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# Y-Axis Rotation from Matrix

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### #1Endemoniada  Members   -  Reputation: 423

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Posted 06 August 2014 - 09:25 PM

Hi guys,

I'm trying to get the y-axis rotation from a matrix.

I found this:

r=atan2(-M._31, sqrt(M._32*M._32+M._33*M._33);

...but I don't know if that's for D3D9 matrices. It's tricky because of the return values of atan2 as well.

I looked at D3DXMatrixDecompose() but I don't know how to get the y-axis rotation from the quaternion.

I'm hoping to get the rotation in [0...twoPi]

Thanks.

### #2Stephany  Members   -  Reputation: 166

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Posted 07 August 2014 - 12:16 PM

Are you just wanting the angle relative to the world? Here are some things that might help you figure it out:

+ You can get the absolute angle between two vectors using acos( dot( v1, v2 ) ).

+ If your world is composed of X=right, Y=up, Z=forward, then your world's forward vector is (0,0,1).

+ The vector you might be interested in is the forward/Z vector of the matrix (M._31,M._32,M._33).

+ If your matrix only contains Y rotations, the positive/absolute angle will be equal to acos( M._33 )

+ You'll know if the angle is flipped (rotating the opposite direction) if M._31 is negative or positive.

+ If your matrix contains other (X/Z) rotations, it may be necessary to *smash* the matrix forward vector to the X/Z plane, then get the angle.

+ To put it simply, smashing the vector would be something like normalize( m._31, 0, m._33 ). But take care that (m._31,0,m._33) isn't zero.

Hope that helps

Edited by Stephany, 07 August 2014 - 12:22 PM.

### #3HappyCoder  Members   -  Reputation: 4428

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Posted 07 August 2014 - 03:14 PM

Let me make sure I understand what you want. Assuming that +y is up, you want to know what angle something is facing. In that case you could just transform the forward vector by the matrix.

transformed = forward * matrix;

then use atan2 on the x and z component of the result. So if +z is forward, the angle would be

atan2(transformed.x, transformed.z);

0 radians would be pointing down the z axis with a positive value rotating towards positive x.

If forward will always be positive z, then you can simply put.

atan2(matrix._13, matrix._33)

If you don't mind me asking, what do you need the angle for? Usually there are more elegant solutions to problems using matrices and vectors than finding angles and using trig functions.

Edited by HappyCoder, 07 August 2014 - 03:14 PM.

My current game project Platform RPG

### #4Endemoniada  Members   -  Reputation: 423

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Posted 08 August 2014 - 12:17 AM

Hey guys,

I didn't think it was a good thing but figured I'd give it a shot. I'm trying to test something that's not working and think it might help.

A typical example is this:


D3DXMATRIX R, T, M;

D3DXMatrixTranslation(&T, 2.0f, 0.0f, 0.0f);
D3DXMatrixRotationY(&R, -D3DX_PI*0.5f);
M=R*T;
M=M*SomeOtherMatrix;
D3DXMatrixInverse(&M, NULL, &M);

float r=GetYRotation(&M); // the helper I want to make

// if it worked properly it would do this:
D3DXMatrixRotationY(&R, D3DX_PI*0.5f);
float r=GetYRotation(&M);
// r would equel pi*0.5



That's all I would be using it for.

Thanks.

Edited by Endemoniada, 08 August 2014 - 12:20 AM.

### #5JohnnyCode  Members   -  Reputation: 982

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Posted 09 August 2014 - 04:08 PM

D3DXMATRIX R
, T, M;

D3DXMatrixTranslation
(&T, 2.0f, 0.0f, 0.0f);
D3DXMatrixRotationY(&R, -D3DX_PI*0.5f);
M=R*T;
M=M*SomeOtherMatrix;
D3DXMatrixInverse(&M, NULL, &M);

float r=GetYRotation(&M); // the helper I want to make

// if it worked properly it would do this:
D3DXMatrixRotationY(&R, D3DX_PI*0.5f);
float r=GetYRotation(&M);
// r would equel pi*0.5

if the M matrix will always be Y rotation matrix, you could derive the parameter it is constructed from just as this : acos (M[0][0]).

But this parameter is not an angle with Y axis - it is actualy an angle in ZX plane - thus making objects revolve around Y axis if transfomed by M.

But if you will incompose also Z and X rotation matricies into the M matrix (multiply the M with them), then you will not be able to use this simple solution. You will need to solve 3 linear equations. I recomend Cramer method for solving them as it does not use division, so you will be zero safe. Following is a method that constructs world rotation and translation from the 3 axises sinuses and cosinuses and position

function ComposeWorld4x3(cx,cy,cz,sx,sy,sz,x,y,z,res,offr)
{
var m1=cy*cz;
var m2=cy*sz;
var m3=-sy;

var m4=sx*sy*cz-sz*cx;
var m5=sx*sy*sz+cx*cz;
var m6=sx*cy;

var m7=cx*sy*cz+sz*sx;
var m8=cx*sy*sz-sx*cz;
var m9=cx*cy;

res[offr]=m1;
res[offr+1]=m2;
res[offr+2]=m3;
res[offr+3]=0.0;
res[offr+4]=m4;
res[offr+5]=m5;
res[offr+6]=m6;
res[offr+7]=0.0;
res[offr+8]=m7;
res[offr+9]=m8;
res[offr+10]=m9;
res[offr+11]=0.0;
res[12+offr]=x;
res[13+offr]=y;
res[14+offr]=z;
res[15+offr]=1.0;
}


the res is a 16 array in column memory layout (res[3]is 4th component of first column) and cx,cy,cz,sx,sy,sz are cosinuses of X,Y,Z and sinuses of X,Y,Z angles respectively

Now if you look at it you may take equations as those:

var m1=cy*cz;

var m9=cx*cy;

var m3=-sy;

you should find out cy through third simple equation m3=-sy such as sqrt(1.0-asin(-1.0*m3)) and use 2 linear equations system only:

M[0][0]=cy*cz;

M[2][2]=cx*cy; where cy is aleary existing constant you computed

this is cramer's rule

Consider a system of n linear equations for n unknowns, represented in matrix multiplication form as follows:

where the n by n matrix has a nonzero determinant, and the vector is the column vector of the variables.

Then the theorem states that in this case the system has a unique solution, whose individual values for the unknowns are given by:

where is the matrix formed by replacing the ith column of by the column vector .

So you would perform b column to be M[0,0],M[2,2] and A matrix to be

[cy,0]

[0,cy]

you then compute x1 and x2 by applying last sentence of quoted Cramer's rule on them both. You will then poses cx,cy,cz and use acos on all three to know angles around all three axises of a rotation matrix

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