Hot Air Balloons
I get to see hot air balloons almost every weekend with mild weather, spring/summer/fall. The thumbnail is a hot air balloon in front of Mount Meeker and Long’s Peak.
I watched a hot air balloon launch fall 2022. The balloon envelope gets packed without the circular cover, visible as a black and yellow “bulls eye” dartboard looking thing in the image above. The handlers start inflating the balloon, then one of the handlers snaps the circular cover in place. It’s a little weird to think that all the hot air stays in place because a mere sheet keeps it there.
What’s up with the shape?
Hot air balloons have an odd shape, which seems to go by the names “inverted teardrop” and “natural shape”. I’ve always wanted to know why this shape works, and how the envelope makers attain it. I’m perfectly capable of understanding whatever explanation there is: I’ve got a degree in Aerospace Engineering. I used to do structural engineering for major aerospace companies. Here it is, 2023, and it’s very difficult to search up any coherent engineering analysis, just some handwaving about reducing heat transfer, and having zero circumferential (“hoop”) stress.
This is the best publication I could find on the “natural shape” of a balloon. The author calculates the “natural shape” using a system of 4 differential equations. I don’t care to spend 6 months understanding that system, so I’m just going to let it lie.
Will it stay in one piece?
I found Cameron Balloons A type envelope figures to try to understand hot air balloons a little better.
The A-160 envelope has these facts:
| Volume | 160,000 ft3 |
| max FAA certified weight | 3200 lb |
| envelope wt | 334 lb |
| height | 73 ft |
| max diameter | 71 ft |
How much pressure does the top of the balloon’s envelope have to withstand to hold up the 3200 lb maximum weight?
We simplify our mental model of the balloon to a cylinder 71 feet in diameter with a flat plate at the top, and open at the bottom, shaped roughly like a large tin can. It’s open at the bottom, like a real hot air balloon, so the pressure on the flat plate at the crown of the balloon is what holds up the weight.
P = F/A = (3200 lb)/(π * (12*35.5 in)2) = 0.006 psi
That’s not a giant amount of pressure.
Using the max diameter is justified because at that location, the meridional stress in the fabric is normal to the imaginary disk that would comprise a flat plate on a cylindrical can shaped balloon.
Cameron Balloons “Caliber” fabric is said to be 1.7 ounce, high tenacity, rip-stop nylon, woven from quality DuPont 6,6 nylon fibers. I found a data sheet for 1.6 ounce ripstop nylon fabric that claims a thickness of 0.0039 inches.
Assuming a spherical balloon crown, which probably isn’t accurate, the stress in the fabric is:
stress = Pr/2t = (.006 lb/in2)(35.5*12 in)/(2*.0039 in) = 327 psi
Assuming a √2 elliptical balloon crown seems more reasonable. Many liquid fueled rockets, Titan launch vehicles, Centaur upper stages, probably others, had elliptical cross section heads on their fuel and oxidizer tanks. This was because a pressure vessel that’s a oblate ellipsoid with a ratio of radius to height of √2 has zero circumferential stress at its equator. That would be no compressive stress or tensile stress. Since a balloon envelope is fabric, it can only carry tensile stress. A √2 elliptical crown is almost certainly a good guess at the shape because of this.
Meridional stress in an elliptical balloon top = Pρ/2t, where ρ is the radius of curvature. ρ varies from a2/b at the very crown to b2/a at the equator. The balloon radius is 35.5 feet, that’s a in this discussion, so b (height of crown above max diameter) is a/√2 = 25.1 feet.
Max ρ = a2/b = (35.5)2/25.1 = 50.2 feet.
Max stress in this shape balloon crown = Pρ/2t = (.006 lb/in2)(50.2*12 in)/(2*.0039 in) = 463 psi
That’s laughably small. Considerations other than membrane stress guide the choice of fabric. Maybe sewn joints reduce the strength, or maybe durability or heat resistance choose the fabric.
Double check the √2 dome assumption
Estimate envelope volume as if it were half an ellipsoid and a cone. The envelope is 71 feet high, b = 25.1 feet, so the conical part is 45.9 feet tall.
Volume of half an ellipsoid = (2*π/3)*25.1*35.52 = 66,250 cubic feet
Volume of a cone = π/3 * 35.52 * 45.9 = 60,575 cubic feet
Total volume is 126,825 cubic feet. That’s not terribly far off the 160,000 cubic feet nominal volume. The extra 36,000 cubic feet can easily be encompassed by the bulging of the conical part of the envelope.
Will It Float?
Air has a density of 0.0752 lb/cu ft at sea level (14.7 psi air pressie), 68F
Density varies linearly with temperature (absolute!) and inversely with pressure. In Windsor, CO, air pressure is about 12.3 psi, and balloons fly in about 50F air. Ambient air density in Windsor will be:
0.0752 * (12.3/14.7) * ((68+460)/(50+460)) = 0.0651 lb per cubic foot
Apparently the hot air is about 225F, so the density of the air in a balloon’s envelope will be:
0.0752 * (12.3/14.7) * ((68+460)/(225+460)) = 0.0484 lb per cubic foot
The 160,000 cubic foot volume of an A-160 would displace .0651*160000 = 10416 pounds of ambient air, and the air in the volume would weigh .0484*160000 = 7744 lb. The net lift is 10416 - 7744 = 2672 lb, about 500 lb less than the FAA max certified load. A gondola, propane tanks and 2 or 3 people are going to weigh a lot less that 2672 pounds, more like 1500 pounds.
Double check the 0.006 psi internal pressure
Knowing the temperature and barometric pressure lets me calculate the density of the air inside the balloon. From that we can estimate pressure in the envelope another way.
Cameron Balloons web page says that 250F is the maximum temperature. So back at sea level, 250F air has a density of:
0.0752 * ((68+460)/(250+460)) = 0.0559 lb per cubic foot
If we assume an upside-down column of hot air a constant density fluid would create a hydrostatic pressure of:
P = ρgh
We’ve got “density” in pounds per cubic foot, so the g (32.2 ft per sec2) is already multiplied in. h will be 71 feet, the height of the envelope.
P = (0.0752 - 0.0559 lb/ft3)(71 ft)(1 ft/12 in)2 = 0.01 psi
That’s not too far off of 0.006 psi
0.01 psi is an outside estimate: the density of the air in the balloon certainly varies from “mouth” at the bottom, which is open, and so is at ambient pressure, to the crown, which is where the hottest, and therefore least dense, air would rise.
Assuming a linear density gradient from mouth to crown:
P = ρgh/2 = 0.005 psi
The maximum 0.01 psi pressure would make the max stress in the envelope fabric:
Pρ/2t = (.01 lb/in2)(50.2*12 in)/(2*.0039 in) = 772 psi meridional stress
That’s still quite low. Fabric wear resistance or quality of joining the gores of the envelope, or how the gondola load is sheared into the envelope matter more than plain membrane strength.
Conclusion
Science triumphs, the A-160 could fly (float) in Windsor, CO on a 50F morning, and not burst at the seams.
Note
The ability to do this sort of examination is due entirely to the ability to look up facts like balloon envelope dimensions and volume, material thickness, air density and the behavior of air when pressure and temperature change, and how the shape of pressure vessels affects the membrane stress.
Any college junior civil or mechanical engineering major should have the knowledge to do this examination. Before The Internet, the facts to fill in the equations used to be much harder to come by. Having those facts makes this examination into something other than an exercise in guessing, makes it somehow more satisfying.
I’m still peeved that the “natural shape” is so hard to calculate.