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superman3275

Just finished my Circular Collision Function!

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Hello all! I have spent today working on a good set of functions to test collision in my game (I'm expecting to finish my character tomorrow) and I'm very happy! The method I used was finding the distance betweent the circles (I used the pythagoreon theorem!) and then seeing if the distance was less than either circles radius. It doesn't work for ovals (Only perfect circles), however the circles can have different diameters. Do you know how I can check collision for Ovals also? I've been thinking through the problem all day, however I've had no luck. Cheers :)!

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google search for ellipse's intersection would give you what you need.

 

basically, all you have to do is find the point on the ellipse closest to the other ellipse(or circle), and it becomes simple distance checks.

 

 

bool EllipseIntersection(float axRadius, float ayRadius, float ax, float ay, float bxRadius, float byRadius, float bx, float by){
 //Get direction+distance toward b
 float cx = bx-ax;
 float cy = by-ay;
 float cSqrt = sqrtf(cx*cx+cy*cy); //Check for 0
 if(cSqrt <=0.0001f) return true; //they are ontop of each other.
 //Normalize c:
 float cnx = cx/cSqrt;
 float cny = cy/cSqrt;
 //Find a's ellipse point toward b:
 float aex = axRadius*cnx;
 float aey = ayRadius*cny;
 float aeSqrt = sqrtf(aex*aex+aey*aey); //Get distance to that point.
 //Find b's ellipse point toward a:
 float bex = bxRadius*-cnx;
 float bey = byRadius*-cny;
 float beSqrt = sqrtf(bex*bex+bey*bey); //" "
 return cSqrt<=aeSqrt+beSqrt;
}

 

 

someone correct me if i did something wrong, i wrote that up just now from memory of the last time i wrote an ellipse-ellipse collision function.

 

edit: tested and fixed an issue, i thought i could test without taking the sqrt, but that's only with circles.

Edited by slicer4ever

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When using Pythagoras for collision detection you can cut the cpu cost from the sqrt by doing the comparison in the second dimension. In other words, rather than using sqrt on (a^2 + b^2) and then comparing it to (c) just compare (a^2 + b^2) to (c^2). Squaring is much cheaper than sqrt'ing and you'll get the same relationship for &lt;, &gt;, and ==.<br /><br />Is that what you were mentioning, slicer? I see your edit there.

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From instinct I'd think that you'd need to calculate the radius along the vector from centerpoint-to-nearestpoint and then just handle it like circle collision.

This is intriguing. How would you store the collider? I'm thinking a centerpoint x/y vector, a width and height and then an angle of rotation... So to find the radius for a given angle you'd subtract the angle of rotation from it (to axis-align the calculation) then...

Hmm...

I think you could apply slicer's equation at that point, though you'd have to also adjust for rotation with the other ellipse if you're doing ellipse-vs-ellipse testing.

I don't know if there would be a good way of reducing the complexity in this case, so you'd definitely want to do a broad-phase test before getting into the heavy math.

Some related links I found:
http://www.gamedev.net/topic/309198-aabb-for-an-ellipse/
http://www.gamedev.net/page/resources/_/technical/game-programming/general-collision-detection-for-games-using-ell-r1026 Edited by Khatharr

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When using Pythagoras for collision detection you can cut the cpu cost from the sqrt by doing the comparison in the second dimension. In other words, rather than using sqrt on (a^2 + b^2) and then comparing it to (c) just compare (a^2 + b^2) to (c^2). Squaring is much cheaper than sqrt'ing and you'll get the same relationship for &lt;, &gt;, and ==.<br /><br />Is that what you were mentioning, slicer? I see your edit there.

yes, i first took the Sq's, but when i was doing a test on the code to check that it was working i realized that i couldn't do that.

 

for example:

ellipse A: x = 0, y = 0, xRad = 5.0f, yRad = 5.0

ellipse B: x = 10, y = 0, xRad = 5.0f, yRad = 5.0

(essentially two circles)

so, their distance squared would be 100

but each radius squared would be 25, and their total is 50, while the two circles are touching, the sq's don't reflect that, however the sqrt's do.

 

and yes, that code's only for axis-aligned ellipse's.

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