Once the weights of your squares are fixed. Your sum turn into a weight sum of the area of intersection of your circle with each of the squares.
The intersection area between a circle and a square can further breakdown and expressed in terms of the area of intersection between a circle with four quadrants. i.e. those set of the form
$$(-\infty, u] \times (-\infty,v ] = \big\{\; (x,y) \in \mathbb{R}^2 : x \le u, y \le v \;\big\}$$
Let $C$ be your circle. For any $(u,v) \in \mathbb{R}^2$, let
$$\Delta_C(u,v) = \text{Area}(\;C \cap (-\infty,u] \times (-\infty,v]\;)$$
For any rectangle $[a_1, a_2] \times [b_1,b_2]$, we have following identity:
$$\text{Area}( C \cap [a_1, a_2] \times [b_1, b_2 ] ) =
\Delta_C( a_1, b_1 ) - \Delta_C( a_1, b_2 ) - \Delta_C( a_2, b_1 ) + \Delta_C( a_2, b_2 )$$
Assume your circle is centered at $(p,q)$ with radius $r$ and $C_0$ is the unit circle, By translation symmetry and scaling argument, we have
$$\Delta_C(a, b) = r^2 \Delta_{C_0}\left(\frac{a-p}{r},\frac{b-q}{r}\right)$$
At the end, we just need to figure out what $\Delta_{C_0}( x, y )$ are. For simplicity of presentation, we will drop the $C_0$ subscript from $\Delta_{C_0}$ from now on.
For any $u \in \mathbb{R}$, let $h(u)$ be the area of intersection between the unit disk $C_0$ with the half plane $(-\infty, u)\times(-\infty,\infty)$. It is not hard to work
out
$$h(u) = \Delta(u,\infty) = \begin{cases}
\pi,& u \in [1,\infty)\\
\pi - \cos^{-1}(u) + u\sqrt{1-u^2},& u \in (-1,1)\\
0,& u \in (-\infty, -1]
\end{cases}$$
To determine the value of $\Delta(u,v)$, one can use following table.
$$\begin{array}{rcc}
\hline
\verb/Condition/ && \Delta(u,v)\\
\hline
u^2 + v^2 \le 1 && \frac{h(u) + h(v)}{2} - \frac{\pi}{4} + uv\\
u \le -1 \vee v \le -1 && 0\\
u \ge 1 \wedge v \ge 1 && \pi\\
u \ge 1 && h(v)\\
v \ge 1 && h(u)\\
u \ge 0 \wedge v \ge 0 && h(u) + h(v) - \pi\\
u \ge 0 \wedge v \le 0 && h(v)\\
u \le 0 \wedge v \ge 0 && h(u)\\
\verb/otherwise/ && 0\\
\hline
\end{array}$$
One look at the conditions in this table one by one.
If $(u,v)$ satisfy a condition, then $\Delta(u,v)$ will be given by the expression at the right. If not, move to next condition.
Finally, here is a picture illustrating in which range, one should use which formula for $\Delta(u,v)$.
I think you're overcomplicating it.
Since the ratio $f$ of sector area to circle area always equals the ratio of $θ$ to 2π radians, we have:
$θ = 2\pi f$ in radians or $360^\circ\cdot f$ in degrees.
If we bisect $θ$, we have two right triangles, the hypotenuse being $R$, and the height, $r$. We want to find $r$ since $h + r = R$. We know that $\cos(θ/2) = r/R$ by definition, so:
$r = R\cos(θ/2)$.
Therefore:
$h = R - r$,
$= R - R\cos(θ/2)$,
$= R\,( 1- \cos(\pi f) )$ in radians,
or $R\,(1-\cos(180^\circ\cdot f)$ in degrees.
For $f = \frac{3}{8}$ and $R = 23.5\,\text{mm}$, $h = 14.5\,\text{mm}$.
Please do follow up in the comments if needing further clarification.
Best Answer
Let $t$ be the angle subtending the coloured segment. Then the area of the segment is $(t-\sin{t})\frac{r^2}{2}$.
You want the area to be equal to $F\cdot \pi r^2$.
The $r^2$ cancels, leading to the equation:
$$t - \sin t - F\cdot 2\pi = 0$$
If you let $f(t)=t - \sin t - F\cdot 2\pi$, then $f'(t) = 1 - \cos t$. Applying Newton's method to this function $f(t)$ gives you the iterative step:
$$t_{n+1} = t_n - \frac{f(t_n)}{f'(t_n)}$$ $$= t_n - \frac{t_n - \sin t_n - F\cdot 2\pi}{1 - \cos t_n}$$ $$= \frac{\sin t_n - t_n \cos t_n + F\cdot 2\pi }{ 1 - \cos t_n}$$
Note that the brackets in the OP are incorrect.
Once you've found a good approximation of $t$ (e.g. $t_{10}$), you can calculate the distance from $O$ to the segment as $y=r\cos \frac{t}{2}$, and then the height of the segment as $h=r-y$.