Can you help me find my baloons?

August 19, 2009 6:21 PM

If I were to release 50 regular helium party balloons, from kingsway drive kitchener ontario canada, whats the likely hood of me finding two or three of them. These balloons were released today around noon. I need to find at least two or three of them. Anyone who could help me out in this endeavor would be amazing. map embed in extended explination.

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The Sooner the better so i can plan out my excursion. If this helps any I believe most of our weather comes out of the north west.

posted by Chamunks to Sports, Hobbies, & Recreation (13 answers total)

You're probably not going to find any of them, except in small pieces.

Research has shown that after Latex balloons are launched, they often rise to an altitude of about five miles where they begin to freeze in the -50 degree Fahrenheit cold. In addition, the strong differential between the gas pressure inside the balloon and the near vacuum outside the balloon at that altitude causes the balloons to expand to the point were they eventually burst. However, because the Latex is frozen, the bursting balloon tears into shreds (the exact scientific term is called "brittle fracture"). These tiny, spaghetti-like pieces then scatter over a wide area as they fall back to Earth where they naturally begin to decay according to the process we discussed earlier.

(source)
posted by jedicus at 6:27 PM on August 19, 2009 [1 favorite]

Next time, laminate some bits of paper with return information and as payload.
posted by phrontist at 6:31 PM on August 19, 2009

Balloon Track is a free windows program that uses publicly available weather data to predict the flight patterns of high altitude weather balloons. I've never had a Windows machine around to use it but if you are clever you should be able to figure it out. I can't speak as to how the flight characteristics of party balloons differs from those of weather balloons but the program might still give you some good hints and tell you what direction to drive.

It looks like weather data for Ontario is available form the sources that Balloon Track can import from but really someone with more experience than I would need to have a look to figure things out.
posted by ChrisHartley at 6:33 PM on August 19, 2009

Regular means latex, right? Not mylar? Almost zero chance, except if you find some that got caught in trees or man-made structures near the release point. If you are trying to guess where they came down after an unobstructed flight, forget it.
posted by planetkyoto at 6:34 PM on August 19, 2009

Response by poster: Its mainly to recover the notes that have been placed within the baloons. I basically need the max range and possibility of finding the notes inside.
posted by Chamunks at 6:40 PM on August 19, 2009

Yours is not the likely hood. Please read this thread before you release the next batch.
posted by weapons-grade pandemonium at 6:49 PM on August 19, 2009

fwiw, I performed this exact experiment 30 years ago and a rancher 8 miles east of King City was kind enough to return the card at his expense.

The flight, about 50 miles.

Winds that day were IIRC normal onshore breezes for the Salinas Valley.
posted by @troy at 6:54 PM on August 19, 2009

Response by poster: Nah Nope just some tickets to give away a 4000$ sony bravia television from leons.
posted by Chamunks at 7:25 PM on August 19, 2009

Response by poster: I would like one of these televisions that would be fantastic.
posted by Chamunks at 7:28 PM on August 19, 2009

Hmmm I did find the winds aloft report for the area today at noon local time -- here it is. It looks like the winds were blowing 247 degrees at 4 knots at 1151ft, drifting to 285 degrees at 29 knots at 10,000ft.

This paper (on the environmental impact of balloon releases) puts the rate of ascent of the typical helium filled 11 inch diameter toy balloon at 5.89 feet per second, with a rupture altitude of 28,000 fet and a total ascent time of 4,753 seconds (80 minutes).

It should be possible to combine that information with the winds aloft report and figure out where your balloons would have landed, or at least where they would have burst. I'll keep working on it. Probably there are so many shaky assumptions built into these calculations that the margin of error is about the size of Lake Ontario.

Please note that I really don't know what I'm doing, so if someone out there has a better idea please step in.
posted by ChrisHartley at 7:45 PM on August 19, 2009

Best answer: I'm going to go to bed but I projected this route for your balloon:
241 0.154333
252 0.154333
248 0.154333
255 0.185200
244 0.185200
240 0.308667
248 0.154333
262 0.185200
270 0.740800
272 0.463000
277 0.555600
282 1.419867
281 0.771667
275 0.802533
274 4.228733
275 0.864267
277 1.790267
280 0.895133
288 2.716267
285 2.778000
281 3.950933
273 5.556000
267 4.938667
263 3.889200
261 2.592800
262 3.889200
263 11.667600
267 4.259600
268 5.802933
269 7.562333
276 7.562333

The first column is the bearing, the second is the distance in kilometers. Start from your starting point, go 0.154333km at a heading of 241, etc. One of the scripts on this page may be helpful. If I didn't have to go to bed I'd put together a script to calculate out the final destination but maybe someone else will chime in.
posted by ChrisHartley at 9:25 PM on August 19, 2009 [1 favorite]

Response by poster: Wow thank you If I find more than one balloon I will make sure you get something out of this. I make no promises obviously but wow thanks.
posted by Chamunks at 10:09 PM on August 19, 2009

Ummm whoops, I believe the the bearing is the direction the wind is blowing FROM, so your balloons would actually move the 180 degrees in the opposite direction.
posted by ChrisHartley at 5:50 AM on August 20, 2009