C++ 欧拉角到四元数然后四元数到欧拉角
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原文地址: http://stackoverflow.com/questions/11103683/
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Euler angle to Quaternion then Quaternion to euler angle
提问by user1466739
I'm using lib glm (http://glm.g-truc.net/) for test quaternion but I've a problem; when I convert euler angle to quaternion then immediatly quaternion to euler angles, my result are totally different from my initial euler angles. Is this normal? Could it be because the rotations are not communative?
我正在使用 lib glm ( http://glm.g-truc.net/) 进行四元数测试，但我遇到了问题；当我将欧拉角转换为四元数然后立即将四元数转换为欧拉角时，我的结果与我最初的欧拉角完全不同。这是正常的吗？可能是因为轮换不是可交流的吗？
Code test:
代码测试：
#include <glm\quaternion.hpp>
#include <math.h>
#define PI M_PI
#define RADTODEG(x) ( (x) * 180.0 / PI )
#define DEGTORAD(x) ( (x) * PI / 180.0 )
int main( void )
{
float RotX = 90.f;
float RotY = 180.f;
float RotZ = -270.f;
if ( RotX || RotY || RotZ )
{
std::cout << "Init: x= " << RotX << ", y= " << RotY << ", z= " << RotZ << "\n";
glm::quat key_quat(glm::detail::tvec3<float>(DEGTORAD( RotX ),
DEGTORAD( RotY ),
DEGTORAD( RotZ )));
glm::detail::tvec3<float> v = glm::eulerAngles(key_quat);
/* // the result is even worse with this code here
RotX = RADTODEG(v.x);
RotY = RADTODEG(v.y);
RotZ = RADTODEG(v.z);
*/
RotX = v.x;
RotY = v.y;
RotZ = v.z;
std::cout << "Final: x= " << RotX << ", y= " << RotY << ", z= " << RotZ << "\n";
}
return (0);
}
Result:
结果：
Init: x= 90, y= 180, z= -270
Final: x= -90, y= -3.41509e-006, z= -90
thank you in advance o/
提前谢谢你o/
回答by Ali
Yes, it is normal. There are 2 waysto represent the same rotation with Euler angles.
是的，这是正常的。有两种方法可以用欧拉角表示相同的旋转。
I personally don't like Euler angles, they mess up the stability of your app.I would avoid them. Plus, they are not very handyeither.
我个人不喜欢欧拉角，它们会破坏应用程序的稳定性。我会避开他们。另外，它们也不是很方便。
回答by Jav_Rock
Have a look at this page. It has everything you need (even some code samples!) for dealing with 3D transformations.
看看这个页面。它拥有处理 3D 转换所需的一切（甚至是一些代码示例！）。
回答by frodo2975
If you end up needing quaternion's to Euler angles, but you need an arbitrary rotation order, I came across a site with conversion code. Sometimes the trick is just finding the right rotation order. (Btw, the orders that have the same letter twice, like XYX, are proper Euler angles, but the ones like XYZ are Tait-Bryan angles).
如果您最终需要四元数到欧拉角，但您需要任意旋转顺序，我遇到了一个带有转换代码的站点。有时诀窍就是找到正确的旋转顺序。（顺便说一句，两次具有相同字母的阶数，如 XYX，是正确的欧拉角，但像 XYZ 的阶数是 Tait-Bryan 角）。
Here's the link: http://bediyap.com/programming/convert-quaternion-to-euler-rotations/
这是链接：http: //bediyap.com/programming/convert-quaternion-to-euler-rotations/
And here's the code:
这是代码：
///////////////////////////////
// Quaternion to Euler
///////////////////////////////
enum RotSeq{zyx, zyz, zxy, zxz, yxz, yxy, yzx, yzy, xyz, xyx, xzy,xzx};
void twoaxisrot(double r11, double r12, double r21, double r31, double r32, double res[]){
res[0] = atan2( r11, r12 );
res[1] = acos ( r21 );
res[2] = atan2( r31, r32 );
}
void threeaxisrot(double r11, double r12, double r21, double r31, double r32, double res[]){
res[0] = atan2( r31, r32 );
res[1] = asin ( r21 );
res[2] = atan2( r11, r12 );
}
void quaternion2Euler(const Quaternion& q, double res[], RotSeq rotSeq)
{
