Cross Section of a Pancake Laying on a Cylinder Consider a circular pancake with some height h laying on the surface of a cylinder.  A cross section taken in the axial direction (RED) will have a flat bottom because there is zero curvature.  A cross section taken in the circumferential direction (BLACK) will have some rounded cross section as it is laying over the full curve of the cylinder.  Now, consider some cross section taken at an angle between RED and BLACK.  

I am trying to understand how I might calculate how much the cross section curves given that angle, and given radius r of the cylinder.  My understanding is that, if the pancake stays the same size, the curvature will decrease with increasing r.  So I am looking for a single expression involving angle and radius that will give me the RED case (zero curvature) and the BLACK case (some curvature) at the two extremes, and also provide a curvature value I can use at some intermediate cross sectional angle.  
Thank you for everyone's help with this problem.  
 A: As clarified by the comments I'm answering the version where the green curve is on the intersection of a plane (tilted by an angle $\theta$) and the cylindrical tube. Here's a pic:

When $\theta=0$ we get the black circle as the intersection. When $\theta=\pi/2$ we get the red line (together with its parallel copy on the opposite side of the cylinder, missing from the pic). When $\theta$ is in-between we something like the green curve in my pic.
That green curve is an ellipse (ask, if you don't see why). Its short axis is a diameter of the cylinder (connecting the two points, where the black and the green curves intersect). Therefore the short semiaxis of this ellipse is $b=r$.
To get the longer semiaxis let's rotate the picture so that we see that black cross section "head on" from the side:

We see that the projection to our viewing plane has a right triangle with one leg the radius of the black circle (clearly of length $r$) and the longer semiaxis $a$ of the green ellipse as the hypotenuse. The angle between those black and green lines is the angle of the plane tilt, $\theta$. Basic trigonometry tells us that $\cos\theta=r/a$
implying that the longer semiaxis of the green ellipse has length $a=r/\cos\theta$.
It is well known that at the endpoint of the short semiaxis the radius of curvature of an ellipse is $$\rho=\frac{a^2}b=\frac{r}{\cos^2\theta}.$$
Therefore the curvature at the desired point is
$$
1/\rho=\frac{\cos^2\theta}r.
$$
As a final check we see that when there is no tilt (black) $\cos\theta=\cos0=1$ and the curvature is $1/r$ as expected. Also, when $\theta=\pi/2$ (red) we get zero curvature.

Catering for the case that we really want to follow a geodesic, i.e. a helix. If we arrange the helix to share the same tangent vector with the green ellipse, we get a parametrization
$$
(x,y,z)=(r\cos t, r\sin t, rt\tan\theta),
$$
IOW a helix with $a=r, b=r\tan\theta$. The radius of curvature of the helix is
$$
\rho=\frac{a^2+b^2}a=\frac{r^2(1+\tan^2\theta)}r=\frac r{\cos^2\theta},
$$
in agreement with the earlier calculation. I read this to mean that at the point of interest the ellipse follows the geodesic so closely that the radii of curvatures agree. In the pic below the blue curve is the geodesic helix.

A: We can consider


*

*the cylinder: $y^2+z^2=r^2$

*the plane: $y=m \cdot x$
and then parametrize the intersection by


*

*$x(t)=\frac1m r\sin t$

*$y(t)=r\sin t$

*$z(t)=r\cos t$
then the limits for $t$ for a fixed $m$ can be found by the condition
$$L=\int_{-t_0}^{t_0}\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dx}{dt}\right)^2+\left(\frac{dx}{dt}\right)^2}dt$$
