layout: default title: Ray Triangle Intersection permalink: /pathtracer/ray_triangle_intersection grand_parent: "A3: Pathtracer" nav_order: 1 parent: (Task 2) Intersections usemathjax: true

Ray Triangle Intersection

Step 1: Triangle::hit

The first intersect routine you will implement is the hit routine for triangle meshes in student/tri_mesh.cpp.

We recommend that you implement the Moller-Trumbore algorithm, a fast algorithm that takes advantage of a barycentric coordinates parameterization of the intersection point, for ray-triangle intersection. We will provide motivation for the algorithm down below, but feel free to look at other resources for ray-triangle intersections.

Down below will be our main diagram that will illustrate this algorithm:

First, we can parameterize any point \textbf{P} in terms of the vertices \textbf{p}_0, \textbf{p}_1 andd \textbf{p}_2 with barycentric coordinates u, v and w:

\textbf{P} = w \cdot \textbf{p}_0 + u \cdot \textbf{p}_1 + v \cdot \textbf{p}_2.

We can simplify this by using the fact that if an intersection point lies within the triangle (which is what we aim to find), then the sum of the barycentric coordinates is equal to 1 and each of the coordinates are non-negative. If any of the coordinates fail to satisfy this property, then we can simply return that there was no intersection. This allows us to simplify the equation from above:

\begin{align*} \textbf{P} &= w \cdot \textbf{p}_0 + u \cdot \textbf{p}_1 + v \cdot \textbf{p}_2 \\ & = (1 - u - v) \cdot \textbf{p}_0 + u \cdot \textbf{p}_1 + v \cdot \textbf{p}_2 \\ & = \textbf{p}_0 + u \cdot (\textbf{p}_1 - \textbf{p}_0) + v \cdot (\textbf{p}_2 - \textbf{p}_0) \end{align*}

On the other hand, we can express any point on the ray with origin \textbf{o} and normalized direction \textbf{d} as a function of time t. Thus, if our ray does intersect the triangle, then we can set these two equations equal:

\textbf{o} + t \cdot \textbf{d} = \textbf{p}_0 + u \cdot (\textbf{p}_1 - \textbf{p}_0) + v \cdot (\textbf{p}_2 - \textbf{p}_0) \textbf{o} + t \cdot \textbf{d} = \textbf{p}_0 + u \cdot \textbf{e}_1 + v \cdot \textbf{e}_2 \implies \textbf{o} - \textbf{p}_0 = u \cdot \textbf{e}_1 + v \cdot \textbf{e}_2 + t \cdot (-\textbf{d}) \implies \begin{bmatrix} \textbf{e}_1 & \textbf{e}_2 & -\textbf{d} \end{bmatrix} \cdot \begin{bmatrix} u \\ v \\ t \end{bmatrix} = \textbf{o} - \textbf{p}_0 = \textbf{s}

Where \textbf{s}, \textbf{e}_1 and \textbf{e}_2 are defined in the initial diagram.

We can now use Cramer's rule where we solve for each variable as the fraction of determinants to solve the above equation:

$$ \begin{bmatrix} u \ v \ t \end{bmatrix} = \frac{1}{(\textbf{e}_1 \times \textbf{d}) \cdot \textbf{e}_2} \cdot \begin{bmatrix} -(\textbf{s} \times \textbf{e}_2) \cdot \textbf{d} \ (\textbf{e}_1 \times \textbf{d}) \cdot \textbf{s} \ -(\textbf{s} \times \textbf{e}_2) \cdot \textbf{e}_1 \end{bmatrix}$$

Once you've successfully implemented triangle intersection, you will be able to render many of the scenes in the media directory. However, your ray tracer will be very slow!

A few final notes and thoughts:

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