# Why don't towers use buoyancy to reduce weight?

Instead of using steel and concrete to build towers why not use mostly light weight, high tensile materials and then suspend the four corners from four giant helium balloons?

The four main cables would be anchored to the ground at the bottom, and held apart at the top by the force of the balloons pushing off each other, as they are pulled inward by the tension required to suspend each floor. So the four corner cables would essentially form a parabola, like the top cables of a suspension bridge, except vertically.

Could a skyscraper be constructed in this way?

How big would the balloons need to be to suspend the building?

• What would happen in the case of a leak? Structures like this are designed to fail slowly, rather than catastrophicly where possible... 🐈 – Jonathan R Swift Mar 1 '18 at 7:10
• What about the other users of the airspace above : helicopter landing pads for emergency etc... – Solar Mike Mar 1 '18 at 7:13
• You are a funny FUNNY guy. First windy day... . – Carl Witthoft Mar 1 '18 at 18:50
• What high tensile material did you have in mind? – paparazzo Mar 7 '18 at 14:38
• Steel cable was what came to mind, like a suspension bridge. Maybe Kevlar checker plates for the floor surface. Unless you could think of better materials. – Lorry Laurence mcLarry Mar 8 '18 at 11:13

The engineering design of towers is not limited in a practical sense by their weight, meaning that the scheme you propose solves no problems in that field. Truly gigantic balloons would be necessary and the cost of a "floating tower" suspended from them would be huge- probably dominated by the price of the helium necessary to fill them and keep them full in the face of leakage.

I would also not want to be in one of them when a windstorm hits because large lighter-than-air structures respond badly to wind shear and turbulence.

I think there are multiple aspects to this question. If I'm not mistaken, your concept would basically mean, that every floor is anchored to those 4 cables in the corners, thus basically "floating" in the air? As a consequence, the upward force generated by those balloons would be (greater) equal to the weight of the floors. Something like this:

In this graphic I have neglected the tension due to the dead load of the cables.

stability and safety

Probably the biggest problem aside from statics would be the stability and safety of the structure; leaks, failure of a balloon, wind, etc.

statics

In normal conditions, Helium has a density of $\rho_{He}\approx0.179kg/m^3$, while air's is around $\rho_{air}\approx1.29kg/m^3$. If we neglect the weight of the hull of the balloon as well, we can come up with a net buoyancy force $f_B$ per cubic meter of Helium. $$f_{B}=(\rho_{air}-\rho_{He})g=10.9N/m^3_{He}$$

If we take a $10m\times10m$ concrete floor, with a thickness of $t=0.2m$, and a density $\rho_c=2'500kg/m^3$, you would already need a balloon volume $$V_{He}\approx45'000m^3$$ which is a sphere of radius $r=22m$. An that is for one single floor.

availability of helium

Though being the most abundant element in the universe, helium is relatively rare on the earth. Currently, most helium is extracted from natural-gas, but those fossil fuel resources are limited. I'm not sure whether we would want to risk losing such a precious resource.

• Oh, no problem (He shortage). We'll use hot-air balloons :-) – Carl Witthoft Mar 1 '18 at 18:51
• Hot air would make your balloons far, far larger and only increase the wind problems. – Chris Stratton Mar 1 '18 at 23:23