Finding the height of an arc with a known arc length and a known width I have a graph that is 5280 units wide, and 5281 is the length of the arc.

Knowing the width of this arc, and how long the arc length is, how would I calculate exactly how high the highest point of the arc is from the line at the bottom?
 A: Consider the complete circle which contains that arc. Let $\theta$ be the central angle (in radians) subtended by the arc, and $r$ the radius of the circle.
Then, the length of the arc is given by $r*θ$ (by the definition of the radian).
Using the right triangle formed by the center, the midpoint of the chord and one of the ends of the arc, we get that the length of the chord is $2r\sin \left( \frac{\theta}{2} \right)$.
Therefore, we get the following equations:
$$\theta  r = 5281 \iff r = \frac{5281}{\theta}$$
$$ 2r\sin \left( \frac{\theta}{2} \right) = 5280$$
Substituting we get:
$$\sin \left( \frac{\theta}{2} \right) = \frac{5280}{5281} \frac{\theta}{2}$$
which cannot be solved analytically. You can solve it numerically, though, for a value of roughly $\theta = 0.067 $, giving $r = 78 821$.
Then (using the same right triangle as before), the height of the arc is given by $ r - r \cos \left( \frac{\theta}{2} \right) =44.2$.
A: This is an old chestnut.  If you assume the arc is circular you get the diagram below.  With $r$ the radius of the circle, we have $$r\theta=2640.5\\r \sin \theta=2640$$  As $r$ will be rather large and $\theta$ small, we can use the Taylor series for the sine.  We need to keep the third order term because the first order one will subtract out.  This gives $$r(\theta-\frac 16\theta^3)=2640\\\frac 16r\theta^3=0.5\\r=\frac 3{\theta^3}\\\theta=\sqrt{\frac 3{2640.5}}\approx 0.03371\\r\approx 78340$$
The height of the arc is $$r(1-\cos \theta)\approx \frac 12r\theta^2\approx 44.5 feet$$

A: Here is another way to approach the problem, albeit it must also use numerical methods.

WLOG, suppose the circle is centered at the origin. Then we define $f(x)=\sqrt{r^2-x^2}$, as the graph of the upper half of the circle. The length of an arc of a graph is given by $$\int_b^a\sqrt{1+\left(\frac{\partial}{\partial x}f(x)\right)^2}dx$$
Then we have:
$$\int_{b}^{a}\sqrt{1+\frac{x^2}{r^2-x^2}}dx=\left.-r \tan ^{-1}\left(\frac{x \sqrt{r^2-x^2}}{x^2-r^2}\right)\right|_b^a$$
Where $a,b=\pm\frac{5280}2$. They you can just use numerical methods like NR to solve for $r$:
$$g(r):=2 r \tan ^{-1}\left(\frac{2640}{\sqrt{r^2-6969600}}\right)-5281
\\r_{k+1}=r_k-\frac{g(r_k)}{g'(r_k)}\\
r\approx 78335.1$$