switch(rotSeq){
case zyx:
threeaxisrot( 2*(q.x*q.y + q.w*q.z),
q.w*q.w + q.x*q.x - q.y*q.y - q.z*q.z,
-2*(q.x*q.z - q.w*q.y),
2*(q.y*q.z + q.w*q.x),
q.w*q.w - q.x*q.x - q.y*q.y + q.z*q.z,
res);
break;
case zyz:
twoaxisrot( 2*(q.y*q.z - q.w*q.x),
2*(q.x*q.z + q.w*q.y),
q.w*q.w - q.x*q.x - q.y*q.y + q.z*q.z,
2*(q.y*q.z + q.w*q.x),
-2*(q.x*q.z - q.w*q.y),
res);
break;
case zxy:
threeaxisrot( -2*(q.x*q.y - q.w*q.z),
q.w*q.w - q.x*q.x + q.y*q.y - q.z*q.z,
2*(q.y*q.z + q.w*q.x),
-2*(q.x*q.z - q.w*q.y),
q.w*q.w - q.x*q.x - q.y*q.y + q.z*q.z,
res);
break;
case zxz:
twoaxisrot( 2*(q.x*q.z + q.w*q.y),
-2*(q.y*q.z - q.w*q.x),
q.w*q.w - q.x*q.x - q.y*q.y + q.z*q.z,
2*(q.x*q.z - q.w*q.y),
2*(q.y*q.z + q.w*q.x),
res);
break;
case yxz:
threeaxisrot( 2*(q.x*q.z + q.w*q.y),
q.w*q.w - q.x*q.x - q.y*q.y + q.z*q.z,
-2*(q.y*q.z - q.w*q.x),
2*(q.x*q.y + q.w*q.z),
q.w*q.w - q.x*q.x + q.y*q.y - q.z*q.z,
res);
break;
case yxy:
twoaxisrot( 2*(q.x*q.y - q.w*q.z),
2*(q.y*q.z + q.w*q.x),
q.w*q.w - q.x*q.x + q.y*q.y - q.z*q.z,
2*(q.x*q.y + q.w*q.z),
-2*(q.y*q.z - q.w*q.x),
res);
break;
case yzx:
threeaxisrot( -2*(q.x*q.z - q.w*q.y),
q.w*q.w + q.x*q.x - q.y*q.y - q.z*q.z,
2*(q.x*q.y + q.w*q.z),
-2*(q.y*q.z - q.w*q.x),
q.w*q.w - q.x*q.x + q.y*q.y - q.z*q.z,
res);
break;
case yzy:
twoaxisrot( 2*(q.y*q.z + q.w*q.x),
-2*(q.x*q.y - q.w*q.z),
q.w*q.w - q.x*q.x + q.y*q.y - q.z*q.z,
2*(q.y*q.z - q.w*q.x),
2*(q.x*q.y + q.w*q.z),
res);
break;
case xyz:
threeaxisrot( -2*(q.y*q.z - q.w*q.x),
q.w*q.w - q.x*q.x - q.y*q.y + q.z*q.z,
2*(q.x*q.z + q.w*q.y),
-2*(q.x*q.y - q.w*q.z),
q.w*q.w + q.x*q.x - q.y*q.y - q.z*q.z,
res);
break;
case xyx:
twoaxisrot( 2*(q.x*q.y + q.w*q.z),
-2*(q.x*q.z - q.w*q.y),
q.w*q.w + q.x*q.x - q.y*q.y - q.z*q.z,
2*(q.x*q.y - q.w*q.z),
2*(q.x*q.z + q.w*q.y),
res);
break;
case xzy:
threeaxisrot( 2*(q.y*q.z + q.w*q.x),
q.w*q.w - q.x*q.x + q.y*q.y - q.z*q.z,
-2*(q.x*q.y - q.w*q.z),
2*(q.x*q.z + q.w*q.y),
q.w*q.w + q.x*q.x - q.y*q.y - q.z*q.z,
res);
break;
case xzx:
twoaxisrot( 2*(q.x*q.z - q.w*q.y),
2*(q.x*q.y + q.w*q.z),
q.w*q.w + q.x*q.x - q.y*q.y - q.z*q.z,
2*(q.x*q.z + q.w*q.y),
-2*(q.x*q.y - q.w*q.z),
res);
break;
default:
std::cout << "Unknown rotation sequence" << std::endl;
break;
}
}
回答by Pawel
Euler -> Quaternion
欧拉 -> 四元数
Extracted from Three.js.
摘自 Three.js。
Here's a piece of code which works for me:
这是一段对我有用的代码：
function eulerToQuaternion(eulerXYZ) {
var c1 = Math.cos(eulerXYZ[0] / 2),
c2 = Math.cos(eulerXYZ[1] / 2),
c3 = Math.cos(eulerXYZ[2] / 2),
s1 = Math.sin(eulerXYZ[0] / 2),
s2 = Math.sin(eulerXYZ[1] / 2),
s3 = Math.sin(eulerXYZ[2] / 2),
x = s1 * c2 * c3 + c1 * s2 * s3,
y = c1 * s2 * c3 - s1 * c2 * s3,
z = c1 * c2 * s3 + s1 * s2 * c3,
w = c1 * c2 * c3 - s1 * s2 * s3;
return [x, y, z, w];
};
function calculate() {
var quat = eulerToQuaternion([document.querySelector('#x').value, document.querySelector('#y').value, document.querySelector('#z').value]);
document.querySelector('#result').innerHTML = quat.join(' ');
}<h3>Euler radians in XYZ order:</h3>
<fieldset>
<label>X:
<input id="x" value="1.5" />
</label>
<label>Y:
<input id="y" value="1" />
</label>
<label>Z:
<input id="z" value="0" />
</label>
<button onClick="calculate()">To Quaternion</button>
</fieldset>
<h3>X Y Z W result:</h3>
<div id="result"></div>