Instructing your Kids

Question: Can anyone cite any regulation that would prevent my child (not infant) from logging time for a private pilots license (i.e. 12 years from now when they are 16 for a solo, 13 years for a license when they are 17)?
Background: I have children that I would like to start instructing and having them log flight training when they fly with me. I am VFR/IFR and medically current, and have a current FAA flight instructors license, therefore my question doesn't involve my own qualifications. Just if there are any known restrictions against the kid logging flight time. To clarify I don't mean both kids logging time on the same flight, just one kid.
I have found none in the CFRs (e.g. 61).
Comment: So this is a 4 year old you want to instruct? I hope there is regulation against that.
Comment: So when they go for a job interview now they have to explain why it took them 13 years to earn a PPL.
Answer: If your child is going to be around airplanes and flying with you for the next 12 years, they're not going to need to log time from back when they are 4 years old. They will have plenty of logged time later in life.
I would suggest that you pick an age, say at least 8 or 10, where your children can start to fill in log books of their own. If they keep an interest in aviation until that time, getting their own log book that they can fill in themselves will be a big step for them to look forward to.

Instructing your Kids


what does Yahoo Finance Open/Close price of bitcoin mean?

Question: I have a question regarding the open and close price of historical bitcoin data. Do they make sense? The bitcoin market is 24/7 do we really have a close or open price. When does yahoo finance record the open/close price of bitcoin? I know their reference time zone is GMT but when do they calculate the open price and when do they calculate the close price?
Answer: Yahoo uses Coinmarketcap.

Coinmarketcap defines it as such:

The market opens at 12:00 AM (00:00) and closes at 11:59 PM (23:59)
UTC time.
For the Historical Tab of a Coin/Token, we take a snapshot when the
market opens and when the market closes.
Data is collected, recorded, and reported in UTC time unless otherwise
specified.

a quick comparison shows that this is indeed the case.

There is no time difference between Coordinated Universal Time and Greenwich Mean Time. However, GMT is a time zone and UTC is a time standard, meaning no country uses it as a local time.
London uses Greenwich Mean Time (GMT) during standard time and British Summer Time (BST) during Daylight Saving Time (DST), or summer time. Therefore, it will not match UTC all the time.
The choice for UTC is natural, because pretty much everything, starting from Unix Epoch is based on UTC as well.
Whenever you have markets that trade 24/7, you need to decide about some artificial snapshot if you want historical data.  You can find some details on what is frequently done in FX in this question.
Answer: It looks like local London time, not GMT, is used. Open/close is marked at midnight for this 24/7 market.
https://finance.yahoo.com/quote/BTC-USD?p=BTC-USD


How do I completely format my disk?

Question: I want to format because my laptop has stopped connecting to my wireless connection and hopefully this will fix it.
Comment: I would suggest that this is perhaps not the right way to approach the problem if getting you wireless working is the *only* benefit from reinstalling. If it is, I suggest starting another question with details of your wireless hardware (the output of `lspci` is often helpful) and fix it rather than ignoring it.
Answer: Take a look at this: How to clean a hard drive before reinstall
Placed here for your convenience:

If you refer to a fresh reinstall,
  when reinstalling you will find
  several options for the usage of your
  hard disk drive, which includes but
  not limits to:
"Side by Side" install, which will
  install your new OS sharing the hard
  disk drive but without losing your
  current installation. This way, GRUB
  will be updated or installed as
  required.
"Use the entire disk" which will
  physically erase the whole disk in
  order to use all its space for the new
  OS install.
If what you wish is to use the hard
  disk drive for a data tasks such as
  BackUp or other data handling, you can
  achieve that by first erasing the
  whole partition and do it from
  scratch. Formatting the disk will also
  do the trick.
Several tools will help you to achieve
  this task in Linux and other OS's.
In Maverick, under
  "system/administration" you will find
  a "Disk Utility", inside of which you
  will find a disk drive and other
  devices tree on the left and its
  characteristics and tool list in the
  right side, where you can find both
  "Delete Partition" and "Format Drive"
  options.
You can do this from a Live Session.
A screenshot is placed here for your convenience.

Addittionally, take in consideration that some wireless devices has stopped working as a part of an update from Maverick. That's my case with a Compaq Presario V3000 (V3617LA) which wireless isn't working anymore even with all my efforts and all the web/forums tricks.
Good Luck!
Comment: Will formatting it make it start off as if it were the first time I had turned it on?
Comment: Kevin: even if you format the drive, there are ways to bring back the data. Check this: http://askubuntu.com/questions/25311/best-tool-to-recover-removed-files/25364#25364 I mean: the information will remain there up to the time that the physical clusters are used by new data. See this: http://ultimateeditionisrael.freeforums.org/viewtopic.php?f=79&t=543 Even so, yes, you will have a better performance if you completely remove both the partition and format the drive. But this task can be achieved at the very moment of the OS install and will have the same performance.
Answer: Completely reformatting your disk is relatively easy in Ubuntu. You simply need to used the disk manager tool that comes installed with Ubuntu desktop. This website has a quick easy tutorial if you are unsure how to do it. Completely Fromat a Disk in Ubuntu
Comment: Note that the author appears to be a member of the above site/self promotion. Also, whilst this may theoretically answer the question, [it would be preferable](http://meta.stackexchange.com/q/8259) to include the essential parts of the answer here, and provide the link for reference.

How do I completely format my disk?


One common index with separate pdf files in separate directories

Question: Sorry for any mistakes, this is my first time here.
I have about 70 different .pdf files, each in its own directory. Each of course has their own index. What I need is one total index into one separate document.
In case it is relevant, I use MiKTeX and memoir document class.
Thanks in advance, Peter.
Comment: Welcome to TeX.SE :-).
Comment: I think it should be possible to open each of the individual `.ind`  files which has been generated during the compilation of the relevant `.tex` file and read the `\indexentry` as a string and storing all such lines in a single file, but what are the merging rules? If index entry A in file X appears on page 17 and in file Y on page 194, listing both (or more page numbers) is wrong, of course, since the total index does not where the original files came from.
Answer: If you're interested in doing this inelegantly and you have .idx files for these .pdfs, this process will work.

Combine the contents of the .idx files into one file, it won't matter if they are out of order.
Dump this text list into Microsoft word or some other program capable of sorting.
Sort it A-Z using the first field (this, I think, will be an ASCII sort just like in Latex's indexing)
You get a list with a bunch of things such as : \indexentry{contents}{11}
Find and replace \indexentry{ with \\, leaving you with \\ contents}{11}. Include a space after \\; it should be \\_.
Find and replace { with whatever you want as the gap between the citation and the page numbers, e.g. \hfill or \dotfill, leaving you with \\ contents} \hfill 11}. Be sure to include spaces around your \hfill or whatever you choose.
Now, replace all } with nothing; make sure the Replace box is empty. That should leave you with \\ contents \hfill 11.
From here, I think it really depends on your indexing terms, that is, whether you have lots of parents (e.g. \indexentry{parent!parent!contents}{11}) or sortkeys (e.g. \indexentry{sortkey@contents}{11}). I would expect it to still work though. You can still find and replace parent terms with nothing so you go from \\ parent!contents \hfill 11 to \\ contents \hfill 11.

Caveat 1: Before doing step 8, eliminate any numbers or symbols being used as sortkeys for the parent and then re-sort the list. That is, if you have something left at this point like \\ 01@Genesis!0202 @2:2}{33} you should Find and Replace 01@ (with nothing). At this stage, that should produce \\ Genesis!0202 @2:2 \hfill 33.
Caveat 2: You may have to do some manual deletions if you do have something like \indexentry{01@Genesis!0202 @2:2}{33}.  If you have sortkeys that differ from reference to reference (0202 in the example I just gave, rather than 01, since 01 sorts the parent and 0202 sorts the child/subitem) those will require manual deletion.

After you've repeated the Find and Replace to deal with more complicated examples, cut and paste it into your tex editor in an environment with the number of columns you need. (I've never done this, but I'm sure it would be straightforward). Stick a \chapter*{Index} on it and that should be it.

Step 8, and its caveats, would only apply to quite complex indices; I would think regular subject indices don't require that kind of  attention.
Comment: You might be interested to read [mark your inline code](http://meta.tex.stackexchange.com/q/863) I tried to do it for you, but I became confused if you mean the 3 backslashed from the sourcecode of you question or the two ones, which are currently displayed in you question.
Comment: I didn't mark it because I figured since I'm talking about working with the text in MS Word it would be confusing to mark it as if it's code. I see now it could be quite the opposite! I'll edit it now.


What is the intuition behind the unit normal vector being the derivative of the unit tangent vector?

Question: I've seen the math, but... It just doesn't make sense to me. How is the slope going to point perpendicular to the vector that is clearly a straight line not going in that direction?
Answer: Take some point $P_1 = C(s_1)$ on a curve $C$. Let $T_1$ be the unit tangent at this point. Now move a tiny distance along the curve, to get to the point $P_2 = C(s_2)$, and let $T_2$ be the unit tangent at this point. Now draw the vector $V = T_2 - T_1$. If $P_1$ and $P_2$ are very close together, $T_1$ and $T_2$ will be almost equal, and $V$ will be roughly perpendicular to both of them (because the triangle with sides $T_1$ and $T_2$ will be almost isosceles and very skinny). Here's a picture.

