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Say I have a cylindrical can of beans. If I try to draw a 'circle' on the label of the can by using a compass and then flatten the label into a plane, will the 3D 'circle' originally on the cylinder label become an ellipse. I tried to solve it using analytical geometry but it turned out to be somewhat difficult. It sure looks like an ellipse but I got an expression for the semi-major axis that contains the arcsin function involving the radius of the compass and that of the cylindrical can. Can anybody provide an intuitive answer please?

Adam Rubinson
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Ad Nazir
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    If the radius your compass is set to exceeds the diameter of the cylinder, the result will certainly not be an ellipse. – Hagen von Eitzen Oct 09 '22 at 21:15
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    I'm not sure why this question is down-voted or close-voted. It seems a reasonable question, although OP should probably include what their attempts to solve it using analytical geometry are, so that we can help assess it. ¶ What you're looking at is the intersection of a cylinder and a sphere centered on the surface of the cylinder; you then "unroll" the cylinder and ask about the shape of the resulting two-dimensional figure. I suspect the result isn't an ellipse after all, but I don't know offhand; I'd have to sit down and figure it out. In the meantime, kindly include your attempt? – Brian Tung Oct 09 '22 at 21:20
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    Sketch of approach. Without loss of generality, let the cylinder be $x^2+y^2 = 1$ and the sphere be $(x-1)^2+y^2+z^2 = r^2, 0 < r < 1$. We then get $z^2 = r^2+2x-2$. To unroll, we let $x = \cos\theta$ and then we have $z^2 = r^2+2\cos\theta-2$ in the $\theta$-$z$ plane—not an ellipse. – Brian Tung Oct 09 '22 at 21:28
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    Sorry, that should be $0 < r < 2$. Let's see if this Desmos link works. – Brian Tung Oct 09 '22 at 21:35
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    I was going to post an answer, but it was the same thing BrianTung suggested. I'll leave this desmos demonstration to play with: https://www.desmos.com/calculator/ojdpff7zec – Merosity Oct 09 '22 at 21:38
  • I originally downvoted as I thought the wording was hand-wavy but now I understand the question better. – Adam Rubinson Oct 09 '22 at 23:13
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    @Merosity The $x$ coordinate also needs to be scaled as $R_{can}$ changes. When $R_{can}$ is very large and $R_{sphere}$ is not, you should get approximately a circle. I inserted one factor of $R_{can}$ in your formula to make that happen: https://www.desmos.com/calculator/ynncpreilu – David K Oct 10 '22 at 00:51

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Brian Tung has the right idea in comments of representing the problem as a sphere-cylinder intersection – scaling the cylinder to have radius $1$ and the sphere representing the compass to radius $r$, the drawn curve when flattened has equation $$y^2=r^2+2\cos x-2$$ which can be rearranged to $$y^2+(2\sin x/2)^2=r^2$$ When $x$ is small, $2\sin x/2\sim x$ and the curve does approximate a circle, but it is never exactly an ellipse. When $r=2$ we get Viviani's curve, which provides intuition on why the curve deviates from ellipticity as $r$ increases: the intersection looks more like a helix.

Parcly Taxel
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