# Length of arc of a semi-circle knowing the height and length

I have a hole I need to cover with plastic using pvc pipe bent into an arc. I have the width and the height of the arc. Height is from the top of the hole to the top of the arc. The width is measured from each side of the hole. I need the length the arc will be so I know how wide the plastic must be to cover it. Assume for the sake of clarity my hole is 22' W and the Arc height would be 5'.

Seems this would be enough information to calculate the length. I can only find formulas based on radius and angle. Both would be unknown because I have no idea how to get those considering the scale. Mind you the arc height is at my discretion.

Assuming that your hole is indeed circular, it can be done like this.

Of course we'll call the radius $r$. Also, I'm using $w$ as half the measured width. In this diagram, $$\sin \theta = \frac{w}{r} \quad\quad\quad \cos\theta = \frac{r - h}{r}$$

One way to combine these expressions is by using the identity $$\sin^2\theta +\cos^2\theta = 1$$

After some simplification, this produces $$r = \frac{w^2 + h^2}{2h}$$

We can also find the angle $\theta$ now $$\sin \theta = \frac{w}{r} \\ \theta = \arcsin\frac{w}{r} \\ \theta = \arcsin\frac{2hw}{w^2+r^2}$$

Combining these together we can find the arclength $l$ $$l = r(2 \theta) \\ l = 2 \frac{w^2 + h^2}{2h} \arcsin\left(\frac{2hw}{w^2 + h^2}\right) \\ l = \frac{w^2 + h^2}{h} \arcsin\left(\frac{2hw}{w^2 + h^2}\right)$$

For $h = 5$ and $w = 11$, $l = 24.915...$

I'll use $s$ for half the width of the hole, $h$ for the height of the arch, and $d$ for the unknown depth of the center of the circle below the level of the hole.

$d^2 + s^2 = (d+h)^2 = d^2 + 2dh + h^2$

$s^2 = 2dh + h^2$

$2dh = s^2 - h^2$

$d = (s^2 - h^2)/2h$

The radius of the circle is $r = d+h = \frac{s^2}{2h} + \frac{h}{2}$. The angle of the half-arc is $\arcsin(s/r)$ or equivalently $\arctan(s/d)$; multiply this (in radians!) by $2r$ to get the length of the whole arc.