Contact Detection Algorithm

Question

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Hi I want a contact detection algorithm to detect collision between 2 ball (2D). I used this algorithm but it doesn't work well, Please Help me.

C#

//in Ball Class:

public bool Collides(Balls otherSprite)
{
    // Check if two Ball collide
    if (this.rectangle.X + this.texture2D.Width > otherSprite.rectangle.X &&
    this.rectangle.X < otherSprite.rectangle.X + otherSprite.texture2D.Width &&
    this.rectangle.Y + this.texture2D.Height > otherSprite.rectangle.Y &&
    this.rectangle.Y < otherSprite.rectangle.Y + otherSprite.texture2D.Height)
        return true;
    else
        return false;
}


if (Balls[0].Collides(Balls[1]))
{
    Vector2 tempVelocity = Balls[0].velocity;
    Balls[0].velocity = Balls[1].velocity;
    Balls[1].velocity = tempVelocity;
}

Posted 3-Mar-13 23:17pm

aloneintherain

Updated 3-Mar-13 23:45pm

Richard MacCutchan

v6

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Richard MacCutchan · Accepted Answer · 2013-03-03T23:52:00

Solution 1

Try using the Rectangle.Intersect Method[^].

Posted 3-Mar-13 23:52pm

Richard MacCutchan

Comments

Yossi Yaari 4-Mar-13 13:21pm

you meant IntersectsWith(...) of course

Richard MacCutchan 4-Mar-13 13:41pm

No, I meant Intersects, but your suggestion is also worth looking at. What I was really aiming for, was to get the questioner to make use of the documentation.

Yossi Yaari 5-Mar-13 0:50am

OK

Sergey Alexandrovich Kryukov 15-Mar-13 14:52pm

In case of rectangular scene, even of the ball is round, not rectangular, it works, if you consider the trivial geometry of the game. My 5. Simple and correct.
—SA

CPallini · Accepted Answer · 2013-03-03T23:56:00

Solution 2

Well, if you want to detect collision between two balls then the condition should be (subject to Sergey's approval :-D).

C#

(XC0-XC1)*(XC0-XC1)+(YC0-YC1)*(YC0-YC1) <= R1*R1+R2*R2+2*R1*R2

That is distance between the two centers is less than (or equal) to sum of their radii

Posted 3-Mar-13 23:56pm

CPallini

Comments

Yossi Yaari 4-Mar-13 13:13pm

I think he means circles or 2D balls.
in any case your solution for calculating distance between centers is correct.

I wonder if it is worth testting Rectangle.Intersect in advance?

CPallini 4-Mar-13 15:56pm

I don't know: Intersect is (hopefully) a fast method. However multiplication too is relatively fast.

Yossi Yaari 5-Mar-13 0:53am

the best reason I think to do so is to minimize chance for error.
if the rectangles don't intersect - their not close enough. so very small chance for significant false positive answer on collision, if there is an error in the calculation.

CPallini 5-Mar-13 3:49am

I would use it for speeding-up the process: if rectangles don't intersect then you may skip other checks.

Yossi Yaari 5-Mar-13 14:21pm

that is of course correct in general.
and here the mathematical solution is very light weight. so I'm not sure how much it will improve performance. as you your self commented.

Andreas Gieriet 5-Mar-13 18:31pm

My 5!
The formula is right. If you want to speed up, you may of course consider to pre-calculate some values (situation permitting). E.g. the right-hand side of the relation might be constant, assuming you have only two circles. And dx*dx + dy*dy is probably faster if you calculate the dx and dy first. So, it reduces to dy = x0-x1; dy = y0-y1; d = r0 + r1; dd = d*d; collision = dx*dx+dy*dy<=dd;
Cheers
Andi

Yossi Yaari · Accepted Answer · 2013-03-04T07:20:00

I think the best solution combines both comments above, and should look something like this:

C#

if( ! Ball1.IntersectsWith(Ball2))
{
   return false;
}

double Ball1CenterX = (Ball1.Right - Ball1.Width)/2;
double Ball2CenterX = (Ball2.Right - Ball2.Width)/2;
double Ball1CenterY = (Ball1.Bottom - Ball1.Height)/2;
//double Ball2CenterY = (Ball1.Bottom - Ball1.Height)/2;    // oops! (Ball1 ?) [MTH: fixed below]
double Ball2CenterY = (Ball2.Bottom - Ball2.Height)/2;

double Ball1Radious = Ball1.Width/2;
double Ball2Radious = Ball2.Width/2;

return (Ball2CenterX - Ball1CenterX)^2 + (Ball2CenterY - Ball1CenterY)^2 
         <= 
        (Ball1Radious + Ball2Radious)^2;   // [MTH: fixed. don't "times 2"]