Rotate a 3D vector

[Cuallito]

How do you rotate a 3D vector by XYZ corrdinates without using matrices or any of that junk?

[kedaman]

It's not junk, it's a very fast way to do transformations. who limited you from not using them?

[Cuallito]

Ok then, how do you rotate a vector by XYZ corridenates around a certain location (with matrices or not ;p)? Also, you do you rotate a XYZ vector by a quaternion? Hey, I've only had Algebra I! Thnx in advance

[kedaman]

here's a tutorial for you:
http://www.manning.com/Lussier/Chapter3.html

[Cuallito]

Ok... Let's see if I got this right: Take a matrix, rotate it by ur xyz corridinates, and then multiply it by ur original vector. It that right?

[kedaman]

that is if you need to do a transformation. X x Y x Z can be done in one step and then transform, any vector you multiply with the result will be rotated in respect to the transformation.

[parksie]

This is in C but the theory's sound...

void BuildRotMatrix() {
	m_pfMatrix[0][0] = 1.0 - 2.0 * (m_qTotal[1] * m_qTotal[1] + m_qTotal[2] * m_qTotal[2]);
	m_pfMatrix[0][1] = 2.0 * (m_qTotal[0] * m_qTotal[1] - m_qTotal[2] * m_qTotal[3]);
	m_pfMatrix[0][2] = 2.0 * (m_qTotal[2] * m_qTotal[0] + m_qTotal[1] * m_qTotal[3]);
	m_pfMatrix[0][3] = 0.0;

	m_pfMatrix[1][0] = 2.0 * (m_qTotal[0] * m_qTotal[1] + m_qTotal[2] * m_qTotal[3]);
	m_pfMatrix[1][1]= 1.0 - 2.0 * (m_qTotal[2] * m_qTotal[2] + m_qTotal[0] * m_qTotal[0]);
	m_pfMatrix[1][2] = 2.0 * (m_qTotal[1] * m_qTotal[2] - m_qTotal[0] * m_qTotal[3]);
	m_pfMatrix[1][3] = 0.0;

	m_pfMatrix[2][0] = 2.0 * (m_qTotal[2] * m_qTotal[0] - m_qTotal[1] * m_qTotal[3]);
	m_pfMatrix[2][1] = 2.0 * (m_qTotal[1] * m_qTotal[2] + m_qTotal[0] * m_qTotal[3]);
	m_pfMatrix[2][2] = 1.0 - 2.0 * (m_qTotal[1] * m_qTotal[1] + m_qTotal[0] * m_qTotal[0]);
	m_pfMatrix[2][3] = 0.0;

	m_pfMatrix[3][0] = 0.0;
	m_pfMatrix[3][1] = 0.0;
	m_pfMatrix[3][2] = 0.0;
	m_pfMatrix[3][3] = 1.0;
}

Where m_qTotal is a rotation quaternion.

[Cuallito]

Well I coded a little sub, but it won't rotate anything:

Function RotateVector(ByRef VectorIn As D3DVECTOR, X As Single, Y As Single, Z As Single) As D3DVECTOR
    Dim Matrix(1 To 2, 1 To 3) As Single
    
    
    'X rotation
    Matrix(1, 1) = VectorIn.X
    Matrix(2, 1) = 0
    Matrix(1, 2) = Cos(X) * VectorIn.Y
    Matrix(2, 2) = -Sin(X) * VectorIn.Z
    Matrix(1, 3) = Sin(X) * VectorIn.Y
    Matrix(2, 3) = Cos(X) * VectorIn.Z
    
    VectorIn.X = Matrix(1, 1) + Matrix(2, 1)
    VectorIn.Y = Matrix(1, 2) + Matrix(2, 2)
    VectorIn.Z = Matrix(1, 3) + Matrix(2, 3)
    
    'y rotation
    Matrix(1, 1) = Cos(Y) * VectorIn.X
    Matrix(2, 1) = -Sin(Y) * VectorIn.Z
    Matrix(1, 2) = VectorIn.Y
    Matrix(2, 2) = 0
    Matrix(1, 3) = -Sin(Y) * VectorIn.X
    Matrix(2, 3) = Cos(Y) * VectorIn.Z
    
    VectorIn.X = Matrix(1, 1) + Matrix(2, 1)
    VectorIn.Y = Matrix(1, 2) + Matrix(2, 2)
    VectorIn.Z = Matrix(1, 3) + Matrix(2, 3)
    'z rotation
    Matrix(1, 1) = Cos(Z) * VectorIn.X
    Matrix(2, 1) = -Sin(Z) * VectorIn.Y
    Matrix(1, 2) = Sin(Z) * VectorIn.X
    Matrix(2, 2) = Cos(Z) * VectorIn.Y
    Matrix(1, 3) = VectorIn.Z
    Matrix(2, 3) = 0
    
    VectorIn.X = Matrix(1, 1) + Matrix(2, 1)
    VectorIn.Y = Matrix(1, 2) + Matrix(2, 2)
    VectorIn.Z = Matrix(1, 3) + Matrix(2, 3)
    
    
    RotateVector = VectorIn
    
End Function

Whats Wrong? BTW, Thx Parksie for the sub, but is there a way I can rotate a vector by a quaternion(luv that word !)without having to make a matrix out of it first? That seems kind of slow...

[kedaman]

looks like you don't care about preserving the matrix, you could use this directly then:

there's no need for quaternions if you don't have combinations of several rotation transformations.

[Cuallito]

Well, I cam up with this:

Function RotateVector(ByRef VectorIn As D3DVECTOR, X As Single, Y As Single, Z As Single) As D3DVECTOR
    Dim Temp(10) As Single
    Dim xx As Single, yy As Single, zz As Single
    
    yy = VectorIn.Y
    xx = VectorIn.X
    zz = VectorIn.Z
    
    Temp(0) = Sin(X)
    Temp(1) = Cos(X)
    Temp(2) = Sin(Y)
    Temp(3) = Cos(Y)
    Temp(4) = Sin(Z)
    Temp(5) = Cos(Z)
    Temp(6) = yy * Temp(5)
    Temp(7) = xx * Temp(4)
    Temp(8) = Temp(5) * xx
    Temp(9) = Temp(4) * yy
    Temp(10) = Temp(3) * zz
    
    RotateVector.X = (Temp(3) * (Temp(8) - Temp(9)) + Temp(2) * Z)
    
    
    RotateVector.Y = (Temp(0) * (Temp(2) * Temp(8) - Temp(2) * _
        Temp(9) - Temp(10)) + Temp(1) * (Temp(7) + Temp(6)))
    
    RotateVector.Z = (Temp(1) * (-Temp(2) * Temp(8) + Temp(2) * _
        Temp(9) + Temp(10)) + Temp(0) * (Temp(7) + Temp(6)))

    
End Function

But it still doesn't work!? Thanks guys for your help.

[kedaman]

don't use arrays when unnessesary. replace that Z with ZZ for the X component expression.

[parksie]

The trigonometric functions in VB (actually, in nearly all programming languages) accept angles in radians. If you use degrees you'll be waaaay out .

To convert from degrees to radians multiply by (pi / 180).

[kedaman]

so if the angle isn't specified by the user, screw degrees and store radians.