Why does this curved wind tunnel rotate, why is it not just curved? A curved wind tunnel has a radial pressure gradient, but this pressure gradient doesn't exist in a real turn, so how do you get zero pressure gradient in a tunnel section?

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    $\begingroup$ It would seem rotation is required to get the outside of the car model to go faster than the inside. One might also consider turning the model on a track with all motion generated by the model, as in real life. $\endgroup$ Commented Mar 24 at 14:15
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    $\begingroup$ @RobertDiGiovanni Flow in curved tunnel is already faster outside then inside... $\endgroup$
    – 22flower
    Commented Mar 24 at 14:30
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    $\begingroup$ Ah, but is it, especially when the apparatus is turning in the opposite direction (viscosity) of flow. I would imagine flow velocity and turn rate would have to be fine tuned to achieve desired results. The design has merit. $\endgroup$ Commented Mar 24 at 15:33
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    $\begingroup$ OK, we'll take the car model out and put a little airplane model in there, OK? Isn't understanding airflow related to aviation? $\endgroup$ Commented Mar 24 at 20:03
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    $\begingroup$ (ctd) Observer onboard test model, on the other hand, sees difference in airspeed from one wingtip to other, just as they would if tunnel were not there at all (discounting the disturbance that would arise from repeatedly flying through own wake.) The whole idea becomes more ingenious the more it has time to "sink in" what is really going on here-- $\endgroup$ Commented Mar 25 at 18:36

4 Answers 4


While superficially similar, this curved windtunnel solves a very different problem from a straight windtunnel. A straight windtunnel is used to study steady flight without needing a very long straight line. A rotating windtunnel is used to shield from wake turbulence from the last pass through the circle.

To simulate straight flight, you could just fly an airplane in a straight line. However, for any meaningful test duration, you would need a very long and straight section of air, and all your instrumentation would need to travel with the plane. Hence, a windtunnel moves air instead of the plane, which is cheaper than a sled on a very long rails.

To simulate curved flight, you could just mount an airplane on a rotating boom, like a merry-go-round, going in circles in stationary air. Unlike straight flight, you 're-use' the same bit of air by just going in circles. However, there is one problem: once you've gone full circle, you enter your own wake, which is not stationary and may contain turbulence. The curved windtunnel from your link is nothing more than a very fancy 'wind shield' that makes sure the rotating plane flies through exactly stationary air (stationary relative to an outside observer), without any wake from the previous pass.

A stationary curved windtunnel would not be the same, because now the air mass needs to curve. This introduces very different phenomena. Perhaps easiest to understand is the pressure gradient, because the air is 'flung' to the outside when going in a circle by centrifugal forces. On top of that, the flow path is disturbed because of Coriolis forces, which are less intuitive but just as real.

  • $\begingroup$ So when tunnel rotate radial pressure gradient is zero? How coriolis forces appear in tunnel? $\endgroup$
    – 22flower
    Commented Mar 25 at 21:37
  • $\begingroup$ When the tunnel rotates, the air is stationary. The radial pressure gradient is then zero, and there are no coriolis forces. Because from the perspective of the air, it's just sitting still, so there are no forces. $\endgroup$
    – Sanchises
    Commented Mar 26 at 6:07
  • $\begingroup$ If tunnel dont rotate, why coriolis force exist if we dont have air movement in radial direction, air particles moves in pure circular path? $\endgroup$
    – 22flower
    Commented Mar 26 at 10:09
  • $\begingroup$ @user707264 That is correct, however: when you put a scale model in the airflow, the air particles no longer go in an exactly circular path. So then they do experience a coriolis force, and the resulting flow field is different. $\endgroup$
    – Sanchises
    Commented Mar 26 at 10:39
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    $\begingroup$ @quietflyer Yes that is correct, if I sit on rotating merry go around, coriolis force is zero but if I try to walk toward center or outward then is not zero.. $\endgroup$
    – 22flower
    Commented Mar 26 at 11:51

Here you'll find his paper about the wind tunnel. Apparently the rotation has been introduced in order to compensate for the Coriolis effect.


Key point from another related answer:

A stationary curved windtunnel would not be the same, because now the air mass needs to curve. This introduces very different phenomena. Perhaps easiest to understand is the pressure gradient, because the air is 'flung' to the outside when going in a circle by centrifugal forces. On top of that, the flow path is disturbed because of Coriolis forces, which are less intuitive but just as real.

