Converting a Bicycle without gears into a Geared one?

Question: I have a racer street bicycle without gears. I want to convert that into a geared one. For that I will need to modify the rear part of the frame where the wheel goes in. Is it advisable to do so?
Comment: it would be nice if you could provide some pictures of your bike, including the detail view of back triangle and dropouts.
Comment: Probably not.  You might be able to install a multi-speed hub, but they're fairly expensive.
Comment: @DavorinRusevljan: Yes, I will upload the pics.
Comment: This is just two gears and it is manual.  http://www.whiteind.com/double-double-system.html
Comment: [See also this question](http://bicycles.stackexchange.com/questions/6590/whats-the-easiest-way-to-add-gears-to-my-single-speed)
Answer: Modifying frame is usually not worth the trouble. Maybe your bike was previously equipped with gears, and it has necessary provisions (derailleur hanger)? If not consider a hub with internal gears. If that is too expensive, consider a new bike.
Comment: Unless it is emotional issue, like you want to have your grandfathers bike with gears no matter what is the cost, than it is either hub with internal gears, or new frame.
Comment: @WedaPashi Just because it didn't come with gears doesn't mean it doesn't meet the necessary requirements. Many bikes are set up and sold as single speeds but easily accept gears.
Comment: No, it was never equipped with gears, so it doesn't have provisions.
Answer: Depends on just what bicycle it is, but its probably possible to add gears in some way. But it may be expensive or complicated.
First thing to check is the frame spacing, ie the width between the rear dropouts. Many singlespeeds use 120mm wide hubs, whereas modern geared hubs are 130mm (road) or 135mm (mountain bike). If it is a steel frame, then it is usually possible to spring it out a bit, up to about 5mm or so, to fit a wider hub.
If the frame is only 120mm, then your options are limited. There are some internal hub gears for 120mm frames, eg from Sram or Sturmey-Archer. But they are fairly expensive, or have a limited range of gears.
If your frame is wide enough for a 130mm or 135mm hub, then you can use a standard road or mountain bike hub with a cassette.
Then you will need to attach the derailleur. If the frame is designed for singlespeed, it won't have a derailleur hanger. But you can bolt one on, depending on what type of droputs you have. eg if you have horizontal dropouts, you could use this DMR Chain Tugs & Mech Hanger.
Also the frame probably won't have guides for the gear cable. You can buy clamp on cable guides, or you could just use zip-ties to hold the cable in place.
Whether all this is worth it is another question. As you will have to buy quite a few parts (wheel, cassette, derailleur, gear levers etc), which could get expensive. Plus quite a bit of work fitting it all. It may be easier just to buy a bike with gears instead.


What happens when the DeBroglie wavelength becomes smaller than the Planck length?

Question: A particle around the size of a grain of salt would have a DeBroglie wavelength close the plank length. Does this mean quantum phenomena become hidden for larger particles since a DeBroglie wavelength smaller than the plank length makes no sense? Does the quantum information become hidden under a sort of Schwartzchild radius similar to when a black hole is formed and thus classical mechanics is all that's left?
Answer: A grain of salt can have a de Broglie wavelength much smaller than its size, and this is part of the explanation for why we don't see macroscopic quantum behavior of the grain of salt. This doesn't really have anything to do with the Planck length, though. Any time the de Broglie wavelength is smaller than our instruments are able to resolve, we'd have a hard time doing anything like a double-slit experiment with that object (for example).
Regarding the Schwarzschild radius and the Planck length:

An object will (presumably) form a black hole if its mass $M$ is all concentrated into a region smaller than its Schwarzschild radius $2GM/c^2$, which doesn't involve Planck's constant.
The Planck length $\sqrt{G\hbar/c^3}$ does involve Planck's constant, of course.

If we choose the mass so that these two scales are roughly equal to each other, then we have $M\sim\sqrt{\hbar c/G}$, which is the Planck mass. According to https://www.physlink.com/education/askexperts/ae342.cfm, a typical grain of salt has a mass of $\sim 6\times 10^{-5}$ grams (subject to lots of variation, of course), which is comparable to the Planck mass $\approx 2\times 10^{-5}$ grams.
An object may have a de Broglie wavelength much less than its Schwarzschild radius, but that doesn't mean it's a black hole or that gravity matters at all. The de Broglie wavelength only indicates the kind of scale that we'd need to resolve in order to observe quantum interference with that object, and has little to do with the object's physical size.
But if we squeezed all of the mass of a grain of salt into a space comparable to the Planck length, then it would become a black hole. At least, that's what classical general relativity would predict, but this is surely pushing classical general relativity beyond its domain of validity, because quantum-gravity effects are presumably essential at the Planck scale, so the true result would probably be even less like "classical physics" than either ordinary black holes or ordinary quantum objects.
Answer: The DeBroglie wavelength $\lambda$ depends on the speed of the particle: $\lambda = \dfrac{h}{mv}$
So a particle that has a mass like a grain of salt can have an arbitrarily large DeBroglie wavelength if it moves slowly enough. And likewise, a tiny particle like an electron can have an arbitrarily small wavelength if it moves fast enough.
Comment: Your answer makes a good point, and I would add this: For the center-of-mass of the grain of salt to have a location that is defined to within $<\Delta x$, it's momentum cannot be defined much more sharply than $\Delta p\sim \hbar/\Delta x$. So it doesn't really have a single de Broglie wavelength, which would be $\lambda=h/p$ if the momentum were exactly $p$ (in which case it would be completely delocalized). Instead, it has a spread of wavelengths, with characteristic wavelength $\lambda\sim h/\Delta p\sim\Delta x$, where $\Delta x$ is normally microscopic for something like a grain of salt.


led driver negative output and GND

Question: If I connect the positive output of a LED driver to anode and cathode of LED is connected to power ground (Negative output of driver is remain unconnected), whether the LED Will work or not?
Comment: I like the "Corrupted Virus" on the lower right corner
Comment: It won't work correctly, because you will be bypassing the feedback resistor that the LM3410 uses to sense the current.  I wouldn't like to predict what would happen.
Comment: Probably not.  But it might work, or it might glow dimly, depending on the design of the LED driver.
Comment: Lm3410 is using as driver. can you say it will work or not
Answer: You would be eliminating the control circuit and the NMOS protection circuit when the Feed Back pin is bypassed.
Either the LED or the NMOS switch will blow.  The resistor is a small ≈4Ω sense resistor not a current limit resistor.

7.3.1 Current Limit
  The LM3410 uses cycle-by-cycle current
  limiting to protect the internal NMOS switch. This current limit does
  not protect the output from excessive current during an output short
  circuit. The input supply is connected to the output by the series
  connection of an inductor and a diode. If a short circuit is placed on
  the output, excessive current can damage both the inductor and diode.
  -- LM3410 Datasheet

led driver negative output and GND


Why are the cardinalities of direct and inverse limits of $\mathbb{Z}/p^k\mathbb{Z}$ different?

Question: The inverse limit of the finite cyclic group $\mathbb{Z}/p^k\mathbb{Z}$ leads to the group of $p$-adic integers $\mathbb{Z}_p$, while the direct limit of $\mathbb{Z}/p^k\mathbb{Z}$ leads to the Pruefer $p$-group $\mathbb{Z}(p^\infty)$. For any given $p$, the $p$-adic integers are uncountable, having the cardinality of the real numbers, while the Pruefer $p$-group is countable, having the cardinality of the rationals (hence the natural numbers). Why is it that the cardinalities of the direct and inverse limits of $\mathbb{Z}/p^k\mathbb{Z}$ are not the same?
Comment: It's because they are different.
Comment: I have never heard of an indirect limit. Perhaps you mean inverse limit?
Comment: Is there a reason, except for the common word "limit", to think they should be the same?
Comment: @DerekHolt Yes, I meant inverse limit, my brain wasn't working when I wrote it up.
Answer: I believe you mean inverse limit for indirect limit. There is no such thing as 'indirect limit' in category theory.
The difference in the cardinalities of the two groups is due to the fact that the constructions of the inverse limit and the direct limit of $\mathbb{Z}/p^k\mathbb{Z}$ use different operations upon the constituent groups. The inverse limit of $\mathbb{Z}/p^k\mathbb{Z}$ involves the categorical product for abelian groups, the direct product of countably infinity many groups of $\mathbb{Z}/p\mathbb{Z}$, which is of cardinality $p^\mathbb{N} \sim |\mathbb{R}|$ or the cardinality of the reals. On the other hand, the direct limit of $\mathbb{Z}/p^k\mathbb{Z}$ involves the categorical coproduct for abelian groups, the direct sum of countably infinity many groups of $\mathbb{Z}/p^k\mathbb{Z}$, which is of cardinality $\sum_{i=0}^{\infty} p^i \sim |\mathbb{N}|$. This is because in the infinite case, the direct sum of abelian groups is only a subset of the direct product of abelian groups, namely those with finitely many non-zero terms.
It can be proven that the underlying set of the Pruefer $p$-group is isomorphic to the natural numbers in a base-$p$ positional notation system, while the underlying set of the $p$-adic integers is isomorphic to the circle group (the real numbers modulo one) in a base-$p$ positional notation system.


