I'm testing some vector reflection and I want to check what happens when a ball collides with a paddle.
So if I have:
Vector2 velocity = new Vector2(-5, 2);
position_ball += velocity;
if (position_ball.X < 10)
{
Vector2 v = new Vector2(1,0); // or Vector2.UnitX
velocity = Vector2.Reflect(velocity, v);
}
then, correctly, velocity is (5,2) after Reflect
, but if I do:
if (position_ball.X < 10)
{
Vector2 v = new Vector2(1,1);
velocity = Vector2.Reflect(velocity, v);
}
then velocity is (1,8) and not (5, -2) that is the solution of reflection equation R = V - 2 * (V . N)
Why is that?
Answer
The reflection equation is R = V - 2 * (V . N) * N
This formula also assumes that N is a normal vector, which it isn't in your case new Vector2(1,1).Length() == 1.414...
Try this instead:
if (position_ball.X < 10)
{
Vector2 v = new Vector2(1,1);
v.Normalize();
velocity = Vector2.Reflect(velocity, v);
}
The reflection of [-5,2] will actually be [-2,5] (or [1,8] if not normalized) and not [5,-2] as you assumed:
R = V - 2 * (V.N) * N
N = [1,1] * 2^-0.5
V = [-5,2]
Since Nx = Ny in this particular case we can refer to either of them as Nc
R = V - 2 * (v.N) * N
= V - 2 * (Vx*Nc + Vy*Nc) * N
= V - 2*Nc(Vx+Vy) * N
= V - 2*Nc(-5+2) * N
= V - 2*Nc(-3) * N
= V + 6*Nc * N
= V + [6*Nx*Nx, 6*Ny*Ny]
= V + [6*Nx^2, 6*Ny^2]
= [6*Nx^2-5, 6*Ny^2+2]
= [6*(2^-1)-5, 6*(2^-1)+2]
= [-2, 5]
(N^a)^b = N^(a*b)
which is used in the above proof (2^-0.5)^2 = 2^-1
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