欧拉到矩阵和矩阵到欧拉的转换 [英] Conversion euler to matrix and matrix to euler

问题描述

我正在尝试使用 .NET/C# 将根据欧拉角描述的 3D 旋转转换为矩阵，然后返回.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:

• 左手系统(x 右，y 上，z 向前)
• 旋转顺序:绕y，绕x俯仰，绕z倾斜
• 使用左手规则(拇指指向+无穷大)为正旋转

我的试用期是:

欧拉转矩阵(为了简化，我删除了 x,y,z 平移部分)

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,
};

矩阵到欧拉

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; }

一定是错的.如果我取 3 个角度，将它们转换为矩阵并将矩阵转换回角度，结果与初始值不同.

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.

我浏览了几个具有不同公式的站点，从 euclideanspace.com 开始，但我现在完全迷失了，找不到正确的计算.我很感激一点帮助.机上有数学家吗?

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

完整的解决方案如下

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 |

乘以您的订单:

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 |

给出反向推导:

tan B = M12/M22

tan B = M12/M22

sin P = -M32

sin P = -M32

棕褐色 H = M31/M33

tan H = M31/M33

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