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This following text (which is page 301 of a book called A History of Aerodynamics: And Its Impact on Flying Machines) seems to say that you must increase air pressure (or similar) to accurately model aerodynamics of a small-scale model:

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I'm semi-literate in Physics (undergraduate-level some decades ago, but I don't readily understand the Wikipedia article). Could you please explain why this phenomenon is so, preferably with an intuitive or verbal explanation, not only mathematical formulae?

And, perhaps more importantly, I think it's saying that if you're using a 10:1 scale model (e.g. a 7 metre model of a 70 metre airplane) you should test it at 10 atmospheres of air pressure to get realistic results. Given that it's 10:1 scale what should the wind speed be in the wind tunnel? Should it be 1000 km/hour, just like the real thing? Or should it be 100 km/hour to reflect the fact that distances are scaled down by a factor of 10?

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The Reynolds number is defined as:

$$ \text{Re} = \frac{\rho U L}{\mu} $$

where $\rho$ is the fluid density, $U$ is a characteristic velocity, $L$ is a characteristic length scale, and $\mu$ is the molecular viscosity. Broadly speaking, it can be thought of as the ratio of inertial (ie. convective) to viscous forces.

The trick in defining a Reynolds number is picking appropriate scales for the problem. $\rho$ and $\mu$ are set by your operating conditions while $U$ and $L$ are chosen by the practitioner. There are conventions to choose of course, for example when testing a wing, the $U$ is the freestream velocity and the $L$ is the chord length (size of the wing normal to the flow).

Physically speaking, when the Reynolds numbers of two configurations are equal, the flow physics is roughly equivalent. I say roughly because it isn't the only non-dimensional number to relate -- there's also Mach number, Prandtl number, etc etc.. But, depending on the physics of interest, matching Re may be good enough.

So, in that context, it becomes easier to understand how scale models work. If you decrease $L$ by a factor of 10, you need to do one of 3 things to keep Re the same -- increase density by a factor of 10, increase velocity by a factor of 10, or decrease viscosity by a factor of 10.

Generally speaking, the only way to change density without changing viscosity is to increase pressure. So, if you want to keep working with an air tunnel for your scale model, you would need to increase pressure by a factor of 10 and that will make the Re match.

Alternatively, you could run the tunnel faster. But if your real velocity is, say, 100 m/s, making it 1000 m/s for your scale model will completely change the physics -- it is now supersonic. So you can't do that.

You could change the working fluid. If you run in a water tunnel, density is roughly 1000 times bigger. This, with the reduction of length by a factor of 10, means we still need to change something else. Viscosity also changes (I don't remember offhand how much), but let's pretend it stayed the same. That means we would also reduce our velocity by a factor of 100.

So, if our real problem is in air at 100 m/s, our made up water tunnel (with fake viscosity) using a 1/10th scale model at 1 m/s should have the same physics. Assuming the physics we want is inertial/viscous forces and not, say, heat transfer.

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  • $\begingroup$ you just beat me to it. Nice answer. $\endgroup$ – ZeroTheHero Feb 21 '17 at 13:52
  • $\begingroup$ I think you're saying that increasing the air pressure keeps the Reynolds number constant, so there's no need to decrease the velocity too. So, for example, a 70 metre airplane at one atmosphere and 1000 kph will behave like a 7 metre model at ten atmospheres and 1000 kph. $\endgroup$ – ChrisW Feb 21 '17 at 13:59
  • $\begingroup$ @ChrisW Yes, exactly. Because increasing pressure by a factor X also increases density by a factor X, which offsets the reduction in L by X. But there's more than one way to do that -- changing medium to something like water and playing with the velocity could also work. But, going from 1000 kph to 10,000 kph alone to offset the change in length would not work -- it drastically changes the physics. $\endgroup$ – tpg2114 Feb 21 '17 at 14:02
  • $\begingroup$ Can you predict what the lift and drag will be, approximately, of the 7 metre model at ten atmospheres and 1000 kph? Will those forces equal the lift and drag experienced by the 70 metre airplane? i.e. about 500 tons of lift? $\endgroup$ – ChrisW Feb 21 '17 at 14:03
  • $\begingroup$ @ChrisW Yes. You would measure the drag and lift of the scale model, which would be plugged into the equations $D = 1/2 \rho U^2 C_D A$ and $L = 1/2 \rho U^2 C_L A$ to find the drag and lift coefficients $C_D$ and $C_L$ ($A$ is the area). These, assuming the shape is scaled the the same, would be the same between small model and full-size. Then you plug in the numbers for $\rho$, $A$, $U^2$ of your full-size configuration and get the drag and lift. $\endgroup$ – tpg2114 Feb 21 '17 at 14:06

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