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1. Mathematics

We want to find the intersections of

  • A ray starting at point \(O\) and having direction \(\vec\Delta\).

  • An infinitely long cylinder with radius \(R\). We choose the cylinder to be aligned with the Z-axis.

A cylinder can be seen as an infinite number of circles stacked upon each other. To find the intersections, we temporarily get rid of the Z-dimension, which reduces the problem to 2D space in which we need to find the intersection of a ray with a circle. We thus have

  • A ray starting at point \(O' = (O_\mathrm{x}, O_\mathrm{y})\) where \(O_x\) and \(O_y\) are the x- and y-coordinates of the original ray’s origin \(O\). The ray’s direction is \(\vec\Delta' = (\vec\Delta_\mathrm{x}, \vec\Delta_\mathrm{y})\).

  • A circle centered at \((0, 0)\) with radius \(R\).

A circle is defined as the set of all points \(P\) at a distance \(R\) from a given center (in our case, \((0, 0)\).) \(P\) must therefore satisfy the following equation:

\[\mathrm{dist}(P, (0,0)) = R\]

The distance can be expressed using the dot product:

\[\sqrt{(P - (0,0)) \cdot (P - (0,0))} = R\]

Squaring both sides gets rid of the square root:

\[(P - (0,0)) \cdot (P - (0,0)) = R^2\]

\(P\) must also lie on the ray, which means it must also satisfy

\[P = O' + \vec\Delta' \cdot t\]

Combining both equations, we get

\[(O' + \vec\Delta' \cdot t - (0,0)) \cdot (O' + \vec\Delta' \cdot t - (0,0)) = R^2\]

Expanding and rearranging yields

\[\vec\Delta' \cdot \vec\Delta' \cdot t^2 + 2 \cdot (O' - (0,0)) \cdot \vec\Delta' \cdot t + (O' - (0,0)) \cdot (O' - (0,0)) - R^2 = 0\]

We simplify this equation by introducing

\[\begin{array}{rcl} a & = & \vec\Delta' \cdot \vec\Delta' \\ b & = & 2 \cdot (O' - (0,0)) \cdot \vec\Delta' \\ c & = & (O' - (0,0)) \cdot (O' - (0,0)) - R^2 \end{array}\]

which turns the equation to

\[a \cdot t^2 + b \cdot t + c = 0\]

This is a standard quadratic equation:

  • Compute the discriminant:

    \[D = b^2 - 4 \cdot a \cdot c\]

  • If \(D < 0\), the ray does not intersect with the circle, and hence neither does it intersect the cylinder.

  • If \(D \geq 0\), there are two intersections with \(t\)-values

    \[t_1 = \frac{-b-\sqrt{D}}{2 \cdot a} \qquad t_2 = \frac{-b+\sqrt{D}}{2 \cdot a}\]

\(t_1\) and \(t_2\) tell us where in 2D space the ray hits the circle. We can restore the Z-dimension and find out the 3D-intersections:

\[ P_1 = O + \vec\Delta \cdot t_1 \qquad P_2 = O + \vec\Delta \cdot t_2\]

The normal at position \(P\) is

\[ \vec{N} = (\frac{P_\mathrm{x}}{R}, \frac{P_\mathrm{y}}{R}, 0)\]

2. Summary

Given

  • A ray starting at point \(O = (O_\mathrm{x}, O_\mathrm{y}, O_\mathrm{z})\) and having direction \(\vec\Delta = (\vec\Delta_\mathrm{x}, \vec\Delta_\mathrm{y}, \vec\Delta_\mathrm{z})\).

  • A infinitely long cylinder with radius \(R\) around the Z-axis.

The intersections can be found using the following steps:

  1. Compute \(O' = (O_\mathrm{x}, O_\mathrm{y})\).

  2. Compute \(\vec\Delta' = (\vec\Delta_\mathrm{x}, \vec\Delta_\mathrm{y})\).

  3. Solve the quadratic equation

    \[ (\vec\Delta' \cdot \vec\Delta') \cdot t^2 + (2 \cdot (O' - (0,0)) \cdot \vec\Delta') \cdot t + (O' - (0,0)) \cdot (O' - (0,0)) - R^2\]

  4. If the quadratic equation has no solutions, the ray does not intersect the cylinder and the algorithm ends here.

  5. Call the two solutions \(t_1\) and \(t_2\).

  6. The hit positions can be found using

    \[P_1 = O + \vec\Delta \cdot t_1 \qquad P_2 = O + \vec\Delta \cdot t_2\]

  7. The normal at an intersection point \(P = (P_\mathrm{x}, P_\mathrm{y}, P_\mathrm{z})\) is

    \[\vec{N} = (\frac{P_\mathrm{x}}{R}, \frac{P_\mathrm{y}}{R}, 0)\]