x F-1 engines.
All-up mission launch mass of this "Saturn XVI" would be 9297 tons.
First 3 stages produce ~9600 m/s of delta-v, taking the beast to a 185km circular orbit. Fourth stage produces ~3400 m/s to send the spacecraft to the moon.
The mass could be brought down substantially with more optimized engines; both the F-1 and H-1 were designed as first-stage engines. Larger nozzle extensions, particularly on the 3rd and 4th stages, would improve the specific impulse, without requiring all-new engine designs. With an "H-1V" producing 320s ISP and an "F-1V" producing 337s, total launch mass could be brought down to 6820 tons or so, with "only" 12 engines on the first stage.
Both my answers are, obviously, merely rough Kerbal-style feasibility estimates.
(Amended tonnages to reflect more conservative fuel tank mass fractions.)
Comment: Great work,calculating how much more bigger would be the Saturn V to have the same payload values.It is my mistake that i haven't specified very well, i was interested to know the configuration and the performance if only 2nd and 3rd stage would be changed, replacing J-2 with RP-1/LOX engines and without changing the 1st stage (that will be the same). How many and what engines would be used for 2nd 3rd(and a possible 4th) stages.What payload values would have now.Probably it would have lower values and would be a shorter rocket.But how much would change the values and how would it look like?
Comment: I will edit my question
Comment: Provided as a separate answer for clarity.
Comment: It is really impressive, about 3 times of propellants mass, dry mass and number of engines for the first stage.
Answer: Yet another approach to the answer is to look at the Soviet N-1 rocket, which evolved in fact to attempt a Soviet version of the Lunar Orbit Rendezvous strategy accomplished by Apollo. The Soviets lacking a good developed hydrogen engine, it was all kerlox.
Their approach took several measures to greatly lighten the spacecraft involved, a version of Soyuz and a lander called LK that relied on a kerosene-oxygen "crasher" to take the LK down from orbital speed to a half kilometer or so above the landing site--and more importantly than losing altitude, took off most of the orbital speed. Thus the LK itself was a single stage unlike the American LM, basically a smaller analog of the LM ascent module with extra fuel for final descent, hovering and landing. (It also had a complete backup engine in case its main engine failed, of the same thrust but with no throttle ability, for abort or as a backup ascent engine). The LK could only handle one cosmonaut, for a much briefer stay on the lunar surface, and upon returning to the Soyuz in orbit (and to board it to descend) the cosmonaut had to spacewalk--no transfer tunnel.
Using more efficient pumped kerosene-oxygen engines than the American pressure fed hypergolics, the Soviet approach required a lighter stack to be launched to the Moon, but even so due to the lower efficiency of kerosene compared to hydrogen, the fourth stage that would serve to send that stack to encounter the moon had to be larger in proportion. Counting the kerosene-oxygen "D" block as a fifth stage, the attempt to get the Lunar mission done with a single launch was very marginal despite the fact that the five-stage rocket outmassed the Saturn V by a considerable margin. To try to lighten it enough to enable the mission they took all sorts of dubious weight-saving expedients, such as removing most of the telemetry. The earliest version of N-1 was not expected to boost more than 45 or so tons to low Earth orbit. Between adding engines, raising the dimensions a bit, and these weight trimming and propellant maximizing expedients, it was to put 95 tons into LEO in the Lunar version that was tested several times. Each attempt led to failure, and had they succeeded in getting a manned Lunar stack on the fourth stage, to inject it on to the Moon, I fear the corner cutting would have spelled disaster sooner or later, probably sooner.
Had they gone over to a two-launch strategy and made fewer heroic efforts to maximize payload and made it somewhat more robust, I think they could have pulled off a Moon landing with margin to spare--indeed the escape/survival modes enabling the crew to return despite major failures might have been far superior to Apollo with a two launch plan. An Alternate History timeline I much enjoyed was written several years ago on this premise.
But looking at the huge size, numerous engines and many stages of the N-1 as built, compared to a much more meagre tonnage to low lunar orbit, illustrates the huge advantage the Americans enjoyed in developing hydrogen upper stages. With kerosene alone, it would have been possible to achieve greater efficiencies than our program accomplished with that fuel mix--the Soviet pumped kerosene engines in their N-1/LK program were all notably more efficient than the best American kerosene engines, though none came close to matching the sheer mighty thrust of the US F-1 engine.
But even so, with markedly superior ISP kerosene engines, the Soviets, or Americans foregoing development of hydrogen engines, would have had to launch a much greater tonnage off of launch pads to achieve comparable results in terms of a moon landing. If the goal were merely to build space stations or the like, the tradeoff is much more reasonable.
But then again, one disadvantage hydrogen burning rockets have is the difficulty of storing the hydrogen for a long time. In space, it is not that difficult to keep LOX liquid and stored for a long time, but hydrogen boils off (or the tanks explode!) Centaur has demonstrated that one can retain enough fuel after several days to be worthwhile, so in principle the Apollo stack might have been braked to low Lunar orbit insertion by such a stage with some savings in weight, but the design was frozen in the early sixties around using nothing but pressure fed hypergolic engines after TLI, and TLI would be initiated just hours after reaching the parking orbit.
So, the drawback of hydrogen boiling off is not a problem if one is using the hydrogen to put something into low Earth orbit or plans to use it pretty soon after that. Thus to argue ker-lox would be better than using hydrogen upper stages one would have to look closely at the structural fixed mass penalties that hydrogen storage imposes. But with good design these are not terrible; look at the STS fuel tank, that massed 36 tons dry but held something like 750 tons of oxygen and hydrogen. Or, the S-IV third stage of the Saturn V (also the second stage of Saturn 1B) which massed 9 tons or so dry, including the engine, but held 120 or more tons of propellant-these are dry weight fractions well under 1/10, while few ker-lox or hypergolic stages actually used were dramatically lower than that.
Another penalty of using hydrogen is that the ratio of thrust to engine weight is also inferior to what a similar state of the art investment can achieve with less energetic but cooler-burning and denser propellants. The Shuttle SSME had a good thrust/weight ratio compared to first generation kerosene or hypergolic engines, but this was accomplished by heroic and expensive to develop, build and maintain methods involving extremely high pressures and temperatures. The tradeoff between efficiency and thrust is inherent in the basic physics; to get a faster exhaust one operates at higher temperatures; to get decent thrust against high sea level atmospheric pressure one uses high pressure; a lower energy per kilogram propellant mix burns cooler and trades off higher mass flow for lower power for the same thrust. If thrust is the goal, it often makes sense to burn something cooler and use simpler, cheaper machinery.
Bottom line is, that initial boosting off the launch pad, the job of a first stage, is a task accomplished by use of massive thrust, and the benefits of using hydrogen are most marginal there, while the advantages of less ambitious fuel mixes in simpler but stronger engines are greatest, especially factoring in easier storage. Vice versa, once raised off the ground, out of the lower atmosphere and boosted to modest speeds that buy time for weaker but more efficient hydrogen based engines to work, at much lower thrusts orbit can be achieved with lower overall upper stage masses and fewer stages. Thus the design of the Saturn V was a good synergy. Note that the first stage is more massive by far than any upper stages--Saturn V had roughly a factor of 5 jump between the third and second stages, but the first stage was something between 3 and 4 times the mass of the entire upper stack, of two stages and a 45 ton Lunar stack all put together. To move this huge total mass engines creating tremendous thrust were required; even if the great challenge of making hydrogen engines that could produce that thrust were met, the drawbacks of storing huge quantities of hydrogen versus compact volumes of kerosene would have offset much of the advantage to be gained by higher efficiency faster exhaust speeds. They economized on these where it counted the most, in the first stage. A mixed stack, despite the problems of managing several types of propellant instead of just two, was the superior way to go.
Comment: This might be a great answer - I don't know yet though. There is nothing wrong with long answers if they are good, but I wonder if you could add a bit of a summary, and maybe some structure via headings to help the reader? Even the "bottom line" alone is over 250 words long!