PHP ajax with jquery slidedown

Question: i have a complicated PHP conundrum. I have a constantly updating table in my database (Lets call it TABLE) with multiple records being added a second, these results will be posted using PHP on a page (page.php) using AJAX.
The maximum amount of posts shown are 1 every three seconds, for which i have a relevancy algorithm. I don't know how to merge the jquery transitions with the constantly updated records from the table (Got by AJAX).The best i came up with isn't much better than a hack, and dosen't accomadate for an infinite number of posts.
Jquery:
$(#5).delay(1000).slideDown(700);
$(#4).delay(4000).slideDown(700);
$(#3).delay(8000).slideDown(700);
$(#2).delay(12000).slideDown(700);
$(#1).delay(16000).slideDown(700);

PHP:
Not required
HTML:

I don't know what to do with AJAX.
Any help would be greatly appreciated,
thanks
Answer: If your question is simply how to slidedown the newest appended div, try this:
HTML:

Note: id attributes starting with numbers are invalid.
jQuery
setInterval(getData, 3000);

function getData() {
    // call ajax, get data.
    // success: function () {...
    var id = parseInt($("#updatePanel div:first").attr("id").replace("div", "")) + 1;

    $div = $("

").attr("id", id)
        .addClass("post")
        .text(dataFromAJAX)
        .css("display", "none") // change this to be a class - this is just to make it clear.
        .prependTo("#updatePanel")
        .slideDown();
}

Example fiddle
Answer: I'm not sure, but is this basically checking for new posts every three seconds, and then displaying them with a slide down?
If so, once you've pulled the new posts with ajax, append/prepend the code to a wrapper container, preset to div to hidden, and then slide it down.
I can help with code if this is what you want but you'll need to post your ajax so I can get a better idea.

PHP ajax with jquery slidedown


How can I sync history of Safari?

Question: Question is in the header. I know that Google Chrome stores his history in sqlite database, and there's impossible to sync Chrome's history. Can I do this with Safari? I didn't find this option.
Comment: Could you elaborate on what you're trying to achieve?
Comment: Well, I would like to synchronize Safari history between computers and maybe mobile devices if possible. So browsers on iMac and MacBook has a same browsing history. Now, each browser has its own history. It would be convenient. When you enter an address or search query - autofill works, and you can do things faster.
Comment: Well, I tried this using Dropbox to sync the History.plist file through symlinks, but it doesn't seem to work how I'd like. If anyone managed to get this working, let me know, I'd really enjoy Safari syncing my browser history throughout the devices I use (Mac mini at work, MacBook Pro at home).
Comment: I tried Dropbox too. Finally, I've considered to cancel this idea.
Answer: Somebody heard you, because this is now baked in. Your iCloud account will sync bookmarks, browser history, and even tabs in Yosemite and iOS 8.
Comment: Yes, that's right. I think this question can be closed as solved.

How can I sync history of Safari?


Is it true that the orbit space of a free finite group action on a CW-complex is also a CW-complex?

Question: Suppose a finite group G acts freely and continuously on an n-dimensional CW-complex X. Then can we conclude that the orbit space of this action is still an n-dimensional CW-complex? (or homotopy equivalent to an n-dimensional CW-complex?)  In particular, we do not assume G acts cellularly on X.
Comment: Here is a suggestion for proving that $X/G$ is homotopy equivalent to a CW-complex in a special case when the complex $X$ is countable and locally finite. Any metrizable ANR is homotopy equivalent to a CW-complex. If the complex $X$ is countable and locally finite, then it is a metrizable separable ANR, and I suspect metrizability and separability are inherited by $X/G$. Now a metrizable separable space that is locally an ANR is globally an ANR, so $X/G$ would then
be a metrizable ANR.
Comment: In seeing whether $X/G$ is homeomorphic to a CW-complex, even the case when $X$ is a smooth manifold is unclear. Indeed, if the $G$-action is nonsmoothable, then $X/G$ would only be a topological manifold, and in general it is unclear to me whether $X/G$ is homeomorphic to a CW-complex. (I think the existence of a CW structure on a topological manifold is unknown in dimension 4 and also for noncompact manifolds in higher dimensions, at least the proof in Kirby-Siebenmann's book is for compact case only).
Comment: @Igor: I think you should write your comments as an answer. Note that if $X$ is metrizable, so is $X/G$ (by averaging the distance function on $X$ under the group action: Sum of distance functions is again a distance function). Local finiteness passes to the quotient, separability too.
Comment: Perhaps you can perform repeated subdivisions on the cell structure of $X$ to arrive at the cellular case? I'm not claiming this is always possible, but it may be.
Comment: I would see if their is some sort of Borel construction. At the very least, you can take the singular chains on $X$, make the group action free with a simplicial Borel construction, then take the geometric realization of that. After this process, you will be in possession of a CW complex with the right weak homotopy type. I suspect that if you began with a CW complex, you will have a homotopy equivalence.
Answer: Lemma If $X$ is a countable locally finite CW-complex and $G$ acts freely and properly discontinuously on $X$, then $X/G$ is homotopy equivalent to a CW-complex.
Proof Any metrizable ANR is homotopy equivalent to a CW-complex
(I am not sure who proved it first but see Theorem 3.6.1 here. Since $X$ is countable and locally finite, it is a metrizable separable ANR. As Misha remarks in comments averaging the metric over the group action implies that $X/G$ is metrizable. Also a countable dense subset of $X$
projects to a countable dense subset of $X/G$. Finally, if a metrizable separable space is locally ANR, it is an ANR (see Borsuk's "Theorey of Retracts", Corollary 10.4, Chapter IV). It follows that $X/G$ is a metrizable ANR as desired.
 Remark In seeing whether $X/G$ is homeomorphic to a CW-complex, even the case when $X$ is a PL manifold is unclear. The difficulty is that it seems unknown which topological manifolds are homeomorphic to CW-complexes (Kirby-Siebenmann prove this for compact manifolds of dimension $\ge 6$ (or maybe $\ge 5$?, but certainly not $4$).
So there might exist manifolds not homeomorphic to CW-complexes but whose finite covers
are PL.
Comment: I wonder when $G$ is finite and the CW-complex $X$ is of dimension $n$, can we choose the CW-complex homotopic to the orbit space $X/G$ to be $n$-dimensional too?
Comment: @Igor: I do not think you need compactness for this. The point is that you can exhaust an open topological manifold $X$ by an increasing sequence of codimension $0$ compact submanifolds with boundary $X_i$. Let $B_i=\partial X_i$. Then, inductively, the handle structure extends from the collar $B_i\times I\subset X_{i+1}$ to the rest of $X_{i+1}$ (all the existence results are relative). Incidentally, the handle decomposition of Kirby-Siebenmann works in dimensions $\ge 6$; it is extended to dimension $5$ by Frank Quinn in "Ends of Maps-III".
Answer: If $G$ (finite or more generally discrete) acts cellularly on $X$, i.e.

if $\sigma$ is an open cell of $X$ then $g\sigma$ is again an open cell in $X$ for all $g \in G$
if $g \in G$ fixes an open cell $\sigma$ (i.e. $g\sigma=\sigma$), then it fixes $\sigma$   pointwise (i.e. $gx=x$ for all $x \in \sigma$)

then $X/G$ is a CW-complex. This follows from Prop. 1.15 and Ex. 1.17(2) of tom Dieck: Transformation Groups
Comment: What I want to know is exactly the case when G does not acts cellularly!
Comment: Yes, but when he answered this, you had not made that clear
Comment: @Igor: 1) Right, the question in its original version makes no assumption on cellularity. Hence there are two case: a) cellular action b) non-cellular action. My answer obviously treats case a). So how do you conclude I misread the question ? 2) If one is only interested in case b) then it's good style to point it out, as mentioned by David. - Anyway, now it's clear and I'm curious about the answer of this interesting problem.
Comment: I am very sorry. I thought the cellularly action case is easy. So I did not mention it at the beginning. But after your answer, I realize I should emphasize where the difficulty of the question lies.
Comment: @David White: your comment seems unfair. If you look in the edits, the first version of the question did not assume the action was cellular; it said "Suppose a finite group G acts freely and continuously on a CW-complex X. Then can we conclude that the orbit space of this action is still a CW-complex? (or homotopy equivalent to a CW-complex?)". In my view this is clearly stated, and it was Ralph who misread the question; why blame the OP?
Comment: Ralph, I should not have used the word "misread"; sorry. I wanted to stress that the question was clearly stated, and one should not blame the OP who did make everything clear, I think, by saying "G acts continuously".
Answer: The 3-sphere gives an example of an action with fixed points. If one takes the solid Alexander horned sphere, then Bing proved that its double is homeomorphic to the 3-sphere. So the quotient of the involution acting on $S^3$ is the solid Alexander horned sphere. However, the solid horned sphere is not homeomorphic to a CW complex. This follows from the answer to this question on the Alexander horned sphere. If the solid Alexander horned sphere were a CW complex, then one could attach the exterior 3-ball to get a CW structure on $S^3$ with the Alexander horned sphere being the boundary of the closure of a 3-cell, which is a contradiction to the other question.
Comment: This action is not free, as requested, the subspace of fixed points is Alexander's horned sphere (the common boundary).
Comment: Ian, does not your involution have fixed points?
Comment: I also missed the assumption that the action was supposed to be free.  However, the question is still a good one without that assumption and Ian's answer is instructive so I hope he leaves it up.
Comment: I missed the freeness assumption, but as Neil suggests I'll leave it up.
Answer: This is not really an answer, but a comment about an interesting special case.  Suppose that $G$ acts smoothly on $S^2$.  By averaging we can choose a $G$-invariant Riemannian metric.  This gives $S^2$ a conformal structure, making it a Riemann surface.  Any Riemann surface homeomorphic to $S^2$ is conformally equivalent to the standard Riemann sphere.  Thus, we can reduce to the case where $G$ acts on $\mathbb{C}\cup\{\infty\}$ by conformal and anticonformal maps, which must have the form $z\mapsto (az+b)/(cz+d)$ or $z\mapsto (a\overline{z}+b)/(c\overline{z}+d)$.  I think it even works out here that the quotient $(\mathbb{C}\cup\{\infty\})/G$ is always either a sphere or a disc.  Thus, one cannot get any local pathology in this context.  This contrasts with other settings where smooth functions can generate topological pathology: for example, any closed subset of $\mathbb{R}^n$, however fractal, can be expressed as the zero set of a smooth function $f\colon\mathbb{R}^n\to\mathbb{R}$.
Along somewhat similar lines, I think one can show that when $X$ is a one-dimensional CW complex with continuous action of a finite group $G$, then $X/G$ is again a one-dimensional CW complex (up to homeomorphism, not just homotopy equivalence).
Comment: Neil: Even more, if $G$ acts topologically freely and properly discontinuously on a topological manifold $X$ of dimension $\ne 4$ then the quotient $X/G$ is homeomorphic to a CW-complex. The same applies if $X$ is merely a simplicial complex of dimension $\le 3$ and $G$ acts topologically.
Comment: Misha: is your $X$ compact? If not, would you give a reference for the first claim, say in higher dimensions?


