I am converting between coordinate systems where Z is the vertical axis in one, and Y in another. I have the rotation values represented as a quaternion.  I see this related issue but it's switching the handedness (DirectX and OpenGL):   How would I go about swapping these values? I'd prefer not to convert to a vector3, modify and convert back again.
I have a quaternion, coming from a system with the following:(intertial sensor)  Right handed. Forward direction: Y axis Right direction: X axis Up direction: Z axis   I need to convert this into a coordinate system that is: (Unreal engine)  left-handed. Forward direction: X axis Right direction: Y axis Up direction: Z axis   I have tried negating the axis, and angle, I have tried switching values, i cannot get this to work. All help greatly appreciated! I am working in C#, with Microsoft.Xna.Quaternion.
std::string formatting like sprintf Same gravatar is displayed for different users in the stackexchange.com , under the Top Network Askers in the right side bar.   The red circled image's source (&lt;img src=) are from   https://www.gravatar.com/avatar/?s=18&amp;d=identicon&amp;r=PG&amp;s=30                                ^   Seems the gravatar id is missing here in the image source.  For an example from the list, the user  gravatar id should be    https://www.gravatar.com/avatar/66c66c4d97a7bc3859a85253f003e1fb?s=18&amp;d=identicon&amp;r=PG&amp;s=30   but it is mapped with   https://www.gravatar.com/avatar/?s=18&amp;d=identicon&amp;r=PG&amp;s=30   which caused the display issue. A zero-dimensional ring is Noetherian? Proposition: Let $A$ be a non-zero ring that is not a field. Suppose $A$ is zero dimensional. Then it is Noetherian.   Proof: Let $p$ be a prime ideal of $A$. If $p$ is not maximal, then $p \subsetneq m$ for a maximal ideal $m$. Hence $dim A \ge 1$, contradiction. Hence every prime ideal is maximal. Consequently, every increasing (or decreasing) chain of prime ideals will be trivial in the sense that it will have length $0$. Hence the ring is Noetherian (and Artinian).  Is there a problem with the above proof? What command can I type into the Terminal so that I can delete all .svn folders within a folder (and from all subdirectories) but not delete anything else? I would like to know if there is any way to add custom behaviour to the auto property get/set methods.  An obvious case I can think of is wanting every set property method to call on any PropertyChanged event handlers as part of a System.ComponentModel.INotifyPropertyChanged implementation.  This would allow a class to have numerous properties that can be observed, where each property is defined using auto property syntax.  Basically I'm wondering if there is anything similar to either a get/set template or post get/set hook with class scope.  (I know the same end functionality can easily be achieved in slightly more verbose ways - I just hate duplication of a pattern) Why aren't there any common words for 'defecating' and 'urinating'? Error entering equation in DSolve If I take the $k \in \mathbb{N}$ power of 10 and mod it by a large prime, I notice that the remainders repeat at some point.  For instance $10^k mod~7$ seems to repeat every $8$th value of $k$.  Given $k$ (the power of ten), and a large prime number $p$, how can I compute the size of the cycle?  I tried writing a python script that computes the first instance of a repeat. However, I don't know if this works since a single instance of a repeat may not necessarily mean a full cycle. When for the first time A new question I answer This haiku appears:     Little slab of meat   In a wash of clear jelly   Now I heat the pan   Such a message is Very strange; is anyone Seeing the same thing?  Have not seen again. Came and went like the wind does! Why I am so blessed? What happens when a familiar from the Find Familiar spell "dies"? I am a student who has taken the basic calculus courses but I am working through Calculus (the fourth edition) by Spivak in my spare time in order to both review the material and to gain a more rigorous understanding of the concepts. I am currently in Chapter 8 where he supplies a proof of the basis of the Intermediate Value Theorem, Theorem 7-1 (pp. 135-136 of the fourth edition). It seems mostly straightforward but I think there is some nuance of it that I don't grasp.  I am including in this post the statement Theorem 6-3 since he references it in the proof of Theorem 7-1 as well as part of Theorem 7-1 itself.     Theorem 6-3 Suppose $f$ is continuous at $a$, and $f(a) &gt; 0$. Then $f(x) &gt; 0$ for all $x$ in some interval containing $a$; more precisely, there is a number $\delta &gt; 0$ such that $f(x) &gt; 0$ for all $x$ satisfying $|x-a| &lt; \delta$. Similarly, if $f(a) &lt; 0$, then there is a number $\delta &gt; 0$ such that $f(x) &lt; 0$ for all $x$ satisfying $|x-a| &lt; \delta$.     Problem 6-16 is also referenced. However,it is the same thing as Theorem 6-3 for one-sided limits. The following proof is the Theorem 7-1 up through the paragraph which I don't fully understand.     Theorem 7-1: If $f$ is continuous on $[a,b]$ and $f(a) &lt; 0 &lt; f(b)$, then there is some number $z$ in $[a,b]$ such that $f(x) = 0$.  Proof: Define the set $A$ as follows:  $$A = \{x : a \le x\le b, \mbox{ and } f \mbox{ is negative on the interval } [a,x] \}.