Finding the Volume of a Slightly non-Rectangular Box

[LightningInAJar]

Homework Statement

This isn't actually for homework. Just my own reference.

Relevant Equations

I realize a rectangle's volume is measure by multiplying length x width x height.

I have a storage tote that has a larger top than bottom. How do I figure out its volume? Is this like a 3D trapezoid?

Can I measure the volume of the rectangle assuming the top is the same as the bottom, then as if the bottom were the same as the top, then just subtract the two?

 

[berkeman]

Homework Statement:: This isn't actually for homework. Just my own reference.
Relevant Equations:: I realize a rectangle's volume is measure by multiplying length x width x height.

I have a storage tote that has a larger top than bottom. How do I figure out its volume? Is this like a 3D trapezoid?

Can I measure the volume of the rectangle assuming the top is the same as the bottom, then as if the bottom were the same as the top, then just subtract the two?

I would carefully measure the inside dimensions of the tote at the bottom and top and make a drawing of the volume with those dimensions. Then I'd just add the volume of the 8 extra triangular pieces to the volume of the smaller box.

But those totes also have a slight curve to their shape, so to get a more accurate number I'd use a garden hose and a bathroom scale

How accurate do you need this volume number to be, and why?

https://m.media-amazon.com/images/I/51ABxdvMwBL._AC_UL400_.jpg

 

[LightningInAJar]

I would carefully measure the inside dimensions of the tote at the bottom and top and make a drawing of the volume with those dimensions. Then I'd just add the volume of the 8 extra triangular pieces to the volume of the smaller box.

But those totes also have a slight curve to their shape, so to get a more accurate number I'd use a garden hose and a bathroom scale

How accurate do you need this volume number to be, and why?

View attachment 316149
https://m.media-amazon.com/images/I/51ABxdvMwBL._AC_UL400_.jpg

Oh, just curious how accurate the sticker is about its volume and to compare similar bins that claim 1:2 volume differences.

 

[BvU]

Put the thing on a scale and fill it (completely) with water !

 

[Mark44]

Oh, just curious how accurate the sticker is about its volume and to compare similar bins that claim 1:2 volume differences.

What does "1:2 volume differences" mean?

 

[FactChecker]

To get the formula for your original question, we can integrate the delta-height times the width and length at each height.
Let $W_b$ and $L_b$ be the width and length at the bottom, and $W_t$ and $L_t$ be the width and length at the top. Let $H$ be the total height and $h$ be the height at a particular height between $0$ and $H$.
The width at height $h$ is $W_b+(W_t-W_b)h/H$. Likewise, the length is $L_b+(L_t-L_b)h/H$.
So the volume of a horizontal slice of thickness $dh$ at any given height, $h$, is $( W_b+(W_t-W_b)h/H )( L_b+(L_t-L_b)h/H )dh$.
The volume integral is: $\int_{0}^{H} ( W_b+(W_t-W_b)h/H )( L_b+(L_t-L_b)h/H ) \, dh$.
We can multiply this out and group the coefficients of different powers of $h$. (I wish I had Maple!)
Then the volume integral is $Volume = \int_{0}^{H} ( A + (B/H) h + (C/H^2) h^2) \, dh$.
Where
$A=W_b L_b;$
$B=(L_b(W_t-W_b)+W_b(L_t-L_b));$
$C=(W_t-W_b)(L_t-L_b)$.
$Volume = A H + B H/2 + C H/3 = (A+B/2+C/3)H$.

 

[LightningInAJar]

What does "1:2 volume differences" mean?

Meaning one bin claims 20gals and the other 10gals.

 

[256bits]

Meaning one bin claims 20gals and the other 10gals.

Is there a fill level for the bins?
20 gals might mean 19.9, or 20.1, depending on how high you fill with water. ( pick your decimal value )
Same for the 10 gal bins.
You will get a roundabout value, but never accurate to a T.
Perhaps the manufacturer has to compensate for ullage ( sounding ) . Who knows what reg's come up at the oddest time. That air space that you see in bottles - that is ullage.
https://www.container-xchange.com/glossary/ullage/

Plus, using water with a scale, you will have to temperature compensate for the density of water, much like buying fuel in the winter/ summer.