Use of article 'the'

Question: Of Rama and Usha the latter is the more responsible.
I think use of the is superfluous before competative degree more.
But my book suggest that the given sentence is absolutely right. Please explain if there is any rule regarding this in English grammar.
Answer: The sentence would be correct either with or without the inclusion of the.
When the is included, the sense is that Usha is the more responsible person; or the more responsible of the two.
When the the is omitted, more simply acts to modify responsible.
Another example might be:

She is more capable.
She is the more capable (person is implied)
She is the more capable administrator.

Both constructions are correct. It depends on exactly what you want to say.
Comment: The expression **the more [adjective]** is used when explicitly comparing two people or things; it can be followed by **of the two**.

Use of article 'the'


Troubleshooting hanged UDP socket in a Python server application code

Question: I am writing a UDP server application that serves as a back end to Teltonika FMB630 car mounted devices.
I already took care of the protocol specifics and decoding, the problem I am facing relates to the UDP socket used.
My UDP server has to send an acknowledgement to the client device upon receiving a message (that is the protocol), however, if I send those ACKs, the server socket stops receiving data after a while.
The server's UDP socket object is passed to an concurrent.futures.ThreadPoolExecutor that fires a function (send_ack) that sends the ACK, however this is not the issue because I tried calling send_ack in the main thread, after receiving data and the same issue occurs.
I suspect the problem is the remote device somehow breaks the connection or the ISP or MNO doesn't route the reply packet (this is a GPRS device) and then the socket.send() method that is used to send the acknowledge, somehow freezes other socket operations, specifically recvfrom_into called in the main thread loop.
I wrote two scripts to illustrate the situation:
udp_test_echo.py :
#!/usr/env/bin python

import socket
import concurrent.futures

def send_ack(sock, addr, ack):
    print("Sending ACK to {}".format(addr))
    sock.connect(addr)
    print("connected to {}".format(addr))
    sock.send(ack)
    print("ACK sent to {}".format(addr))

s = socket.socket(socket.AF_INET, socket.SOCK_DGRAM)
s.bind(("127.0.0.1", 1337))
data = bytearray([0] * 10)

executor = concurrent.futures.ThreadPoolExecutor(max_workers=4)

while True:
    print("listening")
    nbytes, address = s.recvfrom_into(data)
    print("Socket Data received {} bytes Address {}".format(nbytes, address))
    print("Data received: ", data, " Echoing back to client")
    executor.submit(send_ack, s, address, data[:nbytes])

udp_test_client.py:
#!/usr/env/bin python

import socket
import time
import random

def get_random_bytes():
    return bytearray([random.randint(0,255) for b in range(10)])

ip = "127.0.0.1"
port = 1337
s = socket.socket(socket.AF_INET, socket.SOCK_DGRAM)
s.connect((ip, port))

while True:
    stuff_to_send = get_random_bytes()
    print("Sending stuff", stuff_to_send)
    s.sendall(stuff_to_send)
    print("reply: ", s.recvfrom(10))
    time.sleep(0.1)

Running udp_test_echo.py in one terminal and udp_test_client.py in another, we see normal operation but if you Ctrl+C the test client and re run it, you will see that the server doesn't respond until it is restarted.
Is there a way to timeout a specific sending operation from a specific call to socket.send() method without affecting other calls ? (I want my socket.recvfrom_into call to block on the main thread)
If I settimeout on the entire socket object, I am going to have to deal with many exceptions while waiting for data in the main thread and I don't like to have to rely on exceptions for proper program operation.
Answer: The culprit was the socket.connect() call in send_ack, when being called on the server's socket object it causes the socket to no longer be bound and listen on the port specified in the start of the program.
Instead the send_ack function was changed to be:
def send_ack(sock, addr, ack):
    print("Sending ACK to {}".format(addr))
    sock.sendto(ack, addr)
    print("ACK sent to {}".format(addr))

socket.sendto(data, address) uses the existing connection instead of starting a new one.


Are oxidizer to fuel mixture ratios based on mass or volume?

Question: Wikipedia references the ratio of liquid oxygen to RP-1 as 2.56. Is this based on mass or volume? Are they using some typical combustion chamber pressure?
Answer: It's always by mass but there's one additional problem that rockets pretty much never use  stoichiometric oxygen to fuel ratio (O/F) that you'd expect if you calculated molar mass of propellant components an their ideal exhaust product(s). So the source should mention that somewhere if they're using stoichiometric ratios, typical O/F or even quoting some specific system's ideal O/F (and for where, usually given for sea-level and vacuum).
As a note, quoting volumetric ratio wouldn't make sense without also quoting storage pressure and temperature. And in case of mixtures and otherwise non-pure (e.g. non-refined) propellants also their molar composition. And ullage volume. And so on.
Answer: I believe that's the mass ratio. The liquid rocket propellants page on Wikipedia gives 2.77:1 as the mass ratio; the optimum varies with the exact engine design.
There's some interesting discussion of fuel ratios in this old usenet article.
Comment: Ah, old usenet articles - always a find!
Answer: It is a mass ratio as the ratio itself is based on the molecular masses of the fuel and oxidizer. Check out http://www.braeunig.us/space/propuls.htm and there "Combustion and Exhaust velocity" for details.


Use rear facing dual camera in Android

Question: I'm trying to capture using both of the rear facing cameras on my xiaomi mi 5s plus.
When I'm trying to iterate through the id's of the cameras using camera2 api, I only get 2 id's, one for the front facing camera and one for the rear one.
I've looked in the lg development examples (http://mobile.developer.lge.com/develop/sdks/lg-dual-camera/) and they are using 3 ids. When running their example, the app crashes.
The code used for getting the ids is:
<code>CameraManager manager = (CameraManager) getSystemService(Context.CAMERA_SERVICE);
for(String id: manager.getCameraIdList())
    Log.e("test", id);
</code>
Comment: a quick question: is LG G6 -- the successor of LG G5-- supported by the sdk and the examples?
Comment: Take two layouts and follow the camera code on onCreate.
Answer: Capturing images from a dual camera is not natively supported before Android P. Android P now supports multi camera capture -> https://developer.android.com/preview/features#camera

Different vendors provide dual cameras for their Android devices in
  hope to improve the photo quality for average user, more often than
  not, specifically tuned for special conditions like challenging
  illumination or distortions of selfie mode. Each vendor uses
  proprietary technologies to handle dual cameras, and they are not
  interested to disclose the implementation details. The only public
  interface they support is a virtual single camera which is more or
  less compliant with Google specs.

Courtesy of this Stack Overflow answer

Use rear facing dual camera in Android


What does the smile in Railley lips signify?

Question: In the final scene of 12 monkeys, we see James Cole is shot dead and as he said earlier his child copy is present there. Railley sees him and smile comes to her lips. What does it signify? Does it mean she hoped she had not lost Cole and that she is meant to be with him in coming time? Is Cole bound in a never ending time-loop?
Answer: It is more of a wry, fatalistic, and sad smile, as when she sees young Cole at the Airport, she finally puts all the pieces together from what Cole had told her.
In the theater, Cole falls asleep and remembers/dreams seeing her running in the Airport.  He wakes panicked, to find her gone.  He runs into the lobby to see her as a blonde, just as she was in his memory/dream.
He says "It was always you.  In my dream, it was always you."
In a sympathetic response, she replies with "I remember you like this."
Cole becomes confused.  "You do?" he asks.
She says "I felt I've known you before.  I feel I've always known you."
Cole's confusion deepens, and he replies "I'm so scared."
That scene is important, because it shows how she has really turned that corner, in believing and wanting to believe what Cole tells her.  She feels it is more important for him to understand that she is with him.  More importantly, she wants to believe him, to the point of making herself think she has indeed seen him in the past and her dreams.
For Cole, he knows to some degree that his dream is founded in a memory, and he knows he has traveled through time.  So for him, dreaming and remembering her at the airport when he was a kid is more of a revelation when he sees her as a blonde.  What confuses him so, and begins to scare him, is when she not only agrees with him, but tells him that she has always known him as well.  Cole knows she hasn't traveled through time.  Cole knows she couldn't have seen him in her past.  He may even begin to doubt his sanity, and his dreams/memories at this point... for he had previously believed that it was a memory.  However, if what she was saying was true, that she really felt like she had always known him as well, then he might not be remembering things.
So, later, at the airport, when she sees him as a child, the last pieces of the puzzle fall into place for her.  She knows that she is doomed (along with most everyone else) because he was telling the truth all along.  Any lingering doubt she had at that point has washed away.  She knows he was telling the truth in the theater when he said she was in his dream.  She knows that she scared and confused him by telling him she had always known him.  It's an OMG moment for her.  She knows that he survives the coming wave of death, only to be sent back to die before it happens.  The futility of everything they have done settles on her, and it all comes out in that smile.
So, I wouldn't say that the smile is honestly a happy one, except in that she was relieved to know the man she fell in love with wasn't crazy after all.
Comment: What a beautiful and poetic answer. +1 doesn't do it justice, but it'll have to do.
Answer: This smile is the most important shot of the entire film. It indicates that Railey is satisfied knowing that young Cole will go through everything that happened, exactly as it happened before.
With this, it answers the central question of the story. If everything is predetermined, so that the tragic impotence of our actions is already a foregone conclusion, how can we be happy?
She reaches the same conclusion as the 19th century philosopher Friedrich Nietzsche, who writes,

The Heaviest Burden. What if a demon crept after you into your loneliest loneliness some day or night, and said to you: "This life, as you live it at present, and have lived it, you must live it once more, and also innumerable times; and there will be nothing new in it, but every pain and every joy and every thought and every sigh, and all the unspeakably small and great in thy life must come to you again, and all in the same series and sequence - and similarly this spider and this moonlight among the trees, and similarly this moment, and I myself. The eternal sand-glass of existence will ever be turned once more, and you with it, you speck of dust!" - Would you not throw yourself down and gnash your teeth, and curse the demon that so spoke? Or have you once experienced a tremendous moment in which you would answer him: "You are a God, and never did I hear anything so divine!" If that thought acquired power over you as you are, it would transform you, and perhaps crush you; the question with regard to all and everything: "Do you want this once more, and also for innumerable times?" would lie as the heaviest burden upon your activity! Or, how would you have to become favourably inclined to yourself and to life, so as to long for nothing more ardently than for this last eternal sanctioning and sealing? [The Gay Science, 341]

To Nietzsche, however, this eternal recurrence is a challenge we must pose ourselves. Can we say this about our lives? How differently would we live our lives if we made this our primary concern?
Only extraordinary individuals can be content with the idea that their life will recur eternally. Such a person would affirm life in every action they take, which is Nietzsche's highest ideal.
In the case of Railley, it is not so much her own life as that of Cole that is affirmed. Her love for Cole -- not just Cole as an idea, but Cole as he really exists in the world -- makes her happy to see him live, and die, in an eternal loop, forever. She wouldn't change a thing about it, because that would mean changing Cole, and Cole is already perfect as he is.
That is what makes the moment so powerful, I think.


