Balloon-borne camera, part 7: No more helium!

I'm going to go right ahead to some bad news that necessitated a major change in our strategy: Canada has no more helium gas for sale and more won't be available until September. Well, there is some helium around, but it's reserved for important applications like MRI and not for party balloons. This is apparently the result of a 1996 law that states the US government, owner of the world's largest helium reserve, has to sell it off by 2015. I don't really get it, but maybe you can do better by reading this Popular Mechanics article.

The upshot is that the day after we got these amazing pictures:

we learned that it would be impossible to buy enough helium to supply 7 weeks of balloon inflation. So frustrating! The balloons worked so well!

At least we weren't flying blind in trying to find a new strategy. After all, I had adapted kite aerial photography techniques for balloons, so we decided to try to go back to kites. More on that next time!

Interlude: We have a Higgs !/?/*

CERN sent out a press release two days ago announcing the discovery of a new particle! The CERN press release is kind of boring and doesn't have any pictures. The statements from the ATLAS and CMS experiments are exciting and have lots of pictures!

This is my favorite one:

Di-photon (γγ) invariant mass distribution for the CMS data of 2011 and 2012 (black points with error bars). The data are weighted by the signal to background ratio for each sub-category of events. The solid red line shows the fit result for signal plus background; the dashed red line shows only the background.

Here is where I try to explain the graph clearly. It is a histogram, meaning it just counts events. The vertical axis keeps track of how many times an event has happened. The horizontal axis is particle mass*. So, for example, we see about 500 140 GeV particles, and about 1200 110 GeV particles.

What is this particle? We expect the Higgs boson to decay into two photons. The combined energy of the two photons is the same as the energy of the initial Higgs boson, because energy is conserved. This is the signal (S in the plot). We also expect other particle decays to produce two photons. This is the background (B in the plot). This happens very frequently at low energies, and less frequently at higher energies.

If we had just the background, we would expect a smooth decay, but what's going on at 125 GeV? Something is producing lots extra two-photon events! This must be because a new particle is being created at that energy! It produced two photons, so it must be a boson!

Technically, that's all we know so far about the new particle - it's new, it has a mass of about 125 GeV, and it's a boson - but it's very likely the Higgs boson that we've been looking for!

*As I explain in one of the following paragraphs, it's really the combined energy of the two photons but from conservation of energy we know that this must equal the mass of the particle that produced them.

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?)

Thanks, I'm so glad you asked!

It turns out that they are usually made of thin rubber, which is easy to tear if you tie it to the ground with rope. Since we're going to tie it down to the ground with rope, we thought it wouldn't be a good idea. Plus, half the fun was to use a bunch of really colorful balloons instead of just one giant drab one! Moving on, this formula leads to the following conclusions: You need 10-11 17" balloons, 4 24" balloons, or about 1 36" balloon to equal the volume of 30 12" balloons. In the end, we ordered a bunch of 36" balloons because that's what we were able to find online, and also because it would make it really easy to add a lot of lift just by adding a single balloon.