Effects of keeping a guitar in the car

Question: I have a sales job, and am on the road most days. I often find myself with free time between appointments and would like to take advantage of that free time by practicing guitar. I have a 3/4 size acoustic that I think would be easy enough to play in the car. My question is, what effects will leaving the guitar in the hot (or cold) car have on it?
Comment: Well it's going to ruin it, but all guitars get ruined eventually.  I say just keep it there and play it, and when it gets ruined, get a new one.
Comment: Wood might not fare well; probably should look for something carbon-composite. A little electric with a pignose, maybe?
Comment: i keep a jazz bass in my small car, through snow and sun alike - it has suffered exactly zero side effects or harm, it has nitro paint so should be more vulnerable to fluctuations. It might be helpful if the OP stated his location/ climactic situation as this would be inadvisable in say, Alaska, or new Mexico. Check your car insurance for any special conditions of storage.
Answer: My advice is to treat your guitar like a human passenger. If you wouldn't leave a person in your car with the windows rolled up on a particular day, then don't leave your guitar there either. Pick it up and carry it wherever you go.
Avoid leaving your guitar in your car in hot weather. Don't expose your guitar to prolonged high temperature or changes in humidity that your physical body can't stand. Remember that in many climates in the summer if you leave your car parked outside with the windows rolled up, the car, with no air circulation, becomes like an oven. The air in the car will reach temperatures high enough to kill people and animals--higher than 125 degrees Farenheit (52 degrees Celsius) when it is 100 degrees Farenheit outside in direct sunlight. Needless to say this will severely fatigue the wood in your guitar but will also cause expansion and contraction of the metal in the frets and the strings. It will also weaken the glue holding the wooden plates and braces together, causing the guitar to slowly fall apart.
If you want your guitar to play in tune with proper string height (action) you will also find yourself needing frequent adjustments to the truss rod as the neck and the metal truss rod bows or warps positively or negatively every time it is exposed to a large swing in temperature.
When I took my classical guitar anywhere in a car in the summer time in Atlanta, Georgia, I frequently had to deal with nylon strings that snapped from the fatigue of all the expanding and contracting due to the changes in heat.
I do not live in a cold climate, but I have read that in the northern US and in Canada, in the winter time, guitars suffer cumulative damage (particularly with cracked and crazed finishes) when they are taken out of a heated home, carried to a car in freezing weather to be transported, and then brought back into a heated home and taken right out of the case and played. I'm told you need to permit the cold guitar to remain unplayed in the unopened case until it slowly warms up and reaches room temperature. You should also de-tune the strings of the guitar before you take it out in cold weather because the metal strings contract in the cold.
With repeated exposure to extreme heat or cold, the tension on the neck and body from the strings will increase and decrease since metal expands and contracts with temperature changes much more than wood. The bridge, top, body and neck are already under considerable controlled tension from the strings; wide swings in temperature over time will weaken everything in the wood and the glue. In extreme cases with prolonged exposure, the bridge plate could come unglued or simply rip loose from the top, or the headstock could snap, or the neck block, where the neck is glued into the body, could pull away, changing the angle of the neck and rendering the guitar unplayable.
Comment: I came to post this answer and it's already here. The advice I always got is that if you'd be uncomfortable in a particular temperature or humidity, your guitar will be uncomfortable there too. My personal practice is to never leave an acoustic instrument in a car (though I occasionally break this on warm nights after dark when I have a post-gig dinner).
Answer: I occasionally have to do this, with electrics and acoustics. You have a few problems which can occur:
In cold weather, condensation is your biggest enemy - you will get moisture on metal elements and areas of dense wood. For an electric guitar, this will eventually harm your electrics, but on both kinds it will rust your strings faster.
In a car, temperatures can vary over a wide range - this can warp your guitar, and split the wood, as different woods expand and contract at different rates, and it can cause glues to fail.
Tuning will suffer - every temperature change will change tensions etc., but this can be fine.
An acoustic will suffer more than an electric, in almost every case. As @luser said in his comment, a cheap electric will probably be a better bet.
Answer: Keeping it in a case with proper humidification will help somewhat but it will be eventually wear out. Expect warped necks, split tops and lots of tuning. Any extreme temperatures and rapid temperature changes are murder on the wood.
I recommend getting a really cheap guitar with heavy and solid construction and accept that it will wear out over a year or two. Avoid frost and parking in direct sunlight. If it's hot don't leave it in the trunk but on the back seat floor out of the sun with one window a little open.
Answer: As Luser Droog commented, the wood may experience some problems. Cars tend to experience quite vast fluctuations in temperature, and humidity, meaning that parts, particularly necks and bodies, will expand or contract, playing havoc with tuning.
Moisture and humidity may also cause problems, warping wooden parts or causing splits, as well as accelerating the rusting of strings.
As well as this, be aware that leaving a guitar out of a case in a car exposes it to sunlight, causing woods and varnishes to fade over time if prolonged.
Answer: I would advice against placing a wood guitar you like in your car.
Rainsong builds graphite guitars that can handle a beating and temperature/humidity changes.
http://rainsong.com/
Answer: I think all the risks stated above are correct. I have personally seen a lacquer finish cracked by cold temperatures in a car, and a soundboard cracked by dryness in a home.
But you might still want to take those risks depending on how valuable the guitar is, because the benefit of playing might outweigh the cost of a travel guitar or an inexpensive beater. I have a Martin Backpacker that I've taken camping in cold and wet weather, then played next to a hot campfire. I also loaned it to a friend who dropped it and cracked the soundboard. It still serves its purpose. If I were in your situation I would just take the basic precautions of leaving it in the trunk to avoid the most extreme heat and taking it inside overnight during any extreme weather.
The more conservative advice on this page definitely applies to any guitar of significant value. You should also consider the risk of theft, as a musical instrument left in view is a good invitation for a break-in.
Answer: I'm in the same predicament.  I've chosen to buy a piece of junk garage sale acoustic for $40 and keep it in my Tacoma case in my van.  The amount of practice I get in over a lunch break makes it easily worth $40 a year or however often (within reason of course).  Do it!
Answer: Treat it like your child, take it with you and love it as it were your own 'cause it is! So it sits next to you at dinner, so what! A few minutes in the car is one thing but an hour or two is another. Would you leave your child in the car for an hour or two? Definitely get a humidifier; they're like 10 bucks online. Oaisis makes a good one.
Answer: I left my Martin 000x1ae in the truck today. It was cold this morning. However, here in South Carolina, it got up to about 85 degrees and a crack showed up right down the middle of the front. I kinda like it though.
Answer: Some tricks that may help alleviate risks when leaving an instrument in your vehicle are to park in the shade on a warm day, and cover the instrument with a blanket in an attempt to moderate the speed of the temperature changes. A 3/4 size guitar wrapped in a blanket inside a full size guitar case has a much better chance of avoiding serious damage from heat and cold, but there are also limits to the protection provided and prolonged exposure to temperature extremes is certainly risky.
Answer: where do you live? I left a jazz bass in my car for three years with no ill effects whatsoever. The ambient weather and its seasonal range will have a huge effect upon the answer that is best for you

Effects of keeping a guitar in the car


How useful is advisers' advice?

Question: This is one of my first multiplayer games and I've been following my advisers' advice for the most part, but I don't seem to be doing all that well. Is their advice actually good, or should I not be following their advice?
Comment: I have not had the pleasure of playing Civ5 yet, but I highly doubt AI-controlled advice would be useful in a multiplayer scenario, especially considering that all of your opponents also have the same advice being given to them.
Answer: Unit/Building Suggestions
In my experience the best approach to using the advice of the AI is to ask yourself why the AI is suggesting a certain unit or building. A worker could mean you should build more tile improvements around your cities. A military unit could mean you're undefended or need a bigger military. Building suggestions could mean you need more production, or maybe the AI just decided you need more culture.
I would recommend deciding upon a victory condition to strive for at the start of the game, and then working towards that goal. When the AI makes suggestions, take them into account and then decide what will work best for your strategy at this time. As others have pointed out, the AI just makes generic recommendations based on what would probably be a good decision right now, but it does not necessarily coincide with the strategy you have in mind.
Advisor Panel
You can access the advisor panel by using the button in the top right of the screen. You have the Science, Economic, Military, and Foreign advisors. Generally here they just tell you general information that could be useful. Most of their advice on this screen I do not usually use, as most of it you can gather yourself. The most useful information here I find is the military strength of your rivals, which I always consult before making any serious military decisions.
Answer: Generally, you might want to listen to the computer, but only to a certain extent. You have to keep in mind that while you may be going for one strategy, the computer is running calculations for a generic, somewhat-decent move. If you're playing multiplayer, remember that people's brains are always more complex than the computer AI, and as the AI isn't trying to interpret their future moves, only the move that would be best for you in that instant. Computer AI is almost always, without fail, defeated by real people in strategy games, so keep that in mind.
Answer: The AI advice is 100% completely useless. I turned off all of their suggestions.  You should try to understand the game mechanics yourself.  If you try to follow their advice you will build a very generic average civilization.  To get ahead in Civ you must reach for something important (a tech, or massing certain units, or wonders, gold etc.) and neglect some other things as you reach for what you need to get the advantage. If you play on lower difficulties you can follow the stupid advice and still win.
More importantly though, how fun is it to play if you are just clicking buttons you're told to click?
Comment: The advisors are not useless, they're just generic.  They point out problems you have if you haven't been paying attention, and give the obvious solution to them.  When you're just learning the mechanics, having to learn every single thing at the same time can be overwhelming, and the AI advisors will make sure you don't make any huge mistakes.  That said, winning against humans requires more than just following AI advice.
Comment: That's high-level play.  If someone is still listening to the advisers, it's usually a sign that they're not sure what they're doing yet.  The advice you're giving is for people who already know the system, and for those people, the various Civilization forums give wonderful varieties of strategies.  Step 1: walk.  Step 2: run.  The advisers are for people who are still working on step 1.
Comment: I guess it depends on your play style, I think building a generic civ is the worst thing you can do.  It means in the end somebody else will come and easily gain advantages over you.   (If you play against AI below king difficulty I would retract my comment that generic civs and advisors are useless, I should have prefaced my answer.)

