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Asked: 2026-05-11T20:44:46+00:00

I’d like to deflect a ball at an angle depending on where it hits

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I’d like to deflect a ball at an angle depending on where it hits a paddle. Right now, I’m only changing the y coordinate, which results in an uninteresting deflection. It will angle but independent on impact location against the paddle. I’d like something more fun. Speed, momentum, mass and other factors don’t need to be taken into consideration. Just angle depending on impact location of paddle. I’ve read this Not a number error (NAN) doing collision detection in an iphone app but it seems overly complicated for what I’m looking for. Is there a simpler way to calculate the deflection?

The objects are two UIImageViews.

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1 Answer

  1. Editorial Team

    Well, nothing realistic but you could do something so that the outbound angle is only dependent on where on the paddle it hits.

    I have never done any iPhone or objective C coding so I’ll just write up something in pseudo/C code.

    First I’d calculate the speed, which is the length of the speed vector, or:

    double speed = sqrt(velX * velX + velY * velY); // trigonometry, a^2 + o^2 = h^2

    Then we want to calculate the new angle based on where we hit the paddle. I’m going to assume that you store the X collision in impactX and the length of the paddle in paddleLength. That way we can calculate an outbound angle. First let’s figure out how to calculate the range so that we get a value between -1 and 1.

    double proportionOfPaddle = impactX / (double) paddleLength; // between 0 and 1
double impactRange = proportionOfPaddle * 2 - 1; // adjust to -1 and 1

    Let’s assume that we do not want to deflect the ball completely to the side, or 90 degrees, since that would be pretty hard to recover from. Since I’m going to use the impactRange as the new velY, I’m going to scale it down to say -0.9 to 0.9.

    impactRange = impactRange * 0.9;

    Now we need to calculate the velX so that the speed is constant.

    double newVelX = impactRange;
double newVelY = sqrt(speed * speed - newVelX * newVelX); // trigonometry again

    Now you return the newVelX and newVelY and you have an impact and speed dependent bounce.

    Good luck!

    (Might very well be bugs in here, and I might have inverted the X or Y, but I hope you get the general idea).

    EDIT: Adding some thoughts about getting the impactX.

    Let’s assume you have the ball.center.x and the paddle.center.x (don’t know what you call it, but let’s assume that paddle.center.x will give us the center of the paddle) we should be able to calculate the impactRange from that.

    We also need the ball radius (I’ll assume ball.width as the diameter) and the paddle size (paddle.width?).

    int ballPaddleDiff = paddle.center.x - ball.center.x;
int totalRange = paddle.width + ball.width;

    The smallest value for ballPaddleDiff would be when the ball is just touching the side of the paddle. That ballPaddleDiff would then be paddle.width/2 + ball.width/2. So, the new impactRange would therefore be

    double impactRange = ballPaddleDiff / (double) totalRange / 2;

    You should probably check the impactRange so that it actually is between -1 and 1 so that the ball doesn’t shoot off into the stars or something.

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