# Vacuum balloon?

Hi, first time posting here.

I was watching a show about hydrogen/helium balloons and it got me thinking about buoyancy in atmospheric gasses (nitrogen) and what might work better than the common gasses and I considered that since the goal of a balloon is to reduce the internal mass relative to the mass of the medium why not just eliminate it entirely and use a vacuum? I understand ordinary balloon material would not work as it would simply be crushed by atmospheric pressure... so my question to you guys is can you think of any way to maintain a volume of vacuum or near-vacuum while still keeping the object light enough to provide more lift than a traditional helium balloon? If someone loves doing math you could come up with the equation to relate the volume of the vacuum with the lift capacity it would have if it could be maintained in the atmosphere...

Now, I've seen video of a train car (the cylindrical ones that carry liquids) being completely crushed by nothing but atmospheric pressure due to very low internal pressure, so I know this container would have to be fairly robust and we likely are lacking the material to make this feasible, but what about materials that are on the horizon, like something made with carbon nanotubes maybe? Maybe with a reinforced internal structure like a honeycomb structure with thousands of pockets of vacuum?

It would be great if we could create something like this with significant lift capacity to deliver space vehicles high into the atmosphere where they would "launch" from the floating platform or something, since that would be reusable ad infinitum it would reduce the cost to deliver a payload to LEO by a great deal I would imagine.

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Thanks for the link, I read through that thread and came up with another idea in the meantime...

What about using electromagnetism to force metal foil apart to counteract the pressure of the atmosphere? You'd still need a rigid frame but it and the membrane itself would not have to withstand nearly as much pressure, meaning it could be lighter.

Also, one distinct advantage a vacuum balloon would have over a helium or hydrogen balloon is that it would strengthen as it increased in altitude and would find an equilibrium point and float there, where other balloons will eventually burst.

DaveC426913
Gold Member
What about using electromagnetism to force metal foil apart to counteract the pressure of the atmosphere? You'd still need a rigid frame but it and the membrane itself would not have to withstand nearly as much pressure, meaning it could be lighter.
And how much would this electromagnetic generator weigh??

Also, one distinct advantage a vacuum balloon would have over a helium or hydrogen balloon is that it would strengthen as it increased in altitude and would find an equilibrium point and float there, where other balloons will eventually burst.
I pointed this out in the other thread, but it is a small advantage for the short range before the balloon reached equilibirum. And weather balloons have solved this problem anyway. They don't burst.

How about a fairly large mylar balloon of the same material as the party balloons, thin film of aluminum I think, with an internal support grid of something like concrete (very high compression strength, so it can be very thin)?

BTW fun is rarely doing something the obvious practical way.

DaveC426913
Gold Member
How about a fairly large mylar balloon of the same material as the party balloons, thin film of aluminum I think, with an internal support grid of something like concrete (very high compression strength, so it can be very thin)?

BTW fun is rarely doing something the obvious practical way.
You won't need to worry about compressions strength so much as you will torsional forces.

In general, pick a volume for the balloon, say 1000m^3. Find the wight of that much air. Your entire structure, including mylar, support and 1000m^3 of helium (178kg) would have to weigh less than that.

You won't need to worry about compressions strength so much as you will torsional forces.

In general, pick a volume for the balloon, say 1000m^3. Find the wight of that much air. Your entire structure, including mylar, support and 1000m^3 of helium (178kg) would have to weigh less than that.
That would be for a practical solution, as opposed to a demonstration of technology where all I would need to do is make it off the ground.

Thing is, a vacuum balloon is not all that much more bouyant than a hydrogen balloon.
I think this is the show stopper here. There just isn't enough payoff for a complex solution, so unless a vacuum ballon can be cheap and easy, why bother?

DaveC426913
Gold Member
That would be for a practical solution, as opposed to a demonstration of technology where all I would need to do is make it off the ground. No, even just to make a demonstration and make it off the ground your entire setup would indeed have to weigh less than the air displaced.

If you wanted a practical solution beyond that then it would need to weigh significantly less, enough to haul a payload as well. No, even just to make a demonstration and make it off the ground your entire setup would indeed have to weigh less than the air displaced.

If you wanted a practical solution beyond that then it would need to weigh significantly less, enough to haul a payload as well.
Right, but not less than helium, except as a practical, ie boring solution.

DaveC426913
Gold Member
Right. I see your point now. You were referring to your vacuum balloon. It would not have to be more efficient than helium. Simply getting off the ground would suffice to be a proof of concept.

I had this same idea, here some figure about the feasibility of the system:

Note: the idea on this calculation is to see how shape and structural factors go with the characteristic dimension. the value of those factor in turn give information about the feasibility of the design with the current materials.

