# Balloons and lifting gases

Hydrogen is the lightest element, so it's cable of lifting the most weight in out atmosphere (probably not the best terminology there, but you get the picture)

Would hot hydrogen (in the same sense as hot air) be able to lift even more mass? Would a higher or lower density of hydrogen in a ballon lift more? If you could have a balloon which had nothing in it (it was a vacuum inside) would that lift more than a hydrogen balloon?

Basically what are the physics to balloons and lifting?

(really not sure what to tag this, so if someone else could that'd be great)

• In theory the best gas for a balloon is vacuum, but it is hard to make a lightweight-enough container. – user68 Dec 14 '10 at 9:13

## 4 Answers

Would hot hydrogen (in the same sense as hot air) be able to lift even more mass?

Yes. Though I suppose the fire danger goes up, and you certainly can't use a propane burner to warm it...

Would a higher or lower density of hydrogen in a ballon lift more?

Lower density always means higher buoyancy.

If you could have a balloon which had nothing in it (it was a vacuum inside) would that lift more than a hydrogen balloon?

Yes, and this has been proposed in various ways in science fiction literature. The engineering challenge is finding a away to confine the vacuum that is as light as a gas bag so that you don't loose the advantage to extra weight.

In general a volume $V$ of material of density $\rho$ immersed in a fluid of density $\rho_f$ experiences a buoyant force of

$$F_b = gV\rho_f$$

and a weight of

$$W = -gV\rho$$

so the available lifting force is

$$F_l = gV(\rho_f - \rho) .$$

Where the object is floating at the surface of a liquid the buoyant force is modified to reflect the volume of liquid displaced $F_b = g V_d \rho_f$ where $V_d$ is enough to cover the weight of the floating object.

• @Marek: Er...well, I'm not an expert in this business. But...$\rho = \rho(T,P)$ and $P = P(h)$, $T_{atm} = T_{atm}(h)$ where $P$ is pressure, $T$ is temperature and $h$ is height. Those functional relationships may be complicated. In a large balloon I suppose that $\rho$ may vary over the volume necessitating a integration over V. As for control: heat the gas and it expands (i.e. gets less dense), or vent some and V drop proportionately, so you can adjust the lift in either direction. – dmckee Dec 13 '10 at 23:57
• @Noldorin : the use of hydrogen in the Hindenburg was not a mistake. Helium, mined in the US, was basically too strategic to be sold to Nazi Germany at that time, because of potential military use of airships... – Frédéric Grosshans Dec 14 '10 at 11:44
• @Noldorin: It was not a mistake*by scientist/engineers, it was a deliberate action by US gov. forbidding the sell of helium to Nazi Germany. The German engineers wanted to use helium but they could not. To make things clear: I am not accusing the US of any wrongdoing here, because the potential military use of airships at that time was clear, and the belligerent nature of Nazi Germany was obviously to be taken into account. I just want to say that the reason for the lack of helium in the Hindenburg is not because the danger was unknown/neglected by engineers, but because geopolitics of the 30s – Frédéric Grosshans Dec 15 '10 at 16:39
• @Noldorin: Just read the following paragraph on wikipedia to have the story of hydrogen vs helium in the Hindenburg: en.wikipedia.org/wiki/… – Frédéric Grosshans Dec 15 '10 at 16:41
• If you use a closed loop (heat exchanger) system, there is no reason why you could not warm your hydrogen with a propane burner - as long as there is no oxygen in the hydrogen. If there were some low concentration of oxygen, it would soon react, leaving water and pure hydrogen. Not saying it'd be a great idea (the marginal gain is small because it's already so light and it's not without danger), but it is certainly feasible. Really hot hydrogen is used industrially (for example hydrogen firing of Xray targets to drive off carbon contamination) - you just have to be careful of leaks and oxygen. – Floris Jul 31 '14 at 12:53

dmckee's answer is a great not-too-technical description of buoyancy. Read that first. But in case you're interested, I thought I would go into some more detail.

