Unrolled intersection of 2 cylinders

Question

I'm writing a macro with Visual Basic for Autocad. My problem is: 1)I have 2 cylinders with orthogonal vertical axis, so they intersect themselves generating a 3D space curve. Cylinders could have different diameters 2)Now I have to unroll one of them so that the external surface becomes a rectangular with a "hole" (the intersection) Question: what is the equation of the intersection curve in the 2d unrolled plane? It's a sort of ellipse

I need to know it because this "cylinder" is a sort of pressure vessel and the 2d plane is my plate which will be rolled, but I have to cut the exact shape from plate before rolling it.

Thank you

Answer 1

Take a point $P$ on the intersection boundary. Its coordinates with respect to the unrolled vertical cylinder are $PH$ and circular arc $PK$. If cylinder radii are $R$ (vertical) and $r$ (horizontal), and $PH=y$, then $OH=\sqrt{r^2-y^2}$ and arc $PK$ has length $$ x=R\arcsin{\sqrt{r^2-y^2}\over R}. $$ This is then the equation of the intersection curve.

Answer 2

Assuming the axis coincident, we write the equation of one cylindre in cylindrical coordinates $\rho=R$ and plug it in the implicit equation of the second, $x^2+z^2=r^2$, giving

$$R^2\cos^2\theta+z^2=r^2.$$

Now unrolling the first cylindre amounts to considering the coordinates $X=R\theta,Y=z,$ and

$$Y=\pm\sqrt{r^2-R^2\cos^2\frac XR}.$$

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A few cases:

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When $R=r$, we find a sinusoid, as we might have expected,

$$Y=\pm R\sin\frac XR.$$

And when $X$ remains close to $\dfrac{\pi R}2$ (when $r\ll R$), the equation can be written

$$\left(X-\frac{\pi R}2\right)^2+Y^2\approx r^2,$$ which is a circle of radius $r$. And with $r\gg R$, the two straightlines, $Y\approx\pm r$.