Suppose one intersects unit-radius solid cylinders in $\mathbb{R}^3$, with each cylinder axis passing through the origin. For example, two such cylinders produce the Steinmetz solid.

But if we imagine starting with one cylinder with axis aligned along the $x$-axis, and spin that axis about $z$, obtaining an infinite number of cylinders all of whose axes lie in the $xy$-plane, the resulting intersection is a unit-radius ball:

If we set a unit direction vector $u$ along each cylinder axis, the spinning walks the $u$ vector tips around a great circle on the sphere $S$ of directions.

Now imagine an arbitrary closed curve $C$ on $S$. Form the intersection $I(C)$ of the cylinders determined by all the direction vectors lying on $C$. Again all cylinders have unit radius, and all axes are through the origin. My question is:

Q. For which $C$ will $I(C)$ be the unit-radius ball?

This might be straightforward, but I am not seeing it...

An analogous question can be posed in $\mathbb{R}^d$, with each cylinder a line cross a $({d{-}1})$-ball.