我很困惑为什么android的矩阵类的multiplyMV方法似乎顺时针旋转我的矢量坐标,当我认为它是逆时针时。
在此代码中,pos是矢量坐标并设置为< 0.0f,5.0f,0.0f> ,矩阵将围绕Z轴旋转坐标矢量-45度。我希望结果向量坐标在< +,+>中。象限,即< 3.535534,3.535534,0.0> 。但相反,它以相反的方向旋转坐标,< - ,+>象限,即< -3.535534,3.535534,0.0> 。
Matrix4 mtxRot = Matrix4.InitRotateEulerXYZ(0.0f, 0.0f, -45f);
pos.Set(0.0f, 5.0f ,0.0f);
mtxRot.TransformCoordVec(pos);
这是我的Matrix4.InitRotateEulerXYZ
public static Matrix4 InitRotateEulerXYZ(float x, float y, float z)
{
Matrix4 rotMatrix = new Matrix4();
/* XYZ = | cz*cy, sz*cx + cz*sy*sx, sz*sx - cz*sy*cx |
| -sz*cy, cz*cx - sz*sy*sx, cz*sx + sz*sy*cx |
| sy, -cy*sx, cy*cx | */
// Convert from degrees to radians
x = MathHelper.DegreesToRadians(x);
y = MathHelper.DegreesToRadians(y);
z = MathHelper.DegreesToRadians(z);
rotMatrix.GetArray()[0] = MathHelper.Cos(z) * MathHelper.Cos(y);
rotMatrix.GetArray()[1] = -MathHelper.Sin(z) * MathHelper.Cos(y);
rotMatrix.GetArray()[2] = MathHelper.Sin(y);
rotMatrix.GetArray()[4] = (MathHelper.Sin(z) * MathHelper.Cos(x)) + (MathHelper.Cos(z) * MathHelper.Sin(y) * MathHelper.Sin(x));
rotMatrix.GetArray()[5] = (MathHelper.Cos(z) * MathHelper.Cos(x)) - (MathHelper.Sin(z) * MathHelper.Sin(y) * MathHelper.Sin(x));
rotMatrix.GetArray()[6] = -(MathHelper.Cos(y) * MathHelper.Sin(x));
rotMatrix.GetArray()[8 ] = (MathHelper.Sin(z) * MathHelper.Sin(x)) - (MathHelper.Cos(z) * MathHelper.Sin(y) * MathHelper.Cos(x));
rotMatrix.GetArray()[9 ] = (MathHelper.Cos(z) * MathHelper.Sin(x)) + (MathHelper.Sin(z) * MathHelper.Sin(y) * MathHelper.Cos(x));
rotMatrix.GetArray()[10] = MathHelper.Cos(y) * MathHelper.Cos(x);
return rotMatrix;
}
这是我的Matrix4.TransformCoordVec方法
public Vector3 TransformCoordVec(Vector3 vec3)
{
Matrix4.inVec[0] = vec3.X;
Matrix4.inVec[1] = vec3.Y;
Matrix4.inVec[2] = vec3.Z;
Matrix4.inVec[3] = 1.0f; // homogeneousCoord
Matrix.multiplyMV(Matrix4.outVec, 0, this.matrix, 0, Matrix4.inVec, 0);
vec3.X = Matrix4.outVec[0]; vec3.Y = Matrix4.outVec[1]; vec3.Z = Matrix4.outVec[2];
return vec3;
}
非常感谢任何帮助!
