## 06 July 2012

### Balloon-borne camera, part 6: You're gonna need a bigger balloon

Last time, we stuck 30+ balloons together and lifted the camera off the ground! As cool and Up-like as it looked, it just wasn't going to be practical for a single camp counselor to prepare this activity once a week. We decided to look for bigger balloons! Bigger is always better! Now that I've got you excited, be ready for some super-boring math!

How many big balloons would we need? It's not hard to come up with a formula for the volume occupied by 30 12-inch balloons (assuming they actually take a spherical shape when inflated):

$Vol = \frac{4}{3}\pi(\frac{12}{2})^{3} \cdot 30$

The first part is the formula for the volume of a sphere, where 12" -- the size of the balloon -- is the diameter of the sphere and therefore 12/2 is its radius. Then we tack on a 30 for 30 balloons. We want to find out how many balloons of a given size, S, will take up the same volume as 30 12" balloons, which we calculated would be about what we'd need to lift the camera and rig. The volume of large balloons is

$Vol = \frac{4}{3}\pi(\frac{S}{2})^{3} \cdot N$

Since we want the volumes to be the same, we set the equations equal to each other. A little rearranging gives us:

$N = 30\cdot\frac{(^{12}/_{2})^3}{(^{S}/_{2})^3} \rightarrow N = \frac{C}{S^3}$

Here we've put all the constant values into the C, to make clear the dependence on $^{1}/_{S^3}$.

This formula should be valid for lift if the air pressure inside the large balloons is the same as the air pressure inside the smaller balloons, and if all the balloons are more or less spherical once inflated. In any case, we're just going for ballpark figures here so this equation is only a guide. A little online research into balloon suppliers shows that balloons come in a few different standard sizes - 17", 24", 36", and very, very large.

You might be tempted to ask why we don't use super big weather balloons.

(Why don't you use super big weather balloons, Jonathan?)