欧拉转换为基质，基质欧拉 [英] Conversion euler to matrix and matrix to euler

问题描述

我想转换欧拉角度来看描述成一个矩阵3D旋转，然后回来，用.NET / C＃。我的约定是：

I'm trying to convert a 3D rotation described in term of euler angles into a matrix and then back, using .NET/C#. My conventions are:

• 在左手系（x权，Y上，Z向前）
• 为了旋转：绕Y，绕x螺距，银行航向绕Z轴
• 旋转是正使用左手定则（大拇指指向正无穷大）

我的审判是：

欧拉到矩阵（我已经删除了X，Y，Z的翻译部分的简化）

Euler to matrix (I've removed the x,y,z translation part for simplification)

Matrix3D matrix = new Matrix3D() {
    M11 =   cosH * cosB - sinH * sinP * sinB,
    M12 = - sinB * cosP,
    M13 =   sinH * cosB + cosH * sinP * sinB,
    M21 =   cosH * sinB + sinH * sinP * cosB,
    M22 =   cosB * cosP,
    M23 =   sinB * sinH - cosH * sinP * cosB,
    M31 = - sinH * cosP,
    M32 = - sinP,
    M33 =   cosH * cosP,
};

矩阵欧拉

const double RD_TO_DEG = 180 / Math.PI;            
double h, p, b; // angles in degrees

// extract pitch
double sinP = -matrix.M23;            
if (sinP >= 1) {
    p = 90; }       // pole
else if (sinP <= -1) {
    p = -90; } // pole
else {
    p = Math.Asin(sinP) * RD_TO_DEG; }             

// extract heading and bank
if (sinP < -0.9999 || sinP > 0.9999) { // account for small angle errors
    h = Math.Atan2(-matrix.M31, matrix.M11) * RD_TO_DEG;
    b = 0; }
else {
    h = Math.Atan2(matrix.M13, matrix.M33) * RD_TO_DEG;
    b = Math.Atan2(matrix.M21, matrix.M22) * RD_TO_DEG; }

这一定是错误的。如果我把3角，将它们转换成一个矩阵，如果低于起始数值不同的矩阵转换回角度的结果。

It must be wrong. If I take 3 angles, convert them into a matrix and convert the matrix back into angles, the result if different than the intial values.

我已经浏览几个网站有不同的配方，从euclideanspace.com，但我现在已经完全丧失，无法找到合适的计算。我AP preciate一点帮助。是否有一个数学家板载？

I have browsed several sites with different formulas, starting with euclideanspace.com, but I'm now completely lost, and can't find the right computations. I' appreciate a little help. Is there a mathematician onboard?

推荐答案

首先，应该：

sinP = -matrix.M32

编辑：完整的解决方案如下

Full solution follows

我的推导：

Rx(P)=| 1      0       0 |
      | 0  cos P  -sin P |
      | 0  sin P   cos P |

Ry(H)=|  cos H  0  sin H |
      |      0  1      0 |
      | -sin H  0  cos H |

Rz(B)=| cos B  -sin B  0 |
      | sin B   cos B  0 |
      |     0       0  1 |

乘以您的订购：

Multiplied with your ordering:

R = Ry(H)*Rx(P)*Rz(B)
  = | cos H*cos B+sin H*sin P*sin B  cos B*sin H*sin P-sin B*cos H  cos P*sin H |
    |                   cos P*sin B                    cos B*cos P       -sin P |
    | sin B*cos H*sin P-sin H*cos B  sin H*sin B+cos B*cos H*sin P  cos P*cos H |

这给出了相反的推导：

Which gives reverse derivations:

谭B = M12 / M22

tan B = M12/M22

罪P = -m32

谭H = M31 / M33

tan H = M31/M33

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