Make the whole app as subscription

Question: I have read about subscription but all of them is about in app items. Are there apis or methods to make the full app as a subscription? I have also tried in Google console making a new app but there is only FREE and PAID option. I want my app to be subscription like the magazines in Google play. Can anyone provide a reference or is this doable?
Answer: Your subscription would be an IN APP item. You would check if their subscription is active in the app and if not disallow use of the app.
For instance check the subscription in the main activity and if they dont have an active subscription then either send them to the subscription activity or whatever, and if they decide not to subscribe simply exit.
Answer: what you need to do is create a bare bones app that show users what they can accomplish with your app, like if this was an accounting app, they can add only 3 items,etc.
Then have an in-app subscription item that is "unlimited items" ,etc. They buy that.
I'm doing subscriptions for specific features in my app, like multiple accounts is a sub item and I'm planning on adding more, like "admin" etc.

Make the whole app as subscription


Could someone please explain what this Japanese symbol means?

Question:
Hopefully someone is able to let me know what this symbol means. I took a guess that it might be Japanese, but please direct me elsewhere if it is not! Thanks!
Comment: https://en.wiktionary.org/wiki/%E6%A5%BD
Answer: I think that glossing 楽器 as "tool for fun" is a bit misleading. In Chinese the character 楽 (old form 樂) represented both the word for "pleasure" and the word for "music", which had different pronunciations. That's why in Japanese this character is pronounced  らく in words to do with pleasure , such as  行楽 (こうらく) , "a pleasure trip", and がく in words to do with music, such as  音楽 (おんがく), "music".   楽器  (がっき)is literally a "music instrument".
In the Analects, Book 16 Section 5, Confucius is recorded as saying "There are three things people take pleasure in that are beneficial and three things people take pleasure in that are harmful. Taking pleasure in the discriminating study of rituals and music, taking pleasure in talking about the virtues of other people and taking pleasure in having many friends are beneficial. Taking pleasure in extravagance, taking pleasure in idleness and loafing about and taking pleasure in feasting are harmful."  "Taking pleasure in" is  楽 in all cases. "Music" is also  楽. "Taking pleasure in the discriminating study of rituals and music" is:  楽節礼楽 - "take pleasure / discriminate / ritual / music". There are other passages in the Confucian Classics which make word-plays on the two meanings of this kanji.
English also has a word meaning both "pleasure" and "music": "glee", as in "he hugged himself with glee" and "he joined the glee-club". The two things are obviously connected: the purpose of music is to give pleasure.
Answer: It is originally Chinese, but in Mandarin it has been simplified so it looks different. The Japanese variant might however still be present in some parts of the Chinese speaking world, such as Taiwan.
Anyway, from a Japanese perspective, it says 楽, pronounced raku, and means "with ease" or "comfort". You can also add a suffix to the kanji creating the word 楽しい, which is pronounced "tanoshii" and means "fun".
It is also used in various combinations with other characters to create other words, such as 楽器, pronounced gakki. The second part means "tool", so a "tool for fun" is how to say the word "musical instrument" in Japanese.


Stemmer processor not working

Question: I have configured Search api module in my Drupal 8 search. This is the first time I am configuring search api. I am using database search and not solr. My problem is stemmer processor is not working properly. I am not getting intended result for keywords due to stemmer processor issue. Stemmer processor is not stemming the words while indexing but is stemming while search query is executed. For example the word 'cars' is stored in the database as 'cars' itself while indexing but when a search query for 'cars' is executed it stem the word to 'car' and I am not getting the result. How can I fix this.
Answer: It seem to be a bug. I run into the same problem today, so I created an issue on the search api queue Stemmer is not working properly on indexing or on search. Please check that issue and elaborate if necessary.

Stemmer processor not working


Why an electron "rotate" around the nucleus at a speed close to the light one?

Question: Why an electron "rotate" around the nucleus at a speed close to the light one?
I mean where he gets all this energy?
One DOES NOT simply approach the speed of light AFAIK.
Comment: Who said that it does? Non-relativistic QM describes light elements **very** well.
Comment: I close this question as _unclear what you're asking,_ but the question is really closed because it rests on a false premise. In other words, in order to re-open this question, OP (or somebody else?) should add a published reference that backs up his premise.
Answer: (1) if it were a classical system, the correct word is "revolve" rather than "rotate"; the Earth rotates every 24 hours and revolves around the sun (approximately) every 365 days
(2) since it's not a classical system, the notion of an electron revolving around the nucleus is suspect.  Indeed, for S orbitals, the orbital angular momentum is zero.
So, where do you get this notion that your question is based on?
Comment: That's why I put " around rotate.
I don't remember where I have read it...
Comment: @حكيمالفيلسوفالضائع, perhaps some more research is in order then?
Comment: Of course taking the Fourier transform of the position distribution yields a momentum distribution from which you can determine if the semi-classical approximation is adequate to your needs.


Within a Clojure project using deps.edn, where is the package name and version tracked?

Question: Where would one track/store the package name and version string within a Clojure project using deps.edn? For example, within a Rust project, there is the Cargo.toml file where the package name and version string are stored along with other metadata about the package.
Answer: Not there. The deps.edn file is not a comprehensive description of your project, it's a file for specifying what the project's dependencies are, where to get them, what classpath the project has, and how to run it.
Project name and version can be stored there as e.g. JVM options under some alias, but that's... unconventional to say the least.
You'd be much better off creating a completely separate file and using it however you like. An EDN file with a map with :name and :version keys, a properties file with two keys, specially marked sections in your README, an existing package.json, a pom.xml created and updated with clj -X:deps mvn-pom where you keep the version and the groupId/artifactId of your project, and so on.
And also, not every project needs a version and a name. While libraries do need them (and then pom.xml usually makes the most sense), apps often don't.
Comment: `build.clj` is probably the most common place these days for group/artifact/version?


Lumpy green balls on tomato leaf

Question: We live in central Maryland, USA, and pulled this off a tomato plant in our garden this morning (mid-summer). Any idea what these are? Not sure if they're beneficial or parasitic, but I'd say the latter if I had to guess.
Answer: The shape of those looks more like caterpillar droppings (frass) than galls to me, though I've never seen them so green or shiny. Here's an image of tomato hornworm droppings, it looks quite similar to your picture:

If the balls are firmly attached to the leaf then they are galls, if they can be dislodged, then poop is more likely.
Comment: On closer look poop, is more likely.
Answer: Galls, the plants reaction to certain insects. Generally no action is warranted because of minimal harm to the host plant.

Lumpy green balls on tomato leaf


How can I resize the UIImage to specific size

Question: In my app, I am taking picture with camera and displaying it in 320*320 UIView. But as the image resolution is more than that of UIView its kinda look squeezed. Is there any way I can resize that image?
Comment: @Andreas: That is completely different questions. And that one talks about content adjustment.I am more interested in preserving the all the context of the image so even if I do some operations on image like cropping or slicing I won't loose the content of it...
Answer: Here's how you can resize the image while preserving its aspect ratio. The code below is from a category for UIImage:
+ (UIImage*)imageWithImage:(UIImage *)image
              scaledToSize:(CGSize)newSize
{

    float heightToWidthRatio = image.size.height / image.size.width;
    float scaleFactor = 1;
    if(heightToWidthRatio > 0) {
        scaleFactor = newSize.height / image.size.height;
    } else {
        scaleFactor = newSize.width / image.size.width;
    }

    CGSize newSize2 = newSize;
    newSize2.width = image.size.width * scaleFactor;
    newSize2.height = image.size.height * scaleFactor;

    UIGraphicsBeginImageContext(newSize2);
    [image drawInRect:CGRectMake(0,0,newSize2.width,newSize2.height)];
    UIImage* newImage = UIGraphicsGetImageFromCurrentImageContext();
    UIGraphicsEndImageContext();

    return newImage;
}

Comment: if this is a category, then what's the need for the image parameter? might as well use `self` instead
Answer: I do this, but first I enable the photo picker to editable so it is default 1:1 aspect, like so:
imgPicker.allowsImageEditing = YES; //I think this is what you're really looking for

then I resize the image using this method:
-(UIImage*)resizeImage:(UIImage*)image{
    CGSize newSize = CGSizeMake(480.0, 480.0);
    UIGraphicsBeginImageContext(newSize);
        [image drawInRect:CGRectMake(0,0,newSize.width,newSize.height)];
        UIImage* newImage = UIGraphicsGetImageFromCurrentImageContext();
    UIGraphicsEndImageContext();
    return newImage;
}

Comment: I was not looking for imagePickerEditing but its  sure something I did not known. Thanks for sharing..
Comment: Yea, I mean if you're using 1:1 ratio's it's not a bad option if it fits well with your UI. It let's the user scale and position with an interface they're use to.


