flat bag volume

[Unregistered]

How can I determine the volume a bag will hold given its flat dimensions? Say a sandbag 20 x 36"? Is there a formula to estimate volumes of flat bags?

[Robert Fogt]

You would need 3 dimensions to calculate volume.

volume = length * width * height

If each dimension is in inches, the volume will be in cubic inches.

[Robert Fogt]

This seems to be a good time to mention a new calculator I added to the site. It calculates the volume of a rectangular object.

http://www.onlineconversion.com/object_volume_box.htm

[rahmesa@rediffmail.com]

[QUOTE=Robert Fogt]what is the formula for calculating polythene bag volume and the conversion of area to volume

[Robert Fogt]

The conversion for area to volume is:
volume = area * height

To calculate the volume of a bag you need:
volume = length * width * height

[drcruzr]

With all due respect, Robert Fogt does not seem to have the imagination needed to generate a formula to find the volume of a flat bag. Perhaps he never put a sandwich in a baggie. We know you need 3 dimensions to generate a volume. But flat bags are defined by 2 dimensions only. Nonetheless, they get filled all the time. If it was easy, I’d do it myself, but the math gets pretty complex. Let’s start by defining terms. Bags are usually given as Length (L) by Width (W), with the opening on the W side. That makes the circumference (C) of the opening 2W. The diameter of that opening, 2W/Pi, would be the maximum depth of our flat bag. We could use L x 2W/Pi, but both ends are flat. I suspect the formula would be something like (L/W x constant (K)) x 2W/Pi x L. We could, I suppose, fill up a bunch of various sizes of bags and see if a pattern emerges, but a mathematical formula seems much cleaner, don’t you think? Any thoughts? Anyone?

[drcruzr]

How about this: for the case where the length of the bag is longer than the width, so the center can go to a cylinder I'm postulating this formula:
V = ((Wsquared/Pi) x (L-W)) + (1/2 or 1/4 or 1/8 x Wcubed/Pi).

[gubment_cheez]

How about this: for the case where the length of the bag is longer than the width, so the center can go to a cylinder I'm postulating this formula:
V = ((Wsquared/Pi) x (L-W)) + (1/2 or 1/4 or 1/8 x Wcubed/Pi).
so lets review.
((W2/pi)*(L-W))+([.5|.25|.125]*W3/pi) = V
and lets take a bag that is 20W x 20L those are nice even measurements
((202/pi)*(20-20))+([.5|.25|.125]*203/pi)
(127.32395447351626861510701069801 * 0) + (.5|.25|.125]*203/pi)
0 + 1273.2395447351626861510701069801
oooor
0 + 636.61977236758134307553505349006
oooor
0 + 318.30988618379067153776752674503

interesting

[Unregistered]

What if you wanted to know the volume at any time not just the maximum volume?

[drcruzr]

I hate being wrong, but have to admit it when I am. My first shot at a formula to estimate the volume of a flat bag seriously UNDERestimated the actual values I was able to obtain experimentally. I still think it should be close for very long bags, but for bags where the length and width are at all close, it's not too good. I'm still open to suggestions. Does anyone know people who work with flat bags? They must know something we don't...

[gubment_cheez]

it would be even worse for bags that were wider than they were long (those half-ziplock baggies, for example, although in this case you could probably substitute width for height and vice versa)

[Unregistered]

What does the 1/2, 1/4 or 1/8 depend on?

[Unregistered]

I derived a formula based on drcruzr's idea of cylinder approximation. Two assumptions:
1. Circumference of cylinder = 2w and
2. Area of cylinder = area of bag. This will overestimate the actual volume of the bag since some of the bag material is wasted due to crumpling at the corners.

Volume = Wsquared/Pi * (L - W/Pi)
instead of...
V = ((Wsquared/Pi) x (L-W)) + (1/2 or 1/4 or 1/8 x Wcubed/Pi).

I found an article on wikipedia about what's called the "paper bag problem." It gives an approximate formula. I have a 29 x 17 cm bag. (drum roll) Here are the results.

