Japan Data: Postcode/Census

Question: I'm primarily looking for a GIS data source for postal codes in Japan as well as census data for different scales of geography.  Has anyone found a good source of these data sets that is in English?
Answer: Not sure if this is what you had in mind, but there is a CSV file available from the Japan Post web site that has postcodes and their corresponding locations in Romaji (roman characters or English if you wish to think of it in those terms) for all of Japan. It is available from a link on this page:
http://www.post.japanpost.jp/zipcode/dl/roman-zip.html
There is no equivalent English page (that I could find), but if you can identify the kanji '全国一括' (also identifiable by size:(988,276Byte)) it's in the lower right corner of the group of links (the data is provided by prefecture (ken) as well, thus the large number of links), and the file is zipped as: KEN_ALL_ROME.LZH
There are some other attributes included in the file (not LAT/LONG), but I did not research what they are.
If this is enough as is, or not what you wanted, that's fine. However, if you are interested in knowing more about the data set, I may be willing to put a little more effort into looking into it, if asked.
At this point, I am not sure about Census data sets. Sorry.
HTH
Comment: Kurt,
Thanks for your find.  I'm looking for the spatial boundaries ( polygons of the postal code's extent ) with a record tied to each polygon.  This is usually found in a shapefile format.
Answer: JBO,
Naturally, the polygons (shapefile) are independent of language, but you can use the post office file to match the attributes with the romanized data.
The shapefile for postal code areas is PAREA-Zip, which is sold by PAREA (a division of KOKUSAI KOGYO CO.,LTD.), and whose website is http://www.parea.jp/datebase/area_map/#zip
There is an email provided on that page (no English page), and they will likely reply to a succinctly written English message if you are interested in purchasing their data.
For census data, the GIS formats are available from PASCO (presumably not free) in their PFM series (probably PFM25000-P). Their English page describing these datasets is http://www.pasco.co.jp/eng/products/map/
HTH

Japan Data: Postcode/Census


How can I tell what level of difficulty I've completed on a given mission?

Question: In Hitman Absolution, where do I look, or how can I tell which missions have been completed on a particular difficulty level? I played some missions on professional difficulty, and some on normal, but I can't remember which ones I've done on each difficulty setting. I'm trying to get the Absolution achievement, and I don't want to go through each one over again.
Answer: Short answer, you can't. In this case, having a good memory and/or having the foresight of noting down which level you've played at which difficulty is paramount.
You can, however, look at the total score you have for each level. If the score is higher than you'd expect to get from normal mode, than it's very likely that you have finished the level on hard. Remember, in normal mode you get 25% bonus to your total score and in hard mode, 50%. This could actually be quite reliable depending on your playing style. Because under normal mode, many players actually tend to just go all gun-ho on the enemies, whereas in hard mode, you are forced to take the steathier approach, as you take more damage, do less damage to enemies (or they have more hitpoints; irrevelant in this case); enemies are more alert and numerous. So ironically the more meticulous and careful you are when playing on normal mode, the less you are able to tell the differences in score between difficulties.


How does the correlation coefficient differ from regression slope?

Question: I would have expected the correlation coefficient to be the same as a regression slope (beta), however having just compared the two, they are different. How do they differ - what different information do they give?
Comment: if they are normalized, they are the same. but think of what happen when you make change of units...
Comment: I think the top scoring answers to [this Q](https://stats.stackexchange.com/q/22718/136579) (and maybe even [my A](https://stats.stackexchange.com/a/386483/136579) to it where I show that the correlation coefficient can be seen as the absolute value of the geometric mean of the two slopes we obtain if we regress y on x and x on y, respectively) are also relevant here
Answer: Assuming you're talking about a simple regression model $$Y_i = \alpha + \beta X_i + \varepsilon_i$$ estimated by least squares, we know from wikipedia that $$ \hat {\beta} = {\rm cor}(Y_i, X_i) \cdot \frac{ {\rm SD}(Y_i) }{ {\rm SD}(X_i) } $$ Therefore the two only coincide when ${\rm SD}(Y_i) = {\rm SD}(X_i)$. That is, they only coincide when the two variables are on the same scale, in some sense. The most common way of achieving this is through standardization, as indicated by @gung.
The two, in some sense give you the same information - they each tell you the strength of the linear relationship between $X_i$ and $Y_i$. But, they do each give you distinct information (except, of course, when they are exactly the same):

The correlation gives you a bounded measurement that can be interpreted independently of the scale of the two variables. The closer the estimated correlation is to $\pm 1$, the closer the two are to a perfect linear relationship. The regression slope, in isolation, does not tell you that piece of information.
The regression slope gives a useful quantity interpreted as the estimated change in the expected value of $Y_i$ for a given value of $X_i$. Specifically, $\hat \beta$ tells you the change in the expected value of $Y_i$ corresponding to a 1-unit increase in $X_i$. This information can not be deduced from the correlation coefficient alone.
Comment: As a corollary of this answer, notice that regressing x against y is not the inverse of regressing y against x !
Comment: One more point- the correlation coefficient does not have units. The slope does.
Answer: With simple linear regression (i.e., only 1 covariate), the slope $\beta_1$ is the same as Pearson's $r$ if both variables were standardized first.  (For more information, you might find my answer here helpful.)  When you are doing multiple regression, this can be more complicated due to multicollinearity, etc.
Comment: In simple linear regression, as Macro, shows above, $\hat{\beta} = r_{xy}\frac{s_y}{s_x}$. Is there an analogous expression for multiple regression? It seems there isn't for the reason you're getting at with "multicollinearity," but I think you really meant covariance here?
Comment: @Iamanon, try reading: [Multiple regression or partial correlation coefficient? And relations between the two](https://stats.stackexchange.com/q/76815/).
Answer: The correlation coefficient measures the "tightness" of linear relationship between two variables and is bounded between -1 and 1, inclusive.  Correlations close to zero represent no linear association between the variables, whereas correlations close to -1 or +1 indicate strong linear relationship.  Intuitively, the easier it is for you to draw a line of best fit through a scatterplot, the more correlated they are.
The regression slope measures the "steepness" of the linear relationship between two variables and can take any value from $-\infty$ to $+\infty$.  Slopes near zero mean that the response (Y) variable changes slowly as the predictor (X) variable changes.  Slopes that are further from zero (either in the negative or positive direction) mean the response changes more rapidly as the predictor changes.  Intuitively, if you were to draw a line of best fit through a scatterplot, the steeper it is, the further your slope is from zero.
So the correlation coefficient and regression slope MUST have the same sign (+ or -), but will not have the same value.
For simplicity, this answer assumes simple linear regression.
Comment: you indicte that beta can be in $-\inf, \inf$, but isn't there a case-by-case bound on beta implied by the ratio of variance of x and y?
Answer: Pearson's correlation coefficient is dimensionless and scaled between -1 and 1 regardless of the dimension and scale of the input variables.
If (for example) you input a mass in grams or kilograms, it makes no difference to the value of $r$, whereas this will make a tremendous difference to the gradient/slope (which has dimension and is scaled accordingly ... likewise, it would make no difference to $r$ if the scale is adjusted in any way, including using pounds or tons instead).
A simple demonstration (apologies for using Python!):
<code>import numpy as np
x = [10, 20, 30, 40]
y = [3, 5, 10, 11]
np.corrcoef(x,y)[0][1]
x = [1, 2, 3, 4]
np.corrcoef(x,y)[0][1]
</code>
shows that $r = 0.969363$ even though the slope has been increased by a factor of 10.
I must confess it's a neat trick that $r$ comes to be scaled between -1 and 1 (one of those cases where the numerator can never have absolute value greater than the denominator).
As @Macro has detailed above, slope $b = r(\frac{\sigma_{y}}{\sigma_{x}})$ , so you are correct in intuiting that Pearson's $r$ is related to the slope, but only when adjusted according to the standard deviations (which effectively restores the dimensions and scales!).
At first I thought it odd that the formula seems to suggest a loosely fitted line (low $r$) results in a lower gradient; then I plotted an example and realised that given a gradient, varying the "looseness" results in $r$ decreasing but this is offset by a proportional increase in $\sigma_{y}$.
In the chart below, four $x,y$ datasets are plotted:

the results of $y=3x$ (so gradient $b=3$, $r=1$, $\sigma_{x}=2.89$, $\sigma_{y}=8.66$) ... note that $\frac{\sigma_{y}}{\sigma_{x}}=3 $
the same but varied by a random number, with $r = 0.2447$, $\sigma_{x}=2.89$, $\sigma_{y}=34.69$, from which we can compute $b= 2.94 $
$y=15x$ (so $b=15$ and $r=1$, $\sigma_{x}=0.58$, $\sigma_{y}=8.66$)
the same as (2) but with reduced range $x$ so $ b= 14.70$  (and still  $r = 0.2447$, $\sigma_{x}=0.58$, $\sigma_{y}=34.69$)

It can be seen that variance affects $r$ without necessarily affecting $b$, and units of measure can affect scale and thus $b$ without affecting $r$


Dynamics CRM 4.0 Access Audit?

Question: In Microsoft Dynamics CRM 4.0, is there any way to audit what records were viewed by a particular individual at what time without any special plugins? If you need a plugin, can you install the plugin and then look at past data?
Answer: There is no native way to do this, but there are plugins that can be deployed, and it's the only way to grab all data view, even from external applications that are using the SDK.
When plugins are deployed, they are active on particular events. So if you are looking a past data, the plugin will be launched, because the plugin is not aware of the underlying data.

Dynamics CRM 4.0 Access Audit?


security in the standard model → random oracle model?

