segments of a parabolic umbrella I would like to build a
kind of parabolic umbrella. Therefore I am trying to calculate the shape of the fabric parts.
My mathematical approach to this problem was to slice a parabolic surface along it's axis into (infinitly thin) circles and enscribe polygons within these circles.
For an n-sided regular polygon the center angle is given by: $$\gamma=\frac{2\pi}{n}$$
The length of one side of this polygon for a given radius r is: $$s=2 r \sin{\frac{\gamma}{2}}$$
Finally the radius is given by the parabola $$r=\sqrt{\frac{x}{a}}$$ where a is constant.
All this adds up to: $$\frac{s}{2}=\sqrt{\frac{x}{a}}\sin{\frac{\pi}{n}}$$
This is a function that provides the width of a segment for any value of x.
I tried building a paper model using this equation but it did not work out so well. So I doublechecked for mistakes but could not fix it.
Could someone point me in the right direction, please.

Examplary calculations for n=12 and focal point f=3 with $$f=\frac{1}{4a}$$

 A: In your formulas $x$ is the distance measured along the axis of the paraboloid. But when you unfold a single piece of the umbrella on a plane, the middle line (black curve in figure below) is longer than the corresponding part of the axis. To obtain the right equations one must then obtain the length of this middle line, from the vertex of the paraboloid up to $x$.
This is not too difficult: the parametric equation (in space) of the left side of the green piece is given by:
$$
x=r,\quad y=0,\quad z=ar^2,
$$
where the axis of the paraboloid is along $z$-axis and
I used as a parameter the radius $r$ of a section.
The parametric equation of the right side of the green piece is:
$$
x=r\cos\gamma,\quad y=r\sin\gamma,\quad z=ar^2.
$$
The black middle line is halfway between these sides, and its parametric equation is then:
$$
x=r{1+\cos\gamma\over2},\quad y=r{\sin\gamma\over2},\quad z=ar^2.
$$
We can obtain the length of this line when the parameter runs from $0$ to $r$ by computing the integral:
$$
l=\int_0^r\sqrt{dx^2+dy^2+dz^2}=
\cos{\gamma\over2}\int_0^r\sqrt{1+{4a^2t^2\over\cos^2{\gamma\over2}}}dt=\\
=\frac{1}{2} r \cos{\gamma\over2}\sqrt{1+{4a^2r^2\over\cos^2{\gamma\over2}}}
+{\cos^2{\gamma\over2}\over4a}{\ln \left(\sqrt{1+{4a^2r^2\over\cos^2{\gamma\over2}}}
+{2ar\over\cos{\gamma\over2}}\right)}.
$$
The expression above gives then the height of the unfolded piece, while its width is given by
$$
s=2r\sin{\gamma\over2}.
$$

To draw the shape we can set $s=|x|$ and $l=y$ as coordinates in a cartesian plane, and substitute
$\displaystyle r={|x|\over2\sin{\gamma\over2}}$
into the equation for $y$, to eliminate $r$ and
get the final equation:
$$
y
={|x|\over4\tan{\gamma\over2}} \sqrt{1+{4a^2x^2\over\sin^2{\gamma}}}
+{\cos^2{\gamma\over2}\over4a}
{\ln \left(\sqrt{1+{4a^2x^2\over\sin^2{\gamma}}}
+{2a|x|\over\sin\gamma}\right)}.
$$
Here's a plot of this function made with GeoGebra, for $\gamma=45°$ and $a=1$.