How to change height of UITextView from xib programatically?

Question: I have a UITextView with non-editable text inside of it that I have made in a xib, and I need it to automatically change its height (when it's displayed), so that it perfectly fits the text inside of it. I have tried numerous methods but the majority either don't work on ios7 or just dont seem to work for me. I figured it might just be a problem with what I'm seeing. Here's the code that I'm running:
CGRect frame = CGRectMake(0,0,5,5);
storyText.frame = frame;
NSLog(@"%f", storyText.frame.size.height);

I have put this method both in viewDidLoad and viewDidLayoutSubviews but it doesn't seem to work in either of them. The NSLog is indeed printing 5, but this is what i'm seeing when i run the app.

What I am seeing above is exactly what it looks like in my xib. Anyone know why?
Answer: You can solve your issue like that:
dispatch_async(dispatch_get_main_queue(), ^{
    CGRect frame = CGRectMake(0,0,5,5);
    storyText.frame = frame;
});


What is the relationship between bayesian and neural networks?

Question: I'm looking for computationally heavy tasks to implement with CUDA and wonder if neural networks or bayesian networks might apply. This is not my question, though, but rather what the relation between the two network types is. They seem very related, especially if you look at bayesian networks with a learning capability (which the article on wikipedia mentions). At a glance, bayesian networks look at bit like a specific type of neural networks. Can anyone sum up their relationship, and if there is any connection beyond the apparent similarity?
Comment: should migrate to http://stats.stackexchange.com/questions
Comment: http://stats.stackexchange.com/questions/94511/difference-between-bayes-network-neural-network-petri-nets-and-decision-tree
Answer: Bayesian networks represent independence (and dependence) relationships between variables.  Thus, the links represent conditional relationships in the probabilistic sense.  Neural networks, generally speaking, have no such direct interpretation, and in fact the intermediate nodes of most neural networks are discovered features, instead of having any predicate associated with them in their own right.
Comment: Hmmm ... I always had the impression that bayesian statistics was all about discovering not assigning relationships.
Comment: That's right, to an extent, and thus there's a difference between learning the structure of a Bayesian network, and learning a given network's parameters.  It turns out that you can learn several plausible structures, but it's hard to know if you are missing variables. See "Causality" by Judea Pearl
Comment: Actually, the more I think about it, the key point to emphasize is that the nodes and links are connected through Bayes rule, and as such are subject to Bayesian updates upon further data.  Beyond this, I'm not sure to what can appropriately be ascribed as Bayesian.
Answer: It is reported that Bayesian networks are more resistant to the "overtraining" that is seen in some neural networks. In other words some neural networks become so "trained" to the observed measurements used in training that they aren't useful for the general cases.
Answer: Indeed they are. I see a bayesian network as a neural network applying the Baye's Theorem on large scale, but I don't remember details. I know where you can find them and I recommend this book for that.
Comment: This is not true.


Wolverine in both Pacific and European theaters?

Question: If you watch X-Men Origins: Wolverine and then watch The Wolverine, you can find an incongruency in Wolverine's story. In X-Men Origins, you can see that in the World War II scene, Wolverine and Victor fight in Normandy. In The Wolverine, he is a prisoner of war in Japan. What happened? Both movies were made before Days of Future Past.
Comment: _"If you watch X-Men Origins: Wolverine and then watch The Wolverine, you can find an incongruency in Wolverine's story."_ No, you find *several* incongruencies, and wonder why at least one of these two ever got made :-)
Comment: True. But I am wondering how this big one slipped through. The war was over in Japan before more troops got shipped over from Germany. How could they mess that up?
Comment: Wolverine being multiple places at once is often joked to be one of his powers, even (especially) in comics.
Answer: Actually there's no incongruency.
The Normandy landings happened in June 1944, the atomic bombings in August 1945. Unless I'm forgetting The Wolverine stating Logan was a prisoner for a very long time, it's not unreasonable to think he could have been in both theaters.
It's certainly not explicitely explained, but time-wise, it's not physically impossible.
Comment: But based on the technology of the time, moving troops from Germany to Japan would take too long. The war ended before all the logistics were worked out. Also, for Wolverine to become a POW in Japan, he must be in combat in Japan.
Comment: @Neo1009 Moving troops in mass may have been difficult but given the skills that he likely had shown they could have sent him individually to the pacific theater which would have been a lot easier.
Comment: are we sure that both scenes are from only WWII? could it be one in WWI? he was a prisoner a long time as shown in the movie


Add-AppxPackage : 0x80073CF6, ... could not be registered. Merge Failure : error 0x80070003 :

Question: I have a PC running windows 10 enterprise th2 build 1511/10586
The PC is joined to a domain. The issue occurs in a Domain User Account with Standard User permissions.
First - The photo app kept on crashing after about 5 seconds of being launched.
So i attempted to re-register the UWA's on the PC using ....
Get-AppxPackage -AllUsers| Foreach {Add-AppxPackage -DisableDevelopmentMode -Register "$($_.InstallLocation)\AppXManifest.xml"}

(Admin: Powershell)
This did not fix the issue ... so i attempted to do a user re-registration.
Get-AppxPackage | Foreach {Add-AppxPackage -DisableDevelopmentMode -Register "$($_.InstallLocation)\AppXManifest.xml"}

(User Level Powershell)
After this i had lost many application ... Photos, store, video, maps, messenger had become all "Semi" registered. No Icons, and long packages names in the start menu - also none of them would launch.
so i tried the Admin registration again... this time receiving my error for the first time .... Add-AppxPackage : Deployment failed with HRESULT: 0x80073CF6 .... Merge Failure : error 0x80070003 ... etc for the 5 apps. The rest of the UWA's were unaffected.
At this point i followed -> http://www.howtogeek.com/224798/how-to-uninstall-windows-10s-built-in-apps-and-how-to-reinstall-them/ and attempted to reinstalled the apps.
This did not work either. Photo apps didn't even seem to be found by the Remove-AppxPackage ...
Then i tried...
Get-AppxPackage *photos* | Add-AppxPackage

This appeared to work but did not change anything.
I also tried to login with my username (domain administrator) and theses apps all worked (however the name was not correct). I then performed an all user registration again from admin Powershell, which fixed my profiles app names but did nothing to the other user account.
Doing research this issue has been around for a long time. How do i completely remove these applications, sideload the store app and then proceed to reinstall them like they never existed as i would do with a win32? or alternatively what other ways is there to repair a users installation of these applications?
Answer: I had this problem too and it made me go rage for weeks. Your question is 6 months old but maybe it is still of interest. I use Windows 10 Professional with Server 2012 R2 in Active Directory.
The error will show something like "NOTE: For additional information, look for [ActivityId]" and suggesting a command to run. Run this command and see what it shows you.
In my case there were "just" some XML files missing for reasons I still don't understand. I found the XML on another computer and messed up the security permissions to the AppRepository folder to access it. But it worked.
Another solution is to reinstall Windows 10 by creating a new Windows installation media and run Setup.exe inside Windows 10, keeping all Settings and Apps. I don't know if this works for Windows 10 Enterprise.
I came into huge problems when deleting local user profile copies to clean up the system, the main and only reason were these corrupted Apps. In my case, I solved it by getting the XML files and making a clean re-installation of all store apps. I had to try multiple times, as lots of files were missing.


How to remove member from viber group

Question: How can i remove a member of a group which do not have admin. Admin left Viber and i have mistakenly added wrong guy. I and other members tried but we could not remove him. how can i remove him.
Answer: Viber Support page mentions this:

Removing participants from a group
  Participants cannot remove other participants from a group. Each participant can choose to exit the group at any time.

So it seems like your only choice is to exit the group and start a new one.

How to remove member from viber group


I working on sharepoint website

Question: All Pages Scrolling is very slow using all iOS devices. Please provide me fix.
whats is fix for this js or css?
Answer: Followup questions:

Are you using the application or the browser? If so, what type?
What version of SharePoint? (If you are unsure, you can check here)

I working on sharepoint website


Ideas on a nano-material that mutates into another, after acted upon by heat?

Question: This idea is from Hanger 1, the UFO Files: advanced alien technology that ended up on Earth.  Purportedly, a sample of this gold/silver material changed to titanium and molybdenum when heated. Any suggestions on describing the science in a story?
They also unearthed minute pins with precisely wrapped coils (http://www.ancient-origins.net/unexplained-phenomena/ancient-nanostructures-found-ural-mountains-are-out-place-and-time-002046).
Comment: A gold and silver alloy transmutated into titanium and molybdenum? If that is possible it's a technology so advanced it's indistinguishable from magic. Seriously, it's magic.
Comment: Welcome to Worldbuilding, Artidan22, Gold, silver, titanium & molybdenum are elements. They are absolutely unlikely to transform into one another by heating them. Nucleosynthesis (see L.Dutch's answer) can transmute elements inside stars and as a result of a supernova. Just heating won't do that.
Answer: Gold, silver, titanium & molybdenum are elements. They are absolutely unlikely to transform into one another by heating them. Because the only process that transmutes elements is nucleosynthesis. This happens inside stars and involves massive temperatures, pressures and energies. Supernovas are also involved in nuclear transmutation. This is impossible with ordinary heating.
If this is a nanomaterial it's conceivable it might be a substance that resembles gold or silver and on heating changes into something resembles titanium or molybdenum.
This wouldn't involve nucleosynthesis or transmutation. Such a nanomaterial could be a complex composite of ordinary elements and compounds amalgamated together to enable changing from, for example, a gold-like material into titanium-like when heated. Perhaps the faux titanium might change into molybdenum-like substance on being heated to a higher temperature.
Remember these nanomaterials only resemble gold, silver, titanium and molybdenum. While they may behave as if that is what they are, the materials only resemble their elemental counterparts.
You are welcome to devise how and what nanomaterials changes into whatever substance is suitable for your story.
Answer: Such a process exists, and it's normally called nucleosynthesis.
Energetic enough atomic nuclei (Temperature of some million K) hit each other and combine, changing the resulting material.
It normally happens in those objects known as star or supernovae.
Humans do it on smaller scale in particle accelerators, but only with few atoms per time.
Comment: well, thanks, at least it exists...didn't anyone see this on "Ancient Aliens?"  not the best source for facts, but odd ones for sure.
Answer: Realistically, we already have things sort of like that. Others have mentioned nucleosynthesis through fusion/fission, where the elements the material is composed from changes (but this involves immense amounts of energy). But there are other processes that, to a layman, can look like the material becomes totally different.
The easiest example would probably be heat breaking an unstable chemical bond (thermolysis), turning one compound into others. Usually you would get a color change and some part of the material escaping as a gas.
Another way would be a phase transition (note: these are reversible, so if you cool the material back down, it goes back to it's original state), like ice melting into water. A good example would be something like liquid crystals, the different phases of which interact with light differently. Or the lipid bilayer phase transition rapidly making it more fluid and flexible.
For something even cooler, look into low temperature phenomena like superconductivity and superfluidity or even into "negative temperature" systems.
With a fictional nano-material the effects of most of these can be made fairly spectacular while still sounding plausible. Have fun worldbuilding!


