I'm developing a voxel game with OpenGL and want to optimize the number of rendered faces. How can I estimate how many faces at most a modern graphics card of a certain brand is able to compute, while still managing to render at 60fps?
With me going it alone in development I only have limited resources at my disposal i.e one pc with which I develop and test the game. so I only know for certain my game works with that setup.  What options have I got in testing the game on the many different hardware/operating system combinations? Its a windows desktop game.
Prevent external devices being locked to launcher If I connect an external device (such as a USB stick or android smartphone) it will automatically open in nautilus and get a corresponding icon locked to unity launcher. Is there any way to prevent this?  I'm using Ubuntu 14.04 How to rename multiple files by removing the extension? Best practice for Rails App to run a long task in the background? I have a Rails application that unfortunately after a request to a controller, has to do some crunching that takes awhile.  What are the best practices in Rails for providing feedback or progress on a long running task or request?  These controller methods usually last 60+ seconds.    I'm not concerned with the client side... I was planning on having an Ajax request every second or so and displaying a progress indicator.  I'm just not sure on the Rails best practice, do I create an additional controller?  Is there something clever I can do?  I want answers to focus on the server side using Rails only.  Thanks in advance for your help.  Edit:  If it matters, the http request are for PDFs.  I then have Rails in conjunction with Ruport generate these PDFs.  The problem is, these PDFs are very large and contain a lot of data.  Does it still make sense to use a background task?  Let's assume an average PDF takes about one minute to two minutes, will this make my Rails application unresponsive to any other server request during this time?  Edit 2:  Ok, after further investigation, it seems my Rails application is indeed unresponsive to any other HTTP requests after a request comes in for a large PDF.  So, I guess the question now becomes: What is the best threading/background mechanism to use?  It must be stable and maintained.  I'm very surprised Rails doesn't have something like this built in.  Edit 3:  I have read this page: .  I would love to read about various experiences with these tools.  Edit 4:  I'm using Passenger Phusion "modrails", if it matters.  Edit 5:  I'm using Windows Vista 64 bit for my development machine; however, my production machine is Ubuntu 8.04 LTS.  Should I consider switching to Linux for my development machine?  Will the solutions presented work on both? Hello Physics StackExchange!  I will put the TL;DR in the beginning: I need a self contained, relatively hand-holding sequence of textbooks that covers up from the end of Griffith's Intro to QM to relativistic QM/particle physics to QFT.    This will be my first post. I'm a medical physics grad student who majored in biomedical engineering for my undergrad, with serious regrets about not doing math/physics double major in hindsight. Oh well. I've made it my goal to eventually understand QFT.  My background in relevant subjects is Calc 1 to 3 and Diff EQ, 2 part freshman physics, an E&amp;M course that used Griffiths Electrodynamics, a summer of self studying Griffith's QM (got most of the way through). I will also be officially taking a class in QM using Griffith's textbook this next spring. I am aware of the Lagrangian and Hamiltonian approaches to mechanics as well. In addition I have familiarity with special and basic general relativity using tensors and all this. What would be the best way for me to get from where I am now to a strong understanding of QFT autodidactically? I have no aversion to studying pure maths where I need to.  EDIT: My question is much different than simply asking for a handholding explanation of QFT. Generally these textbooks assume a certain amount of physics/math background, and I was asking for a sequence of textbooks leading up to QFT from the end of a specific textbook, Griffith's QM. Did you even read my question? Did you read the thing you linked me to? How to pivot rows into columns MySQL Been looking to pivot rows (progress_check) into columns check 1, check 2 etc... No sum or totals required, just to display the results hopefully... Can anyone help, Thanks Ad Let $A$ be a diagonalizable matrix and $\lambda$ be an eigenvalue of A. Prove that rank($\lambda I - A$) = rank(($\lambda I - A)^2$). Let $A$ be a diagonalizable matrix and $\lambda$ be an eigenvalue of A. Prove that rank($\lambda I - A$) = rank(($\lambda I - A)^2$).  