Question: Can a protocol proved secure in the standard model be considered secure in the random oracle model?
Answer: As correctly pointed out by Ricky Demer it is not necessarily true. However, this implication does not hold for very specific cases. In the case of random oracle gates the existence of the RO changes the functionality of the "scheme", since with RO there are RO-Gates and without there aren't.
In most cases the existence of the RO does not affect the functionality of the scheme, meaning that with and without RO you have the exact same functionality. In these cases this implication holds I think.
The ROM, in comparison to the standard model, provides you with an additional oracle that realizes a truly random function, i.e. a function where the input and output are absolutely unrelated.
If you have proven something secure in the standard model, then you didn't need this additional oracle. The exact same proof will so to speak also hold in the ROM but will not use the oracle.
Even though an adversary might use the RO too, I don't see how this would help him if it doesn't affect the functionality of the scheme, since in this case the scheme and the proof are completely unrelated to the RO.
Answer: Not necessarily.
For example, if there is a public-coin collision-resistant hash family then there is

a (statistical) zero-knowledge argument system (with negligible soundness error)

for NP that uses a constant number of rounds and has a public-coin verifier.

However, in the random oracle model, constant-round public-coin computational zero-knowledge argument systems (with negligible soundness error) only exist for easy problems.
The point is that the simulator must commit to the adversary's code and then give a universal argument about what that code outputs on a given input. $\:$ Such arguments can't handle random oracle gates.
Also, I have not seen any proof that a random oracle does not significantly increase the

power of (fully) non-uniform adversaries, even against non-interactive hardness assumptions.
Comment: Thanks for your answer. I've not really understood.


Contact form or mailto: link?

Question: I'm aware that mailto: links only work for people who use a mail client. Assuming most novice users don't, is it safe to assume that contact forms are only appropriate when the target audience isn't a tech-savvy one?
Answer: Arguments against forms:

Users have to fill in more
information (at least their email
address)
There is more room for
errors. What if the user mistypes the
email address? No way of contacting
him/her.
Losing your internet connection,
pressing a wrong button etc. can
cause the loss of the message (very frustrating)
The mail isn't in your sent mail folder. No way to look up what you sent and when you sent it.

I'm not saying contact forms are wrong. Just some points to consider.
Also: On smartphones your email client is usually an app so there is no problem of mailto: links not working.
Hope that helps, Phil
Comment: Not sure if this has ever been done but what about upon clicking the submit/send button having it cc a copy to yourself? This could eliminate the part about not having a record of it.
Comment: @Matt - Lots of sites cc copies (or at least give you the option to do so) when using the contact form. A prime example is eBay - whenever you contact the seller, you can get a copy of it.
Comment: While the email client may be an app on a smartphone, it doesn't guarantee that it has been configured. I was pretty surprised to find out that a work colleague does not have Mail on his iPhone configured, because he uses Gmail. I think it's likely that a lot, if not most users don't use their phone's default email client.
Comment: @Charles - Good point about eBay, I remember that now!
Comment: Good thoughts here.
Answer: I would say that contact forms are always better to use. The general population is now very comfortable filling out online forms, and the contact form is so common that it should produce no question and what it's use is.
Facebook, Twitter, eBay, etc, have created a standard for it.
In fact, I would go so far as to say that using the mailto: would be scary for users who didn't realize that you could open their default mail client.
...and even more alienating for people who didn't know they had a mail client, when they see it open it.
:)
Comment: Clearly forms are the trend on most sites. However, at least *having* the email address as an option (with or without a mailto link) is probably also worthwhile.
Answer: Go for both. In the end I am sure you will not have just a form in the whole page, or just the mailto link. So give people the possibility to choose. The form shouldn't be scary for a person that wants to contact you. So filling in 3 fields shouldn't be a problem.
Answer: This depends on the nature of your site and how you interact with your customers.
The email address should be with the full contact details - address and telephone numbers, maybe photos of the people on the customer service team.
The contact us form is definitely the way to go, not least because you can ask people questions to filter their enquiry to a given department.
The design of the contact us form is important too - you need to supply meaningful confirmation of their enquiry in email, e.g. 'thanks for your enquiry we will be writing you back soon' text.  Also set the mouse focus to the first text box so they don't have to click on it. Also consider putting in some javascript over-ride for when they press return - some people do this to move to the next box and end up submitting the form.
For many small businesses 'contact us' is the second most important page on their website - that is where the leads come from. You can record anything and everything you can about their computer, IP address and anything else in $_SERVER. From this information you can then see where your customers come from. Depending on the business this may be helpful to know.
Comment: Yes.  Do actually publish an email address.  This gives people the option of using their webmail should they wish.
Answer: Native mailto: links have real strong advantages. They are obvious, and are summed up in the word native.
But the Web form has a strong advantage too. It is handy to use when I don't have a ready-to-use e-mail client at hand. Or when I don't want to launch it. I am on the Web, I stay on the Web. Quick. No hassle.
Have both.
The good approach is the one chosen by Bare Bones Software, for example. A native contact e-mail link, and a contact Web form. And they tell their snail mail address too, and phone number, and fax number. This is good. And send to the user a confirmation of his/her message. So that s/he have his/her message in his/her mailbox.
Comment: +1 for providing both solutions and letting the user choose what he prefers.
Answer: What if instead of a mailto: or a contact form we just provide the e-mail address, like we provide the facebook or twitter page?
It just don't open your client email (you have to go by yourself).
If you are on the internet your email may be already opened, so it would be fast.
You avoid the problems cited by Phil.
Comment: Scraping bots. Your mail can easily go to a lot of unwanted places.

Contact form or mailto: link?


What does Character development actually mean?

Question: I keep hearing the terms "character development" and "characterisation", but I'm still not sure what they mean? How does one "develop" a character?
Answer: Character development can refer to either the task of sitting down and creating a character (working out their appearance, history, mannerisms, and so on), or it can refer to the change a character undergoes during the course of a story.
In the first instance, the idea is to create a fictional person, complete with flaws and weaknesses, history, mannerisms, hopes, fears, someone that is often even more real than people you meet in real life. This is (generally) a person you will use in your story, whether as a protagonist or antagonist. It's like you getting to know someone so well that you know absolutely everything about them. Just how much you need to know depends on how important the character is to the story, but generally speaking, the more you know about the character, the better able you'll be able to write about them in relation to your story.
In the second instance, character development describes the change an individual undergoes through the course of a story as a result of the conflict or conflicts that person encounters. Think of the character Scrooge from A Christmas Carol, who begins as a crotchety, tight-fisted, greedy old man, but over the course of the story, various "conflicts" force him to change his ways so that, by the end, he becomes a better man, generous and kind.
Comment: The two are often used interchangeably, meaning "fleshing out the character" - both in designing them and in showing them to the audience. Remember the Han Solo and Greedo scene in Star Wars? Its impact on the whole story was minor but it was essential to develop the character of Han, as a ruthless, practical mercenary with no qualms, and subsequent re-release where Greedo shot first caused fan outrage as it totally reversed characterization of Han Solo (not to mention it made the scene far less plausible.)
Comment: I'm aware that technically "characterization" is the process of actually describing a character, but regarding OP's question, wouldn't the first scenario you described fall more into the characterization category?
Comment: @DeVil - Yes, the first scenario falls under characterisation, but "character development" can be used to mean "characterisation", depending on context. Example: "I'm going to spend some time on character development, and work out his history."
Comment: This is the 3rd best answer Im giving you today Craig! Well done!
Comment: The most effective character development is that which is shown, not told, in my opinion (and hopefully in general). The Han and Greedo scene mentioned above is a good example of this. You learn a great deal about Han Solo through this scene. He's in debt, he's on the run, and he cares for little but himself (or so he thought...).

Characters can often be developed on the fly, as if you are learning the character as you write your story. For some characters you may need only start with little more than a single sentence and everything else can be improvised from that.


What is wrong with a high-speed landing for an ultralight or GA aircraft?

Question: Is it okay to land an ultralight or a general aviation plane at a speed higher than the recommended landing speed for such plane? Will there be any damage if the descent rate is very low?
Comment: [Related](https://aviation.stackexchange.com/q/2485/62). Higher speed landings are common in training, e.g. to demonstrate the effects of a tailwind, or to simulate a flaps failure.
Answer: Keeping too much speed in an approach in any airplane can be risky depending on how much runway you have as it all has to bleed off before you can stop. It's a great way to end up in a hedge. Extra airspeed means you'll either float in ground effect for a very long time, or if you somehow manage to get the wheels down you won't get much braking action as the wheels won't have much weight on them on a tricycle configuration, in a tailwheel configuration you will have braking action but you'll have to be gentle or you'll nose in. In fact, trying to hold down an airplane that is far over rotation speed is tricky, and it can all go wrong. There's no good reason you'd want to try this unless you had some sort of condition in the airplane that necessitated it.
Landing fast isn't good for your tires, the extra ground speed will cause more wear on initial contact as the wheel has to spin up that much faster. Your brakes could overheat or at least would wear out faster as well. Other than that I can't think of any particular damage or wear.
Comment: *"Other than that I can't think of any particular damage"* when thinking of ULF you have to keep in mind that they usually have low safety factors and pushing your gears to the ground at high spead may cause damage to them especially if the runway is gras or similar.
Comment: Sure, @TimothyTruckle, but that's no different from the risk of any other landing if it's mishandled. You have a greater risk of a hard landing if you land fast, but landing fast doesn't automatically mean you land hard.
Answer: You can, in theory, fly a faster landing speed than the default Vref of 1.3*Vso.  But it's wasteful in the round out and potentially very dangerous on short field landings.
When an airplane lands, the pilots needs to enter the round out with only enough energy to arrest the descent and establish the airplane in the correct attitude for touchdown approx 1-2 ft above the runway.  The airplane should be then quickly be exhausted of energy and set down on its landing gear, ideally within the first 1/3 of the runway length.  Excess speed here combined with the dramatic reduction in induced drag due to ground effect will cause the aircraft to 'float' down the runway until it slows enough and touches down.  Attempting to force the airplane onto the runway with forward elevator pressure is hazardous as the PA-28 in the video below shows.  A fundamental rule of flying airplanes is never force an airplane to land if it does not want to.  Go around and try again.

