How to add friction/damping to circle to circle prediction formula?

Question

I've been working on determining the time at which two objects would collide, if they would at all. I had referenced this thread which was very helpful: Circle-Circle Collision Prediction

(Oax + t*Dax - Obx - t*Dbx)^2 + (Oay + t*Day - Oby - t*Dby)^2 = (ra + rb)^2

I was able to solve that for t but now I'm wondering how I would best apply drag/friction to it.

If ball1 is traveling at a fixed rate of 9, and ball2 is traveling at 7... this works great. Problem is that if their velocities will decrease over time.. so if we're covering a lot of ground, ball1 will end up traveling at 4 and so the collision result could change.

For example: Ball1 is traveling down a slight slope, gaining speed. Ball2 is going straight and losing speed.

I'm assuming that the part of this formula needing adjustment would be the t*Dax, t*Dbx, t*Day, and t*Dby. These appear to be linear, right? So I would want to replace these with something representing my damping.

There would probably be multiple things determining my damping such as: surface friction and gravity.. or could reference easing equations such as Robert Penner's equations in this file http://gsgd.co.uk/sandbox/jquery/easing/jquery.easing.1.3.js

Thanks!

Edit:

Looked up some projectile motion formulas and was playing around with d = V₀t - ½ at² . Was wondering if combining that would be the trick, but not sure where/how it would be applied. I considered the following...

(O₁x + (t·D₁x - a₁t²) - O₂x - (t·D₂x + a₂t²))² + (O₁y + (t·D₁y - a₁t²) - O₂y - (t·D₂y + a₂t²))² = (r₁ + r₂)²

Update (thanks to the post by willywonkadailyblah)

After working with it a bit, here's a quick Javascript example of it working. You can find references to Vector classes and Quartic solvers online.

// p1 = c1 + (v1 * t) + 0.5 * a1 * (t ^ 2)
// p2 = c2 + (v2 * t) + 0.5 * a2 * (t ^ 2)

var p1c = new Vector(10,  50),
    p1v = new Vector(10, -10),
    p1a = new Vector( 5,   2),

    p2c = new Vector( 50,  10),
    p2v = new Vector(-10,  10),
    p2a = new Vector( -9,   0),

    r1  = r2 = 15,
    p1, p2,
    oc, ov, oa,
    A, B, C, D, E,
    q, r, s, t, u;

    var quadraticAnswers = solveQuadratic(p1c, p1v, p2c, p2v, r1);
    var quarticAnswers = solveQuartic( p1c, p1v, p1a, p2c, p2v, p2a, r1, r2 );

    console.log( "Quartic", quarticAnswers );
    console.log( "Quadratic", quadraticAnswers );
    console.log( "Best Quartic", quarticAnswers.bestAnswer );

    function solveQuartic(p1c, p1v, p1a, p2c, p2v, p2a, r1, r2) {
        // delta p
        oc = p1c.subtract(p2c);

        // delta v
        ov = p1v.subtract(p2v);

        // delta a
        oa = p1a.subtract(p2a);

        //
        A = oa.magnitude().pow(2).multiply(0.25);
        B = ov.multiply(oa);
        C = oc.multiply(oa).add( ov.magnitude().pow(2) )
        D = ov.multiply(oc).multiply(2);
        E = oc.magnitude().pow(2);

        q = A.x + A.y;
        r = B.x + B.y;
        s = C.x + C.y;
        t = D.x + D.y;
        u = E.x + E.y - Math.pow(r1 + r2, 2);  // hidden because of the issue with radius not adding up

        // the quartic/cubic/quadratic solver
        var ret = solveEquations(q, r, s, t, u);

        if (oa.x + oa.y < 0) {
            ret.bestAnswer = ret.x4 || ret.x3 || ret.x2 || ret.x1;
        }
        else {
            ret.bestAnswer = ret.x2 || ret.x1 || ret.x4 || ret.x3;
        }

        return ret;
    }

VectorJS works well with this, but you have to manually add a pow method. https://evanw.github.io/lightgl.js/docs/vector.html

Here's a quartic/cubic/quadratic solver for JS. You'll need to do a fair amount of modification on the source to get it to work with your app or to translate it into other languages. http://www.freewebs.com/brianjs/ultimateequationsolver.htm

I was going to post a C++, but figured this would be easy for a lot of people and many people use JS nowadays.

Answer 1

You wanted to simulate surface friction / gravity, which most models at this scale assume to be constant.

SUVAT equations for the positions of each ball:

Need distance between centers < sum of radii - take square magnitude (self dot-product):

Let

And then all you need to do is to solve this quartic equation to find the smallest positive real root. Easy!

Well OK. Here are two sites to help you:

• https://en.wikipedia.org/wiki/Quartic_function
• https://math.stackexchange.com/questions/785/is-there-a-general-formula-for-solving-4th-degree-equations

If you don't want to do this, then a fast iterative method such as Newton-Raphson will do; contrary to time-step based methods these don't limit the accuracy of the answers.