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This question already has an answer here:

Reading this question reminded me of an idea I once had, and I wonder whether it's feasible.

Imagine a hollow ball made of something strong, like steel. Imagine there is nothing inside; only vacuum.

Is there any size of ball and any available material such that, if you made a ball like this, it would float like a helium balloon? In other words, that the mass of the ball itself plus the (0) mass of the vacuum would be less than the mass of the air it displaces, making it bouyant?

I suspect there is no material strong and light enough, but I don't know how to do the calculations.

(It's fun to imagine these being used in fantastical bridges or airships, though I suppose a bridge would be unwise; you wouldn't want it to rise or sink when the weather changes.)

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marked as duplicate by DavePhD, BMS, Kyle Kanos, JamalS, Brandon Enright Nov 19 '14 at 19:47

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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According to the research paper linked here: http://www.researchgate.net/profile/Weicheng_Cui/publication/222221948_An_overview_of_buckling_and_ultimate_strength_of_spherical_pressure_hull_under_external_pressure/links/53f1a2950cf26b9b7dd0da3c

The pressure difference which can be held by a sphere of any particular material is a function of (t/R), where t is the thickness and R the radius.

In other words the thickness of the material required to withstand a given pressure (here 1 atmosphere) is proportional to the radius of the sphere.

Combined with the fact that the surface area grows with the square of the radius, the mass of the sphere will be proportional to the radius cubed.

The mass of the air missing from the sphere - and hence the lift - is also proportional to radius cubed.

So if you find a material which is strong enough and light enough to work, it will work equally well on all scales.

The yield strength equation in the above link gives a linear relation between the yield strength of the material, the thickness-radius ratio and the withstandable pressure. And the calculation of the mass of the sphere will provide an inverse relation between the thickness-radius ratio and the density of the material. Combining these equations should give a minimum ratio of yield strength to density for a material capable of making a vacuum balloon. I haven't completed the calculation but I expect steel's ratio will be an order of magnitude too small and that exotic materials will be required.

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An interesting question, but I think this would be absolutely impossible for the following reason.

Vacuum chambers, particularly large ones, need thick strong walls to prevent air pressure making them collapse. The larger the volume the greater the problem.

Pressure = force over area; so force = area times pressure.

$F=pA$

For area of $10\ \mathrm{m^2}$, force is equal to $1\,000\,000\ \mathrm N$ equivalent to the weight of a mass of $100\,000\ \mathrm{kg} = 100\ \mathrm t$

So the bigger you make the vacuum baloon the worse the problem with air pressure pushing on the walls needing the walls to get thicker.

Normally with hot air ballon/airship/zepplin/blimp etc. the larger you make them the more lift you get until you can lift whatever you need to. From IanF1's excellent answer it is clear that there is no benefit to increase the volume for this type of baloon.

Further edit after good comment from IanF1. If the structure is lighters than the air it displaces, then the larger the structure the more lift it will generate and the more it will be able to carry. My point above is that for a normal ballon the weight of the container will be proportional to the area, but the lift will be proportional to the volume. So as the volume is increase the ratio of the lift due to the volume of hot air or helium to the weight of the container will get larger and larger, whereas for the vacuum filled ballon it will remain constant.

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    $\begingroup$ You make a good point - making a balloon larger won't increase its ability to lift itself, but (as long as it can lift itself) it will increase how much other stuff it can lift. $\endgroup$ – IanF1 Nov 19 '14 at 18:31
  • $\begingroup$ @IanF1 - that is a good point too! - I will edit. $\endgroup$ – tom Nov 19 '14 at 21:09

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