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DISCLAIMER: I'm new to this (specific) site, and I couldn't find any other Stack site that matches with my question.

I'm working on a mobile project and am hanging cardboard "medallions" on it. I am calculating the torque created by each side to balance them out.

However, the medallions are really light, so I can't just weigh them: instead, I plan to find the volume (area calculated using free online tool Sketch and Calc, height measured with ruler) then multiply it by cardboard's density.

But that's where I run into problems: I don't know the density of cardboard. I was wondering if someone could help me out.

I'm referring to the corrugated cardboard with two flat linerboards, with a flute in between them. Specifically, the type of cardboard Amazon uses with its boxes (the type of cardboard I'm using).

I would like the density in kg/(cm^3) (the medallions, along with being light, are also small).

After searching on the internet, I found (corroborated by two sources) that the density of cardboard is from 0.6-0.7 kg/m^3. Does this make sense?

DUO Labs
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    Make 10 then weigh all at once, onthe small electronic scales should work. If not just hang them and adjust their distance from the pivot... – Solar Mike Dec 20 '19 at 22:05
  • @SolarMike Yes, I was going to just adjust by eye, but I have to make a report of the weights used (it's for school). Also, all I have is a big kitchen scale, the ones usually used for weighing a person (and I don't have that much cardboard) – DUO Labs Dec 20 '19 at 22:07
  • Weigh the box first then divide by the total area - even a complete box... – Solar Mike Dec 20 '19 at 22:09
  • @SolarMike An empty box, yes? And don't you mean "divide by volume"? I don't know if I have a complete box though... perhaps, but is there another way if that fails? – DUO Labs Dec 20 '19 at 22:10
  • Well if the thickness is constant then you can work with the area used ie so many grams per cm^2... think of how paper is rated... – Solar Mike Dec 20 '19 at 22:16
  • You need someone *in your market* to weigh an empty Amazon box on a gramscale and tell you the box dimensions. I tried to do it, but my scale is gone, and I'm not in your market. – Harper - Reinstate Monica Dec 20 '19 at 22:45
  • @Harper-ReinstateMonica What do you mean "in my market"? – DUO Labs Dec 20 '19 at 22:53
  • @QuoteDave Cardboard boxes are heavy and bulky and Amazon doesn't ship boxes all over the world. They use boxes made in your local market, e.g. the Spanish warehouses go to Spanish cardboard box makers and buy cardboard boxes there. That has no reflection on Indian or US boxes. – Harper - Reinstate Monica Dec 21 '19 at 00:52
  • It may be my lack of understanding, but I don't see the relevance of the cardboard's density to any measurement of torque. Isn't what you're looking for is the "dynamic weight" of the cardboard? And don't air currents and positioning affect the torque as much, much or more than static weight? – gnicko Dec 21 '19 at 02:49
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    Just balance it with pennies tapped to the back. Do all of your calculations and then make a point about how all of that goes out the window in the real world: *Verify in Field*. – Mazura Dec 21 '19 at 03:20
  • @GregNickoloff Since I can not measure the weight exactly, I will take the density and multiply by the volume (which I can easily find) to find the weight. – DUO Labs Dec 21 '19 at 03:28
  • Yes, but there is enough inaccuracy built in to that to make it rather useless. There are too many variables to cover. Can you weigh a 1 x 1 square of the cardboard and multiply out a "calculated weight" more easily? – gnicko Dec 21 '19 at 03:35
  • @GregNickoloff Like I said: either 1) I don't have enough cardboard, or 2) It'll be too light. – DUO Labs Dec 21 '19 at 03:51
  • @QuoteDave - Unfortunately, if nothing in the equation changes, you have no chance for a solution. Perhaps you could take a cardboard sample to your school and use a balance in one of the science labs to measure the sample. You could also buy a cheap postal scale from an office supply shop and weigh a cardboard sample that way. Then just multiply out a weight for the actual pieces. I think that this will be much more "workable" for you than trying to figure out the density of the cardboard you're using and the volume of the pieces with enough precision to make the trouble worthwhile. – gnicko Dec 21 '19 at 04:01
  • @GregNickoloff I'm actually surprised that this information isn't readily available from popular cardboard manufacturers. – DUO Labs Dec 21 '19 at 04:30
  • @GregNickoloff However, I did find that Amazon does offer that kind of information, so I'm using that as a basis. – DUO Labs Dec 21 '19 at 04:47

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Calculating the density of the cardboard seems like extreme overkill here.

Just take a representative piece of your cardboard to the local postal service retailer (like a FedEx or UPS store) or even to a government post office and ask them to weigh it with their sensitive scale. If they ask why just tell them you may need to know for future shipping purposes.

Jimmy Fix-it
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  • Given that my local UPS is an hour away, I don't think I'll be going there on a whim. – DUO Labs Dec 20 '19 at 23:22
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    OK, if you want to be snarky, scales that weigh down to 1/100th of a gram can be had from Amazon starting at $5, just buy one. Your question didn't indicate that your endeavor is actually just a whim... – Jimmy Fix-it Dec 20 '19 at 23:48
  • Cool, I'll buy one. Thanks! – DUO Labs Dec 20 '19 at 23:49
  • Though searching on the internet, I got that the density of cardboard being between 0.680-0.700 (verified by two sources) kg/m^3. Does this make sense as being accurate? – DUO Labs Dec 21 '19 at 02:00
  • I imagine it would depend, as almost all things do, on the specifics. There's more than one kind of corrugated cardboard. Some are made out of thicker paper, some have bigger flutes in the corrugation, some have different thicknesses of the facing boards. If accuracy is key, I'd not guess. I'd obtain a piece of the desired board, cut a measured piece out of it, measure the thickness, and weigh it. Direct observation exceeds estimation, every time. – Edwin Buck Dec 21 '19 at 03:29