Editing Ray tracing (graphics) (section)

===Example===
[[File:Parametric surface illustration (trefoil knot).png|thumb|[[Trefoil knot]], created with a [[parametric equation]] and ray traced in [[Python (programming language)|Python]].]]

As a demonstration of the principles involved in ray tracing, consider how one would find the intersection between a ray and a sphere. This is merely the math behind the [[line–sphere intersection]] and the subsequent determination of the colour of the pixel being calculated. There is, of course, far more to the general process of ray tracing, but this demonstrates an example of the algorithms used.

In [[vector notation]], the equation of a sphere with center <math>\mathbf c</math> and radius <math>r</math> is

:<math>\left\Vert \mathbf x - \mathbf c \right\Vert^2=r^2.</math>

Any point on a ray starting from point <math>\mathbf s</math> with direction <math>\mathbf d</math> (here <math>\mathbf d</math> is a [[unit vector]]) can be written as

:<math>\mathbf x=\mathbf s+t\mathbf d,</math>

where <math> t</math> is its distance between <math>\mathbf x</math> and <math>\mathbf s</math>. In our problem, we know <math>\mathbf c</math>, <math>r</math>, <math>\mathbf s</math> (e.g. the position of a light source) and <math>\mathbf d</math>, and we need to find <math> t</math>. Therefore, we substitute for <math>\mathbf x</math>:

:<math>\left\Vert\mathbf{s}+t\mathbf{d}-\mathbf{c}\right\Vert^{2}=r^2.</math>

Let <math>\mathbf{v}\ \stackrel{\mathrm{def}}{=}\  \mathbf{s}-\mathbf{c}</math> for simplicity; then

:<math>\left\Vert\mathbf{v}+t\mathbf{d}\right\Vert^{2}=r^{2}</math>

:<math>\mathbf{v}^2+t^2\mathbf{d}^2+2\mathbf{v}\cdot t\mathbf{d}=r^2</math>

:<math>(\mathbf{d}^2)t^2+(2\mathbf{v}\cdot\mathbf{d})t+(\mathbf{v}^2-r^2)=0.</math>

Knowing that d is a unit vector allows us this minor simplification:

:<math>t^2+(2\mathbf{v}\cdot\mathbf{d})t+(\mathbf{v}^2-r^2)=0.</math>

This [[quadratic equation]] has solutions
:<math>t=\frac{-(2\mathbf{v}\cdot\mathbf{d})\pm\sqrt{(2\mathbf{v}\cdot\mathbf{d})^2-4(\mathbf{v}^2-r^2)}}{2}=-(\mathbf{v}\cdot\mathbf{d})\pm\sqrt{(\mathbf{v}\cdot\mathbf{d})^2-(\mathbf{v}^2-r^2)}.</math>

The two values of <math>t</math> found by solving this equation are the two ones such that <math>\mathbf s+t\mathbf d</math> are the points where the ray intersects the sphere.

Any value which is negative does not lie on the ray, but rather in the opposite [[Line (mathematics)|half-line]] (i.e. the one starting from <math>\mathbf s</math> with opposite direction).

If the quantity under the square root (the [[quadratic equation#Discriminant|discriminant]]) is negative, then the ray does not intersect the sphere.

Let us suppose now that there is at least a positive solution, and let <math>t</math> be the minimal one. In addition, let us suppose that the sphere is the nearest object on our scene intersecting our ray, and that it is made of a reflective material. We need to find in which direction the light ray is reflected. The laws of [[Reflection (physics)|reflection]] state that the angle of reflection is equal and opposite to the angle of incidence between the incident ray and the [[surface normal|normal]] to the sphere.

The normal to the sphere is simply

:<math>\mathbf n=\frac{\mathbf y- \mathbf c}{\left\Vert\mathbf y- \mathbf c\right\Vert},</math>

where <math>\mathbf y=\mathbf s+t\mathbf d</math> is the intersection point found before. The reflection direction can be found by a [[Reflection (mathematics)|reflection]] of <math>\mathbf d</math> with respect to <math>\mathbf n</math>, that is

: <math>\mathbf r = \mathbf d - 2(\mathbf n \cdot \mathbf d ) \mathbf n.</math>

Thus the reflected ray has equation

: <math>\mathbf x = \mathbf y  + u \mathbf r. \, </math>

Now we only need to compute the intersection of the latter ray with our [[field of view]], to get the pixel which our reflected light ray will hit. Lastly, this pixel is set to an appropriate color, taking into account how the color of the original light source and the one of the sphere are combined by the reflection.