Cylinder approx: 2.17 L
Wikipedia formula: 1.98 L
Actual volume of bag: 2.15 +/- 0.05 L

I'll be doing this experiment again to confirm this. I use bags at work that are made out of a plastic that does not stretch significantly. Also, I sometimes use square bags which presumably would not follow this formula as well.
-SK

[Unregistered]

This is an interesting question. After reading everyone's answers and thinking of bags of seed, I would say a bag would have the cylinder shape except for the top inch and bottom inch for a 50 pound bag of seed.

You would really have to pack in the sand to get the cylinder shape and you may not get the top sewed shut.

When the bag is laying on it's side, it's base area appears as an oval excluding the top and bottom inch.

RB

[Unregistered]

I derived a formula based on drcruzr's idea of cylinder approximation. Two assumptions:
1. Circumference of cylinder = 2w and
2. Area of cylinder = area of bag. This will overestimate the actual volume of the bag since some of the bag material is wasted due to crumpling at the corners.

Volume = Wsquared/Pi * (L - W/Pi)
instead of...
V = ((Wsquared/Pi) x (L-W)) + (1/2 or 1/4 or 1/8 x Wcubed/Pi).

I found an article on wikipedia about what's called the "paper bag problem." It gives an approximate formula. I have a 29 x 17 cm bag. (drum roll) Here are the results.

Cylinder approx: 2.17 L
Wikipedia formula: 1.98 L
Actual volume of bag: 2.15 +/- 0.05 L

I'll be doing this experiment again to confirm this. I use bags at work that are made out of a plastic that does not stretch significantly. Also, I sometimes use square bags which presumably would not follow this formula as well.
-SK

Can you post how you got the Wiki equation (en[dot]wikipedia[dot]org/wiki/Paper_bag_problem)to work? I keep getting 2319.0673136863, and I don't know if that's cubic liters or what!

[JohnS]

Can you post how you got the Wiki equation (en[dot]wikipedia[dot]org/wiki/Paper_bag_problem)to work? I keep getting 2319.0673136863, and I don't know if that's cubic liters or what!

The Wiki formula gives you the cubic form of whatever length units you use. If centimeters, then cm³. You will separately have to do any conversion to liters ( 1 L = 1000 cm³).

[Unregistered]

Thanks John. Just what I need to get the formula embedded into code!

[Unregistered]

Short answer: numerical modeling (oh, the irony!). Long answer: I'm going to assume not to care about actually closing the bag at first, or that gravity would distort its shape during filling. Since a bag is most full when it is most round, and most bags are thin sheets, the bag cross-section at the opening is a circle, and a line or extreme oval at the bottom. Thus the area or cross-section at any distance along the bag is going to be an oval, so the area equation to use is that of an oval; it necessarily includes the area of a circle. However, the area isn't constant all the way down the bag since clearly the bottom of the bag essentially has no area. What is constant is the circumference of the particular area. Therefore, the area will be tied to the circumference of the oval, which is constant. The task is then to integrate along the length of the bag all of the infinitesimal volume slices starting with oval axes which are equal (a circle) and continuing by decrementing one axis to zero (a line) along the way (while keeping the other constant at the bag width). You could imagine that the decrementing of the axis length could be varied to obtain different cross-sections combinations, like holding the bag in a circular cross-section for a finite distance. In such cases, the integration would be done along sections of the bag length, thereby segmenting the volume integration. However, the more cylinder-like the bag is forced to be, the more it will deviate from an oval cross-section at the bottom. To see this, take a bag and stuff the widest cylinder you can in it and observe what form the bottom takes: unfilled triangles. No, I'm not going to bother with the numerical analysis of this unless I am bed-ridden for some time.

[Unregistered]

Even though this thread is dead, Since I found this via google search, I thought a usefull response would be heplful. Perry's Chemical Engineers' Handbook has exactly what you are looking for. There are details sufficnet that you can determine how much of any material you can fit in a bag with know flat dimensions. Or you can calculate the size bag you need based on how much you want it to contain.