But, by the definition of derivative,
$$
N = \frac{dT}{ds} = \lim_{s_2 \to s_1} \frac{T_2 - T_1}{s_2 - s_1}
$$
and this limit will also be roughly perpendicular to $T_1$.
Comment: Can we conclude that every fixed length vector is a curve of a circle or part of a circle?
Answer: The dot product of the unit tangent vector with itself is of course equal to 1.
$$ \mathbf{T} \cdot \mathbf{T} = \|\mathbf{T}\|^2 = 1^2 = 1. $$
Take the derivative of both sides, and remembering the product rule,
$$ \frac{d}{ds} ( \mathbf{T} \cdot \mathbf{T} ) = \frac{d}{ds}(1) $$
$$ \mathbf{T} \cdot \frac{d \mathbf{T}}{ds} + \frac{d \mathbf{T}}{ds} \cdot \mathbf{T} = 0 $$
$$ 2 \mathbf{T} \cdot \frac{d \mathbf{T}}{ds} = 0 $$
or
$$ \mathbf{T} \cdot \frac{d \mathbf{T}}{ds} = 0. $$
That is to say, the derivative of the unit tangent vector is perpendicular to the unit tangent vector, i.e. it's a normal vector.
The essence of a derivative is the approximation of functions by linear equations: $\mathbf{T} (s + \delta s) \approx \mathbf{T} (s) + \delta s \mathbf{T}' (s) $.
Performing this vector addition tip-to-tale, the three vectors $\mathbf{T} (s + \delta s),\space \mathbf{T} (s) \space \text{and} \space \delta s \mathbf{T}' (s)$ form the sides of an isosceles triangle with two legs of length 1 and an arbitrarily small base with length proportional to $\delta s$. As $\delta s$ goes to zero, the angle between the two legs goes to zero, and the base angles go to right angles.
Comment: @David H:  $\Vert T'(s) \Vert \ne 1$ in general; instead, we have the curvature $\kappa = \Vert T'(s) \Vert$.
Comment: See, I understand that. I could go through the math, but I want to be able to visually see or verify this, which I'm having a hard time doing (probably because I might be conceptually wrong in what I think a derivative in this case is).
Comment: @RobertLewis Wow, of course. I must have the dumb.
Comment: @David H:  s'all good, we caught that grouder!
Answer: The unit tangent vector gives the instantaneous velocity.  But unless you go in a straight line forever, you will turn.  Suppose you turn left.  The unit tangent vector still points forward at any given moment, but it is turning left -- its derivative is leftward.  The unit normal points left, to indicate the direction that the tangent is changing.
Comment: @David No. In the case of straight-line motion, the unit tangent vector is a constant vector, so its derivative is the zero vector.
Comment: If you do go in a straight line forever, then what happens? Is our unit normal vector still the derivative? I believe so... So you wouldn't be turning left or right, which means the tangent isn't changing, but what would happen? Is another way to think about this the fact that the unit normal is the double derivative of our vector function, which tends to show the concavity. In this case, the concavity is showing the general trend, which is that the graph is going left or right? I still don't quite see it. If we were to treat our first derivative as a new vector function, that seems to break.
Comment: @DavidH, indeed. Then, the unit tangent vector on non-straight-line motion would definitely not be a constant vector. If we were to treat our unit tangent vector as our vector function of interest, then wouldn't the derivative of it be a tangent vector, which is tangent to it? Or, perhaps, I'm thinking of the tangent vector incorrectly.
Comment: @David See the answer I just submitted. Does that help?
Answer: First of all, the unit normal vector $N$ is in general not the derivative of the unit tangent vector $T$.  What is true is that the derivative of $T$ with respect to the arc length along the curve $s$ is a vector $T'(s)$ normal to $T(s)$ in the sense that $T'(s) \cdot T(s) = 0$, and if $T'(s) \ne 0$, so that $\Vert T'(s) \Vert \ne 0$, we can define the unit normal $N(s)$ by the equation
$N(s) = \kappa^{-1}T'(s), \tag{1}$
where
$\kappa = \Vert T'(s) \Vert \tag{2}$
is the magnitude of $T'(s)$. (1) is then equivalent to
$N(s) = \kappa T'(s), \tag{3}$
the way (1) is usually written; it is one of the Frenet-Serret formulas which are the standard differential geometric means of addressing the properties of curves in $\Bbb R^3$.  It is very important, in utilizing these formulas, to take derivatives with respect to $s$, the arc length of the curve $\gamma(s)$ under consideration.  Otherwise, though we will still have $T'(s) \cdot T(s) = 0$, since $T(s) \cdot T(s)$ is constant, implying $T'(s) \cdot T(s) = \frac{1}{2}(T(s) \cdot T(s))' = 0$, $\kappa$ will not necessarily reflect the true rate of bending of the curve $\gamma(s)$.  This is perhaps most easily seen via the observation that, if $r$ is another parametrization of $\gamma$, then along $\gamma$, $\frac{d}{dr} = \frac{ds}{dr} \frac{d}{ds}$.  This means, for example, that
$\frac{dT}{dr} = \frac{ds}{dr} \frac{dT}{ds} = \frac{ds}{dr} \kappa N; \tag{4}$
if we accept $\Vert \frac{dT}{dr} \Vert$ as the definition of curvature, it's value can vary extensively with the choice of $r$, and its geometrical meaning is obscured.  Taking the curve parameter as arc length $s$ standardizes the definition of $\kappa$ by isolating it from the somewhat arbitrary choices of possible parametrizations of $\gamma$.
In the above, the unit vector $N$ gives the direction of bending of the curve $\gamma$, and the curvature $\kappa$ gives its magnitude.  Intuition can be developed for this, in terms of elementary physics, by imagining one driving in a car at a fairly high but constant rate of speed.  If the vehicle is traveling at say, $100 \,\text{km/hr}$, we can "unitize" this rate by defining a new distance measure, the umpla:  $1 \, \text{umpla} = 100 \, \text{km}$; thus "unit speed" in umpla-units is
$100 \, \text{km/hr}$; quite rapid, actually.  We imagine the car as turning while maintaining this speed; then the car is undergoing an acceleration in a direction perpendicular to its motion; the magnitude of this acceleration is $\kappa$ and its direction is $N$.  Now imagine, if you will, for any $s_0$ a circle of radius $\kappa^{-1}(s_0)$ centered at $\gamma(s_0) + \kappa^{-1}(s_0)N(s_0)$, where we hold $s_0$ fixed.  This circle touches the curve $\gamma(s)$ at $\gamma(s_0)$.  A particle traveling at unit speed on such a circle has velocity vector $T(s_0)$ at $\gamma(s_0)$ and will experience a centripetal acceleration, in the direction $N(s_0)$, given by the classic formula $\frac{v^2}{r}$ with $v = 1$ and $r = \kappa^{-1}(s_0)$, exactly the same as the acceleration $\kappa N$ of $\gamma$ at $s = s_0$; the greater $\kappa(s_0)$ is, smaller $r$ will be, and vice versa; tight turns produce large such accelerations; you can almost feel yourself thrown against the door on a tight right.
Answer: An example from physics: gravity pulls Earth towards the sun, and the earth travels in a (vaguely) circular orbit. So, the attractive force acting on the earth is always a radial vector (pointing towards the center of the orbit), yet the velocity vector that Earth travels with is always tangent to the circle traced out by the orbit.
Force is related to acceleration, and acceleration is the derivative of velocity.
Comment: Ah, and we could generalize this for all curves such that if it isn't perfectly circular, then we are still going towards any one direction at one time (it just so happens that the unit normal vector won't always go towards the origin in a non-perfectly circular orbit... perhaps). Why is this different from derivatives in calc 1 though, where we're used to derivatives explicitly being slope? I feel as though it's odd that we'd come across this.
Comment: Actually, is it the fact that we're dealing with vectors, and vectors could be anywhere we want them to be, so the derivative is still the slope, but that doesn't necessarily have to look anything like what the tangent looks like. They're different functions, related to one another through the slope. In vectors, it just so happens that the slope of our tangent vector will always be orthogonal to the tangent vector. Yes? If not, then our slope of the tangent vector is orthogonal to the original function we are worried about.
Answer: If you have a curve in $\mathbb{R}^3$, you can translate each unit tangent vector to this curve so that its foot point starts at the origin in $\mathbb{R}^3$.
When you view all of these unit tangent vectors with foot points now starting at the origin, the collection of head points produces a curve in $S^2$.
Now, let's say your curve in $\mathbb{R}^3$ is $c(t)$ and that the unit tangent field to your curve is $T(t)$, that is, $$T(t) = \frac{\dot c(t)}{|\dot c(t)|}.$$
Now, the vector $T'(t)$ is found by computing the velocity vector to the curve $T(t)\in S^2$. Since the velocity vector to any curve that lives in $S^2$ must be tangent to $S^2$, we must have $T(t)$ and $T'(t)$ are perpendicular.
Answer: This can be very confusing. A simple explanation is as follows: The (principle) unit Normal vector (dT/dt) is always orthogonal to the curve and, therefore, also to the unit Tangent vector T, and is contained in the vector subspace defined by the two unit vectors T and N. The acceleration of a particle on a curve is partioned between the unit Tangent and the unit Normal vector. Thus, acceleration = kT + mN. For example, a stright line has k=0, and there is  no acceleration in the normal direction. However, the unit Normal vector to the line is still well defined. The point is that the unit Normal vector is part of a coordinate system that may be used to define a component of acceleration.
Comment: "The unit normal vector to the line is still well defined". Really? How?


Creating prime ideals in rings

Question: Are there different ways to create prime ideals in a ring other than taking quotients? I recently came across a construction of a prime ideal in a Noetherian ring $A$ given in the book on Algebraic Number Theory by James Milne as follows. Take $c$ belonging to $A$, where $c$ is not zero and not a unit in $A$. Construct the $A$-module $M = A/(c)$ and choose an $m \neq 0$ in $M$ such that $Ann(m) \subset A$ is maximal (We can do this because A is Noetherian). This particular construction helped in proving a theorem about local, Noetherian and integrally closed rings (that they are PIDs). I wanted to know if you have come across any more such constructions.
Comment: There are a few exercices in the second chapter of Eisenbud's book "Commutative algebra" that show that ideals that are maximal with respect to some properties are prime.
Comment: The construction you mention is called "associated primes", and is one of the basic tools in commutative algebra. I suggest that you look at some introductory book in the subject (e.g. Atiyah-MacDonald).
Comment: There are many proofs in commutative algebra in which an ideal, maximal with respect to some property, is shown to be prime. For a general examination of this phenomenon, see Lam
and Reyes, J. Algebra 319 (2008), no. 7, 3006--3027.
Comment: I'm not sure if it's a construction, exactly, but minimal prime ideals have a similar flavor and are often important.
Answer: I'll turn my comment into an answer. Indeed, there are many proofs in commutative algebra in which an ideal, maximal with respect to some property, is shown to be prime. "Of course the best known and probably most important is Krull's result that an ideal maximal with respect to missing a multiplicatively closed set is prime. Also, an ideal maximal with respect to not being principal, invertible, or finitely generated or an ideal maximal among annihilators of nonzero elements of a module is prime." In the paper listed below, the authors prove a metatheorem, the Prime Ideal Principle. They define the notion of a family of ideals $F$ being Oka or Ako. If $F$ is Oka or Ako, then any ideal maximal with respect to not being in $F$ is prime.
Lam, T. Y.; Reyes, Manuel L. A prime ideal principle in commutative algebra.  J. Algebra  319  (2008),  no. 7, 3006--3027.
Comment: Wow, this seems quite interesting. Thank you for the answer, will go through this paper :)