Why do massive rapid-rotating stars move bluewards off the zero-age main sequence (ZAMS)?

Question: I'm reading a well-cited paper by Szecsi et al. 2015 titled "Low-metallicity massive single stars with rotation". The paper can be found here. I was surprised that figure 5 depicted an HRD in which rapid rotators of 500 km/s evolved bluewards on the HRD. I didn't know any star could move anything but rewards when leaving the ZAMS. Other rotators below that 500 km/s move in a more typical fashion. Can anyone explain what is going on here?
Comment: You need to explain what the lines on the graph represent.
Comment: The paper attributes this to chemically homogeneous evolution caused by rotational mixing. Did you look at the Maeder (1987) paper it references?
Comment: @ ProfRob: I updated the plot, thanks. I see that the mixing causes it to be more homogenous but I'm not sure how that changes the temperature. It's not obvious to me.
Answer: The blueward evolution is a simple consequence of the homology relations for main sequence evolution.
If the star is rotating very fast, then this induces strong mixing of the interior, such that the star can be treated as chemically homogeneous. As main sequence hydrogen burning proceeds, the entire star is gradually turned into helium, with a consequent increase in the mean molecular weight $\mu$.
Therefore, the evolution can be thought of as a sequence of models with similar mass, but increasing $\mu$.
The luminosity evolution is
$$L \propto \mu^{7.5} M^{5.5} R^{-0.5}\ .$$
The strong $\mu$ dependence comes about through the strong dependence of opacity on $\mu$ for stars in radiative equilibrium.
On the other hand, the radius is only weakly dependent on $\mu$. That is because the central temperature depends on $\mu M/R$ (from hydrostatic equilibrium), but the strong temperature dependence of the CNO cycle, means the central temperature hardly increases as burning continues. As a result
$$ R \propto \mu^{2/3} M^{4/5}\ .$$
If we assume a blackbody, then $T_{\rm eff}\propto L^{1/4}R^{-1/2}$ and thus
$$ T_{\rm eff}\propto \mu^{1.54} M^{0.98} R^{-0.12}\ .$$
If we start off with 10% He by number, then $\mu=0.56$. But if we end with 80% He, then $\mu=1.2$.
On the face of it, this would appear to more than double the surface temperature during the main sequence evolution. The effect will be mitigated by mass loss and the disproportionate effect on opacity contributed by whatever metals are present (these are not zero metallicity stars).


IIS Management Console is not present in web management tool

Question: IIS Management console is not present under web management tool. Two options are there:

IIS 6 management compatibility
IIS Management Scripts & Tools.

I have checked both, but it did not work. Please help.
Answer: I'm assuming you're using Windows 7 since you specified IIS 7.
The IIS Management Console should be available under Control Panel → Administrative Tools.
If it's not there, verify you have IIS installed under Control Panel → Programs and Features → "Turn Windows features on or off".


How to add custom option to wp_dropdown_categories?

Question: I need to add a custom option to wp_dropdown_categories. Now, the whole wide web world haven't gotten a solution for that ... therefore I ask for a solution here … as I can't really imagine that the developers of WordPress have not considered that this would not ever be needed, right?
Comment: What do you mean with 'custom option'? An extra argument that can be passed to the function? What do you want it to achieve?
Answer: You can do like this
  add_filter( 'wp_dropdown_cats', 'add_yet_another_option', 10, 2 );

  function add_yet_another_option( $output, $r ){
    $output = str_replace( '</select>','',$output ); // remove closing select tag
    $output .= '<option class="level-0" value="some-value"> Some Custom Options </option>'; // custom option.
    $output .= '</select>'; // add closing select tag
    return $output;
  }

But why would someone put custom option on the category dropdown ?
Comment: It's a client asking for it ... and as you probably already know: clients are stupid ; ).

But thanks, I'll check it out.


Pages not loading on PHP 5.4.4 w/ APC - Apache 2.4.2

Question: I'm hoping someone can help me with this, as I've kinda hit a wall. Here's the situation:
I'm currently setting up a knowledge base using MediaWiki. I'm using XAMPP to aggregate my tools. The site itself is working fine, and coming along nicely.
I'm now trying to set up a cache to speed things up. I've opted to use PHP APC, as it is recommended for MediaWiki. The troubles started here.
Current versions are:
PHP: 5.4.4
Apache: 2.4.2 (apache2handler)
SQL: 5.5.25a
XAMPP: 1.8.0
MediaWiki 1.19.1
I am currently using Windows, so the first issue I had was finding the correct php_apc.dll file for my PHP. I tried numerous versions unsuccessfully (with log errors stating the .dll cannot be found). I eventually was able to secure a version which seems to work, APC 3.1.14 for PHP 5.4, downloaded from http://dev.freshsite.pl/php-accelerators/apc.html. It is the TS version, as this option is enabled on my PHP.
Now, with this presumably correct version of APC, Apache is now able to start fine. phpinfo() also shows the APC extension as available. The problem is now that no pages on the wiki will load. Precise situation is that:

All pages on the Wiki do not load (IP/mediawiki/*)
Pages on XMAPP itself do seem to load correclt (IP/XAMPP/*).
Other utilities are not able to load either, (e.g. phpMyAdmin).

No error messages seem to be created either in apache logs or php logs. Now, I am by no means an expert on this.
I have confirmed that environment variables in Windows TMP and TEMP both point to the same folder (C:\WINNT\TEMP), on which I gave "everyone" full access rights.
Here is an image showing the phpinfo() for APC that I can see:
https://i.stack.imgur.com/hYHwP.png
Any ideas?
edit: I forgot to mention: there actually are 0kb sized files that seem to be related to APC being created under c:\winnt\temp. For instance, one of them is called ".apc.a02684".
Answer: This has come up previously on the MediaWiki Support Desk.  MediaWiki apparently doesn't run under Apache 2.4.


What is the objective for League of Legends? Are there Quests?

Question: Been wanting to play but I don't know if its worth playing. What is the objective of the game?
Comment: Very nearly down-voted this question for a lack of research effort, but actually the [LoL new player guide](http://euw.leagueoflegends.com/learn/new_user_guide) doesn't do a good job of explaining basics of the game to those who are unfamiliar with MOBAs. :/
Answer: League of Legends is a MOBA, Multiplayer Online Battle Arena. You battle with your team to fight the enemy team to destroy towers guarding the enemy base, and then destroy said base. An average match of League of Legends lasts for about 20-50 minutes, though some matches may exceed this.
In the new playstyle, Dominion, your team tries to capture a majority of the 5 capture points on the map while defending the points you own. The team who can keep a majority of the points longest wins. These matches usually don't last more than 25 minutes.
There are (currently) 95 champions that any person can play, after unlocking them with real money or in-game points, each with several abilities and all with their own stats. By working with your team you can choose a character to support others, kill the enemy team, distract the enemy team or keep the enemy from acting.
Comment: Dominion mode was removed years ago, so this answer is dated (as the answered date reflects=)).
Answer: Forgot to mention 3v3, a game format very similar to 5v5 however it usually takes less time. And of course the number player is limited. There are no quests to say in the 5v5 but there are in dominion
Answer: At the time this question was asked and answered, there was not a quest system.  Now there is.  Usually it just consists of "First Win" which gives an XP reward and resets daily.
Occasionally there are events which almost always come with some sort of quests as well.  These can result in XP and in game money as well as shards for champions and skins, emotes, icons and more.
Some of the events are PVE or mix but some are PVP only.
Recently (at least in North America), they also added a rewards system for watching the professional events (LCS in NA).
As noted in comments above, Dominion map has been removed.
There are several rotating game maps to keep things interesting which stay for a limited period of time.  They are announced slightly in advance and don't seem to have a regular pattern except (maybe ironically) around April Fools.
All of the above are 5v5 except for a very few PVE events which are 5 vs the map and one rotating event which was 3v3.
Not mentioned by anyone yet either is the ARAM game mode which is 5v5 PVP with a single lane where everyones champions are picked randomly from their pools rather than normally you can pick which champion you want.
Otherwise, much of the upvoted information remains the same today.
Answer: There are many different game modes, Dominion, 5 V 5, 3 v 3, Co-op, Custom Games. The objective changes based on the game style you want to play however a couple of them are similar with some small difference.
Dominion - Capture Points and prevent the other team from capturing points. It's basically a circle and you control you champion to reach certain points. Your team starts at 500 life as well as the other team. Killing enemy champions and capping points reduce the health of the other team.
5v5, 3v3 - this is played against other players, you are playing a chosen champion with 4/2 other teams and their champions. You move down lanes and take out defending towers so that you can move on to the enemy base like area and destroy their main "life" known as the Nexus. Doing this all the while the enemy is doing the same to you. So think of it like playing capture the base as a kid with laser tag guns. Make your way to the fort, taking out all that you can. If you're hit, you go back to base and revive.
Co-Op - is a bit decieving, This is based off of 5v5 and dominion. You play with 4 other real players however, you do face AI/bots instead of other human players. Same concept as the other games but just a little less competitive. You can say it's a learning arena
Custom Game - same as all the other game types just you are more so allowed to establish the rules/game type and who plays.
Comment: I don't think Co-Op is very deceiving, since the game mode is explicitly stated as "Co-Op Versus AI" in the game. :)
Comment: It is not less competitive either.  It's far more competitive against computer programs, bots that gain xp and gold based on time as well as how they do in game.  It is only less competitive against other human beings.  To say that only human beings can be competitive is a bit arrogant.
Comment: Dominion mode was removed years ago, so this answer is dated (as the answered date reflects=)).
Answer: The objective is to have fun :)
LoL is a purely match-based PvP game (there are bots, but they're none too smart), so each match pretty much stands on its own. Riot does have a fairly comprehensive lore attached to the world, and each character, but the player doesn't experience any kind of story. The game isn't any kind of RPG, just an endless supply of deathmatches :)
You do have a single meta-character (called a summoner) which gains XP with each game, and there's a level cap of 30. However, each level only grants a fairly small increase to character power (in comparison to most RPGs, for instance) and it doesn't take much playing to get to the maximum level.
The characters you actually play each game with have a fixed skill-set, and they level up, gain gold items and money entirely within each match, and are reset at the start of each match. You can play with any of the 'free' characters (a small selection of characters which changes each week), or characters you've bought with in-game currency or real money. You can also buy runes (which are the closest thing LoL has to gear) with in-game currency, and attach them to your summoner. They confer small bonuses which apply to whatever character you select when actually playing a game.
For most people, the main reason for playing is to earn more in-game currency which will let you buy new characters. Each character has a unique skillset and playstyle, and although most players will develop an affinity for a particular character, its nice to be able to pick from a stable of characters if you feel like mixing it up.
Answer: I'd like to say its about PVP and effectively killing your opponent but like the others said, you have to destroy the base.
Don't get me wrong though, you can be that player who specifically kills enemy champions.
Answer: I know this will get a lot of hate. However, have you ever played DoTA? The Riot team, the creators of League of Legends, worked on the DoTA mod from WarCraft III. Almost the same concept, however the gameplay is a bit more lenient on new players. You have to capture points and defend your own.