So we keep the air stationary, and we move the test model around in a circle.

(See 2:04 in the video to see the model moving in a circle.)

As seen by an outside observer, the air inside the tunnel should actually not be moving at all.

More specifically, this should be true of the air at the location where the test vehicle would normally be, at any given instant, as seen by an outside observer, in a test run where no test vehicle is actually present.

So the fan should not pump air through the tunnel at high speed. The fan should only provide enough force to prevent the test model, and the ducting itself, from pulling the air inside the tunnel along with it as the whole apparatus whirls around in a circle.

What does this really mean? What is meant by the statement at 1:42 in the video where the narrator states that "the axial fan is required to produce the required pressure overrides to overcome the losses within the system"?

The three-dimensional geometry involved in the airflow inside the rotating ductwork is complex. What does it really mean to say that the airflow approaching the moving vehicle at any given instant has zero velocity as seen by an outside, stationary observer?

From the point of view of someone in the test vehicle, we can imagine that we have a tube coming down from above that is magically constantly "unbending" somewhere out in front of us to continually provide a new segment of horizontal (curving) tube full of air (air that is in motion relative to us, but stationary relative to the outside world) for us to pass through. So air that was formerly above us, is constantly being brought down to our level. We pass rapidly through that airmass, which is then released out of the open end of the tunnel behind us. Since the tunnel walls move along with the test model, there is clearly always a non-zero airflow velocity relative to any given point on the wall of the wind tunnel. Looking at the apparatus as a whole, air is constantly being pumped in through the top and exiting at the bottom. Yet the instantaneous velocity of a parcel of air as it is approached by the moving test vehicle is always zero as seen by an outside observer. Admittedly, this is a bit of a mind-bender!

Here's yet another take on the whole problem: A rotating ductwork apparatus of the shape described in the video is one way to take the airflow from a fan and decelerate it to zero velocity, as seen from the viewpoint of an external, stationary observer.

Could we accomplish the same thing while leaving the walls stationary?

  • Initial response: No, because if the model is simply spinning around inside a donut-shaped chamber, you have no way for the fan to exert a force on the air inside that chamber in any particular direction to keep the air inside the chamber from being dragged along with the moving vehicle. For the fan to do any good, it has to be inside a duct with a beginning and an end, one portion of which is bent into an arc of the appropriate curvature. If you try to spin a model inside something like that, you'll soon hit a wall. The only way to spin a test vehicle around inside something like that, is to rotate the tunnel along with the test vehicle.

  • But, actually... Possibly. The tunnel could be an enclosed donut (torus) with stationary walls, with the test model whirling around inside, supported by an arm protruding through a slot in the inner wall of the tunnel, or possibly attached to a little motorized or magnetically-propelled cart that runs around the tunnel on a track like a toy train. On the opposite side of the tunnel, supported in the same fashion and moving around the tunnel at the same speed, we could have a fan that blows just hard enough to decelerate the air in the tunnel to zero windspeed as seen by an outside observer. The purpose of this fan is to prevent the test model, supporting arms, and any other moving apparatus inside the tunnel from eventually dragging all the air in the tunnel along with it. Straightening vanes behind the fan could smooth out any vortices or other turbulence and ensure that the air downwind of the fan was truly stationary as seen by an outside observer. One significant advantage of this arrangement would be that since the tunnel walls are stationary relative to the air in the tunnel, especially the air immediately in front of the moving test model at any given instant, we wouldn't have wall friction to contend with in our quest to create a uniform of chunk of air with zero windspeed (as seen by an outside observer) immediately in front of the moving test model. But we do have the complication of devising a method to drive the test model (and the fan) around inside the enclosed tunnel, or if we are mounting them on whirling arms with a central pivot point, we'll need a slot in the inner wall of the tunnel for the arms to protrude through.

Some other key points from the video:

  • In a turn, the airflow is different than when travelling linearly.

  • In a turn, the airflow is curved, as seen from the vehicle's perspective.

  • This means that the flow is faster over the "outside" side of the vehicle than over the "inside" side of the vehicle.

  • This also means that the angle of the (free-stream) airflow approaching the front of the vehicle is different than the angle of the (free-stream) airflow leaving the rear of the vehicle.