How useful is advisers' advice?


Leaving wi-fi on all the time

Question: Does it highly affect the stand-by time to leave wi-fi open all the time on my phone?
Answer: Yes, I would imagine leaving WiFi on all the time would drain the battery faster. Even if you're not actively connected to an Access Point (AP), the phone is still constantly scanning all the APs in the vicinity of where you are and that takes energy.
Answer: The general rule is to turn off what you are not using.
Answer: This has been my experience with HTC's Evo:
WiFi uses less power compared to 3G when actively downloading/uploading large amounts of data.  Even when doing a lot of browsing.  However, unlike 3G, the WiFi radio never goes to sleep.  So when you aren't actively downloading data, it's best to turn it off.  This is unlike GPS and Bluetooth where if they aren't being used, they don't use much power at all.
Answer: Turning off anything that isn't in use will generally increase battery life.  Also remember that the biggest drain on the battery is generally the display.  Setting your brightness to the lowest level (or at least auto) can do a lot to save your battery life.  Also, if you typically connect to WiFi at home, but don't have access at your work, there are apps that can automatically turn off the WiFi radio (among other things) in the morning, then turn it back on in the late afternoon/early evening.  I use Advanced Volume Mode Scheduler to do this.
Answer: Wifi uses significantly less power then 2G or 3G and in fact uses less while connected and inactive than when it is searching for an AP. So I leave it on all the time and my battery lasts longer.
Comment: That is only half the story. Wifi uses significantly less power when actively transmitting data since it transmits lower range signals; but maintaining a Wifi radio without transmitting any data uses much more battery than maintaining a 3G radio.
Answer: As JamesRyan pointed out, leaving WiFi connected and inactive uses less battery than if it has to keep searching for an AP.
I use a HTC Desire and initially had my WiFi set to sleep after 15min, but then realised that every time I unlocked my phone it would wake the WiFi, and in turn it would start searching for an AP. Afterwards, I set WiFi to never turn off - this actually resulted in more battery power left at the end of the day (even with the occasional sync of my accounts - email, twitter etc).
Answer: By my experience it seems that if WiFi is connected to a network then leaving it on will improve standby time as all the periodic data access that an Android phone does will go through WiFi. And accessing data over WiFi is more battery efficient than 2G/3G/LTE.
On the other side, if WiFi is not connected to a network then it will constantly search for one and decrease the standby time.
So what I try to do is to leave WiFi on whenever I think I will be around a known WiFi AP for some time but I leave it off the rest of the time. The drawback is that it forces me to change the setting at least a couple times per day.
Answer: Yes, it decreases standby time because WIFI is one of the more significant things draining the battery. As long as WIFI is enabled, your phone is constantly scanning for new networks and "listening" on TCP ports.
Additionally it will be sending data from applications, requesting updates, etc.; and if you have the Play Store set to automatically update apps then it will also be downloading and installing apps, which also significantly drains the battery.

Leaving wi-fi on all the time


Why are "some" letters silent in English?

Question: There are many such words that we all know about, but please explain why the makers of the English language made up words with silent letters?
Comment: You talk about the English language as if a group of people sat down one day and decided on all the rules. We all agree that had this been true, they should have stuck to their day jobs. Words, in particular, have appeared in the English lexicon from many different sources - extremes perhaps being classical Greek and youtspeak. There is no governing body (the French like to think they have one for their language), and there is often dissension over what are 'allowed' words, and how they should be pronounced (and spelt). Many anomalies have appeared over time - and many remain.
Comment: @ShivamPatel: *everyone* thinks that their mother-tongue is spoken "exactly as its written", simply because they have fully internalized that particular language's spelling-to-pronounciation mapping.
Comment: There are silent letters in every language. There are silent letters in your mother tongue.
Comment: @RegDwigнt there are no silent letters in my mother - tongue its "Gujarati" and its exactly spoken as we write!!! Even complicated words in Gujarati novels are pronounced as they are written
Comment: Just a quick web search brings up that visarga is silent in Gujarati. It used to be pronounced as /h/, but now no longer is. The Wikipedia article lists a whole range of other features of the language that are not represented in writing. In fact it flat-out says that the very alphabet is actually not an alphabet but an abugida, so that vowel notation is completely secondary by design.
Comment: [Related](https://english.stackexchange.com/q/594992/2085).
Answer: This is a very big topic which cannot really be answered in a few lines, but part of the answer is that some letters which we don't pronounce now were once pronounced in the past. The pronunciation of words containing them has changed, but their spelling hasn't.
If you're interested in the history of English spelling, I recommend 'Spell It Out' by David Crystal.
Answer: What @Edwin Ashworth says is true; however, there have been people who have sat down and tried to standardize English, and some were responsible for adding or unnecessarily maintaining spelling anamolies:
"Caxton is credited with standardising the English language (that is, homogenising regional dialects) through printing. This facilitated the expansion of English vocabulary, the regularisation of inflection and syntax, and a widening gap between the spoken and the written word. However, Richard Pynson, who started printing in London in 1491 or 1492, and who favoured Chancery Standard, was a more accomplished stylist and consequently pushed the English language further toward standardisation.[citation needed]
It is asserted that the spelling of "ghost" with the silent letter h was adopted by Caxton due to the influence of Flemish spelling habits.[7]"
from http://en.wikipedia.org/wiki/William_Caxton
Early dictionaries were also aimed at standardization, and of course the authors were responsible for deciding how things should be spelt:
"To the end of establishing an English that could serve the complex needs of education, the Elementarie ends with a list of 8000 "hard words". Mulcaster does not define any of them, but attempts to lay down a standard spelling for them at a time when English lacked universal standardized spellings. Besides making movements toward spelling rules for English (such as the role of the silent e in vowel length in such pairs as bad and bade), the list represents a call for English to have its first dictionary, to gather "all the words which we use in our English tung … out of all professions, as well learned as not, into one dictionarie, and besides the right writing, which is incident to the Alphabete, [the lexicographer] wold open vnto us therein, both their naturall force, and their proper use." "
from http://en.wikipedia.org/wiki/Richard_Mulcaster
Note the above quote also points out that not all silent letters are arbitrary, as in the case of silent -e ending to indicate a long vowel. Think also "far/fare" "car/care" "sit/site" -- although of course this rule doesn't always hold true: "have", "live" [the verb], "give", etc.
Answer: A language (any language) wasn't built in a day. They're also not stand-alone.
Many factors affect the way words are pronounced. Some are:

The land in which the words are frequently used.
The other languages spoken by the people.
The standard/basic phonetic sounds made by the letters.
The standard rules that letters make when used with others.
The language from which an English word is derived.

Here are some examples of words/names related to various lands...

The name John (in UK) -> Yohan/Yowan -> Io Wan (Belgium)
Joseph (British) -> (pronounced) Yoseph in Austria -> Yusuf (Middle East)
The very letter 'j' is pronounced 'y' in Germany and Austria.
British  anti as in 'tip' while Americans say 'anti' as in 'type'
Sugar is from Arabic (Sukkar)

And examples of standard prnunciations:
Sounds created by 'i_e' where the middle blank is a letter.
kite, bite, site - they all end with the same sound.
But this is not always the case. My point is that certain letter combinations together have a certain standard sound with exceptions of course.
So the list goes on...
And this is why you'll find words with some letters not prnounced.


algorithm to calculate the control points of a cubic Bezier curve

Question: I have all points where my curve pass through, but I need to get the coordinates of the control points to be able to draw the curve. How can I do to calculate this points?
Answer: When what you already have is a set of points where your curve must pass through, Bézier is not what you want; you should be using a parametric (cubic) spline.
Assuming that your points $\mathbf{P}_i=(x_i,y_i)$, $i=1\dots n$ represent a general curve (as opposed to a function $f(x)$, for which simply applying the usual cubic spline algorithm on your points suffices), Eugene Lee proposed centripetal parametrization to generate suitable parameter values $t_i$ associated with your $\mathbf{P}_i$. The prescription for generating the $t_i$ (in its most general form) is

where $\left\| \cdot \right\|$ is the (Euclidean) length, $e$ is an adjustable exponent in the interval $[0,1]$, and $t_1=0$. (A usual value for $e$ is 0.5, but $e=1$ is sometimes used as well.)
From this, one applies the usual cubic spline algorithm to the sets $(t_i,x_i)$ and $(t_i,y_i)$, from which you now have your parametric spline. (The periodic spline is recommended for closed curves, and the "not-a-knot" spline for all other cases.)
The MATLAB Spline Toolbox has support for this parametrization scheme, though it shouldn't be too hard to write your own implementation.
Answer: It depends how many points you have.
If you have only 4 points, then you will need only one Bezier cubic segment. Suppose your known points (through which you want the curve to pass) are $Q_0, Q_1, Q_2, Q_3$. First you have to choose 4 parameter values $t_0,t_1,t_2,t_3$ to assign to these points. The centripedal approach described in the other answer is good. In simple situations, where your points are fairly equidistant, a much simpler choice is just $(t_0,t_1,t_2,t_3) = (0, 1/3, 2/3, 1)$. Then you have to solve a system of 4 linear equations, as follows. Suppose $P_0,P_1, P_2, P_3$ are the (unknown) control points of the Bezier curve, and denote the curve by $C(t)$. Then we want $C(t_i) = Q_i$ for $i=0,1,2,3$. This means
$$ \sum_{j=0}^3{b_j(t_i)P_j} = Q_i$$
for $i=0,1,2,3$, where $b_0, b_1,b_2,b_3$ are the Bernstein polynomials of degree 3. Solve these equations to get $P_0,P_1, P_2, P_3$. If you always use $(t_0,t_1,t_2,t_3) = (0, 1/3, 2/3, 1)$, then the coefficient matrix of the linear system is fixed, and you can just invert it once (exactly), and save the answer. This gives you a simple formula relating the $P$'s and $Q$'s.
If you have more than 4 points, then you can use a Bezier curve with degree higher than 4. The calculations are analogous to those shown above. But a better approach is to use a "spline" that consists of several Bezier cubic segments joined end-to-end. There are many ways to compute splines. One of the easiest (and the most popular amongst graphics folks) is the Catmull-Rom spline. It gives you simple formulae for the control points in terms of the given points $Q_i$.


