Average hot air balloons have a diameter of about 18 m. The largest out there is roughly 32 m.

My question is, are there any reasons other than fuel concerns that would prevent having a larger balloon, say around 50-60 m as a lifting body? I.e. problems with air mixing, pressure differentials, etc in such a large volume?

  • $\begingroup$ @PhilSweet, oh, whew! You had me quite concerned. How would I generate one of those spreadsheets? I'm not an Engineer. I'm a software developer doing simulations in my head based on observation and using simplified geometry so as to cancel out extra forces and simplify the math. $\endgroup$ – Adrian Feb 8 '20 at 0:29
  • $\begingroup$ I've got the answer back up and hopefully it's sorted a bit better. $\endgroup$ – Phil Sweet Feb 8 '20 at 0:47

You need to figure out how to land a large balloon in conditions with no wind, without the passengers ending up underneath the collapsed balloon.

In windy conditions, you also need a larger landing area to avoid damaging the collapsed balloon, and at some size point you will need machinery to pack it for transportation because of its own weight.

The largest passenger carrying balloon (as of 2017) was around 40m tall, which gives some idea how much ground area is required to operate it, remembering that balloons are not steerable so you don't have the option of landing at a purpose-built site.

Also, the only real purpose of having a bigger envelope is to carry more load, and eventually you hit design problems trying to make a completely flexible structure like a balloon strong enough.

  • $\begingroup$ This balloon will be tethered to the ground so it will alwayshave a landing spot that can accommodate it. Your first comment about wind and last comment about envelope strength is something I hadn't fully considered though. $\endgroup$ – Adrian Feb 7 '20 at 18:24
  • $\begingroup$ You need to edit that constraint into the question rather than bury it in the comments. $\endgroup$ – Transistor Feb 8 '20 at 10:56

As the volume of a hot air balloon is increased, so increases its surface area- through which the burner heat is lost. Since the volume increases faster than the area for a spherical balloon, more volume means more lift and lower heat losses for that amount of lift. However, more volume also means more fuel burned to fill the balloon to stay aloft, and beyond a certain balloon size the burners have to be running continuously to maintain lift. That determines the practical maximum size of a hot air balloon for a standard set of burners.

This also means that in general, a balloon is designed for a certain payload specification. This sets the balloon volume and hence the fuel burn rate, and from that comes the burner capacity in BTU's.

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    $\begingroup$ But wouldn't that would also be the case for several smaller balloons containing the same volume? A single sphere would have the least surface area compared to many adding up to the same volume. $\endgroup$ – Adrian Feb 7 '20 at 14:12
  • $\begingroup$ You are right. I will edit my response. $\endgroup$ – niels nielsen Feb 7 '20 at 17:38
  • $\begingroup$ The balloon will be tethered, so power is not an issue as the power can come from a ground station. That's why I was saying to ignore fuel concerns. $\endgroup$ – Adrian Feb 7 '20 at 18:16
  • $\begingroup$ OK, got it. -nn $\endgroup$ – niels nielsen Feb 7 '20 at 21:35

What you need to do is look closely at the physics of the inflated envelope and how the shape is determined. Let's limit the variables and hold the temperature differential constant as we make the balloon bigger. As the envelope height dimension grows, the pressure differential between the inside and outside grows. So that if the envelope tension is the same, the Gaussian curvature will increase. That's not what you want, you want the curvature to decrease in order to make the balloon bigger. So now you have to decrease increase the envelope tension in proportion to the square of the size change in order to maintain dimensional similitude. You want most of the envelope tension to come from the payload support tendons, and little from the hoop tension.

So you have to solve a fairly complicated but reasonably well conditioned set of equations that account for -

  1. Envelope weight (lbs/sqft) as a function of size and temperature difference.
  2. Envelope shape as a function of construction details, payload, temperature, and size.
  3. Envelope thermal performance wrt fuel system cost and weight.

You can definitely set this up in Excel and use the add-in solver to optimize the envelope parameters.

As sanity check, if we start with a balloon with 20 tendons at 25# tension each, and double the linear scale, we now need 40 tendons to maintain panel size, and need to tension them at 100# to keep similarity with the original. Thus lift (total tension) goes with volume, as we would expect for similar shapes.

  • $\begingroup$ So the envelope isn't actually load bearing but the tendons are? I'm assuming that is what the lines are going from the top off the balloon to the base are? Now why did you quadruple the tension on these tendons as opposed to just doubling for similarity? $\endgroup$ – Adrian Feb 8 '20 at 5:38
  • $\begingroup$ Quadrupling: If the pressure on a membrane doubles, which it will since the height of the congruent patch has doubled, you need to double the tension to keep the curvature the same. But you don't want to keep the curvature the same, you need to cut it in half to maintain similitude, so you have to double the tension again. $\endgroup$ – Phil Sweet Feb 8 '20 at 11:38
  • $\begingroup$ Note that the weight of the tendon system grows faster than the volume. The total cross section of the tendons is the cube of the smaller version, and the length of each tendon is double, so the tendon weight goes as scale factor to the 4th power. $\endgroup$ – Phil Sweet Feb 8 '20 at 11:46
  • $\begingroup$ The total cross section of the tendons is the cube of the smaller version, Wait, what? Shouldn't that be a factor of 16? Double the number of tendons of double length of quadruple diameter? $\endgroup$ – Adrian Feb 8 '20 at 14:14
  • $\begingroup$ To keep the stress the same, if you cube the payload, you have to cube the cross section area of the tendons. And they are twice as long, so a 4th power law for weight. There is a large class of problems where the structural weight goes as scale factor to the 4th power. Here, part of the structure, the tendons, scales to the 4th, and the envelope weight, as a first approximation, scales to the 3rd. The key is that you can't scale the stress in the fibers. The stress has to stay the same. $\endgroup$ – Phil Sweet Feb 8 '20 at 17:30

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