Calculating segment length on circle I'm building a physical machine and I'm trying to figure out a geometrical problem.
The machine is composed by a cylinder, and the wall of this cylinder is composed by many wooden boards, each of them must be spaced of a value X. What should be the number of wood slats needed, and of which width?
I tried to solve the problem calculating the circonference of the circle and working on it like a straight line. Then I've split the line by using the amount of space I would have to keep between the wooden boards, but then I got stuck.
Graphical rappresentation:

Any clues? Thanks!
UPDATE
Thanks for the comments. In fact, the only data I have is the radius of the circle and the space I need to leave between the wooden boards. I don't have the number of woods. Do I need to brute-force it with an algoritm?
ps: I've updated the image, now it should be more clearer. 
 A: The limit number of wood slats is that there's an upper bound. 
If the spacing is $X$ and the radius is $r$ then of course the spacing and assume there's $n$ wood slats, then the spacing alone would take $nX$ of circumference while the total circumference is $2r\pi$. To make this work we must have $2r\pi \ge nX$  or $n \le 2r\pi/X$.
This is assuming that the spacing and the wood slats are properly curved.
After selecting $n$ we could calculate the width of the slats since the total circumference would be $nX + nY = 2r\pi$ which means that $X+Y = 2r\pi/n$.
A: As Paul comments, $2\pi r = n(x + y)$, or
$$
2r = \frac{x + y}{\pi/n}.
\tag{1}
$$
This equation expresses your wish to have some (sufficiently large) number $n$ of "board-and-space units" filling the circumference of a circle of radius $r$. "Sufficiently large" means "enough boards that you can pretend the circle is a regular $n$-gon." If you want more accuracy (for $n \geq 3$), use $x + y = 2r\sin(\pi/n)$, or
$$
2r = \frac{x + y}{\sin(\pi/n)}.
\tag{2}
$$
instead.
Since only the space width $x$ is known on the right, there is no unique solution to your problem: For each integer $n \geq 3$, equations (1) and (2) each have a unique solution, i.e., a board width $y > 0$ satisfying the circumference constraint.
What you actually do will depend on whether you'd rather have a specific number of board-and-spaces, or a specific width of board (so you can use standard lumber).


*

*If you fix $n$, then $y$ is uniquely determined, and you'll probably need to cut lumber to width.

*If you want the largest number of slits while making the fewest cuts/wasting the smallest amount of lumber, you can do this: Plug $x$ and $y$ into (1) or (2) (depending on how much accuracy you need), and find the first (i.e., smallest) integer value of $n$ making the right-hand side larger than the left. This tells you how many boards you need. Now plug the known values of $x$ and $n$ into the corresponding equation, calculate $y$, and cut the boards to that width.
For example, if you want to build a cylinder of radius $24$ inches with $1$ inch gaps using $5$-$\frac{3}{4}$ inch boards, then
$$
\frac{2\pi r}{x + y} = \frac{48\pi}{6.75} \approx 22.34,
$$
so you'll have $n = 23$ units. Using equation (2), the boards must be cut to
$$
y = 2r\sin(\pi/n) - x = 48\sin(\pi/23) - 1 \approx 5.5\text{ inches.}
$$
Keep in mind that this discussion makes implicit assumptions:


*

*The boards are flat (rather than having arcs of circles as cross-sections), and the "radius" is the distance from the center of the cylinder to the inside edge (not the inside center line) of a board.

*The gap width $x$ is much smaller than the circumference $2\pi r$.