Creating prime ideals in rings


Help with finding a specific passage in the Talmud

Question: I'm currently involved in a research project about literacy in the Ancient Mediterranean, and I keep finding mentions of a passage in the Talmud referred to as [Ketubot 8:11 32c]. I checked Sefaria and other online Talmuds, but I can't find any that contain a passage with that numbering in Ketubot. Have I missed something fundamental about how the Talmud is ordered?
Answer: It took a bit of detective work but I think it refers to the 8th chapter of Massekhet Ketubot in the Jerusalem Talmud, halacha (law) 11, see the original here.
If I google Ketubot 8:11 32c the way you have it, I find multiple references (e.g., bottom of here) to the verse from Deuteronomy 32:47 which is the one referred in the gemara above (the reference 32 appears in the margin of the gemara).
See the Sefaria version of the gemara here.
Puzzle solved I think.
(I am less clear how the gemara relates to the content of the article from where you took the reference. Not even sure this is the correct reference. Baba Batra 21a seems much more interesting re literacy in the Ancient Mediterranean. You will tell us I guess.)
Comment: The Venice edition of the Yerushalmi came with 4 columns per page. That's what 32c references: page 32 column 3. The Vilna edition used a different pagenation so people nowadays usually stick to Chapter and Halakha like 8:11, but this source was being extra clear I guess. Here's the original back side of page 32 in the Venice edition https://commons.wikimedia.org/w/index.php?title=File:Yerushalmi-Venice-1523-HB45298.pdf&page=359 You'd want the right hand column, which is the third on the page (two on front, two on back).
Comment: Thanks! The part that I need seems to be in 50b, but that put me close enough to find it. One thing I still can't understand is why it would consistently be marked as 32c, even though every Talmudic numbering system I've seen only includes a and b.
Bava Batra 21a is indeed of interest (especially the parts about class sizes, attitudes towards teachers, etc.), but I want to find sources from across multiple tractates, so I still have some digging to do.


Does private key belong to one public key in Bitcoin?

Question: More address belong to one wallet in Copay. I have a 12-word mnemonic to my old Copay wallet, but not the private key. If I would infer the private key of the 12-word could I reach all the public adresses with that private-key and start withdrawal?
Comment: If you were to translate the mnemonic into the seed value it represents & feed that through the correct (I'm guessing bech32) PRNG algorithm, then yes, you could generate all of the private keys to spend the btc in all of the addresses in your Ciopay wallet.
Answer: Each private key corresponds to one public key, but a mnemonic key contains the master private key which describes how each private key is generated.
If you want to withdraw from it, I would use Electrum. It can show you the addresses and you can export individual private keys.
Comment: While each private key *technically* only corresponds to 1 public key, the compressed and uncompressed versions of that pubkey correspond to 2 very different wallet addresses. ... might be worth noting.
Comment: Also, and more importantly, the mnemonic passphrase IS NOT for a "private key," but instead is for a "seed value" that is fed into a PRNG (pseudo-random number generator) that derives a list of private keys. The seed value neither is a private key, nor describes how to generate any. It is simply a "bookmark" that sets the PRNG to start "reading" on the right "page" of a near-infinitely long "book" of potential keys it can generate.
Answer: I don't know about your other question, but the public key and private key always work in a pair.  So one public key and one private key have to work together.  It is a one to one relationship.


What material is used for the PCB traces?

Question: Are the traces of a PCB always made of copper,  we can use another conducting material?
Also, is there is another conducting material that can be used in the cladding layer instead of copper, or do we always use copper clad?
Comment: It's just to satisfy my curiosity. Thank you
Comment: Why would you want anything else? Everything else is either way too expensive or a bad conductor, can't solder to it, or magnetic *or a combination thereof). Or are you just curious? I've not seen anything other than copper clad. I suppose you could use silver or gold for $$$. There's aluminum but you can't solder to it due to its passivation layer. I'm not sure you can easily solder to it even if you plate it. And being able to etch any of these other materials is a different matter entirely.
Comment: Sometimes PCB traces may appear to be another material, as they are coated after etching with a less readily oxidised metal - solder, nickel or even gold. But they are still copper under the plating.
Answer: There are also "aluminum trace PCBs" (to differentiate from "aluminum base PCB", which use aluminum as a heat-sinking backplane) but they are more expensive than copper PCBs.
It serves a very niche market: superconducting traces for cryogenic electronics, transparent PCBs for X-ray systems, PCBs with long radiation length materials for nuclear beam applications.
Pro: very good for aluminum wire bonding.
Con: it is difficult (although possible) to solder to aluminum using fluxes based on fluoroborate. However, it is also possible to have copper plated into the aluminum to make it easily solderable.
Answer: For extreme low-cost low-power applications as pocket calculators carbon traces printed on plastic sheet are a thing. They also function as keyboard contacts.
The high resistance of carbon isn't a problem as the currents needed for keyboard matrices and LCDs are in the 1 µA range.
Answer: The PCB process generally assumes copper conductors.
This is somewhat by definition because all the equipment is designed around a copper etching and plating process.
And the technology - the common set of equipment, standards, practices, and knowledge - also generally assumes copper process.
This is the "technology brand" known as printed circuit boards aka PCBs, for now at least.
The reasons for this are described in the comments, copper conductors are superior for many applications.
That being said, other conductors can be printed/plated/etched and otherwise attached to various substrates.
For example, here is a silver plated process with ENIG finish on ceramic substrate produced by Remtec
They certainly look like PCBs to me, but the manufacturer does not refer to them as such.
Answer: I have seen flexible printed circuit heaters that used nichrome instead of copper. However, these are pretty specialized and are designed for a single purpose: to act as electrical resistance heating elements. They are common, but specialized in the sense that they are only really good for one purpose: heating.


Can I rename my craft/ folder?

Question: I have one site on my server that's currently using Craft, and I'd like to setup a second one to use Craft as well. Can I rename the craft folder, and then set the $craftPath variable to point to it?
Answer: Absolutely! In fact, you've outlined the steps needed to do so...

Rename craft folder
Change $craftPath value in index.php

That's it!
Comment: Thanks Lindsey - I thought I could, but it telling me it couldn't connect. I didn't read the error message all the way through (one of my big faults, I just automatically assume...), and it was having a problem connecting to the database. Typed in my user name wrong.  Whoops.

Can I rename my craft/ folder?


How to display server answer waiting with a minimal visual difference to the interface

Question: We have users complain to a visual difference between normal application view and waiting for a server response (that happens, actually, quite often, after pressing the button, changing the focus in the data grid, creating a new row, deleting a row, etc.). Time waiting is not so huge, up to 3 sec in most cases. The request is to avoid the visual difference (as far as this loading state appears quite often, it makes users irritated) but at the same time - to prevent any data from being changed (as while waiting the server respond, all application is not available for interactions). Can someone assume if it's possible - to lock data changing + not to have this overlay with a spinner? (The example shows the current realization)
Answer: Usually if data is loaded inside the grid (async, RXJS), it should be still valid to edit the filters (just keep in mind to have some debouncing).

Try to focus the visual difference just on the part, which will be updated, in the given situation it's the grid itself.
Also try to look for telerik's kendo UI grid, or syncfusion data-grid. They do have online samples, you could show your clients, so in the end they could decide the final look.
Consider speeding up gird - under 3s is slow (despite being industrial standard);  response below 1sec is acceptable, around 300ms is great. Try use paging or if the data is not too much, load it into memory!
Answer: You should limit yourself to only protecting fields and controls that actually matter, and don't try to shortcut the work by having a blanket mask.
For example, your "search" filter doesn't matter if the user changes something. It's not live data. So there is no need to disabled this area of the application.
If you are refreshing the grid of results then you should clear the grid initially and then show the loading graphic over the grid control only.
With regards to the "Search" buttons, etc. you can decide how they should behave based on what you want to happen. For example, you might want to disable the "Search" button while a search is in progress, and then re-enable it once the search is complete.


Why are Navier-Stokes equations needed?

Question: Can't we picture air or water molecules individually? Then, why are Navier-Stokes equations needed, after all? Can't we just aggregate individual ones? Or is it computationally difficult, or inefficient to generate the picture of whole flow?
Comment: It is unclear to me what you mean by Navier-Stokes "theorem". If by that you mean the [Navier-Stokes _equations_](http://en.wikipedia.org/wiki/Navier%E2%80%93Stokes_equations), which describe hydrodynamics well in appropriate regimes, then these are not theorems but rather physically motivated equations of motion whose solutions describe well physical flows in appropriate regimes. While a direct computation of a flow from first principles, molecule-by-molecule, is indeed possible in principle, the sheer number of bodies makes it unthinkable for any macroscopic scale.
Answer: Consider a standard volume of $1\textrm{ m}^3$ of air. This contains on the order of $10^{25}$ molecules of O2 and N2.
If you needed to simulate or explain the physics occurring in that volume of air, would you want to model $10^{25}$ molecules and all the interactions between them or, say, 100x100x100 cells based on the Navier-Stokes equations?
Theoretically, it is possible to simulate every fluid flow ever by tracking every single molecule. But direct simulation of turbulence using the Eulerian Navier-Stokes equations requires $Re^{9/4}$ grid points and is thus totally impractical for Reynolds numbers larger than a few thousand. So simulating something with $10^{25}$ things to track is completely impossible.
Comment: Just recording the position coordinates as doubles would take 1000000000000000 GB of RAM.
Comment: A quibble, but a cubic metre contains more than a mole (6.023e23 molecules). At STP a mole of an ideal gas is 22.4 litres.
Comment: I don't think numerical simulation was the reason Navier and Stokes derived their equations.
Comment: Just to add to this correct answer: even if you *could* simulate every molecule, you would gain little by it unless for some reason you knew the initial position of every molecule to start with. Even then, it would only help you to predict the future behaviour of that particular volume of fluid at that particular time - it wouldn't tell you anything about what would happen if you repeated the same experiment again, but with different molecules that weren't in exactly the same starting places.
Comment: @Bernhard No, numerical simulation was not the reason the equations were derived. But, they were also derived at a time when kinetic theory was non-existent, and so the derivation reflected the Eulerian description of the problem they could observe. The question asked why use NS over a molecular approach, which whether you are simulating or trying to analyze the equations, it's quite difficult to handle the number of molecules that would need to be tracked to do anything on the macroscopic scale.
Comment: @Skilvvz It wasn't so much about turbulence as it was a domain size for something of practical interest, that people can relate to conceptually. Turbulence was a second example to highlight that NS can be hard enough, Boltzmann would be even harder.
Comment: You're right, it's been a long time since I had to think about those relationships. I'll fix it
Comment: Why would we need that much air to have turbulence? You should mention the minimum amount where NS equations are relevant.
Answer: As apparent from the other answers, the continuum approach of the Navier-Stokes equations makes life easier, by reducing the degrees of freedom. For example, Poisseuile flow (laminar flow between flat plates or in a pipe), is easily solved using the Navier-Stokes equations, but I have no clue how to solve it using just molecular behavior (on the back of an envelope that is).
If you calculate the Knudsen number of a specific problem, which is just the molecular mean free path divided by some macroscopic length scale, you can see if the continuum approach is a valid for the considered system.
Answer: Apart from the experimental and numerical difficulties to track each particle in a macroscopic piece of matter, there are physical reasons.
Theoretically, we can show that in the hydrodynamic regime the exact equations of motion for the collection of atoms/molecules reduce to the Navier-Stokes equations. This is shown in statistical mechanics.
Due to complex molecular effects, including molecular chaos, most degrees of freedom self-cancel and only some few dynamical modes survive. Those modes are collective --i.e. they describe the fluid as a whole-- and robust --i.e., they survive in macroscopic scales of space and time--.
That is, even if you were to trace the individual motion of each particle and next average to describe the collective motion of the whole fluid, you would see that your description is practically indistinguishable [*] from that given by solving the Navier-Stokes equations.
[*] The differences are of the order of the inverse of the size of the system and vanish for a macroscopic system.
Comment: "Theoretically, we can show that in the hydrodynamic ..." I believe this paragraph is wrong at the current time. I thought you can derive NS from the Boltzmann equation, so techinically you can derive NS to a certain extent only for gases, but not for liquids. Or do you know statistical NS derivation for liquids?
Comment: @Yrogirg: There is a kinetic theory for dense fluids as well, but is not of the Boltzmann type, of course. Moreover, you can obtain the hydrodynamic equations without using any kinetic equation, for instance using the Mori formalism for the microscopic dynamical variables.