Anje Falkenrath and Madness order

Question: Let's say I activate Anje Falkenrath to discard a card with madness. The madness card goes to exile instead. Do I get to draw a card before I decide to cast the madness spell? Do I get a choice to draw or cast first?
Answer: The rulings below the card in Gatherer state:

If you discard a card with madness to pay the cost of Anje's activated ability, you'll draw a card last. You may untap Anje before or after you choose whether to cast the discarded card, but not after you've chosen to cast it but before that spell resolves.

(emphasis mine)
Why? The first object on the stack is the 'draw a card' ability, and you pay for it by discarding the card. That causes both the triggered ability part of Madness (see below) and Anje's last ability to trigger, and you may choose in which order you put them on the stack:

702.34a Madness is a keyword that represents two abilities. The first is a static ability that functions while the card with madness is in a player's hand. The second is a triggered ability that functions when the first ability is applied. "Madness [cost]" means "If a player would discard this card, that player discards it, but exiles it instead of putting it into their graveyard" and "When this card is exiled this way, its owner may cast it by paying [cost] rather than paying its mana cost. If that player doesn't, they put this card into their graveyard."

Those abilities resolve first, while the 'draw a card' ability is still on the stack, and therefore drawing the card always happens last.
If you expect it makes a difference in your decision to pay the Madness cost or not, consider putting the Madness ability on the stack first, followed by Anje's untap ability. The untap ability resolves first, which gives you the opportunity to activate Anje's second ability again, and the card draw from that activation will happen before you have to decide to pay the madness cost for the first discarded card. (Then again, maybe you expect the next card draw to matter, so I'm not sure it will make much of a difference.)
Comment: I'm not 100% sure but I think that if you put the Madness ability on the stack first, so Anje's untap ability resolves last, you could activate Anje a second time to draw a card before having to choose whether to pay the Madness cost or not.
Comment: Yes, except that you said "untap ability resolves last" when you meant resolves first. If you had several madness cards you could have multiple in exile with their triggered abilities on the stack at the same time.
Comment: Thanks for catching that (and verifying that my idea should work, though its value is debatable).

Anje Falkenrath and Madness order


Calculating the parameters of binary systems?

Question:
If I know the masses of the two bodies (they have equal mass), how do I calculate the two ellipse eccentricities, length of the apoapsis and speed at the apoapsis, that make stable system, as in this picture.
Answer: You can't. There are an infinite number of stable orbits that are possible depending on the angular momentum and total energy of the system - which must be negative for a bound orbit.
If you specify the eccentricity (there is only one eccentricity value for the orbit) and orbital period then Kepler's third law will give you the semi-major axis, from which, the apoapsis can be calculated. Or if you specify the apoapsis then the eccentricity can be calculated. If you wish to consider the centre of mass frame then each body executes an ellipse with the same eccentricity (see here for example) and with equal semi-major axes (if the masses are equal), where their sum is equal to the semi-major axis of the system.
The speed at apoapsis can be obtained from the vis viva equation once the orbital parameters are established.
Comment: @DomahidiPéter all bound two body orbits are "two ellipse systems" since circles are just a special case of an ellipse. The only other options are two hyperbolas for unbound, and two parabolas for the borderline zero total energy case (zero velocity at infinity). Because the ratio of their distances to their center of mass must always be (by definition) inverse to their ratio of mass, they have no choice. They are always opposite each other relative to their center of mass. Whatever Fred Astaire does, Ginger Rogers has to backwards, and in high heels!
Comment: This is work for "two ellipse" system?
Comment: The parameters of a binary orbit include *one* eccentricity and *one* semi-major axis. @DomahidiPéter
Comment: I mean orbit system like in picture


Can a connected USB device read all data from the USB bus?

Question: Is it possible for a (malicious) (hardware) USB device to access all the data that is transferred through the USB bus and then read/store this information, essentially sniffing all transferred data? Or is the USB bus switched and only sends data to the correct recipient, i.e. only allowing the intended recipient to read it?
Answer: Most likely yes, but it depends
Much like PATA, SCSI, and Ethernet devices, USB devices don't directly connect to the computer. They connect to a Host Controller that manages all signaling and communication. All ports are connected to something called a Root Hub, and to each Root Hub you may connect other hubs and subsequently more hubs.
Each of these hubs have multiple downstreams and exactly one upstream. "What does that mean?" you ask. Well, it means that whatever data sent by the hub is sent to all child hubs and devices, while data sent by the hubs and devices are only sent "upwards" to Root Hub.
So, if a number of devices are connected to ports that lead to the same Root Hub (they're all controlled by the same Host Controller), then any of the devices can sniff the data only in the direction Computer -> Device.
In my laptop, for example, the ports on the right side are controlled by a Host Controller, and the ports on the left side are controlled by another Host Controller.

Meaning that data sent to any device on the right side can be sniffed by any device on the right side, but not devices connected to the left side.
I remember a colleague of mine modifying a USB stick to prevent it from ignoring data sent to other devices. So if you plug that modified USB stick to a computer, you can capture all the files copied to other USB sticks on the same computer.
Update:
@Polynomial's comment made me question the whole answer, since my information is based on my colleague's description. To be sure, I tried to find some reference. I dug in the USB specs and I found this:

  ...
  In the downstream direction, hubs operate in a broadcast mode. When a
  hub detects the start of a packet on its upstream facing port, it
  establishes connectivity to all enabled downstream facing ports. If a
  port is not enabled, it does not propagate packet signaling downstream.

Also, a TOTAL PHASE KB article seems to agree

USB 2.0 works through a unidirectional broadcast system. When a host
  sends a packet, all downstream devices will see that traffic. If the
  host wishes to communicate with a specific device, it must include the
  address of the device in the token packet. Upstream traffic (the
  response from devices) are only seen by the host or hubs that are
  directly on the return path to the host.
Comment: I think it's to do with the differentiation between *control* and *data* packets, which can both be broadcast and unicast data. The spec doesn't explicitly prevent either from being broadcast, but by default the data packets *should* be unicast. I'd imagine one of the USB sticks designated itself master, and the other slave, and relayed the unicast traffic horizontally.
Comment: I'm pretty sure this isn't true. The host controller manages 1:1 connections to each of the child devices back to the root USB controller. These connections may not be full-duplex, but as far as I'm aware they aren't shared buses. They're physically separated and don't do broadcast.
Comment: *"Each of these hubs have multiple downstreams and exactly one upstream [...] it means that whatever data sent by the hub is sent to all child hubs and devices, while data sent by the hubs and devices are only sent upwards to Root Hub."* - this is definitely incorrect. USB uses unique addressing for downstream and horizontal traffic, and round-robin polling to determine who gets to send upstream traffic back to the root host. There's no downstream broadcast.
Comment: @Polynomial Hmmm.. My understanding of this comes from my colleague's explanation of the whole process. But I've seen it with my own eyes how data copied to one USB stick are mirrored on the other. Although, now that you say it, I _could_ be wrong because the demonstration was on two USB sticks connected to an external USB hub. Do external USB hubs behave differently?
Answer: The answer by Adnan is correct. However to complete the picture, for superspeed mode in Usb3.0, the packets downstream are routed by the hub instead of broadcast unlike usb2.0.
Do note that the usb3.0 hub consists of the usb2.0 portion and a separate superspeed portion. The usb2.0 portion of the superspeed hub operates in broadcast mode as before.
In short, the contents you write to a drive running in superspeed mode cannot be snooped.


are RTS/CTS packets channel dependent?

