This planet-like balloon started out as two elastomer sheets, heat-sealed together into a spiraling tube. As the balloon was inflated, it changed from flat to a saddle-like shape. With more air, the pressure inside increased, triggering an instability that caused the middle of the balloon to bulge. As inflation continued, the central bulge expanded, unbonding layer after layer of the seal. Even late in inflation, the balloon maintains hints of its original shape in the form of a ring around the Jovian bulge in the middle. (Image credit: N. Vani et al.)
Tag: balloons
High-Altitude Balloon Flight
Tangentially fluids-related, but SpaceWeather has a fun video of a high-altitude helium balloon bursting. Although this balloon carried a space-related payload, it’s the same type of set-up used for weather balloons. With only a few basic assumptions, it’s possible to do some neat calculations on the buoyancy, loading capacity, and behavior of such balloons.
Giant Water Balloon Physics
Playing with a giant water balloon and high-speed cameras is like a giant experiment in surface tension, right up until the tensile strength of the balloon comes into play. The rippling in the balloon is reminiscent of the motion of droplet breakup or impact on superhydrophobic surfaces. (submitted by Daniel B)
Reader Question: Hot Air Balloon Physics
lazenby asks:
and boyancy in air? is the lifting capacity of a hot air balloon equal to the modulo of the weight of the air in the balloon with the weight of the same volume of air outside the balloon?
for that matter, does the lift of a big helium weather balloon decrease as air pressure, and so weight of the air outside the balloon, drops? and is this exactly counterbalanced by the lessening density of the helium in the balloon?
all of these things keep me awake.
Hopefully you won’t be sleepless much longer. Buoyancy in air follows the same principles as buoyancy in water. Determining the lifting capacity of a balloon is a matter of determining how heavy the balloon can be before the buoyant force is equal to the weight. See the free body diagram and little derivation below to see what the maximum payload mass is for a helium balloon. You can click on the picture to enlarge it.
The second part of your question raises some interesting points. As a balloon’s altitude increases, the atmosphere around it gets colder and less dense, all of which should reduce the buoyant force. At the same time, the balloon itself expands to equalize the pressure inside and outside of the balloon, which should increase the buoyant force. (At some point the pressure drops sufficiently that the tensile strength of the balloon material is unable to cope with that expansion and the balloon bursts, but we’ll ignore that here.) For this problem, we’d want to know what payload the balloon can carry without losing lift, and, with a couple assumptions, that’s pretty easy to figure out. I’ve done that derivation below.
The real key to the calculation is assuming that the helium in the balloon maintains the same temperature as the air outside. Since balloons rise slowly, this seemed a more reasonable assumption than imagining that the balloon remains warm compared to its surroundings. That calculation is doable as well but requires more than a couple lines, unfortunately! Thanks for your questions!