A graphic example of why floating is risky on short field landings comes from this one of a Piper Aztec attempting to land at Gustaf III on St Bart's.  Pilot was too high and 10-15 kts too fast.  The float used up critical runway the pilot needed for braking and by the end was unable to stop with the runway available.
Comment: @gsamaras - I suspect that the pilot just wanted to get the plane on the ground. And y'know what? HE DID!
Comment: @gsamaras Yes, they should have. This is a perfect example of most of the [Classic Five Hazardous Attitudes](http://aviationknowledge.wikidot.com/sop:hazardous-attitudes).
Comment: Couldn't the Piper just go around (after seeing that it floated way too much)? Or maybe it didn't had enough power to?
Comment: Yes the PA-28 should have made a go around
Comment: Maybe it should be a question on its own, but at the St Barts airport (2nd video), why don't planes approach from the other direction. Looks safer to me (easier to use the whole runway), but what do I know?
Comment: The can fly the Rwy 28 approach but the winds were favoring Rwy 10 that day.
Answer: Depends.
With tricycle gear airplanes the limitation is mainly pitch attitude in the landing flare so you don't come down nosewheel first, which is very bad on the nosewheel and ultimately on the airplane.  So you need to be slow enough that the mains will touch first with landing flap.  But beyond that it's just wear and tear on the tires and brakes.
With taildraggers it's a different story.  If you do wheel landings (tail high touchdown), which is my preference, you can, in theory, fly the thing on to the runway at any speed you want within reason, limited only by what the tires can take and possibly propeller clearance if going too fast.  In fact a normal wheel landing is done quite a bit faster than a 3 point, touching down in a relatively level attitude maybe 10 mph faster than a full stall 3 point landing.  You actually fly the airplane down to about 1 or 2 feet above the surface and push the stick to plant the mains when doing a wheelie.
Answer: I don't think there's anything inherently dangerous about landing a light aircraft faster than the prescribed speed, however you're unlikely to be very successful. The most likely outcome is just a continuous float down the runway until the airspeed drops off anyway.
Landing an aircraft is more or less stalling the wings in a controlled manner, very close to the ground. This causes the wings to no longer be taking the weight of the aircraft, and gives you the ability to use wheel brakes.
Answer: Someone told me once:
When landing a plane the stall horn should sound just as you touch the ground, and this is basically how you land: Gradually reducing lift from your wings to the point where the plane's weight fully rests on it's wheels. If you go too fast there might be too much lift to touch down!
Landing at a low descent rate won't damage the airplane at all, but the higher ground speed required to do so might, as others have pointed out.
While you can attempt to land at high speeds, the lift still produced by the wings can make it hard to stay on the ground and keep the plane intact at the same time. Gliders, which are obviously very light, use airbrakes to reduce lift/increase drag drastically when descending and landing.

ASK 21 glider with deployed speedbrakes. Source: Wikimedia Commons
For planes without any airbrakes a fast approach most likely results in a long float above ground due to the lift, or a long roll on the ground, for which you may not have the room on short airfields, as mentioned by others already.

I have a real life example of a high-speed approach incident
(Recollection from memory - might not be 100% accurate)
A couple years ago a pilot flying our club-owned ultralight arrived back at the airfield. They wanted to get it down quickly, my guess is because of gusting winds, but unfortunately their airspeed was way too high for landing. I do not know whether this was due to the winds or pilot error.
Instead of going around, or using the generous runway length to slow down further, they wanted to touch down early. The plane was still floating above ground because of it's high speed, so they pushed the nose down in an attempt to land it, which is almost instinctive, but also very wrong to do. The plane bounced off the nose gear two or three times before it collapsed. Prop and nose hit the ground, slowing the plane down rather quickly after that.

Aftermath of the described incident, picture mine
Nobody was hurt in the crash, but the plane was trashed. Trainees at the airfield learned a valuable lesson that day: Why you should land at appropriate speeds.
Answer: If a tricycle-geared airplane attempts to land with too much speed, the nose wheel touches first, which has all sorts of undesirable dynamic consequences.
The position of the nose wheel gives it plenty of leverage, and if it touches with any force (or the ground has any roughness), it will pitch the airplane nose-up, and the airplane has sufficient extra speed to balloon as the angle of attack increases. Once it leaves the ground, however, the aerodynamic controls reassert themselves, and the pilot is holding the elevator in the position that led to the high-speed touchdown in the first place. Unless he does the right thing at this point, which is to hold off until the speed has decreased (or go around), the nose wheel will come down again, probably harder than before (especially if the pilot tries to force the airplane down), and the cycle will repeat.
Furthermore, if the rotation of the airplane causes the main wheels to touch, this will immediately bring the nose wheel down again, hard. This can be seen in the first of Carlo Felicone's videos, where this starts happening after the second or third bounce, and after about six more increasingly violent and rapid bounces, the nose wheel collapses.
Your problems are not over if you manage to get through this phase. Now you are "wheelbarrowing", with most, or all, of the contact force being borne by the nose wheel, and any bump on the ground before you have slowed down to a correct landing speed can start the above cycle of bouncing. Furthermore, with the point of contact being well ahead of the center of gravity, you are unstable in yaw, and the slightest swing, uncorrected side-wind or drift could start a groundloop. It is basically the same reason as why taildraggers are more at risk of groundlooping, but with a longer moment arm.
Answer: I like to view landing as a critical transition period between two distinct regimes of vehicle control. The way that you handle an aircraft is incredibly different in these two regimes and many landing accidents have failing to completely transition from the flight phase to the ground-handling phase as a primary factor.
The whole idea behind the flare is to have the wheels touch the ground at nearly the same instant that the wings stall when the aircraft is in the proper landing attitude at the correct speed. And though landing incorrectly will often put you in an attitude that is detrimental to the aircraft, I would argue that its design and gear location are driven by this purpose rather than the landing attitude being driven by the aircraft/gear design.
One operation where this is quite visibly apparent is in landings with very high crosswinds. Since the method for handling an aircraft in a crosswind are particularly different in the two regimes, the maneuvers to transition between the regimes at landing are necessarily very pronounced. Sometimes you will even feel like you have to force the aircraft into the transition in extreme conditions.
If you have experience riding motorcycles, I find it quite similar to the difference between low-speed handling where you simply turn the direction you want to go, and where because of wheel's gyroscopic procession you actually counter-steer at high speed.


The Action of the inverse of a shifted matrix on a vector

Question: This question concerns a matrix $A\in\mathbb{C}^{n\times n}$, whose inverse we know the action of on a particular vector $v\in\mathbb{C}^{n}$. If we know that $A^{-1}v = u$, is there any way to express $(A+\alpha I)^{-1}v$?  We have that $\alpha \in \mathbb{C}$ but it is not bounded in magnitude and $A$ is any matrix.
Answer: The only case in which you have enough information is if $u$ is a scalar multiple of $v$.  If $u = t v$ with $t \ne 0$, i.e. $A v = v/t$, then $(A + \alpha I) v = (1/t + \alpha) v$ so
either  $(A + \alpha I)^{-1}$ doesn't exist or $(A + \alpha I)^{-1} v = (1/t + \alpha)^{-1} v$.  If $u$ is not a scalar multiple of $v$, you don't have enough information to determine $(A + \alpha I)^{-1} v$.
Comment: Also known as "eigenvector"?
Answer: By adding a constant identity to $A$, you increase all of its eigenvalues by $\alpha$.
I rewrite your transformation as
$$(A+\alpha I)^{-1}v = X(\Lambda+aI)^{-1}X^{-1}v,$$
where $X$ is a set of eigenvectors for $A$ and $\Lambda$ is the eigevalues diagonal.
We have that $u = X\Lambda^{-1}X^{-1}v$, or
$$X\Lambda X^{-1}u = v.$$
We can obtain
$$(A+\alpha I)^{-1}v=X(\Lambda+aI)^{-1}\Lambda X^{-1}u=X(I-a(\Lambda+aI)^{-1}) X^{-1}u=u-aX(\Lambda+aI)^{-1}X^{-1}u$$
due to $\frac{\lambda_i}{\lambda_i+\alpha}=1-\frac{\alpha}{\lambda_i+\alpha}$.


How to check if a hotfix is needed

Question: I've got a customer who's said they need a hotfix (https://kb.sitecore.net/articles/574334). They're on 8.2 Update 3, but the notes for what download to get only goes through 8.1. So I don't know if the fix was integrated somewhere down the line. Is there any way to check beyond reading the release notes for each and every version from 8.1 initial through 8.2 Update 3?
Comment: Not exactly the same but similar question here: https://sitecore.stackexchange.com/questions/4269/how-can-i-find-out-if-a-support-dll-is-still-needed-when-upgrading
Comment: Ask Sitecore support...
Comment: I've put a request into them as well, but I didn't know if someone here knew of a reference site I wasn't aware of.
Comment: What makes the client think they need that specific hotfix?
Answer: You could read all the release notes, and it might be in there. But if you really want to be sure, the best thing to do is ask Sitecore Support. I would assume they have a system that can check these things better than going through release notes ;)
Of course, while waiting for an answer you could still read the notes but that might turn out as a waste of time...
As answered here before (How can I find out if a support dll is still needed when upgrading?), after upgrading a solution with a bunch of support dlls this "solution" is also the best approach.
I realise this might not be a good solution, but as long as we don't have the tools to easily search for this information and Support probably/hopefully does, we need to bother them with these questions...
Answer: What I have done is to open the fix dll's with Reflector, and noticed that it's fixes code from ParseGenericSearchEngine from Sitecore.Analytics.
Then I've compared code for ParseGenericSearchEngine from Sitecore.Analytics, from Sitecore 7.5 and Sitecore 8.2 update 3, and I've discovered that Sitecore 8.2 update 3 has the same code as in Sitecore 7.5 for ParseGenericSearchEngine.
So, I could say that the fix its needed, but as @Gatogordo mentioned before, Sitecore Support should be the one to give you a definitive clear answer.

