There are competing effects at work. But no, the surface area is not a good gauge of the pressure differential. Paradoxically, increasing surface area is compatible with decreasing pressure differential.Swamp Thing said:Isn't it the case that the inflated surface area of the balloon is a measure of the pressure differential? And the area seems to be such that the balloon would normally burst?
Bandersnatch said:I think this calls for a bit lower-level approach:
For a spherical balloon, the tensile stress in the rubber is $$\sigma=\frac{R}{2h}\Delta P$$where h is the present thickness of rubber. For incompressible rubber, h is related to the initial thickness and radius at low pressure by: $$h=\left(\frac{R_0}{R}\right)^2h_0$$hutchphd said:The curvature and tensile strength matter.
For a curved surface the Young-Laplace equation $$\Delta P=-\gamma (\frac 1 R_1 +\frac 1 R_2)$$ gives Pressure differential in terms the principal radii of curvature and the surface tension
Edit: I put in the minus sign
It is and the maximum exterior 'suck' you can get is when ambient pressure is 0Bar. At that stage, the internal air will have expanded somewhat and what started as 1Bar will have reduced considerably so the pressure difference will be a lot less than -1Bar. Simple Boyle's Law would suggest that an increase in internal volume of 4 would produce a pressure difference of 0.25Bar - and so on. At some stage the envelope will be stressed but not enough to rupture it.Swamp Thing said:the differential (guage) pressure between inside and outside the balloon is what matters
The embarrassing question is why a condom should ever be required to stretch that far! Not by me, your honour.kuruman said:The darn thing filled the entire available volume without bursting until