The barrage balloons of WW-II had to be anchored very carefully to avoid having them go into the stratosphere. As the balloon expands (which it does when it goes higher) it will become more buoyant, not less so there is a positive feedback loop in there which usually ends in destruction unless you take precautions. Such anchoring requires very long cables, which makes them a bad choice to defend against jets.

So, the risks are non-zero and if one were to get sucked into a jet engine (especially the payload portion) the mayhem would be considerable, but they are so small that a 'notice to airmen' suffices unless you're operating very close to an airfield when you launch.

What is interesting about this incident is how far the balloon came down from where it was launched, it must have travelled for a long time, maybe even circumnavigated the globe more than once before landing.

I have no experience with balloons at all, but even a plowed field feels like a sidewalk to me when I run into it at the same speed under canopy. I leave a sizable mark, and I have seen friends break femurs, pelvises, etc at the end of a similar descent.

For flexible envelopes this is more or less inevitable (balloon rises->atmospheric pressure drops->balloon expands->density decreases->balloon rise etc), for more rigid envelopes it is a balance that may work out in favor of the balloon staying in one piece (oscillating in altitude as it cools down/heats up again with the day/night cycle), or it may burst depending on the pressure differential. Most of them are pretty flimsy.

From: http://www.stratostar.net/faq.html

"When a balloon is filled on the ground with lift gas (helium or hydrogen), it can range in size from 2.5 ft to 8 ft in diameter. During the balloon's flight it will grow more than 4 times the diameter and upto 83 times the volume measured at launch, until it can't strech any more and will burst! A high-altitude weather balloon filled with 268 cu/ft of helium will have a diameter of about 8 ft at sea level, but as the balloon climbs through the atmosphere it will expand to 35ft in diameter and will have a volume of 22,449 cu/ft before it pops."

I've worked on a token ring network as recently as 2006. Thankfully, part of the project was migrating off of it.

I don't expect to hear a postmortem from Google, but I'd be astonished if this wasn't a malfunction of some sort-- These balloons almost certainly have an emergency cut-down device of some sort capable of safely and rapidly returning the payload to Earth.

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.

See also:

http://weatherjackwilliams.com/answers-weather-balloons-and-...

edit: extensively edited after avoid3d spotted a crucial error in the math, I had dropped 6 orders of magnitude.