How to check if a hotfix is needed


Answering old questions

Question: Should I answer old questions (over a year), even if they have been already answered? I think that this practice is correct, but since in many forums it is discouraged I'm not sure.
Comment: The _primary_ goal of SE is not to help the OP (though of course we want to). It is to create Q&A pages  that others will find useful. Answer away!
Comment: @Dilaton: It's the other way around: A useful Q&A page helps a lot of people learn and understand -- not just the 10 people who see it when it comes up on the main page, we're talking about the hundreds who come from Google. It doesn't matter if the OP isn't around to benefit, that's just one less person from many who can learn from the post.
Comment: Yep, I once had a related [question](http://meta.physics.stackexchange.com/q/4082/2751).
Comment: @Manishearth I thought Physics SE is for people to help each other learning, doing, and understanding physics, and a useful Q&A page is what comes out of it as a result ...? This is how I see it :-). Cheers
Comment: @Dilaton: Huh? Whatever you just said is irrelevant to this discussion. Yes, SE was created for this purpose, to help a wider audience than just the OP. They noticed that posts on most traditional forums would be quite localized and would help _one_ person -- the OP. So they built a network where the posts are meant to help many, many others. That's why we have a "too localized" close reason. Sure, we focus on helping the OP as well, but while writing an answer you must keep in mind that you want the maximum number of _other_ visitors to benefit from it. That's all.
Comment: @Manishearth I think this is the SE owner/stuff point of view. Nevertheless, it is important to have a nice coherrent physics community (including knowledgeable long term users) who likes to learn and do physics together to establish a nice helpful atmosphere and make people stay. Otherwise, there will be just transient users coming in but disappearing soon. With just transient users it can still work if the site is just meant for laypeople to find answers to very basic and everyday life questions, but not if the site is meant to feature higher level Q&A, for which experts are needed, too.
Comment: The layperson/non-expert issue is irrelevant here.
Comment: Unless you aren't duplicating other answers (with or without knowledge), it's always good to add something useful (what makes your post unique). I'd point Ben Crowell. He's an active member answering *both old & new* questions. Of course, he adds interesting things to neighbor posts and get votes ;-)
Answer: Stack Exchange sites are not forums, and many of the rules which are conventional among forums don't apply here. The rule against answering old questions is one of them. So yes, you can and should answer old questions when you feel you have something to contribute, just as you would do with new questions. (Even if a question already has an accepted answer, don't let that stop you from posting another answer of your own.)
In fact, one of the guiding principles behind the design of the Stack Exchange software is to allow old questions to receive new answers.
Answer: I think we should answer old questions because most of the time your new answer will definitely add more information to the existing answers and the new answer will give a different way of explaining which also is useful.

Answering old questions


Scrum PO (product owner) requesting status everyday! Help!

Question: We do daily scrum in which the PO does not attend. But still later PO comes and ask the status and whole team responds. So we are sort of having double daily scrums
Comment: Is your board up to date and clear, can you form an impression of what's going on from looking at it?
Comment: Can you move the daily scrum to a moment where the PO can attend? Can the Scrum master intercept the PO, explain the importance of focus and walk through the board together, only asking individual members for additional clarification if needed?
Answer: John MP already has a good answer (ask the Product Owner (PO) what s/he actually needs, and then discuss options to satisfy these). However, there are two thing I'd like to add:
The Daily Scrum is not a Status Meeting
It's concerning that a whole-team status update meeting is being considered redundant with the Daily Scrum. Likewise the suggestion that the PO could just attend the Daily Scrum to get such an update. The purpose of the Daily Scrum is for the Development Team to collaborate; to keep them on the same page and help each other out (or at least, discover opportunities to do so, after the meeting). The Development Team should not be reporting to anyone (not the PO, not the Scrum Master, not a senior dev, etc.), and turning it into a status meeting is likely to violate this.
The Purpose of the Scrum Board is to provide a Status Overview
The Scrum Board should be the ideal tool to solve this. Its purpose is to provide an overview of the status of the Sprint, both to the whole Scrum Team (including the PO) and possibly even directly to external stakeholders. I see three possibilities: either the Scrum Board is not being maintained correctly, the PO is not educated in its use, or the PO has specialized needs that cannot be completed via looking at a board. The first two are problems that need to be addressed immediately. For the third, as previously suggested, you'll need to work with the PO - there has to be a better way to fulfill the PO's needs than a daily all-hands meeting.
Answer: Sounds like the PO herself is creating an impediment for the team by way of inefficient meetings (i.e. unnecessarily duplicating efforts). I think it calls for a discussion where the SM, in a one on one with the PO, can explain the inefficiency and suggest options. One option being moving the meeting to a time where the PO can attend.
Another option is to understand what the PO is truly looking for. As one of the roles of the PO is to work with the business stakeholders, it's possible that she is keeping an eye out for high-level updates that impact the business. If so, the SM should be more than capable to email anything of relevance that falls into the area after the meeting (assuming the PO cannot attend). For instance, a blocker that's specifically connected to a business aspect.
But coming back to my first point: I'd like to think that almost everyone is professional and open to communication. Just discuss the issue with the PO and find the best option that suits her needs without jeopardizing the team's time.
Answer: The Scrum Guide
"The Scrum Master enforces the rule that only Development Team members participate in the Daily Scrum." This is because "The Daily Scrum is a 15-minute time-boxed event for the Development Team to synchronize activities and create a plan for the next 24 hours."
"Significant aspects of the process must be visible to those responsible for the outcome. Transparency requires those aspects be defined by a common standard so observers share a common understanding of what is being seen."  In this case the Sprint Backlog artifact is the aspect of the process in question.
One thing not desired is an absent Product Owner.  It is good for the person filling that role to be asking questions and providing input.  It is bad for that person to be seeking status updates.
This is a good topic for the Sprint Retrospective event.  Discuss with the entire Scrum Team about the problem (not the person) and discuss options for improvement.  Scrum requires the team to inspect and adapt to provide value.


Where did that recursion question go?

Question: That recursion question «Is recursion not something programmers know?» (10k link, img) suddenly disappeared. Why and what happened to it?!
Comment: obligatory reference to [Where did that recursion question go?](http://meta.programmers.stackexchange.com/questions/7469/where-did-that-recursion-question-go#comment22387_7469)
Answer: GlenH7, Amon, and I were the three delete votes on it.  A closed question that is at -3 can be deleted by users that have 20k rep without having to wait for two days.
This question was attracting significant discussion in the comments.  The question post had 13 comments, the posts had 10, 14, and 4 comments on them (and showed no sign of stopping or slowing down).
The question had a number of reopen votes on it, which would have likely made it (and I will point out that it would likely get closed again - neither Glen, Amon nor I had cast close votes on it in the first round).  The question itself was (I believe) a rant about the lack of experience of php coders, an opinion poll about what to do when interviewing a candidate who can't do recursion for a Fibonacci series.  The answers were veering off into critiquing the interview question and the applicability of recursion in today's computing environment.
The combination of these things (the discussion in comments, the off topic nature of the question, that the question was likely to get reopened and reclosed (with probably a few more answers jumping into the bandwagon)) strongly suggests that it is going to become even more contentious with time. At best (or worst depending on opinion), it would be one of those ugly historical locks (historical significance on a 1 day old question?).
Discussions are indeed fun and interesting - but they are a poor fit for the Stack Exchange Q&A format. I would love to banter about this in chat, or see what anecdotes occur on Quora or Reddit for it... but it is very tough to get it to fit in the Q&A format as asked and written.
Deleting the question gives people who have 10k reputation (and the OP) the opportunity to try to edit it into a question that should be reopened without the drama associated with bumping the question and repeated "leave closed" reviews.  In its current form, however, it should remain closed and ultimately be deleted as it isn't a good fit for the Q&A format. That it was done sooner than later was to try to reduce the poor experiences for the people who chance to answer it in the reopen/reclose cycle and reduce any associated drama.
Comment: @BenAaronson [the second answer the question got](http://i.stack.imgur.com/hBGDk.png). The answer is tangent to the question being asked - and there are *lots* of tangents to the question (including critiquing the question itself). The reason these are showing up as answers (and getting up votes) is that question itself doesn't have a possible concrete answer - every answer to it *must* be an opinion and everyone has one (and then people vote according to if they *agree* with the statement rather than if it is a good answer that solves the problem and we get a bike shed).
Comment: It's also "Unclear what you're asking," since the OP never specified what they meant by, "No one has been able to answer it." It's not clear whether that means no one gave *any* answer (sounds unlikely) or if no one gave exactly the desired answer. For all we know, he rejected answers that used an `$n < 1` check instead of `$n < 2`. (Unless that's clarified in comments, but it should've been edited into the question if so.) That's a some hay on top of a mountain, of course, but it is one more reason for it to go.

Where did that recursion question go?


(When) do people use Nydus worm?