 

[FactChecker]

If you have one of each, fill the smaller one up twice and see if it all fits in the larger one.

 

[berkeman]

Meaning one bin claims 20gals and the other 10gals.

Can you post links to each of these totes please? I'm guessing that you are misreading the specs. You didn't work for NASA on the Mars missions, do you?

 

[FactChecker]

You didn't work for NASA on the Mars missions, do you?

Ouch! That was cold! :-)

 

[hutchphd]

Are these bins similar in shape? Then it is easy.

 

[LightningInAJar]

One bin is 18gal and taller, the other is 10gal and shorter but longer. They are Sterilite brand storage bins so they aren't really for holding water, nor do I want to waste any.

I measured the first at about 16" to 18" long, 12" to 14.5" wide, and about 15" tall compensating for the lid.

The second bin I measured 20.75" long, 14.5" to 16.5" wide, and about 5.25" tall.

I calculated that the smaller bin is maybe only 90% of what it should hold assuming that the 18gal one does in fact hold that much. I think tomorrow I might cheat and draw these shapes out using Sketchup and use the volume tool since the tops and bottoms are slightly different sizes.

 

[FactChecker]

I measured the first at about 16" to 18" long, 12" to 14.5" wide, and about 15" tall compensating for the lid.

Using the equations of post #6, I get that this is 3385 cubic inches.

The second bin I measured 20.75" long, 14.5" to 16.5" wide, and about 5.25" tall.

This is 1688 cubic inches.
The ratio is 1:2.
Water is 231 cubic inches per U.S. liquid gallon. 3385 would be 14.7 gallons and 1688 would be 7.3 gallons.

 

[LightningInAJar]

Using the equations of post #6, I get that this is 3385 cubic inches.

This is 1688 cubic inches.
The ratio is 1:2.
Water is 231 cubic inches per U.S. liquid gallon. 3385 would be 14.7 gallons and 1688 would be 7.3 gallons.

Those are the volumes I got also. Weird that it would be labeled as 10gal and 18gal if one is actually evenly double the other.

 

[hutchphd]

Unless you can provide good error bars for your calculated volumes it is neither weird nor not weird
It is simply an estimate of undetermined quality, and therefore of limited utility.

 

[FactChecker]

Those are the volumes I got also. Weird that it would be labeled as 10gal and 18gal if one is actually evenly double the other.

I also thought that the exact 1:2 ratio was surprising (3385/1688=2.005), but I do think it's largely coincidental. It doesn't surprise me that the volumes are exaggerated as much as possible. That's advertising. I'm sure that their volume is at least up to overflowing.

 

[LightningInAJar]

Well thanks all for the formulas.

 

[Mark44]

I measured the first at about 16" to 18" long, 12" to 14.5" wide, and about 15" tall compensating for the lid.

The second bin I measured 20.75" long, 14.5" to 16.5" wide, and about 5.25" tall.

Missing the shorter length of the second bin.

As an alternative to the formula @FactChecker showed in post #6, you can get a reasonable approximation by multiplying the average of the two lengths, the average of the two widths, and the height. Doing this for the first bin, I get 17 * 13.25 * 15 = 3378.75 cu. in. This is smaller than the value shown by FactChecker by 6.25 cu. in., which represents an error of about 0.2%.

What I'm doing is using the formula for the area of a trapezoid with parallel bases of $b_1$ and $b_2$, and a height of h. The area of such a trapezoid is the height times the average of the two bases, or $A = h\frac{b_1 + b_2}2$.

 

[FactChecker]

Missing the shorter length of the second bin.

I took that to mean that those sides were straight up and the top and bottom measurements were equal.

As an alternative to the formula @FactChecker showed in post #6, you can get a reasonable approximation by multiplying the average of the two lengths, the average of the two widths, and the height. Doing this for the first bin, I get 17 * 13.25 * 15 = 3378.75 cu. in. This is smaller than the value shown by FactChecker by 6.25 cu. in., which represents an error of about 0.2%.

Good point. The linear average is much closer than I would have assumed. I knew that the true equations were nonlinear, so I never thought to compare my answers to the linear average.