Getting x1 and y1 after glrotate
Author
Does anyone know how to get the object's coordinates after glrotate? Thanks in advance.
You can use the glGet*() family of functions with either GL_MODELVIEW_MATRIX or GL_PROJECTION_MATRIX to get the matrix and multiply the matrix yourself manually, depending on which transformed coordinate set you want.
Author
I'm still a bit lost. So I can do this
glTranslatef(Obj.x1,Obj.y1,0)
glRotatef(90,0,0,1)
glTranslatef(-Obj.x1,-Obj.y1,0)
then use
glGetFloatv(GL_MODELVIEW_MATRIX, matrix)
but I'm not sure where to go from there.
glTranslatef(Obj.x1,Obj.y1,0)
glRotatef(90,0,0,1)
glTranslatef(-Obj.x1,-Obj.y1,0)
then use
glGetFloatv(GL_MODELVIEW_MATRIX, matrix)
but I'm not sure where to go from there.
Quote:but I'm not sure where to go from there.Google 'matrix vector multiplication'. The code will depend on matrix majorness and vector convention. In OpenGL, you have:
[0 2][x] = [0*x+2*y][1 3][y] [1*x+3*y]
Author
Augh I'm still having trouble with this! I should've stayed in math my 12th grade year. Ah I'm sure I'll figure out something with the info given. Thx.
If you want a quick solution, there's a dirty trick I used sometimes when I was bored to do the math:
I supply the coordinates as a (dummy) light position, GL automatically transforms it with the current MV matrix and stores it, and then I retrieve the new coords with glGetLight. It's an awful hack though, so better work with your math, it's very simple.
...float pos[]={3.24,5.12,8.457,1};//w=0: Vector, w=1: PositionglMatrixMode(GL_MODELVIEW);glPushMatrix();//Manipulate matrix here...glLightfv(GL_LIGHT0,GL_POSITION,pos);glGetLightfv(GL_LIGHT0,GL_POSITION,pos);glPopMatrix();...I supply the coordinates as a (dummy) light position, GL automatically transforms it with the current MV matrix and stores it, and then I retrieve the new coords with glGetLight. It's an awful hack though, so better work with your math, it's very simple.
Quote:Original post by mikeman
It's an awful hack though, so better work with your math, it's very simple.
it may be a hack, but i like it. very creative.
Author
These functions have some strange effects. Just to make sure we're on the same page, this is what it does right? I can't have the right collision detection if I don't get this right.
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