Deeper meanings of barycentric subdivision

Question: I just want to ask if there is any deeper motivation or clear geometric "sense" behind the barycentric subdivision. Some friend asked me about this a few months ago, looking back the section at Hatcher, I still feel quite confused. I remember one friend told me combinatorically one can do this from posets back to posets, but this does not give me any way to "understand" it properly. In some books (Bredon, for example), the author use excision property as one of the axioms, I'm wondering "where they came from, why they make any sense?".
Comment: Poincare duality in terms of dual cell subdivision might be interesting for you.
Comment: I think the OP has seen singular homology, and has seen the proof of the excision property in Hatcher's book.
Comment: There is more to life (as an algebraic topologist) than simplicial homology. In particular fractals probably (depending on your understanding of this word) have homology groups. The keyword is singular homology which generalises simplicial homology (and other types of homology) to arbitrary topological spaces.
Comment: While we are at it, the cubical definition and subdivision in Massey's "Singular homology theory" might be of interest, just to see that you don't need to use simplexes.
Comment: Hi. I know Poincare duality. But I don't believe one can define singular homology successfully for fractals for the simple reason they they are not continuous objects. So maybe I should search on this before raising the question.

The second part got deleted by Daniel's request. I don't enough on this to "edit", so I delete it.
Comment: I didn't mean to imply that there is a Poincare duality for fractals. In fact I know almost nothing about fractals. All I wanted to say was that there is an interesting proof of Poincare duality in terms of dual cells, and that it might interest you.
Comment: Oh, it is not for you. Maybe I should divide that into paragraphs. It's the second commenter.
Answer: There can be many reasons for subdividing simplices, barycentrically or otherwise.
For a simplicial complex (triangulated space) there are the simplicial homology groups. These are known to be isomorphic to the singular homology groups, therefore (1) invariant under homeomorphism, and in particular (2) invariant under (not necessarily barycentric) subdivision. Before the invention of singular homology, I believe that (1) was unknown. Fact (2) was a key part of the theory. Subdivision is important simply because even if your space is made out of simplices you will sometimes care about subsets which are only unions of simplices after you cut the space up finer.  In simplicial homology, excision is an easy algebraic fact, stemming from the fact that when a complex is a union of two subcomplexes then every simplex is in one or the other (or both).
In singular theory, as you know, invariance under homeomorphism is a triviality but excision requires some work. The point is that when a space is a union of two open sets then (bad news) not every singular simplex is in one or the other but (good news) simplices can be systematically replaced by combinations of smaller simplices to show that this does not matter. This is where subdivision is used, and there is no reason it has to be barycentric. It's like with the fundamental group: you might explore a space by using maps of a standard unit interval into it, but in proving the Seifert-Van-Kampen Theorem you might want to subdivide that interval into little pieces.
Barycentric subdivision also rises in PL (piecewise linear) topology in one other specific technical way that has nothing much to do with homology: regular neighborhoods. In a finite simplicial complex $K$, the smallest neighborhood of a given subcomplex $L$ that is itself a subcomplex does not in general have $L$ as a deformation retract, but this becomes true if you first barycentrically subdivide twice.
And in the interplay between categories and simplicial constructions barycentric subdivision turns up in various ways.
ADDED in response to Hatcher's answer and its comment thread:
Yes, there is a way of extending to all $n$ the pattern that begins: cut a segment in half, cut a triangle into four equal pieces using midpoints of edges ... It is sometimes called "edgewise subdivision", I believe. It may be realized for simplicial sets as follows: A simplicial set is a functor $\Delta^{op}\to Set$ where $\Delta$ is the category of standard nonempty ordered finite sets; its subdivision is obtained by composing with (the opposite of) the functor $\Delta\to\Delta$ which takes an ordered set to two copies of that set laid end to end. This leaves the realization unchanged. Applied to a standard $n$-simplex, it gives a certain subdivision with $2^n$ pieces. If $n>2$ then the pieces are not all the same shape. If $n=3$ you get a tetrahedron cut into four scaled-down models of itself sitting in the corners and four more whose union is an octahedron; these four all share an edge, the only internal edge that there is. It's not immediately clear to me what diameter estimate is available for the pieces.
This can be generalized so that you now cut an edge into $k$ equal pieces and a triangle into $k^2$ congruent pieces (almost half of which are upside down) and in general cut an $n$-simplex into $k^n$ pieces. This $k$-fold edgewise subdivision plays a role in the area of cyclic homology and related things: when a simplicial set $X$ has the kind of extra structure that makes it a cyclic set (a suitable action of a cyclic group of order $m$ on the set $X_{m-1}$ for all $m>0$) then its realization has an action of the circle group, and to make the action of the subgroup of order $k$ appear as a simplicial action you can do the $k$-fold edgewise subdivision described above.
There is also another edgewise subdivision. In this one the $1$-simplex is cut in half as before and the $2$-simplex is cut into four pieces in the following way: join the middle vertex to the midpoint of the opposite side, and join that midpoint to the midpoints of both of the other sides. This construction corresponds to the functor $\Delta\to\Delta$ that takes an ordered set to two copies of the same laid end to end but with the order reversed in one copy.
See also my answer to the recent MO question "Endofunctors of the Simplex Category".
The second edgewise subdivision that I described can be used to analyze the relationship between two definitions of algebraic $K$-theory: Quillen's $Q$-construction is essentially a subdivision of Waldhausen's $S$-construction.
Comment: Thanks! This really helps. I need sometime to finish reading this.
Comment: The topological invariance of simplicial homology groups seems te be known before the invention of the singular homology. For example, Alexander's paper [Combinatorial Analysis Situs](https://www.ams.org/journals/tran/1926-028-02/S0002-9947-1926-1501346-5/S0002-9947-1926-1501346-5.pdf) already covered this topic. As a very old textbook, Alexanderoff's book *Elementary Concepts of Topology*, with a preface written by Hilbert in 1932, also covers this topic (which might be the first book related to algebraic topology that I read).
Answer: In the other replies there has been some mention of alternative methods for subdivision besides barycentric subdivision, but these are rarely encountered in algebraic topology. What are some of these other methods, in fact?  Preferably they should be natural and canonical, not based on random choices. I dimly recall seeing somewhere (in a paper of Quillen or Segal?) a subdivision method generalizing the simple idea of subdividing a triangle into four triangles by adding new vertices at the midpoints of the three edges, but the generalization to higher dimensions isn't obvious.  Does anyone know a reference for this?  Another approach might be to use the canonical subdivision of an n-simplex into n+1 cubes, one at each vertex of the simplex, then subdivide each cube into small cubes in the obvious way, then subdivide the small cubes into simplices in some natural way.  This seems a bit cumbersome, however.
A drawback of barycentric subdivision is that it takes some work to show that sufficiently many iterations of barycentric subdivision produce arbitrarily small simplices. It would be nice to have a subdivision method for which this was obvious.
Comment: That's one of the ways in which singular homology is easier with cubes than with simplices: there is a natural way to subdivide a cube and it is obvious that sufficiently many iterations produce arbitrarily small cubes.
Comment: I'm pretty sure this discussion is about what's usually called (Segal's) edge-wise subdivision, and it seems to come up quite a bit actually.  A google search turns up a bunch of references.
Comment: There are two kinds of edgewise subdivision. See my edited answer to this question. I think that Segal's is the second one that I mention there.
Comment: I would say that other subdivisions are not so rare, in the sense that while individually each subdivision might not appear in many different places, needing an alternate subdivision of some sort is not unusual.  Immediately I can think of a generalization of edgewise subdivision used by Bokstedt, Hsiang and Madsen (I now see that Goodwillie mentions it in his answer) and a decomposition of a simplex into products of lower-dimensional simplices used by McClure-Smith (first case: a 2-simplex as the union of a product of 1-simplices (that is, a square) and two 2-simplices).
Comment: Thinking it over, I see that these edgewise subdivisions do make things smaller in the sort of way one needs. But the story is rather different from the barycentric one. In his book Allen Hatcher shows that the diameter of an $n$-simplex shrinks by at worst some definite amount depending only on $n$ (in fact $\frac{n}{n+1}$), no matter what metric (inherited from an ambient Euclidean space) is used. This fails for at least one kind of edgewise subdivision of a $3$-simplex (where new vertices are placed at midpoints of edges); the ratio of diameters can be arbitrarily close to $1$.
Comment: Yet it's not hard to see that in the $k$-fold edgewise subdivision of an $n$-simplex every little simplex has diameter bounded by $\frac{D}{k}$ where $D$ is the diameter of a cube containing the big simplex. And edgewise subdivision repeated $m$ times is $2^m$-fold edgewise subdivision.
Comment: I don't really understand how it could be arbitrarily close to 1, but the $\frac{D}{k}$ bound makes sense. This is second time I got answers from the book authors. I'm glad Hatcher is still alive.
Comment: By arbitrarily close to $1$ I mean: you can have a $3$-simplex in $\mathbb R^3$ with diameter $1$ such that when you subdivide it into $2^3$ simplices in this way then some $1$-simplex of the subdivision  has diameter $>1-\epsilon$. This $1$-simplex connects the midpoints of two opposite edges. This does not contradict the D/k bound. Iterated barycentric subdivision creates some very distorted simplices; iterated edgewise subdivision does not.
Comment: Or maybe it can distort them a lot. But somehow, if you do it $m$ times, the diameter of a little simplex is bounded by $\frac{D}{2^m}$ for a constant $D$ even though you don't get this by working out a bound for what happens when you do it one time.
Comment: The oldest reference that I am aware of generalizing the subdivision of an equilateral triangle into four equilateral triangle is Freudenthal (1942) and the decomposition into four triangles which share a common vertex is the book of Whitney (1957) in Appendix 1, §1.
Answer: You can think of a simplex as a finite ordered list (i.e., the vertices).  The simplices of its barycentric subdivision are the lists of subsets of the first list, ordered by inclusion.
Answer: The following paper includes a very detailed and elegant description of how to construct edgewise subdivisions that subdivide a $d$-simplex into $k^d\cdot d$-simplices all of the same volume and shape characteristics, for every integer $k\geq 1$
Edgewise Subdivision of a Simplex
H. Edelsbrunner and D. R. Grayson
DISCRETE AND COMPUTATIONAL GEOMETRY
Volume 24, Number 4, 707-719
Answer: This is really just a comment/question for Tom (or for any
other knowledgable topologist), but it has got far too long.
It's also an attempt made from a position of almost
complete ignorance to (re)construct an alternative
foundation for singular homology :-)

Is the following a correct description of his first "edgewise decomposition"
  of the $3$-simplex?

