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How big would a hollow rigid object need to be to float, (not in water but in air) if all of the air was vacuumed out and the container sealed?

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3 Answers 3

up vote 3 down vote accepted

Another way to look at this is to solve for what weight/area you are allowed to use. Assume a spherical craft. Surface area goes up in proportion to r^2. Volume (which will generate your lift) is proportional to r^3. If you do the math, you find that you are allowed to use (r x 0.4)kg/m^2, or about (r x 0.88) lbs/m^2. So the problem gets easier to solve as the craft gets bigger.

The force experienced by a totally evacuated body at sea level, would be about 14.7 psi, which is about 11.4 ~TONS~ per m^2. (Don't you just love mixing metric and English units? I do...)

At r = 1m, you can only use 0.88 lbs of material for skin and structure to hold back that massive force (good luck with that).

At r = 10m, you get 8.8 lbs/m^2. That equates to a (1/17) inch thick aluminum skin and no internal supporting structure. ---> Still a no-go.

At r = 100m, you are allowed 88 lbs/m^2. ... O.K., now we are talking. 1/8 inch aluminum plate weighs about 19lb./m^2. That leaves 69 lbs./m^2 for ridigization of the skin panels and internal structure. Something that approximated a monocoque structure (like an egg shell: able to take tremendous pressure with no deep internal bracing) might work. As an engineer, I'd be willing to take on that challenge; but... that is a H-U-G-E craft! OVER two football field lengths in diameter, and weighing in at about 5 million kg before evacuation!

Maybe someone would be able to do it for r = 50m, which would be a budget of 44lbs./m^3. My guess is that technical feasibility lies somewhere in the design area of 50m < r < 100m.

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Since air weighs about 1.2 kilograms per cubic meter, if you could make your Magdeburg hemispheres strong enough and they enclosed one cubic meter and weighed less than 1.2 kilograms, they ought to float like a helium balloon

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The big problem is making the container strong enough. It has to withstand that 15 pounds per square inch air pressure and not collapse. –  Jim Graber Apr 15 '11 at 1:31
    
you can edit your answer to add additional details rather than adding comments –  Nick T Apr 15 '11 at 2:36

Just to flesh out Jim's answer.

Assume you made a cube (makes the math easier!) 1m x 1m x 1m
Aluminium has a density of 2700kg/m^3
And you need the cubeto have a mass of less than 1.2Kg
- so can use (1.2/2700) m^3 of material

The surface area is 6m^2 then you can have a thickness of (1.2/2700)m^3 / 6m^2 = 70um
Or about half the thickness of a hair!

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