Ray-Plane Intersection

Ray-Plane Intersection

A plane is defined by the equation: Ax + By + Cz + D = 0, or the vector [A B C D]. A, B, and C, define the normal to the plane. If A2 + B2 + C2 = 1 then the unit normal Pn = [A B C]. If A, B, and C define a unit normal, then the distance from the origin [0 0 0] to the plane is D.

A ray is defined by: R0 = [X0, Y0, Z0]

Rd = [Xd, Yd, Zd]

so R(t) = R0 + t * Rd , t > 0

To determine if there is an intersection with the plane, substitute for R(t) into the plane equation and get:

A(X0 + Xd * t) + B(Y0 + Yd * t) + (Z0 + Zd * t) + D = 0

which yields:

t = -(AX0 + BY0 + CZ0 + D) / (AXd + BYd + CZd)

= -(Pn· R0 + D) / (Pn · Rd)

First compute Pn· Rd = Vd. If Vd = 0 (incident angle, q = 900) then the ray is parallel to the plane and there is no intersection (if ray in in the plane then we ignore it). If Vd > 0 then the normal of the plane is pointing away from the ray. If we use one-sided planar objects then could stop if Vd > 0, else continue.

Now compute second dot product V0 = -(Pn· R0 + D) and compute t = V0 / Vd . If t < 0 then the ray intersects plane behind origin, i.e. no intersection of interest, else compute intersection point:

Pi = [Xi Yi Zi] = [X0 + Xd * t Y0 + Yd * t Z0 + Zd * t]

Now we usually want surface normal for the surface facing the ray, so if Vd > 0 (normal facing away) then reverse sign of ray.

So algorithm analysis:

1. Compute Vd and compare to 0: 3 "*"s, 2 "+"s, 1 comparison.

2. Compute V0 and t and compare to 0: 3 or 4 "*"s, 3 "+"s, 1 comparison.

3. Compute intersection joint: 3 "*"s, 3 "+"s.

4. Compare Vprd to 0 and reverse normal: 1 comparison.

Total = 10 "*"s, 8 "+"s, 3 comparisons.

Example calculation:

Given a plane [1 0 0 -7] ( plane with x = -7)

Ray with R0 = [ 2 3 4 ], Rd = [ 0.577 0.577 0.577]

Compute Vd = Pn · Rd = 0.577 > 0 , therefore the plane points away from the ray