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- Thread starter mascasa002
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it is not a constant

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as volume goes down, pressure goes up, hence the non-constant comment from Curl

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sophiecentaur

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Starting with 1 atmosphere in the unloaded strut, if the area of the cylinder is A (in m

It doesn't follow Hooke's Law, of course, but you can predict the force needed for a given displacement. Actually, because of the Law involved, you can measure a bigger range of forces using an 'air spring' than a steel coiled one.

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nasu

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If you have a gas of volume V and want to compress it by [tex]\Delta V[/tex], you need an increase of pressure

[tex]\Delta p = -B \frac{\Delta V}{V}[/tex]

B can be found for both isothermal or adiabatic processes and for air is of the order of [tex]10^{-5}[/tex] Pa.

If you apply this to a cylinder of length L and area A,

then

[tex] \Delta p = F A[/tex]

and

[tex] \frac{\Delta V}{V}=\frac{\Delta L}{L}[/tex]

Then

[tex] \frac{F}{\Delta L}=- \frac{B A}{L}[/tex]

and assuming a constant B for small compressions, you could say that the term on the right hand side is the elastic constant of the cylinder.

It depends on the geometry too, not only on the properties of the material. Same as for a real spring.

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