# Ray Tracing 005: Adding a Sphere

Episode 4 introduced a ray class. Now it’s time to let the ray hit the first object: a sphere.

The equation of a sphere centered at the origin is

$x^2 + y^2 + z^2 = R^2$

which is essentially derived from Pythagoras’ Theorem extended to three dimensions.

For any $$(x,y,z)$$, if $$x^2 + y^2 + z^2 = R^2$$ then $$(x,y,z)$$ is on the sphere, otherwise it’s not.

The more general equation of a sphere centered at a point $$(x_0,y_0,z_0)$$ is

$(x-x_0)^2 + (y-y_0)^2 + (z-z_0)^2 = R^2$

This can also be expressed as a dot-product:

$$\begin{pmatrix} x \\ y \\ z \end{pmatrix} - \begin{pmatrix} x_0 \\ y_0 \\ z_0 \end{pmatrix} = \begin{pmatrix} x-x_0 \\ y-y_0 \\ z-z_0 \end{pmatrix}$$ $$\begin{pmatrix} x-x_0 \\ y-y_0 \\ z-z_0 \end{pmatrix} \cdot \begin{pmatrix} x-x_0 \\ y-y_0 \\ z-z_0 \end{pmatrix} =$$ $$(x-x_0)^2 + (y-y_0)^2 + (z-z_0)^2 = R^2$$

For center $$C = (x_0,y_0,z_0)$$ and a point $$P = (x,y,z)$$, we can also write:

$$\begin{pmatrix} P - C \end{pmatrix} \cdot \begin{pmatrix} P - C \end{pmatrix} = R^2$$

The ray $$p(t) = A + t*B$$ intersects the sphere where $$p(t) = P$$.

If the ray hits the sphere, there is a t for which p(t) satisfies the sphere equation.

$$\begin{pmatrix} p(t) - C \end{pmatrix} \cdot \begin{pmatrix} p(t) - C \end{pmatrix} = R^2$$ $$\begin{pmatrix} A + t*B - C \end{pmatrix} \cdot \begin{pmatrix} A + t*B - C \end{pmatrix} = R^2$$

I’m looking for t, and A, B, C and R are constants. I can rearrange the equation as follows:

$$\begin{pmatrix} t*B + (A-C) \end{pmatrix} \cdot \begin{pmatrix} t*B + (A-C) \end{pmatrix} = R^2$$

Let $$oc = (A-C)$$ and rearrange:

$$t^2 * (B \cdot B) + 2 * t* (B \cdot oc) + (oc \cdot oc) = R^2$$

Now let $$a = (B \cdot B)$$ and rearrange:

$$a * t^2 + 2 * t * (B \cdot oc) + oc \cdot oc - R^2 = 0$$

Let $$b = 2 * (B \cdot oc)$$:

$$a * t^2 + b * t + oc \cdot oc - R^2 = 0$$

Then, let $$c = (oc \cdot oc) - R^2$$ to arrive at the normal form:

$$a * t^2 + b * t + c = 0$$

This is a quadratic equation that can be solved for t with the Quadratic Formula:

$$t = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a}$$

The square root portion of this equation can either be positive (meaning there are two real solutions), or zero (meaning there’s one real solution), or negative (meaning there are no real solutions).

This relates directly to the geometry. If there are two real solutions, the ray intersects the squere at two points. If there’s exactly one real solution, it touches the sphere at one point on the surface. If there are no real solutions, then the ray doesn’t touch the sphere at all.

The new function hit_sphere() implements the math derived in this episode. The only difference is that I only care about the root and whether it’s greater than 0. I don’t need a precise solution here. All rays that miss the sphere will result in the background gradient of the previous episode.

The sphere is located in the center of the screen with a radius of 0.5. If the ray intersects this sphere, the color function will just return the color red for now.

// sphere-example.cpp
bool hit_sphere(const vec3& center, float radius, const ray& r) {
vec3 oc = r.origin() - center;
float a = dot(r.direction(), r.direction());
float b = 2.0 * dot(oc, r.direction());
float discriminant = b*b - 4*a*c; // for quadratic equation
return (discriminant > 0); // all we care about is whether the root is not 0
}

vec3 color(const ray& r) {
if (hit_sphere(vec3(0,0,-1), 0.5, r)) return vec3(1,0,0); // just return red if sphere was hit

vec3 unit_direction = unit_vector(r.direction());
float t = 0.5f * (unit_direction.y() + 1.0f);
return (1.0f - t) * vec3(1.0f, 1.0f, 1.0f) + t * vec3(0.5f, 0.7f, 1.0f);
}


## Javascript and Canvas

Here’s also the same example in Javascript.

Please check my repository at github.com/celeph/ray-tracing for the complete code. For more details, check out Peter Shirley’s excellent book that inspired this episode: “Ray Tracing in One Weekend” and his original code and book repository.