Question: I'm not a Zerg player, but I've never seen Nydus worm used against me either. When do people actually use it?
Comment: This [4-part commentary](http://www.youtube.com/watch?v=FVbXyZpsUYA) by HDStarcraft showcases good use of the Nydus network.
Comment: I've been playing SC2 since the day it was released, am a gold-level player, and I still don't know what the Nydus worm does.  (Just transports units?  Can it be built anywhere?  Does it require a drone?)
Comment: Blueraja: http://starcraft.wikia.com/wiki/Nydus_network Basically you can build a Nydus Network once you have a Lair. The Nydus Network building gets an ability to spawn a Nydus worm anywhere you have vision. The Network and all Worms then act as one giant transport. Ground units can enter and exit the system from any Network or Worm, thus allowing near-instant transport wherever you can place a Worm.
Answer: Nydus worm can be very effective in many different use cases.
Combat: All you need for a nydus worm deployment is line of sight which you can easily achieve with an overlord/overseer.  This can be an effective way to get through a walled in opponent without actually taking out their wall.  Nydus worms also generate creep so you could even setup spine crawlers if you so wish.
Another way to use it in combat is to distract your opponent. Because the nydus worm can be heard anywhere, they usually go into panic mode and will send their army back to their base or will be otherwise distracted in scouting for it at which point you can attack with your army without actually using the worm.
Its also important to note that when a nydus worm node is destroyed the units inside are unharmed unlike a warp prism, medivac, overlord.  Unless you take out the whole network.
Economy:
This is a bit tricky and requires you to be on top of things micro wise.  But you can use a nydus worm to mine out an undefended high yield without actually building a hatchery there.
Setup the nydus worm in your base right next to your hatchery and set the rally point to your hatchery then shift click back to the node.  Then where you've opened the nydus worm at the high yield, set its rally point to the minerals then back to the worm.
Hot key your nydus worm's mineral node to a key and the base node to a key.  Load it up with a bunch of drones then every 20 seconds (i forget the exact timing) select the opposing nydus node hotkey and press 'D'. this will unload your drones, and depending on which side they unloaded they will either start mining or they will return the goods back to your hatchery.
Obviously this use is very complex and requires you to remember to unload and at which node to unload at (have the drones just mined or just returned gathered minerals?).  You also lose the added benefit of having more larva and another hatchery.
Comment: You'll know when you hear it.  Sounds like a giant worm screaming as it busts through solid ground.
Comment: minor quibble: "line of sight" not "line of site"
Comment: what kind of sound does it make?
Answer: 1v1 is risky at this point in time because of the time it takes to build and the noise it makes.  It is also easily killed.
However, in team games, a nydus can be very effective which is where I personally will use it.  It is an agreed upon strat with my teammates, usually someone's observer provides the line of sight.  I've used it to get a quick mineral line attack when the opponents push out, and I've also used it to provide a smaller secondary attack to draw attention away from the main forces of my allies while they attack.
Answer: The main uses have been posted, but another use is simply as an intra-base tram system. On big maps, it can be very handy to shuttle armies around very quickly using the Nydus network. You can even use it to ship your attacking army up to the frontlines, and send in reinforcements quickly like how Protoss use Warp Gates.
Another use is expansion defense. If you are a good little Zerg and have lots of expansions, you can save resources on defenders by simply adding a Nydus worm exit or two to each expansion. Then you can keep a few dozen Hydralisks in the system, and pop them out if they are needed. It's not as good as static defenses since the units exit one at a time, but it can help protect isolated expansions.
Comment: You could even use it to *create* expansions. Want to get to an island but no time for the overlord transport research? Nydus worms to the rescue!
Comment: Very true. They are affordable, can pop anywhere, and generate creep. And apparently they burrow through the space-time continuum since as you mentioned they can pop up on islands in space...
Answer: It's a way to get across the front defenses of someone, or used as a surprise attack (like to hit the mineral line of the main base or something). It also gets units across the board very quickly.
One reason why you might not have it hit you very often, is maybe you properly defend your borders, so an overlord can't get in your base anywhere to drop the worm.
Answer: As Mark already said, Nydus worms can be used to get around the enemy frontline. A tactic i use sometimes as Zerg is to send an overlord into the back of his base (if possible a little out of sight) and spawn the nydus worm there. Then i send all units to attack his base. As soon as there is too much resistance i simply retreat into the nydus worm and spawn it somewhere else, i.e. at one of his expansions.
You can repeat this until the enemy is either dead or attacks your base out of desperation. Just remember to retreat before you lose too much and if the enemy can't muster enough defenses, just send more reinforcements through the nydus worm and finish his base off.
This works best with hydralisks as often the first resistance you encouter will be his air units, as they travel a straight line into his base while his ground units have to go around through the choke.
The advantage of nydus worms over overlord drops is:

Nydus worms can't be shot down by air defense
You can send reinforcements almost instantly
You can retreat easily without having to worry about your overlords getting shot
Answer: Nydus + Overlord/Overseer dropping creep is a very effective combo for a common backdoor strategy. Most Zerg players will look for an exposed part of an enemy's base and then simply transport units to the enemy base. If used strategically, spawning the nydus and then removing the creep from the overlord will minimize visibility from the enemy (Zerg buildings don't generate much [if any] creep, thus this could be unnoticed from an amateur player.

(When) do people use Nydus worm?


Make sourdough starter without discarding

Question: I understand the point of discarding half the sourdough starter as the yeast develops so that you don't have exponential amounts of starter. Why not start with a half ounce of flour and just add the same weight as what's in the container? (I plan to try this.) On the fifth day, feeding once a day, there would be 8 ounces of flour.
Comment: There are many uses for superfluous starter: Pancakes, crumpets, whatever. It's half flour and half water, more or less, so just substitute it into your normal recipes.
Answer: While I have successfully begun a starter with only 10 grams of flour at the outset, I think I agree with Stephie that doing a little more is more efficient.  I used to do very wasteful starters until I happened upon this site back in 2002 or something.  (Amazingly, it's still there.)  You can read the details there, but basically the guy who did it designed his initial starter regimen based on the amount of flour he'd eventually use to bake bread (i.e., with no waste).  (It also has a way to get to a lower pH earlier which helps the starter along while inhibiting bad stuff; this was years before the internet discovered the "pineapple juice solution.")
But I think this depends a LOT more on your feeding regimen, especially in the early days, as well as the flour.  I've written a lot about the variables in my other answers here on sourdough related questions, so I'm not going to repeat it all here.
Briefly, though, YES, it is possible.  It can be just as reliable as using more flour and a lot less wasteful.  Even if you don't do a true exponential growth model for your starter, you can also be significantly less wasteful by keeping the starter relatively small until it's established.  There's no reason to be feeding and throwing away a cup or more of flour every time for every feeding, particularly if the starter isn't even growing yet.
Also, the same applies to maintaining your starter.  If you want to do detailed calculations, I've explained them here.  When I first started with sourdough, I used to produce way too much of it and throw it away.  Now I generally save just a tablespoon or two, feed it a little, and put it away for the next batch and build up from there when necessary.
Answer: I won't say your idea doesn't work, but there are a few points to consider:
A "ripe" sourdough contains a balanced culture of yeast and bacteria (lactobacillus) that can keep fungi and other unwelcome bacteria under control.
Let me use the political landscape of today's US as an example:
You have two big parties in slightly varying majorities - like in a sourdough that may be a bit more yeasty or a bit more acidic depending on the circumstances. But together that system keeps extreme political groups that may topple over the whole system from rising to power - like in your sourdough.
Now if you take a very small sample and especially if your two parties (yeast and bacteria) aren't fully established yet, even a comparatively small number of extremists (mold, other bacteria) can endanger the whole system.
So if you want to start a sourdough from a tiny amount of flour, you need to work very carefully and cleanly to avoid contamination and still might have too few yeast spores etc. in your initial amount of flour to get to that stable system going. Technically, it is possible, though.
Larger amounts of the initial flour/water mix increase the probability of successfully getting a sourdough going - most sources I know suggest around 50g / 2oz of flour as minimal initial amount for home bakers. While I respect your frugality in not wanting to discard anything, a few spoonfuls of flour are pretty cheap and if you have to start over a few times, you'll gain nothing.
Comment: Great analogy. Makes sense. I'll try it and avoid contamination. Figured out an easy use of starter discards: rolled out, trimmed, and toasted flatbread cooked in a toaster.
Answer: In the beginning of a starter's life, the point is to develop yeast and friendly bacteria, NOT develop lots of starter. If you don't discard, you end up with way too much, and it is not wasteful because tossing about half keeps everything at a manageable amount. You don't want too little either because as was mentioned, there needs to be enough there to keep it fresh and good. The pioneers coming access America in covered wagons had no refrigerators yet their starters did good because they were populated with wenough yeast and bacteria to stay that way. Everyone suggests about a cup of starter, a cup of fur and either 1/2 cup or 2/3 cup water, and this has worked well for me. I know my answer may have seemed a bit contradictive, but too little or too much should be avoided.
Answer: Starter becomes acidic over time.  You need to discard and replenish to keep acid under control.  Too much acid inhibits yeast growth...bad for your loaves!
Comment: Maybe baking soda can help. So much experimenting to do!
Comment: @motorbaby baking soda is unnecessary, and certainly not traditional for sourdough starter.
Answer: You need the flour to feed the bacteria. The amounts given in the starter-process mean that the colony of bacteria will be happily fed and able to grow until the discarding and filling up.
Your starting-amount will simply not keep the colony happy and fed for the required time. Also, with that small a mass, chances are the starter would dry out until the next "feeding".
Comment: Sorry, no. In a new / beginning starter there are only few yeasts and bacteria present, starvation is not a problem.
Comment: Then mine must have died of other causes... poor thing! ^^
Comment: happens, sadly so :-(
Comment: My home is very dry. If I put fabric over the starter container, the top of the starter dries out (and I don't use synthetic-estrogen-containing plastic wrap). I use a mason jar and put a lid on top with a loose ring; that keeps it moist while allowing gases to escape.
Answer: It's not clear from the question whether the OP asks about the starter culture or the mother starter. The answers so far seem to pertain to the starter culture; being small and done only once, the quantity of discard from this process is not very meaningful. Discard from repeated uses of the mother starter - the starter used to raise and flavor dough for baking - is done frequently with larger amounts, so it is the more meaningful question.
I bake bread at home for one person, so the amount of mother starter I use each time does not vary greatly. By "bread" I mean loaves, muffins, pizza, rolls, etc. Knowing what I'm going to bake next, I replace the amount of mother starter just removed to make a dough with a slightly larger quantity of flour-plus-water than I will need for my next recipe, to account for starter "lost" to cleanup, etc. This process ensures that I am using about 50g - 100g of *mother starter" each time, leading to repeatable re-growth of the mother starter.
The SCOBY in a sourdough starter is much more forgiving that some would have you think. Remember: Nature is a mother - all of her creatures want to live and are likely to find a way to establish themselves in your starter as long as you follow basic rules of cleanliness and don't expose them to chemical challengers.
Comment: Please clarify the difference between your mention of starter vs mother. Should be inherently clear that the purpose of the question is to "make" starter, not maintain it from something that has already been created. Thanks.


What is the motivation for a vertex algebra?

