I have a camera module with a lens viewing angle of 185° and a focal length of 1.8.  But as far as I can tell, this camera module is not classified as a fish-eye lens.  The working distance is specified from 14cm->infinity.  And while I understand that the horizontal FOV (mm) can be calculated from the lens viewing angle and desired working distance, I am confused at how this 184° viewing angle would effect the outputted image (would it be fish-eye?  Is the working distance too small for the fish eye effect to be noticeable?)  Thanks for any clarification.
What is the relationship between the focal length and field of view for an equal-area ("equisolid") projection fisheye lens?  Alternatively, what is the relationship between the 180-degree image circle size and the focal length for such a lens?  Ultimately I want to find the field of view of a Tokina 10-17 mm fisheye zoom at focal lengths other than 10 mm on APS-C ().
Change footer in beginning of chapters I want the footer of all pages to be "Page N°X" on the right. I used this:  \usepackage{fancyhdr} \pagestyle{fancy} \fancyhf{} % clear all header and footer fields \fancyhead[R]{\leftmark} \fancyfoot[R]{\footnotesize Página N$^\circ$ \thepage\ } \renewcommand{\headrulewidth}{0pt} \renewcommand{\footrulewidth}{0pt}   It works on every page but the chapter beginning ones. How to make a good glow effect around a sun? Why do metals only glow red, yellow and white and not through the full range of the spectrum? Intellisense in IntelliJ no longer working I used to be able to color code each folder in Finder.   I know I can use tags in Mavericks, but is there a way to restore the ability to actually color code folders?  Thanks. Capistrano for Django Save site as template using client object model The relativistic energy-momentum equation is:  $$E^2 = (pc)^2 + (mc^2)^2.$$ Also, we have $pc = Ev/c$, so we get: $$E = mc^2/(1-v^2/c^2)^{1/2}.$$  Now, accelerating a proton to near the speed of light, I get the following results for the energy of proton:  0.990000000000000   c =&gt;    0.0000000011    J =      0.01    TeV  0.999000000000000   c =&gt;    0.0000000034    J =      0.02    TeV  0.999900000000000   c =&gt;    0.0000000106    J =      0.07    TeV  0.999990000000000   c =&gt;    0.0000000336    J =      0.21    TeV  0.999999000000000   c =&gt;    0.0000001063    J =      0.66    TeV  0.999999900000000   c =&gt;    0.0000003361    J =      2.10    TeV  0.999999990000000   c =&gt;    0.0000010630    J =      6.64    TeV 0.999999999000000   c =&gt;    0.0000033614    J =      20.98   TeV  0.999999999900000   c =&gt;    0.0000106298    J =      66.35   TeV  0.999999999990000   c =&gt;    0.0000336143    J =      209.83      TeV  0.999999999999000   c =&gt;    0.0001062989    J =      663.54      TeV  0.999999999999900   c =&gt;    0.0003360908    J =      2,097.94    TeV  0.999999999999990   c =&gt;    0.0010634026    J =      6,637.97    TeV  0.999999999999999   c =&gt;    0.0033627744    J =      20,991.10   TeV    If the LHC is accelerating protons to $7 TeV$ it means they're traveling with a speed of $0.99999999c$.  Is everything above correct? Uniform integrability of a backward submartingale Let $\{\mathcal{F}_n\}_n$ be a decreasing sequence of sub-$\sigma$-fields of $\mathcal{F}$($\mathcal{F}_{n+1}\subset\mathcal{F}_n$) and let $\{X_n\}_n$ be a backward submartingale($E[|X_n|]&lt;\infty$, $X_n$ is $\mathcal{F}_n$ measurable and $E[X_n|\mathcal{F}_{n+1}]\geq X_{n+1}$).  If we know $l=\lim_{n\rightarrow\infty}E[X_n]&gt;-\infty$, how to prove that $\{X_n\}_n$ is uniformly integrable? Many thanks!  I have an idea, it is enough to show:    $\sup_nE[|X_n|]&lt;\infty$;  $\forall\epsilon&gt;0$, $\exists\delta&gt;0$ s.t. $\forall A\in\mathcal{F}$ such that $\mathbb{P}(A)&lt;\delta$, we have $\sup_nE[1_{A}|X_n|]&lt;\epsilon$.    For $|X_n|=2X_n^+-X_n$, we have  $$E[|X_n|]=2E[X_n^+]-E[X_n]\leq 2E[X_0^+]-l\leq 2E[|X_0|]-l$$  which implies $\sup_nE[|X_n|]&lt;\infty$.  Secondly, $\forall A\in\mathcal{F}$, we can suppose without loss of generality that $A\in\mathcal{F}_0$.  $$E[1_{A}|X_n|]=2E[[1_{A}X_n^+]-E[[1_{A}X_n]$$  Does someone have an idea for this term? Thanks a lot..... Posting only once every 40 minutes even for experienced users? I have low rep on Physics &amp; Woodworking, but not so on Stack Overflow.  I just tried to ask &lt; 40 minute interval on those two and got hit with the :    I posted once on Woodworking, then ~5 minutes later tried to post on Physics, and was denied because of the 40 min rule. So apparently the system looks cross-network to assess this rule.  It seems the system is looking at my lowest cross-network rep when applying this rule rather than highest rep? What is the point of this very conservative approach? It would seem there is very little chance that a user with high rep on ANY site is likely to spam or abuse the lower-rep sites. And obviously this discourages participation across the network.  