# c++ – How to perform a fast ray cylinder intersection test?

First we’ll subtract the cylinder center from the ray origin, to translate our whole problem to be centered at (0, 0, 0).

Case 0: Inside

First, check if the ray origin is inside the cylinder from the start.

$$text{origin}_x^2 + text{origin}_z^2 leq text{radius}^2 quad text{and}quad -frac {text{height}} 2 leq text{origin}_y leq frac {text{height}} 2$$

If so, it’s conventional to return no hit (we only count hits going into the cylinder). Though you could instead invert the following cases to check for a hit exiting the cylinder.

Case 1: Above / Below

If our shifted origin is above $$frac {text{height}} 2$$ or below $$-frac {text{height}} 2$$, then we check whether its direction takes it closer to the cylinder:

$$text{origin}_y times text{direction}_y < 0 ?$$

If it does not, we know the ray is too high/low to ever hit the cylinder, and we can early-out with no hit.

Otherwise, we can advance the ray to the time $$t$$ when it crosses the plane of the cylinder’s top (or bottom):

$$t_{text{plane}} = -frac {text{origin}_y} {text{direction}_y}$$

The ray’s position at that time is $$vec p_1= vec {text{origin}} + t_text{plane} cdot vec {text{direction}}$$

If that point is inside the cylinder’s radius,

$${p_1}_x^2 + {p_1}_z^2 leq text{radius}^2$$

Then we have a hit against the top/bottom cap of the cylinder, with a vertical collision normal.

If not, we can proceed as though the ray had started somewhere beside the cylinder, rather than above/below:

Case 2: Beside

Next we’ll check if the ray hits the circle. For that we want to know the earliest positive time $$t$$ at which…

$$||vec {text{origin}} + t cdot vec {text{direction}}|| = r\ (vec {text{origin}} + t cdot vec {text{direction}})^2 = r^2\ t^2 cdot text{direction}^2 + t cdot 2(vec {text{origin}} cdot vec {text{direction}}) + text{origin}^2 – r^2 = 0$$

You can solve this with quadratic formula. Choose the smallest positive solution, and call it $$t_text{circle}$$. If there is no positive real solution, return no hit.

Then, as we did with case 1, find the position at that time: $$vec p_2= vec {text{origin}} + t_text{circle} cdot vec {text{direction}}$$

Now we check whether that’s on our cylinder, or if it missed:

$$| {p_2}_y| < frac {text{height}} 2 ?$$

If the hit is within this height range, then we have a hit, with collision normal parallel to $$vec p_2$$. Otherwise, we miss the cylinder and return no hit.

If you do find a hit, add the cylinder center back to the hit position $$vec p_1$$ or $$vec p_2$$ to get the hit position in world space.