Parametric equation of a non-wrapping circle on the surface of a cylinder

Question

I have a cylinder of radius $R$, and I wish to draw a circle of radius $r$ on its surface that does not wrap around it. So, the centre will lie somewhere on the surface of the cylinder, and the whole of the circle will lie on the surface as well.

Can someone help me to find a parametric representation?

Answer 1

Credit goes to @achillehui for giving me the answer. I am just elaborating on what he said, and clearing up why the scaling by $R$ is needed.

• The general way to go about finding the parameterisation of a circle on a cylinder (or to a sphere or a cone or any mappable surface) is to find a mapping from the plane to the cylinder, and then apply this mapping to a circle on the plane.

Quoting @achillehui from above:

Assume cylinder has z-axis as symmetry axis. You can parametrize the surface as $$(u,v)↦(x,y,z)=(R\cos(\frac{u}{R}),R\sin(\frac{u}{R}),v)$$. To draw a circle on it, compose circle's parametrization $$t↦(u,v)=(r\cos t,r\sin t)$$ with that of cylinder. You get $$t↦(x,y,z)=(R\cos(\frac{r \cos t}{R}),R\sin(\frac{r \cos t}{R}),r \sin t)$$. If you want circle looks like a circle, make sure you use the right scaling factor for $u,v$ (the two $R$ above).

• The composition of mappings is what I said at the beginning. What really piques my interest is the re-scaling by $R$. @achillehui explains this using a Taylor expansion around $(u,v) = (0,0)$ for the large $R$ limit. Thus, the unscaled parametrisation $(u,v)↦(x,y,z)=(R\cos(u),R\sin(u),v)$ gives $(u,v)↦(x,y,z)=(R,Ru,v)$, so that we have a mapping from circle to ellipse: $$u^2+v^2=1 \: (\rm{circle}) \mapsto (y/R)^2 + z^2 =1 \:\rm{(ellipse)}.$$

Scaling this by $R$ gives $(u,v)↦(x,y,z)=(R,u,v)$ so that we have: $$u^2+v^2=1 \:(\rm{circle}) \mapsto y^2 + z^2 =1 \:\rm{(circle)}.$$

• I have a different argument, more intuitive for me. It is that the first unscaled mapping is not an isometry, whereas the second is. Only isometries preserve the geodesic curvature $\kappa_g$ (see: Why the geodesic curvature is invariant under isometric transformations?), so that circles ($\kappa_g =1$) remain circles under the mapping.

This can be seen clearly by looking at the action of the mapping on the line $u=\rm{const.}$, which maps onto a circle concentric with the cylinder's axis. For the unscaled map, the interval of length $2\pi$ is mapped onto a circle of length $2\pi R$, which clearly involves a scaling by factor $R$. Dividing by $R$ thus ensures isometry.

------------------------------------------- IMAGES ------------------------------------------------

• Unscaled parametrisation (with $R=1$): $$t↦(x,y,z)=(\cos(r \cos t),\sin(r \cos t),r \sin t), \:t \in [0, 2\pi]$$

• Scaled parametrisation: $$t↦(x,y,z)=(R\cos(\frac{r \cos t}{R}),R\sin(\frac{r \cos t}{R}),r \sin t), \:t \in [0, 2\pi]$$