Say I want to rotate the other points around the center point, and I don't have opengl to push, rotate, and pop... how can I do it? [Edited by - ScottC on August 2, 2005 4:52:12 AM]

You could use matrices like the APIs but just code it all yourself, it's fairly easy. Just search google for 2D Rotation Matrices or just 3D ones if thats what you need and also google for matrix maths.

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^^Linkified^^

ace

Quote:Original post by ace_lovegrove
You could use matrices like the APIs but just code it all yourself, it's fairly easy. Just search google for 2D Rotation Matrices or just 3D ones if thats what you need and also google for matrix maths.

EDIT:



^^Linkified^^

ace

the sites I find are confusing.

You need to grasp a few concepts before this will work:
- matrix multiplication (click)
- matrix/vector multiplication
- transformation matrices (click here)

Let us assume a 2D rotation. Furthermore simplify by rotating around the origin first so C = (0,0). We shall rotate Phi angles in radians. Clicky2 gives the rotation matrix for this:

 cos Phi   sin Phi-sin Phi   cos Phi

Just fill in the values and you have a simple rotation matrix.

Transformation matrices are applied to a coordinate by multiplying the coordinates vector with the matrix. If the previous matrix is R and we want to rotate the point P we compute P' = P R. See Clicky1.

Now transformation can, besides rotations, also be translations. We could build a matrix that transforms some point P 5 units up along the Y axis. Thus, we could also construct a transformation matrix that would move the origin (0,0) to the desired center point C = (Cx,Cy). Now the last step is that we can combine two transformation matrices using multiplication.

We had a rotation matrix R and we just introduced a translation matrix T. If you want to rotate around C you can simply compute the transform matrix M = T R and transform your points with this.

This will probably sound complicated but I suggest you read out the given links and try some examples on paper. Feel free to ask if you are stuck.

Illco

It is not difficult at all.
You can do that on tree steps, all convened into a single step
Say the origin is p(x0, y0)

The first thing you do is to translate everything to the origin

x’ = x – x0
y’ = y – y0

in matrix form this is equivalent to

[x’ y’ 1] = [x y 1] * [1 0 0]
[0 1 0]
[-x0 -y0 1]


now you rotate each point by the angle you want


x’’ = x’ cos(a) - y’ sin (a)
y’’ = x’ sin(a) + y’ cos (a)

in matrix from this is

[x’’ y’’ 1] = [x’ y’ 1] * [cos(a) sin(a) 0]
[-sin(a) cos(a) 0]
[0 0 1]

finally you translate everything back to where it was


x’’’ = x’’ + x0
y’’’ = y’’ + y0

and in matrix form this is:


[x’’’ y’’’ 1] = [x’’ y’’ 1] * [1 0 0]
[0 1 0]
[x0 y0 1]

Now calling the tree matrices inv(T), R, T, and concatenating all tree equation you get

P’’’ = P * inv(T) * R * T

The expression :

inv(T) * R * T

Is very common in linear algebra and it is called similar transformation. You can develop the algebra by expanding the expression and save few operations here and there, by I recommendation you do the matrix multiplies in some matrix class and it will guaranty correct result not matter what the situation is.

Quote:Original post by Anonymous Poster
It is not difficult at all.

ironically this is the only thing I could understand...

Basically rotation math is always done around the origin. (0, 0) So if you want to rotate something around this point, you make sure it's coordinates are relative to this, probably with a translation if it's not in local space. Do your rotate, then translate it to the point you want it.

Well I am sorry I confused you. What GodBeastX is exactly what I said.
If you know a least the concept of matriices and matrix multiplication it should be clear. If you do not know that but you have a matrix class this is what you do

Matrix RoationAboutAPoint ( Point P, Angle A)
{
Matrix invT = TranlationMatrix (P.Scale(-1));
Matrix R = RotationMatrix (A);
Matrix T = ranlationMatrix (P);

Matrix mat = invT * R * T;
Return mat
}

then you multiply each point by the matrix mat.