# 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!

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!

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

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

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

posted by blue_beetle at 11:55 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

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

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

posted by DarkForest at 1:38 PM on July 17, 2007

*"Divide into?"*

posted by rhizome

posted by rhizome

As in: 3 into 12 goes 4 times.

posted by blue_beetle at 2:16 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

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

This thread is closed to new comments.

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