I'm working on a project and I have a question. How does the volume and pressure of the balloon affect the time the hovercraft hovers above ground?
This relates to my earlier question where I described the experiment in more detail.
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I actually added this to the answer to your other question but will repeat it here...
If the balloon is bigger, the time that the toy can hover will increase - by a surprisingly large amount. Using the result from my other answer that pressure (and thus flow rate) scales with $1/r^2$, and volume scales with $r^3$, then time (which is the time it takes for the balloon to deflate) will scale with $r^5$.
Proof:
From flow rate: $$\frac{dV}{dt} \propto \frac{1}{r^2}$$ From equation for volume of balloon: $$\frac{dV}{dr}=4\pi r^2\\ $$ Combining: $$\frac{dr}{dt} = \frac{\frac{dV}{dt}}{\frac{dV}{dr}} \propto r^{-4}$$ Integrating: $$t \propto r^5$$
In other words, a bigger balloon will allow for much longer floating time, assuming that the flow rate is proportional with the pressure (and that the pressure of the fully inflated balloon is still large enough to keep the craft floating). That's an interesting result I was not expecting.
Following a conversation with Mike Dunlavey in the comments, if the flow rate goes with pressure squared (as it might for a simple aperture), then the answer changes: the time for the balloon to empty would go with the seventh (!) power of radius.
I expect that experiment will give the answer, and would urge you, once you have built the project, to report back on your findings. It is probably easiest, given modern technology, to just constrain the hovercraft from floating about (surround it with three regularly space pins into the surface below) and film it against a background of graph paper. Estimate the size for each frame (or just find the frame where the radius is "one square smaller") and plot the log of the result. The slope will help you determine the correct value of the exponent.