The buoyant force on a submerged object (e.g. a balloon submerged in air) is equal to the weight of the displaced fluid,

$$F_b = \rho_f g V$$

as dmckee said. The physical origin of this force is actually the pressure difference between the top and bottom surfaces of the floating object. Pressure in a fluid at a certain height is related to the depth of the fluid above that height by

$$P(z_2) - P(z_1) = \rho_f g (z_2 - z_1)$$

that is, density of fluid times gravitational acceleration times height difference. If you have a rectangular box whose top and bottom surfaces are flat, then it's pretty easy to calculate the buoyant force as the pressure differential times the area of those surfaces,

$$F_b = (\Delta P)(A) = \rho_f g \Delta z A = \rho_f g V$$

For an irregular shape, you'll have to do an integral of some sort. For example, I once wrote a blog post which discusses, in part, deriving the buoyant force (and weight) from the minimization of potential energy, and that method can be easier to apply to irregular objects. (There are also a couple of interesting applications, even if you don't care about the math.)

You can also take into account variations in density (or gravitational acceleration) over the size of the balloon by doing an integral. But, according to the US standard atmosphere model, the density of the atmosphere takes about $20\ \mathrm{km}$ to drop off to near zero, which corresponds to a fraction of a percent change over the height of a typical hot air balloon (a few tens of meters). That fraction of a percent is generally negligible, so you're pretty safe just using a single value for the density.

However, you can't neglect differences in density between vastly different altitudes. Remember that the buoyant force on the balloon is equal to the weight of the amount of fluid displaced. As you go higher, the density of the air drops, which means the balloon displaces a lower mass of air. Therefore, as the balloon rises, the buoyant force drops. Eventually it reaches a height at which the buoyant force exactly balances out the weight of the balloon (and basket), and the balloon levitates at that level. As has been said in the comments, for a controlled balloon, the operator can adjust the level by either heating the gas inside the balloon (thus making it expand and displace more air) or by letting some gas out (thus making the balloon contract and displace less air).

• +1 - Very nice answer. One point: it isn't obvious to me that buoyancy decreases with height. The density of air decreases with height, but so does absolute pressure. So if you have a balloon, the balloon will expand as it rises. In expanding, it takes up more volume, and this would increase its buoyancy. If the temperature were constant, then using the ideal gas law, if the density of the air halves, the pressure halves, and a balloon doubles in size, and buoyancy is constant. – Mark Eichenlaub Dec 14 '10 at 4:45
• @Mark: Depends on the balloon. High altitude balloons are designed to take advantage of exactly that effect and can rise to the neighborhood of 100,000 ft (30,000 m) elevation (I had a friend on a balloon borne CMB experiment back before COBE). The kinds of balloons people ride in have little give in the bags and won't go that high. – dmckee Dec 14 '10 at 5:35
• @Mark: good point. I was assuming that the balloon size would be roughly constant due to tension in the fabric, although now that you mention it I'm not sure whether that's true for normal hot air balloons. (Certainly you're right for weather balloons and the like) I'll expand on that point if/when I have time. – David Z Dec 14 '10 at 5:38
• You will find that many high altitude balloons are "slack bags" - in other words, they are designed so they will expand as the external pressure drops. See for instance flightopportunities.nasa.gov/media/uploads/88/… – Floris Jul 31 '14 at 12:49

Approximate mean atomic mass of air: 29
Approximate mean atomic mass of hydrogen: 2

So with both gases at the same temperature and pressure, you are getting $\frac{27}{29}=93\%$ of the theoretical lifting capacity.

You would have to heat the hydrogen to about 200 C in order to get to 95% efficiency. It would be much better to work on making the surrounding/supporting structure lighter than to heat the hydrogen - the apparatus plus fuel for the heating is likely to weigh much more than you gain in lift.

A vacuum balloon is a possibility, but I doubt it can provide more lift than a hydrogen balloon: you need a rigid shell to prevent implosion of the vacuum balloon, and my feeling is the shell will be too heavy. However, while vacuum balloons cannot compete on lift, they can have better altitude control: you can bleed air in to lose altitude and pump out air to gain altitude.

Together with my coauthor, I proposed some designs of vacuum balloons made of commercially available materials, such as boron carbide and aluminum honeycombs (US patent application 20070001053 (11/517915), or http://akhmeteli.org/wp-content/uploads/2011/08/vacuum_balloons_cip.pdf ). The main problem is so-called buckling (loss of stability). It is possible, but difficult, so nobody has built a vacuum balloon so far, as far as I know, although the idea of vacuum balloon is centuries old.