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?
[Math] Finding the height of an arc with a known arc length and a known width
geometry
Related Solutions
If $r$ is the radius then you have presumably found, as Isaac did, that $l=r\theta$ and $\frac{c}{2}=r \sin(\theta/2).$ You could eliminate $\theta$ to get $$2r \sin\left(\frac{l}{2r}\right) = c.$$ That you can solve for $r$ using numerical methods.
In addition to the answer pointed out by @n_b I would like to show how to reach there:
The final result:
$$\LARGE\boxed{R=\frac H2+\frac{W^2}{8H}}$$
float cos_t = (R-H)/R;
P_x=(rect.width/2);
P_y=-R.cos_t;
Process:
Without using calculus(actually it would never be solved with calculus) and only using trigonometry:
The component of R along $\theta$ is $R.\cos\theta$ and perpendicular to it is $R.\sin\theta$
Now remaining component is $R-Rcos\theta=R(1-\cos\theta)$
Now we have: $$ W=2.R.\sin\theta\implies\sin\theta=\frac W{2R}\\ H=R(1-\cos\theta)\implies\cos\theta=1-\frac HR=\frac{R-H}R $$ Using the identity $\sin^2\theta+\cos^2\theta=1$ $$\left(\frac W{2R}\right)^2+\left(\frac{R-H}R\right)^2=1$$ $$\frac{W^2}{4R^2}+\frac{R^2+H^2-2RH}{R^2}=1\\\text{ I assume you are familiar with }(a\pm b)^2=a^2+b^2\pm 2ab$$ $$\frac{W^2}4+R^2+H^2-2RH=R^2$$ $$\frac{W^2}4+H^2=2RH$$ $$R=\frac{W^2}{8H}+\frac H2$$ It is easy to see that center is at the midpoint of width. Also y-coordinate is $-R.\cos\theta$
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Best Answer
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$.