How to add an event to a different user's calendar

Question: Is there a way to upload the iCalendar (calendar file) to a different person's Google Calendar directly?
I shouldn't be able to view the other person's calendar but I should be able to upload calendar events directly (Note: I do have the permission to upload to his calendar).
How should I achieve this?
Does the other person need to provide me any rights/permissions in Google Calendar to authorize such activity?
Answer: Try this:  This uses Jquery and a small plugin to generate add to Calendar and If you wish even using ical/vcal to users desktop calendars
Works like a charm
https://github.com/tardate/jquery.addtocalendar


Is it possible to see Fraunhofer lines with amateur equipment?

Question: Is it possible to see Fraunhofer lines with amateur equipment?
Would it be possible (with reasonable effort) to identify elements or is this hard?
Answer: Yes!  See: http://www.skyandtelescope.com/skytel/beyondthepage/121557614.html
Answer: Certainly. Spectroscopes are available to fit most telescopes, or you could make your own using a diffraction grating.
Answer: Some will and some wont - it all depends on your instrument/device and your source (solar vs. stellar), and whether you are looking visually or doing photography. Newton missed them completely, Wollaston saw them in a kind of fleeting manner and thought they were divisions lines between the colors (which means he didnt see much!), but Fraunhofer being the far more skilled practicioner/investigator saw and catalogued hundreds, he was able to calculate the wavelength of d1-d2 sodium lines using Young's forumla for diffraction (he was using a grating vs. a prism!) ... so seeing the Fraunhofer lines is not automatic and the new highly taunted very expensive low-order diffraction gratings no being sold to amateur astronomers reveals very little in the way of stellar spectroscopic details -
so stick to looking for the solar Fraunhofer lines after build a cd-spectroscope ... and
build it exactly as told! Then you probably will see several dark Fraunhofer absorption lines ... or better yet look for bright Fraunhofer EMISSION LINES in your room lights after building a "diffraction grating spectroscope". The Fraunhofer lines are all the same for a given element and wavelength, whether absorption or emission, and the emission versions are much easier to see!
Answer: There are lots of cool "make your own spectroscope" projects online.  Most use a CD as the grating.  They aren't for scientific purposes but are fun and educational.  You'll be able to see the spectral differences between sunlight and incandescent lights and others.  You can make them from a computer print-out on card stock, or use a cereal box!


