1. Tracing Light in a Virtual World

2. Creating Randomness and Accumulating Change

If you recall, the very high level algorithm for path tracing is:

- From each pixel, fire a ray out into the scene from the eye.
- If the ray hits an object
- Use the material properties to accumulate attenuation
- Bounce the ray in a new direction
- GOTO 2
- If the ray hits a light
- Multiply the light and the attenuation
- Add the result to the accumulation buffer
- GOTO 5
- If the ray doesn't hit anything
- GOTO 5
- GOTO 1 until sufficiently converged

The first post in the series covered step 1 and the second post covered step 3b. This post is going to cover how to test if a ray hits an object in the scene (aka, 2 and 3).

### Ray - Sphere Intersection

Foreword: Scratchapixel has an excellent lesson covering ray-object intersections. Much of what I say below is based on their lessons. Much props to the authors.

struct Sphere { float3 Center; float RadiusSquared; };

A ray can be represented as a vector representing its origin and a unit vector representing its direction:

struct Ray { float3 Origin; float3 Direction; };

Let's look at a picture of the most common type of ray - sphere intersection:

\[\begin{align}

\overrightarrow{R_{o}} &= ray \: origin \\

\overrightarrow{R_{d}} &= ray \: direction \\

\overrightarrow{S_{c}} &= sphere \: center \\

S_{r} &= sphere \: radius \\

\overrightarrow{P_{0}} &= first \: intersection \: point \\

\overrightarrow{P_{1}} &= second \: intersection \: point \\

t_{0} &= distance \: from \: \overrightarrow{R_{o}} \: to \: \overrightarrow{P_{0}} \\

t_{1} &= distance \: from \: \overrightarrow{R_{o}} \: to \: \overrightarrow{P_{1}} \\

\end{align}\]

We would like to find $\overrightarrow{P_{0}}$ and $\overrightarrow{P_{1}}$. Mathematically, they are defined as:

\[\begin{align}

\overrightarrow{P_{0}} &= \overrightarrow{R_{o}} + t_{0} \overrightarrow{R_{d}} \\

\overrightarrow{P_{1}} &= \overrightarrow{R_{o}} + t_{1} \overrightarrow{R_{d}} \\

\end{align}\]

We already know $\overrightarrow{R_{o}}$ and $\overrightarrow{R_{d}}$, so we just need to find $t_{0}$ and $t_{1}$. In order to do so, let's define a few new variables.

And then define $t_{0}$ and $t_{1}$ in terms of $t_{ca}$ and $t_{hc}$:

\[\begin{align}

t_{0} &= t_{ca} - t_{hc} \\

t_{1} &= t_{ca} + t_{hc} \\

\end{align}\]

Now we can begin solving for our unknown variables. To start, let's look at the right triangle formed by $t_{ca}$, $\overrightarrow{L}$, and $d$.

We can solve for $t_{ca}$ by using the definition of cosine and the dot product:

\[\cos\left ( \theta \right ) = \frac{adjacent}{hypotenuse} = \frac{t_{ca}}{\left \| \overrightarrow{L} \right \|}\]

\[\overrightarrow{m} \cdot \overrightarrow{n} = \left \| m \right \| \left \| n \right \| \cos \left ( \theta \right )\]

\[\begin{split} ie: \end{split} \qquad \qquad

\begin{split}

\overrightarrow{R_{d}} \cdot \overrightarrow{L} &= \left \| \overrightarrow{R_{d}} \right \| \left \| \overrightarrow{L} \right \| \cos \left ( \theta \right ) \\

&= \frac{\left \| \overrightarrow{L} \right \| t_{ca}}{\left \| \overrightarrow{L} \right \|} \\

&= t_{ca}

\end{split}\]

$\overrightarrow{R_{d}}$ is a unit vector. Therefore, $\left \| \overrightarrow{R_{d}} \right \| = 1$ and cancels out. Then, if we replace $\cos \left ( \theta \right )$ with its definition, we can cancel $\left \| \overrightarrow{L} \right \|$ from top and bottom. Thus, we're left with just $t_{ca}$.

Using the Pythagorean Theorem and a trick with the dot product, we can solve for d:

\[\overrightarrow{m} \cdot \overrightarrow{m} \equiv \left \| \overrightarrow{m} \right \| ^{2}\]

\[\begin{align} t_{ca} \: ^{2} + d^{2} &= \left \| \overrightarrow{L} \right \| ^{2}\\

t_{ca} \: ^{2} + d^{2} &= \overrightarrow{L} \cdot \overrightarrow{L} \\

d^{2} &= \overrightarrow{L} \cdot \overrightarrow{L} - t_{ca} \: ^{2}

\end{align}\]

To solve for $t_{hc}$, let's look at the triangle formed by $S_{r}$, $t_{hc}$, and $d$.