BALOON LIFT:
The lift force depend on the weight of the air, that the baloon volume occupy, so:

g [m/s3] = gravity
RoAir [kg/m3] = air density at a given altitude and temperature
BV [m3] = Baloon Volume

Lift [N] = g [m/s2] * RoAir [kg/m3] * BV [m3]

The Baloon volume will be the cube of the characteristic dimension multiplied by a geometrical constant which account for the baloon shape

d[m] = characteristic dimension
Kv  = volume shape factor (for a sphere is equal to Pi*4/3)

BV [m3] = (d [m] ^3) * Kv 

BALOON WEIGHT:
It's very difficult for a sphere to withstand compression (it will deformate and crush), so the best way to make a light baloon is probably to apply a thin surface on a structure with high compressive strength over weight ratio. in this case, the baloon weight will be given by two component:
-The surface (that withold the vacuum)
-The structure (that withstand the compressive force)

BALOON SURFACE WEIGHT:
The baloon weight will depend on the baloon surface area, the surface thickness and the surface material density (some sort of plastic may be the best)

g [m/s2] = gravity (9.8)
BS [m2] = Baloon surface area
T [m] = Surface thickness
RoSurface [Kg/m3] = Baloon surface density (for rubber is 1200)

BSW [ N ] = g [m/s2] * BS [m2] * T [m] * RoSurface [Kg/m3]

Again i can generally express the surface area the square of the characteristic multiplied by another shape factor

d[m] = characteristic dimension
Ksa  = surface area shape factor (for a sphere is equal to Pi*4)

BS [m2] = (d [m] ^2) * Ksa 

BALOON STRUCTURE WEIGHT:
This is probably the most limiting part of the design, i will compute the Crushing Force that the baloon feel, and express the weight of the structure as a fraction of this force. (example. if the Structure Weight constant need to be 1/1000, mean that a baloon structure of 1N of weight, need to withstand 1000N of compressive force from the surface)

The Crushing Force the structure need to withstand is the baloon surface area multiplied be the pressure difference

DeltaP [pa] [N/m2] = difference in pressure between outside and inside of the baloon
BS[m2] = baloon surface area

CF [N] = BS [m2] * DeltaP [pa]

The structure weight will be the crushing force multiplied by a structure constant, which account for the structure shape and material

CF [N] = Crushing Force felt by the baloon structure
Kx  = structure weight over resistence ratio

BXW [N] = CF [N] * Kx 

FINAL CONSIDERATION:
The fly limit happen when the baloon weight equal the lift (the weight need to be lower for pratical porpouse, so this give the lower bound of the shape and structural constant)

LIFT [N] = BaloonSurfaceWeight [N] + Baloon Structure Weight (BXW) [N]

Doing some math and semplification and resolving in respect to the Kx constant:

g [m/s2] = gravity (9.8)
T [m] = Surface thickness
RoSurface [Kg/m3] = Baloon surface density (for rubber is 1200)
Kx  = structure weight over resistence ratio
d[m] = characteristic dimension
Ksa  = surface area shape factor (for a sphere is equal to Pi*4)
DeltaP [pa] [N/m2] = difference in pressure between outside and inside of the baloon
Kv  = volume shape factor (for a sphere is equal to Pi*4/3)

Kx  = g [m/s2] * { d [m] *RoAir [Kg/m3] *Kv  - T [m] *Ka  *RoSurface [kg/m3] } / DeltaP [pa]

Kx of 1/100 is easier to obtain then 1/1000, the higher Kx, the better.

This formula tell that:

1) Bigger baloon have advantage over small baloon since d (the characteristic dimension) is on the numerator (it make sense, since the lift force increase with volume but crushing force increase only with the square of d).
This does not account for the difficulty to achieve let's say Kv = 1/100 with a structure that weight 1N or one that weight 10N)

2) Shape that have higher volume over dimension ratio is better (it make sense since moving more air, increase the lift) (example: cube)

3) Shape that have lower surface over dimension ratio is better (it make sense since i lower the crushing force exerted on the structure) (example sphere:)

4) Point 2 and 3 go in opposit direction, in the design you need to balance them in order to not have a bottleneck (highest value for the difference)

5) Higher gravity make thing easier, the baloon would work better on jupiter then earth (make sense since the gravity will influence more the liquid than the structure weight)
the same can be said for higher liquid density (example: water better than air)

6) Having lower surface density or lower thickness (aka better surface material) is better (obvious, but if the formula get at least the obvious thing right there is chance for it to be right also)