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Old Sep 27, 2002, 08:39 PM
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Kings Park, New York, USA
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Why does nose drop in turns on flying wings?

On a zagi or Slope Slayer, why does nose drop when plane is banked?

On a 3 axis aeroplane, I can see where rudder can act like down elevator once banked.

Ron parigoris
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Old Sep 27, 2002, 09:10 PM
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You got it - left rudder (for example) holds the tail down ("nose up") in a right bank.

If you have a CPU TX, you can decrease the "nose drop" in a flying wing by programing your turns with mostly down elevon deflection in the high wing (i.e., down right elevon deflection for a left turn- hardly any up left elevon.

That will differentially increase the drag on the upper wing with respect to the lower, and have the effect of opposite rudder in a conventional aircraft.
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Old Sep 27, 2002, 11:07 PM
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Palmdale, CA
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Unless you the pilot do something to prevent it, ALL airplanes will drop the nose in a turn.
Banking a plane reduces the amount of lift holding it up. It must go down!
With a conventional airplane, there's more than likely an unobserved back-stick input.
A wing lacks a good horizontal reference such as a fuselage, so the nose goes down much more before the flier notices.
Differential as Jube suggests will diminish the effect, but not eliminate it.
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Old Sep 28, 2002, 02:29 PM
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In a conventional aircraft you don't use the rudder to control the nose up or down in a coordinated turn, you use the elevator. The rudder is used to keep the turn coordinated.(Overcome the yaw created by difference in lift between left and right wing.)

As you enter the turn (starts to bank) the correct deflections are:
1. Apply aileron in the direction of the turn.
2. Almost at the same time as you apply the aileron you start to give some rudder in the same direction.
3. As the aircraft starts to bank smoothly increase elevator to keep the nose at the same attitude. (Or slightly higher.)
4. When you have the desired bank angle, ease off on the aileron deflection, at the same time as you decrease rudder deflection. In order to maintain the same bankangle you gonna end up with having to actually put in some aileron in the OPPOSITE direction as you are turning in.

This mean that established in a turn the controls should be: Back pressure on the elevator, rudder in the direction of the turn and aileron in opposite direction of the turn.

This is true for real conventional aircraft, and as I understand it should also be true for model aircraft as well.

BUT the big question is how this works on a Wing, and that I don't know.

Tompa
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Old Sep 28, 2002, 03:21 PM
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Kings Park, New York, USA
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Still don't get it.

Hey guys. Thx for the input, but i still don't get it.

I know it happens, i know how to compensate for it (on a flying wing differential), still don't know why?

"Banking a plane reduces the amount of lift holding it up. It must go down!"

sorry I still don't get it, we are talking nose pitching down in turns, needing up elevator to compensate.

OK any aeroplane in a 60 degree bank and holding altitude, will pull 2 Gs, matters not weight or speed or size, lets think bout this, you haul back on stick to pull gs and pull nose through turn. without pulling back nose drops and does not turn. Nope still can't figure why nose drops.

Flying full sized 3 axis, i kinda sortta had in the back of my mind that the rudder was kinda sortta acting as elevator when banking (correcting for adverse yaw) and needed elevator to compensate.

Thx. for any help with dense matter between sholders.

ron Parigoris
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Old Sep 28, 2002, 04:47 PM
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Gotland, Sweden
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It is hard to explain without drawing pictures, and my skill in drawing pictures on computers are very limited (read zero). But I give it a try.

An aircraft in steady flight without any acceleration has in regard to this question three forces acting on it.
1.Weight acting vertically down
2 Lift acting up, perpendicular to the wing span
3 Lift from the tail, acting perpendicular to the stab. DOWN

This three forces each create a momentum and sum of these must be zero, or the aircraft will tilt nose up or down.

If you increase bank, without adding elevator (increasing angle of attack) you are keeping the same total lift from the wing but tilting it, so the vertical part counteracting the gravity decreases. At the same time you also tilt the negative lift from the tail, decreasing the vertical part of the down force from the tail. But the weight is unchanged, still acting straight down.

If you to make it "simple" now calculate the momentum from the position were you have the lift, you can disregard the lift and just look at the weight and the down force from the tail. (To do it this way is not totally correct, but the effect gonna be the same.)

In this situation you have
1.Not changed the down force from the weight, and the momentum is also unchanged.
2.Decreased the vertical part of the down force from the tail, thus decreased the momentum trying to keep the tail down.

The sum of the momentum are no longer zero, the nose down momentum is larger, tilting the nose down.

In short you tilt the force from the tail, but not the weight. It is always gonna be straight down.

I don't know if anyone can make any sense of this explanation, I should add a picture if I only knew how.

Tompa
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Old Sep 28, 2002, 05:51 PM
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Banjul
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Tompa,

Well done (and WITHOUT graphics)!!!

VP
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Old Sep 29, 2002, 02:19 AM
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Kings Park, New York, USA
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Hey Tompa

Thx for the explination.

I thought and drew on some napkins and what you said makes sense.

appriciate the help.

Ron Parigoris
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Old Sep 29, 2002, 08:05 AM
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Andy W's Avatar
Marietta, GA
Joined Jun 1999
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Here it is with pictures.
..a
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Old Sep 30, 2002, 07:29 AM
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Another way to look at it is in view of the fact that center of pressure is behind the center of gravity in a stable airplane.

So, the airplane acts like a "dart" - it aligns itself with relative airflow.

When you turn an airplane on its side, the relative airflow vector shifts in an upward direction as the airplane falls, and the nose goes down to align itself with the airflow.
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