# How to ray-intersect a transformed geometry, without actually transforming it: a geometric illustration

$\newcommand{\mvec}[1]{\mathbf{#1}}\newcommand{\gvec}[1]{\boldsymbol{#1}}\definecolor{eqcol2}{RGB}{114,0,172}\definecolor{eqcol1}{RGB}{172,0,114}\definecolor{eqcol3}{RGB}{0,172,114}\definecolor{eqcol4}{RGB}{230,190,120}$The goal of this article, is to describe how we can transform some geometry with a transformation matrix, and then compute the intersection between the transformed geometry, and some ray. However, we want to do this without actually transforming the geometry, since this can be expensive.

Let us say we have a ray $\mvec{p}(t) = \mvec{o} + \mvec{d}t$, with origin $\mvec{o}$ and direction $\mvec{d}$, and we want to find its intersection with some geometry, by finding the value of $t$ for this intersection.

For an example, consider the intersection between above ray, with the given values of $\mvec{o}$ and $\mvec{d}$. and the green geometry. The ray certainly intersects this geometry, and the value of $t$ is $t=2$, since the ray is heading straight towards the wall of the geometry, and the distance between $\mvec{o}$ and the wall of the geometry is 2. Let us now translate this geometry, 4 units upwards on the $y$-axis, by applying the following matrix: \begin{equation*} T = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 4 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}, \end{equation*} then we get this geometry:

This translated geometry is intersected by the ray with origin $\mvec{o}=(-4,4)$ and direction $\mvec{d}=(+1,0)$, with again $t=2$. However, is there a way we can calculate this exact intersection, without having to translate the geometry at all. The geometry might be very complex, and have millions upon millions of triangles, so transforming the whole geometry might be expensive. Also, the geometry might also be storing some kind of spatial data structure for fast ray-intersection queries(like a bounding-volume hierarchy), which would also have to be updated if we transformed and updated the geometry vertex data. If we found a way to intersect the translated geometry, without actually having to translate it all, these costs would become zero.

So, is there a way to intersect the geometry with the ray, without having to transform the geometry at all? Yes, and it is actually rather simple: instead of transforming the geometry, we apply the inverse of the transform on the ray. And then, we intersect this inverse transformed ray, with the original geometry. A small example should convince the reader that this is a valid approach. The inverse of the translation matrix $T$ we just used, is \begin{equation*} T^{-1} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & -4 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}, \end{equation*} If we apply this matrix to the ray with properties $\mvec{o}=(-4,4)$, $\mvec{d}=(+1,0)$, we arrive at the following situation:

Now as we can observe, we can intersect this inverse transformed ray, with the untransformed geometry, and by doing so, we arrive at our desired result $t=2$.

One subtle detail here, is that we must make sure to apply this inverse transformation on both the ray origin, and direction. To understand this, let us go through a slightly more subtle example. In this example, the transformation matrix is \begin{equation*} M = \begin{bmatrix} 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 4 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 & 4 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} , \end{equation*} This is a transform that first translates +4 units on the $x$-axis, and then does a 90 degrees counter-clockwise rotation about the origin $(0,0)$. If we apply this transform to a geometry, we arrive at the below situation

We want to intersect the above ray with the original geometry instead of the transformed geometry, and so we begin by inverse transforming the ray origin. The inverse of the transform is \begin{equation*} M^{-1} = \begin{bmatrix} 0 & 1 & 0 & -4 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \end{equation*} now apply this to the ray origin: \begin{equation*} \begin{bmatrix} 0 & 1 & 0 & -4 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} -3 \\ +4 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ +3 \\ 0 \\ 1 \end{bmatrix} \end{equation*} And after doing this transform, we obtain:

And we can see that the ray obviously will not intersect the original geometry in this configuration. Because we also need to inverse transform the ray direction, so let us do this: \begin{equation*} \begin{bmatrix} 0 & 1 & 0 & -4 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} +1 \\ 0 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ -1 \\ 0 \\ 0 \end{bmatrix} \end{equation*} And by doing this, we end up with

Now the ray does intersect the geometry, and the point it does intersect, is the same point it would have intersected, on the transformed geometry. We highlight these two intersection points in blue for clarity

As long as we remember to inverse transform both the ray origin and direction, we can intersect a transformed geometry, without actually having to transform it. It even works if the transform matrix has a scaling component to it. We must just make sure to not normalize the ray direction after inverse transforming, and we will get the correct result. The one requirement to using the technique is that the transform matrix is invertible. If it is not invertible, then we are not able to inverse transform the ray to begin with.