Furthermore, an improvement at least would be to warn the user ASAP so they aren't stuck with a nicely worded question in their browser window only which could be forgotten, lost, etc. As it is you only get the warning when you click [Post Your Question]. I can't destroy blocks in Minecraft Creative Mode (singleplayer) The  page says that Box-Jenkins is a method of fitting an ARIMA model to a time series. Now, if I want to fit an ARIMA model to a time series, I will open up SAS, call proc ARIMA, supply the parameters $p,d,q$ and SAS will give me AR and MA coefficients. Now, I can try different combinations of $p,d,q$ and SAS will give me a set of coefficients in each case. I select the set with the lowest Akaike information criterion.  My question is: where in the above procedure did I use Box-Jenkins? Am I supposed to use Box-Jenkins to come up with initial estimates of $p,d,q$? Or did SAS use it internally somehow? I am superimposing CGI over video footage, and have run into a problem in making materials for objects that act as 'shadow catchers'.  Essentially I want a shader that is shadeless—like an emitter—but catches shadows and light from other objects. This is so I can reconstruct objects in footage so other objects in the scene interact convincingly with them.  Here's an overcomplicated diagram to illustrate my point:   (In this particular scenario, the problem could be glossed over by using RGB curves, but that is definitely not the case with more complicated scenes)  Isn't this a common situation? How to the real guys do it? Thank you.  (I am welcome suggestions of other ways of doing this.) I have some stuff in settings.py that I'd like to be able to access from a template, but I can't figure out how to do it. I already tried  {{CONSTANT_NAME}}   but that doesn't seem to work. Is this possible? String answer1 = "yes"; if (answer1 == "yes") gives false. Why? What is the relationship between symmetry and degeneracy in quantum mechanics? Let me remind you about the following classical  examples in quantum mechanics.     Example 1. Bound states in 1-dim potential V(x).   Let $V(x)$ be a symmetric potential i.e.   $$V(x) = V(-x)$$   Let us introduce the parity operator $\hat\Pi$ in the following way:   $$\hat\Pi f(x) = f(-x).$$   It is obvious that    $$[\hat H,\hat\Pi] = 0.$$   Therefore, for any eigenfunction of $\hat H$ we have:   $$\hat H|\psi_E(x)\rangle = E|\psi_E(x)\rangle = E\hat\Pi|\psi_E(x)\rangle,$$   i.e. state $\hat\Pi|\psi_E(x)\rangle$ is eigenfunction with the same eigenvalue. Is $E$ a degenerate level? No, because of linear dependence of $|\psi\rangle$ and $\hat\Pi|\psi\rangle.$   Consider the second example.     Example 2. Bound states in 3-dim a potential $V(r)$. Where $V(r)$ possesses central symmetry, i.e. depends only on distance to center.   In that potential we can choose eigenfunction of     angular momentum $\hat L^2$ for basis    $$|l,m\rangle,$$   where $l$ is total angular momentum and $m$ - its projection on chosen axis (usually $z$). Because of isotropy eigenfunction with different $m$ but the same $l$ correspond to one energy level and linearly independent. Therefore, $E_l$ is a degenerate level.      My question is if there is some connection between symmetries and degeneracy of energy levels. Two cases are possible at the first sight:   Existence of symmetry $\Rightarrow$ Existence of degeneracy Existence of degeneracy $\Rightarrow$ Existence of symmetry     It seems like the first case is not always fulfilled as shown in the first example. I think case 1 may be fulfilled if there is continuous symmetry. I think the second case is always true. Homework question on whether two quotient spaces are homeomorphic Isn't $HOME supposed to be the location of the user's home directory? I'm having trouble with a web application I'm writing: The web application, which runs as user nobody, needs to launch an inkscape process to manipulate some SVG files. Because this runs as user nobody, there is no home directory. What I am doing is creating a temporary directory under /tmp and trying to set that as the home directory via export HOME=/tmp/someUniqueId. I then set a few things in that "temporary" home directory, like a symlink .fonts to the folder for the font files to use.  I know this works for finding the ~/.fonts directory. But when I launch Inkscape, all I get is a message that Inkscape could not create the .inkscape directory:    Doesn't $HOME set the location of my home directory? Why is Inkscape still trying to access / as my home directory? What is the word that is used to describe the oversimplification or non-academic discussion of a complex issue?   Typically, I'm looking for a word like pedantic, or puerile, except that the word covers the fact that the discussion is particularly non-academic. I know that such a word exists since I have used it before, but it has skipped my mind. Ubuntu 14.04 Wireless not working: no network interface. Atheros [168c:003e] device I just installed Ubuntu 14.04.3 LTS after installing windows 10 in my new VN7-571G and the wireless is not working. It never had: even during the installation it was not recognized. I run a script to gather the data , which is . lspci -knn | grep Net -A2 output 03:00.0 Network controller [0280]: Qualcomm Atheros Device [168c:003e] (rev 20)         Subsystem: Lite-On Communications Inc Device [11ad:0804]  The OS behaves like it has no wifi card. I do not see wifi settings in network settings I'm not a linux power user, so I do not know what to do to solve the issue. What should I do to make it work?