Some time ago I needed to solve analytically the intersection of a ray and a cone. I was surprised to see that there are not that many resources available; there are some, but not nearly as many as on the intersection of a ray and a sphere for example. Add to it that they all use their own notation and that I lack math exercise, after a bit of browsing I decided I needed to write a proof by myself to get a good grasp of the result.

So here goes, the solution to the intersection of a ray and a cone, in vector notation.

- We define a ray with its origin $O$ and its direction as a unit vector $\hat{D}$.

Any point $X$ on the ray at a signed distance $t$ from the origin of the ray verifies: $\vec{X} = \vec{O} + t\vec{D}$.

When $t$ is positive $X$ is in the direction of the ray, and when $t$ is negative $X$ is in the opposite direction. - We define a cone with its tip $C$, its axis as a unit vector $\hat{V}$ in the direction of increasing radius, and $\theta$ the half angle between the axis and the surface.

Any point $X$ on the cone verifies: $(\vec{X} – \vec{C}) \cdot \vec{V} = \lVert \vec{X} – \vec{C} \rVert \cos\theta$ - Finally we define $P$ the intersection or the ray and the cone, and which we are interested in finding.

$P$ verifies both equations, so we can write:

$$

\left\{

\begin{array}{l}

\vec{P}=\vec{O} + t\vec{D} \\

\frac{ \vec{P} – \vec{C} }{\lVert \vec{P} – \vec{C} \rVert} \vec{V} = \cos\theta

\end{array}

\right.

$$

We can multiply the second equation by itself to work with it, then reorder things a bit.

$$

\left\{

\begin{array}{l}

\vec{P}=\vec{O} + t\vec{D} \\

\frac{ ((\vec{P} – \vec{C}) \cdot \vec{V})^2 }{ (\vec{P} – \vec{C}) \cdot (\vec{P} – \vec{C}) } = \cos^2\theta

\end{array}

\right.

$$

$$

\left\{

\begin{array}{l}

\vec{P}=\vec{O} + t\vec{D} \\

((\vec{P} – \vec{C}) \cdot \vec{V})^2 – (\vec{P} – \vec{C}) \cdot (\vec{P} – \vec{C}) \cos^2\theta = 0

\end{array}

\right.

$$

Remember the mouthful earlier about $\hat{V}$ being in the direction of increasing radius? By elevating $\cos\theta$ to square, we’re making negative values of $\cos$ positive: values of $\theta$ beyond 90° become indistinguishable from values below 90°. This has the side effect of turning it into the equation of not one, but two cones sharing the same axis, tip and angle, but in opposite directions. We’ll fix that later.

We replace $\vec{P}$ with $\vec{O} + t\vec{D}$ and work the equation until we get a good old quadratic function that we can solve.

$$

\require{cancel}

((\vec{O} + t\vec{D} – \vec{C})\cdot\vec{V})^2 – (\vec{O} + t\vec{D} – \vec{C}) \cdot (\vec{O} + t\vec{D} – \vec{C}) \cos^2\theta = 0

$$

$$

((t\vec{D} + \vec{CO})\cdot\vec{V})^2 – (t\vec{D} + \vec{CO}) \cdot (t\vec{D} + \vec{CO}) \cos^2\theta = 0

$$

$$

(t\vec{D}\cdot\vec{V} + \vec{CO}\cdot\vec{V})^2 – (t^2\cancel{\vec{D}\cdot\vec{D}} + 2t\vec{D}\cdot\vec{CO} + \vec{CO}\cdot\vec{CO}) \cos^2\theta = 0

$$

$$

(t^2(\vec{D}\cdot\vec{V})^2 + 2t(\vec{D}\cdot\vec{V})(\vec{CO}\cdot\vec{V}) + (\vec{CO}\cdot\vec{V})^2) – (t^2 + 2t\vec{D}\cdot\vec{CO} + \vec{CO}\cdot\vec{CO}) \cos^2\theta = 0

$$

$$

t^2(\vec{D}\cdot\vec{V})^2

+ 2t(\vec{D}\cdot\vec{V})(\vec{CO}\cdot\vec{V})

+ (\vec{CO}\cdot\vec{V})^2

– t^2\cos^2\theta

– 2t\vec{D}\cdot\vec{CO}\cos^2\theta

– \vec{CO}\cdot\vec{CO}\cos^2\theta

= 0

$$

Reorder a bit:

$$

t^2((\vec{D}\cdot\vec{V})^2 – \cos^2\theta)

+ 2t((\vec{D}\cdot\vec{V})(\vec{CO}\cdot\vec{V}) – \vec{D}\cdot\vec{CO}\cos^2\theta)

+ (\vec{CO}\cdot\vec{V})^2 – \vec{CO}\cdot\vec{CO}\cos^2\theta

= 0

$$

There we go, we have our $at^2 + bt + c = 0$ equation, with:

$$

\left\{

\begin{array}{l}

a = (\vec{D}\cdot\vec{V})^2 – \cos^2\theta \\

b = 2\Big((\vec{D}\cdot\vec{V})(\vec{CO}\cdot\vec{V}) – \vec{D}\cdot\vec{CO}\cos^2\theta\Big) \\

c = (\vec{CO}\cdot\vec{V})^2 – \vec{CO}\cdot\vec{CO}\cos^2\theta

\end{array}

\right.