Question: RTS and CTS packets in IEEE 802.11 are for dealing with hidden node problem. If I enable this feature on channel 1, would other nodes in other channels(e.g 2) receive and respect RTS/CTS packets?
Comment: Did any answer help you? if so, you should accept the answer so that the question doesn't keep popping up forever, looking for an answer. Alternatively, you could provide and accept your own answer.
Answer: No. Nodes on other channels are not listening to you. That's one reason why you should always use non-overlapping channels (1,6,11)

are RTS/CTS packets channel dependent?


Interesting: How to achieve this product catalogue effect

Question: Can anybody let me know how to achieve this effect with Magento.
This effect based on catalogue or folder wise structure of products.
The main catalogue or folder shows the image itself alongwith total number of products inside it.
After clicking on any of the folder all the products related to that can be seen.
This is even based on layered navigation based.
It is very much convenient for customers to browse the collection.
They can even find their product once they come back again.
The standalone link of those products is also there on the site by which direct browsing of those products can be possible from under their main categories.
I am providing the links of both way.
1.
http://www.utsavfashion.in/store/variety.aspx?cat=salwar
Below is the link under main category salwar kameez (where all the products can be seen and can be filtered using multiple checkbox layered navigation filter in the left side)
2. http://www.utsavfashion.in/store/items.aspx?cat=salwar
I am sure everybody will like this and want this in their own site.
Even developers and programmers would like to use and merge this feature in their Themes and Magento default features.
Regards
Answer: After a bit browsing, I found that there are few extensions out there in the market for multi-select checkbox layered navigation filter that can be used for point number 2
However, I am yet to find out the option for point number 1


MySQL Workbench Database Migration - data truncates after apostrophe

Question: I'm using MySQL Workbench Data Migration Wizard to migrate an MS SQL Server R2 database to MySQL database. The schema and data are migrating properly but for the fields having apostrophe, the data after apostrophe is truncated.
Here is an example:
Original data in a field(in SQL Server):

It has come to the Association's attention that some crew members are being required to perform administrative/office duties whilst on airport reserve.

As members would all be aware, your Enterprise Agreement states that Airport Reserve is duty time spent by a flight attendant at the airport awaiting assignment to a flight or duty (Clause 10.1)

However, in the past it has been agreed between the Association and the Company that checking and loading of an aircraft would be an appropriate requirement and should a flight attendant offer to assist the Company in performing other duties within their capabilities, that is also acceptable, but cannot be required.  The Association's position on this particular issue has been long standing and, in a time where administrative positions have been reduced and with a staff freeze in place, we can see no reason to reconsider this.

Returned data (in MySQL):

It has come to the Association

As you can see, everything after the apostrophe is missing.
Please help me.
Answer: Could be the migration tool interprets the data as incomplete quoted string and discards hence that part. Better file a bug report in MySQL bug system to get this fixed.


Distances in General Relativity, near massive bodies

Question: I've been studying the basics of General Relativity, and my question is: does it make sense to say that near a black hole (or any massive body), distances increase, the way on a topographical map the contours get closer and closer near a steep hill? So that an object escaping from near the event horizon is actually having to travel thousands of times more than the 'real' distance to reach a destination far away.  This seems to jive with the idea that an infalling object (from an outsiders perspective) seems to travel an infinite distance, fading away but never crossing the horizon, and also with the Shapiro time delay phenomenon.  But I want to ask in case this is a false analogy. And if it is true, how would that "stretched distance" (i.e. all the local distances on the trajectory added together) relate to the distance seen by an outsider?  How could the outsider ever say "I am 100 million miles from a black hole," if any test particle would actually travel many times greater distance to traverse it?
And how could a distant observer measure distance to a black hole anyway?
How would a distant observer measure a valid distance to any object, if not by sending and receiving a light ray?
Comment: While your topological map analogy is correct to describe the gravitational length contraction, the total distance to the horizon measured by a fee falling observer is finite and only somewhat larger than the coordinate distance. The "infinite distance" premise of your question is incorrect.
Comment: @RC_23 You can however get a notion of the diverging coordinate distance in the coordinate system of a hypothetical observer hovering arbitrarily close to the horizon.
Comment: Related: https://physics.stackexchange.com/q/21319/123208 & https://physics.stackexchange.com/a/170506/123208 Also see my description of Schwarzschild coordinates here: https://physics.stackexchange.com/a/552874/123208
Comment: @safesphere So in the case of AdS spacetime, light emitted outward from r=0 returns to r=0 in some finite time as measured by the observer at r=0. Does that mean that AdS space is finite?
Comment: @Razor This question is about the Schwarzschild spacetime. If you are interested in the AdS spacetime, please feel free to ask a separate question.
Answer: You can form $\int ds$ along a radial line at some given $t$ and thus create a notion of ruler distance. The ruler distance from the horizon is finite. You can see this also by looking at Flamm's paraboloid: distances along the paraboloid are finite.
However the time required for a particle to move outwards from the horizon is infinite.


What is the total amount of database queries Joomla does on a request with cache disabled?

Question: Thats about my question, I want to know how many db queries is done on a request when cache is disabled.
Thanks :)
Comment: The answer is **42**.
Answer: Enable the debug mode from Global configuration.
This will show you all queries that were logged during the last request.
Last request mean, if there are multiple redirection, then it will show the queries of last redirection(request).
Answer: That depends on the backend code (server side application code). Nobody will be able to tell you this (except the one that wrote the backend code).
Example: I use my browser to call a website.
In order to display the site i do not know if there is a db-query involved at all to create the HTML code of the page.
But it might be that on the backend some very expensive db operations take place. The user will never know. Only the persons involved in the backend process might be able to find out how many queries are necessary.


Is employee's 401(k) contribution still a tax deduction for employer through "wages paid"?

Question: Small business owners can claim wages paid to employees as tax deduction. If one of his employees would elect to contribute $15K of his salary into 401(k) account then would the employer still be able to claim the portion contributed to 401(k) as tax deduction (i.e. on full wages)? If so, then wouldn't a single 401(k) contribution be tax deduction twice - once on employer's books and once on employee's books?
Comment: The employee and employer each still pay FICA and Medicare on that 15000.
Answer: It is not a net deduction on the employee's books, because it is counted as income first and then removed from income when contributed to the 401(k), just as when the employee contributes to a traditional IRA. That is, the portion of wages that the employee contributes is neutral, neither increasing nor decreasing the employee's taxes.
EDIT: Yes, when an employee contributes part of their wages to a 401(k), those wages remain fully deductible by the employer:

You paid your employees wages but they chose to defer a portion of those wages into the 401(k). However, you still  get a deduction for the full wage expense.

But I am trying to convey that there is not "tax relief twice for a single contribution". If the employer starts offering a 401(k) and the employee starts contributing part of their existing wages to it, the employee's taxes go down but the employer's taxes don't change. If the employer raises the employee's wages and the employee contributes that raise to the 401(k) -- keeping take-home pay the same -- then the employer's taxes go down but the employee's income taxes don't change. The contribution never generates a full net deduction for both parties at once.
Comment: nanoman can confirm, but I believe the clear point is that wages are a deduction from an employer's income (i.e. an expense), but once paid to the employee, what they do with it is irrelevant. Unless, of course, the 401(k) is matched, then the 'matching' part is another expense/deduction for employer.
Comment: It seems I did not use the term "tax deduction" correctly. Though can you affirmatively answer whether 401k contribution still gives benefit of full wage tax deduction to employer? It seems like could be a no-brainer for small business owners to offer 401k to their employees because there is tax relief twice for a single contribution.
Answer: Well, "deduction" is a bit imprecise. For the employee, it's an adjustment to income; as far as the IRS is concerned, it's money that the employee hasn't gotten yet. For the employer, it's an expense; it's part of the gross revenue, but not part of the net income. When lay people say "deduction", they usually are thinking about below-the-line deductions, and this is such to neither employee nor employer.
Another point to understand is that this isn't tax-free, it's tax-deferred. This is money that isn't going into the employee's pocket. It's going into a special account, and if the employee wants to spend it, they have to pay taxes on it.
So just how many incomes and deductions there are is to some extent a matter of accounting. If you want to treat it as a deduction to both employee and employer, you have to treat it as income to both. For instance, suppose I run a tutoring center, and a client pays me $30, and I then pay an employee $20, and the employee defers $5 of that into a 401(k) plan. That $5 can be analyzed as being a deduction for both me and my employee, but then it has to be analyzed as an income for both of us. If the employee then withdraws it from their 401(k) plan, that has to be considered a separate income under this analysis. That is, under this analysis, this $5 started out in the client's pocket. Then it went to me (gross revenue for me), then it went to my employee's 401(k) plan (income for them), then it went from the 401(k) account into the employee's bank account. Three income events, two deduction events.
We can also treat only the employee actually having access to the money as an income event for them, in which case we have two income events, and one deductions. Or we can not include it in my income, in which case it's two income events and one deduction again. We can do both, in which case one income event, and no deductions. Or we could go the other way. Maybe I contract to a school, and the school has $35 in funding that they used to hire my tutoring business. Then we could analyze it as four income events (the school gets funding, they pay me, I put the money in the employee's 401(k) account, the employee withdraws it), and three deductions.
No matter how you analyze it, there will be one more income event than deduction. Adding the employee to the analysis doesn't make the contribution any more effective; it can add a deduction to the analysis, but only if it also adds an income event.


How is the exposure of this histogram?

Question: Is exposure correct for this luminance histogram?
What does this say about the dynamic range of the scene?
Comment: Answer depends on what the subject is: whether it was a zebra, or a gray cat. Why do you ask - can you give us some context?
Answer: That is what's known as a histogram.  The exact meaning requires more context.  It could either represent the light intensity from left to right across the image or, based on what the plot looks like, it more likely indicates the amount of the image that falls in any particular luminance (brightness).
Assuming it is the later, which it appears to be, then the dynamic range of the image is pretty limited as the majority of the image is not making use of the brighter or darker portions of the color space (the total amount of colors that could be produced).

How is the exposure of this histogram?