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New to the site so excuse any clumsy writing:

I am attempting to write an equation that finds the surface area along a pyramid as you move up and down the height of it. I know the initial areas of the top and bottom and thought I could use the Frustum pyramid equation to find the volume: \begin{align} V=(\frac{h}{3})*(A_1+A_2+\sqrt{A_1*A_2}) \end{align} And then move up and down between there to determine the area of the top (as the bottom stays constant). This train of thought has limited me in finding an equation that would help to determine the new surface area of the top as I change the height... thoughts?

Edit, I am looking to find the surface area of water as it drains from a rectangular hole... here is an example of use:

pond layout

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  • $\begingroup$ Could you elaborate a little further on the problem? What do you mean by " along a pyramid ", " as you move up and down the height of it " and " there "? By the way, it should be ... $A_1+A_2$... @ EngineeringSam $\endgroup$ – Deniz Sargun Feb 7 '16 at 23:05
  • $\begingroup$ Thanks for the correction. Further explanation: I have a rectangular hole in the ground that is shaped like a pyramid. So, knowing the top area of my pond and the bottom area of my pond, I want to know the water's surface area at any given time as it infiltrates. Thanks in advance for the help. $\endgroup$ – EngineeringSam Feb 10 '16 at 21:46
  • $\begingroup$ I've added an image with a pond to show what I am speaking about $\endgroup$ – EngineeringSam Feb 10 '16 at 21:58
  • $\begingroup$ Another good exercise is to show that the top surface area could also be calculated via the differential volume added to the pyramid shape. $\endgroup$ – Deniz Sargun Feb 12 '16 at 19:52
  • $\begingroup$ lol, I fell short on that one... I'm still stuck in that way of thinking. Have you been able to do so? $\endgroup$ – EngineeringSam Feb 14 '16 at 0:40
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If by "new surface area of the top as I change the height" you mean "area of a cross-section at a given height $y$ through a frustum of fixed height $h$ with areas $A_1$ and $A_2$ at top and bottom", then the formula you're looking for is $$A_y = (\sqrt{A_1} + \dfrac y h (\sqrt{A_2}-\sqrt{A_1}))^2$$ where $y$ varies from $y=0$ at the end with area $A_1$ to $y=h$ at the end with area $A_2$.

To see where the formula comes from (I just derived it), assumed WLOG we have the frustum of a square pyramid. The side-length of a cross section varies linearly from $\sqrt{A_1}$ to $\sqrt{A_2}$ as $y$ varies from $0$ to $h$ - the side length is the expression on the RHS being squared.

Example: a cone frustum of height $h$ with area $A_1=\pi r_1^2$ on the bottom and $A_2=\pi r_2^2$ on the top.

  • Reasoning via geometry: the radius varies linearly: at height $y$ above the base the radius is $r=r_1+\dfrac{y}{h}(r_2-r_1)$, so the cross-section area at height $y$ is $A_y=\pi(r_1+\dfrac{y}{h}(r_2-r_1))^2$.
  • By the formula: $A_y=(\sqrt\pi r_1 + \dfrac y h(\sqrt \pi r_2-\sqrt\pi r_1))^2 = \pi(r_1+\dfrac{y}{h}(r_2-r_1))^2$

For your case of a pyramid with rectangular cross-section, i.e. a rectangle $a\times b$ on the bottom and rectangle $ka \times kb$ on the top, when you muddle through all the algebra you get the same result.

