I need help resolving a geometric hydrology related question for a storm drain pipe. The pipe is 6.91 ft x 5.35 ft. The equipment at the site is giving me real-time depth of water data, and with the equation I got from a very helpful user on this website I now can calculate the corresponding width of water. I now need to determine the arc length at any given water depth, which I can then add to the width already calculated to determine the entire perimeter of the submerged cross section. how do I use the data I've got to determine arc length? I have attached a sketch. pic

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    $\begingroup$ The length of an elliptic section is related to elliptic integral (hence the name), which usually cannot be expressed in terms of fundamental functions (polynomial, exponential, logarithmic, trigonometric), You may need to find a numerical approach. $\endgroup$ – Quang Hoang Oct 19 '15 at 18:34
  • $\begingroup$ Can you live with an approximation? $\endgroup$ – John Alexiou Oct 19 '15 at 19:20
  • $\begingroup$ Most likely I can, depending on how off the approximation is. what are your thoughts ja72? $\endgroup$ – user281838 Oct 19 '15 at 19:33

Consider an approximation. Start by using the half angle $\psi$ that spans the arc


Note that $\tan \psi = \frac{x/2}{depth}$. I also used a dimensionless parameter for the shape equal to $\epsilon = \frac{h}{\sqrt{w^2-h^2}}$ where $w$ is the known width and $h$ is the known height. This created the arc length integral as

$$ \ell = \int \limits_{0}^{\psi/2} \sqrt{w^2-(w^2-h^2)\cos^2 \psi}\,{\rm d}\psi $$

$$ \ell = \frac{w}{\sqrt{1+\epsilon^2}} \int \limits_0^{\psi/2} \sqrt{\epsilon^2 + \sin^2 \psi}\,{\rm d} \psi $$

I plugged the integral into Wolfram Alpha to give me the approximation

$$ \ell = (w) \psi + \left( \frac{h^2}{24 w}-\frac{w}{24} \right) \psi^3 - \left( \frac{h^4}{640 w^3}-\frac{h^2}{960 w}-\frac{w}{1920} \right) \psi^5 + \ldots O(\psi^7)$$


For any parametric curve $(x(t),y(t))$ the length is $$\Delta \ell = \int \limits_t^{t+\Delta t} \sqrt{ \left(\frac{{\rm d}x}{{\rm d}t}\right)^2 + \left(\frac{{\rm d}y}{{\rm d}t}\right)^2}\,{\rm d} t $$


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