Why are Navier-Stokes equations needed?


What does the switch inside the fire do?

Question: There's a fire in a room and I used my mega douser to put out the fire, I went inside and pressed the action button and I heard something change inside the dungeon, can somebody tell me what it did ?
Answer: Congratulations, you've unlocked the secret room that will take you to the Oubliette.  It's a secret level between the first and second floors.  Look for a new room on your map.
If you can't find a weapon that fires something wet, you can also roll a water barrel into the room to put the fire out.  This takes a bit more work, but you can import a barrel from another room if you're so inclined.
It takes two keys to unlock the trap door down to the Oubliette, so I hope you saved up!  If not, there's always next time...
Do note that falling down the trap door means you'll leave this level completely, so if you want to, say, kill the boss first for a chance at guns/hegemony credits, do so beforehand.
Comment: Just to add on to the answer, as of version 1.0.8 there will always be a water barrel in the fireplace room if you've opened the secret passage on previous runs. (You still have to roll it over to the fireplace though, so be careful while fighting in this room to prevent it from accidentally popping.)
Comment: Does this button have to be hit each run, or just once total?
Comment: @Powerlord, you've got to put out the fire and hit the switch every run.
Comment: Yeah, I found this out when I tried it. So, of course the one time I opened it, I ended up dying against the Chamber 1 bosses because I kept mistiming dodges while trying to charge a very slow charging gun.


What is the gender of an aircraft?

Question: I know that a ship is always referred using "she", but what about aircraft? What should we use when we're referring to aircraft? Is it the same for automobile?
Comment: Lipis, no. Ships are not *always* referred to as *she*. It's traditional for people who work with them but, in general, everyday English, that's rare.
Comment: It depends. You need to look under the rear luggage bay to know for sure.
Comment: I Googled "she car" to get [this](http://www.phrases.org.uk/bulletin_board/18/messages/78.html).
Comment: For cars it's been my observation that men tend to choose female names and women choose male names.  My SIL called her old pickup "Fred".
Comment: If you listen to Tracey Curtis-Taylor who is a pilot, you'll hear her telling the journalist that she calls her plane "he".
http://www.bbc.co.uk/programmes/b07fg1x4 listen to the discussion about the gender of the plane starting @ minute 02:33 at the beginning of the programme.
Answer: The most common way to refer to an airplane is it.
If one were to decide to infer gender onto an aircraft, female would be the obvious choice for airships (eg Zeppelins), and by extension probably the logical one for smaller aircraft like planes.
However, I would advise against doing any such silly thing. Getting rid of gender for inanimate objects is one of the main features of English I like.
For automobiles, there really isn't a standard. The most common usage is again "it". My mom had an odd tendency to name her cars, but they were generally neutral names like: "The Blue Blump". Probably the most popular personified cars in media are Herbie the Love Bug (male), Christine (female), and the Cars cartoon series (both).

Not gonna rewrite this more than a decade later, but it got poked, and on a reread, I realized I do know of some examples for aircraft. The two planes that dropped the first 2 atomic bombs were named "Enola Gay" (female) and "Boxscar" (neuter). The Wright Brothers' first plane was "The Spirit of St. Louis" (neuter). There's a list of B-17 bomber names here. Almost all are female, several are neuter, but I do see one "King Bee".
Comment: Funny, it is something that I miss when speaking English. :)
Comment: Looking at news sources, both general and aviation-specific, "it" is almost always used. Example: [CNN](https://edition.cnn.com/travel/article/inside-boeing-new-777x/index.html): "One of the most noticeable things about the experimental plane, aside from the fact that it's filled with monitoring equipment, is just how huge it is."
Comment: I agree! But it's something that men (well, many of them anyway) love to do, unfortunately there's no way to stop them...
Comment: "*There's a lady who's sure. All that glitters is gold. And she's buying a stairway to heaven...*" ~ Led Zeppelin, "[Stairway to Heaven](http://www.youtube.com/watch?v=w9TGj2jrJk8)".
Comment: I'll wait few more days before deciding on which one should be the correct answer :)
Comment: @StuartF - As the pronoun, I'd agree. However, *if* the plane is named, and the name is being used in the conversation as well, one would imagine the proper pronoun for the gender of the name could be used. FWIW, the Enola Gay website seems to be using "it" when a pronoun is used though.
Answer: Aircraft are usually referred to as "she", just like ships. Automobiles follow the same principle.
Comment: I think you mean *usually referred to as "she"* rather than "he". I'm be pretty sure that aircraft are much less likely than ships to be assigned any gender at all. Just by the by, I personally find that plural *aircrafts* rather odd here - I think I might follow it up with a question!
Comment: You're right, my "aircrafts" is ungrammatical. Slipped. Sorry!
Comment: I asked anyway, because there's definitely increasing usage where it's **not** unintentional.
Comment: FumbleFingers, you made a good point. Vehicles of any kind, do not need to have a gender, in English. Because of that, if a gender is chosen, it doesn't have to be one or the other.
Comment: By whom are they referred to as "she"? All the websites I've looked at from news to discussion forums show "it" is generally used.
Answer: As a pilot, the aircraft is definitely a she. Look after her, keep her in balance and she'll reward you; she'll work with you to get you there smoothly and comfortably and she'll becomes an extension of your hands.
Each aircraft has a personality of her own; it's easy to personify different idiosyncrasies as different personality types. A slight miss on throttle-forward might be an emphasyemic cough, an out-of-balance rudder might be a wooden leg; these things become what you love in an a/c.
Answer: If an aircraft is given a gender, it is normally female, along the same lines as for ships and for much the same reasons (complex creatures with very fickle attitudes toward their men, that are nonetheless objects of affection).
The anthropomorphism of aircraft is rarer than for ships, but still very common especially in the military, where bombers and even fighter aircraft have traditionally been given a name and "nose art" for good luck (more commonly as simply something for the lonely pilots and ground crew to ogle). The bigger the plane, the more likely it is to be anthropomorphized. Race cars are also anthropomorphized by their drivers from time to time.
Comment: Right; engender - to cause to exist or develop. We are creating and/or assigning human qualities such as gender where there are none. However, "anthropomorphizing", though more of a mouthful, is also more exact and less ambiguous.
Comment: http://www.merriam-webster.com/dictionary/engender
Answer: Things that contain people are feminine. Ships, cars, aircraft, countries, cities.
Of course "it" is always at least as correct.
Comment: Interesting concept!  +1 for the generalization.  But you got to be careful extending it.  Under this rule, a house is a "she".  Not sure that works.
Comment: Cyberherbalist, that's a good point. Deciding a gender for something that is not living, is unnecessary in English and just becomes ridiculous and bizarre.
Answer: As I read the question: 'WHAT IS the gender of an aircraft? What should we use when we're referring to aircraft?...', NOT, '...the question was basically whether one should refer to an aircraft as "she" as we do a ship...'. To continue, I think 'nose art' had a lot to do with it during WWII, where 'airships' often had female artwork (and attendant names) directly under the pilot's side of the fuselage as an 'affectionate' anthropomorphism to remind the crew of females they'd left behind, dreamed of, or had the characteristics of women within a plane's often sensitive controls during take-off, cruise and landing. Having said that, there were also not a few aircraft with male (often chauvinistic, comical or downright dirty) artwork and names (depending on their necessarily rigorous maneuverability). Still, I think it comes back to the finesse of the 'delicate handling' of aircraft. They are sensitive to the touch, respond with only a modicum of manipulation during flight, and with just the right amount of direction, bring her crew and passengers into a comfortable repose whether her '_heels' are lifted-up or placed gently on the ground after an exhilarating trip - alleviating pent-up stress and with a great deal of satisfaction.
~N'est ce pas?
Comment: Great detail, but the question was basically *whether* one should refer to an aircraft as "she" as we do a ship, and you haven't quite given that answer! Could you edit to include it?

What is the gender of an aircraft?


Fit a equilateral triangle on three arbitrary parallel lines with an edge and compass

Question: How can you fit a equilateral triangle on three arbitrary parallel lines with an edge and compass?
Comment: IIRC this was a Math Olympiad problem (IMO or USAMO) from the 1970s or 1980s. I can remember working on it as a practice problem when I was a senior in high school in spring 1991.
Answer: Well, I'm late to this, but I've been obsessing over this far too much to back out now.  So: Label your parallel lines $a$, $b$, and $c$ from bottom to top.  Construct a line $d$ with a "positive slope" that crosses $b$ such that the top right angle of their intersection is $60^\circ$.  Extend the line so that is crosses $c$.  Call the intersection of $d$ and $c$ point $A$.  Construct another line $e$ such that it lies to the left of $d$, is parallel to $d$, and is the same distance from $d$ as $a$ is from $b$.  Call the intersection of $e$ and $b$ point $B$.  Draw an arc centered at $B$ and having radius $AB$, and crossing $a$.  Call the intersection of the arc with $a$ point $C$.  $\triangle ABC$ is your triangle.

