I am currently making a space game. The player's ship is stationary at world origin while things come towards it. However, the player is supposed to move towards the next destination.  To create the illusion of movement, I have particles moving backwards while my ship remains stationary. However, the particles seems to get lost visually during complex battle scenes.  Is there any other way to give the illusion of movement where there is an absence of stationary reference points?  Here are some screenshots of the game       The referenced question that has been identified as a duplicate does not answer my question. It is somewhat unclear what some of the implementations are. I seek a tried example that is used in other games if possible.  Most of the implementations suggested is more relevant to a first person perspective. I am looking for 3rd person effect where the player cannot "Visually Feel" the effect through a jolt.
I'm making a game where you see a spaceship (in 3rd person) travelling at high-speed between planets (it takes 2-3 seconds move between them).  What effect could I use to convey the effect of extreme speed? I already know about the usual effects, change in FOV, white lines around the subjects, etc..   I'm asking about some things that are more space-like, I mean even if it's not how it would be perceived in real physics, I need it to look like it's fast and it's in space.  Don't hesitate to ask for clarification, my English is nowhere near being perfect, maybe I didn't convey my thought perfectly.
How do I draw the graph of inequalities in the coordinate plane? I want to graph the following inequalities (done in desmos) in latex. I need the code. Do I have to worry about copyright issues for code posted on Stack Overflow? I was wondering about the copyright of code snippets from answers. Let's say you ask a question and the solution consists of ten lines of code, which are given in an answer. Are there any restrictions about the use of that code?   It seems that the content of this page is under the . So does one have to attribute the author if code is used in an open source project? Can it be legally used in closed source projects at all?  I guess ten lines of code are not really problematic, but is there a critical number of lines? Let $A \subset B$ be a finite ring extension. Fix a ring homomorphism $\nu : A \to \Omega$, where $\Omega$ is an algebraically closed field.   I want to show there exists a non-zero homomorphism $v : B \to \Omega$ of $A$-modules.  I want show that the space $V$ of all the homomorphisms $v$ in (i) is a non-zero finite dimensional vector space over $\Omega$. Show that the algebra $B$ acts on this space. Show that eigenvectors of this action correspond to homomorphisms of algebras $\lambda : B \to \Omega$ that extend the homomorphism $\nu$. Deduce that the set of such homomorphisms $\lambda : B \to \Omega$ is finite and non-empty.    We can assume that $B = Ab_1 + ... + Ab_n$ where $b_1,...,b_n \in B$. The chose $v(b_i)=\omega_i \in \Omega-\{0\}$ and then define $$ v( a_1 b_1 + ... + a_n b_n ) = \nu(a_1) v(b_1) + ... + \nu(a_n) v_n = \nu(a_1) \omega_1 + ... + \nu(a_n) \omega_n \ .$$ This is an $A$-module homomorphism. Am I correct? For the second part, that $V$ is not-zero follows from the first part. That it is finite dimensional follows from the fact that $v_i(b_j)=\delta_{i,j}$ (Kronecker's delta) with $i=1,...,n$ spans $V$ and hence $\dim V \le n &lt; \infty$. I define the action of $B$ over $V$ to be $$b \bullet v = \left( x \mapsto v(b)v(x) \right) \mbox{  for all } b,x \in B \ .$$ This seems to be the natural action to define. Am I correct? However, I haven't managed to solved the rest (the eigenvector stuff and the relation to $\nu$). How can I re-enable backspace navigation in Chrome? Let $f(\sin x)$ be a given function of $\sin x$.   How would I show that $\int_0^\pi xf(\sin x)dx=\frac{1}{2}\pi\int_0^\pi f(\sin x)dx$? Why is "rm" so much slower than deleting files via the File Manager? Can spells or spell-like abilities do non-lethal damage? I'm working on a Paladin/Warlock combination, based off of , and as part of this, I noticed Miniman's remark that Repelling Blast won't be that useful, because as a Paladin, you want people up close and personal when you're in a combat situation, not further away.  However, as a Paladin, your creed sometimes leads you to not want to engage in full-blown combat with a potential adversary -- you may want to drive them off or incapacitate/capture them instead.  With this, I wonder: can Repelling Blast, or other spells and spell-likes (such as invocations), be made to do non-lethal damage?  If so, I'm thinking that it'd be useful in exactly that sort of situation -- halting someone/pushing them away without having to wade into a melee with them or blow them to smithereens. Can someone explain me this  please I am a bit confused?  