If we spin an air balloon, the centrifugal force would cause it to expand, since it is directed outwards. Increased volume of the balloon would increase the buoyant force. Can this be used to noticeably increase the lift of the balloon?

  • $\begingroup$ I love the idea. How do you intend to spin it up and keep it spinning? $\endgroup$ Aug 28, 2021 at 17:25
  • $\begingroup$ Lets say its a blimp and there are two connection points between the baloon part and the useful weight part, which has some sort of engines that spin it up. In this case it would be spinning horisontally $\endgroup$
    – user369070
    Aug 28, 2021 at 17:33
  • $\begingroup$ What do you mean by an "air balloon"? A balloon filled with air has no lift: it is under pressure from the rubber and thus denser than the atmosphere. Anyway, wouldn't its axial dimension lessen? $\endgroup$ Aug 28, 2021 at 18:31
  • 1
    $\begingroup$ To increase the volume by spinning would require a stretchable material (like rubber). It is my understanding the the material of a normal hot air balloon is not stretchable. $\endgroup$
    – R.W. Bird
    Aug 28, 2021 at 18:58

1 Answer 1


To see if the idea has merit, let's try to quantify the magnitude of the effect. Simplifying assumptions:

  1. The balloon has the shape of a cylinder. Let $R$ denote the radius, and let $H$ denote the height along the symmetry axis.

  2. The balloon spins around the cylinder's symmetry axis. Let $v$ denote the tangential velocity of the curved surface, so that the angular velocity is $v/R$.

  3. The balloon's mass $M$ is evenly distributed over the round face of the cylinder, so the top and bottom caps have negligible mass.

  4. Stretching the material to increase the cylinder's radius does not require any force, except for the force required to overcome the ambient atmospheric pressure, but assume that the cylinder's height (along the symmetry axis) is incompressible.

  5. The balloon is empty (vacuum), or at least the air pressure inside the balloon is negligible compared to the outside atmospheric pressure $P$, so that the cylinder's spin is the only thing that balances the outside atmospheric pressure.

From assumption 1, the balloon's volume is $$ V=\pi R^2 H. \tag{1} $$ Assumption 5 says that the outside pressure needs to balance the centripetal force: $$ Mv^2/R = P\times 2\pi R\times H, \tag{2} $$ where the right-hand side is the pressure times the area of the curved surface (assumption 4). Combine equations (1)-(2) to get $$ Mv^2 = 2PV. \tag{3} $$ The bouyant force is $\rho gV$, where $\rho$ is the outside air density and $g$ is the acceleration of gravity. For the balloon to float, this needs to exceed the gravitational force $Mg$ on the balloon, so we need $$ \rho Vg > Mg. \tag{4} $$ Combine (3)-(4) to get this result for the tangential velocity: $$ v > \sqrt{\frac{2P}{\rho}}. \tag{5} $$ For perspective, compare this to the speed of sound of an ideal gas: $$ c = \sqrt{\frac{\gamma P}{\rho}} \tag{6} $$ where the adiabatic index $\gamma$ is $\sim 1.4$. The conclusion is under the assumptions listed above, the balloon's tangential velocity must exceed the speed of sound in order for the (empty) balloon to stay afloat.

That's if the spin is the only thing counteracting the outside atmospheric pressure. In practice, we could modify assumption 5 by filling the balloon with a gas (like helium or hot air) whose pressure-to-density ratio is higher than the outside air, and we should modify assumption 4 by using a realistic material. I'd recommend making those adjustments to the calculation before submitting the patent application.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.