Why does this work?  Here's a hint: Start with an equilateral triangle $\triangle ABC$.  Draw any line through $A$, and then draw a parallel line through $B$.  Now identify the center $O$ of $\triangle ABC$.  You have now have a triangle and two lines.  Rotate the entire configuration $120^\circ$ around $O$.  What do you get?  Rotate another $120^\circ$.  What do you get now?
Comment: Anyway it's a nice solution.
Comment: Beautiful. Both answers are good, but yours is better explained (with scaffolding attached), so I have accepted this one. Thank you !
Comment: Thanks!  BTW, a variation would be to construct a third line $f$, parallel to $d$ and the same distance from $d$ that $b$ is from $c$.  The intersection with $a$ would give point $C$.  This might be conceptually simpler, but requires much more work than just drawing $\angle ABC$.
Comment: "Construct a line dd with a "positive slope" that crosses bb such that the top right angle of their intersection is 60∘" -- this isn't a valid compass and straightedge construction, at least not directly. (though you can if you break it down into a few more steps) Neither is constructing a line "e" the same distance from "d" as "a" is from "b".
Comment: @JasonS  I skipped details to keep my post from getting longwinded.  Both constructions can be accomplished with compass and straightedge.
Comment: @JasonS  Thanks!
Answer: I've found a different invariant that solves this problem: two circles bisecting a perpendicular between any two parallels cut each other at points that are exact middles of the side opposite to the vertex on the third parallel.
If you want more explanations and sketches please go to:
http://romanyandronov.elementfx.com/pse/ryapserac03.html
Sorry for providing the link - it's a small part of a bigger thing. In my articles I'm interested not in the answers themselves but rather in the ways one can find them. You may find other things of interest there.
This is an eleven-step construction, but I think that number can be reduced.
Outline: random point on any parallel, perpendicular through it, two circles bisecting any segment between any two lines, line through their intersection point and a point on the remaining parallel, perpendicular to that last line locates two remaining vertexes:
Comment: Your website is absolutely cool!
Answer: For brevity, I will not go into details about how to construct perpendiculars and 30-degree angles using straightedge and compass. Same for doubling lengths. Translating those operations to compass-and-straightedge primitives is left to the reader. Given the three parallel lines and a vertex C arbitrarily fixed on one of those lines, I will find the two locations of another vertex, named E and E', that are suited for the fit, and I will not lose words about the rest because that should be obvious.
In C, construct a line perpendicular to the three parallel lines, and name the new intersection points A and B. Now have a look at the attached figure, which is symmetric with respect to the perpendicular line AC, and assume that all filled triangles are equilateral. In this demonstration, |AC| >= |AB| >= |BC| for tidyness, but it turns out that algebraically there is no such restriction for the scheme to work.

Rotate the kite CEFE' 60 degress around E to find that it is congruent to DEE'G. Therefore |DG|=|DE|=|DC|, so |CG|=2|CB|. Using this you can locate G and then branch off CG in G at angles of +/-30 degrees, intersecting with the parallel line through A to find E resp. E'.
Comment: @CravingSpirit: How I came up with the solution? Brace yourself. I actually calculated a suitable rotation angle (ECA in the attached figure) using trigonometry. Then I tried to interpret the result geometrically, which led me to define the point G by |CG| = 2|CB| and assert that the angle EGA is 30 degrees, in other words, that the triangle GEE' must be equilateral. This suggested that a simple geometric proof was within reach, and I found it via congruence resp. rotation by 60 degrees. You see, the first way found is rarely presentable. The figure has been made with Geogebra.
Comment: Could you please explain a bit how you came up with the solution (i.e the scaffolding) ? Thanks!
Comment: What do you mean a suitable rotation angle, and how did you calculate it? Thanks~
Answer: My previous construction took 11 steps to accomplish. Here's an 8-step one. My reasoning is here:
http://romanyandronov.elementfx.com/pse/ryapserac04.html
Invariant this time - an equilateral triangle built on the outer parallels locates two vertexes one of which is on the inner parallel:

In the sketch above triangle AGH locates A and C. That observation lead me to the following 8-step construction (the construction lines are numbered):

Algebraically (C() is circle, L() is line):
$$O_1, O_2 \in L_1$$
$$C(O_1, O_1O_2)$$
$$C(O_2, O_2O_1) \cap C(O_1, O_1O_2) = A, A_1$$
$$L(O_1, A) \cap L_2 = B, L_3 = C$$
$$C(O_1, O_1C) \cap L_1 = D$$
$$C(B, BD) \cap L_3 = E$$
$$L(B, D)$$
$$L(D, E)$$
$$L(E, B)$$
Answer: I'm not going to go into the details of a ruler-and-compass construction, but I'll indicate the main idea, which is very simple.
Referring to the figure in the question, $CDE$ being equilateral is equivalent to saying that $D$ is obtained by rotating $C$ by an angle of $60^{\circ}$ about $E$.
Therefore, start by picking an arbitrary point $E$ on the third line. A suitable point $D$ can (and must) be obtained by rotating the first line by $60^{\circ}$ about $E$ and letting $D$ be the intersection of the second line with the rotated line. You get two triangles for a given $E$ this way, one by rotating clockwise, the other counterclockwise.
Comment: This is beautiful! A true aha moment, and much simpler than the other solutions.
Answer: Here is a simple 10-step construction. Looking at the diagram given in the OP, denote the uppermost line by $l$, the middle line by $m$, and the lowermost line by $n$.
On $l$ choose a point $A$ and construct a line $t_1$ through $A$ which forms a $60$ degree angle with $l$ and has negative slope. Let $t_1$ intersect $m$ at $B$. Through $B$ construct a line $t_2$ which forms a $60$ degree angle with $m$ and has positive slope. Let $t_2$ intersect $n$ at $C$. Using a compass mark point $D$ on $m$ such that $AD=AC$ and $D$ is on the opposite side of $AC$ as $B$. I claim triangle $ACD$ is equilateral.
Note that $\angle ABC=120$, so it suffices to prove that quadrilateral $ABCD$ is cyclic. We will do this by showing that $\angle ADB=\angle ACB$. By Law of Sines on triangle $ABD$, we have $AB/AD=\sin(\angle ADB)/\sin(60)$. By Law of Sines on triangle $ABC$, we have $\sin(\angle ACB)=(AB/AC)\sin(120)$. But $AC=AD$ and $\sin (120)=\sin(60)$, so combining the equations give $\sin(\angle ADB)=\sin(\angle ACB)$. $\angle ADB$ is less than $120$ and $\angle ACB$ is less than 60, so their sum is less than $180$. Thus $\angle ADB=\angle ACB$, as desired.
Sorry for the lack of a diagram.
Answer: Here is a very simple construction:
Suppose the parallel lines are called a,b,c. A transversal crosses a at A, b at B and c at C. Let XYZ be an arbitrary equilateral triangle. Find point E on XY so that the ratio EX over EY is the same as BA over BC. From X and Y draw lines parallel to ZE. This is almost the answer. Now, we only need to proportionally blow it up or down.
Comment: This is an question which already has a well-accepted answer. You are not contributing anything new.
Answer: Given 3 parallel lines A B C (top to bottom), construct a large equilateral triangle between lines A and C where the bottom triangle leg is on line C and the top triangle vertex is on line A. Mark the bottom right point of the triangle point c1.  Also mark the intersecting points of lines B and the left leg of the triangle point as b1.  Set the compass point on point c1 and the compass lead on point b1.  Now draw an arc that intersects line A, at new point a1. Use the straight edge and draw the new equilateral triangle between points b1, c1, and a1.
I figured this out in 1965 and made a proof to confirm validity, but have long forgotten the proof.
Regards, Ray McKaig


What does it feel like to ride a mtb that has a bent frame

Question: What are the signs and symptoms of a bent mtb frame? Or bent forks? What does it feel like when riding it?
Comment: If you post pictures of your bike, especially any part that might be bent, we might be able to confirm and offer solutions
Comment: I thought I had somehow badly buckled my rear wheel.  The bike was wobbling significantly and pedaling felt spongy.  Turns out my frame was broken.  The point is if you think it's bent you should check for that by examining the frame not going by how it feels to ride.
Comment: Same experience as @EricNolan with a broken frame at night - so I didn't notice the frame was broken in front of me and I suspected that the problem was in the front wheel, the fork or the stem. However that wobbling started when one frame tube was already broken, so not actually an early warning sign. At most it warned that a few seconds later a second tube would broke and I would end with two separated half bicycles (and fortunately no worse consequences).
Answer: Don't count on any warning signs.
A few possible signs of impending frame/fork failure or a bent frame which are all pretty non-specific:

Creaking
Soft feeling when pedaling or going over bumps (especially if the rear is damaged)
Strange steering behavior if a bent/cracked fork or frame changes the steering angle slightly
Veering to one side
Shifting not working correctly or it changes gear when putting a lot of force on the pedals because the frame is bending/twisting and changing effective cable length

It's a good idea to look over your frame and fork occasionally and check for any cracks or bends.
Comment: Can confirm that a frame close to failure in the rear triangle can change gears on you unexpectedly
Comment: Yes when I'm pedaling hard there's a wierd grinding crunching like sensation as though my pedals are attached to a mill in sandy gravel. Then my gears in the rear start to slip. My back tire seems to want to slip when riding on dirt, gravel or anything other than smooth concrete even though its a mtb. Plus I've gone through 3 rear tires. They're straight and true when put on bike but before long they wobble until its so severe I have to replace it once again
Comment: @Alison: This could be many things, not necessarily a damaged frame. Maybe it's just a combination of worn pedal or crankset bearings and the wheel being out of true. If you have rear suspension it could also be worn bushings.
Answer: We can infer from Jobst Brandt's article that the bicycle will not track straight and will always veer off to one side if ridden no-hands if frame or fork are out of alignmnt.
See: https://sheldonbrown.com/brandt/align-fork.html
See: https://yarchive.net/bike/frame_bending.html
His repair suggestions are for steel bikes and shouldn't be attempted on alu, carbon or titanium.
Answer: FRAME
I broke a bike (several bikes) at the seatpost junction.

This started as an odd-noise - an occasional tappy sound, and a general lack of speed where everything felt sluggish, like there was brake drag.  Then I felt a clonk and the bike settled - I thought the seatpost was slipping down.  This resulted in a long walk to work.

