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 Control Surface Efficiency
 Jaylink From United States of America, joined Jun 2006, 44 posts, RR: 0Posted Sun Jul 8 2007 20:01:57 UTC (8 years 11 months ago) and read 2622 times:

 Does anyone know where I could find detailed information on the physics of control surfaces? For example, at 20 knots, even a full extension of elevators will have no measurable effect. Then, sometime before Vr, the control surfaces will become usable, but sloppy. Then, past V2, they will become crisp and responsive. So, let's say (for example) I'm in a 737-700, and I pull back all the way on the yoke immediately upon starting my takeoff roll. How can I determine at what speed the nose wheel will leave the ground with full elevator extension? Presumably this will be before the aircraft would begin to rotate with only a partial extension. Thanks.
 SilverComet From Mauritius, joined Apr 2007, 85 posts, RR: 0 Reply 1, posted Mon Jul 9 2007 00:12:33 UTC (8 years 10 months 4 weeks 1 day 19 hours ago) and read 2600 times:

 Quoting Jaylink (Thread starter):How can I determine at what speed the nose wheel will leave the ground with full elevator extension?

As far as I know, no such speed is defined in the books. For a particular aircraft (in your case the 737), this speed will largely depend on the weight distribution (position of aircraft CG w.r.t main landing gear). The further forward the CG the greater the downforce required on the elevator to lift the nose. This means a higher speed is required with a full elevator deflection. I don't even think the charts provided in the aircraft performance manuals allow you to work out that speed given the position of the CG.

In my opinion the only way you can find out is to work out the numbers on your own knowing:

1. distance between elevator and MLG
2. distance between aircraft CG and MLG
3. OAT and pressure (read: density)
4. surface area of elevators
5. some basic knowledge of levers and moments

Downforce on elevators will then be (ideally) equal to: (1/2) x density x area x speed squared x K;

where K is some constant that depends on the shape of the elevators and their angle of attack, among other things.

All that being said, the closest defined speed to what you are looking for is the minimum unstick speed, Vmu, which is the minimum speed at which the aircraft will get off the ground with the elevators fully deflected and the wings at the maximum AoA possible on ground (that means the tail is scraping the runway). This speed can be worked out from the performance charts given the weight of the aircraft and the outisde conditions (temperature, pressure).

 Jetlagged From United Kingdom, joined Jan 2005, 2620 posts, RR: 25 Reply 2, posted Mon Jul 9 2007 19:13:16 UTC (8 years 10 months 4 weeks 1 day ago) and read 2541 times:

 Quoting SilverComet (Reply 1):All that being said, the closest defined speed to what you are looking for is the minimum unstick speed, Vmu

Actually the speed Jaylink is trying to determine is defined, as Vmr (minimum rotation speed).

As well as the five things SilverComet listed there are some others you would need to take account of:

Aircraft mass (CG position alone is not much use).
Stabiliser trim angle.
Approximate Cl v AOA for the stabiliser.
Stabiliser area.
Zero lift pitching moment of the wing.

Stabiliser angle is a key parameter, because if this is mis-set the aircraft may not rotate at all. It will certainly affect Vmr.

So there are lots of veriables involved, and much guess work.

As for the basics of control surfaces, there's some basic stuff here:

http://www.grc.nasa.gov/WWW/K-12/airplane/rotations.html

Basically, control surfaces get more effective as airspeed increases, in fact their effectiveness increases with speed squared, but reduces with air density. This explains why it can be next to nothing at 20 knots, yet very significant at 100 knots.

 Quoting SilverComet (Reply 1):Downforce on elevators will then be (ideally) equal to: (1/2) x density x area x speed squared x K

(1/2) x density x speed squared is the air flow's dynamic pressure, multiplying by the area gives a force. K would vary according to elevator deflection. For simplicity it could be assumed to be a linear relationship.

 Quoting Jaylink (Thread starter):Presumably this will be before the aircraft would begin to rotate with only a partial extension.

Correct. At half deflection you would need roughly twice as much dynamic pressure to get the same pitching moment, so about 41.4% more speed, i.e. square root 2 times the Vmr for full elevator deflection.

 The glass isn't half empty, or half full, it's twice as big as it needs to be.
 SilverComet From Mauritius, joined Apr 2007, 85 posts, RR: 0 Reply 3, posted Mon Jul 9 2007 20:08:30 UTC (8 years 10 months 4 weeks 1 day ago) and read 2534 times:

 Quoting Jetlagged (Reply 2):Actually the speed Jaylink is trying to determine is defined, as Vmr (minimum rotation speed).

Hmm I didn't know that; I stand corrected.

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