# How much lift does the average latex helium filled party balloon produce?

How much lift does the average helium filled party balloon produce? (not including any extras like ribbon string)

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The mass difference of the air it would have contained and the Helium it does = the volume of the balloon multiplied by the difference in density of the helium and air.

Suppose the balloon is spherical and 12" in diameter (physicists can only do the arithmetic for spherical objects, and preferably in a vacuum) .

That gives it a volume of $\frac43 \pi r^3$ so annoyingly mixing units, $\frac{4}{3} \pi\, 0.15\:\mathrm{m^3} = 0.014\:\mathrm{m^3}$

Air has a density of $1.2\: \mathrm{kg \cdot m^3}$ at room temperature and pressure and Helium $0.176\: \mathrm{kg \cdot m^3}$ So your balloon has a lifting capacity of $0.014 \cdot (1.2-0.176) = 0.014\:\mathrm{kg}$

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"(physicist's can only do the arithmatic for spherical objects, and preferably in a vacuum)" This is only true in the sense that you have to do calculus to do it. (Meaning correct that it's not arithmetic) If you wanted to do that you would generate a model with a differential equation, integrate over the curve of the line (Times two, if you're balloon is symmetrical (A reasonable assumption, assuming uniform density of the latex)) and then do what the previous answer said. Do not walk away thinking there is something so simple that cannot be calculated by a physicist! And I'm only an engineer –  user20587 Feb 4 '13 at 9:56
The helium is under higher pressure than the air, being squeezed by the rubber, but this probably makes a negligible difference. OTOH, I'm not so sure it's OK to neglect the weight of the rubber itself. Is it much less than 14g? I don't know. –  Steve B Feb 5 '13 at 15:38
Don't you have to subtract the mass of the balloon skin itself? –  b_jonas Dec 7 '13 at 10:10
@b_jonas - the question is how much lift. But yes you need to subtract the envelope from the "useful" payload –  Martin Beckett Dec 7 '13 at 20:21