Recently one of our coworkers travelled back to India for a month to get married. Obviously a vacant office for a month was just too much to pass up so we decided to fill his office from floor to ceiling with balloons. The idea of the prank was to fill every square inch of his office with balloons. In order to budget for such an endeavour, we needed to determine the amount of balloons it would take to fill up his office.

Our solution was, given the radius of the average balloon, place the balloons on top of each other and calculate how many balloons it would take to fill up such a space. Given that the size of the office was 9′ x 12′ x 9′, we calculated that approximately 1000 balloons would be needed to fill his office. While we realized that the balloons would not be stacked neatly into rows, we assumed that there would be some space between the balloons and this would by assuming the balloons were stacked, we would take this space into account.

Unfortunately budget was a limiting factor in this prank and we were only able to purchase 720 balloons. Approximately 30 balloons perished due to one member of our team blowing them up too large but we ended up placing approximately 690 balloons in the office. With 690 balloons, the office was slightly less than half full. While the prank was effective, I wondered how in the future such estimations can be made better.

So the problem comes down to how to determine the number of balloons required to fill up a given volume. This is called a “Fermi Question” because there is no exact answer. Variations in balloon size will guarantee that there is no exact answer.

There are lots of ways to solve this problem, but since using balloon dimensions did not have much success I decided to start with the volume of an actual balloon. I picked one balloon that was of “average” size and traced an outline of it. The first thing I noticed that, although the balloon was 11″, it was only blown up to 10″. This is likely because we had too many “casualties” from balloons blown up too much, so we made them a bit smaller on purpose. To calculate the volume I traced the balloon on graph paper and assumed that the balloon was symmetrical and created by a curve rotated about the x axis. I took a representative 16 points from the curve and used a curve fitting program to give me a best fitting polynomial. I then integrated the square of the polynomial times pi to obtain the volume of the balloon. Unfortunately due to the error from tracing, the curve fitting program, and calculations I threw the result out because I felt the error was too large. I could have written a program to reduce the error but I obviously have more serious things to do.

So I decided to estimate the volume of the balloon by assuming that the balloon is close the a sphere shape, which given some squishing is almost true. The volume of the balloon therefore came to approximately .30 sq ft.

If balloons were packed into the office in a way so no air was left between them, approximately 3,240 balloons would fit. Of course items are already located in the office, such as desks, bookshelves, boxes, and computers so the number must be less than this. Here we can rely on empirical evidence. Knowing that 1600 balloons would likely have been enough to fill the office based on the space left after placing 700 balloons in the office. A good formula for balloon coverage would be.

(Volume of room) / (volume of balloon) * 1/2

Of course I don’t feel entirely comfortable with just placing an arbitrary “adjustment factor” in there without sound mathematics. While surely we must adjust for the existing objects in the room, how can we best calculate the amount of space between the balloons?

Some obvervations are in order. When a balloon is dropped and allowed to find its own position, it tends to lay on one side with the bottom of the balloon touching the floor. If further balloons are placed on top of this balloon, it might be reasonable to assume that the balloon will move to its side from the pressure. Balloons placed on top of the bottom layer of balloons, however, will likely lay vertically or at an angle as the first balloon originally was. Given this model, a simulation could likely be created that would give a better estimate of the space between balloons. Of course static electricity could interfere with this, and I’m not sure what effect that would have.

If anyone has more insights into this I would appreciate the feedback.

Great! But where is the picture ? There is no such thing as http://nisd on the internet..

Your problem is closely related to <a href="http://mathworld.wolfram.com/SpherePacking.html">sphere packing</a>. If you assume that your balloons are round, you can use results for sphere packing to come up with a more appropriate fudge, er, <em>adjustment</em>, factor. Since you won’t be placing the balloons carefully, you might want to assume a random packing, which has a packing density of 0.6 to 0.65. If you’re even more concerned about underfill, you might want to use 0.76 (the greatest known packing density of unifrom ellipsoids).

I’d probably use 0.76.

Hey, not all of us are viewing on interal servers!!!

Great post btw.

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