Ray tracing: Computing ray-sphere intersections

03 Sep 2018
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Part 1: Ray tracing: Computing ray-circle intersections
Part 2: Ray tracing: Computing ray-sphere intersections

Whereas the previous post was about 2D ray-circle intersections, this one is about 3D ray-sphere intersections. We'll extend the circle standard equation to a sphere, and switch from scalars to vectors for defining both line and sphere. With vectors, we can generalize and perform intersection calculations for any dimension in a succinct manner.

Sphere offset from origin

Except for the added third dimension, the equation of a sphere centered at a position \(c\) is identical to that of the circle. Its equation still follows from the Pythagorean theorem: $$(x - c_x)^2 + (y - c_y)^2 + (z - c_z)^2 = r^2$$ With the sphere centered at origin \((0,0,0)\), the \(c\) terms disappear and the equation becomes $$x^2 + y^2 + z^2 = r^2$$

Vector standard form of a line

The \(y = ax + b\) definition of a line in 2D doesn't generalize to 3D. In 2D the line equation expresses that a change in one dimension may cause a change in at most one other dimension. In 3D, however, a change in one dimension may cause a change in two other dimensions.

Instead, a line in 3D is expressed as a starting position \(p_0\) and an amount of movement \(t\) in some direction \(p_1\). Statements about position and direction are best expressed in terms of vectors and operations on vectors: $$\vec p = \vec{p_0} + t\vec{p_1}$$ Different values of \(t\), for time, result in a line described by a fixed origin and a position vector pointing at different points on the line (we consider a position vector and a point synonymous). Different values of \(t\) has the effect of scaling the position vector along the line, allowing us to get to any position on the line.

In ray tracing we're only interested in the half-line where \(t > 0\). That half-line represents a ray traveling from the camera into the scene, and that's why we refer to it as a ray instead of a line. \(t = 0\) would be the position of the camera and \(t < 0\) a ray traveling out back of the camera.

As an example, we can find the parametric equation of the line passing through \((1,-1,4)\), the position of the camera, in the direction of \((3,5,-1)\), a scene object. We do so by substituting the two vectors into the vector equation for a line: $$\begin{eqnarray*} \vec p = (1,-1,4) + t(3,5,-2)\\ \vec p = (1,-1,4) + (3t,5t,-2t)\\ (x,y,z) = (1+3t,-1+5t,4-2t) \end{eqnarray*}$$ For ray-sphere intersections, we're after values of \(t\) describing points of intersection. Once we have those values, we can substitute those into the parametric equation of a line to get the \((x,y,z)\) coordinate of intersections.

Line and sphere intersections, scalar edition

As an example, suppose we have the parametric equation of a line \((10 + 2t, 5 + t, 2)\) and the equation of a sphere \(x^2 + y^2 + z^2 = 9\) on standard form. As with 2D intersections, we're after values for \(x\), \(y\), and \(z\) that when substituted into both equations make those equations equal. We can put in what we know about the unknowns of one equation into the other: $$\begin{eqnarray*} (10 + 2t)^2 + (5 + t)^2 + (2)^2 = 9 & & \textrm{\{substitute in line equation\}}\\ (10 + 2t)(10 + 2t) + (5 + t)(5 + t) + 4 = 9 & & \textrm{\{expand squares\}}\\ 100 + 20t + 20t + 4t^2 + 25 + 5t + 5t + t^2 + 4 = 9 & & \textrm{\{expand parenthesis\}}\\ 5t^2 + 50t + 129 = 9 & & \textrm{\{simplify and arrange in \(t^2\), \(t\), and constants order\}}\\ 5t^2 + 50t + 120 = 0 & & \textrm{\{move 9 to left side to arrive at the standard form of quadratic equation\}} \end{eqnarray*}$$ We went from one sphere equation with three unknowns, \(x\), \(y\), and \(z\), and ended up with a quadratic equation with one unknown \(t\). Solving for \(t\) means applying the same method as previous: $$\begin{eqnarray*} t = \dfrac{-b \pm \sqrt{d}}{2a} \textrm{, where \(d = b^2 - 4ac\)}\\ d = 50 - 4(5)(120) = 100\\ t_1 = \dfrac{-50 + \sqrt{100}}{2(5)} = \dfrac{-50 + 10}{10} = \dfrac{-40}{10} = -4\\ t_2 = \dfrac{-50 - \sqrt{100}}{2(5)} = \dfrac{-50 - 10}{10} = \dfrac{-60}{10} = -6\\ \end{eqnarray*}$$ Substituting these \(t\) values into the parametric vector equation of the line, we arrive at the points of line-sphere intersection: $$\begin{eqnarray*} \vec{p_1} = (x_1,y_1,z_1) = (10 + 2(-4), 5 + (-4), 2) = (2,1,2)\\ \vec{p_2} = (x_2,y_2,z_2) = (10 + 2(-6), 5 + (-6), 2) = (-2,-1,2) \end{eqnarray*}$$

For additional details on calculating line-sphere intersection, refer to this video.