Let our standard tetrahedron have vertices (0,0,0), (1,0,0), (1,1,0)
and (1,1,1) in $\mathbb{R}^3$. So it consists of the $(x,y,z)$
with $1\ge x\ge y\ge z\ge0$. We split this into eight small tetrahedra.
To avoid fractions, let's first double the size so we split the tetrahedron
with vertices (0,0,0), (2,0,0), (2,2,0) and (2,2,2) into tetrahedra
with vertices:

(0,0,0), (1,0,0), (1,1,0) and (1,1,1),  (*)
(1,0,0), (2,0,0), (2,1,0) and (2,1,1),  (*)
(1,0,0), (1,1,0), (2,1,0) and (2,1,1),
(1,0,0), (1,1,0), (1,1,1) and (2,1,1),
(1,1,0), (2,1,0), (2,2,0) and (2,2,1),  (*)
(1,1,0), (2,1,0), (2,1,1) and (2,2,1),
(1,1,0), (1,1,1), (2,1,1) and (2,2,1),
(1,1,1), (2,1,1), (2,2,1) and (2,2,2).  (*)

Then the starred tetrahedra are those in the "corners" of the
large tetrahedron while the other four share the "internal"
edge from (1,1,0) to (2,1,1).
How can this be generalized? Take as the standard $n$-simplex $\Delta_n$
that defined in $\mathbb{R}^n$ by the inequalities
$1\ge x_1\ge x_2\ge\cdots\ge x_n\ge0$. Then the unit cube can be
decomposed as $n!$ copies of $\Delta_n$ obtained by coordinate
permutations. Then $\mathbb{R}^n$ itself can be decomposed into
copies of $\Delta_n$ by translating the decomposition of the unit
cube by vectors in the integer lattice. Call this our standard
decomposition of $\mathbb{R}^n$.
We can now decompose our standard simplex $\Delta_n$ into $k^n$ congruent
simplices each similar to $\Delta_n$. Again it's more convenient
to scale $\Delta_n$ by a factor of $k$ and then decompose into
simplices congruent to $\Delta_n$. But $k\Delta_n$ is
a union of $k^n$ simplices in the standard decomposition of $\mathbb{R}^n$,
and this does it.
What I haven't worked out yet, is if there an analogue of
the barycentric chain maps $S$ and $T$ (in Hatcher's notation)
and what those should be. If so then this decomposition would provide
an alternative approach to the excision theorem in singular homology.
There would be a couple of advantages to this over the classical approach.

The standard simplex $\Delta_n$ lives in $\mathbb{R}^n$
rather than $\mathbb{R}^{n+1}$.
Repeated subdivision of the standard simplex results in simplices
similar to the original, as opposed to repeated barycentric subdivision
which produces a plethora of different-shaped simplices. The technical
advantage here is that we immediately see what the diameter
of the simplices in our subdivision are, rather than have the bound
of the factor $n/(n+1)$ occurring in the barycentric subdivision
which takes some effort to prove.

On the other hand, there are disadvantages: for instance,
most of the faces of $\Delta_n$ are not congruent to $\Delta_{n-1}$.
Is there some reference where these ideas are fully worked out?
Answer: I won't address the second part of your question, since I'm not sure it's well-defined.  But the first (as I understand it, "where does barycentric subdivision come from?") is a good question.
If you've read Hatcher's proof of the excision theorem, you'll remember he defines, for an open cover $\mathcal{U}$ of $X$, the chain complex $C^\mathcal{U}(X)$ to be the subcomplex of $C(X)$ given by singular simplices whose images are contained in an element of $\mathcal{U}$.  He shows that the inclusion $C^\mathcal{U}(X)\to C(X)$ is a homotopy equivalence using barycentric subdivision---and excision, in the second form he states it, is obvious enough for the homology of $C^\mathcal{U}(X)$.
So there are two questions:

(1) What is the motivation for introducing the complex $C^\mathcal{U}(X)$?
(2) Why do we want barycentric subdivision to prove the homotopy equivalence?

Question (2) is easy enough -- we don't actually need barycentric subdivision, we just need something like it.  We want to be able to send an arbitrary simplices $\sigma$ to sums of simplices contained within elements of $\mathcal{U}$, such that the sum in question is homologous to $\sigma$ (i.e. the boundaries cancel out).  The obvious thing to do is to apply the Lebesgue number lemma, so we need some way of making simplices smaller by some definite factor.  Furthermore, we need the map in question to be a chain map (to commute with the boundary map), which means it has to be built up inductively -- $\partial S=S\partial$ means that restricting the subdivision of an $n+1$-simlex to its faces must give the same subdivision as simply subdividing the faces.  Barycentric subdivision is an obvious way to do this.  You can motivate it yourself by trying to come up with a subdivision satisfying these two criteria for $1$-simplices and $2$-simplices; I bet you'll come up with barycentric subdivision.  But it's by no means necessary -- there are any number of similar subdivisions.  (You might for example define singular homology via cubes; then there's an extremely obvious subdivision!)
Question (1) is much deeper.  Of course, excision in the second form Hatcher gives it suggests an approach like this -- but the intuition is a bit more exciting than that.  What this approach means is that homology can be computed "locally"!  Introducing singular cohomology, this idea leads directly to sheaf cohomology -- motivationally, if not historically.
Comment: Hi, to be honest, I don't think your answer is really related to what I'm asking about. These thoughts are what I already have in mind when I wrote down the question in here, not the "deeper thoughts" I'm looking for. But thanks for commenting and answering.
Comment: Could you give an explicit description of "Introducing singular cohomology, this idea leads directly to sheaf cohomology -- motivationally, if not historically."? Thanks.
Answer: Not sure whether this is an answer to the question but thought of sharing what I feel makes the idea of barycentric subdivision very natural.
On metric spaces one has the crutch of the notion of distance to make sense of what is "small" and then things like Lebesgue Number help one get covers with as small open sets as possible. This notion of smallness gets hard to emulate on general topological spaces and here barycentric subdivision helps get across that by exploiting together the benefit of having a notion of smallness in Euclidean space and the notion of continuity. If something is small in the euclidean norm (as available on the simplex) it will map to something small in an arbitrary topological space under a continuous map.
Further barycentric subdivision sort of very optimally exploits the notion of convexity unlike say thinking in terms of cubes. I am not sure how to make it precise but using cubes instead of tetrahedrons in 3D will bring in many more maps than necessary.
Homology in some sense depends on lower amount of information than a priori it seems to need. Like instead of all continuous maps from the simplex to the space if one takes only those maps which are non-degenerate on some face of the simplex, even then one gets the same homology theory. (This is what Massey does). This kind of think might get harder to see with  out barycentric subdivision.
I am not sure what sense it would make if one needed to do infinite barycentric subdivision since in that "limit" one will land up looking at maps from 0 volume subsets of euclidean space to the topological space and won't these simply not allow non-trivial continuous maps from them? I am not sure how to make it precise.
In "reasonable" topological spaces for any open cover chosen on it, the corresponding cover gotten on the simplex has a Lebesgue number which is a finite number. Hence one can do a finite number of barycentric subdivisions till each piece has diameter less than this and hence a finite number of barycentric subdivisions should be enough.
It would be illuminating if someone can explain why this might fail for things like fractals. And if it does then how does one go around it?
Also the notion of barycenter coincides with the idea of "center of mass" if one can attach equal masses to all the vertices of the simplex. This gives quite an intuitive help. I guess should be cases where this identification is more tangibly fruitful.


Prove that, for any set of consecutive integers with an odd number of terms, the sum of the integers is always a multiple of the number of terms.