Question: The mathematical definition of a vertex algebra can be found here:
http://en.wikipedia.org/wiki/Vertex_operator_algebra
Historically, this object arose as an axiomatization of "vertex operators" in "conformal field theory" from physics; I don't know what these phrases mean.
To date, I haven't been able to gather together any kind of intuition for a vertex algebra, or even a precise justification as to why anyone should care about them a priori (i.e. not "they come from physics" nor "you can prove moonshine with them").
As far as I am aware, theoretical physics is about finding mathematical models to explain observed physical phenomena. My questions therefore are:
What is the basic physical phenomenon/problem/question that vertex operators model?
What is the subsequent story about vertex operators and conformal field theory, and how can we see that this leads naturally to the axioms of a vertex algebra?
Are there accessible physical examples ("consider two particles colliding in an infinite vacuum...", etc.) that illustrate the key ideas?
Also, are there alternative, purely mathematical interpretations of vertex algebras which make them easier to think about intuitively?
Perhaps people who played a role in their discovery could say a bit about the thinking process that led them to define these objects?
Comment: The main problem for mathematicians who want to learn about QFT is that there is no universal definition of what a QFT is. I think Wightman axioms are a big help to start. The first chapter of Kac's "Vertex algebras for beginners" shows how the notion of vertex algebra arises naturally from Wightman axioms.
Comment: But is there a "fundamental example" that captures what a QFT is supposed is to do (or be)? I would be happy just to hear a story about a particular physical system, in a way that illustrates the important features of QFT.
Comment: You ask what physical problem vertex operators model but you actually give the answer yourself! :-) They can be used to answer questions about "two particles colliding in an infinite vacuum". A pair of strings coming from infinity, interacting "once", and going off to infinity, say, sweep out a surface that is topologically a sphere with 4 tubes sticking out out of it. String theory is (sort of) conformally invariant and this surface is conformally a Riemann sphere with 4 punctures in it. Vertex operators arise when studying quantum fields on Riemann spheres in the vicinity of these punctures.
Comment: Addendum to Qiaochu's comment:

Also see http://ncatlab.org/nlab/show/conformal+field+theory and its reference Martin Schottenloher: "A mathematical introduction to conformal field theory", for a pre-vertex-algebra introduction to conformal field theory.
Comment: Have you seen http://ncatlab.org/nlab/show/vertex+operator+algebra ?
Answer: Vertex algebras precisely model the structure of "holomorphic one-dimensional algebra" -- in other words, the algebraic structure that you get if you try to formalize the idea of operators (elements of your algebra) living at points of a Riemann surface, and get multiplied when you collide.
Our geometric understanding of how to formalize this idea has I think improved dramatically over the years with crucial steps being given by the point of view of "factorization algebras" by Beilinson and Drinfeld, which is explained (among other places :-)  ) in the last chapter of my book with Edward Frenkel, "Vertex algebras and algebraic curves" (second edition only). This formalism gives a great way to understand the algebraic structure of local operators in general quantum field theories -- as is seen in the recent work of Kevin Costello -- or in topological field theory, where it appears eg in the work of Jacob Lurie (in particular the notion of "topological chiral homology").
In fact I now think the best way to understand a vertex algebra is to first really understand its topological analog, the structure of local operators in 2d topological field theory.
If you check out any article about topological field theory it will explain that in a 2d TFT, we assign a vector space to the circle, it obtains a multiplication given by the pair of pants, and this multiplication is commutative and associative (and in fact a commutative Frobenius algebra, but I'll ignore that aspect here). It's very helpful to picture the pair of pants not traditionally but as a big disc with two small discs cut out -- that way you can see the commutativity easily, and also that if you think of those discs as small (after all everything is topologically invariant) you realize you're really describing operators labeled by points (local operators in physics, which we insert around a point) and the  multiplication is given by their collision (ie zoom out the picture, the two small discs blend and look like one disc, so you've started with two operators and gotten a third).
Now you say, come on, commutative algebras are SO much simpler than vertex algebras, how is this a useful toy model?
well  think about where else you've seen the same picture -- more precisely let's change the discs to squares. Then you realize this is precisely the picture given in any topology book as to why $\pi_2$ of a space is commutative (move the squares around each other). So you get a great model for a 2d TFT by thinking about some pointed topological space X.. to every disc I'll assign maps from that disc to X which send the boundary to the basepoint (ie the double based loops of X), and multiplication is composition of loops -- i.e. $\Omega^2 X$ has a multiplication which is homotopy commutative (on homotopy groups you get the abelian group structure of $\pi_2$). In homotopy theory this algebraic structure on two-fold loops is called an $E_2$ structure.
My claim is thinking about $E_2$ algebras is a wonderful toy model for vertex algebras that captures all the key structures. If we think of just the mildest generalization of our TFT story, and assign a GRADED vector space to the circle, and keep track of homotopies (ie think of passing from $\Omega^2 X$ to its chains) we find not just a commutative multiplication of degree zero, but a Lie bracket of degree one, coming from $H^1$ of the space of pairs of discs inside a bigger disc (ie from taking a "residue" as one operator circles another). This is in fact what's called a Gerstenhaber algebra (aka $E_2$ graded vector space). Now all of a sudden you see why people say you can think of vertex algebras as analogs of either commutative or Lie algebra (they have a "Jacobi identity") -- -the same structure is there already in topological field theory, where we require everything in sight to depend only on the topology of our surfaces, not the more subtle conformal geometry.
Anyway this is getting long - to summarize, a vertex algebra is the holomorphic refinement of an $E_2$ algebra, aka a "vector space with the algebraic structure inherent in a double loop space", where we allow holomorphic (rather than locally constant or up-to-homotopy) dependence on coordinates.
AND we get perhaps the most important example of a vertex algebra--- take $X$ in the above story to be $BG$, the classifying space of a group $G$. Then $\Omega^2 X=\Omega G$ is the "affine Grassmannian" for $G$, which we now realize "is" a vertex algebra.. by linearizing this space (taking delta functions supported at the identity) we recover the Kac-Moody vertex algebra (as is explained again in my book with Frenkel).
Comment: For a nice exposition of the topological 2d picture there is Kock's book: http://www.mat.uab.es/~kock/TQFT.html
Answer: The original motivation for vertex algebras is explained briefly in the original paper http://www.jstor.org/stable/27441 as follows. For any even lattice one can construct a space $V$ acted on by vertex operators corresponding to lattice vectors. More generally one can write down a vertex operator for every element of $V$. These vertex operators satisfy some complicated relations, which are then used as the definition of a vertex algebra. In other words, the original example of a vertex algebra was the vertex algebra of an even lattice, and the definition of a vertex algebra was an axiomatization of this example.
This was motivated by my attempt to understand Igor Frenkel's work on the Lie algebra with Dynkin diagram the Leech lattice.  Frenkel had constructed a representation of this Lie algebra using vertex operators acting on the space $V$, and I was trying to use his work to understand its root multiplicities. I did not use any insights from conformal/quantum/topological field theory or operator product expansions when defining vertex algebras (as implied by some of the other answers), for the simple reason that I had barely heard of these concepts and had almost no idea what they were.
This is not all that helpful for understanding what a vertex algebra really is. A better view is to regard them as something like a commutative ring with an action of the (formal) 1-dimensional additive group. In particular any such ring is canonically a vertex algebra. The difference between rings with group actions and vertex algebras is that the "multiplication" of a vertex algebra from $V\times V$ to $V$ is not defined everywhere: it can be thought of as a "rational map" rather than a "regular map" in some sense. More precisely if we write $u^g$ for the action of a group element $g$ on $u$, then the product $u^gv^h$ is not defined for all group elements $g$ and $h$, but behaves formally like a meromorphic function of $g$ and $h$, which in particular may have poles when $g$ and $h$ are trivial. Making sense of this informal idea produces the definition of a vertex algebra. (For more details see the unpublished paper http://math.berkeley.edu/~reb/papers/bonn/bonn.pdf.) This means that vertex algebras behave like commutative rings: for example, one should define modules over them, tensor products of modules, and so on.
Answer: The answers here have focused on the mathematical aspects of VOAs and the motivation coming from QFT, the specialization to Conformal Field Theory, and then the further specialization to two-dimensional holomorphic CFT. Two-dimensional CFT's did arise from string theory where the fields on the 2d world-sheet of the string define a CFT. However there is another important part of the story which has not been mentioned and is more directly tied to physical phenomenon and that is the theory of critical phenomenon. Many systems, such as water-ice, magnetic systems and so on undergo phase transitions as a thermodynamic parameter such as temperature is varied. Typically these are first order transitions, meaning that there is a latent heat associated to the transition. Sometimes one can vary an additional parameter and find a line of first order transitions which terminates at a second order transition. The second order transition is characterized by fluctuations on all scales: the theory becomes scale and conformally invariant at that point. It also turns out that the behavior of thermodynamic quantities as one approaches the critical point are characterized by numbers called critical exponents which are universal for systems with the same symmetry structure. These exponents are related to what are called the conformal dimensions of operators in CFT and they are directly measurable in the lab for a variety of systems. One important tool which was used in the study of critical phenomenon is the operator product expansion or operator algebra of K. Wilson and L. Kadanoff. There is a huge literature on this. Here is a reference to an early paper on the operator algebra for the Ising model: http://prb.aps.org/abstract/PRB/v3/i11/p3918_1 . VOA's are a rigorous mathematical formalization of this kind of algebraic structure. For someone who wants to learn about CFT starting from a particular physical system (or at least a mathematical idealization of a physical system) the Ising Model is a good place to start.
Answer: Speaking as a string theorist, I would say Victor Kac's book "Vertex Algebras for Beginners" followed fairly closely what physicists originally thought. So you might want to have a look at it.
Answer: Rex asked:
"*As far as I am aware, theoretical physics is about finding mathematical models to explain observed physical phenomena. My questions therefore are:
What is the basic physical phenomenon/problem/question that vertex operators model?*"
Please have a look at the nLab at the pages that have already been mentioned:
ncatlab.org/nlab/show/vertex+operator+algebra and
ncatlab.org/nlab/show/conformal+field+theory
There is also a lot of material about QFT there.
Here is a quick and dirty and oversimplified introduction to some aspects of QFT:
In the Wightman framework we can think of quantum fields as operator valued distributions, that map test functions living on a given spacetime to (maybe unbounded) operators. Essentially selfadjoint operators represent observables that can be measured, in principle, by some experiment or device. Physicsists are mostly interested in systems that have interactions. We can think of elementary particles as localized excitations of quantumf fields that are solutions to specified wave equations (in the distributional sense). In order to have any interactions, these wave equations should have non-linear terms (actually you can define the notion of "free field" (a field with no interactions whatsoever) this way: A free field in the sense of physicists is a quantum field that is a solution of a linear equation).
Non-linear terms entail products of quantum fields, which are undefined, in general, because products of distributions are undefined, in general. The history of QFT is about the struggle of physicists to dodge this problem in one way or another. One way to dodge this problem is to introduce, as an axiom, so called "(associative) operator product expansions". This is a way to formulate, as an axiom, (handwaving:) that the "severity of the singularity of products of distributions" is under control.
operator product expansion
An operator product expansion (OPE) for a family of fields means that there is for all fields and all $z_1 \neq z_2$ a relation of the form
$$
\psi_i(z_1) \psi_j(z_2) \sim \sum_{k \in B_0} C_{ijk} (z_1 - z_2)^{h_k - h_i - h_j} \psi_k(z_2)
$$
Here $\sim$ means modulo regular functions. (See ncatlab.org/nlab/show/conformal+field+theory for an explanation of the notation.)
In a handwaving way, this axiom says that we assume we know something about the kind of singularity of the product of two fields, as their support comes closer and closer, and it is not so bad.
A rigorous interpretation of an OPE would interpret the given relation as a relation of e.g. matrix elements or vacuum expectation values.
An OPE is called associative if the expansion of a product of more than two fields does not depend on the order of the expansion of the products of two factors.
Warning: Since the OPE has no interpretation as defining products of operators, or more generally the product in a ring, the notion of associativity does not refer to the associativity of a product in a ring, as the term may suggest.
The axioms of vertex operator algebras are an axiomatization of OPEs. In this sense there is reason to expect that vertex operator algebras will play a key role, on one way or another, in a rigorous construction of interacting QFTs in four dimensions.
Answer: Name-dropping here, but I can claim that Richard Borcherds discussed vertex algebras with me, before anyone knew what they were (not that I was able to be of any help). At that stage they were purely algebraic objects, with some kind of "Jacobi identity". At some point he said he had a definition as Lie algebra in an internal sense in a category - this idea was binned as not in fact useful. Much later he said something about a relationship with the Wightman axioms, but I gather that isn't really watertight when it comes down to it. So I doubt whether the emergence of an axiomatic theory can be "rationally reconstructed" without doing some damage to the history.
Comment: I'll join in the name-dropping.  Over a period of years, I saw Borcherds give several talks on vertex algebras, and my impression was that he was putting a lot of energy into massaging the definition so that it seemed mathematically inevitable.  I believe he began one talk with some words such as "the aim of this talk is to make the theory of vertex algebras trivial".  I also remember him emphasizing that the theory of vertex algebras is not an algebraic theory (in the sense of universal algebra).  So although I have no answer for Rex, I have reason to hope there are good answers out there.
Answer: I'll take a more physical POV. Although, we don't have any mathematically concrete definition of QFT in general, but we can define 2d Conformal Field theory (this is attributed to infinite symmetries and exact solvability), other that CFTs we can also define Topological QFTs (Atiyah).Greame Segal proposed a geometric definition of CFT.
In Conformal field theory we deal with vertex operators analog to operators in QFT, we can write a Taylor like expansion of two vertex operators, k.a Operator product expansions (OPEs) which gives the QFT analog of two fields interacting. All these notions are captured by the axiomatization based on Vertex Algebras.
To see a better picture,look at the classic paper of Belavin, Polyakov, and Zamolodchikov, where an algebraic approach of CFT was proposed.
Recently Kapustin and Orlov proposed a more general definition of Vertex algebras and they showed the relation between their algebraic definition and Segal's geometric one.
Comment: @J Verma: I have been wondering if there were rigorous algebraic constructions which capture both the holomorphic and anti-holomorphic sectors of a conformal field theory. Is the paper you mentioned by Kapustin and Orlov (I guess CMP Vol 233, No 1 (2003), 79-136, right?)
the answer to this question?
Answer: This may hold you off until a REAL answer comes along.  I found something resembling intuition on this subject when I first studied Quantum Field Theory.  I'd recommend looking there before trying to tackle the many-headed beast that is conformal field theory.  Many (but not all) of the VOA axioms can be seen in the properties of a QFT.  As far as a "mathematician-friendly" place to learn about QFT goes I remember liking the book by Ryder.