Line and sphere intersections, vector edition

Above we established the standard vector equation for a line. For a sphere centered at \(c\) with radius \(r\), its corresponding standard vector equation is as below. In a moment, we'll show why that's the case: $$(\vec p - \vec c) \cdot (\vec p - \vec c) = r^2$$ \(\vec p\) represents every position vector on the sphere and \(\cdot\) is the dot product of two vectors. The term product in general refers to the result of multiplication, but unlike the single product definition for scalars, vectors have both dot product and cross product. We're only interested in the dot product, defined as the sum of the products of the corresponding entries, making the result a scalar, not another vector.

In the following equations, whether scalar product or dot product is meant can always be inferred based on whether both operands are scalars or both are vectors. Strictly speaking, we don't need the \(\cdot\) symbol, but it's convention.

At first sight, the standard vector equation of a sphere looks nothing like its scalar counterpart. But note the dot product's definition: $$\vec u \cdot \vec v = {u_x}{v_x} + {u_y}{v_y} + {u_z}{v_z} = r^2$$ If \(\vec{u} = \vec{v}\), as is the case for the vector sphere definition, expanding the dot product turns it into the scalar standard equation of a sphere. So while at first sight the dot product and sphere seem unrelated, defining the sphere in terms of the dot product is a handy shortcut: $$\vec u \cdot \vec u = {u_x}{u_x} + {u_y}{u_y} + {u_z}{u_z} = u_{x}^2 + u_{y}^2 + u_{z}^2 = x^2 + y^2 + z^2 = r^2$$ As with the scalar solution to computing intersections, where we substituted the line equation into the circle equation, we can replace \(\vec p\) in the sphere equation by the vector equation of the line (don't be confused by the use of \(\vec p\) in both definitions. Those are different \(\vec p\)s): $$(\vec{p_0} + \vec{p_1}t - \vec c) \cdot (\vec{p_0} + \vec{p_1}t - \vec c) - r^2 = 0$$ Through rewriting and simplification, the goal is to turn this complicated looking equation into a quadratic one and solve for \(t\). For an actual line and sphere, \(\vec{p_0}\), \(\vec{p_1}\), \(\vec c\), and \(r\) are known quantities, leaving \(t\) as the only unknown:

