How many balloons would it take to fill a room?
July 17, 2007 10:56 AM   Subscribe

Packing efficiency of 9" balloons?

I know how many play pen balls fill a room. I also know the cost ($8,272.6 for a small bathroom). So how about balloons?

How would one model the number of 9" balloons required to fill a given volume? What's the proper term for "balloon," anyway (the closest I've found is "prolate spheroid")? I don't expect a solution to the minimal packing problem, just a well-reasoned estimate that assumes random placement. Oh, and these balloons will be filled the old-fashioned way.

Thank you!
posted by nilihm to Science & Nature (13 answers total) 1 user marked this as a favorite
 
Since all you want is an estimate, model them as spheres and then use a packing density of 0.6.

So number of balloons= 0.6*VolofRoom/VolofBalloon

Even an optimal ellipsoid packing won't get above 75%.
posted by vacapinta at 11:10 AM on July 17, 2007


It depends on how efficiently they're packed. Prolate spheres with random orientation should average out about the same as spheres. If the balloons are 9" x 10", just calculate for 9.5" spheres.
posted by weapons-grade pandemonium at 11:14 AM on July 17, 2007


I laughed when I saw this because I have a mathematician friend who has devoted a good portion of his life to sphere packing problems. I know that's no help. Just sayin'.
posted by MarshallPoe at 11:26 AM on July 17, 2007


Cube the diameter, divide it into the volume of the room. Done and done.
posted by blue_beetle at 11:55 AM on July 17, 2007


"Divide into?"
posted by rhizome at 11:57 AM on July 17, 2007


A practical solution might be to pack a large cardboard box with balloons and extrapolate.
posted by weapons-grade pandemonium at 12:08 PM on July 17, 2007


It is definitely way more efficient to deflate the balloons before seeing how many of them you can cram into a given space. Trust me, I learned the hard way.
posted by Caper's Ghost at 12:13 PM on July 17, 2007


A practical solution might be to pack a large cardboard box with balloons and extrapolate.

The problem with that is that the cardboard box would have to be significantly larger than the balloons for meaningful extrapolation. The box itself introduces "edge effects" which would force a type of packing (e.g. a lattice packing) which would not happen in say a large room. All these packing densities are thus limits which ignore these effects.
posted by vacapinta at 12:16 PM on July 17, 2007


just a well-reasoned estimate that assumes random placement.

If you just want a well-reasoned estimate, and not an exact answer, just model the balloons as spheres. It should be close enough, and as a bonus you can use the applet on the page you linked to in your original question.
posted by DevilsAdvocate at 12:41 PM on July 17, 2007


Since balloons are significantly deformable (unless really filled to their limits), I'd expect to be able to get significantly more balloons into a given space than spheres. Done right, you might be able to fill almost all the space with balloons.
posted by DarkForest at 1:38 PM on July 17, 2007


"Divide into?"
posted by rhizome


As in: 3 into 12 goes 4 times.

posted by blue_beetle at 2:16 PM on July 17, 2007


How are you going to do the compression DarkForest?
posted by Mitheral at 2:40 PM on July 17, 2007


blue_beetle, that would give you a packing of about .52, less than the sphere packing.

Mitheral: Maybe DarkForest will use a vacuum? Or get out of the room and push?
posted by Monday at 5:08 PM on July 17, 2007


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