Using the Pythagorean Theorem again, we can solve for $t_{hc}$:

\[\begin{align}

t_{hc} \: ^{2} + d^{2} &= S_{r} \: ^{2} \\

t_{hc} &= \sqrt{S_{r} \: ^{2} - d^{2}}

\end{align}\]

Whew! We made it! Using the above equations, we can find the two intersection points. Let's look at some code that implements the above and walk through some special cases / early outs.

/** * Test for the intersection of a ray with a sphere * * @param ray The ray to test * @param sphere The sphere to test * @param normal_out Filled with normal of the surface at the intersection point. Not changed if no intersection. * @return The distance from the ray origin to the nearest intersection. -1.0f if no intersection */ __device__ float TestRaySphereIntersection(Scene::Ray &ray, Scene::Sphere &sphere, float3 &normal_out) { float3 L = sphere.Center - ray.Origin; float t_ca = dot(L, ray.Direction); // Ray points away from the sphere if (t_ca < 0) { return -1.0f; } float d_squared = dot(L, L) - t_ca * t_ca; // Ray misses the sphere if (d_squared > sphere.RadiusSquared) { return -1.0f; } float t_hc = sqrt(sphere.RadiusSquared - d_squared); float t_0 = t_ca - t_hc; float t_1 = t_ca + t_hc; float nearestIntersection; float normalDirection; if (t_0 > 0 && t_1 > 0) { // Two intersections // Return the nearest of the two nearestIntersection = min(t_0, t_1); normalDirection = 1.0f; } else { // Ray starts inside the sphere // Return the far side of the sphere nearestIntersection = max(firstIntersection, secondIntersection); // We reverse the direction of the normal, since we are inside the sphere normalDirection = -1.0f; } normal_out = normalize(((ray.Origin + (ray.Direction * nearestIntersection)) - sphere.Center) * normalDirection); return nearestIntersection; }

The algorithm can be summarized as follows:

- Calculate $t_{ca}$
- If $t_{ca}$ is negative, $\overrightarrow{R_{d}}$ is pointing away from the sphere. Thus, there can not be an intersection
- Calculate $d^{2}$
- If $d^{2}$ is greater than $S_{r} \: ^{2}$, the ray misses the sphere.
- Calculate $t_{hc}$
- Calculate $t_{0}$ and $t_{1}$.
- If $t_{0}$ and $t_{1}$ are both positive, the ray starts outside the sphere and intersects it twice. Choose the closest of the two intersections
- If either $t_{0}$ or $t_{1}$ is negative, the ray starts inside the sphere and intersects it on the way out. Choose the positive intersection.
- They both can't both be negative, since that would mean the ray is pointing away from the sphere, and we already checked for that.
- See following picture.

### Ray - Plane Intersection

A plane can be represented using any point on the plane, and a unit normal vector from the plane.struct Plane { float3 Point; float3 Normal; };

\[\begin{align}

\overrightarrow{P_{p}} &= plane \: point \\

\overrightarrow{P_{n}} &= plane \: normal

\end{align}\]

Let's look at the intersection from the top.

We would like to find $d$. We can start by taking the dot product between $\overrightarrow{R_{d}}$ and $\overrightarrow{P_{n}}$. We get $\cos \left ( \theta_{1} \right )$, since both $\overrightarrow{R_{d}}$ and $\overrightarrow{P_{n}}$ are unit vectors.

\[\begin{align}

\overrightarrow{R_{d}} \cdot \overrightarrow{P_{n}} &= \left \| \overrightarrow{R_{d}} \right \| \left \| \overrightarrow{P_{n}} \right \| \cos \left ( \theta_{1} \right ) \\

&= \cos \left ( \theta_{1} \right )

\end{align}\]

Let's look at the triangle formed by $\overrightarrow{P_{n}}$ and $\overrightarrow{L}$, where $\overrightarrow{L}$ is the vector between $\overrightarrow{P_{p}}$ and $\overrightarrow{R_{o}}$.

If we take the dot product between $\overrightarrow{P_{n}}$ and $\overrightarrow{L}$, we get $a$.

\[\begin{align}

\overrightarrow{P_{n}} \cdot \overrightarrow{L} &= \left \| \overrightarrow{P_{n}} \right \| \left \| \overrightarrow{L} \right \| \cos \left ( \theta_{2} \right ) \\

&= \frac{\left \| \overrightarrow{L} \right \| a}{\left \| \overrightarrow{L} \right \|} \\

&= a

\end{align}\]

Finally, let's look at the two triangles formed between $\overrightarrow{P_{n}}$, $\overrightarrow{R_{d}}$, and the plane itself.

