# Thread: Hight of pyramid, at half it's volume?

1. ## Hight of pyramid, at half it's volume?

A classic pyramid with square footprint, hight h, sidelength a and volume V.
Like seen, eg. in Square Pyramid Calculator

Original question:
Assume it is a hollow container, with infinitively thin walls. It's filled with a volume of water = V/2, ie, it is half full (by volume). How far up from the bottom would the water reach, expressed in units of h, ie. How deep will the water be inside the pyramid, compared to the total hight of the pyramid.

My question:
Can this be solved in general, ie. Without knowing any specific values of h and a? Is the depth of water always a specific (constant) fraction of h. Or is the answer depending on the slant angle of the pyramid surface?

For completeness. The original question mentions specific values of h and a, I can't remember the values, but lets say h=300m and a=200m, though this is of cause not relevant for my question...

2. ## Re: Hight of pyramid, at half it's volume?

you can try h' = 0.707*h and a' = 0.707*a

3. ## Re: Hight of pyramid, at half it's volume?

Originally Posted by MikeHammer
A classic pyramid with square footprint, hight h, sidelength a and volume V.
Like seen, eg. in Square Pyramid Calculator

Original question:
Assume it is a hollow container, with infinitively thin walls. It's filled with a volume of water = V/2, ie, it is half full (by volume). How far up from the bottom would the water reach, expressed in units of h, ie. How deep will the water be inside the pyramid, compared to the total hight of the pyramid.

My question:
Can this be solved in general, ie. <--- yes
1. I assume that a is the sidelength of base square of the pyramid.

2. The volume of the pyramid is calculated by:

$V_1 = \frac13 \cdot a^2 \cdot h$

You are looking for a pyramid with $V_2 = \frac12 \cdot V_1$ with a' as the sidelength of the square and h' as the height of the smaller pyramid.

3. Use proportion: $\dfrac ha=\dfrac{h'}{a'}~\implies~a'=\dfrac ah \cdot h'$

4. Now you know:

$\frac13 \cdot (a')^2 \cdot h' =\frac12 \cdot \frac13 \cdot (a)^2 \cdot h$

$\left(\frac ah \cdot h' \right)^2 \cdot h' =\frac12 \cdot (a)^2 \cdot h$

$(h')^3 = \frac12 \cdot h^3~\implies~h'=h \cdot \sqrt[3]{\frac12} = \frac12 \cdot h \cdot \sqrt[3]{4} \approx 0.7937\ h$

5. Plug in this value into the equation of #3 to determine the value of a'.

4. ## Re: Hight of pyramid, at half it's volume?

Dear earboth

A million thanks. I could not have hoped for a more complete and fulfilling answer.

Mike

5. ## Re: Hight of pyramid, at half it's volume?

Having looked closer on your calculations... I can follow you clearly as far as 3., but in 4. I get confused.

Originally Posted by earboth
4. Now you know:

$\frac13 \cdot (a')^2 \cdot h' =\frac12 \cdot \frac13 \cdot (a)^2 \cdot h$

$\left(\frac ah \cdot h' \right)^2 \cdot h' =\frac12 \cdot (a)^2 \cdot h$
until here it's clear ... but how do you get from that to this next line?

Originally Posted by earboth

$(h')^3 = \frac12 \cdot h^3~\implies~h'=h \cdot \sqrt[3]{\frac12} = \frac12 \cdot h \cdot \sqrt[3]{4} \approx 0.7937\ h$
Can you please spell that out in more details for me.

Regards Mike

6. ## Re: Hight of pyramid, at half it's volume?

Originally Posted by MikeHammer
Having looked closer on your calculations... I can follow you clearly as far as 3., but in 4. I get confused.

until here it's clear ... but how do you get from that to this next line?

Can you please spell that out in more details for me.

Regards Mike
1. Expand the bracket:

$\left(\frac ah \cdot h' \right)^2 \cdot h'=\frac12 \cdot (a)^2 \cdot h~\implies~\frac{a^2}{h^2} \cdot h'^2 \cdot h'=\frac12 \cdot (a)^2 \cdot h$

2. Collect like terms and multiply through by $h^2$

$a^2 \cdot h'^3=\frac12 \cdot (a)^2 \cdot h^3$

3. Divide through by $a^2$

$h'^3=\frac12 \cdot h^3$

4. Now calculate the 3rd root as shown in the last line of my previous post.

7. ## Re: Hight of pyramid, at half it's volume?

Thanks again. Now even I got it :-)

Mike