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In the movie Transporter 3 a submerged car is floated to the surface by filling a large bag with air from the tyres.

I know that movies are about the worst places to get examples of physics in action, and my first thought was that if the air from the tyres was enough to inflate the bag and lift the car then wouldn't the air do that while still in the tyres? But then I wondered: is compressed air less buoyant than air at atmospheric pressure?

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  • $\begingroup$ movies are one of the best places to get examples of bad physics in action! $\endgroup$
    – gregsan
    Commented Sep 24, 2013 at 7:27
  • $\begingroup$ @mods: add "popular-science" (then delete this) $\endgroup$
    – Dave
    Commented Sep 24, 2013 at 13:53

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Typical air pressure is about $15psi$, typical car tire pressure ratings are $\approx 30psi$. So the air will go from a pressure of about $45psi$ to about $15psi$, so the air in the tires will expand by a factor of something like $3$ (I'm assuming that the car is in shallow enough water that we can ignore it's pressure).

My ROM for the volume inside the tires is something like $\pi \times 1m \times 0.1m \times 0.2m \approx 0.06 m^3$ (one meter diameter, 10cm by 20cm cross section). Using all four tires gives us a compressed volume of $\approx 0.25m^3$.

Multiplying this by the expansion factor gives us something like $0.75m^3$.

The density of water is $1000kg/m^3$, so the overall bouyancy is $(750kg)\times g\approx7500N$. (The density of the air itself is about $1kg/m^3$ so I've ignored it in computing the buoyancy force.)

Although this calculation yields a result that gets to the order of magnitude, this result is too small by a factor of at least $2$ and probably $4$ to actually float the car. Possibly under special circumstances, larger wheels and a very light car, this could be done.

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  • $\begingroup$ what's a rom? rough order of magnitude? $\endgroup$ Commented Sep 24, 2013 at 2:15
  • $\begingroup$ @NowIGetToLearnWhatAHeadIs yes, ROM stands for rough order of magnitude; my intended use here is "as a crude approximation". $\endgroup$
    – Dave
    Commented Sep 24, 2013 at 2:28
  • $\begingroup$ Be careful to convert all pressures used to absolute pressure, before calculating expansion factors. If atmospheric pressure is 15 psi and the tire gauge pressure is 30 psi, then the expansion factor is 3;$$EF = \frac{30+15}{15}=3$$ $\endgroup$
    – DJohnM
    Commented Sep 24, 2013 at 3:35
  • $\begingroup$ @User58220 : thanks. I had assumed that gauges measure absolute pressure. updated. $\endgroup$
    – Dave
    Commented Sep 24, 2013 at 15:50
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The weight of fluid displaced by an object is the bouyant force on the object. Whether the fluid is displaced by a vacuum (in a light, rigid container) or compressed air makes no difference, the bouyant force is the same.

Compressed air will have a greater downwards weight, so a bouyant bottle of compressed air has less nett bouyancy than an evacuated one: the upwards bouyant forces are the same in both cases, but the compressed air weighs the bottle down, whereas the vacuum doesn't.

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Bouyancy is simply about relative density - the density of an ambient medium and the density of an object within that medium.

In the scenario described, there is a fixed quantity of air in the tires. To do their normal job, the tires will be inflated to (perhaps) 30 psi or so over ambient air pressure (out of the water). Released from the tires, but still contained in a bag, you'll still have the same quantity of air (mass will be the same), but at a lower pressure it will occupy a larger volume. The total system (car, tires, and bag) will not change in mass, but will displace more water, so will have lower density, which means greater bouyancy.

Movie science must always yield to drama and the needs of the story; in the movies, there's certain to be more than enough air in the tires to float the car; in the real world, probably not. It would have the effect depicted (would increase bouyancy), but almost certainly not enough to achieve the result appearing on screen.

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  • $\begingroup$ Recently became a diver - and bouyancy, air pressure, density, have become a fascination. I guess then the logic follows the deeper down the less boyant my air becomes (not considering the tank). $\endgroup$
    – terary
    Commented May 7, 2016 at 9:35

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