$$  &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Clearly $A \ne \emptyset$, since $a$ is in $A$; in fact, there is some $\delta &gt; 0$ such that $A$ contains all points $x$ satisfying $a \le x &lt; a + \delta$; this follows from Problem 6-16, since $f$ is continuous on $[a,b]$ and $f(a)&lt;0$. Similarly, $b$ is an upper bound for $A$ and, in fact, there is a $\delta &gt; 0$ such that all points $x$ satisfying $b-\delta &lt; x \le b$ are upper bounds for $A$; this also follows from Problem 6-16, since $f(b) &gt; 0$. &nbsp;&nbsp;&nbsp;&nbsp;&nbsp; From these remarks it follows that $A$ has a least upper bound $\alpha$ and that $a &lt; \alpha &lt; b$. We now wish to show that $f(\alpha) = 0$, by eliminating the possibilities $f(\alpha) &lt; 0$ and $f(\alpha) &gt; 0$. &nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Suppose first that $f(\alpha) &lt; 0$. By Theorem 6-3, there is a $\delta &gt; 0$ such that $f(x) &lt; 0$ for $\alpha - \delta &lt; x &lt; \alpha + \delta$. Now there is some number $x_0$ in $A$ which satisfies $\alpha - \delta &lt; x_0 &lt; \alpha$ (because otherwise $\alpha$ would not be the least upper bound of $A$). This means that $f$ is negative on the whole interval $[a,x_0]$. But if $x_1$ is a number between $\alpha$ and $\alpha+\delta$, then $f$ is also negative on the whole interval $[x_0,x_1]$. Therefore $f$ is negative on the interval $[a,x_1]$, so $x_1$ is in $A$. But this contradicts the fact that $\alpha$ is an upper bound for $A$; our original assumption that $f(\alpha) &lt; 0$ must be false.    The point I don't understand is in the final paragraph of the excerpt. Why does Spivak split the interval $[a,b]$ up into the subintervals $\lbrack a,x_0 \rbrack$ and $[x_0,x_1]$? Isn't it true that, because $\alpha$ is a least upper bound of $A$ and $f$ is continuous on $[a,b]$, it must be true that if $a \le x &lt; \alpha$ then $f(x)&lt;0$? I understand the need to pick out the $x_1$ but not the $x_0$. Especially perplexing is the statement "there is some number $x_0$ in $A$ which satisfies $\alpha - \delta &lt; x_0 &lt; \alpha$ (because otherwise $\alpha$ would not be the least upper bound of $A$." Wouldn't it be true that if there were some $x_0$ such that $a \le x_0 &lt; \alpha$ and $f(x_0) \ge 0$ that $x_0$ would be an upper bound for $A$, contradicting that $\alpha$ is the least upper bound of $A$?  Thanks for any help. Are PDO prepared statements sufficient to prevent SQL injection? Is there a C++ decompiler? Purge unused data (e.g. particle system, or groups) by script How can I completely remove a objects (e.g. particle system or groups used in particle systems) in blender using python?  The problem is this: I have a script, with which I create a scene (e.g. adding some normal objects and some group of objects with a particle system). For a research study I would like to repeat that that several times, without restarting blender. So I can create a scene, delete everything, e.g. with a script like this:  import bpy scn = bpy.context.scene for obj in scn.objects:     obj.select = True bpy.ops.object.delete()   If I rerun the script to recreate the scene with objects using particle system, the latter is not shown in the scene anymore (only the placeholders of the particle system). Now I'm not sure where exactly the problem lies, but I guess not everything is immediately deleted, running the script above. Only the startup of blender helps (but that's not what I want) or pushing the purge button. The purge button seems to start the command:  bpy.ops.outliner.orphans_purge()   But I can't get this piece of code to run in the console. It says that "the context is incorrect". Is there a way to purge the unused data by python script? Thanks a lot for any help. Although I have installed the OpenLayers plugin in QGIS 2.4, I cannot see it in the list when I click on Plugins.    Does anyone else have this issue? How does light speed up after coming out of a glass slab? How do I efficiently iterate over each entry in a Java Map? If I have an object implementing the Map interface in Java and I wish to iterate over every pair contained within it,  what is the most efficient way of going through the map?  Will the ordering of elements depend on the specific map implementation that I have for the interface? I have customers who want to buy large amount of flash memories but he wants to copyright all files inside it.   The flash will contain files that will be delivered for his customers but they're not allowed to copy or delete any of the files the only allowed permission is to ready the files.   I searched and found many solutions but all have on function it WriteProtect the USB though I can't add files to Drive at all but I can delete or copy any file in the drive.   Is there any way to protect the flash memory so I can't copy data out of it? In discussing , I noted that while the statement:     Any vector space has a basis   is equivalent to the axiom of choice, I wondered if the statement that:     Any vector space either has a finite basis or an infinite set of linear independent vectors   was weaker than the axiom of choice. It feels weaker - it feels like you can inductively define a countably infinite set of linearly independent vectors without full choice, but I'm not sure if that is possible with ZF alone, or requires all of the Axiom of Choice, or is weaker but requires some form of choice. Input image strip without stretching to fit output frame When you add an image strip to the Video Sequence Editor, it stretches/shrinks your image to fit the full output frame.  How can I add an image at its original resolution?  For example, if I add a 60x60 pixel image, it should appear very small in the final 1080p render.