FORKS I had an 80s road bike which turned out to have bent forks.  It wasn't visible unless you laid the fork crown flat and the measured the height of the dropouts from a flat reference surface.  The rim brakes were about centered when assembled.
The symptom was if the bike ever hit a bump like a stone, stick, crack or crossing a ramp then the bike felt like it had "stepped to the right"   If it was a MTB you'd wonder if the front had slipped off a rock, requiring some effort to hold your balance.  Not ideal if pedalling at full road speed, and I got that thing up into the 40 km/h range.
I did repeated checks looking for cracks and other issues like this, but the only way it was diagnosed was that one dry day I rode through a shallow puddle then onto dry road.  I glanced back (perhaps I was about to turn across traffic?)  and noticed that the bike was leaving two distinct wheel tracks of water on the dry road.
Upshot was that under a sudden load, the fork acted like a spring, loading up one side and throwing the front of the bike to the right.
You can just make out that the for tines are bent back (relative to the headtube) in this photo, and that the wheel is turned slightly less than the fork tines/legs.

Also notice the rust on the top tube and down tube, around 2 inches/50mm aft of the head tube.  This is characteristic of a frontal impact that has tweaked both tubes and allowed moisture to the steel underneath.

Here's another bike with a frontal impact.  This one was straight, but pushed back.
This one rode correctly, but it actively wanted to turn to either side.  So it required the rider to keep both arms in tension, holding the wheel in the middle.  None of this casual self-centering or hands-free riding.  If you let go, it would throw you at any speed.

SEATPOSTS This happened at least three times, always due to a hard landing and far too much seatpost stickout.  It feels like the saddle clamp has failed and the saddle has rotated so the nose is upward.

MEASURING Bicycle frames are symmetrical.  So you can use a string pulled taught from one rear dropout, up and around the head tube, and back down to the other rear dropout.  As long as the string isn't fouling cables or chain, then the horizontal gap to the frame should be the same on both sides of the bike, within a millimetre is fine.
Suspension can also throw off this measurement, if the bushings, bearings, pivots are worn/sloppy.  Any joint should have rotational movement, but not sideways movement.  Is it possible you're feeling slop in suspension?
Comment: I was expecting an answer from you on this one ;)
Comment: I actually disagree with the claim that bicycle frames are symmetrical. Sometimes, they are not, like in Cannondale Ai.
Comment: @MaplePanda I seem uniquely qualified to talk about broken bikes don't I :)  MTB experience is a bit light though.
Comment: @Criggie That would be Cannondale's AI offset. The owner is not guaranteed to know about it—it's only 6mm so not blatantly obvious from inspection, and shops are unlikely to mention it during the sale.
Comment: @juhist true - the right hand chainstay is frequently dented for clearance from the chainrings.  Sometimes the stay takes a different path, higher or lower. I believe there's also a rear wheel standard for MTB where the right hand dropout is further outboard from centerline than the left dropout, but its pretty rare and unusual and the owner would know that.  The vast bulk of frames display symmetry across the centerline.
Answer: The front tire is very close to the down tube, or may even touch it periodically.
This happened to me long tone ago after the hard forward hit. The workshop said that frame is not safe to ride and not repairable.


How can I match an existing stain color on exterior shutters? The bottom half has been faded by sprinkler hitting them

Question: Two of my 14 shutters have water damage on the lower half. I want to sand the bottom half and re-stain and seal these two but do not know the color originally applied. Is there a way to color match without removing the shutters and carrying them to the paint store?
Answer: Matching stains on wood is a tricky business.  A paint store should have samples of stained wood you can take home to compare, but differences in wood species can throw off the color of a stain.  It's not likely that the store will be able to "color-match" the stain even if you brought them a shutter - solid colors are easy to match, but transparent ones are nearly impossible, due to the color variations found in the grain of the wood.  If you do decide to try that route, most stores will need a sample of the color to be matched that's about the size of a quarter or slightly larger - which you'll have to extract from your existing shutters.
Another issue you'll face is that stains and wood finishes tend to darken over time as they age and are exposed to UV light.  So, in the off-chance you do find a stain that matches, it may be significantly darker than the existing wood in a few years.
To be perfectly honest, your best bet would be to paint the shutters a solid color, or sand them down and re-stain them with all-new stain.  A solid color will be easy to retouch as the finish is exposed to the elements.
Source: I worked for a big-box retailer for about 5 years, mostly in Flooring, but I also worked in Paint when we were short-handed.  Between the two, I've probably answered most of the common questions about repairing pre-finished wood.
Comment: Thanks Chris!  I sanded down the wood that was damaged and used the stain "chips" from the store that I know the builder bought all of the paint and stain from. I bought the stain that was closest to the color that remains on the shutter. I figured if it didn't work, I can just sand the whole things and restain them with the new stain. I will have to do all of them, but that will look even better.


Is it advisable to drop certain levels of a categorical variable?

Question: Let's say that I have one categorical variable with six levels, and I then create five indicator variables in order to represent the six levels. If two of the five variables are insignificant, then do I drop these two? I assume not, but I was not sure. I was thinking that it might be better to test the full (all five variables) versus the reduced (just the three significant variables)  model and, if that was not significant, then just leave all five of the variables in. I was not sure what to do. Oh and I meant for this to be in the context of fitting a logistic regression model.
Comment: It seems to me that you're mixing two distinctly different actions: dropping **levels** of categorical variables and dropping **variables** themselves. A clarification would be helpful.
Comment: is it ok to just drop some of the levels?
Comment: Please see my answer. Just after posting it I've discovered @Ben Kuhn's answer, which seems to be very nice (+1), if a bit short.
Answer: The latter approach (comparing two models with and without the five variables and decide if you should keep them as a set) is better.
The problem with dropping the indicator is that you'll change the p-values of the remaining levels as well, as you're shifting the intercept (aka the reference group.) Given a model:
$y = b_0 + b_1 Lv2 + b_2 Lv3 + b_3 Lv4 + b_4 Lv5 + b_5 Lv6 + \epsilon$
The intercept represents the mean of $y$ for group $Lv1$. Now, if we drop, say, the last two terms:
$y = b_0 + b_1 Lv2 + b_2 Lv3 + b_3 Lv4  + \epsilon$
Because you only drop the variable and not the cases, subjects in levels 5 and 6 need a place to go: notice that your intercept is now picking the groups $Lv5$ and $Lv6$ as well, representing the mean $y$ for levels 1, 5, and 6.
So, two major points: 1: your reference group can change and such change is not always sensible. 2: you may be surprised to see the significant results you wish to keep may be gone, due to the reference group mean has also changed.
Comment: I might be wrong, but I've had an impression that the OP is interested in dropping factor levels as opposed to variables themselves. Nevertheless, your answer is nice (+1).
Comment: I just re-read your answer and I'm taking my words about the levels back - sorry about misreading the text. I probably need a cup of coffee right away :-).
Answer: You should leave all five indicator variables in. Dropping predictors because they are non-significant leads to biased estimates for regression coefficients and inflated p-values.
A good reference that discusses this at length is Frank Harrell's Regression Modeling Strategies. You can find a summary of the problems with dropping insignificant features in section 4.3 there.
Comment: What if you're only interested in prediction, and don't care about the p-values?
Answer: Here's my two cents. I can't say with full certainty, but, I guess, it very much depends on a model and data. If I understand this answer correctly, @gung advises to test your model(s) after dropping all and then some levels. However, the details on how exactly to perform the testing are rather fuzzy (at least, to me). Perhaps, he will be kind enough to expand on that for beginners like me.
You may also find relevant and useful this course notes document on logistic regression (in R) by Professor Christopher Manning (Stanford University). Among other things, he describes dropping whole categorical variables (factors in R terminology) and manipulations with categorical variable levels, such as collapsing several levels into a single one and other manipulations, as well as the impact of those actions on quality of regression models and interpretations of analysis results.
Comment: My point was only that explaining the tests isn't the same thing as saying you ought to use them for factor manipulation (a subset of model selection, as you say) - which, given the context of this question, an unwary reader might be liable to conclude. (And everything said in gung's model selection answer applies whenever you pick the "best" model from many based on a measure of fit - whether you look for the "best" by dropping whole predictors or collapsing levels of categorical predictors.)
Comment: @gung is explaining there how to perform tests of null hypotheses that the expected response is the same for all levels, or for any two levels, not advising model selection based on the results of these tests. Given his answer [here](http://stats.stackexchange.com/questions/20836/), I'd imagine he'd be quite circumspect about advising that. Your link shows *how* to collapse levels based on the results of hypothesis tests, but offers no motivation or justification for doing so (though domain knowledge is being used as a constraint).
Comment: @Scortchi: Thank you for your comment. I didn't say anything about model selection in my answer. In regard to gung's answer in this thread, well, it maybe clear to more experienced people, like you, but to me it's a bit fuzzy. Not sure why you reference here gung's model selection answer, which is nice, but IMHO not directly relevant to this discussion's topic.
Comment: What else is dropping/leaving in factor levels or whole factors (in this question) referring to but model selection?
Comment: @Scortchi: Well, yes, factor manipulation _can be_ considered as model selection, but, more accurately, as **subset** of it (as model selection might consider other _aspects_ of models). Therefore, my points are: 1) model selection is a _much larger area_ than factor manipulation; 2) gung's model selection answer doesn't touch on the _factor manipulation aspect_, which is the topic of this question.
Comment: @Scortchi: Now I understand what you mean. Point taken. I will need to read on that (and more) at some point later. With my (non-statistics) Ph.D. dissertation defense coming next month, needless to say that I'm a little distracted to further improve my statistical knowledge at the moment... :-).