They are similar triangles, so we can solve for $d$ using the Law of Similar Triangles and the definition of cosine:

\[\cos \left ( \theta_{1} \right ) = \frac{b}{c}\]

\[\begin{align}

\frac{a + b}{c + d} &= \frac{b}{c} \\

c \left ( a + b \right ) &= b \left ( c + d \right ) \\

ac + bc &= bc + bd \\

ac &= bd \\

d &= \frac{ac}{b} \\

d &= \frac{a}{\cos \left ( \theta_{1} \right )}

\end{align}\]

Let's look at the code implementation of the above and walk through it.

/** * Test for the intersection of a ray with a plane * * @param ray The ray to test * @param plane The plane to test * @param normal_out Filled with normal of the surface at the intersection point. Not changed if no intersection. * @return The distance from the ray origin to the nearest intersection. -1.0f if no intersection */ __device__ float TestRayPlaneIntersection(Scene::Ray &ray, Scene::Plane &plane, float3 &normal_out) { float cos_theta1 = dot(plane.Normal, ray.Direction); // If cos_theta1 is greater than -epison, // the ray is perpendicular to the plane or points away from the plane normal if (cos_theta1 > -1.0e-6f) { return -1.0f; } normal_out = plane.Normal; float a = dot(plane.Normal, plane.Point - ray.Origin); return a / cos_theta1; }

The algorithm can be summarized as follows:

- Calculate $\cos \left ( \theta_{1} \right )$
- If $\cos \left ( \theta_{1} \right )$ is greater than or equal to zero, the ray is perpendicular to the plane, or faces away
- In the code, we use a small episilon to account for floating point innacuracies
- Calculate a
- Return $\frac{a}{\cos \left ( \theta_{1} \right )}$

### Testing Them Out

In order to test the code out, I added on to the eye ray code we created in the first post. After creating the ray, I tried to intersect it with an object in the scene. If it hit, I did some basic Lambertian shading with a hard-coded directional light. If it didn't, I colored the pixel black. Here's the results:Very nice! If I do say so myself. I don't have a plane in the scene because I didn't implement the plane intersection code until I had skipped ahead to the path tracing itself. (I got too excited. Ha ha!) So you'll have to take my word for it that the code does work. Below is the kernel code used to create the image above:

__global__ void PathTraceKernel(unsigned char *textureData, uint width, uint height, size_t pitch, DeviceCamera *g_camera, Scene::Sphere *g_spheres, uint numSpheres, uint hashedFrameNumber) { int x = blockIdx.x * blockDim.x + threadIdx.x; int y = blockIdx.y * blockDim.y + threadIdx.y; // Create a local copy of the camera DeviceCamera camera = *g_camera; // Global threadId int threadId = (blockIdx.x + blockIdx.y * gridDim.x) * (blockDim.x * blockDim.y) + (threadIdx.y * blockDim.x) + threadIdx.x; // Create random number generator curandState randState; curand_init(hashedFrameNumber + threadId, 0, 0, &randState); // Calculate the first ray for this pixel Scene::Ray ray = {camera.Origin, CalculateRayDirectionFromPixel(x, y, width, height, camera, &randState)}; // Generate a uniform random number //float randNum = curand_uniform(&randState); // Try to intersect with the spheres; float closestIntersection = FLT_MAX; float3 normal; for (uint i = 0; i < numSpheres; ++i) { float3 newNormal; float intersection = TestRaySphereIntersection(ray, g_spheres[i], newNormal); if (intersection > 0.0f && intersection < closestIntersection) { closestIntersection = intersection; normal = newNormal; } } float3 pixelColor; if (closestIntersection < FLT_MAX) { float attentuation = max(dot(normal, make_float3(0.70710678118f, 0.70710678118f, -0.70710678118f)), 0.0f); pixelColor = make_float3(0.846, 0.933, 0.949) * attentuation + make_float3(0.15f, 0.15f, 0.15f); } else { pixelColor = make_float3(0.0f, 0.0f, 0.0f); } if (x < width && y < height) { // Get a pointer to the pixel at (x,y) float *pixel = (float *)(textureData + y * pitch) + 4 /*RGBA*/ * x; // Write out pixel data pixel[0] += pixelColor.x; pixel[1] += pixelColor.y; pixel[2] += pixelColor.z; // Ignore alpha, since it's hardcoded to 1.0f in the display // We have to use a RGBA format since CUDA-DirectX interop doesn't support R32G32B32_FLOAT }

### Conclusion

There we go! We now have everything we need to do path tracing. The next post will be just that: creating a basic path tracing kernel. I've already implemented it in code, so it's just a matter of how quickly I can write up the post. (teaser picture). Keep an eye out!The code for everything in this post is on GitHub. It's open source under the Apache license, so feel free to use it in your own projects. The path tracing code is already pushed to the repo, if you're feeling antsy, you can give that a look.

As always, feel free to ask questions, make comments, and if you find an error, please let me know.

Happy coding!

-RichieSams