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  • $\begingroup$ As mentioned in my response above, I am looking to find the surface area of water as it drains in a rectangular hole. I know the top and bottom areas to start with. I cannot assume a square... since it will be built rectangular. Thanks for the help! $\endgroup$ – EngineeringSam Feb 10 '16 at 21:48
  • $\begingroup$ I've added an image with a pond to show what I am speaking about $\endgroup$ – EngineeringSam Feb 10 '16 at 21:58
  • $\begingroup$ @EngineeringSam: As long as the top and bottom areas are the same shape (i.e. scaled versions of each other), the formula I wrote works. The square pyramid example is just there because it's easy to see the connection between the formula and that example. I'll add an example with a cone frustum as an illustration. $\endgroup$ – Frentos Feb 10 '16 at 23:45
  • $\begingroup$ Yup, that further explanation helped! I see now where my train of thought went wrong... trying to compare the area through volume and not just using reasoning that the area would increase uniformly as the height increased. Thanks! $\endgroup$ – EngineeringSam Feb 12 '16 at 18:31
  • $\begingroup$ can't we just find the difference in A2 and A1.... THEN multiply it by (the height up the pond divided by the total height)... Then add this value to A1? ...why root the terms then square them? $\endgroup$ – EngineeringSam Feb 12 '16 at 19:49
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Given an upside down rectangular frustum right pyramid with top and bottom surface edge lengths $a_0$, $b_0$, $a_2$ and $b_2$ and height $h_3$ ( nomenclature to be clear in the following argument ), let us extend it to a pyramid of height $h_0$. Regarding it as a pond filled with water, let us determine the lateral surface area, $l_4$, of the volume of water as a function of the height of water, $h_4$.

Call the lateral surface area of the extended part as $l_2$ and the lateral surface area of the pyramid formed by the volume of water and the extension as $l_1$, then we have the following.

$$ l_4= l_1-l_2 $$

Now, for a rectangular pyramid of base side lengths $a$ and $b$ and height $h$ the lateral surface area, $l$, is computed as below.

$$ l= 2 \thinspace \left( \frac{1}{2} \sqrt{h^2+ \frac{b^2}{4}} \thinspace a+ \frac{1}{2}\sqrt{h^2+\frac{a^2}{4}} \thinspace b \right) $$

$$ = \sqrt{h^2+ \frac{b^2}{4}} \thinspace a+ \sqrt{h^2+ \frac{a^2}{4}} \thinspace b $$

Then,

$$l_4= \left( \sqrt{h_1^2+ \frac{b_1^2}{4}} \thinspace a_1+ \sqrt{h_1^2+ \frac{a_1^2}{4}} \thinspace b_1 \right) - \left( \sqrt{h_2^2+ \frac{b_2^2}{4}} \thinspace a_2+ \sqrt{h_2^2+ \frac{a_2^2}{4}} \thinspace b_2 \right) $$

and due to the similarity arguments in the triangles formed

$$ h_2= \frac{h_3}{a_0-a_2} \thinspace a_2. $$

Further, for any length parameter $\alpha$,

$$ \frac{\alpha_1}{\alpha_2}= \frac{h_1}{h_2} $$

$$ = \frac{h_2+h_4}{h_2} $$

$$ = 1+ \frac{h_4}{h_3} \thinspace \frac{a_0-a_2}{a_2}. $$

Let us denote this parameter by $k$. Note that it depends linearly on the water level $h_4$. Thus,

$$ l_4= (k^2-1) \left( \sqrt{h_2^2+ \frac{b_2^2}{4}} \thinspace a_2+ \sqrt{h_2^2+ \frac{a_2^2}{4}} \thinspace b_2 \right) $$

where each term can be written in the given variables. Another point to note is that the second multiplier is constant in the water level and by the $k^2$ term the lateral surface area of the volume of water is quadratic in the water level, as expected.

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  • $\begingroup$ Nice approach! This would work in the instance I want to apply it... The universal approach given by Frentos above might be the one I use as my default application because not all ponds are created equal :o) Thanks for the solution! $\endgroup$ – EngineeringSam Feb 12 '16 at 18:34
  • $\begingroup$ @EngineeringSam. Thank you :) Great that you have found an answer. Just note the difference between the two results. Frentos has computed the top surface area of the water whereas I have come up with the lateral surface area of the volume of the water. These are two different answers to two different questions. I can now see that you were looking for the expression for the top surface area in which case Frentos' s answer is both true and intuitive and generalisable. And you have still not corrected the expression $ ...A_1+A_2...$ in the question statement. $\endgroup$ – Deniz Sargun Feb 12 '16 at 19:50
  • $\begingroup$ thanks.... It's up to date! $\endgroup$ – EngineeringSam Feb 12 '16 at 19:57

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