Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have for some time tried to create a formula to guess or find the weight of a roll of aluminum foil wound onto a core. Since my I have not been able to solve this issue I want to ask this question here and hopefully be able to solve it this time.

The initial data I have is the following:

Core Diameter = 76,2 mm
Material Thickness = 108 microns
Roll Length = 751 m
Width = 486 mm

I use a formula to find the outer diameter (converted to cm for this example)

R = \sqrt{L×t/π + r2}
R = \sqrt{75100*0,0108/π + (7,62/2)^2}
R (Outer radius) = 33,0267 cm

Am I using the correct values and if I have these values, can I calculate the weight of a given roll?


After reading the question again i noticed that i forgot to mention a few things in my question. Sorry about that.

The roll is made out of aluminum oxide foil wound onto a core of PVC material which is hollow in the middle. Outer diameter of the core is 91mm but inner diameter (hollow part) is 76.2mm. Weight of the core is 1.64 KG.

John's answer pointed my in the right direction (sadly math is not my strongest area) and found the way to calculate the weight. However i feel like i'm not doing it correctly, partly because i have to remove the core so that the calculation is correct.

Known limitations to the calculation:

  • Thickness is not always correct. Depending on foil type it's from 110µm to 115µm and on a few 85µm (i have a table for that depending on material).
  • Roll can be tightly wound or in rare cases less well wound. This may affect the weight.

I hope that by calculating the weight i can predict with a small margin of error if the roll length or roll width is incorrect (basically catch larger discrepancies)

So my revised question is: Am i trying to solve an impossible scenario? If not, am i on the right track with this or am i missing something in my calculations? Is there some other way of doing this?

The following image is a screenshot of the calculation i made in excel. The data is from an actual roll that has been worked. Length might be a few meters less or more but i know for sure that the weight is correct.

My Calculation

share|cite|improve this question
John's calculation is fine. He is correct the winding doesn't matter. It will just change the roll OD. You can just take the core weight as a given and add it in at the end. Do you believe the density value you have? Presumably the original is $3.95$ g/cm^3 Use it to calculate the foil weight that you should have and compare with the actual. – Ross Millikan Dec 10 '13 at 22:27
His formula is good. By using his formula i get 172.59 kg. which leaves me to believe that the density is wrong. To match the correct weight the density has to be around 2.5g/cm3. In this example the roll is used to produce capacitors so the foil is etched and later formed by running it through a current and is bathed in chemicals to hold a certain voltage. The capacitance is measured in µF/cm2, can this be a factor in the density? – dansige Dec 10 '13 at 23:29
Etching can make the surface rough, removing material, and reducing the effective density. In reality the thickness is less than you think. The density you are using of $3.95$ is correct for crystalline aluminum oxide, but etched material often has pores that reduce it considerably. This makes it not a reliable calculation. – Ross Millikan Dec 11 '13 at 0:40
up vote 3 down vote accepted

I don't think the roll matters.

The mass will be $m = \rho V = \rho l w t$, where $\rho$ is the density of the aluminum foil. ($V$ is volume, $l$ length, $w$ width, and $t$ thickness.)

You have the length, width, and thickness already.

To get the appropriate density, you may need to take an appropriate sampling of the foil and measure the mass. If there's variability in there -- thickness, porosity, roughness, etc. -- then you'll need to take enough samples of sufficient size, from appropriately differently places from the large roll, to bring the measurement error down to something you can live with.

share|cite|improve this answer
In practice, the density of the foil as rolled will be lower than the standard density of aluminum. This uses the known values, which should give you the best result. If you use the rolled dimensions you need to estimate the packing density. – Ross Millikan Dec 6 '13 at 23:24
Thank you John. When i re-read my question i noticed i did not describe my problem correctly. But using your answer i managed to solve my problem in a way. I will edit my question later today. – dansige Dec 10 '13 at 17:24
My pleasure. I added some text about measuring the density of the foil. My assumption is that you are making this wrapped core yourself and that you have long sheets of aluminum to play with. If not, then maybe the manufacturer can give you tolerances. (My background is experimental physics, so I vaguely remember doing some of these things. :) ) – John Dec 11 '13 at 20:11
Your assumption is correct, although i'm not the guy who is wrapping it to the core so yes if i need to i can cut some samples. Since i have a few specs for the different foil types i can get the orginal thickness and more. So i think that if i can calculate the density loss due to process (based on the solution and how long it has to work on the foil) i could probably find a way to finish the calculation. But i think i need to ask more people at work :) – dansige Dec 12 '13 at 0:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.