# “Unrolling” a 3d wedge

I have a parametric equation that describes a particular intersection in 3d space. I'd like to flatten this by deforming the object without stretching its length, as if this were the outline of a sticker, and I was making it flat again. Clearly this is not possible for a general curve, and this is not possible for even the curve in question except along one of two axes. (It is possible for this curve because it is the intersection of two right cylinders, and it's possible to "unroll" the length of a right cylinder into a rectangle.)

I had considered using the arc length formula dx' = $\sqrt{dx^2 + dz^2}$. However, this will only give a positive value for dx' and dx has clearly positive and negative parts.

What is the proper way to perform this operation, and what is this called?

If you're interested or it's relevant, the equation of this curve is:

$$\begin{split} x(t) &= b \cos(t) \\ y(t) &= b \sin(t) \\ z(t) &= \sqrt{a^2 - b^2 \sin^2(t)} \end{split}$$

Which is the intersection of the cylinders:

$$\begin{split} x^2 + y^2 &= b^2 \\ x^2 + z^2 &= a^2 \\ \end{split}$$

Where $b < a$. And the plane I'm attempting to unroll is the one defined by the second of those two equations.

• the square root from the radical is only considered positive by convention - mathematically its both +/- valued, a branched function. I suspect d_x' = r*d_theta may help – f5r5e5d Mar 25 '18 at 3:22
• I included only the positive surd because I graphed only the positive branch ;) – OmnipotentEntity Apr 2 '18 at 22:29

Since we only have to unroll in the $x$-direction, we can fix some $y_0$ and look at the plane $y=y_0$. Let $x_0$ be half the length of the curve defined by the intersection of your surface and this plane, and $f(x_0)$ be half the width of the strip defined by the cylinder with radius $b$ and this plane. Here is a picture of the plane: From this, we get $f(x_0)=\sin(\varphi)a=\sin(\frac{x}{a})a$.
Geometrically, if you unroll your surface, you unroll a part of the circle in this picture so that you get a 1-disc of radius $x_0$, and to get back from $x_0$ to the value of the x-coordinates of the border of this disc before unrolling, you have to apply $f$. So $f$ doesn't model your operation of unrolling, but the inverse to it!
Why that way around? Because now we can combine it with our global equation for the cylinder with radius $b$. Namely, $x^2+y^2=b^2$ has to be satisfied for the not unrolled curve, so $f^2(x)+y^2=b^2$ is our condition for the unrolled curve, or spelled out: $\sin^2(\frac{x}{a})a^2+y^2=b^2$.
Finally, you might want to add the secondary condition $|x|<a\frac{2\pi}{4}$ to ensure that you only unroll from the top half of the cylinder, and only get solutions which make sense according to the geometric reasoning above. (But you can also find nice geometric interpretations for the solutions you get without this secondary condition, and which stay valid even for $b\geq a$.)