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How much difference between internal and external pressure can a balloon withstand?

how would the value be calculated?

I found an equation:

http://library.thinkquest.org/C003758/Function/laplacelaw.htm

But, how does that relate to a material?

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    $\begingroup$ A theoretical spherical balloon of perfectly uniform material is different from a real-world balloon, which is asymmetric and almost certainly has weak spots/imperfections where the first breach will occur. $\endgroup$ – Carl Witthoft Jan 2 '14 at 12:30
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For a spherical balloon (which is a reasonable model to start with) the tension in the material is $\sigma = 0.5 P R /\delta$ where P is the pressure difference between inside and outside, R is the radius, $\delta$ is the thickness. The material will break at some critical value of the tension which is called the tensile strength of the material. For example, for rubber the tensile strength is $\sigma_c$=15e6 Pa $\approx$150 atm, so a rubber balloon with R=1 and $\delta$=1 mm (in the inflated state) can withstand the excess inside pressure $P=2 \sigma_c \delta/R$=0.3 atm. See more info and numerical values of the tensile strength for some common construction materials on http://en.wikipedia.org/wiki/Tensile_strength.

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  • $\begingroup$ You maybe should note that $R$ and $\delta$ are the radius and thickness of the balloon in the inflated state, in which the rubber has deformed elastic. This leads to as decrease in pressure when you inflate the balloon, because the radius changes differently than the thickness when inflating the balloon. $\endgroup$ – fibonatic Jan 3 '14 at 11:13
  • $\begingroup$ Agreed, maybe nylon would be a better example because rubber is so elastic. Or maybe stick with a cylindrical balloon to make sure $R/\delta$ stays constant? $\endgroup$ – Maxim Umansky Jan 3 '14 at 16:42
  • $\begingroup$ Thank you! :-) And this only applies if the material is isotropic correct? $\endgroup$ – user2218665 Jan 5 '14 at 0:54
  • $\begingroup$ For anisotropic situation, if the tensile strength of balloon skin is different for two directions within the plane of the skin (this is the plane where the tension is) then the smaller value will define the break point. $\endgroup$ – Maxim Umansky Jan 5 '14 at 16:07

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