## Sunday, November 29, 2015

### Macy's Thanksgiving Day Parade Balloons

We just finished up with buoyancy, but are still talking about fluids, which I was thinking about while watching the Macy’s Thanksgiving Day Parade, and the physics of the parade balloons. The balloons would have a force of gravity pulling them down, and a force of buoyancy from the air pushing it up and allowing it to float (assuming it is a sunny and windless day).
ΣF=ma=0=FB-Fg
FB=mg
ρVg=mg
The balloon (specifically the Big Bird balloon) is filled with 12,000 cubic feet of helium, which has a density of 0.164 kg/m3, making the balloon weigh about 2,000 N (ignoring the weight of the balloon). In this case, the balloon is not accelerating in the y direction, so the forces must equal zero, meaning that the buoyancy force must be equal to 2,000 N. In the x direction, there would be a forward force by the force of the humans pulling the balloon, and the drag force of the balloon.
ΣF=ma=Fpulling-FD
ma=Fpulling-(½ρv2CDA)

For this case, the drag would have to be a less force than the force of the people pulling the balloons, since the balloon should be accelerating forward. Assuming the drag coefficient is .47, the density of the air is 1.225 kg/m3,the balloon has a radius of 6.5 m, and the balloon is traveling at 1 m/s, the drag force would be 32.2 N. So, the force of the people would have to be above this, on a windless day. Adding wind would drive the system out of whack, which is why there are actually many issues with the Macy’s Thanksgiving Day Parade, as seen in this video!