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Is there an equation in cylindrical coordinates for an ellipse (tilted at some angle) on the surface of a right circular cylinder of radius r? For simplicity, I envision the cylinder to be coincident with the x-axis.

I am aware that the cylinder could be "unwrapped" into a plane, which would result in the ellipse becoming a sine curve. I am just not sure how that information ties into cylindrical coordinates.

EDIT: I have realized that I am looking for a parametric equation. For an ellipse on the surface of a cylinder of radius r which has a certain angle of inclination, is there a parameterization where I can calculate the axial coordinate seperately from the azimuth angle for a given t?

Thank you.

Ellipse on a cylinder

Thetravellingfool
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  • You need a pair of implicit Cartesian equations to describe a curve in 3D. If you want a single equation, it’s not too hard to come up with a parameterization of the curve. – amd Nov 28 '18 at 21:58
  • That is a very good point; In my mind I am looking for a parameterization of the curve that takes into account the angle of inclination, but I failed to put that in my original question. For a given cylinder radius, I believe I need two parameters: an expression for the axial coordinate and an expression for the azimuth. Could you help me with that parameterization of the ellipse? – Thetravellingfool Nov 28 '18 at 23:35
  • You can choose a cylindrical coordinate system $(\rho,\phi,z)$ such that the ellipse is $\rho = r, z = r \tan\alpha \sin\phi$, where $\alpha$ is the angle between the plane normal and cylinder axis. You’ve already accepted an answer, but I can expand on this if you like. – amd Nov 30 '18 at 22:36

2 Answers2

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For greater simplicity, is better that cylinder axis $=$ $z$-axis. Let be $R$ the cylinder radius, $ax + by + cz + d = 0$ the plane containing the ellipse. Then, $r = R$ and $$a R\cos\theta + b R\sin\theta + cz = d.$$

Martín-Blas Pérez Pinilla
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  • I think that it’s worth emphasizing that this is a pair of implicit Cartesian equations. Without the constraint $r=R$, you get an intersecting-looking surface, not an ellipse. – amd Nov 28 '18 at 21:57
  • @amd, correct. The ellipse is the intersection of the cylinder $x^2 + y^2 = R^2$ and the plane $ax+by+cz+d=0$. – Martín-Blas Pérez Pinilla Nov 29 '18 at 15:39
  • That was meant to be “interesting,” not “intersecting.” My fingers have minds of their own at times. – amd Nov 29 '18 at 19:27
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As stated before assume the (normalized) equation of the plane is

$$ \frac{a x + b y + c z}{\sqrt{a^2+b^2+c^2}} = d $$

and the parametric equation of the cylinder

$$ \pmatrix{x & y & z}= f(\varphi,z) = \pmatrix{ R \cos\varphi, R\sin\varphi, z} $$

Where the two intersect you have your ellipse in cartesian coordinates

$$ z(\varphi) = \frac{d \sqrt{a^2+b^2+c^2}-R (a \cos\varphi+b \sin\varphi) }{c} $$

or

$$\vec{r}_{\rm curve}(\varphi) = \pmatrix{x\\y\\z} = \pmatrix{ R \cos\varphi \\ R \sin\varphi \\ \frac{d \sqrt{a^2+b^2+c^2}-R (a \cos\varphi+b \sin\varphi) }{c} } $$

Now let's find the properties of this ellipse.

The center of the ellipse is at $\vec{r}_{\rm cen} = \pmatrix{0 & 0 & \frac{d \sqrt{a^2+b^2+c^2}}{c}} $

The ellipse in polar coordinates is

$$ r(\varphi) = \| \vec{r}_{\rm curve}-\vec{r}_{\rm cen} \| = \sqrt{R^2 + \frac{R^2}{c^2} \left( \frac{a^2+b^2}{2} + a b \sin(2 \varphi) + \frac{a^2-b^2}{2} \cos(2\varphi) \right)} $$

This allows us to find the major and minor radii

$$ \begin{aligned} r_{\rm major} & = R \frac{ \sqrt{a^2+b^2+c^2}}{c} \\ r_{\rm minor} & = R \end{aligned} $$

The principal axes of the ellipse are on

$$ \varphi = \frac{1}{2} {\rm atan}\left( \frac{2 a b}{a^2-b^2} \right) + n \frac{\pi}{2} $$

John Alexiou
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