It's from a texbook I am reading (block cipher companion)  it says that:      each weak key has  $2^{32}$ fixed points $m$ where $\operatorname{DES}_w(m) = m$ ( $w$ is a DES weak key). In other words encrypting a fixed point gives exactly the same point. To see this consider the $2^{32}$ ciphertexts after 8 rounds of encryption for which the two 32 bit halves are equal. For a weak  DES key, the round keys of the eighth and ninth rounds are equal. Thus the intermediate texts after seven rounds and after nine rounds of encryption will be equal. Since the round keys of the 7th and the 10th rounds are equal, the intermediate texts after 6 rounds and after 10 rounds of encryption will be equal. Continuing like this it follows that the plaintext will be equal to the ciphertext.    Now did not quite get :   Why are there $2^{32}$ fixed points?  And why after the eighth round of encryption the two halves must be equal? What does the author mean by the intermediate texts exactly? And how are they being equal?   P.S.: This question is not asking for what the weak keys are. I am asking for "fixed points" not exactly weak keys. For this reason I think it is not a duplicate of any other question. There is one similar question to what I've asked for which I have checked and saw that no satisfactory answer has been given there either.  Here is the similar question with no proper answers: Is PCA still done via the eigendecomposition of the covariance matrix when dimensionality is larger than the number of observations? I have a $20\times100$ matrix $X$, containing my $N=20$ samples in the $D=100$-dimensional space. I now wish to code up my own principal component analysis (PCA) in Matlab. I demean $X$ to $X_0$ first.  I read from someone's code that in such scenarios where we have more dimensions than observations, we no longer eigen-decompose the $X_0$'s covariance matrix. Instead, we eigen-decompose $\frac{1}{N-1}X_0X_0^T$. Why is it correct?  The normal covariance matrix is of size $D\times D$, each element of which tells us the covariance between two dimensions. To me, $\frac{1}{N-1}X_0X_0^T$ is not even of the correct dimensions! It is $N\times N$ matrix, so what would it tell us? Covariance between two observations?! The mediation package in R returns results in which:   Average Direct Effect (ADE) (the unmediated effect) is not statistically significant The Average Causal Mediated Effect (ACME) (the effect of the mediator alone) is positive and statistically significant The Total Effect (ADE+ACME) is larger than the ADE and ACME alone, positive and statistically significant   Does this indicate that the variable being mediated can only act through the mediator DV? I'm not quite certain as to how this would be interpreted.  Thank you for any help. For $a,b,c,d \in \mathbb{R}$, is $\sqrt{a^2+b^2} + \sqrt{c^2+d^2} \geq \sqrt{(a+c)^2 + (b+d)^2}$? I have tried to show this using the fact that $\sqrt{x+y}\leq\sqrt{x}+\sqrt{y}\; $  for non-negative $x,y$ but haven't been able to arrive at a proof. To get rid of dust that had gathered near the fan, I put a vacuum cleaner to the air intake (with the computer turned off).  This removed the dust.  It also caused the fan to rotate at a fairly high speed. Quite faster than normal speed, judging from the sound.  Do I risk destroying stuff by doing this? Why is $\frac{1}{\frac{1}{0}}$ undefined? I am planning on doing a fresh install of Ubuntu and want to know what is the correct way to install ATI Catalyst Video Driver?      There are multiple valid answers for this question spanning over several versions of Ubuntu. For your convenience an index of each below: How to get a list of applications installed in Ubuntu dash? In Gnome, we have applications grouped up based on their type (like - games, accessories, et al). But in Unity, we need to manually type in the Unity to get the App.  Is there any keyword/utility to display the applications in GUI?  Ubuntu Version : 14.04 Hardware : Dell Inspiron 3437 Cannot create custom config entity fails because ID is blank jquery's append not working with svg element? Do universities hire graduates from lesser universities Recently, someone analyzed  and found that over half of the professors at the top 51 universities graduated from a top 10 university.  have also brought this up. From my personal observations, most schools do hire graduates from better ranked schools.  Is this because of the competitive job market? We have so many good applicants, we have to narrow it down some how! Or is it simply that these schools produce the most PhDs? Has this always been the case? How rare are exceptions to this? I know of a few people who graduated from a top 75 school and got hired at a top 50 school. But what about bigger gaps? The top 10 schools seem to just swap graduates, do they ever hire from a 50+ ranked school?  Update 2018: I have accepted a tenure-track position at a top 75 department at an R1 university immediately after graduating from an unranked department at an R2 university. It does happen! This may or may not generalize to other fields and countries. Where to put my code: plugin or functions.php? Given a ring $R$ (with unity), prove that if $a \in R$ is nilpotent, then $a+1,a-1$ are both invertible.   Suppose $a^n=0$. Then $1=1-a^n=(1-a)(1+a+\ldots+a^{n-1})$, so $1-a$ is invertible.  If $n$ is odd we can use $1+a^n=(1+a)(1-a+\ldots+a^{n-1})$.  But what if $n$ is even?