$$

From there, you know the drill: calculate the determinant $\Delta = b^2 – 4ac$ then depending on its value:

- If $\Delta < 0$, the ray is not intersecting the cone.
- If $\Delta = 0$, the ray is intersecting the cone once at $t = \frac{-b}{2a}$.
- If $\Delta > 0$, the ray is intersecting the cone twice, at $t_1 = \frac{-b – \sqrt{\Delta}}{2a}$ and $t_2 = \frac{-b + \sqrt{\Delta}}{2a}$.

But wait! We don’t have one cone but two, so we have to reject solutions that intersect with the shadow cone. $P$ must still verify $\frac{ \vec{P} – \vec{C} }{\lVert \vec{P} – \vec{C} \rVert} \vec{V} = \cos\theta$, or simply, if $\theta < 90°$: $(\vec{P} – \vec{C})\cdot\vec{V} > 0$.

Note that there is also the corner case of the ray tangent to the cone and having an infinity of solutions to consider. I’ve completely swept it under the rug since it doesn’t matter in the context I was, but if it does to you, you’ve been warned about it. Also remember to check the sign of $t$ to know whether $P$ is in the direction of the ray. You may need to determine which of $t_1$ or $t_2$ you want to use, which depends on your use case. For example is your ray origin inside or outside of the cone?

Now for a little sanity test, let’s consider the corner case $C=O$, where the ray origin is the tip of the cone (thanks Rubix for the suggestion!). We have $b=0$ and $c=0$ thus $\Delta=0$ and $t=\frac{-b}{2a}=0$ which is the expected result.

I also tried the cases $\theta=0$ and $\theta=\pi/2$, but expanding $\Delta$ proved too tedious to proceed to the end. So this is left as an exercise, as they say. :)

Finally, to demonstrate that the result is indeed correct, here is a glorious ray traced cone scene on ShaderToy:

I hope this can prove useful to others too.

Oh, and Happy New Year by the way!

You forgot to recommend verifying that t is positive. Also the pi constant of your shader is wrong.

Very good point, I should expand on it a bit. Thanks for spotting the error too; it’s fixed.

Randomly passing by looking for some unrelated info, saw you’re not escaping your functions in LaTeX! I.e. you should use “\cos” instead of “cos”, else it will render cos in italics, as a product of 3 variables c, o and s.

Sorry, this is a major pet peeve of mine, one sees it everywhere! :”D

Hi Thomas, thanks for pointing it out; I’ve fixed it.

Hi Julien,

Very nice write-up ! I’ve implemented that six years ago in the context of raytracing curves.

But the best way to raytrace curves is found in the paper from Koji Nakamaru and Yoshio Ohno which I believe is widely used nowadays. This way you don’t need to discretized the curve in small segments (facing camera) or small cones, etc.

Cheers!

Hi Guillaume,

Thanks, I didn’t know that reference; I will have a look.

Thanks for putting this together — just the form I was looking for too.

Also, there is one small error: when you solve for the quadratic form for a, b, and c, in the equation for b, everything to the right of 2 should be inside a bracket. I.e., 2 multiplies everything there not just the first term. Thankfully this was easy to spot as you were very thorough in your derivation.

Thanks again!

Oh, you’re right! Thank you for spotting this, I’ve fixed it.

I’m not sure if this is my own problem but the demo looks all black to me?

I think ShaderToy switched to a newer version of WebGL or something, and that broke some of the shaders. That might be related. I haven’t had time to look at it unfortunately (the shader looks fine on my machine).

Hello,

Thank you for providing your proof. How do you go about defining your cone in relation to this proof?

Thanks!

Hello,

See the point #2: the cone is defined by $C$, $V$ and $\theta$. You can use a different definition too, as long as you can derive it to express those three elements.

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Thanks, I’ve implemented this however seem to be having a problem or a special case perhaps. I’m getting solutions where, $a = 0$, i.e. $(D.V)^2 = \cos^2(\theta)$.

Specifically for my implementation, my $\theta$ angle defining the cone is the as the angle from vertical of my ray direction vector?

Is this a case where my ray is lying tangent to the cone?

Hello Ben,

Thank you for reading. I am fairly busy recently and won’t have time to look at your question in a timely manner.

If you haven’t figured the solution to your problem yet, I invite you to ask the question on https://computergraphics.stackexchange.com/ where I’m sure you will get good support.

Thanks for your reply. I’ve found a simple work around of adding a tiny insignificant value (for my use) to the cone angle, θ . This fully eliminates my problem and ensures the solution is always solvable.

Greatly appreciate your detailed post.

Thanks for this! When I put it into my ray tracer, I got a double cone. Not hard to solve the problem (if v dot p < 0 discard), just thought I'd mention.

Yes absolutely: by elevating the equation to square, the sign information is lost. See the remark after the three determinant sign cases.