Why does the melting point get lower going down the Alkali Metal Group with increase in atomic number?

Question: Why does the melting point get lower  going down the Alkali Metal Group with increase in atomic number?
Answer: You have to understand that melting point is related to the bonds in a metal between the metal cation and the 'sea of electrons'. With an increase in atomic number, you have an increase in electron shells.
As the radius of atoms get larger down the group, you could say that the force holding them together is spread over the greater area and hence, the metal cations are more weakly bonded. And coming back to my first point, when the bonding is weaker, the metal's melting point will decrease.
Answer: As you go down any group, there are extra electron shells (for example Lithium has 2 shells, Francium has 7) which cause electron shielding. Because the force of attraction must extend further and through these shells to the outermost shell, the force becomes less so there is less strong bonding occuring.
Thus, it is easier to break these less strong bonds, and because the melting point is a measure of the point at which all the bonds holding a metallic structure together are broken, it will decrease because this point is reached a lot sooner as the bonds are not as strong.
Answer: melting points of all group (i) elements is dependent on the strength of the metallic bond. In metallic bonding, the group (i) cations in the metallic lattice are attracted to the delocalised electrons. Down the group, the number of delocalised electrons and the charge on each cation remains the same at +1 but the cationic radius increases so the attraction between the cations and the electrons in the lattice get weaker down the group and so does the strength of the metallic bond
Answer:

Why does the melting point get lower going down the Alkali Metal Group with increase in atomic number?

Ok, you're looking for a generalization of what happens to the melting point as you go down a column in the periodic table.
Think of the metal atoms as cations $\ce{M^+}$ and anions $\ce{M^-}$ packed into a crystal like NaCl. However for the metals think of the metal atoms switching + and - charges at a high frequency.
So as the atoms get bigger, the energy that is holding the crystal together gets smaller because the "ions" are further apart.
Note that this simple explanation doesn't explain "real" chemical bonding. So there are other layers of truth under this simple model.

Shell structure of an atom
Different crystalline lattices and hence energy "within" the crystal
Ionic , covalent, and metallic bonding

So the "real" truth is gleamed from the interactions of simple models into a more complex system.
Answer: You have to know that elements with high lattice energy has high melting and boilng point.And since lattice energy is inversely related with radius of atoms it increases down a group and hence the melting point increases.
I hope this will help you!
Comment: The alkali metals are not ionic solids.
Comment: This doesn't follow. Atomic radius increases down the group so according to you lattice energy should decrease and so melting point should decrease.
Comment: @bon although this answer doesn't make much sense, the melting point does indeed decrease down the group..


how to move origin point to center of object

Question: I'm wanting to move the origin (widget) to the center of this object. The object (mesh and bones) was appended from a previous animation, and it looks the same in my animation.
  I have tried shift ctrl. alt C > origin to center of mass. Also I noticed that when I shift S cursor to selected, the cursor goes to the widget (origin).
  I have a feeling this happened because I moved the object with it's root bone rather than the x,y,z arrows. The Root bone is sticking out out the center of the objects back
  My animation seems to be working fine with the origin being offset as it is. Just out of curiosity, when I render the animation could the origin being offset like this, hurt anything?
Comment: *"I moved the object with it's root bone"* don't translate rigged objects independently on what they deform. If translating do so in Object mode for both object so their local coordinates remain to be the same. You can stick to resulted origin point as long as it works normally, though it can cause troubles if moved further away in the scene (possible precision errors while showing geometry)
Comment: Possible duplicate of: [How to recenter an object's origin?](https://blender.stackexchange.com/questions/14294/how-to-recenter-an-objects-origin)
Answer: I think your problem is this one:
Your armature has an origin, like any object. If you move and animate the whole armature in Object mode the origin will follow, because its position is linked to the object. But if you go within the object in Pose mode and begin to move the bones, it won't affect the origin position in any way, because the origin is linked to the armature position, not to the bones position within the armature, so with some animations the bones can move far away from the origin.
Generally, when you animate in Pose mode, you don't move the root bone, so this bone will stay close to the origin. But if you begin to move it, for a reason or another, it may also move far away from the origin.
If you want to see what is the real position of the bones within your armature, relatively to the origin, go in Pose mode, select all your bones and press altR, altG and altS, or in the Properties panel > Data > Skeleton, press Rest Position, or switch to Edit mode... you'll come back to the rest position and you'll see that the bones will probably come back close to the origin, so you'll probably won't find any reason to move the origin position.
Now if, in rest position, the origin position doesn't please you, which in my opinion is probably not your problem here, you can change it: Switch to Edit mode, then, like for any other object you need to put the 3D cursor where you want the origin to be, then shiftctrlaltC> Origin to 3D Cursor.
Answer: If you just want to move the object's origin (pivot) to the geometric mean, this option is available in the Object -> Set Origin menu. This is similar to how it was in 2.79 except now the origin has its own menu, whereas before it was under Object -> Transform.
If you want to move the cursor someplace and then snap the origin to the cursor, the same shortcut for the cursor menu in 2.79 also applies in 2.80. Press SHIFT+S and choose "Cursor to Selected" (where your transform pivot option is already set to "Median"). You can then set an object's origin to the cursor location.
Comment: I haven't heard "widget" used to describe it for Blender. An object's *origin* is a small yellow dot; The *3d cursor* is an open circle with a crosshair through it. Blender provides options for transforming (translation/rotation/scale) about an object's origin, the 3d cursor, or various means (averages). Setting an object's origin to the geometric mean is an option in the Object -> Set Origin menu. Before this I would use the 3d cursor menu to first set the cursor to the object's mean, and then set the origin to the cursor location. Let me know if you can't find the options
Comment: I usually set transformations to pivot about the object's *origin* instead of the *cursor* so I don't have to keep moving the cursor again every time I move the object.
Comment: You were right. I selected all the bones and saw that the xyz widget was in the center of the character. When the mesh is selected the xyz widget is not centered. I noticed too that when I go to edit mode the mesh jumps to the xyz widget.  I did try something; I selected my mesh and shift s >cursor to selected, the cursor went to the offset xyz widget.   My animation is almost finished. Is there a way to get the mesh's xyz widget to the mesh?  Also is the origin the same thing as the xyz widget? After reading xyz widget a bunch of times you can probably tell I don't know.
Comment: This worked great for me and it works when you select multiple items too, Blender will set the origin to all of them respectively.