nonasymptotic complexity results

Question: I recall hearing about a result, or maybe a cluster of results, in some area of complexity theory, probably algebraic, to the effect that there are known, specific, short formulas whose minimal derivation is known to be exceedingly long. Or perhaps it is a specific function that requires an exceedingly deep curcuit. The "philosophy" seemed to be that such examples would threaten to make the older asymptotic question obsolete: "Who cares about asymptotics if the constants are huge." Can anyone, given these hints, describe ( and direct me to) the result I overheard?
Comment: Honestly I think that a separate CS theory site could be a mistake.  It should be fine to ask the question here and there should at least be a tool to move questions.
Comment: Actually, I posted a second question a few minutes after this one, the second on homotopy-theory and geometry. Careful consideration will reveal that the two questions are nearly identical. This illistrates the impossiblility of defining computer science as seperate from mathematics.
Comment: Probably this question will be best addressed at cstheory.stackexchange.com. I recall that kind of result, but almost certainly it is not in the book Algebraic Complexity Theory by Bürgisser et al.
Comment: Greg, I hope your skepticism won't prevent you from visiting from time to time :)
Answer: The general fact surrounding some of the other answers is
the following:
Every computably axiomatizable consistent theory $T$
containing trivial arithmetic admits very short theorems
requiring extremely long proofs.
The basic fact is that there can be no total computable
bound on the length of the proof required. To see this,
suppose that there is a computable total function $f$ such
that whenever statement $\psi$ is a statement of size at
most $n$ provable from $T$, then there is a proof of size
at most $f(n)$. In this case, the question of whether $T$
proves $\psi$ will be decidable, since we can simply
inspect all proofs of length $f(n)$, where $n=|\psi|$ and
check if any of them are proofs of $\psi$. But it is
impossible that $T$ proves $\psi$ is decidable, since we
could then produce a consistent completion of $T$, by the
usual process of completing a theory, which is effective
for decidable theories. This would produce a computable
complete consistent theory of arithmetic, in contradiction
to the incompleteness theorem.
One can make the conclusion fairly concrete via the halting
problem. If $T$ is true and contains some trivial
arithmetic, then $T$ will prove all true instances of the
assertion program $p$ halts on input $m$, and prove no
false instances. If there were a total computable function
$f(p,m)$ such that whenever $p$ halted on input $m$, then
there was a proof of this of length at most $f(p,m)$, then
we could decide the halting problem: on input $(p,m)$,
compute $f(p,m)$ and then look at all proofs of that length
and determine if there is a proof of halting or not.
In conclusion:
Theorem. For any computably axiomatizable true theory $T$ and for any total computable
function $f$, there is a program $p$ and input $m$ such
that $T$ proves that $p$ halts on input $m$, but there is
no proof of this in fewer than $f(p,m)$ steps.
Since as we all know, the computable functions can grow
quite outrageously, this means that for any theory there
will be short theorems that require extremely long proofs.
Answer: This is a result due to Gödel in his 1936  paper "Über die Länge von Beweisen" (On the lengths of proofs), republished in English translation in volume 1 of his collected works. Roughly speaking, he shows that there are short theorems that have extremely long shortest proofs in certain formal systems, but much shorter proofs in more powerful systems. Harvey Friedman found some rather dramatic explicit examples, described by Smoryński in The varieties of arboreal experience.
Comment: C. Smoryński says, in the article Richard cites,: "There are many incompleteness and independence results in modern logic. [...]
The exciting thing [in 1985] about the Paris-Harrington theorem [...]
is that it is of the form $\forall k \exists n A(k,n)$
and is undecidable for a different reason."  And then my authority to access this document is (frustratingly!) exceeded.
Comment: There is an integer n such that if T1,...,Tn is a sequence of trees with Ti having cardinality at most 10+i, then some tree can be homeomorphically embedded in a later one.

Friedman shows that this is provable in Peano arithmetic, but the shortest proof has length at least 222...  where the tower of exponent has height 1000
Answer: Perhaps you were told about

Larry J. Stockmeyer, Albert R. Meyer: Cosmological lower bound on the circuit complexity of a small problem in logic. J. ACM 49(6): 753-784 (2002)

From the abstract: "An exponential lower bound on the circuit complexity of deciding the weak monadic second-order theory of one successor (WS1S) is proved. Circuits are built from binary operations, or 2-input gates, which compute arbitrary Boolean functions. In particular, to decide the truth of logical formulas of length at most 610 in this second-order language requires a circuit containing at least $10^{125}$ gates. So even if each gate were the size of a proton, the circuit would not fit in the known universe.
This demonstrates a specific function for which all inputs of size 610 cannot be determined within the physical universe. The proof is essentially a diagonalization argument, but is very carefully executed to take all constants into account.
(Note this is actually based on an asymptotic result, but it took some work to get a concrete lower bound out of it.)
Comment: Thank you. I think this is what I had heard. Many of the other answers were quite interseting as well.  Mike
Answer: Are you by any chance referring to examples like Haken's proof that the pigeonhole principle requires exponentially long resolution proofs ?
Comment: G. Turán, "Resolution proofs of generalized pigeonhole principles"
[Theoretical Computer Science
Volume 62, Issue 3, December 1988, Pages 311-317], from the Abstract:
"We extend results of Haken to give an exponential lower bound on the size of resolution proofs for propositional formulas encoding a generalized pigeonhole principle."
Comment: Thanks, Joe. I didn't know about the more recent reference.
Answer: It's clear from the space and time hierarchy theorems in computational complexity, and more specifically non-deterministic time hierarchy theorems, that there are statements with extremely long proofs in any formal system that: (1) can describe Turing machines or cellular automata or similar; (2) can describe large numbers; and (3) has proofs that can be checked in a reasonable amount of time, such as polynomial time or even exponential time.  The non-deterministic hierarchy theorem says that $\text{NTIME}(f(n)) \ne \text{NTIME}(g(n))$ if $f(n)$ and $g(n)$ are reasonably separated functions.  A fairly precise version of this result is due to Cook, but if $f(n)$ and $g(n)$ are far enough apart, you can conclude that there is a separation just by comparison to deterministic time.  The corollary then is that you can turn instances of such a slow computational problem into a claim which cannot have a short proof; if it did have a short proof, you could find it quickly to answer the computational question.
For example, for each fixed $n$, you can take a question of the form, "can the following unit squares with teeth and notches added tile a square of size $A(n)$, the Ackermann function?"  The answer is always provably yes or provably no, because it is a finite question.  You can use a diagonalization argument in the sense of complexity theory to show that sometimes the proof has an Ackermann-like length.
Actually Mike sent me his question by e-mail, but I didn't think of this answer then and suggested posting the question here.  I don't know if this is the result sought, or merely a  similar result.  I also don't see a way to convincingly argue that asymptotic results are obsolete.  It is true that because some computational problems are asymptotically slow, you can make a sequence of computational problems that are all asymptotically fast, but with a worse and worse constant factor.
Answer: Establishing that minimal proofs (or "derivations") of certain statements or theorems are long, is the prime target of the area called proof complexity.
Concerning examples given in prior answers, like Haken's size lower bound on proofs of the pigeonhole principle, they deal with propositional logic only. Therefore, they are asymptotic results of the form: "there exists a constant $ 0<\epsilon<1 $ such that any resolution refutation of the propositional pigenohole principle $ PHP_n$ must be of length (i.e., number of steps) at least $ 2^{n^\epsilon}$"; Where $ PHP_n \;$,  $ n=1,2,\ldots ,$  is an infinite family of propositional contradictions (expressing the $ n+ 1 $ to $ n\;  $ pigeonhole principle). So propositional proof complexity does not seem to answer the question about non-asymptotic lower bounds.
On the other hand, lower bounds on first order proofs (having quantifiers, i.e., not propositional logic) are not necessarily asymptotic, and if this is what you are looking for, the only thorough survey I know of (dealing with both propositional and non-propositional) proof complexity is:  Pavel Pudlak: The lengths of proofs, in Handbook of Proof Theory, S.R. Buss ed., Elsevier, 1998, pp.547-637, available here
(There is also an older book by Orevkov on non-propositional proof complexity: [1993] Complexity of Proofs and Their Transformations in Axiomatic theories, vol. 128 of
Translations of Mathematical Monographs, American Mathematical Society, Providence,
Rhode Island.)
Answer: I found a citation to this paper by Joel Spencer: "Short Theorems with Long Proofs" [Amer. Math. Monthly 90(6) 365-366 (1983)].
But I do not have access to it at the moment.  Certainly seems relevant from the title!
Edit.  But apparently, not according to michael's comment.
The abstract of this paper by Marian Mrozek,
"Inheritable Properties and Computer Assisted Proofs in Dynamics"
says

Since there exist short theorems having arbitrarily long proofs (a consequence of G"odel's
  incompletness theorems, see [4, 6, 23]) ...

Reference [4] is Spencer's paper.  So I guess this just confirms that this is common knowledge,
without providing the explicit example you seek. Apologies for the distraction!
Comment: Thank you for the reference, but the author only remindes the reader of Godel's incompleteness theorem. No example is produced with which requires a long proof ( within some system.)

nonasymptotic complexity results


Are there v-brakes with long reach between the fork boss and the pad?

Question: I'm planning to upgrade my rims from 305mm to 349mm. However, it seems that most v-brakes have a max distance of 1" from the fork boss to the center of the pad. Are there v-brakes that have a reach of say, 48mm?
Comment: Gidday and welcome to SE Bicycles.  Good first question; thank you for including relevant measurements etc straight off.
Answer: Assuming you do mean reach as distance from the center bolt (for mounting a caliper brake) to the middle of the brake shoe:
For a center mount brake (e.g. many caliper brakes), the reach is determined by the length of the brake arms, and you can get different reaches easily by swapping to a longer or shorter armed brake (or make a drop bolt allowing you to use a shorter reach brake for when you need longer reach).
However, brakes mounted on bosses like V-brakes have their "reach" determined by the position of the bosses in addition to the brake. This can vary from frame to frame (or fork to fork) a bit, but usually not beyond the 10-15 mm or so allowed for brake pad adjustment. The exception is probably the Paul's Moto BMX V-brake:

which is sometimes used for 26" -> 700c conversions and might be enough for your purposes if you can find one.
Another way to do the swap is with a brake post conversion adapter -- a piece of metal which mounts at the brake posts to provide brake posts further up.
Something like this (commonly used for 26 inch -> 700c swaps -- one for these purposes is likely a bit too big for you):

Plus a V-brake with sufficient adjustment.
Another option I've seen (which I wouldn't recommend) is people taking a rectangle of metal, mounting the brake pad at the top of the rectangle, then attaching the bottom of the metal to where the brake pad is normally mounted. This likely results in a good loss of braking power, plus you have to make sure the mount is very secure.
Comment: Never knew that adapter existed. Interesting. We get some requests for things such as this, moreover people wanting to convert this to that, etc at the shop I work at. Good to know.
Comment: Its a bit of a hard part to find (though Xtracycle among others used to make them). Financially, unless you have the parts already laying around, you're usually better off just replacing the bike than putting 700c wheels in 26" frame.
Comment: Do note that if you use the Paul V-brake the "mechanical advantage" of the brake levers will change, changing the available force and travel of the brakes for a given lever motion.
Comment: Indeed, I think the manual of those V-brakes tells you to switch to short pull levers for some configurations.
Comment: @Batman Thanks for the response! I was already looking at the rectangular  v-brake extensions you mentioned but it seems that the u-shaped adapters are better.


Nikon sg-3ir more robust D.I.Y. alternative?

