SDL rotation

[DrChill]

I want to know how to rotate a shape, I declare my shapes like this:

shape foo[]={
     0,0,//vertex 0
     100,100,//vertex 1
     0,100 //vertex 2
};

I was wondering how to rotate it?

(I am using vertices so that when I move to 3d it will be easier(hopefully))

Also while I was looking up rotation I ran into this: http://en.wikipedia.org/wiki/Vertex_%28geometry%29#Definitions

[Bazzy]

Rotate all the vertices around it's center using sine/cosine functions

[helios]

So what does this have to do with SDL?

[DrChill]

@Bazzy
This what I had before:

void rotate(shape* poly, double radians){
	//count the number of vertices ...
	int numVert;
	int i = 0;
	while(poly->vertices[i]->x != NULL) ++i;
	numVert = i;

	i = 0;
	while(i < numVert){
		int x, y ;
		x = poly->vertices[i].x * cos(radians) - poly->vertices[i].y * sin(radians);
		y = poly->vertices[i].x * sin(radians) + poly->vertices[i].y * cos(radians);
		poly->vertices[i].y = y;
		poly->vertices[i].x = x;
		++i;
	}	
}

@helios
Actually it has very little to do with SDL, I guess I wanted to say I was using graphics.

[helios]

All you need to do to rotate a point is provide a pivot point. Then you convert the rectangular coordinates to polar in relation to that point, add the rotation to the angle component, then convert back to rectangular.

[DrChill]

Ok, so first conversion from rectangular to polar:

r = √ (x2 + y2)
r = 2*(x*2 + y*2)

θ = atan( y / x )
theta = atan(y/x)

Is that correct so far?

[helios]

Uh... You do realize that x^.5 is not the same as 2*x, right? And that formula is completely wrong, anyway.
r=(x1-x2+y1-y2)^.5
t=atan2(y1-y2,x1-x2)

Determining an angle from two sets of coordinates is such a common operation that atan2() exists just to do that. It even handles the cases when !x. If !x && !y, the function fails.

[manojr]

The discussion of rotation is incomplete without quaternion.

[DrChill]

Sorry, I am horrible at geometry, thankyou.

[DrChill]

Sorry, I am horrible at geometry, thankyou.