Are there minimal pairs between normal length and long vowels in English?

Question: Are there minimal pairs between vowels of normal length such as a and vowels of long length such as aː?
Answer: Australian English has true phonemic length distinctions.
Some examples:

ferry /feɹi/ vs fairy /feːɹi/
Manning (name) /mænɪŋ/ vs manning /mæːnɪŋ/
Answer: Yes, though it depends on the variety. For example in Scottish English, vowel length does not have minimal pairs since it is determined by "Aitken's law".
In Standard Southern British English (SSBE) however there are minimal pairs for example:

For /ɪ/ vs. /iː/: bit vs. beat; lick vs. leek; hit vs. heat etc.
For /æ/ vs. /ɑː/: cat vs. cart (though this wouldn't hold up in rhotic accents)

Also note the qualitative difference in the short and long vowels, it's not just a vowel length difference, so some might argue that it's not actually vowel length which forms the minimal pair but the quality of the vowel.
Hope this helps.
Answer: A classic example is that virtually all dialects distinguish [bid] "bid" and [bi:d] "bead". Also, dialects often distinguish [bit] "bit" and [bi:t] "bid". A "minimal pair" is a pair of words whose phonetic valued differ only in the presence of A vs. B in the phonetic output. To rephrase, "minimal pair" is a property of phonetic outputs, not underlying phonological analysis. It is a well-known problem with the concept of "minimal pair" that the minimality of the pair is crucially determined by the granularity of the transcription, so while [bid] vs. [bi:d] is "okay" as a phonetic transcription, [bɪd] / [bi:d] is even better (and also not minimal w.r.t. length), and [bɪˑt] / [bɪˑjt] is best (though perhaps dialect dependent). Then we have the contrast [bɪˑt] "bid", [bɪt] "bit", [bɪˑjt] "bead" and [bɪjt] "beat/beet". The underlying contrast in the vowels is length, which is realized phonetically via diphthongization. The underlying consonant contrast is voicing, realized phonetically as lengthening  of the vowel.
My "informants" of Australian English had a vowel quality contrast to go with their length contrast. The Wiki article says "some speakers" and the data are notated with a "citation needed" flag indicating that there is a diphthongization difference to go with the assumed length difference. This can only be resolved with some concrete acoustic evidence. Fletcher et al. look into the "hud / hard" difference, unfortunately their measurements only give a single generic formant value of "the vowel". They do show that there is a large correlation between vowel choice and duration, and a small and speaker-dependent difference in formant value.
In my west coast dialect, there are minimal pairs like "whore" [hɔɹ] and "horror" [hɔ:ɹ], "tear" [tɛɹ] and [tɛ:ɹ]. To validate the claims about transcription, we'd have to have an independent way to establish how much of the speech signal is [ɹ] and how much is vowel. If the rhotic is longer in the orthographically bisyllabic forms, thus we really have [hɔɹ] vs. [hɔ:ɹˑ], then this would not be a minimal pair, it would be like the case of "beat" and "bit", with length and quality co-dependent.
Comment: What is [tɛ:ɹ], a west coast pronunciation of terror or tearer?
I know hardly anything about US accents.
Comment: It was 'terror', and I'm not sure about 'tearer' (not a common word).
Answer: Assuming minimal pair means contrasting phonetic forms, here is one from my midwestern American dialect: [tʰæ̃j̃k] "tank" versus [tʰæ̃:j̃k] "tannic".
Answer: Non-rhotic varieties of English have a number of phonemic contrasts in which vowel duration is pretty important. It might be a primary cue to the vowel contrast in some varieties, as exemplified above. Perhaps English will develop more durational minimal pairs.
For example, the vowel in "hut" is often a short version of the vowel in "heart" or STRUT vs. HEART in lexical sets, though this is not conveyed in traditional transcriptions.
Pairs distinguished by duration (alone, or primarily) are increasingly met for
"piss" vs.  "pierce" (KIT vs. NEAR in closed syllables)
"led" vs. "laird" (DRESS vs. SQUARE in closed syllables)
Answer: Scottish English has short vs. long in near-minimal pairs like "need" vs. "kneed", "brood" vs. "brewed", distinguished purely by vowel duration. This is pretty much only possible for the vowels /i/ and /u/ before a /d/, where the /d/ is either tautomorphemic (for the short one) or a suffix (for the long one). There are also some words where a long vowel occurs unexpectedly, like "dude" and "lewd".
Given that the distribution of the vowel duration is mostly predictable from the status of the /d/, strictly speaking the short-long duration difference is not phonemic, and is one aspect of the so-called Scottish Vowel Length "Rule".