Question: I was recently looking at the nikon sg-3ir to cover the flash on my nikon d3300 so I can use it as a flash trigger. However for what they are,  the sg-3ir units are quite expensive and I was wondering why nobody had just attached a piece of infrared filter to their camera? Also the build quality of these units is quite poor as some people on Amazon have mentioned. I am also aware that these are becoming quite rare now - at least in the U.K
Comment: If you think $12 is expensive, bird photography might not be your best choice of hobbies. Just sayin'.
Comment: Exactly what does this have to do with bird photography?
Comment: @MichaelClark I'm not saying that. What I'm saying is it is expensive for what it actually is. Most Amazon reviews say that it breaks easily therefore it becomes costly to replace. Also it is much more expensive in the UK. And also I'm a kid - I can't spend too much money.
Comment: How do you know that, in fact, nobody *has* just just attached a piece of infrared filter to their camera?
Comment: @inkista you can use them for fill in flash for somewhere remote. There are some examples  of this on YouTube.
Answer: Actually, some people have just used exposed/developed film as a filter, but in these digital days, that stuff's harder to come by. Infrared filter gels are probably as or more expensive than the device you're looking at.
But generally speaking, most people never need this device for two reasons.
First, light from the pop-up flash rarely registers in off-camera flash photography, because the optical master/commander signal is done at a low power level. Because of the inverse square law, used at normal subject distances, this light tends not to register much if at all in the final image, compared to the main flash burst.  And if you're shooting at macro distances, then you're more likely to be using a ring-light setup or some other on-camera combination of flash.
Secondly, optical systems have weaknesses. They require line of sight and low ambient light levels to be really reliable.  They work great in studio conditions; not so great out of doors in bright sunlight without bounce surfaces around. For these reasons, most off-camera flash shooters use radio triggers.  And you can choose to have a "headless" (i.e., a unit with no flash lamp) unit as your on-camera transmitter. This won't emit any light at all and doesn't use the pop-up flash, so eliminates the need to make the master/commander signal non-visible.
Comment: Thanks - I might try some gels as they will be more robust than the build  quality of the sg-3ir. I have also looked at some ttl triggers but these are a bit more of an investment. Thanks for your help.
Answer: It passes IR and filters out the visible portion of the pre-flash sequence. The IR is required to trigger off-camera flash units. The visible pre-flash can cause humans to blink and I suppose cause birds to fly away.


Find files by creation date on Windows Server 2003

Question: What's the easiest way to find all files on a server (Windows Server 2003) created on or before a certain date? (BTW, Windows Search has been and is currently disabled.)
Comment: possible duplicate of [Fastest way to search for file attributes on windows?](http://serverfault.com/questions/11018/fastest-way-to-search-for-file-attributes-on-windows)
Comment: Can't test, but perhaps 'dir /o:d /t:c'  would be of use, although I vastly prefer Powershell if you have it installed. dir help: https://technet.microsoft.com/en-us/library/cc755121.aspx EDIT: I just saw the age of this question..
Answer: robocopy is your friend with the /L switch:
<code>-------------------------------------------------------------------------------
   ROBOCOPY     ::     Robust File Copy for Windows
-------------------------------------------------------------------------------

  Started : Sun Oct 16 06:17:46 2011

              Usage :: ROBOCOPY source destination [file [file]...] [options]

             source :: Source Directory (drive:\path or \\server\share\path).
        destination :: Destination Dir  (drive:\path or \\server\share\path).
               file :: File(s) to copy  (names/wildcards: default is "*.*").

::
:: Copy options :
::
                 /S :: copy Subdirectories, but not empty ones.
                 /E :: copy subdirectories, including Empty ones.
             /LEV:n :: only copy the top n LEVels of the source directory tree.

                 /Z :: copy files in restartable mode.
                 /B :: copy files in Backup mode.
                /ZB :: use restartable mode; if access denied use Backup mode.
            /EFSRAW :: copy all encrypted files in EFS RAW mode.

  /COPY:copyflag[s] :: what to COPY for files (default is /COPY:DAT).
                       (copyflags : D=Data, A=Attributes, T=Timestamps).
                       (S=Security=NTFS ACLs, O=Owner info, U=aUditing info).

           /DCOPY:T :: COPY Directory Timestamps.

               /SEC :: copy files with SECurity (equivalent to /COPY:DATS).
           /COPYALL :: COPY ALL file info (equivalent to /COPY:DATSOU).
            /NOCOPY :: COPY NO file info (useful with /PURGE).

            /SECFIX :: FIX file SECurity on all files, even skipped files.
            /TIMFIX :: FIX file TIMes on all files, even skipped files.

             /PURGE :: delete dest files/dirs that no longer exist in source.
               /MIR :: MIRror a directory tree (equivalent to /E plus /PURGE).

               /MOV :: MOVe files (delete from source after copying).
              /MOVE :: MOVE files AND dirs (delete from source after copying).

     /A+:[RASHCNET] :: add the given Attributes to copied files.
     /A-:[RASHCNET] :: remove the given Attributes from copied files.

            /CREATE :: CREATE directory tree and zero-length files only.
               /FAT :: create destination files using 8.3 FAT file names only.
               /256 :: turn off very long path (> 256 characters) support.

             /MON:n :: MONitor source; run again when more than n changes seen.
             /MOT:m :: MOnitor source; run again in m minutes Time, if changed.

      /RH:hhmm-hhmm :: Run Hours - times when new copies may be started.
                /PF :: check run hours on a Per File (not per pass) basis.

             /IPG:n :: Inter-Packet Gap (ms), to free bandwidth on slow lines.

                /SL :: copy symbolic links versus the target.

            /MT[:n] :: Do multi-threaded copies with n threads (default 8).
                       n must be at least 1 and not greater than 128.
                       This option is incompatible with the /IPG and /EFSRAW options.
                       Redirect output using /LOG option for better performance.

::
:: File Selection Options :
::
                 /A :: copy only files with the Archive attribute set.
                 /M :: copy only files with the Archive attribute and reset it.
    /IA:[RASHCNETO] :: Include only files with any of the given Attributes set.
    /XA:[RASHCNETO] :: eXclude files with any of the given Attributes set.

 /XF file [file]... :: eXclude Files matching given names/paths/wildcards.
 /XD dirs [dirs]... :: eXclude Directories matching given names/paths.

                /XC :: eXclude Changed files.
                /XN :: eXclude Newer files.
                /XO :: eXclude Older files.
                /XX :: eXclude eXtra files and directories.
                /XL :: eXclude Lonely files and directories.
                /IS :: Include Same files.
                /IT :: Include Tweaked files.

             /MAX:n :: MAXimum file size - exclude files bigger than n bytes.
             /MIN:n :: MINimum file size - exclude files smaller than n bytes.

          /MAXAGE:n :: MAXimum file AGE - exclude files older than n days/date.
          /MINAGE:n :: MINimum file AGE - exclude files newer than n days/date.
          /MAXLAD:n :: MAXimum Last Access Date - exclude files unused since n.
          /MINLAD:n :: MINimum Last Access Date - exclude files used since n.
                       (If n < 1900 then n = n days, else n = YYYYMMDD date).

                /XJ :: eXclude Junction points. (normally included by default).

               /FFT :: assume FAT File Times (2-second granularity).
               /DST :: compensate for one-hour DST time differences.

               /XJD :: eXclude Junction points for Directories.
               /XJF :: eXclude Junction points for Files.

::
:: Retry Options :
::
               /R:n :: number of Retries on failed copies: default 1 million.
               /W:n :: Wait time between retries: default is 30 seconds.

               /REG :: Save /R:n and /W:n in the Registry as default settings.

               /TBD :: wait for sharenames To Be Defined (retry error 67).

::
:: Logging Options :
::
                 /L :: List only - don't copy, timestamp or delete any files.
                 /X :: report all eXtra files, not just those selected.
                 /V :: produce Verbose output, showing skipped files.
                /TS :: include source file Time Stamps in the output.
                /FP :: include Full Pathname of files in the output.
             /BYTES :: Print sizes as bytes.

                /NS :: No Size - don't log file sizes.
                /NC :: No Class - don't log file classes.
               /NFL :: No File List - don't log file names.
               /NDL :: No Directory List - don't log directory names.

                /NP :: No Progress - don't display percentage copied.
               /ETA :: show Estimated Time of Arrival of copied files.

          /LOG:file :: output status to LOG file (overwrite existing log).
         /LOG+:file :: output status to LOG file (append to existing log).

       /UNILOG:file :: output status to LOG file as UNICODE (overwrite existing log).
      /UNILOG+:file :: output status to LOG file as UNICODE (append to existing log).

               /TEE :: output to console window, as well as the log file.

               /NJH :: No Job Header.
               /NJS :: No Job Summary.

           /UNICODE :: output status as UNICODE.

::
:: Job Options :
::
       /JOB:jobname :: take parameters from the named JOB file.
      /SAVE:jobname :: SAVE parameters to the named job file
              /QUIT :: QUIT after processing command line (to view parameters).
              /NOSD :: NO Source Directory is specified.
              /NODD :: NO Destination Directory is specified.
                /IF :: Include the following Files.
</code>
Comment: ROBOCOPY unfortunately does not seem work with file creation date but rather only with Last Modified date and Last Accessed date.


What would be appropriate for forge and foundry?

Question: I'm trying to translate a text I've written where we explain forge and foundry but not in the literal blacksmithing sense.  What is the appropriate word in Spanish?
Comment: I think that you have to add more context. Those words are related specifically and literally to blacksmithing in English and so are their equivalents in Spanish, forja and fundición. The verb forjar can be used in the same "metaphorical" way than forge as create (something) strong.
Comment: Just to add to @RubioRic comment the Wikipedia disambiguation page has links to three separate entries for foundry: metal foundry, type foundry, semiconductor fabrication factory. So clarifying which one you want would be beneficial.
Answer: Forge would be translated Forja
Foundry would be translated Fundición
In the first you heat something to make it malleable to be able to shape it, and in the second you completely melt the material to then put it into a mold and give it shape.
Answer: Forge

a furnace or a shop with its furnace where metal is heated and wrought
a blacksmith's workshop; a smithy.

Since this seems to be both the workshop and the "oven" used for the smith's trade, I would translate as Fragua

f. Fogón en que se caldean los metales para forjarlos, avivando el fuego mediante una corriente horizontal de aire producida por un fuelle o por otro aparato análogo.

f. Taller donde está instalada la fragua.

See from the second definition that a fragua is the place (workshop) where the foundry is installed. My guess is that in this question's context1 that workshop would be the  foundry. Now, it would not be a "foundry" in the literal sense, since you are not melting metal into a liquid, pouring it into a mold etc.
You could use in this context herrería for this "foundry" (which by description seems to be more like a smithy).

f. Taller en que se funde o forja y se labra el hierro en grueso.

f. Taller de herrero.

f. Tienda de herrero.

So, for the context give in the question I would translate

forge as fragua (fragua as a furnance)
foundry as herrería (to distinguish from the fragua/taller)

1 A foundry is a factory that produces metal castings.But I understand from the question that you want to change the "scope" of this place from "factory" to something more like "a few blacksmiths", meaning a smithy.