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How do you arrive at the formula

area = (AB/4)[arccos(1 - 2h/A) - (1 - 2h/A)sqrt(4h/A - 4h2/A2)]

from this website http://www.had2know.com/academics/ellipse-segment-tank-volume-calculator.html?

Any ideas will be most helpful thanks.

Rob Wilkinson
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1 Answers1

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Let $a$ and $b$ be the ellipse semiaxes, with axis $a$ cut by the segment base at $\bar x=a-h$ from the center. Then segment area is: $$ area=2\int_{\bar x}^{a}y\,dx=2ab\int_0^{\bar t}\sin^2t\,dt =ab\left(\bar t-\sin\bar t\cos\bar t\right), $$ where I have used the parameterization $x=a\cos t$, $y=b\sin t$ and $\bar x=a\cos \bar t$. From the last equality it follows that $\cos\bar t=\bar x/a=1-h/a$: plug this (and the easily derived expressions for $\sin\bar t$ and $\bar t$) into the above formula and you are done.

EDIT.

With the data given in the comment below, we have:

$$ a=A/2,\quad b=B/2, $$ and in addition, from $\cos\bar t=1-h/a=1-2h/A$ we get $$ \bar t=\arccos(1-2h/A),\quad \sin\bar t=\sqrt{4h/A-4h^2/A^2}. $$ Plugging that into my formula above we finally obtain $$ area={AB\over4}\left[\arccos\left(1-{2h\over A}\right)-\left(1-{2h\over A}\right)\sqrt{{4h\over A}-{4h^2\over A^2}}\,\right], $$ which is exactly the formula you quoted.

You might have got confused by my choice of $x$ as "vertical" axis, but of course that does not change the final result.

enter image description here

Intelligenti pauca
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  • Thanks for your reply. It seems that you have integrated with respect to the x axis, so dx, but should you be integrating with respect to y axis, so dy. I say this because h is the height of the liquid and that changes in the y axis. Please note A and B are overall height and width of ellipse in y and x axis. so if you take semi axis then it would be A/2 and B/2. I'm also wondering how you get cos in formula, I have found this from Dr Math http://mathforum.org/library/drmath/view/61443.html and it has sin in formula. Still unsolved problem for me, hopefully you can help. – Rob Wilkinson Nov 28 '17 at 13:01
  • At the moment I can't remember the precise formulation of your question, and the link you provided is not working anymore: could you please update your question and better explain what you need? – Intelligenti pauca Nov 28 '17 at 15:06
  • OK, I will make it clear for you. The question is how do you prove that the equation for area of an ellipse segment = (AB/4) [arcos(1 - 2h/A) - (1 - 2h/A) sqrt(4h/A - 4h^2/A^2)] in a liquid tank, where height of tank is A in y axis, width of tank B in x axis, height of elliptical segment h (which is also height of liquid in y axis) and arccos is in radians and not degrees – Rob Wilkinson Dec 01 '17 at 06:38
  • Aretino, Just a little bit extra added on that I couldn't before I'm wondering how you get cos in equation and arrive at this equation. Should you be integrating with respect to y axis because height of liquid changes in y axis?That had2know link I gave you doesn't work anymore maybe they have changed their website or doing maintenance, but the Dr Math link for Fuel Left in an Ellipsoidal Tank does work and that works it out from ellipse equation y^2/a^2 + x^2/b^2 = 1, but it has sin in final equation. – Rob Wilkinson Dec 01 '17 at 06:39
  • See my edited answer. – Intelligenti pauca Dec 01 '17 at 10:10
  • Hi Aretino, I am almost there with this question on Ellipse Segment area of tank. There are just a few more things that I need help to understand. I am a bit confused by your limits of integration for $$ area=2\int_{\bar x}^{a}y,dx, where a is the top limit and $\bar x is the bottom limit. Also I'm just wondering how you get \sin\bar t=\sqrt{4h/A-4h^2/A^2}. I can see that $\sin\bar t=\bar y/a. Once I have this understood I can put the answer up. – Rob Wilkinson Dec 09 '17 at 06:18
  • I added a picture, hope it helps. As for $\sin\bar t$ you should remember that for any angle $\theta$ this identity holds: $\sin^2\theta=1-\cos^2\theta$. – Intelligenti pauca Dec 09 '17 at 10:29