JayTracer Mathematical Background

Contents 1 Background 1.1 Notation 1.2 Algebraic Properties of the Dot Product 2 Jaytracer 2.1 Ray 2.2 Plane 2.3 Sphere

Background

Notation

DEFINITION:                                                         EXAMPLE:
all scalars are in normal letters                                   t
all vectors are in capitals                                         V
the coordinates of vectors are named after the vector, suffixed
    with x, y and z                                                 V = [Vx, Vy, Vz]
dot product of vectors X and Y                                      XY
dot product of vector X with itself                                 X^2
length of vector X = sqrt(X^2) = sqrt(Xx^2 + Xy^2 + Xz^2)           |X|
expressions in text areas are enclosed in single quotes when it
    is not obvious for the reader, what is math and what isn't      we see that 'X = tV', which...

Algebraic Properties of the Dot Product

I'll just remind you of the algebraic properties of the dot product:

     AB = AxBx + AyBy + AzBz
     AA = |A|^2 = sqrt( Ax^2 + Ay^2 + Az^2 )^2 = Ax^2 + Ay^2 + Az^2
     AB = BA
 A(B+C) = AB+AC
  (rA)B = r(AB) = A(rB)

Jaytracer

Ray

a ray (a directed line) is given by the following equation.

    X = S + tV

S is the starting point and V is the direction.

Plane

a plane is given by the following equation:

    N (X - P) = 0
N is the normal and P is a point on the plane. The plane is given by all vectors PX, who are
orthogonal to N (e.g. their dot-product is 0).

The intersection between the ray and the plane is the point(s) where the free vectors X are equal.
In the ray X is given by S + tV, which we substitute for X in the plane equation:

    N (S + tV - P) = 0
     NS + tNV - NP = 0
               tNV = NP - NS
               tNV = N(P-S)

Easily solved for t:
                     N(P-S)
                 t = -------
                       NV

If NV = 0, the plane normal and the ray are orthogonal, which means that the ray and the plane are
parallel, i.e. no intersection.

Sphere

a sphere is given by the following equation:

    (X - C)^2 = r^2

C is the center and r is the radius.

The intersection between the ray and the sphere is the point(s) where the free vectors X are equal.
In the ray X is given by S + tV, which we substitute for X in the sphere equation:

    (S + tV - C)^2 = r^2

We introduce D = S - C, the vector from the sphere center to the ray starting point, which gives us:

    (tV + D)^2  = r^2

This can also be written as 'sqrt((tV + D)^2) = r' or '|tV + D| = r' which makes sense: the ray plus
the vector from the sphere center to the ray starting point and the radius of the sphere must have
equal length.

We expand:

    (tVx + Dx)^2 + (tVy + Dy)^2 + (tVz + Dz)^2 = r^2
    tVx^2 + Dx^2 + 2tVxDx  +  tVy^2 + Dy^2 + 2tVyDy  +  tVz^2 + Dz^2 + 2tVzDz  = r^2

'Dx^2 + Dy^2 + Dz^2 = D^2' and 'VxDx + VyDy + VzDz = VD' which leaves:

    tVx^2 + tVy^2 + tVz^2  +  D^2 + 2tVD = r^2

also, tVx^2 + tVy^2 + tVz^2 = t^2V^2, which leaves:

          t^2V^2 + D^2 + 2tVD = r^2
    V^2t^2 + 2VDt + D^2 - r^2 = 0           (2nd degree polynomial of t)

Which we solve for t:
                  _______________________          /             _________  \
        -2VD +- \/(2VD)^2 - 4V^2(D^2-r^2)         |      -B +- \/B^2 - 4AC   |
    t = ---------------------------------         |   x = ----------------   |
                     2V^2                          \              2A        /
                  _______________________
        -2VD +- \/4(VD)^2 - 4V^2(D^2-r^2)
      = ---------------------------------
                     2V^2
                   _____________________
        -2VD +- 2\/(VD)^2 - V^2(D^2-r^2)
      = --------------------------------
                     2V^2
                 _____________________
        -VD +- \/(VD)^2 - V^2(D^2-r^2)
      = ------------------------------
                     V^2

If the discriminant '(VD)^2 - V^2(D^2-r^2)' is < 0, there are no solutions, i.e. no intersections.
Otherwise, there are 1 or 2 solutions.