Comments on: Dynamic Circle/Circle Collision Detection in ActionScript 3

By: James

James — Thu, 03 Mar 2011 12:28:37 +0000

For those that are interested, I have solved the problem above by adding vDiff.multiply(0.9); just after vDiff is calculated in the detect function.

I now have a new question: I have a problem when the two circles are added whilst overlapping each other. A collision is detected, but there are insufficient velocities to move them apart, so they just sit there.

I have attempted several solutions in terms of re-positioning the circles, but this causes problems when you add more than two circles to the mix.

The ideal solution would be for the circles to gradually move apart from one another, but sadly my maths abilities are not good enough to figure that one out.

Hopefully someone can help…?

By: James

James — Thu, 03 Mar 2011 10:07:31 +0000

Would you mind explaining how you would change the code to allow for decremental velocities?

For example, imagine friction comes into play, and the velocity of each circle will reduce by 0.1 each frame.

How would you account for that?

By: drew

drew — Tue, 16 Mar 2010 14:25:32 +0000

@Patrick Corander
if you look at the derivation above, you’ll see that the equation equals at^2 + 2bt + c = 0. the generic form of this equation is at^2 + bt + c = 0, so in this case we know that the second term happens to be a multiple of 2. here’s how i solved for t:
at^2 + 2bt + c = 0
at^2 + 2bt = -c
t^2 + t(2b/a) = -c/a
t^2 + t(2b/a) + b^2/a^2 = -c/a + b^2/a^2
(t + b/a)^2 = -c/a + b^2/a^2
t + b/a = sqrt( (b^2 – ac)/a^2 )
t + b/a = sqrt( b^2 – ac )/a
t = (-b +/- sqrt( b^2 – ac ))/a

By: Patrick Corander

Patrick Corander — Tue, 16 Mar 2010 07:37:24 +0000

This has been bugging me for a few days so I need to ask :p. How do you actually know that b is multiple of 2 here? And if you know that b is multiple of 2, how do you actually reduce the equation at this state (-b +/- sqrt(b^2 – ac))/a?

cheers

By: drew

drew — Sat, 02 Jan 2010 14:26:28 +0000

It is the exact same equation, where it is known that b is a multiple of 2. Because of this, the solution reduces to (-b +/- sqrt(b^2 – ac))/a.

EDIT: I’ve added a derivation of this equation- hopefully that will add some clarity!

By: Michael B Paulson

Michael B Paulson — Sat, 02 Jan 2010 07:18:10 +0000

The quadratic formula is
//since your only doing the lesser of the two

X = (-b – sqrt(b^2 – 4ac)) / 2a

On the example you are missing the “4ac” but have “ac”

So that is what MultiHunter is saying.

By: drew

drew — Tue, 29 Dec 2009 15:29:48 +0000

i don’t know what you’re saying, Multihunter. ax^2 + 2bx + c = 0 is a quadratic equation… its roots are (-b +/- sqrt(b^2 – ac))/a

By: Multihunter

Multihunter — Tue, 29 Dec 2009 06:31:33 +0000

But that’s not the quadratic formula!
This is the quadratic formula:
http://en.wikipedia.org/wiki/Quadratic_equation#Quadratic_formula

By: drew

drew — Fri, 12 Oct 2007 19:56:12 +0000

Hi Elliot, It's just an arbitrarily defined constant meant to leave room for numerical rounding errors. If the dotted relative velocity (in this case defined as the difference between 2 and 1) is less than 0, the circles are moving apart from each other. Because this is just a computationally inexpensive early-out for the algorithm, it's a bit safer to leave a margin for error. This sort of constant is most often named EPSILON. Here's the whole source in case you missed the link above.

By: Elliott Partridge

Elliott Partridge — Fri, 12 Oct 2007 18:03:32 +0000

I’m trying to port this code to C++, and I’m wondering what Collision.OMEGA is. Is that some constant? Thanks for the guide!!