finding the length of a semicircle?? (1 Viewer)

mmmm. · Jun 22, 2021

Hello, i am doing some diy and need to figure out how long the pvc pipe should be if I want the height of the semicircle to be 75 if the diameter is 90 cm. From the image, I want to find the length of the red

Screen Shot 2021-06-22 at 3.07.30 pm.png

CopperPike · Jun 22, 2021

191.45 cm

CM_Tutor · Jun 27, 2021

CopperPike said:
191.45 cm

How did you get this answer?

This is not a simple problem as the curve is not a semicircle - it looks like half an ellipse, in which case the formula for the circumference is not in any HSC course, nor are methods to calculate it. In fact, I'm not sure that there is any closed form for the circumference known.

For anyone wondering, for an ellipse

$\cfrac{x^2}{a^2} + \cfrac{y^2}{b^2} = 1$

the circumference can be approximated by the following formula derived by Indian Mathematician Srinivasa Ramanujan and reported in 1914:

$C \approx \pi\bigg[3(a + b) - \sqrt{(3a + b)(a + 3b)}\bigg]$

In this case, with $a = 45$ , $b = 75$ , and the distance sought $= \cfrac{C}{2} \approx 191.45239795...\ \text{cm}$ with an error that's no more than about 1 in 4,000.

A more precise approximation, also due to Ramanujan, is:

$C \approx \pi(a + b)\left(1 + \cfrac{3h}{10+\sqrt{4 - 3h}}\right) \qquad \text{where} \qquad h = \cfrac{(a - b)^2}{(a + b)^2}$

We have $h = \cfrac{(a - b)^2}{(a + b)^2} = \cfrac{(45 - 75)^2}{(45 + 75)^2} = \cfrac{(-30)^2}{(120)^2} = \cfrac{1}{16}$ , and so:

$\begin{align*} C &\approx 120\pi\left(1 + \cfrac{\cfrac{3}{16}}{10+\sqrt{4 - \cfrac{3}{16}}}\right) \\ &\approx 120\pi\left(1 + \cfrac{3}{16\left(10+\sqrt{4 - \cfrac{3}{16}}\right)}\right) \\ &\approx 120\pi\left(1 + \cfrac{3}{16\left(10+\cfrac{1}{4}\sqrt{4 \times 16 - 3}\right)}\right) \\ &\approx 120\pi\left(1 + \cfrac{3}{160+4\sqrt{61}}\right) \\ &\approx 30\pi\left(4 + \cfrac{3}{40+\sqrt{61}}\right) \\ &\approx 30\pi\left(4 + \cfrac{3\left(40-\sqrt{61}\right)}{40^2 - 61}\right) \\ &\approx 30\pi\left(4 + \cfrac{3\left(40 - \sqrt{61}\right)}{1539}\right) \\ &\approx 30\pi\left(4 + \cfrac{40 - \sqrt{61}}{513}\right) \\ &\approx 30\pi\left(\cfrac{4 \times 513 + 40 - \sqrt{61}}{513}\right) \\ &\approx \cfrac{10\pi}{171}\left(2092 - \sqrt{61}\right) \\ \text{Thus}: \qquad \cfrac{C}{2} &\approx \cfrac{5\pi\left(2092 - \sqrt{61}\right)}{171} \qquad \approx 191.45249147...\ \text{cm} \end{align*}$

and our "less precise approximation" was accurate to about 6 sig fig. The error in the more precise figure is somewhere around or below 1 in 2²⁰, so it's reliable to somewhere around 6 decimal places / 9 significant figures.

finding the length of a semicircle?? (1 Viewer)

mmmm.

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CopperPike

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