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wayman

Volume Of Air In Bellows Of Given Measurement (Math Intensive)

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Posted (edited)

Being curious about how much greater the volume of air in the expanded bellows on a 7-inch concertina is compared with the bellows on a 6-1/4-inch concertina, I tried to come up with the mathematical formula and see whether there was a really fast way to estimate / calculate this. I would love it if someone similarly math-inclined on a Monday morning could look at this and confirm or correct the below.

 

I assume the number of folds is the same for the bellows of the two hexagonal concertinas being compared.

 

I assume that volume reduction due to bellows folds for any given bellows (as a percent of the volume with no folds) is roughly the same as for any other bellows (to the point of being unimportant here) because to do otherwise would be even more folly than this already is.

 

And this is specific to hexagonal concertinas. Those with Aeolas or Edeophones will have to offer me significantly more tea before I consider how to reduce that math into an easy comparison algorithm.

 

Then

 

let N = "size of a regular hexagonal bellows, as measured perpendicular to the flat sides"

let S = "length of a side"

let A = "area of a cross-section of the bellows"

let V = "volume of the bellows"

 

 

With the above assumptions

 

V2/V1 ~= A2/A1

 

S = (N * sqrt(3)) /2

A = (3 * sqrt(3) * S^2) /2

 

And thus

 

A = 9/4 * sqrt(3) * N^2

 

So for two concertinas which measure N2 and N1 where N2 = X * N1

 

A2 = X^2 * A1

 

Given the example common bellows sizes N2=7 and N1=6.25

 

X = 1.125

A2 ~= 1.25 * A1 and V2 ~= 1.25 * V1

Edited by wayman

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Way too complicated! But the right answer. The linear measurements are increased by a factor of 7/6.25 = 1.12; so the area measurement is increased by the square of this: 1.12^2 = 1.2544; or about 25%. That's all you need. But as I said in the thread where this was started, if the bellows folds are also deeper then the extension of the bellows will also be increased so the potential volume of air in the bellows will be even more than 25% greater.

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Yes, very simple calculation "all other things being equal". ((Bigger size/smaller size) squared).

 

However, all other things aren't equal because, amongst other things:

 

1) The depth of the bellows folds may vary between instruments.

2) The thickness of the bellows material and hinges may vary between instruments.

3) Not all of the air is useable because there is still air in the bellows when they are fully closed.

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When I was still teaching math this would have made an interesting problem. First I would have like some theoretical justifications as you have done. Then I would encourage the students to develop a system to actually measure the difference without any damage to the instruments.

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When I was still teaching math this would have made an interesting problem. First I would have like some theoretical justifications as you have done. Then I would encourage the students to develop a system to actually measure the difference without any damage to the instruments.

 

Remove the bellows. In their compressed state sitting flat on the bottom of a large cooking pot, carefully see what volume of dried rice fills them. Then repeat while someone holds the bellows open with one end still held down flat in the pot. For more accuracy, perhaps use quinoa or cous-cous :) Then remove the bellows from the pot, add water, simmer, steam some vegetables... ;)

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don't forget to clean the bellows out after or in a fast session I can see rice seeds bouncing off the walls

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don't forget to clean the bellows out after or in a fast session I can see rice seeds bouncing off the walls

 

It's a way to play concertina and shakey-egg at the same time! ^_^

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