The hint they provided is that for any invertible matrix B, rank(AB) = rank (A) and rank(BA) = rank(A). Is it true that: $a+bi$ is prime in $\mathbb{Z}[i]$ if and only if $a^2+b^2$ is prime in $\mathbb{Z}$ Getting server name during servlet initialization Theoretical Proof of Pauli's Exclusion Principle How to choose capacitor for an IC I'm a beginner in electronics from the software industry. With some self-taught things, I'm trying to implement some basic Arduino circuits. My confusion is about capacitors mainly. My understanding about capacitors is that they act as power storage for few seconds or milliseconds.  I found that most IC's must have capacitors connected to their pins.  My confusion is how to find out which pins need a capacitor, and how to find the correct capacitor for a circuit or a capacitor for an IC.  Finally, why are capacitors necessary in a circuit in such situations? AMD GPU Material in "material preview mode" and "rendered mode", is entirely gray as background. Only outline is displayed Best way to build a Plugin system with Java How would you implement a Plugin-system for your Java application?  Is it possible to have an easy to use (for the developer) system which achieves the following:   Users put their plugins into a subdirectory of the app The Plugin can provide a configuration screen If you use a framework, is the license compatible with commercial developement? What happens when a Wizard loses Intelligence after preparing his spells? A Wizard prepares a number of spells equal to his Wizard level + his Intelligence modifier.  But what happens if he loses Intelligence (the most obvious way I can think of is unattuning to a Headband of intellect and having a base Intelligence score of less than 18) after preparing his spells? Does he...   Lose access to the first [INT mod lost] spells he prepared? Lose access to the last [INT mod lost] spells he prepared? Randomly lose access to [INT mod lost] of the spells he prepared? Keep access to all his prepared spells (but will be able to prepare fewer spells after his next long rest, unless he regains the lost intelligence)? I would like to know the purpose of nohup these days. I ask because I read this  Nohup is short for “No Hangups.” It’s not a command that you run by itself. Nohup is a supplemental command that tells the Linux system not to stop another command once it has started. That means it’ll keep running until it’s done, even if the user that started it logs out. The syntax for nohup is simple and looks something like this: nohup sh your-script.sh &amp; Notice the “&amp;” at the end of the command. That moves the command to the background, freeing up the terminal that you’re working in. Nohup works with just about any command that you run in the terminal. It can be run with custom scripts as well as standard system commands and command line utilities.  With using linux distributions such as SLES 11.4, or RHEL/CentOS 7.x, I can  remote log in to linux over the network via SSH run the simple program below, or any other, doing ./a.out &amp; log out of linux; type exit in the putty SSH terminal my sample program below, or any other including bash, csh, or tcsh shell scripts, run to completion just fine without the need for nohup typing exit in putty to close an SSH session, I would think that qualifies as a logout?  Is there any purpose or value with nouhup today?  Is it a kernel &gt;= 3.x thing? For however long I can remember (kernel &gt;= 2.6) I've never used nohup and have always just use &amp; with no problems. Can someone give me a practical scenario where nohup would be used? #include &lt;stdlib.h&gt; #include &lt;stdio.h&gt;  /*     sample C code, to print numbers to a file named zz.tmp     the numbers 0 to 29 over 30 seconds */  int main ( int argc, char *argv[] ) {    FILE *fp;    int i;     i = 0;    fp = fopen(&quot;zz.tmp&quot;, &quot;w&quot; );     while ( i &lt; 30 )    {       fprintf( fp, &quot;%d\n&quot;, i++ );       sleep( 1 );    }    fclose( fp );    return 0; }  mysleep.sh #!/bin/bash  sleep 1000  On Centos 7.7, optiplex pc login with keyboard &amp; mouse, my account is bash. I do nohup ./mysleep.sh and then control-z.  When typing jobs I see [1]+ Stopped nohup ./mysleep.sh A ps -ef | grep mysleep shows that process id existing, but if I close the terminal window via X in upper right corner, that process goes away so it seems nohup is not working? How to remove comma between taxonomy terms? I style my taxonomy terms as boxes, so I don't need comma between them.  Could somebody give me any advice? On the uniqueness of the Riemann-Christoffel tensor According to Section 6.2, Gravitation and Cosmology by Weinberg, the Riemann-Christoffel tensor is the only tensor that can be constructed out of the second (or lower) order derivatives of the metric tensor and is linear in the second order derivatives. The reasoning behind the same goes like this:   In a class of frames where $\Gamma^{\lambda}_{\mu\nu} =0$, the transformation rule for $\dfrac{\partial \Gamma^{\lambda}_{\mu\nu}}{\partial x^{\kappa}}$ involves an inhomogeneous term which is symmetric in $\mu$, $\nu$, and $\kappa$. Thus, if one is to construct a tensor which is a linear combination of the first order derivatives of the Christoffel symbol then the only way to do so is by eliminating the inhomogeneous part of the transformation and this could be done only by making the combination explicitly antisymmetric in $\mu$ and $\kappa$. Since in these frames, $R^{\lambda}_{\mu\nu\rho}$ $ = \dfrac{\partial \Gamma^{\lambda}_{\mu\nu}}{\partial x^{\rho}} -  \dfrac{\partial \Gamma^{\lambda}_{\mu\rho}}{\partial x^{\kappa}} $, $R^{\lambda}_{\mu\nu\rho}$ is the only tensor that can be formulated using the second (or lower) order derivatives of the metric tensor and is linear in the second order derivatives.  