How to calculate the rotation resulting from ball bounce

Question

Hey all, still working to incorporate more physics simulation into my game (mentioned HERE). Now having the ball successfully and quite realistically bouncing a surface it hits, I wanted to make the ball spin off.

Now this seemed simple enough, I would calculate the rotation speed for the ball every time-tick and then rotate the ball by that much degrees upon drawing.

I knew that the speed of rotation would depend, upon impact, on:

1. Initial Speed of the ball hitting the surface (magnitude of ball velocity vector)
2. Friction co-efficient of ball & surface (constants for simulation)
3. Angle between ball impact velocity vector and the surface normal (approximated by dot vector value of impact and exit velocity vectors. 1 meaning high spin, -1 meaning no spin, and everything else relatively in between)

so, multiplying all of the above together, and making sure they were then transformed to the range 0-1, and multiplied by Max rotation speed, the ball seemed to respond in rotation speed as was expected. Except for one thing, it was always rotating clock-wise (because of positive values)

So my question is this:

1. What do you think of my method? I know it is not the most accurate, but it is good enough for simple simulation. Any easier ones?
2. If this method is fine, then what am I missing? How should I know when ball should rotate clock/anti-clock wise?

thanks for any comments

Answer 1

Your method is nice, because it's very simple. One thing you might need is dependency on previous spin on the ball, which you do not take into account. The spinning ball represents rotational energy, so a realistic simulation would probably have to conserve it along with the other energies.

However, if the ball is not rotating upon impact, I can't imagine a situation in which it begins rotation against the direction of the incident angle. That is, "clockwise" or "counterclockwise" should be relative to whichever side of the normal the incident angle is.

I think simply multiplying the result by the original x-direction vector (+1 if traveling left to right, -1 if traveling right to left) should do it.

Edit: You can use the cross-product for this. Incident cross normal provides a vector in the Z direction only (if we are on the 2D x-y plane). Look at the z-element: if it is positive, the ball's approach should cause it to spin clockwise. If it is negative, the ball should spin counterclockwise.

Answer 2

First get the surface tangent from the surface normal: t = (ny, -nx)

Then you can get the velocity component along the surface as vt = v dot t.

Now you can calculate the rotation of the ball: w = |(normal * r) cross vt|, where r is the radius of the ball.

Here I assume the ball has no rotational inertia and begins to spin instantly at the speed it would if it was to roll along the surface. You can use a coefficient of friction to make it more realistic and, if you want, take into account the rotational inertia of the ball.

Answer 3

Ókay, this might sound stupid but you're not using the dot-product of the ball vector and the surface normal and just doing an arccos to calculate the angle are you? Because then the angle would be positive whether it was positive (up to 90 degrees) or negative (ditto) as cosine is symmetrical around 0.
If this is the case then instead of using the normal of the plane, use the plane direction itself and subtract 90 degrees from the angle, so 0 to 180 would become -90 to +90 degrees (or -half PI to +half PI if you're radially inclined).