Circumference and Piece Length of the Da Vinci Bridge Cheers guys,
I want to build a Da Vinci Bridge as a rose arch. I want to plant it on two pillars, that are d units apart. Problem is, I need to figure out the length of each individual wood piece l, so that the whole bridge spans this given distance of d. I know the bridge would approximate a circle, but each piece of wood overhangs a freely choosable distance of s. It rests on another piece of wood with a given thickness b.
I tried to go over the angle phi that each piece of wood would form in dependence of b, but I can't get there.
I know this is an unusual question, but since I've thought about it for quite some time now, I really want to know the answer for the answers sake itself. I honestly feel it should not be that hard, I just can't really see the "trick".
A badly drawn sketch of the problem:

Every help is appreciated!
 A: If you want to build a five-piece bridge, then the third piece must be horizontal and angle $\phi$ must then be half the angle formed by the first piece with ground ($\phi=20°$ in figure below).
The horizontal distance from the left point of the first piece to the midpoint of the third one must be $d/2$, leading to the equation (I took $s$ as the distance from the vertex of angle $\phi$ to the end of the piece, I'm not sure that is what you intended):
$$
(l-s)\cos2\phi+(l-2s)\cos\phi+{l-2s\over2}={d\over2}.
$$
You also ask for the relation between $\phi$ and $b$: if $b$ is the side of the square inscribed in an isosceles triangle with base $l-2s$ and base angles $\phi$, then it is not difficult to see that
$$
\tan\phi={2b\over{l}-2s-b}.
$$

A: Is it like this general arrangement of Chinese Rainbow Bridge? without overhang $s$? Sometimes  circular wooden logs are filled into the small triangles. Possible to design as intersection of two displaced regular polygons.
woodenbridgeplans.com/home.html

