Subject: intersection of perpendicular cylindrical surfaces

I've encountered this situation in the course of normal architectural business affairs, and my atrophied math abilities no longer serve:

Please consider two right circular cylinders, perpendicular one to the other, and of unlike radii in a 3 dimensional Cartesian space with mutually perpendicular x,y,z axes. If one cylinder is centered on the y axis with radius ra, and the other on the z axis with radius rb, then the expression for the first surface would be x2 + z2 = ra2, y = any number. Likewise, the second cylinder's surface would be x2 + y2 = rb2, z = any number.

It is my intent to define the curve at the intersection of these two cylindrical surfaces. From sketching the conditions it appears that this intersection resembles an ellipse folded about its minor axis.

My approach was to solve one of the two surface expressions for one variable and substitute that result into the other expression. So, taking the second expression and solving for x2 = rb2 - y2, and then pitching that back into the first gives rb2 - y2 + z2 = ra2. Rearranging, z2 - y2 = ra2 - rb2, the right hand side of the equation being a constant. This seems to represent some sort of hyperbolic paraboloid, and not at all what I expected.

Could you please render some guidance, and maybe recommend a good thorough text in which to find recipes for this sort of thing?

Thank you for your time

Hi Charlie,

The curve is easily drawn, but it is not easily described in elementary terms -- it is "twisted" in the sense that any plane intersects it in a finite number of isolated points; and it is "smooth" in the sense that you can find the direction of its tangent vector at any point. It comes in two disconnected pieces. You can get a feeling for what a piece looks like by setting a rubber band on a cylindrical object (or by running your car over a bicycle wheel).

Your calculations show that by looking at the curve from a point far out on the x-axis (or equivalently, by projecting the picture into the yz-plane), you see a hyperbola (namely z2 - y2 = ra2 - rb2). Looking at it from a point on the y-axis you see a circle, and from the z-axis you see a circular arc. If you have a computer that draws curves in 3-dimensions, have it draw -- for t between -ra and ra --

(t, sqrt(rb2 - t2), sqrt(ra2-t2)) together with (t, sqrt(rb2 - t2), -sqrt(ra2-t2)). Then you can view it from any angle. It's just a nice, smooth, mildly distorted circle floating in space!

When rb = ra the intersection of two cylinders is more familiar -- the shape of the intersection (of the solid cylinders) is common in domes of public buildings. The boundary curve is no longer disconnected -- it consists of two congruent ellipses in perpendicular planes, attached at the ends of their minor axes.

Go to Math Central