As has been explored in other ASE questions and answers, these effects are extremely relevant to airplane flight dynamics as well. For example, at least in the special case where rate of curvature of flight path, yaw rate, and pitch rate are all constant, the "aerodynamic damping" effect on pitch rate and yaw rate can be also viewed as a change in local angle-of-attack (e.g. of the horizontal stabilizer, vertical stabilizer, etc) due to the free-stream relative wind curving in accordance with the curvature of the flight path.

Some other links relating to the curvature of the free-stream "relative wind" in turning (circling) flight:

Other ASE questions or answers:

What is airflow direction in turn? (Answer) -- see especially the related links at end

How is a "turning polar" constructed? (Answer)

What has happened to make me experience negative G with the control stick FULL AFT near the top of a loop? (Answer)

External links:

Section 6.1.10 "Long-Tail Pitch Effect" from John S. Denker's "See How It Flies" website

Section 8.10 "Long-Tail Slip" from same website

"Spiral Stability and the Bowl Effect" series from "Model Aviation" magazine, by Blaine Beron-Rawdon-- pertains to stability and control of rudder-controlled rc sailplanes-- the "bowl" refers to the curving lines of the free-stream relative wind in circling flight

  • $\begingroup$ @Sanchises -- On further reflection, I now agree completely w/ the other related answer aviation.stackexchange.com/questions/104426/… . I just didn't fully see it, till I finished typing this answer! $\endgroup$ Commented Mar 25 at 16:15
  • $\begingroup$ I'm using "free-stream airflow", "free-stream relative wind", and "relative wind" interchangably here. To mean the instantaneous airflow direction that would be seen at any given point on the vehicle, if the physical presence of the vehicle were not having any influence on the airflow direction. The curvature I'm referencing is entirely due to the the fact that the trajectory of the vehicle is curved. So the vehicle is rotating as well as linearly translating, and at any given instant two randomly chosen points on the vehicle are likely not moving through the air in the same direction. $\endgroup$ Commented Mar 26 at 15:07
  • $\begingroup$ So the focus is on the direction of the undisturbed airflow, nothing to do with eddies, turbulence, etc. Could possibly explain that a little more clearly and/or use more consistent terminology throughout the answer. $\endgroup$ Commented Mar 26 at 15:08
  • $\begingroup$ "Could we accomplish the same thing while leaving the (TUNNEL) walls stationary? Initial response: No, because if the model is simply spinning around inside a (STATIONARY) donut-shaped chamber," $\endgroup$ Commented Mar 26 at 17:57

An automobile with front steering pivots on its rear wheels. This may cause a significant amount of side lift in front from a change in the relative wind, making it more difficult for the tires to hold the turn.

A aircraft may have some side lift at the nose in a turn, but this is controlled by rudder and elevator (wing). In a properly coordinated turn, the effects will be minimal. However, in uncoordinated flight, effects may also be significant.

flow curvature is achieved in the absence of a pressure gradient

This is why the apparatus rotates. The outside wall turns away from the flow and the inside wall turns towards it. This (ingeniously) means the airflow from the diffuser frame of reference travels in more of a straight line (minimizing pressure gradient), whereas from the test model perspective it is curved. (This model is actually traveling in a circular path through straight air flow!).

Awareness of lift created by a change in relative wind (by certain shapes) helps us understand how a rounded fuselage can actually promote spin tendencies in aircraft once they have "gone flat", having the relative wind passing over the fuselage at a proper angle to generate a lifting force in the spin direction.

  • $\begingroup$ RE "an automobile with front steering pivots on its rear wheels." -- actually I'd suggest that for any given position of the steering wheel, the car tends to adopt a position where front and rear wheels are all "cutting" across the curving line of the actual trajectory at equal and opposite angles, i.e. the curving line of the instantaneous trajectory is striking the outside walls of the front wheels and thus creating a nose-to-inside torque, and also is striking the inside walls of the rear wheels and thus creating a rear-to-outside/nose-to-inside torque. $\endgroup$ Commented Mar 25 at 22:47
  • $\begingroup$ Said torques are expended to overcome any aerodynamic "damping" or aerodynamic resistance to yaw rotation. Also to supply torque to increase yaw rotation rate. If aerodynamic damping in yaw is minimal, and yaw rotation rate is constant rather than increasing, so zero net torque is required, then all wheels can essentially be fully in line with the curving arc of the instantaneous trajectory, regardless of which wheels are actually being "steered" by the steering wheel. Above theory is for CG approx midway between front and rear wheels, may need to be slightly modified if this is not the case. $\endgroup$ Commented Mar 25 at 22:50

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