我试图通过分解矩阵来确定矩阵中保持的X,Y,Z角。我正在使用.net 4.5 c#。
我创建了一个测试来检查以下内容:
我得到以下结果: X:0.5 Y:0 Z:0
我在期待: X:0.45 Y:0 Z:0
测试代码
Quaternion quatDecomposed;
Vector3D translation;
Matrix4x4 rot = Matrix4x4.RotationAroundX(45);
rot.DecomposeNoScaling(out quatDecomposed, out translation);
我创建了自己的Matrix4x4,Vector3D和Angle3D结构,如下例所示。
我的Matrix4x4围绕x方法旋转如下:
public static Matrix4x4 RotationAroundX(double degrees)
{
// [1, 0, 0, 0]
// [0, cos,-sin,0]
// [0, sin,cos, 0]
// [0, 0, 0, 1]
// convert degrees to radians.
double radians = DoubleExtensions.DegreesToRadians(degrees);
// return matrix.
var matrixTransformed = Matrix4x4.Identity;
matrixTransformed.M22 = (float)Math.Cos(radians);
matrixTransformed.M23 = (float)-(Math.Sin(radians));
matrixTransformed.M32 = (float)Math.Sin(radians);
matrixTransformed.M33 = (float)Math.Cos(radians);
//return matrix;
return matrixTransformed;
}
我的分解无缩放方法如下:
public void DecomposeNoScaling(out Quaternion rotation, out Vector3D translation)
{
translation.X = this[1, 4];
translation.Y = this[2, 4];
translation.Z = this[3, 4];
rotation = new Quaternion(new Matrix3x3(this));
}
我希望得到的是Matrix4x4中包含的角度,我这样做如下:
Angle3D angles = new Angle3D(quatDecomposed.X, quatDecomposed.Y, quatDecomposed.Z);
有人能发现我做错了吗?我真正想要解决的是ZYX顺序中矩阵4x4的欧拉角。
提前致谢!
答案 0 :(得分:0)
不应该是矩阵的最后一行是“1”?
[1 0 0 0]
[0 cos -sin 0]
[0 sin cos 0]
[0 0 0 1]
(最后一行最后一列应为1)
答案 1 :(得分:0)
以防万一其他人需要知道,这就是我直接从矩阵中得到欧拉角的方法:
public static Angle3D GetAngles(Matrix4x4 source)
{
double thetaX, thetaY, thetaZ = 0.0;
thetaX = Math.Asin(source.M32);
if (thetaX < (Math.PI / 2))
{
if (thetaX > (-Math.PI / 2))
{
thetaZ = Math.Atan2(-source.M12, source.M22);
thetaY = Math.Atan2(-source.M31, source.M33);
}
else
{
thetaZ = -Math.Atan2(-source.M13, source.M11);
thetaY = 0;
}
}
else
{
thetaZ = Math.Atan2(source.M13, source.M11);
thetaY = 0;
}
// Create return object.
Angle3D angles = new Angle3D(thetaX, thetaY, thetaZ);
// Convert to degrees.;
angles.Format = AngleFormat.Degrees;
// Return angles.
return angles;
}