Make action only when connected

Question: Is there any way to trigger an action as soon as I get connected to a network with a specific network interface i.e. local area connection 2? The OS is winXP.
Thanks in advance
Answer: I think some NIC software might do that. I think I saw something about this in my dell NIC software.
Answer: I'm fairly certain there's no XP functionality to launch an app based on plugging in the cable.
This is straying into Stack Overflow territory, but it looks as if there are definitely WMI events that you can grab onto - though I'm uncertain about whether you can get to these without churning out some code.  Here's an example.
Answer: Definitely worth asking this on Stack Overflow as well, because it's will have to be a programatic solution.
Answer: There are no built in applications that can do this that I'm aware of and I haven't come across any general purpose third party apps that do anything reusable.
If you have a developer handy you can roll your own. I've built a number of applications ( Windows 2000\XP so this should work for you) that use the IPHelper API's NotifyRouteChange function to register a callback that reacts to any change in the routing table. This has proven to be very reliable and rapid as it responds to things like connecting\disconnecting a LAN or WLAN as soon as the adapter's IP-address becomes usable. This is generally a couple of seconds after a cable is plugged in, at most.  It is (obviously) limited to IP networks though.
The System Event Notification System (SENS) has mechanisms designed to be used for this purpose (specifically the ISENSNetwork::ConnectionMade Method) but my experience with this and other SENS functionality is that they can take minutes to trigger. This appears to be intentional as SENS is intended to provide applications with mechanisms to adjust their behavior to network state changes over time (e.g. degradation of link quality), rather than to provide instantaneous reactions such as a response to a new connection being made.
This probably does belong over on StackOverflow though.

Make action only when connected


Why can't noncontextual ontological theories have stronger correlations than commutative theories?

Question: EDIT: I found both answers to my question to be unsatisfactory. But I think this is because the question itself is unsatisfactory, so I reworded it in order to allow a good answer.

One take on contextuality is to develop an inequality on measurement outcomes that is satisfied for any ontological noncontextual theory, and see that it is violated by quantum mechanics.
Another take would be to assume an algebraic structure and see that if one restricts the observable algebra to be commutative, the expected values of certain operators are restricted to lie in a given range, whereas if we allow non-commutativity the range is greater.
These approaches coincide? I've seen plenty of works that assume that it does, but without discussing it; in particular there is this paper by Tsirelson that states (in the particular case of Bell inequalities) that both approaches are equivalent, but without proving it. Is it too obvious?
At first sight, an ontological noncontextual theory is much more general than some theory embedded in a C*-algebraic framework. Why then can't it generate stronger correlations than theories with commuting algebras of observables?
Can one find a more direct connection between non-commutativity and the violation of a contextual inequality?
Answer: Well, a reasonable approach to local hidden variables is to require commutativity of operators on different space-like separated systems. This is pretty straight forward to motivate, since otherwise you are essentially using non-local operators, which can be seen via the relevant transform instead as a non-local hidden variable theory with local operators.
You might want to check out Scott Aaronson's paper exploring the consequence of this when taken as an axiom together with some other desirable properties of hidden variable theories (Phys. Rev. A 71, 032325).
I'm not too sure it makes sense to talk about commutativity beyond this, since we care about post-measurement outcomes it is not clear that the domain of the operator should contain its image, and so multiplication, and hence the commutator, isn't necessarily defined.
Comment: I'm afraid you didn't understand my question. Commutativity of operators on different space-like separated system is respected by QM (or AQFT, for that matter). I'm talking about commutativity of the whole algebra. In fact it may not make sense to talk about commutativity in the strict sense; I've never seen anyone construct an operator algebra to model a hidden-variable theory. But if we only care about the measurement outcome, not the post-measurement state, we have commutativity in a trivial sense, by substituting the observable by its predefined outcome, which is just a real number.
Comment: @Mateus: I think I did understand, which is what my last paragraph attempts to address.
Answer: Non-commutativity of operators ensures that in general we can't construct a joint probability distribution over the observables that we model using those operators. In some states and for some choices of non-commuting operators we can construct joint probability distributions, for example the vacuum state and coherent states of a quantized simple harmonic oscillator generate a positive-definite Wigner function for position and momentum, which can be taken to be a probability distribution. Of course that possibility falls apart when one considers almost any superposition of coherent states, say. The Wigner function is not positive-definite in the general case, making the interpretation of the Wigner function as a probability distribution in the special cases quite tendentious.
Conversely, if we have a commutative algebra of operators we can construct a joint probability distribution over the any subset of the observables in any state over the algebra. One could take this property as a somewhat plausible definition of classicality.
For the technical basis of this, I like best two short papers, John Baez, Letters in Mathematical Physics 13 (1987) 135-136, and Lawrence J. LANDAU, PHYSICS LETTERS A, Volume 120, number 2 (1987), which put remarkably little interpretation in the way of the mathematics, but there is a substantial literature that has tried to get at this relationship in some sort of clear way.
A literature that gives an alternative way into the relationship between non-commutativity and measurement, and that focuses on the relationship between quantum theory and classical probability theory in a way that I find helpful, albeit not conclusive, is the positive-operator valued measure approach, which is well represented by the book by Paul Busch, Marian Grabowski, and Pekka J. Lahti, Operational Quantum Physics, Springer, 1995. Searching the literature or the ArXiv for anything more recent by any of these three authors will give you something enlightening to read. To my taste, Paul Busch is always worth reading.
As far as physicality is concerned, classical physics models measurements as not affecting other measurements, so that joint probability distributions over multiple measurements are possible. In the presence of any finite level of noise ---there always is noise, everywhere (only the thermal component of the noise goes away when one is close to absolute zero, the Lorentz invariant quantum component of the noise is not controllable)--- the uncontrolled nature of the noise is something that has to be accommodated by our models of our measurements. Quantum theory accommodates the non-trivial effects of joint measurements on each other by introducing non-commutativity of the operators that are used to model the measurements, whereas classical physics models the non-trivial effects of joint measurements on each other by modeling the measurement apparatus. Contextual models are precisely models that include the measurement apparatus, or the complete experimental apparatus, in the extreme case the whole universe, not just a putative measured system.
That's somewhat bashed out. Hope someone finds it congenial.
Answer: I don't think there is a more direct connection between noncontextuality and noncommutativity for a couple of reasons. Firstly, there are noncommutaive sets of observables and states that can be simulated by a noncontextual model. Think of the Kochen-Specker model for a qubit for example. Secondly, to really answer the question of what a violation of some inequality means, you should not assume that the data you collect in the experiment is necessarily produced by quantum theory (in particular, we don't do this for Bell's inequalities). Now, there are plenty of operational theories that are contextual (in Rob Spekkens sense) but that do not have a C*-algebraic structure, e.g. the theory wherein the state space is a square. Unless you can define what it means for measurements in these theories to be "commutative", which seems unlikely because they do not have an algebraic structure, then it is clear that the relationship between commutativity and noncontextuality breaks down in this context.
Comment: Good point. It would be interesting to find an operational definition of commutativity. But allow me to be picky: I think my answer is clear on relating commutative theories and noncontextual models; what it lacks is a relation between non-commutativity and contextuality.
Answer: They can't, because ontological noncontextual theories are not more general than commuting subsets of quantum mechanics. In a nuthsell, commutative quantum mechanics is just classical probability theory, and the question whether there exists an ontological noncontextual model for quantum mechanics is precisely the question if it can be reduced to classical probability theory.
To see this, one needs Spekkens' operational definition of contextuality: an ontological noncontextual model is one where each state $\rho$ is represented by a probability distribution $\mu_\rho(\lambda)$ on an ontological space $\Lambda$, and each POVM $\{E^k\}$ by a probability distribution $\xi_{E^k}(λ)$. Then the probability of outcome $k$ will be given by $$\int \mathrm{d}λ\, \mu_\rho(λ) \xi_{E^k}(λ).$$
Now, if all $E^k$ commute, I can write $ρ$ in a basis in which they are diagonal. Then $$\mathrm{tr} (\rho E^k) = \sum_n \rho_{nn} E^k_{nn},$$ that is, the only part of $\rho$ who'll play a part in the expected value is its main diagonal, and that's just a probability distribution. If we do the identifications $\lambda \mapsto n$, $μ_ρ(λ) \mapsto ρ_{nn}$, and $\xi_{E^k}(λ) \mapsto E^k_{nn}$, we have an embedding of an ontological noncontextual model into commutative quantum mechanics.
In the particular case of nonlocality, one could see directly that a diagonal $\rho$ is always separable, and thus admits a local ontological model. The converse path, to construct a separable $\rho$ and commutative algebra from a local ontological model is essentially the same as above.
That said, I don't think my last question is answered; I'd still like to see a more direct, physical connection between non-commutativity and contextuality.