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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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    $\begingroup$ A hovercraft with a balloon? It's unclear what you are talking about. $\endgroup$ Mar 15, 2015 at 17:52
  • $\begingroup$ From what little I've thought about this, and the discussions below, I'm guessing the bottom surface of the puck could be dished somewhat. It's not necessary, or maybe even desirable, for the air layer to be the same thickness throughout. The thickness that matters is where the air escapes to the outside, at the rim. (Real hovercraft have a "skirt" around the edge.) $\endgroup$ Mar 16, 2015 at 21:33

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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.

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  • $\begingroup$ Pressure proportional to flow rate? Usually pressure goes as flow-rate squared (at low velocity). Like in an airplane - double the speed, quadruple the lift. $\endgroup$ Mar 15, 2015 at 17:56
  • $\begingroup$ @MikeDunlavey - using hyperphysics.phy-astr.gsu.edu/hbase/pfric.html#veff flow rate scales with pressure difference. I am ignoring the small effects of density change given we are talking balloons here so that is a second order effect at best. $\endgroup$
    – Floris
    Mar 15, 2015 at 18:01
  • $\begingroup$ That's for viscous laminar flow in a tube. Flow through an orifice with a pressure drop is different. There, if you double the velocity, you double the momentum change per parcel of air, but you also double the number of parcels. (Density change does not enter, as long as the Mach number is low.) $\endgroup$ Mar 15, 2015 at 18:08
  • $\begingroup$ @MikeDunlavey - there is a tube connecting the balloon to the rest of the system... as long as that tube provides most of the resistance to air flow it will give a flow rate proportional to pressure. If not, then the relationship gets even crazier. Of course we have all seen how balloons deflate more quickly as they get smaller... Perhaps I will do some measurements and report back. $\endgroup$
    – Floris
    Mar 15, 2015 at 18:10
  • $\begingroup$ OK, the OP made it clearer what the situation is. It looks to me like the main pressure drop is in that thin film from center to edge. I would suppose it is mainly viscous, but the velocity would have to decrease with radius (assuming a flat bottom surface), so I'm back to square 1 :) $\endgroup$ Mar 15, 2015 at 18:13

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