Finding different surface areas along a frustum pyramid

Question

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

Answer 1

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.

Answer 2

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.