What is the name of the "circle" drawn on a cylinder('s curved) surface with a compass?
If we unroll the cylinder, we get a slightly elongated circle. Is there anything known about this type of curve?
This discussion is all I could find about it: https://uk.comsol.com/forum/thread/149872/how-to-draw-a-circle-on-a-cylinder-surface
Let's say the cylinder is an infinite cylinder of radius $\frac{1}{2}$ around the $z$-axis, and you set your compass at $(\frac 12, 0, 0)$ with radius $r$. Then the spatial curve is the intersection of the cylinder with a sphere of radius $r$ around the compass pin.
Then all points on the spatial curve satisfy the two equations $$x^2+y^2 = \frac{1}{4}$$ (since the curve lies on the cylinder) and $$\left(x-\frac 12\right)^2+y^2+z^2=r^2$$ (since the curve lies on the sphere). Intersection of two conics is a quartic curve in $\mathbb{R}^3$.
By subtracting these equations from one another, we get the equation $z^2 = r^2 + \frac{1}{2} - x$.
Now let's unroll the cylinder via the parametrization $(u,v) \mapsto (\frac 12\cos 2u, \frac 12\sin 2u, v)$
Then the plane curve equation in $(u,v)$ coordinates is $v^2 + \frac 12 - \frac 12 \cos 2u = r^2$.
Using the trigonometric identity $\frac 12 - \frac 12 \cos 2u = (\sin u)^2$, we get a nice alternative description $(\sin u)^2 + v^2 = r^2$ of the plane curve in question.
This implicit curve is trigonometric in one coordinate and polynomial in the other, so I'm not particularly sure whether it's useful at all, but we can still have a look at how it behaves.
For small $r$, the curve approximates a circle, since $\sin u$ approximates $u$ for small $u$. Or I should say that it approximates a row of circles, since we didn't limit $u$.
For $r = 1$, the curve is actually a union of plain old cosine curves $v = \pm \cos u$. It's the unrolling of Viviani's curve.
For $r > 1$, the curve is a union of two periodic cosine-like curves.