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我有一个4x4矩阵(行 - 主)类和一个四元类,我试图提供转换两种表示之间转换的转换方法。将矩阵结果转换为四元数,并将值反转?
这是用于从四元数转换成矩阵,其中_2是System.Math.Pow我的转换功能:
/// <summary>
/// Converts the quaternion to it's matrix representation.
/// </summary>
/// <returns>A matrix representing the quaternion.</returns>
public Matrix ToMatrix()
{
return new Matrix(new double[,] {
{
_2(W) + _2(X) - _2(Y) - _2(Z),
(2 * X * Y) - (2 * W * Z),
(2 * X * Z) + (2 * W * Y),
0
},
{
(2 * X * Y) + (2 * W * Z),
_2(W) - _2(X) + _2(Y) - _2(Z),
(2 * Y * Z) + (2 * W * X),
0
},
{
(2 * X * Z) - (2 * W * Y),
(2 * Y * Z) - (2 * W * X),
_2(W) - _2(X) - _2(Y) + _2(Z),
0
},
{
0,
0,
0,
1
}
});
}
这些是从矩阵转换成四元数我的两个转换函数。请注意,在考虑X旋转时,它们都不起作用。
/// <summary>
/// Converts the matrix to a quaternion assuming the matrix purely
/// represents rotation (any translation or scaling information will
/// result in an invalid quaternion).
/// </summary>
/// <returns>A quaternion representing the rotation.</returns>
public Quaternion ToQuaternion()
{
/* http://www.euclideanspace.com/maths/geometry/rotations/conversions/matrixToQuaternion/index.htm
*/
double tr = this.m_Data[0, 0] + this.m_Data[1, 1] + this.m_Data[2, 2];
if (tr > 0)
{
double s = _N2(tr + 1) * 2;
return new Quaternion(
(this.m_Data[2, 1] - this.m_Data[1, 2])/s,
(this.m_Data[0, 2] - this.m_Data[2, 0])/s,
(this.m_Data[1, 0] - this.m_Data[0, 1])/s,
0.25 * s
);
}
else if ((this.m_Data[0, 0] > this.m_Data[1, 1]) && (this.m_Data[0, 0] > this.m_Data[2, 2]))
{
double s = _N2(1 + this.m_Data[0, 0] - this.m_Data[1, 1] - this.m_Data[2, 2]) * 2;
return new Quaternion(
0.25 * s,
(this.m_Data[0, 1] + this.m_Data[1, 0])/s,
(this.m_Data[0, 2] + this.m_Data[2, 0])/s,
(this.m_Data[2, 1] - this.m_Data[1, 2])/s
);
}
else if (this.m_Data[1, 1] > this.m_Data[2, 2])
{
double s = _N2(1 + this.m_Data[1, 1] - this.m_Data[0, 0] - this.m_Data[2, 2]) * 2;
return new Quaternion(
(this.m_Data[0, 1] + this.m_Data[1, 0])/s,
0.25 * s,
(this.m_Data[1, 2] + this.m_Data[2, 1])/s,
(this.m_Data[0, 2] - this.m_Data[2, 0])/s
);
}
else
{
double s = _N2(1 + this.m_Data[2, 2] - this.m_Data[0, 0] - this.m_Data[1, 1]) * 2;
return new Quaternion(
(this.m_Data[0, 2] + this.m_Data[2, 0])/s,
(this.m_Data[1, 2] + this.m_Data[2, 1])/s,
0.25 * s,
(this.m_Data[1, 0] - this.m_Data[0, 1])/s
);
}
}
/// <summary>
/// This is a simpler form than above, but doesn't work for all values. It exhibits the
/// *same results* as ToQuaternion for X rotation however (i.e. both are invalid).
/// </summary>
public Quaternion ToQuaternionAlt()
{
double w = System.Math.Sqrt(1 + this.m_Data[0, 0] + this.m_Data[1, 1] + this.m_Data[2, 2])/2;
return new Quaternion(
(this.m_Data[2, 1] - this.m_Data[1, 2])/(4 * w),
(this.m_Data[0, 2] - this.m_Data[2, 0])/(4 * w),
(this.m_Data[1, 0] - this.m_Data[0, 1])/(4 * w),
w
);
}
现在我的测试套件有一个简单的测试,像这样:
[TestMethod]
public void TestMatrixXA()
{
Matrix m = Matrix.CreateRotationX(45/(180/System.Math.PI));
Assert.AreEqual<Matrix>(m, m.ToQuaternion().ToMatrix(), "Quaternion conversion was not completed successfully.");
}
这是结果我从测试套件获得:
Expected:
{ 1, 0, 0, 0 }
{ 0, 0.707106781186548, -0.707106781186547, 0 }
{ 0, 0.707106781186547, 0.707106781186548, 0 }
{ 0, 0, 0, 1 }
Actual:
{ 1, 0, 0, 0 }
{ 0, 0.707106781186547, 0.707106781186547, 0 }
{ 0, -0.707106781186547, 0.707106781186547, 0 }
{ 0, 0, 0, 1 }
你会注意到,矩阵中的两个值被倒置。我已经测试了它,并且在每次来回转换(如.ToQuaternion()。ToMatrix())时,这些字段都是反转的。即如果我做四元数/矩阵转换两次,我得到正确的矩阵。由于正确值和结果之间的差异非常简单,我假设它是一个简单的东西,就像一个负号出现在错误的地方,但由于我不是矩阵和四元数学的专家,我无法找到问题所在。
有人知道数学有什么问题吗?