$$\begin{eqnarray*} (\vec{p_0} \cdot \vec{p_0} + \vec{p_0} \cdot \vec{p_1}t - \vec{p_0} \cdot \vec c) + (\vec{p_1}t \cdot \vec{p_0} + \vec{p_1}t \cdot \vec{p_1}t - \vec{p_1}t \cdot \vec c) + (-\vec c \cdot \vec{p_0} - \vec c \cdot \vec{p_1}t + \vec c \cdot \vec c) - r^2 = 0 & & \textrm{\{expand dot product into 3 * 3 + 1 terms\}}\\ \vec{p_0} \cdot \vec{p_0} + \vec{p_0} \cdot \vec{p_1}t - \vec{p_0} \cdot \vec c + \vec{p_1}t \cdot \vec{p_0} + \vec{p_1}t \cdot \vec{p_1}t - \vec{p_1}t \cdot \vec c - \vec c \cdot \vec{p_0} - \vec c \cdot \vec{p_1}t + \vec c \cdot \vec c - r^2 = 0 & & \textrm{\{remove parenthesis included as a visual aid\}}\\ \vec{p_0} \cdot \vec{p_0} + \vec{p_0} \cdot \vec{p_1}t - \vec{p_0} \cdot \vec c + \vec{p_1}t \cdot \vec{p_0} + t^2(\vec{p_1} \cdot \vec{p_1}) - \vec{p_1}t \cdot \vec c - \vec c \cdot \vec{p_0} - \vec c \cdot \vec{p_1}t + \vec c \cdot \vec c - r^2 = 0 & & \textrm{\{factor out \(t\) in term where it appears twice\}}\\ t^2(\vec{p_1} \cdot \vec{p_1}) + t(\vec{p_0} \cdot \vec{p_1} + \vec{p_1} \cdot \vec{p_0} - \vec{p_1} \cdot \vec c - \vec c \cdot \vec{p_1}) + \vec{p_0} \cdot \vec{p_0} - \vec{p_0} \cdot \vec c - \vec c \cdot \vec{p_0} + \vec c \cdot \vec c - r^2 = 0 & & \textrm{\{factor out \(t\) in terms where it appears once\}}\\ t^2(\vec{p_1} \cdot \vec{p_1}) + t\vec{p_1}(\vec{p_0} + \vec{p_0} - \vec c - \vec c) + \vec{p_0} \cdot \vec{p_0} - \vec{p_0} \cdot \vec c - \vec c \cdot \vec{p_0} + \vec c \cdot \vec c - r^2 = 0 & & \textrm{\{factor out common \(\vec{p_1}\) in \(t(...)\) part\}}\\ t^2(\vec{p_1} \cdot \vec{p_1}) + t\vec{p_1}(2\vec{p_0} - 2\vec c) + \vec{p_0} \cdot \vec{p_0} - \vec{p_0} \cdot \vec c - \vec c \cdot \vec{p_0} + \vec c \cdot \vec c - r^2 = 0 & & \textrm{\{group similar element inside \(t(...)\) part\}}\\ t^2(\vec{p_1} \cdot \vec{p_1}) + 2t \vec{p_1} \cdot (\vec{p_0} - \vec c) + \vec{p_0} \cdot \vec{p_0} - \vec{p_0} \cdot \vec c - \vec c \cdot \vec{p_0} + \vec c \cdot \vec c - r^2 = 0 & & \textrm{\{apply distributive law inside \(t(...)\) part: \(ab - ac = a(b - c)\) and remove extra parenthesis\}}\\ t^2(\vec{p_1} \cdot \vec{p_1}) + 2t \vec{p_1} \cdot (\vec{p_0} - \vec c) + \vec{p_0}^2 + \vec{c}^2 - 2\vec{p_0} \cdot \vec{c} - r^2 = 0 & & \textrm{\{group similar elements in constants part\}}\\ t^2(\vec{p_1} \cdot \vec{p_1}) + 2t \vec{p_1} \cdot (\vec{p_0} - \vec c) + (\vec{p_0} + \vec c)^2 - r^2 = 0 & & \textrm{\{apply binomial square law to constants part: \(a^2 + b^2 - 2ab = (a - b)^2\)\}}\\ \end{eqnarray*}$$ Or in terms of the standard form of a quadratic equation: \(at^2 + bt + c\), from above we get \begin{eqnarray*} a = \vec{p_1} \cdot \vec{p_1}\\ b = 2\vec{p_1} \cdot (\vec{p_0} - \vec c)\\ c = (\vec{p_0} + \vec c)^2 - r^2 \end{eqnarray*} Because the dot product of two vectors results in a scalar, once we substitute in the known elements (\(\vec{p_0}\), \(\vec{p_1}\), \(\vec c\), and \(r\)) and simplify, we're left with a quadratic equation.

Continuing with the example from above in which we have \(\vec{p_0} = (10,5,2)\), \(\vec{p_1} = (2,1,0)\), \(\vec c = (0,0,0)\), and \(r^2 = 9\): $$\begin{eqnarray*} t^2((2,1,0) \cdot (2,1,0)) + 2t(2,1,0) \cdot ((10,5,2)-(0,0,0)) + ((10,5,2) + (0,0,0) \cdot ((10,5,2) - (0,0,0)) - 3^2 = 0\\ t^2(5) + 2t(2,1,0) \cdot (10,5,2) + ((10,5,2) \cdot (10,5,2)) - 9 = 0\\ t^2(5) + 2t(2,1,0) \cdot (10,5,2) + (129) - 9 = 0\\ t^2(5) + 2t(25) + 120 = 0\\ t^2(5) + t(50) + 120 = 0\\ 5t^2 + 50t + 120 = 0 \end{eqnarray*}$$ which shows that the scalar and vector editions yield identical results.

Carrying out the simplification steps requires the memorization of rules. Or having a site such as Symbollab.com assist in identifying a rule by name and carry out the simplification.

For additional details on vector math and the parametric equations of a line, refer to this video. And for an intuitive description of the dot product, refer to this one. For a brush-up on vector algebra, refer this video. For an approachable introduction to linear algebra in general, the Essence of linear algebra video series is highly recommended. For book length coverage, Linear Algebra (available for free) is definitely worth a read.


In computer graphics, we almost always want equations expressed in terms of vectors. This gives us the ability to treat the \(x\), \(y\), and \(z\) components of a vector as a single unit, making equations shorter and clearer. Similarly, defining and applying vector operations in code make it shorter and less error-phone. Like in object oriented programming when we create a class to group related data and operations.

Having now derived the ray-sphere intersection equation, we have the mathematics in place to create a first ray tracer; one that shoots rays from the camera, through each pixel, and into the scene where it potentially intersects with a sphere. That'll be the subject of the next post.