iPhone Pong Advanced Deflection Angle

Question

Hi, I am currently developing a simple Pong game for the iPhone. Currently using CGRectIntersectsRect for the collision detection and as for the deflection of the ball when it hits the paddle, I just multiply the ball velocity with -1 (therefore reversing the direction of the ball).

What I am trying to do is to make it so that when the ball hits the paddle, it checks whether how far is the ball from the center of the paddle, and increases the deflection angle the further the ball is away from the center of the paddle. (E.g. In this case, the ball will be deflected back at 90 degrees no matter where it came from, as long as it hits the center of the paddle)

How am I suppose to do that?

Any help would be greatly appreciated.

Thank you.

Answer 1

What you have given us are reference points (centre and edge). What we need are a reference line from which to measure the new angle. Additionally, what you are saying is not consistent and thus does not make sense.

I am guessing that what you are asking is a way to calculate the outgoing angle such that it is only a function of where it hit on the paddle. If it hit the paddle centre, then irrespective of the incoming angle, it will bounce off at an angle of 90 degrees to the paddle. If it hit the paddle edge, then irrespective of the incoming angle it will bounce off at an angle of 45 degrees to the paddle.

If so, then the following should do it (it is not the only way).

Assumption: The paddle shape is a rectangle.

• Let L be the length of the paddle.
• Let K be a constant such that L / (2 * K) = 1 / sqrt(2).
• Let D be the distance from the centre of the paddle (may be + or -).

theta = pi/2 - asin (D / K);

This should give you an angle relative to the paddle.

Hope this helps.