Question: For example, the sum of 1, 2, and 3 (three consecutives -- an odd number) is 6, which is a multiple of 3.
Answer: Suppose that you add up the consecutive integers $n,n+1,\dots,n+m$; call the sum $S$. Now add up the numbers again, but in reverse numerical order, and arrange the sums like this:
$$\begin{array}{c}
n&+&(n+1)&+&\ldots&+&(n+m-1)&+&(n+m)&=&S\\
(n+m)&+&(n+m-1)&+&\ldots&+&(n+1)&+&n&=&S\\ \hline
(2n+m)&+&(2n+m)&+&\ldots&+&(2n+m)&+&(2n+m)&=&2S
\end{array}$$
Each column must have the same sum: each time you move one column to the right, the top number increases by $1$ and the bottom number decreases by $1$, so the total remains the same. There are $m+1$ columns, so the bottom line can be summarized as
$$(m+1)(2n+m)=2S\;,$$
and $$S=\frac{(m+1)(2n+m)}2=\underbrace{(m+1)}_{\text{nr. of terms}}\cdot\frac{2n+m}2\;.$$
Since the number of terms is odd, $m+1$ is odd. But $S$ is an integer, so $(m+1)(2n+m)$ must be divisible by $2$. What does that tell you about $2n+m$?
Answer: Formula for sum of consecutive integers: $S = \frac{1}{2}n(f + l)$ where $S$ is the sum, $n$ is the number of terms, $f$ is the first term and $l$ is the last.
$S$ is an integer. So, if $n$ is odd, then $\frac{1}{2}(f + l)$ has to be an integer [Reason: $\frac{1}{2}(f + l)$ can either be an integer or half integer, but half integer is not an option]. So $n$ divides $S$.
Answer: Hint: Suppose there are $2t+1$ terms. Let $m$ be the "middle" term. Argue that the sum is $m$ times $2t+1$.
I suggest doing it by the "Gauss" method (which is older than Gauss by millenia) fpr summing an arithmetic progression. Go outwards from the middle. The two neighbours of the middle term add up to $2m$. So do the next two terms as you go outwards on both sides. And the next. And so on.
Remark: The same proof works for any arithmetic progression of integers with an odd number of terms.
Answer: Hint $\rm\,\ m = 2n\!+\!1\:$ consecutive integers form a complete set of remainders mod $\rm\:m,\:$ so they are congruent mod $\rm\:m\:$ to the complete set of remainders $\rm\:0,\pm1,\pm2,\ldots,\pm n,\:$ which have sum $\equiv 0.$
Remark $\ $ This provides an arithmetical interpretation of the reflection around the middle term mentioned in some other answers. $ $ Here it becomes the $ $ negation $ $ reflection $\rm\ x \,\to\, -x \,\ (mod\ m),\:$ by which we can pair up each element with its negation, forcing them to cancel out of the sum. Such use of reflections (or involutions) to pair-up terms frequently proves handy, e.g. see prior posts here on Wilson's theorem (in groups), esp. this one to start.


Final Patent Filing Procedure

Question: What are the key elements needed to prove Concept after getting a provisional patent filing accepted in order to complete a final filing documents?
Answer: You do not need to prove the concept by documenting experiments unless it involves biology or possibly chemistry. You do to have the solution in hand, requiring no further invention. But it can all be on paper.
A patent application, to have any real value, needs to show that you are "in possession of the invention", and needs to "teach someone skilled in the art how to make and use the claimed invention without undue experimentation". It does not need to show a particularly efficient or production-ready implementation. You
can't  hold back critical details of whatever you think is the best way to do it. The patent application does not need to be down to the detail of manufacturing drawings as long as you are not hiding the "secret sauce".
Although a provisional application does not need the same formality of a non-provisional application, it does need to
contain this same level of content or it might not do you as much good as you think it does if you end up needing to rely on it to establish a priority date.

Final Patent Filing Procedure


what does it mean to be cheerleading a project when you are the boss?

Question: I've heard the expression "the boss is cheerleading a project" where the boss is not really managing the project, but is just encouraging their people for the project to happen. Where does this come from and why would the boss be the cheerleader when he/she is still the superior?
Answer: Probably the boss himself/herself also has responsibilities towards the project in question, but rather than working on those and mitigating the risks, probably she/he is simply cheerleading without really solving any problems. That's my interpretation.
Comment: Yes, I think this is an important distinction to make. The boss has backed off of direct management for a project, and maintains involvement only through encouragement, not problem-solving.
Answer: Cheerleader here is used in the sense

an enthusiastic and vocal supporter
Answer: Philoto gave a definition.  As for why a boss would do this, inspiring your team (getting them excited about the project) usually works better than just ordering them to do it because you're the boss.
Answer: A "cheerleader" is a member of a group that is "adjunct" to a sports team that is leading the spectators in cheers for the team.
A boss' job is to motivate the team to get the job done. Sometimes, this result can be achieved by being "bossy," but with a competent, self-motivated group, cheering the team on can be at least as helpful. This sometimes occurs when the boss is, or feels less technically competent than the team members. Eighty-three year old Warren Buffett, chairman of Berkshire Hathaway, plays "cheerleader" with his top executives.


Compling KUTE traffic generator 1.4 on Fedora Core 5

Question: I am trying to install KUTE traffic generator on Fedora Core 5 kernel version 2.6.15.
When I try to complile I get this error message:
[root@dhcppc128 kute-1.4]# make
make modules
make[1]: Entering directory `/kute-1.4'
make -C /lib/modules/2.6.15-1.2054_FC5smp/build SUBDIRS=/kute-1.4 modules
make: *** /lib/modules/2.6.15-1.2054_FC5smp/build: No such file or directory. Stop.
make: Entering an unknown directorymake: Leaving an unknown directorymake[1]: *** [modules] Error 2
make[1]: Leaving directory `/kute-1.4'
make: *** [all] Error 2

Also I get errors when trying to patch.
I have been told I need to install the kernel headers, and I have tried installing the kernel headers and kernel devel from the DVD:
I have checked the kernel version and it is: 2.6.15-1.2054_FCSsmp i686
I have tried the kernel-smp-devel.2.6.15-1.2054_FC5.i686 RPM but it gives an error message.
rpm -i kernel-smp-devel-2.6.15-1.2054_FC5.i686.rpm
warning: kernel-smp-devel-2.6.15-1.2054_FC5.i686.rpm: Header V3 DSA signature: NOKEY, key ID 4f2a6fd2
error: Failed dependencies:
/usr/sbin/hardlink is needed by kernel-smp-devel-2.6.15-1.2054_FC5.i686

Installing the kernel-devel.2.6.15-1.2054_FC5.i686 RPM does work but I still get the same problem with compiling, even after a reboot.
I am guessing I need to install the SMP version?
P.S. I am running it on a P4 3.0 Ghz with 1 Gig RAM.
Answer: Fedora Core 5 is prehistoric, on its kernel you won't ever be able to build new modules. The in-kernel API is far from static.


What is the best way to learn how to develop drivers?

Question: At some point in my career I would definitely like to get involved in some embedded programming.  I was thinking that starting out with developing some drivers would be a good place to develop my abilities in the regard.  I'm open to doing so on either windows or linux platform (eventually will want to probably attempt both).  What tools would you recommend that I look into to get me started on this project?  Any other resources that you have found to be helpful (i.e. books, websites, programs, etc)?  Any advice/recommendations? Thanks!
Comment: When you Googled for books on writing linux drivers, what did you find?  Please list books you found with specific questions on those books.   For example, what's wrong with this book? http://oreilly.com/catalog/9781565922921
Answer: If you want to do embedded programming, get some hardware!
For a first introduction, I highly recommend getting an Arduino. I got an Inventor's Kit for Arduino from Sparkfun. It contains a lot of hardware and easy to follow instructions for creating lots of projects. It walks you through assembly on a breadboard, writing the software, and making LEDs blink and motors turn. Lots of fun! Arduino is very expandable, with lots of shields (add-on boards) that give your Arduino extra functionality, like an SD Card reader, GPS and more.
If you want a little more advanced board, the PIC32 Ethernet Starter Kit has a 32-bit microcontroller and USB and Ethernet interfaces. Load the sample projects from the free IDE and you have a web server on a little board. The PIC32 is a powerful device and you can use this kit to experiment with it.
Both of these boards load over USB, no separate programmers needed. They both have all the software and tools free to download.
Comment: Doing embedded programming is indeed awesome and a lot of fun. Keep in mind though that it's different than writing device drivers. When writing for Arduino, all you're worried about is the Arduino itself. When doing device drivers, a lot of your time and energy is spent interfacing with the kernel and userland on the computer. This is something that Arduino doesn't prepare you for.
Answer: The source code of existing drivers is the obvious one.
Other than that, your best bet is probably reading the technical specification of some devices and trying to write one. I'd suggest a cheap network adapter (if such a thing still exists) as it is quite challenging but you're less likely to screw your entire computer up after a failed attempt. You don't need to write a full driver, a standalone program that can do a couple of things by using low-level magic will do.
The other trail you can follow is to concentrate on the driver-ness and create a real driver for a non-hardware device, like a virtual disk drive.
You can then put the two together. Well, if you still fancy it, because you might easily have had enough at this point.
Answer: Check out the book Linux Device Drivers. I'm reading it at the moment and it seems to be quite good (can't say way too much though, at roughly page 50 now).
Other than that, program, and as the source code is available for existing drivers, use it!
Answer: I know that it may seem obvious at this point, but device driver development and embedded systems are both areas where one has to know something about computer architecture and digital design.  One cannot write a device driver for a new device unless one can read and comprehend data sheets and understand timing diagrams.  Embedded systems is an area where one has to be half computer scientist/half electrical engineer.
Answer: Theirs a few online blogs that talk about people writing USB drivers for "unknown" USB devices, it's pretty cool.
I think when I found them I typed into google: "making a linux usb driver"
Answer: I started because my job required it so I started reading hardwards and OS specs.
Seriously though. I need to write a driver. So I learned the hardware API, then the OS driver specs. Got the sample source code of drivers and (in my case it was the Windows NT DDK)  development kit and started coding and learning.
The trick I found  was to actually write one and hone it to do what I wanted which was to link to a 16bit process. It was tough learning process which I beleive would have be harder if I actually didn't have to produce commercial driver.
In your case I'd definitly recommend getting some programmable hardware like the Arduino perhaps the Rasbery PI and start making the hardware obey your command.
Caveat: I'd already done Assembler for years so the machine interface wasn't new to me.
Answer: the following link gives a brief introduction about developing device drivers in windows platform
http://msdn.microsoft.com/en-us/library/windows/hardware/ff554690(v=vs.85).aspx
but before starting driver development you should have full knowledge about the hardware that your are interfacing with the computer.In embedded systems the main communication channel between two components is established with help of RX/TX ports(RX for receiving, TX for transmitting).This even holds true in case of hardware and a PC.Any hardware connected to PC will continuously communicate with each other.Consider if we send 01 to a computer from hardware.we can program a specific task in driver development.conversely if we want switch an LED in the hardware then we need write a program such that on extrapolating information from computer the hardware should switch the light on.
the above link describes this topic far better than me take a look at it.