已修复
InitRotateEulerXYZ和我的Quaternion ToMatrix()方法需要进行转置,以便逆时针旋转正角度。以下是更正的方法。
Quaternion.ToMatrix
/**Converts a quanternion to its equivilant matrix form**/
public Matrix4 ToMatrix()
{
// First, lets check if we need to re-normalize our quaternion
if(normalRegenerationCount <= 1000)
{
Normalize();
}
float x2 = x * x;
float y2 = y * y;
float z2 = z * z;
float xy = x * y;
float xz = x * z;
float yz = y * z;
float wx = w * x;
float wy = w * y;
float wz = w * z;
Matrix4 result = new Matrix4();
// This calculation would be a lot more complicated for non-unit length quaternions
// Note: The constructor of Matrix4 expects the Matrix in column-major format like expected by
// OpenGL
result.Set_11(1.0f - (2.0f * (y2 + z2)));
result.Set_12(2.0f * (xy + wz));
result.Set_13(2.0f * (xz - wy));
result.Set_14(0.0f);
result.Set_21(2.0f * (xy - wz));
result.Set_22(1.0f - (2.0f * (x2 + z2)));
result.Set_23(2.0f * (yz + wx));
result.Set_24(0.0f);
result.Set_31(2.0f * (xz + wy));
result.Set_32(2.0f * (yz - wx));
result.Set_33(1.0f - (2.0f * (x2 + y2)));
result.Set_34(0.0f);
result.Set_41(0.0f);
result.Set_42(0.0f);
result.Set_43(0.0f);
result.Set_44(1.0f);
return result;
}
Matrix.InitRotateEulerXYZ
public static Matrix4 InitRotateEulerXYZ(float x, float y, float z)
{
Matrix4 rotMatrix = new Matrix4();
// Convert from degrees to radians
x = MathHelper.DegreesToRadians(x);
y = MathHelper.DegreesToRadians(y);
z = MathHelper.DegreesToRadians(z);
rotMatrix.matrix[0] = MathHelper.Cos(z) * MathHelper.Cos(y);
rotMatrix.matrix[1] = (MathHelper.Sin(z) * MathHelper.Cos(x)) + (MathHelper.Cos(z) * MathHelper.Sin(y) * MathHelper.Sin(x));
rotMatrix.matrix[2] = (MathHelper.Sin(z) * MathHelper.Sin(x)) - (MathHelper.Cos(z) * MathHelper.Sin(y) * MathHelper.Cos(x));
rotMatrix.matrix[4] = -MathHelper.Cos(y) * MathHelper.Sin(z);
rotMatrix.matrix[5] = (MathHelper.Cos(z) * MathHelper.Cos(x)) - (MathHelper.Sin(z) * MathHelper.Sin(y) * MathHelper.Sin(x));
rotMatrix.matrix[6] = (MathHelper.Cos(z) * MathHelper.Sin(x)) + (MathHelper.Sin(z) * MathHelper.Sin(y) * MathHelper.Cos(x));
rotMatrix.matrix[8 ] = MathHelper.Sin(y);
rotMatrix.matrix[9 ] = -(MathHelper.Cos(y) * MathHelper.Sin(x));
rotMatrix.matrix[10] = MathHelper.Cos(y) * MathHelper.Cos(x);
return rotMatrix;
}
答案 0 :(得分:1)
通过在InitRotateEulerXYZ
中将X和Y设置为0作为参数,然后乘以(0,5,0,0),简化您创建的矩阵:
| cos(θ) sin(θ) 0 0 | | 0 | | 5*sin(θ) |
|-sin(θ) cos(θ) 0 0 | | 5 | | 5*cos(θ) |
| 0 0 1 0 | | 0 | = | 0 |
| 0 0 0 0 | | 0 | | 0 |
代入θ=-π/ 4,得到的矢量由下式给出:
| 5*sin(-π/4) | | -5*sin(π/4) |
| 5*cos(-π/4) | | 5*cos(π/4) |
| 0 | = | 0 |
| 0 | | 0 |
这正是您所看到的答案。所以,我认为这不是Android矩阵乘法例程的问题。您应该重新检查矩阵乘法的推导,或者只选择其中一个already derived at another site并相应地调整您的代码。
或者,您可以使用内置的setRotateEulerM方法并让系统为您生成它。您只需要确定矩阵乘法的顺序与您的意图一致,因为有多种方法可以生成这些矩阵。
另一件需要考虑的事情是,使用Euler Rotation矩阵通常是广义旋转的一种不好的方法,因为它们受到像gimble lock这样的事情的影响。您可能需要考虑其中一个Axis/Angle methods,例如Rodrigues' method,或使用quaternions。如果您希望使用其中一个,那么Android库似乎有native implementations of axis/angle routines。