Ray tracing: Computing ray-sphere intersections

03 Sep 2018

Whereas the previous post was about ray-circle intersections with scalars, this one is about ray-sphere intersections with vectors. We'll extend the circle standard equation to a sphere, and switch to vector notation for defining both sphere and line. Using vectors, we can generalize and perform intersection calculations for any dimension. In a sense, a vector hides details about its scalar components, letting us operate on the vector as a whole.

Sphere offset from origin

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

Vector standard form of a line

Defining a line in 3D isn't as straightforward as in 2D. We don't something similar to \(y = ax + b\). In 2D, the line equation expresses the fact that a change in one dimension causes a change in one other. In 3D, a change in one dimension causes a change in two others which is hard to express in a similar form.

Instead, in 3D a line is expressed as a starting position and an amount of movement in some direction. Statements about position and direction are best expressed mathematically with vectors and operations on vectors: $$\vec p = \vec{p_0} + t\vec{p_1}$$ Plugging in different values of \(t\), for time, result in a line described by the same origin but a position vector pointing at a different point 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 and thus by varying \(t\) we can 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 call it a ray instead of a line. \(t < 0\) would describe a ray traveling out back of the camera.

As an example, we can find the parametric equation of the line passing through \((1,-1,4)\) in the direction of \((3,5,-1)\). We do so by plugging the two vectors into the vector equation for the 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 any scalar \(t\) we get any position vector on the line. For ray-sphere intersections, we're after the some values of \(t\) describing points of intersection. Once we have those values, we plug those into the parametric equation of a line to get the \((x,y,z)\) coordinate.

Line and sphere intersection, scalar edition

As an example, suppose we have the parametric equation of the line \((10 + 2t, 5 + t, 2)\) and the equation of a sphere in standard form \(x^2 + y^2 + z^2 = 9\). As with 2D intersection, we're after values for \(x\), \(y\), and \(z\) that when plugged into both equations make them equal. That implies we can plug 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{\{plug 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 rearrange in \(t^2\), \(t\), constants order\}}\\ 5t^2 + 50t + 120 = 0 & & \textrm{\{move 9 to left side to arrive at the standard form of a 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, namely \(t\). Solving for \(t\) means using 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*}$$ Plugging these values into the parametric vector equation, we get the points of line-sphere intersection: $$\begin{eqnarray*} \vec{p_1} = (x_1,y_1,z_1) = (10 + 2(-4) = 2, 5+ (-4) = 1, 2) = (2,1,2)\\ \vec{p_2} = (x_2,y_2,z_2) = (10 + 2(-6) = -2, 5 + (-6) = -1, 2) = (-2,-1,2) \end{eqnarray*}$$

For a commentary on line-sphere intersection, see here.

Line and sphere intersection, vector edition

We already established the standard vector equation for a line. For a sphere centered at \(c\) with radius \(r\), its corresponding standard vector equation is $$(\vec p - \vec c) \cdot (\vec p - \vec c) = r^2$$ \(\vec p\) is position vectors on the sphere and \(\cdot\) is the dot product of two vectors. The term product in general refers to the result of a multiplication. Unlike the single definition for the product of scalars, vectors have two kinds of product: dot product and cross product.

In the equations which follow, scalar product or dot product may be inferred from the context. We don't need the \(\cdot\) symbol, but it's convention. Like operator overloading in programming, if both operands are vectors, it means dot product, if one operand is a vector, it means scaling by that scalar, if both operands are scalars, it means regular multiplication.

At first sight, the sphere vector equation look nothing like its scalar counterpart. But note the definition of the dot product: $$\vec u \cdot \vec v = {u_x}{v_x} + {u_y}{v_y} + {u_z}{v_z} = r^2$$ With this definition, if \(\vec{u} = \vec{v}\), which is the case for vector sphere definition, expanding the dot product turns into the original standard equation for a sphere. So while it may look mysterious, defining the sphere in terms of the dot product is correct: $$\vec u \cdot \vec u = {u_x}{u_x} + {u_y}{u_y} + {u_z}{u_z} = u_{x}^2 + u_{y}^2 + u_{z}^2 = x^2 + y^2 + z^2 = r^2$$ In a similar way to how we plugged in the line equation into the equation of the sphere above, we can replace \(\vec p\) by the vector equation of the line because where they're the same must be where the line intersects with the sphere: $$(\vec{p_0} + \vec{p_1}t - \vec c) \cdot (\vec{p_0} + \vec{p_1}t - \vec c) - r^2 = 0$$ Through rewriting and simplification, the goal is to turn this complicated looking equation into a quadratic one and solve it 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 square 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 in previous step\}}\\ \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 in constants part: \(a^2 + b^2 - 2ab = (a - b)^2\)\}}\\ \end{eqnarray*}$$ or in terms of the standard form of a quadratic equation: \(at^2 + bt + c\) \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*} Note that calculating the dot product of two vectors results in a scalar, not a new vector. So even though the above equation may look complicated, once we plug in known vectors and simplify, we're left with a quadratic equation of the form \(at^2 + bt + c\).

I struggled quite a bit with the simplifications because I don't have all possible rules memorized. For that case Symbollab.com is helpful in that given an equation, it can identity the rule by name.

Continuing the scalar example above in which \(\vec{p_0} = (10,5,2)\), \(\vec{p_1} = (2,1,0)\), \(\vec c = (0,0,0)\), 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 of ray-sphere intersection yield identical results.

For a commentary of the vector and parametric equations of a line, see here. Similar, see here for intuitive explanation of the dot product. For a brush-up basic vector algebra, see here.


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 makes the code shorter and less error-phone. Thinking of a vector like this isn't much different than in object oriented programming when we create a class to group related data and operations on those data.

With the equation derived in the vector edition section, we have the mathematical tools required to create a first ray tracer; one that shoots rays from the camera, through each pixel, and into the scene, coloring the pixel on ray-sphere intersections.

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