Math question - Collision Detection Tutorial

the way I see it for swept spheres vs segments is like this.

Doing a swept sphere vs segment collision test is equivalent to doing a ray vs a cylinder test. All you need, is to have the sphere centre at a distance of exactly the sphere radius.

The equation of a infinite cylinder is

((P - O) x H)^2 = R^2

O is a point on the cylinder core (say vertex V1 of segment)
H is the direction of the cylinder (say H = V2 - V1)


the ray equation for the sphere centre is

P = C + D.t

C is the centre of the sphere, D is either the displacement or the velocity of the sphere, it does not matter which way you do it.

plug in C + D.t in the infinite cylinder equation and you get

((C - O) x H + t.(D x H))^2 - R^2 = 0

it is an equation in the form

(X + t.Y)^2 - R^2 = 0

X = (C - O) x H
Y = (D x H)

expand the equation and you get

X^2 + t.(2.(X.Y)) + t^2.(Y^2) - R^2 = 0;

and this equation is in itself a second order equation in the form

a.t^2 + b.t + c = 0

where

a = Y^2 = (D x H)^2
b = 2.(X.Y) = 2 * (D x H) . ((C - O) x H)
c = X^2 - R^2 = ((C - O) x H)^2 - R^2


you can do the same system for two sphere colliding, or a ray vs. a sphere intersection (the tests are equivalent)

equation of the surface of a sphere : (P - O)^2 = R^2

Thank you!

That helps me a lot

btw, you need to normalise the segment direction.

or ((P - O) x H)^2 = R^2 will not be meaningful.

say E is your un-normalised direction (E = V2 - V1), while H is the direction of the segment, and as length of 1.

E = H . |H|


H = E / |H|

(P - O) x H = (P - O) x (E/|H|) = ((P - O) x E) / |H|

so

((P - O) x H)² = ((P - O) x E)² / |H|²

and H.H = |H|²

so ((P - O)xH)² = ((P - O) x (V2 - V1))² / (V2 - V1)²

so in the end, if you have a segment [V1, V2], the equation for the capped cylinder around [V1, V2] is

((P - O) x (V2 - V1))² / (V2 - V1)² = R²

=> ((P - O) x (V2 - V1))² = R² . (V2 - V1)²
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plug in your ray equation in there, solve second order equation, there you go