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0 pointsjacquesm11y ago0 comments

That's an interesting back of the envelope question to work out. Something with volume occupied by all aircraft aloft at any given moment, volume of all balloons at any given moment, the total volume of the shell in which commercial air traffic takes place.

I think we can safely forget about the chances of spyplanes hitting balloons, the volume of space versus the number of spyplanes would make that a non-issue, even if there were a lot more balloons.

So 35000 feet (11 km give or take) would be a reasonable upper limit. Let's assume the worst and start from 0, you have a shell above the earths surface up to 11 km above it, which has an approximate volume of: 510.1 million km x 11 = 5610 million cubic kilometers.

That's a lot of space. Every cubic kilometer is 10^9 cubic meters, so 5.6x10^18 cubic meters.

I don't know how many aircraft are typically aloft, but let's say it's 20,000 craft and they're all of the very largest variety (say A380, or Boeing dreamliner). They're approximately 60 meters long, and 6 meter in diameter, so that's 1700 cubic meters, let's double that to include the wing volume, so 3400 cubic meters.

We have 20,000 of them, they're all aloft at the same time, so all the planes take up approximately 68,000,000 cubic meters.

Now for the balloons, they're 10 meters in diameter, worst case they are 50 meters high or so (instrument package dangling below the balloon, assuming a cylinder with a radius of 5 meters and a height of 50), so about 4000 cubic meters. ('assume a spherical cow of uniform density').

So how big is the chance that one balloon intersects in all of space with the volume of all the aircraft given that both have all of the atmosphere to play cat and mouse in?

68,000,000 / (5.6x10^18) = 0.000000000012 (the chance that any given cubic meter is part of the space occupied by an aircraft) multiplied by 4000 (the number of cubic meters in a balloon) is about 0.000000048. So that's pretty small but non-zero, multiply by the number of balloons aloft at any given time, but keep in mind that most of the factors here were taken very pessimistic (as in, favouring the collision). The calculation also totally ignores the relative speeds of the two types of vehicles, ascent speed of balloons, the time factor, ability to manoeuvre and so on.

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falcolas11y ago

The problem is that jetliners don't have all that space you calculated to play around in. They actually have a few well defined corridors and altitudes in which they can operate - the straight lines drawn from one VOR (effectively equivalent to a major airport) to another at 1,000 foot intervals.

Detours are costly due to time and coordination (air traffic control, other aircraft), and reacting to seeing a balloon and moving the aircraft isn't that easy when you're traveling at 300+ MPH in an aircraft which turns like a cargo ship. And that's assuming you can even see the balloon in time to react in the first place.

And that's just the commercial jetliners. Private jets go higher and faster (about 50,000' and 700mph), while GA aircraft fill the skys below 14,000'.

Granted, this still leaves a lot of room in between these major aircraft corridors, but if a balloon should ever intersect with one of them, it's going to cause havoc, even if there's never an actual balloon/aircraft incident.

toomuchtodo11y ago

> They actually have a few well defined corridors and altitudes in which they can operate - the straight lines drawn from one VOR (effectively equivalent to a major airport) to another at 1,000 foot intervals.

Until NextGen (ADS, etc) finishes its rollout and everyone flies direct instead of on Victor airways and VOR to VOR.

falcolas11y ago

True, but even then there will still result in "air highways" from airport to airport.

toomuchtodo11y ago

I would agree with this. Still, the number of "air highways" would increase drastically.

avoid3d11y ago

You lost 6 orders of magnitude to this IMHO:

5610 million km^3 = 5.61 * 10^3 * 10^6 km^3 = 5.61 * 10^9 km^3 = 5.61 * 10^9 * 10^9 m^3 = 5.61 * 10^18 m^3

But you said the resulting volume was = 5.61 * 10^12 m^3.

I think you forgot that it was 5610 million km^3.

jacquesmOP11y ago

Ah! I knew there was something terribly wrong there somewhere (and probably that's not the only thing). Thank you very much, I'll edit the comment to fix it.

utopkara11y ago

Keep in mind that balloons are much cheaper than planes to build and operate. Worse yet, the populated areas on earth are only a fraction of the whole surface area. Expect balloons to make dense clusters rather than evenly distributed over the sky. Needless to say, there will be more flights to more populated areas as well. Increasing the likelihood of collision to a level that the pilots will need to constantly monitor the skies will increase fatigue and cause indirect problems. The cost of all this will be felt by the aviation industry, and will be passed on to us.

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falcolas11y ago

And that's just the commercial jetliners. Private jets go higher and faster (about 50,000' and 700mph), while GA aircraft fill the skys below 14,000'.

toomuchtodo11y ago

Until NextGen (ADS, etc) finishes its rollout and everyone flies direct instead of on Victor airways and VOR to VOR.

falcolas11y ago

True, but even then there will still result in "air highways" from airport to airport.

toomuchtodo11y ago

I would agree with this. Still, the number of "air highways" would increase drastically.

avoid3d11y ago

You lost 6 orders of magnitude to this IMHO:

5610 million km^3 = 5.61 * 10^3 * 10^6 km^3 = 5.61 * 10^9 km^3 = 5.61 * 10^9 * 10^9 m^3 = 5.61 * 10^18 m^3

But you said the resulting volume was = 5.61 * 10^12 m^3.

I think you forgot that it was 5610 million km^3.

jacquesmOP11y ago

Ah! I knew there was something terribly wrong there somewhere (and probably that's not the only thing). Thank you very much, I'll edit the comment to fix it.

utopkara11y ago

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