# Wind as a dynamic problem

1. Sep 10, 2015

### Cristiano

When I see a paper where it is explained the wind effect on an airplane, I always see a kinematics problem.
Now I'm interested in calculating the wind effect starting from the force that the wind exert on the aircraft surfaces.
I know that F= 0.5 * rho * V2 * A * C and I also know that if we have 10 m/s head wind, the airplane decelerate of exactly that speed, but when I put that force in my program, I obtain a much smaller deceleration.
Please, could someone explain?

2. Sep 10, 2015

### DEvens

Hello Cristiano. Welcome to the forum.

You put the 10 m/s into what program? You put it in how? And with what context? You know that a 10 m/s head wind causes the plane to decelerate that much under what conditions?

It is very hard to explain what some unknown computer program did under unknown conditions that don't seem to compare to some other unknown conditions.

3. Sep 10, 2015

### Cristiano

I post a question and you answered with additional 4 questions?
I reformulate my question in a plainer form.

An airplane is cruising at 100 m/s; then it enters into a 10 m/s head wind; its speed becomes 90 m/s.
Please, using forces, could somebody demonstrate why?

4. Sep 10, 2015

### DEvens

Sure. The plane is at aerodynamic equilibrium at a relative air speed of 100 m/s. That means that forces balance at this speed. The 10 m/s ground speed of the head wind means that the ground speed of the plane must adjust so that it still has a 100 m/s air speed. Otherwise, forces will not balance. 100 minus 10 is 90.

You seem to think you are asking something else. Maybe you could expand on what that something else is?

5. Sep 10, 2015

### Cristiano

You wrote: "The 10 m/s ground speed of the head wind means that the ground speed of the plane must adjust so that it still has a 100 m/s air speed.", which is easily demonstrated by adding the velocity vectors, but I can only use force vectors and hence, instead of write Vplane - Vwind, I must write Fplane - Fwind, in other words, the wind is an additional drag for the plane. How can that be written?

6. Sep 10, 2015

### rcgldr

The deceleration related to entering a head wind is only temporary. From the plane's perspective, the relative wind speed increases from 100 m/s to 110 m/s and it the plane slows down asymptotically until the relative wind speed is back to 100 m/s (assuming no throttle changes).

In a real world situation, the plane would climb upwards (increased lift) and assuming positive pitch stability, it would pitch upwards, which would slow it down faster than what just the increase in aerodynamic drag would do, then the plane would return to level flight at a somewhat higher altitude.

7. Sep 10, 2015

### Cristiano

Good! I need to write something (formulas/algorithm) to calculate what you said.

8. Sep 10, 2015

### Staff: Mentor

If the plane were to keep its attitude, it should be easy to see from that equation that a 10% increase in speed results in a 21% increase in drag.

Of course, as said, the interaction in real life is far more complicated, so what else you want to include is determined by how sophisticated you want the model to be.

9. Sep 10, 2015

### cjl

Of course, if the plane keeps altitude, this is not in fact true, since in the higher wind speed, the lift coefficient must be reduced to maintain the same lift (and thus not gain altitude). When the lift coefficient is reduced, the induced drag will also be reduced. Depending on the details of the airplane, the total drag may go up very nearly 21% (if the induced drag was already low and the total drag was dominated by profile drag), or the total drag may rise much less, or even decrease (if the airplane is close to stalling and overall drag is dominated by induced drag). If the total drag decreases (the airplane was close to stall) and the control surfaces are adjusted at all times to maintain level flight, the aircraft may never return to its original speed - it may continue to fly faster than it was prior to the perturbation.

10. Sep 10, 2015

### rcgldr

At 100 m/s == 360 kph, most aircraft (at least sub-sonic aircraft) will be at cruise speed, where parasitic drag is the dominant factor. A propeller's or high bypass jet engine's thrust is going to be reduced if it's rpm and pitch stay constant when the aircraft first encounters a head wind. The mass versus drag is going to affect the rate of deceleration.

11. Sep 10, 2015

### Staff: Mentor

Agreed, but I said "attitude", not altitude. I purposely presented the simplest possible case.

12. Sep 10, 2015

### cjl

Ahh. I misread that. Sorry.

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