How to better manage spam in Gmail Apps

Question: I am a long term Gmail Apps user and use catch-all behaviour in my Gmail account. I have only 1 email address connected and get about 600 spam messages per 30 days, nearly 100% of which are marked as spam automatically. Therefore they don't interfere with the operation of my Inbox.
I also publsih my email address on the Internet widely - social media, mailing lists, forum, live blogging etc and this has not affected the amount of spam at all.
I am also subscribed to a huge amount of notification services and mailing lists, causing maybe around 6000 emails to hit my email and be filed automatically
My first name is German and very unusual for an English speaker - I explain in a moment why I think this may be important.
I use the web interface and "Report Spam" systematically.
I have a client on Gmail apps. They have a brick and mortar business with 12 email addresses and catch-all turned on as well - because Gmail doesn't support mail aliases.
The catch-all account doesn't have the mailing lists and notification services subscribed, but receives around 52000 spam in 30 days.
The client is not throwing his email address around as I am and the few "commercial unrelated" email pieces he receives are from other brick&mortar businesses that he probably has frequented at least once.
However the client has a very common English first name as his email address.
The spam he receives is about 60% bounce notification, ie. caused by the domain name being used as a sender, however rarely does it actually use a common English first name, it is always a random text. About 100 messages per day make it into the Inbox.
The client uses Outlook, so they do not report Spam to Gmail.
Knowing some basic spam filtering techniques, the above comparison is counter-intuitive to me. Is the difference really all down to spam reporting?
A shortcut way to solve this will be to set up a filter that archives all mail that is not addressed to the person or the aliases. Mid term I hope "reporting spam" will solve the problem.
Any other comments anyone can make?
Answer: Responding to :

The client uses Outlook, so they do
  not report Spam to Gmail

.
Gmail's spam filters work the same for a person using the web interface or a client like Outlook. I mean, the spam filters will tag a message as spam / non-spam before sorting it into the appropriate folder.
From this page, moving a message to the [Gmail]/Spam folder performs the same action as clicking the REPORT SPAM button - so there is in fact a method to report spam via an email client.

How to better manage spam in Gmail Apps


How do I create pull quotes in Google Docs?

Question: Can I install an app to make pull quotes in Google Drive? Or can I somehow create them using clever tricks?
Answer: I've done some research and came to the conclusion that it can't be done in Google Docs. This thread in the Google Product Forum shows all kinds of negative energy around this topic.
One user suggested taking a screenshot of a Google Spreadsheet that can be imported into the Google Doc: Jack Kleeman
UPDATE
Here's my own attempt:
I've included the Google-Documents file, with table color set black: Pull Quotes
Comment: Thanks for the insights! As a former engineer for Google Docs, I was already depressed over having to ask this question ... and that list thread made me, well, more depressed. Nice workaround! :D
Answer: There is a very very limited work around for this.
You set your quote in a table in your document by inserting a 1x1 table where you want the quote to go. Set the background of the table and the background of the text to the same color, and adjust the table's column width if necessary.
This screenshot shows what I was going for. It doesn't allow you to align the table to fit inside of another column of text, but it is at least a good way to insert a quote and set it apart from the formatting of the rest of the document.


SP2010 psconfig fails trying to configure farm after installing language packs

Question: I'm using the AutoSPInstaller script. This is what I run:
Start-Process -FilePath $PSConfig -ArgumentList "-cmd upgrade -inplace v2v -passphrase FarmPassPhrase -force -wait" -NoNewWindow -Wait

But I get a bunch of errors:
Performing configuration task 3 of 4
Upgrading SharePoint Products...
Failed to upgrade SharePoint Products.

Total number of configuration settings run: 3
Total number of successful configuration settings: 2
Total number of unsuccessful configuration settings: 1
Successfully stopped the configuration of SharePoint Products.
Configuration of SharePoint Products failed.  Configuration must be performed before you use SharePoint Products.

The PSCDiagnostics Log has the following errors:
03/03/2011 09:35:25  8  ERR            Exception:
Microsoft.SharePoint.Administration.SPUpdatedConcurrencyException: An update conflict has occurred, and you must re-try this action. The object SPUpgradeSession Name=Upgrade-20110303-113635-252 was updated by NA\myID, in the PSCONFIG (3644) process, on machine SHAREPOINT.  View the tracing log for more information about the conflict.
at Microsoft.SharePoint.Administration.SPConfigurationDatabase.StoreObject(SPPersistedObject obj, Boolean storeClassIfNecessary, Boolean ensure)
at Microsoft.SharePoint.Administration.SPConfigurationDatabase.Microsoft.SharePoint.Administration.ISPPersistedStoreProvider.PutObject(SPPersistedObject persistedObject, Boolean ensure)
at Microsoft.SharePoint.Administration.SPPersistedObject.BaseUpdate()
at Microsoft.SharePoint.Upgrade.SPUpgradeSession.Update()
at Microsoft.SharePoint.Upgrade.SPUpgradeSession.ContinueOnLocalThread(Guid id, Boolean consoleOutput)
at Microsoft.SharePoint.Upgrade.SPManager.ContinueSessionOnLocalThread(Guid id, Boolean consoleOutput)
at Microsoft.SharePoint.PostSetupConfiguration.UpgradeTask.Run()
at Microsoft.SharePoint.PostSetupConfiguration.TaskThread.ExecuteTask()
...
03/03/2011 09:35:25  1  ERR      Post setup configuration was not run successfully when using the command line task driver module

Does anybody have any ideas how I can fix this?
Comment: If the problem is AutoSpInstaller relater, you should ask the question on Codeplex not here
Comment: Have you retried running psconfig or viewing the tracing log to see the conflict as per the note in the diagnostic log?
Answer: I used the following commands and managed to successfully run the upgrade
PSCONFIG.EXE -cmd upgrade -inplace b2b -force -cmd applicationcontent -install -cmd installfeatures


Bittering hop contribution to flavor

Question: How much do bittering hops contribute to flavor?  Does a hop addition at the beginning of a 60m boil result in the destruction of most that hop's flavor compounds?
Answer: The 60 minute addition does indeed contribute to flavor.  Its not much but its noticeable.
Especially the higher alpha acid varieties.  I have brewed several single hops beers with a single shot at 60 minutes, and you can definitely pick out some flavor differences between hop varieties.  The beers aren't all simply bitter with no flavor when compared side by side.
However, to really get appreciable flavor additions you still should use some mid boil hops.
I tend to like 10 minutes for max flavor effect.  I don't think you can actually quantify flavor though.
Of course taste and aroma are so closely related mechanically in your nose and mouth its impossible to not have something super dry hopped and thing that its also got a big flavor addition.
I've added a load of hops only at flameout, with a single 60min addition.  The nose is amazing but you'd swear there were mid boil hops added too.
Answer: Bittering hops contribute little to flavor - most of the flavor compounds have been boiled off - or at least, that's the common wisdom. To maintain flavor, hops should be boiled for round 20-30 mins, or dry hopped.
You could do an experiment - boil the same amount of hops in a quantity of water for different lengths of time - 60 mins, 45 mins, 30 mins, 15 mins and 5 mins to see the affect on flavor and aroma.
Although utilization is lower with water, that's not what's important here, and using water rather than wort will be sure your solute has no appreciable flavor or aroma that might mask that from the hops.

Bittering hop contribution to flavor


How do we explain accelerated motion in Newtonian physics and in modern physics?

