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我有一個4x4矩陣(行 - 主)類和一個四元類,我試圖提供轉換兩種表示之間轉換的轉換方法。將矩陣結果轉換爲四元數,並將值反轉?
這是用於從四元數轉換成矩陣,其中_2是System.Math.Pow我的轉換功能:
/// <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
}
});
}
這些是從矩陣轉換成四元數我的兩個轉換函數。請注意,在考慮X旋轉時,它們都不起作用。
/// <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
);
}
現在我的測試套件有一個簡單的測試,像這樣:
[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.");
}
這是結果我從測試套件獲得:
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 }
你會注意到,矩陣中的兩個值被倒置。我已經測試了它,並且在每次來回轉換(如.ToQuaternion()。ToMatrix())時,這些字段都是反轉的。即如果我做四元數/矩陣轉換兩次,我得到正確的矩陣。由於正確值和結果之間的差異非常簡單,我假設它是一個簡單的東西,就像一個負號出現在錯誤的地方,但由於我不是矩陣和四元數學的專家,我無法找到問題所在。
有人知道數學有什麼問題嗎?