I was disappointed with this one! I need hard numbers. What is the buoyancy to drag ratio?
drag coefficient should come from a simple sphere, Cd = 0.47
The buoyancy is dictated by the ratio of densities, but we have several complications.
- The expansion affect the Helium temperature, which impacts its density
- The balloon itself has some weight
I would still prefer to ignore these factors. So the basic idea is:
The density difference is (1.097 kg/m^3). The base density of air is (1.3 kg/m^3). We know gravity, and now I've defined everything else. The remaining unknowns are the velocity and the radius of the balloon. For that, we'll have to add one more bounding equation which takes into account the mass of the falling person. I'll use 100 kg for this, to be the standard human. This is a fairly large human, so it can either represent a nerd carrying extra equipment, or a marine with no extra equipment.
The force balance is simply:
Plug in the expression from above and we'll have an expression with v and R. This is what we want, because R can be adjusted to obtain a certain value of v.
Now that's better. Here is the Google search string to get the velocity with a radius of 2 meters:
Code: Select all
sqrt( (2*(9.8 m/s^2)*(100 kg)-(9.8 m/s^2)*(8/3)*Pi*(2 m)^3*(1.09 kg/m^3) ) / ((1.3 kg/m^3)*0.47*Pi*(2 m)^2) ) = 12.7 m/s
Now here is the velocity for the full range of radii.
At around 2.7 meter radius, the thing just stops. The above equation goes imaginary because it is no longer falling - it is floating. Here is a comparison of the drag and the buoyancy.
This is the juicy stuff I wanted to see. As the balloon gets bigger, the drag component drops off and the buoyancy gets larger. There's little impact from the buoyancy up until a radius of about 1 meter. After that, the cross-over happens fairly fast. But in a volume-based view it would be a good bit more gradual. Ignore the parts beyond 2.6 meters, it's junk. The thing is moving up after that point.