[mpatel]

How can I determine the volume a bag will hold given its flat dimensions? Say a sandbag 25 x 35"? Is there a formula to estimate volumes of flat bags?[/QUOTE]

[IPT Packaging Lab, Brazil]

We made such experimental study for bags that are closed by open mouth constriction, as garbage bags. Perry´s Handbook does not have this kind of bag.
Our results were published in the ABNT NBR9191, the brazilian national standard for garbage bags specification. It shows this relation to eight bag sizes ranging from 39cm x 58cm (= 15 liters) to 115cm x 115cm (=240 liters). The volume for other sizes can be inferred from those by interpolation or extrapolation. [embalab@ipt.br]

[Unregistered]

If we need to find the contact surface area, should we just use:
vol/area? where area = L xW?

[JohnS]

If we need to find the contact surface area, should we just use:
vol/area? where area = L xW?

Basically, on either inside or outside there are two LxW surfaces, when used as a bag.

[alpal]

I found this online while looking for the same answer to: How to calculate the volume of a "pillow bag" with known length and width.

technikpackagingdotcom
To calculate the product volume for a pillow bag with a given bag width and cutoff length the following formula applies:

V=(CSquared/4pi) x H x 0.35

V = Product Volume
C = Circumference of bag = 2 x bag width
H = Cutoff length (Height) of bag
0.35 = 35% usable bag volume

However, try seperating a gallon size zip lock back full of packing peanuts into quart size zip lock bags. You will only fill up three quart size bags with the contents of the gallon size bag.....Usable bag volume seems to vary with dimensions especially with varying densities of product.

[Unregistered]

You would need 3 dimensions to calculate volume.

volume = length * width * height

If each dimension is in inches, the volume will be in cubic inches.

area of a carry bag which dimensions is as 15"*12"*5"

[Unregistered 1]

Hi All,

My theory may be overly simple but in the few test that I have done it seems to be fairly accurate.

Given that I have a bag, that is 50cm x 40cm, the area is 0.2m.
The bag when measured held 20 litres.......

This has worked with all other sized bags,

please note this is with all 4 sides closed, and filled through a hole that is sealed, and the hole is at the top.

Cheers

[Unregistered]

Thought this might be helpful:

To calculate the product volume for a pillow bag with a given bag width and cutoff length the following formula applies:

V = (c^2/4pi) x H x .35

V = Product Volume (cm³)

C = Circumference of bag = 2 x bag width (cm)

H = Cutoff length (Height) of bag (cm)

0.35 = 35% usable bag volume

technikpackaging(dot)com/Bag_Sizing.doc

[Unregistered]

I thought this might help anyone trying to validate new/different equations: I went on a site called sandbagstore.com and clicked on some of their offerings. In the description, they say that a 14x26 bag holds approximately 50 lbs. I did a little more digging and found that 864 cubic in.= 50 lbs of sand

Hope this helps save someone the time/effort of measuring large amounts of sand....
ko

[Unregistered]

I asked this question once with a group of friends of mine. The reference was from a written reference of "a cloth sack that 2 feet wide and 4 feet high." After debating what the volume might be for quite some time we called another friend in to help us out. The friend in question is has a Ph.D in physics.

His response got us looking in a totally different direction. He said that the surface area of the sack (4x4' or 16') used as the given surface area of a sphere would find a very close approximate of the volume of a CLOSED sack (like a ziplock bag).

We decided to put the theory to the test using a quart-sized ziplock bag. The test was done using a syringe (poking 2 holes right next to each other near the seal of the bag). We filled the bag using water through the syringe and got it as close to full as we possibly could. We then emptied the bag into a beaker and measured the volume of water. Based on the volume of water we compared it to the math we did using the surface area of the bag (not including the lip above the seal).

The results were very close. The sphere, mathematically, has a slightly larger volume than the sack. We figured that the result would be even closer once gravity and more precise measurements were accounted for.

Although we wrote everything down while doing this test I no longer have all the notes we took as it was a couple years ago. For anyone that is interested this formula may be of help.

X= (WxL)*2, Surface area of the flat, empty sack. Then:
R= √(X/(4 x 3.14)), The radius of a sphere. Then:
V= 4/3*3.14*R³, The volume of a sphere (and thus the volume of the sack when closed)