I think the presented argument can only suffice to prove that $R^{\lambda}_{\mu\nu\rho}$ is the only tensor that can be formulated from the first derivatives of the Christoffel symbol and is linear in them. I can't figure out why this suffices to assert that $R^{\lambda}_{\mu\nu\rho}$  is the only tensor that can be constructed out of the second (or lower) order derivatives of the metric tensor and is linear in the second order derivatives.  Edit: As the metric is covariantly constant, $\dfrac{\partial g_{{\mu}{\nu}}}{\partial x^{\rho}} = \Gamma^{\kappa}_{{\mu}{\rho}}g_{{\kappa}{\nu}}+\Gamma^{\kappa}_{{\nu}{\rho}}g_{{\kappa}{\mu}}$.  Therefore,   $\dfrac{\partial^2 g_{{\mu}{\nu}}}{\partial x^{\xi}\partial x^{\rho}} = \bigg(\dfrac{\partial \Gamma^{\sigma}_{{\mu}{\rho}}}{\partial x^{\xi}} + \Gamma^{\sigma_1}_{{\mu}{\rho}}\Gamma^{\sigma}_{{\sigma_1}{\xi}}\bigg)g_{{\sigma}{\nu}}+\bigg(\dfrac{\partial \Gamma^{\sigma}_{{\nu}{\rho}}}{\partial x^{\xi}} + \Gamma^{\sigma_1}_{{\nu}{\rho}}\Gamma^{\sigma}_{{\sigma_1}{\xi}}\bigg)g_{{\sigma}{\mu}} + \bigg(\Gamma^{\sigma}_{{\mu}{\rho}}\Gamma^{\sigma_1}_{{\nu}{\xi}}+\Gamma^{\sigma}_{{\nu}{\rho}}\Gamma^{\sigma_1}_{{\mu}{\xi}}\bigg)g_{{\sigma}{\sigma_1}} $  Now, if I want to show that the Riemann-Christoffel tensor is the only (non-trivial and independent) tensor that can be formulated out of the linear combinations of the second order derivatives of the metric tensor then I can equivalently show that the above expansion can be expressed linearly in terms of the Riemann-Christoffel tensor. But I am stuck over how to do it. Also, I am a little bit unclear as to what a linear combination means in this context. The coefficient of the second order term can be only a scalar constant or can it be the metric or can it be the metric that is getting summed over any of the indices as well? How to prove this decomposition of sum of squares? I am reading a paper   It has a mathematical step shown as  $$\sum_{i=1}^{n}(x_i-\mu)^2=n(\mu-\overline x)+\sum_{i=1}^n(x_i-\overline x)^2$$  Here the $\mu$ is unknown parameter and $\overline x$ is the known mean of $\mathbf x=(x_1, x_2, x_3, \dots,x_n)$  How do I demonstrate this identity? How do I create a 50x3 cell array from a 50x3 matrix. Keep just getting a 1x1 cell array with a 50x3 submatrix? How does the integral $\int_{D_C} e^{ia z}P(z)/Q(z)\,\mathrm{d}z$ blow up. In my book I have a theorem that goes something like the following     Let $P(x)$ be $Q(x)$ polynomials such that $\deg(Q) \geq \deg(P) + 2$.   Then \begin{align*}  \int_{-\infty}^{\infty} \frac{P(x)}{Q(x)}  e^{iax} \,\mathrm{d}x  = 2 \pi i \frac{a}{|a|} \sum_{k=1}^{m} \mathrm{Res}\left[  \frac{P(x)}{Q(x)} , z_k \right] \end{align*} where $a$ is a real   constant and $z_1,\,\ldots\,,z_m$ are the singularities to $P(x)/Q(x)$ in   the upper halfplane if $a&gt;0$ and the lower halfplane if $a&lt;0$.   I am having a bit of problems understanding this. I know the reason why we have to switch contour is that $e^{iz}$ blows up. But I really can not see why or how it blows up.   Take the canonical example for why this theorem is useful  $$   J = \int_{-\infty}^\infty \frac{e^{iz}}{z^2+1} $$    Now the clue here is to show that the integral along the curve tends to zero, and then use the residue theorem. To show that the integral tends to zero I did this \begin{align*}  \left| \int_{C_1} \frac{e^{iz}}{1+z^2} \,\mathrm{d}z\right|  \leq \int_{C_1} \left| \frac{e^{iz}}{1+z^2} \right| \,\mathrm{d}z   \leq \sup_{z = R e^{i \theta}} \left( \frac{1}{\left|1+z^2\right|} \right)       \int_{C_1} |e^{iz}| \,\mathrm{d}z \end{align*} where the $ML$-inequality was used in the last inequality  On the circle with radius $R$ we have $|e^{iz}|=R$, and we can use the inequality $|a+b|\leq|a|-|b|$ to simplify further \begin{align*}  \left|\int_{D_R} \frac{e^{iz}}{1+z^2}\,\mathrm{d}z \right|  \leq \sup_{z = R e^{i \theta}} \left( \frac{1}{|z|^2-1}\right)  \int_{D_R} R \,\mathrm{d}z  \leq \pi \frac{R}{R^2-1} \end{align*} which tends to zero as $R \to \infty$ as wanted. but if one instead had $e^{-iz}$ then the theorem states that one has to use the lower half plane. But I do not see where my calculations err if one persists in using the contour in the upper half plane? What goes wrong, and why does it go wrong? $|e^{-iz}|$ should still be $R$ on the semi-circle.   How can one formally show that the integral diverges when picking the contour as a semi circle in the lower half plane? My headphones are not playing voices, but only the music