How to calculate the volume of plastic bags

[JenniferMurphy]

This is really more of a math (geometry) question, but a lot of people here are pretty good at math, so I'm hoping...

I am trying to calculate the volume of various plastic bags used to line trash bins. I know how to calculate the volume of regular solids like cylinders, cubes, rectangular solids, and cones. I don't think these bags are not any of those shapes. They typically come in rolls so they lay out flat before being opened. When opened, the top assumes the shape of the bin, usually a circle or a rectangle. The bottom remains sealed. My first thought was that the cross-sectional area would go from the maximum at the top down to zero at the bottom, so I'd just need to average that over the entire height.

In one roll of bags that I have now that claim to hold 2.6 gallons, each bag is 16" wide and 17" deep. Here's the table of calculations:

Kitchen Scrap Bins.xlsx
	C	D	E	F
5	Value	Units	Quantity	Formula
6	16.00	in	Width of bag	C6: 16
7	17.00	in	Height of bag	C7: 17
8	32.00	in	Circumference	C8: =C6*2
9	10.19	in	Diameter at top	C9: =C8/PI()
10	5.09	in	Radius at top	C10: =C9/2
11	81.49	in sq	Area at top	C11: =PI()*(C10^2)
12	0.00	in sq	Area at bottom	C12: 0
13	40.74	in sq	Average area	C13: =(C11+C12)/2
14	692.64	in cu	Volume in in^3	C14: =C13*C7
15	3.00	gal	Volume in gallons	C15: =CONVERT(C14,"in^3","gal")
Sheet6

Cell Formulas
Range	Formula
C8	C8	=C6*2
C9	C9	=C8/PI()
C10	C10	=C9/2
C11	C11	=PI()*(C10^2)
C13	C13	=(C11+C12)/2
C14	C14	=C13*C7
C15	C15	=CONVERT(C14,"in^3","gal")

My calculations came up with 3.0 gallons, which is not far off from 2.6.

But then I got to thinking. These are very flimsy plastic bags. They are not hard plastic that goes from a circle at the top to a line at the bottom. When filled with water or pretty much anything else, they will bulge out to the maximum cross-sectional area almost all of the way to the bottom, unless stopped by the side of the bin. So the maximum volume would be close to double the above calculation, or almost 6.0 gallons, which is more that twice what is stated on the box.

Can anyone shed any light on this?

Tomorrow, I'm going to see if I can find a bin that is close to the maximum area at the top and then fill a bag with water to see just what it can hold.

 

[jorismoerings]

Hi,
Take a look at this calculation we're using. Be aware this is a theoretical volume calculation which doesn't include parameters for the strength of the bag or does take into account how stretch the material of the bag has. The calculation does take into account a certain amount of wasted space of the bag at the top as there would need to be a way to seal or close the bag.

Book1
	A	B	C
1	Conversion table
2	Liter to US Gallon	1	0,264172
3	CM to Inch	1	0,393701
4
5	Volume Calculation
6		Inch	--> CM
7	Width	16	40,64
8	Length	17	43,18
9	Closure space	2	5,08
10
11
12		Liters	US Gallon
13	Volume	9,347367	2,469313
Sheet1

Cell Formulas
Range	Formula
C7	C7	=B7/C3
C8	C8	=B8/C3
C9	C9	=B9/C3
B13	B13	=((C7^2/PI())*(C8-(C7/2)-C9))/1000
C13	C13	=B13*C2

The calculation comes close to the 2.6 gallons as stated on the box.

 

[JenniferMurphy]

Hi,
Take a look at this calculation we're using. Be aware this is a theoretical volume calculation which doesn't include parameters for the strength of the bag or does take into account how stretch the material of the bag has. The calculation does take into account a certain amount of wasted space of the bag at the top as there would need to be a way to seal or close the bag.

The calculation comes close to the 2.6 gallons as stated on the box.

Wow! That is close.

I can't follow the formula, which is,

Can you explain the derivation?

Here's my implementation. If I reduce the closure to 1.62", I get 2.6 exactly.

Kitchen Scrap Bins.xlsx
	B	C	D	E	F
4	R/C	C	D	E	F
5	5	Value	Units	Quantity	Formula
6	6	16.00	in	Width of bag	C6: 16
7	7	17.00	in	Length of bag	C7: 17
8	8	1.62	in	Closure space	C8: 2
9	9	32.00	in	Circumference	C9: =Wid*2
10	10	10.19	in	Diameter at top	C10: =Cir/PI()
11	11	5.09	in	Radius at top	C11: =Dia/2
12	12	81.49	in sq	Area at top	C12: =PI()*(Rad^2)
13	13	0.00	in sq	Area at bottom	C13: 0
14	14	40.74	in sq	Average area	C14: =(AreaTop+AreaBot)/2
15	15	601.38	in cu	Volume in in^3	C15: = ((Wid^2)/PI()) * (Len - (Wid/2) - Close)
16	16	2.60	gal	Volume in gallons	C16: =CONVERT(VolCubIn,"in^3","gal")
Bag Vol 2

Cell Formulas
Range	Formula
C4:F4	C4	=col()
C9	C9	=Wid*2
C10	C10	=Cir/PI()
C11	C11	=Dia/2
C12	C12	=PI()*(Rad^2)
B5:B16	B5	=ROW()
C14	C14	=(AreaTop+AreaBot)/2
C15	C15	= ((Wid^2)/PI()) * (Len - (Wid/2) - Close)
C16	C16	=CONVERT(VolCubIn,"in^3","gal")

Named Ranges
Name	Refers To	Cells
'Bag Vol 2'!AreaBot	='Bag Vol 2'!$C$13	C14
'Bag Vol 2'!AreaTop	='Bag Vol 2'!$C$12	C14
'Bag Vol 2'!Cir	='Bag Vol 2'!$C$9	C10
'Bag Vol 2'!Close	='Bag Vol 2'!$C$8	C15
'Bag Vol 2'!Dia	='Bag Vol 2'!$C$10	C11
'Bag Vol 2'!Len	='Bag Vol 2'!$C$7	C15
'Bag Vol 2'!Rad	='Bag Vol 2'!$C$11	C12
'Bag Vol 2'!VolCubIn	='Bag Vol 2'!$C$15	C16
'Bag Vol 2'!Wid	='Bag Vol 2'!$C$6	C15, C9

 

[JenniferMurphy]

I did my best to fill one of the 16x17 bags with water. I used a 4-cup measuring cup. I was able to pour 9 4-cup loads into the bag before I couldn't keep it from spilling. That's 36 cups (9x4), which is 2.25 gallons (9x4/16). That pretty much in line with 2.6 gallon cvapacity on the box.

The shape of the bag when filled with water was definitely not a cylinder. It was more like a squashed sphere, which, I just learned, is called an oblate sphere.

The formula for the volume of an oblate sphere is,

As best as I could measure, the dimensions of my bag filled with water were about 11 x 6.

That would make the volume about 1.65 gal. But I could not fill it all the way up. I had 2-3 inches of bag in my fingers. Herfe are a few possible measurements.

And the bag was not really a perfect poblate sphere. It was wider at the bottom than at the top, maybe something like this:

All in all, the 2.6 volume claim is probably correct.