We have a sphere and a circular cylinder. Let the sphere center be $O$ and radius $R$, and the cylinder axis $a$ and radius $r$.
I solved the specific case intersection graphically on 2 planar projections, but I would like to have a general understanding classification of cases, hence this question. So is this correct?
- If a cylinder has its axis through the sphere's center, $O\in a$, the intersection is planar (a circle), if the intersection exists ($r\le R$). Here in both projections $O_1\in a_1$ and $O_2\in a_2$.
- If the cylinder's axis in one projection (e.g. plan) passes through the projection of the center but not in the other (e.g. front), or $O_1\in a_1$ but $O_2\notin a_2$ , or in other words the cylinder position is "through the sphere middle" but not through the center, here I'm not sure. E.g. in my specific example the cylinder is horizontal, $r<R$, and $a$ is directly below $O$. In $projection_1$ the intersection is a line (looks planar), in $projection_2$ it is an ellipse. Which looks like a projection of a circle. However, if the section is planar, then a diagonal section of a cylinder should be an ellipse. But a planar section of a sphere is a circle. Could it be that it just looks planar due to inaccuracy of construction? I have similar looking result in another example too.
- If $O_1\notin a_1$ and $O_2\notin a_2$, then intersection is two non-planar "rings"
- A special case when cylinder touches the sphere, it looks like a 3d figure-8
Best Answer
It is way easier to think when a plane section of a cylinder, that is an ellipse having ratio between the axis that depends on the angle between the cutting plane and the axis of the cylinder, is also a plane section for a sphere i.e. a circle. There is no inaccuracy in point $(2)$: in order to have a planar intersection, the center of the sphere has to lie on the axis on the cylinder.