I'm wondering which offers more resistance: pulling an object at some speed through air, or holding the object steady against wind at the same speed.

I think initially people would think same resistance.

Then I thought that the air that is flowing has probably been compressed under it's own speed (or rather, in order to get to it's speed), and therefore offers more resistance. Also, the wind may be colder (thus denser) if it's the wind I'm used to, but disregarding temperature influence, do you think that my theory is correct? Can we quantify the resistance gained by windspeed?

Thanks in advance.

  • $\begingroup$ Wind isn't colder. It only feels colder due to evaporation. IMO, the answer to this question depends upon the method of generation of wind due to the pressure issue. $\endgroup$ – Manishearth Mar 17 '12 at 5:05
  • $\begingroup$ Ah true. I'm thinking of the usual method of wind generation. $\endgroup$ – user420667 Mar 17 '12 at 6:28

One might say, to be specific: a perfect wind with constant density $\rho$, pressure $P$, velocity $v$ produces the same effect on a body at rest, as on a body, which is moving with velocity $v$ in a still medium of pressure $P$ and density $\rho$. Should you want to specify any other physical parameters, they should be taken the same for the media in both cases.

The statement is guaranteed to be true by galilean invarince. You see a body moving with velocity $v$ in a medium, you start to move yourself with velocity $v$, and you see the body standing still and a wind blowing with velocity $v$. The same physics - just a different reference frame.

  • $\begingroup$ Is it also true for turbulent flow? In the direction normal to the movement of the object, I can imagine that the turbulent fluctuations are different. Is this what people working with wind tunnels neglect? $\endgroup$ – Bernhard Mar 17 '12 at 16:24
  • $\begingroup$ It is only about switching of the reference frames. If your wind is not perfect, and would possess turbulences itself without the body being there, then it is not the same case as a body moving through a still medium. In other words, the wind should be such that if you were moving along with it, it would look for you like a still medium. $\endgroup$ – Alexey Bobrick Mar 17 '12 at 16:51
  • $\begingroup$ So in practice, the windtunnel assumption is not completely correct. As there is not such thing as perfect wind, due to the non-linearity of the process. $\endgroup$ – Bernhard Mar 17 '12 at 18:04
  • $\begingroup$ Perfectly speaking, it is not the same. However, it is pretty easy to calculate out the wind in an empty windtunnel(or measure), and have a control over your conditions. $\endgroup$ – Alexey Bobrick Mar 17 '12 at 18:44
  • 2
    $\begingroup$ You need two thermodynamic variables to completely specify a thermodynamic state for a given composition. Pressure and density give you, say, the temperature, internal energy and whatever else you would wish. $\endgroup$ – Alexey Bobrick Mar 17 '12 at 22:32

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