OT: would anyone happen to know how to calculate bouncing a ball off an angled box?

[Erax]

OK, I can explain this to you for one simple situation and you'll have to work it out for the rest.

Let's say that the ball travels along the X-axis, therefore its trajectory is at 0 degrees (at 90 degrees it would be going 'up' the Y-axis). And let's say further that the wall it's going to bounce off of intercepts the X-axis at some point.

OK, now we are going to measure the wall's angle of inclination starting at the X-axis and ending at the wall's 'upper' segment. This angle is therefore being measured in a counter-clockwise direction. Let's call its value A.

Since the ball's trajectory is at 0 degrees, the angle between it and the lower segment of the wall is also equal to the wall's inclination, A. Now it's going to rebound at an angle to the wall's upper segment that is equal to this inclination, therefore to the X-axis itself it is an angle of 2A.

Some examples : Wall at 45 degrees - ball rebounds at 90 degrees (goes up).

Wall at 30 degrees - ball rebounds at 60 degrees.

Wall at 90 degrees - ball rebounds at 180 degrees (trajectory doubles back on itself).

Wall at 135 degrees - ball rebounds at 270 degrees.

The trick is, A is always between 0 and 180; outside these bounds the wall has 'flipped over' and you still have A between 0 and 180.

That's the best I can do from here. I'm not too good at explaining things without drawing lots of diagrams.

[narf poit chez BOOM]

thanks to the both of you.

% 360

modulo

that's for making sure it's still between 0-360, right?

anyone know how to figure out what side it hit when the box is rotated? i've got an idea, but it's clunky, rotating the whole thing, ball and box, back to 0 while keeping there relative positions and then back.

[ September 13, 2003, 00:36: Message edited by: narf poit chez BOOM ]