Stumbled upon this paper in relation to Vacuum Balloons:
Paper
Specifically:
"The maximum lift attainable from a vacuum balloon is equal to the
weight of the displaced air. This is the maximum theoretical lift (MTL); in
real cases, the lift will be reduced by the weight of the structure and that of
any residual air (no real "vacuum" is perfect), and actual calculations will
need to allow for non-spherical structures, variations in atmospheric pressure
with the weather, altitude and so on. Nevertheless, it is clear that it is feasible
to build vacuum balloons in which a large fraction of the maximum theoretical
lift (MTL) can be obtained.
Above I have shown that the MTL for a sphere of radius 2.5m would be
around 80kg. Doubling the radius to 5m multiplies the MTL by 8, to a value
of 640kg. With a radius of l0m, the MTL is already over 5 tonnes.
Finally, consider the Graf Zeppelin II, the largest airship built. This ship
had a rigid metal framework and was 803 ft. long, about 250m. A vacuum
balloon with a diameter of 250m would have a MTL of over 8000 tonnes,
comparable to an ocean freighter, and certainly well above the toy range"
8000 tonnes of MTL from a 250M spherical vacuum balloon in a perfect scenario still leaves ample potential for a workable MTL to weight ratio in a
none spherical object made up of super light/super strong materials in the real world.
If anyone has the time this might be a fun math problem:
If doubling the spherical radius of a Vacuum Balloon multiplies the MTL by a factor of 8.
What is the size to MTL ratio calculation in a 3 dimensional triangle and at what size would it become "useful" for Humans lets say 10 tonnes for
arguments sake?
(Not phrased very well- dont even know how to describe a triangular delta shape!)
edit on 12-9-2014 by Jukiodone because: (no reason given)
edit on 12-9-2014 by Jukiodone because: (no reason given)