Looking for a cylinder / axis-aligned box volume intersection test

Question

Given:

• An arbitrary circular cylinder (defined by startpoint, endpoint, and radius). An Infinite cylinder is acceptable as well, as long as it passes through those points and has the same radius.
• An axis-aligned box (defined by its minimum and maximum point)
• The box being completely inside the volume of the cylinder should also count as an intersection

To test if the cylinder and the box intersect.

Does such a test exist?

EDIT: I found this on the mathmatics stack exchange, though at the moment I do not follow it. If there is an example of it in code form that would be fantastic.

Answer 1

As mentioned the problem of cylinder box intersection reduces to the problem of line-parallelogram intersection, and line segment-cylinder intersection.

I will assume that you know how to intersect a line and a parallelogram. Here is a sketch for solving the line segment-cylinder intersection:

The canonical equation of an elliptic cylinder aligned with the $Z$ axis and having center $(c_x, c_y, c_z)$ is given as: $$\frac{(x-c_x)^2}{a^2} + \frac{(y-c_y)^2}{b^2} = 1$$ The parametric form of a ray is given as: $$\vec{r}(t) = \vec{o} + t\vec{d}$$ Then by plugging r into the canonical equation one gets: $$\frac{(o_x + td_x -c_x)^2}{a^2} + \frac{(o_y + td_y -c_y)^2}{b^2} = 1$$ Solve the quadratic equation for $t$ and you have your intersection. If the discriminant is negative then there is no intersection. To get a rotated cylinder, the easiest way is to apply the inverse rotation to the ray (the equation may be modified however too). Note that for a line segment, you simply need to check whether $t \in [t_0, t_1]$, where $\vec{a} = \vec{o} + t_0\vec{d}$ is the first point of the line segment, and $\vec{b} = \vec{o} + t_1\vec{d}$ is the second one.

EDIT: To clarify you further questions in the comments:

1) a,b in the denominator are the radii along the $X$ and $Y$ axes respectively. In the trivial case where $a=b=r$ we have the canonical equation for a circular cylinder: $$(x-c_x)^2 + (y-c_y)^2 = r^2$$

2) For the intersection between the cylinder axis and the box a rotation is only required if your box is OOBB and you want to use your AABB intersection code.

3) The solution of the quadratic equation is trivial (note, I renamed $a,b$ to $r_1,r_2$ to further emphasize that they are the radii):

$$\frac{(o_x + td_x -c_x)^2}{r_1^2} + \frac{(o_y + td_y -c_y)^2}{r_2^2} = 1$$ $$r_2^2(o_x + td_x - c_x)^2 + r_1^2(o_y + td_y -c_y)^2 = r_1^2r_2^2$$ $$(r_2^2d_x^2+r_1^2d_y^2)t^2 - 2(r_2^2d_x(c_x-o_x) + r_1^2d_y(c_y-o_y))t + r_2^2(c_x-o_x)^2 + r_1^2(c_y-o_y)^2 - r_1^2r_2^2 = 0$$

Let us substitute: $$A = (r_2^2d_x^2+r_1^2d_y^2)$$ $$B = (r_2^2d_x(c_x-o_x) + r_1^2d_y(c_y-o_y))$$ $$C = r_2^2(c_x-o_x)^2 + r_1^2(c_y-o_y)^2 - r_1^2r_2^2$$

Then you have the quadratic equation:

$$At^2 -2Bt + C = 0$$ $$D = B^2-AC$$

If $D<0$ there is no intersection between the ray $\vec{o}+t\vec{d}$ and the cylinder, otherwise:

$$t_1 = \frac{B-\sqrt{D}}{A}, t_2 = \frac{B+\sqrt{D}}{A},$$

give you the two intersections (it is one intersection if $D=0$).

Now let your segment vertices be $\vec{v}_a = \vec{o}+t_a\vec{d}$ and $\vec{v}_b = \vec{o} + t_b\vec{d}$. If $t_a \leq t_1 \leq t_b$ then you have an intersection at $t_1$, if $t_a \leq t_2 \leq t_b$ then you have an intersection at $t_2$, otherwise there is not intersection.

Answer 2

"Intersection of Box and Finite Cylinder", David Eberly, Geometric Tools, Redmond WA 98052

Answer 3

I ended up finding a fairly simple, fairly performant solution.

1: Create a transformation matrix that rotates the cylinder to one of the axis (x, y or z)

(That part is non-trivial)

2: Apply the transformation matrix to the cylinder

3: Apply the transformation matrix to the eight corners of the axis-aligned bounding box.

(Applying a transformation matrix is not too hard, it looks like this:)

newPoint[0] = (original[0] * xform[0][0]) + (original[1] * xform[1][0]) + (original[2] * xform[2][0] + xform[3][0]);
newPoint[1] = (original[0] * xform[0][1]) + (original[1] * xform[1][1]) + (original[2] * xform[2][1] + xform[3][1]);
newPoint[2] = (original[0] * xform[0][2]) + (original[1] * xform[1][2]) + (original[2] * xform[2][2] + xform[3][2]);

4: Get the distance of the corners to (0,0,0). Keep in mind that depending on your method of determining distance (IE, using a 3D distance formula) you may need to zero out the value of the point on the axis the cylinder was rotated to. If the cylinder was rotated to the Z axis, you would need to zero out the Z value for that point.

5: If any of the points report a distance less than the radius of the cylinder, you know there must be a volume intersection.

6: Special case: If the cylinder has a diameter small enough, it could fit inside of the bounding box completely, thus there would be no points for which their distance < cylinder radius. I am unsure of the maximum circle that fits within the projection of a box while not encompassing any points. My best guess is just to check if the diameter is < than any of the box's edge length, or check if the diameter is less than the diagonal of the box, which may be safer. Not sure.

7: If you want the cylinder to have a maximum length, that's not too bad either. Just take one point of the cylinder, a vector representing its direction and its length, and do origin + (dir * length) to get the endpoint of the cylinder. Transform it as well, and make sure any points you have fall within the Z of that transformed endpoint.