Question: Maybe my question will seem stupid, but I am not a physicist so I have some problems understanding a classic Newtonian experiment: in the bucket experiment, why does he have to introduce the absolute space in order to explain the accelerated motion of water in the bucket. Also i read that in modern physics (Einstein) the accelerated motion of the water in the bucket is explained by the action of gravitational field. How does this work exactly? Other sources say that the accelerated motion remains a dilemma even for modern physics, if so why is that? Sorry that my question is a bit too broad, but i really have a problem understanding this topic so thank you in advance for your answers.
Comment: Related - http://physics.stackexchange.com/questions/3986/newtons-bucket
Comment: I've removed the 'experimental-physics' tag since the Bucket argument is a Gedankenexperiment and not a real experiment at all.
Answer: Newton's bucket is a thought experiment, not a real experiment (although it is also easy to do). The point of the experiment is that the notion of "rotating" or "not rotating" is not measured by the relative distances between objects, but by the relative motion of objects relative to a space-time structure.
When a bucket is rotating, the distance between any two atoms is the same in Newtonian mechanics, so that the geometrical relationship between any two particles is no different than when it is at rest. But when it is rotating, the water is pushed out by a centrifugal force, and when it is at rest, there is no centrifugal force.
Newton concludes that the centrifugal force is a property of the motion of the bucket relative to a notion of absolute non-accelerated space. This conclusion is already present in the structure of Newton's mechanics, where the accelerations are physical forces, and they are only physical forces relative to an inertial frame--- a description of space where the coordinate axes are given at any time, so that the axes can rotate and move at a constant velocity relative to another inertial frame, but they can't accelerate. This description is not completely circular, because it is asserting the existence of rotation and Galilean symmetry, and Newton's laws of motion have this symmetry.
In Special Relativity
Newton's bucket is clarified somewhat in special relativity. When viewed in space-time, with time considered vertical, the motion of points on a stationary bucket make vertical lines that trace out a cylinder in time. When you rotate the bucket, the trajectories of the atoms of the bucket make spirals in time. Even in Euclid's geomety, spirals are different from straight lines, in that straight lines have no curvature and are minimum distance paths between two points.
In special relativity, the reason that a rotating bucket is different from a non-rotating one is obviously pure geometry. The time-paths of the atoms in the rotating case are curved, while the time-paths of the non-rotating bucket atoms are straight. The relative distances between points are not just the relative distances in space, but in space-time, and these are not invariant under rotation.
So extending the notion of distance to space-time makes the choice of reference frame no different from the choice of coordinate axes in geometry. This removes a lot of the philosophical bite of Newton's bucket, because the only reason you expected things to be the same is because of the three dimensional distances were preserved. When you see that the four-dimensional distances are not preserved, why would you expect that everything must stay the same?
Mach's Principle and General Relativity
But for Einstein, this was still not enough. The issue left is that the notion of geometry, the notion of perpendicularity and distance, is something that is acting on material objects, but is not acted upon. This is inconsistent with Newton's third-law philosophy, anything that acts must in turn be modified by its action.
That this is unnatural was pointed out by Ernst Mach. Put Newton's bucket outside in a field at night. When the bucket is rotating, the distant stars are whirling around the bucket in its frame, while when it is still, the distant stars are still. Why should the two notions of rest coincide?
In other words, Mach is asking why the local notion of "at rest, no rotation" should coincide with the cosmological notion of "at rest, no rotation". By asking this question, he is implicitly allowing the local notion of "no rotation" to vary from point to point, and then he is asking "what dynamical law makes the no-rotation frame at any one point coincide with the global notion determined by far away matter?"
Mach concludes that the matter determined the local frame. He concluded that if Newton made the bucket "Many leagues thick", that there would be some effect between the water and the bucket which would make the water follow the bucket motion, so that the centrifugal force would be relative to the bucket, and not relative to the distant stars.
This idea was uppermost on Einstein's mind when formulating GR. Einstein realized that the observed force of gravity allows for an ambiguity in the notion of rest, since it is impossible to distinguish between acceleration and a local gravitational field. So he introduced a dynamical geometry, whose dynamics would produce gravity, and so on to GR, this story is well known.
Frame Dragging
When GR was finished, Einstein asked again about the bucket. If you place a bucket in space-time, there would be a gravitational analog of a magnetic force from the walls of the rotating bucket on the water inside. When the bucket is rotating relative to the local notion of no-rotation, this force will pull the water outward a tiny amount. The force is consistent with Mach's interpretation, when the bucket is rotating and the water is not rotating along, the gravitomagnetic force pulls the water out into an immeasurably tiny parabola.
If the bucked is made more massive, the effect is enhanced. In the limit that the bucket is about to collapse to a black hole, the exterior decouples, so that the water follows the bucket only. In this way, General Relativity incorporates Mach's principle. For a bucket sufficiently massive, the water's shape will be determined by its motion relative to the bucket, not relative to the distant stars.
This effect is called "Frame dragging", and it is a well known prediction of GR which has either already been observered or is soon to be observed.
Cosmological Mach's Principle
But there is a second notion of Mach's principle, that the motion of the bucket is relative to distant matter. This notion seemed for a long time to be incompatible with GR.
There are bogus arguments in the literature regarding this. One argument is that the Kerr solution is a counter-example to Mach's principle, because it describes a black hole rotating in empty space-time, and what's the black hole rotating relative to?
Of course this is nonsense, since you can scale solutions to become arbitrarily small in classical GR. So if you have any rotating object, rotating relative to distant matter, you can scale the solution to become infinitesimal, and send the objects it is rotating relative to to infinity. Then the object will rotate relative to boundary conditions at infinity.
The real question is for situations where there are no boundary conditions. For Einstein, this meant a closed compact universe, like a sphere. In this case, he spent much time trying to establish that Mach's principle would hold precisely, so that the rotation of objects would be relative to distant objects.
The problem is that he needed a stable spherical universe first! A static spherical universe didn't work as a solution of GR. So Einstein considered a universe with a uniform dust of matter all over, and added a cosmological constant to stabilize the universe, and studied this solution as a model cosmology.
But this cosmology is unstable too. The matter will form black holes, the black holes swallow each other, and eventually, one of the black holes becomes cosmological in size. The biggest black hole turns inside out to become a cosmological horizon, and eventually swallows all the other black holes, leaving only a uniform deSitter space.
The deSitter space is the universe of inflation--- it is constantly spreading out, in a process that can be thought of as matter constantly falling into the cosmological horizon that surrounds any observer. The cosmological horizon stays a certain distance away from any observer, and this is the only stable universe of positive cosmological constant.
In deSitter space, the cosmological horizon is a form of matter--- it is continuously linked to the largest black hole in the Einstein static universe. Further, once deSitter space is empty, this horizon cannot rotate, it can only rotate if there is something in the interior (deSitter space is unique). This means that deSitter spaces, like Einstein static universes, also obey Mach's principle, so long as the notion of matter is expanded to include cosmological horizons.
Modern Answer
The modern answer is that Newton's bucket is rotating mostly relative to the cosmological horizon, and partly relative to the distant stars. If all the stars and the cosmological horizon were rotating, we would rotate with them, so that there would be no centrifugal force (this is difficult to even say in GR, because there is no coordinate independent way to set the distant stars and cosmological horizon in a rotating motion, although one can do it for the stars alone, using Godel's universe, so long as the horizon becomes inconsistent).
This is the modern view of Mach's principle. This point of view is obvious given modern holographic physics, but is not in the literature. It is a complete resolution to the problems raised by Mach's principle, so that there is no mystery left. Mach's principle in this point of view, is just a stunted classical precursor to the holographic principle. It is asserting that all motion is to be measured relative to distant horizons.
Comment: +1 (I love your long'isch answers. I feel there are hardly any physics questions bordering philosophical considerations, which can be answered in a short statement without misunderstandings) What I don't see is how a limiting process, as in boundaries at infinity, can have a meaning regarding relative motion. If they are in such a distance, that any changes of their configuration will never propagate to the position you're actually interested in, then it's like they are not really object at all. Also, I'd love if you write a (blog?) post about the holographic principle. Wikipedia doesn't help.
Answer: Because motion is generally regarded as relative. In other words, you cannot tell whether sth. is moving or not without a reference object. However, since in the bucket experiment the dynamical effect of rotation and no rotation is distinct without any reference object, he introduces absolute space so motion is no longer relative, but absolute.

Also i read that in modern physics (Einstein) the accelerated motion of the water in the bucket is explained by the action of gravitational field.

This is wrong. Despite the name "relativity", Einstein's theory is about the absoluteness of spacetime. Rotation is still absolute; something either rotates or not, without reference.

Other sources say that the accelerated motion remains a dilemma even for modern physics.

I don't know any dilemma in this.

Response to Ron Maimon's answer :
Mach's principle and general relativity
It is true that Mach's principle inspired Einstein to discover his famous general relativity, but Mach's principle is in fact incompatible with general relativity. Mach himself rejects general relativity for it admits absolute spacetime.
The corollary of Mach's principle about the bucket experiment is that the centrifugal force water experiences results from the dragging of distant star swirling around. This poses a serious question: how does the motion of stars light-years away affects the  water instantly?
In addition, the well-known result of Thirring has shown that while rotating shells (as a model of distant stars or a big and massive enough bucket that drags the water with gravitation itself) do have effects the same as centrifugal forces and Coriolis forces, they also exert axial forces with the same order of magnitude, without a fictitious force counterpart.
Fictitious forces and gravitational field
The famous equivalence principle states that fictitious forces and gravitation are indistinguishable locally. This gives people false impression that fictitious forces are actually a product of gravitation. But no, while "true" gravitational field is manifestation of spacetime curvature, fictitious forces are a result of non-zero spatial curvature depending on how you slice spacetime into spatial and temporal part. No matter how you rotate the bucket and choose the coordinate system, a flat spacetime is a flat spacetime, and no gravitation is present.
Holographic principle
Holographic principle is more of a conjecture than an established theory. It is "a property of quantum gravity and string theories" as said by Wikipedia. And it is well-known that none of quantum gravity and string theories have ever been experimentally verified and they are constantly evolving, unlike general relativity. So while it is certainly an interesting way to think about bucket experiment, the holographic perspective is by no means the accepted "modern view".
Comment: "This is wrong", No it isn't! In General Relativity, with a massive enough bucket, Newton's experiment doesn't work. The boundary is when the bucket is just about to collapse to a black hole. As you make the bucket thicker, there is less and less centrifugal force, and the water responds to the bucket more. The dilemma for modern physics is resolved by holography.
Answer: My problem with Mach's Principle -- that the water in the bucket experiences centrifugal force (really its inertial tendence to fly off on a tangent to the center of rotation) due to its angular velocity relative to the mass of the rest of the universe -- is that it takes less force to spin the water-filled bucket than it would to spin the universe itself. Therefore, we can be sure it is the bucket which is spinning and not the universe. Thus, both the bucket and the universe enjoy what Einstein called "priviledged states": one "moving" and the other "at rest".


Measure Snapping with ArcGIS Javascript API

Question: I would like to know if the ArcGIS Javascript API has built in snapping functionality for the measure tool.  I am trying to create a measure tool which only measures distances along the edge of roads.  Is there anything built-in or would I need to create this functionality by linking into ArcObjects?
Comment: Alternatively, what would I need to do to create this functionality?
Answer: Not sure it did when the question was first asked but it sure does now.
Check the sample:
http://help.arcgis.com/en/webapi/javascript/arcgis/jssamples/#sample/widget_measurement
Answer: I think you'll need to write your own custom tool.  I don't see a method in the REST API that returns a point and feature ID that is closest to a specified point.  You could use a query request, but then you'd have to use javascript to loop through the returned features to find the closest point on the closest feature relative to the point where the user clicked.  Personally, I think it would be easier to write a REST based SOE, but only because my C# is better than my javascript.
Answer: Based on my observation of how the Silverlight SDK works (which has snapping), I believe you will have to write your own snapping implementation. I don't think there's anything on the server that will really help you.
You will have to retrieve the geometries associated with the road(s) you want to measure via the FeatureServer. Then on the measure tool, you'll have to enable a mouse listener to search the vertexes of the visible geometries.


