# How can you calculate air resistances at different speeds?

I've read that at 50mph air resistance to an average car is the equivalent of driving through water and at 80mph it's the equivalent of driving through oil.

I can't find any references online to back up these figures. Is there a relatively simple way to explain the calculation to verify these figures, or are they incorrect?

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While there are useful calculations one can try and do, usually numerically, fluid dynamics is enormously complicated and these figures must always come down to experiment, and particularly to wind-tunnel measurements of these drag forces. I don't really know where one would look for such figures, though. –  Emilio Pisanty Oct 16 '13 at 19:33
I don't understand (the body of) the question. Please clarify. Do you want to compare air resistance with rolling resistance through fluid covered roads? (At what speed?) –  Glen The Udderboat Oct 16 '13 at 20:54
@GuyThreedee, you may want to clarify the body of your question to match the title. Are you just wanting the relationship between air resistance and speed? –  user29350 Oct 16 '13 at 21:33
Hi, I'd like to know whether the air resistance examples in the body of the question are approximately correct, and, if so, how can you calculate such examples. Thanks. –  Guy Threedee Oct 18 '13 at 11:38

If I understand your question correctly, one aspect that you seem to be asking about is the relationship between and object's speed and the associated air resistance (as in your title) - which is often referred to as 'drag'.

The Physics Hypertextbook chapter Aerodynamic Drag, generally relates drag, defined as:

The force on an object that resists its motion through a fluid is called drag ($R$). When the fluid is a gas like air, it is called aerodynamic drag (or air resistance).

as being proportional to the square of the object's velocity ($v$) as:

$$R \propto v^2$$

So, you can see that as you get fast, the effect of drag increases faster.

This relationship is from the drag formula:

$$R = 0.5\rho CAv^2$$

(where $\rho$ is the fluid density, $C$ is the drag coefficient and $A$ is the surface area affected)

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This parameterization works over a pretty wide range of velocities for reasonably simple objects. Breaks down nearing mach 1, of course and at very low speeds. –  dmckee Oct 31 '13 at 8:37
@dmckee that is very true, I forgot about that point - but in a general sense, it is a good estimation. –  user29350 Oct 31 '13 at 8:38

The resistance to flow, can be "naively" said as proportional to the velocity. It is in fact proportional to the integral over the area times the velocity gradient. The constant of proportionality turns out to be viscosity for Newtonian fluids. The exact calculations are tedious and depends on the nature of the problem.

To answer your question, you could say that it is proportional to the ratio of $\mu v$ where $\mu$ is the viscosity of the fluid and $v$ is the relative velocity under consideration. Now $\mu$ is an experimentally determined quantity and it varies between what "oil" are you referring to.

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That is only for low Reynolds number. For things like autos, aircraft, etc. Reynolds number is high, and drag is proportional to the square of velocity. –  Mike Dunlavey Oct 21 '13 at 16:01