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Awesome. Here are a couple questions. Is it possible to send a balloon up and not have it burst. Just float around for a few days? How do you calculate the ba

Awesome. Here are a couple questions. Is it possible to send a balloon up and not have it burst. Just float around for a few days? How do you calculate the balloon size And helium needed for a given payload?

jgrahamc 16y ago
Yes, it is. That's what the CU Spaceflight people are doing with BallastHalo and there have been autonomous cross-Atlantic flights I believe. The balloon manufacturers give lift information and then you simply fill the balloon and measure the lift.
- tocomment 16y ago
  So how do you keep it from bursting? Or would reading about ballasthalo explain that? I wonder what the actual formula is to calculate the correct balloon size? It seems like it might be some complicated differential equation. Here's why: Say I want to lift 10 lbs. So I figure out I need X volume of helium. So I need a balloon weighing 1 lb to hold the helium. Now I need a balloon big enough to lift 11 lbs instead of 10, repeat, repeat.
  - whatusername 16y ago
    repeat, repeat, http://en.wikipedia.org/wiki/Zeno%27s_paradoxes http://en.wikipedia.org/wiki/Zeno%27s_paradoxes
  - jacobolus 16y ago
    > might be some complicated differential equation If by “complicated differential equation” you mean either “surface area is approximately proportional to radius squared so this could be solved by 8th graders” or “just add 20% margin and it’ll probably work perfectly well”.
    - tocomment 16y ago
      You want to use a 20% margin for spaceflight/stratosphere access? I'd rather find the right equation.
      - jacobolus 16y ago
        We’re talking about a balloon that we want to have rise to the stratosphere in 2-3 hours. If the balloon rises at 4 meters/second, or 7, instead of 5, it’s really not the end of the world. Just pick a size, calculate (rather trivially) how fast it’s going to rise, and then change the amount up or down. If it really must be exactly 5 meters/second (or whatever), the computation is still at about the level of a slightly-above-average 8th grader.
      - rbanffy 16y ago
        A 20% margin would reduce space-access costs immensely. We use exotic alloys, explosive fuels and precision machining because we need to reduce weight to the structural limits. If we hadn't, space access would be cheaper and safer. Sadly, Earth has just enough gravity to ruin our day.

jgrahamc 16y ago

Yes, it is. That's what the CU Spaceflight people are doing with BallastHalo and there have been autonomous cross-Atlantic flights I believe. The balloon manufacturers give lift information and then you simply fill the balloon and measure the lift.

tocomment 16y ago
So how do you keep it from bursting? Or would reading about ballasthalo explain that? I wonder what the actual formula is to calculate the correct balloon size? It seems like it might be some complicated differential equation. Here's why: Say I want to lift 10 lbs. So I figure out I need X volume of helium. So I need a balloon weighing 1 lb to hold the helium. Now I need a balloon big enough to lift 11 lbs instead of 10, repeat, repeat.
- whatusername 16y ago
  repeat, repeat, http://en.wikipedia.org/wiki/Zeno%27s_paradoxes http://en.wikipedia.org/wiki/Zeno%27s_paradoxes
- jacobolus 16y ago
  > might be some complicated differential equation If by “complicated differential equation” you mean either “surface area is approximately proportional to radius squared so this could be solved by 8th graders” or “just add 20% margin and it’ll probably work perfectly well”.
  - tocomment 16y ago
    You want to use a 20% margin for spaceflight/stratosphere access? I'd rather find the right equation.
    - jacobolus 16y ago
      We’re talking about a balloon that we want to have rise to the stratosphere in 2-3 hours. If the balloon rises at 4 meters/second, or 7, instead of 5, it’s really not the end of the world. Just pick a size, calculate (rather trivially) how fast it’s going to rise, and then change the amount up or down. If it really must be exactly 5 meters/second (or whatever), the computation is still at about the level of a slightly-above-average 8th grader.
    - rbanffy 16y ago
      A 20% margin would reduce space-access costs immensely. We use exotic alloys, explosive fuels and precision machining because we need to reduce weight to the structural limits. If we hadn't, space access would be cheaper and safer. Sadly, Earth has just enough gravity to ruin our day.