Native-style scrolling on mobile devices

Question: What does native scrolling refer to on mobile platforms and tablets ? Does it refer to the inertia feel of a scrolling list ? The ability to pull down the list for refresh ? Or both. I'm having a hard time finding a true answer. And where can I find a list of devices/OS that support these feature ?
Comment: I think that it refers to scrolling with a **swipe finger motion**, as opposed to scrolling with a keyboard or with a scrollbar like on desktop/web apps.
Answer: "Native scrolling" is controlled by the core scrolling engine of the native operating system.
"Native-style scrolling" closely mimics the feel of native scrolling including speed, momentum, bounce-back, response, control, appearance, etc.
"Native scrolling" tends to be used as shorthand for both, hence your confusion.
Update: To answer your last question: every device or operating system that scrolls content has its own native scrolling. In computing systems, "native" refers to the basic, built-in functionality.
Answer: "Native scrolling" simply means using the scrolling mechanisms (APIs) that are built into the OS.
Using JavaScript in a web app you can process swipe events and perform a scroll operation.  If the scroll operation does not call the operating systems scroll APIs to perform the scrolling it will probably not feel exactly like a native app's swipe-scrolling (speed, inertia, etc.). This is not native scrolling, and is undesirable because it doesn't feel like most apps (but it's sometimes unavoidable with the current state of mobile web technology).


How can I redirect a call, incall

Question: I want to get a call from somebody and after talking with him redirect that person to another number, on my charge, like call-centers do. Is that possible and if that's so, there is any application that can do that?
Thanks
Comment: I have never heard of this, the only simple solution is to forward the number and get that person to dial the number. Unfortunately AFAIK Android does not do this!
Answer: I guess you could do this using the "Conference Call" trick on android (natively!).
The screenshots below are for vanilla Froyo (Android 2.2) and may vary with your version/dialer app.
During your call (Don't hang up the call), after talking to that person, press the "Menu" button on your android. Press "Add Call" from the menu.

Dial and call the number you wish to "redirect" for him. (on your charge! (which is what you asked for right?) Means you will lose money for this call as if you are calling it)
Now press menu button again when call connects. Press on "merge call".

Finally, Press the menu button again and click on "mute" to make them think you have terminated your call. (you mute yourself but you can hear the conversation between him and the "redirected" call)

Remember, while doing these, you will be charged for 2 calls simultaneously!
Comment: Just remember that not many networks/carriers support conference calls.
Answer: The feature you're looking for is known as call transfer. It's pretty common on office PBX and VoIP systems, but is virtually never seen with mobile phones. The existence of such a feature would require some intelligence in both the device and cooperation from the carrier's equipment, both of which are generally absent. As far as I know, there is nothing in the Android source code to support this feature.
In short, no, it's not really possible.
Answer: While it may not be exactly what you're looking forward to, Google Voice has something like this.
If a caller calls your google voice number, all the phone you have registered ring at once. So you pick up the call on your home phone, then decide you want to go out. Google Voice lets you press 4, and it starts ringing the other phones again (without the other caller hearing anything). All you do is pick up the cell phone, and your call is transferred seamlessly.
I only post this in case someone is looking for this specific solution. Otherwise, you'll have to host your own PBX (like Asterisk) and set it up yourself. Android (well, specifically the cell carrier) doesn't have this option.

How can I redirect a call, incall


Balance between "right tool for the job" and familiarity

Question: So when choosing what language to use for a project, in an ideal world the language is chosen because it's the right tool for the job.  However, I often prefer to use a language that I am fluent in rather than one I would have to learn or that I am only conversational in.  Of course language fluency also entails knowledge of the applicable libraries in the language.  Just because I really like a fairly general-purpose language like Java doesn't mean I should always use it, but at the same time it doesn't mean I should break out something like Perl every time there's some text processing to be done.  How does one find the balance here?
Answer: Wow that is a VERY hard question when taken out of the world of theory and into the world of production.
In Theory
Simple. Always use the best tool for the job, and just learn what you need to.
In Practice
Not only is there the question of your fluency there are a host of other business questions that need to be asked before you can answer this :

Cost of purchasing the "correct tooling"
Cost of supporting this  - people need to be trained
Cost of learning curve
Integration cost with other products ( now and into the future )
... etc

Outside of the theory there are serious ramifications for your technology choice.
Now I am not saying don't pick the correct tool - just make sure the correct tool will be able to break even on its cost implication.
If this is a personal project - always use the "correct" tool - so when you are faced with this decision in the business context you can make a better informed call.
Comment: It's not 'simple' in theory.  Just what does **best** really *mean*? What are the criteria?
Comment: @whatsisname best is subjective and depends on your environment, budget, time frame ... - but in the spirit of a home project it would be the case of trying out a technology that was design to solve that problem. Eg Erlang for distributed, Perl for text manipulation - then you can make your own judgement.
Comment: +1: When all of the factors have been combined, the _correct_ tool may not end up being the _best_ tool - few people get that, choose the best tool, and suffer the consequences.
Comment: One thing I'm more and more sure of is that Java is mostly not the right tool for the job. There are just so many better alternatives. Don't get me wrong. It was a great language back in the '00 and lots of people got used to it but for me it's not the BEST tool (but still a tool in some cases I use, just not the best) for my job anymore. Not for tinkering, not for big data not for web.
Answer: I think familiarity is not given enough credit. Your familiarity very well influences what the right tool is. You have to use the tool to complete your project. Furthermore, using something you enjoy can without a doubt exceed any shortcomings in it's fitness because you'll be more excited to get work done and you'll have better results.
http://headrush.typepad.com/creating_passionate_users/2006/08/when_the_best_t.html
Answer: This isn't really resolvable except as a business question. However, a lot of business questions are made only looking at short-term numbers, which is a mistake with things like this.
My general approach:

If it's a small or short-term thing, always write it in the familiar tools.
If it's a big, long-term thing, look at the cost-benefit tradeoff of learning a new tool.
If you aren't sure, treat it as a short term thing until you have evidence that it's a long-term thing. Then go and look at the decision again.

Three things to keep in mind as you think about cost and benefit: One, people in a hurry tend to short-change the future. Two, maintenance costs are the lion's share of the costs for any successful system. Three, good developers like learning things, and keeping your developers happy is a good long-term investment.
Answer: Great question! As whatsisname said in his answer, "familiarity is not given enough credit." A different tool, different framework, different language could be a lot better than what you're accustomed to using, and you'd still be a lot less productive with it first time around as you learned the ropes.
I've been working for some years as an ASP.NET developer in digital agencies where we have a mix of big projects, small projects, tight projects, well-padded projects, etc. What we try to do, to expand our skills, is look for "soft targets", smaller projects which don't have painfully tight and hard deadlines, and use them as an opportunity to use new tech which might be superior. .NET 2.0, 3.5, 4.0, ASP.NET MVC, Linq to SQL, Entity Framework - all of them, I've used for the first time in such a project.
If you can take your opportunities like that, then hopefully, you'll be ready with a bigger suite of options to choose the right tool without suffering a lack of familiarity. Just like in Julio's example: they found a target where they could add Ruby to their repertoire, and now going forwards they can choose between Java and Ruby.
But if the deadline is short and solid and the project is important, I'd recommend you stick with the familiar tools. Something different might be better suited, but in projects like that, it's all about risk.
Answer: This depends on a few things:
1. How good you are at learning new languages or tools.
If you are a quick study, the barrier to learning new languages or tools is lower. This gives you the opportunity to add another tool to the toolbox.
2. How language/tool independent you make your work environment.
If your workflow is highly tool dependent, then the barriers to learning different languages are higher. If you are wedded to a particular IDE, then switching languages involves significantly more than just learning a language, since editing text will surely frustrate you.
Someone using vim or emacs does not have this problem. All they have to do is learn the new language.
3. Business reality
Learning new tools/languages takes time. That time has a cost. But that cost has the potential to be an investment that pays back more than the initial expense. Also, an ungainly solution will usually take longer to implement and be harder to maintain. If it's anything larger than a small project, and the tools in my existing toolbox don't seem to suit the problem, I will research which tools DO suit the problem.  I've also invested in an environment to suit a generalist approach, learning to use vim as my chosen editor.
Another thing - what is the shortest distance between two points? If someone else has written something that nearly does what I want to do, it is usually quickest to just modify that to suit my needs.
Answer: If there's a new language you are curious about, and you (and the business) can afford it, why not give yourself a couple of weeks to a month to explore it?
This is how I learned ruby. My coder partner had 7 years experience with Java. I had 11 years of Java experience. Neither one of us knew anything about ruby, only that we wanted to try it.
I convinced him and the rest of the company to try ruby for a month (this was going to be a 6-8 month project). Worst case, we would start over this time using Java.
Luckily after a week we were hooked, so it turned out okay. Perhaps you can try something similar? See if you can build something from scratch in some other language, but making it clear to the business why you're doing it, and at least as importantly, what plan B is in case the experiment fails.
Answer: There is obviously no single answer to this question that will apply to all situations.  But here is an aspect that I don't think has been mentioned yet.  If you are a developer you should also take your own marketability into account.  If language X is chosen for your project, how will that look on your resume?  It may be a good idea to choose a language that you are not familiar with so that you have a reason to learn it, expand your intellectual horizons, and make your skill set more attractive for future employers.
Answer: I'd say that familiarity with a language is also an aspect of "being a right tool for a job". I can't imagine a situation where i.e. Prolog would be a right tool for the job for me given my complete ignorance of the language.
Answer: My own version is to "use the right tool available to me" for the job. Being "available" means that I am able to use it, not just that I can buy/get the compiler and//or runtime.
In almost any real life scenario, given a problem you have very limited time to solve it. I don't believe you can truly learn a new language in a very short time. Learning a language means actually reading books, going through other people's code, understanding how it works and the philosophy behind by it. One could just read a tutorial off the web (which is perfectly good as a starting point), and get hacking. But that would lead to terrible code and you would probably be better off writing better code in a language you know in possibly much shorter time.
Despite the merits of the "right tool for the job" proverb, most popular languages are actually general purpose. They may have strengths in certain areas, and not be as good as others in other areas, but they can get most jobs done. Not knowing a language practically means that tool isn't available to you.
I am not saying just learn one (or even two or three) languages and use them for all projects and don't learn anything else. It is important to learn other languages, to acquire more tools to add into your toolbox. But faced with a problem, it is better to stick with the tools you know rather than add to your worries by using unfamiliar technologies. But keep learning other languages so that the next time the choice is easier.