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Let's say you have a scale model that is 1:10th the size of the real thing. You want to test this in a wind-tunnel.

Do you also scale down the windspeed 10x? Why or why not? I would think it has to be scaled down in order to get the right results for things like propeller tip speed.

Edit: I saw this question. It could be a duplicate, except I did not see an answer to my concern there. I want to know if windspeeds are scaled down in a windtunnel, whereas that one asks about how windtunnels are used in general and doesn't seem to mention any scaling of windspeeds.

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In order to achieve the same flow conditions, you need to match several similarity numbers. The most important one expresses the ratio of inertial forces to viscous forces within a fluid and is called Reynolds number (Re):

$$\text{Re} = \frac{v\cdot l\cdot\rho}{\mu}$$ Nomenclature:
$\kern4mm v\kern6mm$flow speed
$\kern4mm l\kern7mm$characteristic length along flow path, like wing chord
$\kern4mm \rho\kern6mm$fluid density
$\kern4mm \mu\kern6mm$dynamic viscosity of the fluid

The next one on the list is the ratio between flow speed and the speed of sound in the fluid and is called Mach number (Ma): $$\text{Ma} = \frac{v}{a}$$ Nomenclature:
$\kern4mm a\kern6mm$speed of sound

Others depend on the nature of the flow - if oscillating vortex shedding is involved, the Strouhal number needs also be observed, with elastic oscillations the Froude number or with heat transfer the Prandtl number.

In your case of a 1:10 model, the flow speed needs to be increased tenfold in order to match the Reynolds number. However, this will bring you in conflict with the Mach number, because the faster flow will probably already be supersonic. This problem is discussed at length in this and the linked answer, however, knowledge of the influence of flow speed is implicitly presumed.

Normally, in a wind tunnel test the flow speed is too low for correct scaling (which means that the viscous forces are higher in proportion to the inertial forces), and the consequences of this mismatch need to be covered by correction factors.

There is no need to scale forces correctly. Wintunnel models were hung on wires which were connected to scales to measure forces and now are mounted on a sting, the bending of which is captured by strain gages or an internal balance. Only the models used in free-fall tunnels for spin research will create as much lift as they weigh. The forces measured are used to compute coefficients, dimensionless numbers which are normalized for speed and size. When multiplied with the speed and size of the original, the (corrected) coefficients will yield the actual forces and moments.

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  • $\begingroup$ Another related Wikipedia article is the one on similitude. $\endgroup$
    – user14897
    Commented Jun 2, 2018 at 0:30
  • $\begingroup$ Highly counter-intuitive. Woulda thought windspeed should reduce 10x too, like a 5 m chord meant for 100 m/s scaled down to 50 cm and 10 m/s, so the wind is moving 20 chord lengths / s in both cases. However, I see in the Lift equation, area would fall 100x, and v^2 would also fall 100x... Whereas mass would fall 1k times based on volumetric density, but Lift falls 10k times. Ouch. Your model would drop like a stone. I guess there are too many factors and they can never all be scaled correctly. Anyway, glad I asked this and glad it got re-opened and answered here. $\endgroup$
    – DrZ214
    Commented Jun 2, 2018 at 5:17
  • $\begingroup$ @DrZ214: There is no need to scale forces correctly. Wintunnel models were hung on wires which were connected to scales to measure forces and now are mounted on a sting, the bending of which is captured by strain gages. Only the models used in free-fall tunnels for spin research will create as much lift as they weigh. $\endgroup$
    – Peter Kämpf
    Commented Jun 2, 2018 at 6:13
  • $\begingroup$ Yeah but if ailerons are just changing the camber of the wing, that's modifying lift, so you would get the wrong impulse or torque. Maybe I'm not understanding the actual purpose of windtunnels. I always imagined they tested the control surfaces and lift of the whole airframe at different angles of attack...at least, before the supersonic advanced stuff. $\endgroup$
    – DrZ214
    Commented Jun 2, 2018 at 12:08
  • $\begingroup$ @DrZ214: Yes, of course they test the control surfaces as well. They measure the forces before and after deflection and compute a dimensionless coefficient from the results. This coefficient is normalized for wind speed and size, so it can be transferred to the larger and faster original. $\endgroup$
    – Peter Kämpf
    Commented Jun 2, 2018 at 17:45
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It really depends on what data you need and the conditions you need it for.

Let's say you want to test a scale aircraft model in a wind tunnel in order to predict the lift forces on the full-sized aircraft during flight.

$Lift = q S C_L$

The dynamic pressure $q = \frac{1}{2} \rho v^2$

$S$ is the planform area

$C_L$ is the lift coefficient

$\rho$ is the air density

$v$ is the air velocity

What you really want to test for is the lift coefficient - you can use that to calculate the lift force during flight for a given dynamic pressure and planform area.

The lift coefficient will vary with Reynolds number and Mach number - these are the quantities you want to match.

So, you will not scale down the windspeed 10x - but it will be different from the flight-scale windspeed because the ambient conditions are different between the wind tunnel and flight at an altitude. The windspeed will be changed to give the correct Reynolds & Mach numbers.

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