# Is there a relationship between control surface deflection and a particular turn radius or bank angle?

I am making a project which includes automation of aileron, rudder and elevator deflection.

Is there a quantifiable relation which can tell me the amount of deflection required for any radius of turn if the other parameters are put constants.

• There is no direct or quantifiable relation between aileron deflection and a given radius of turn. – J Walters Feb 14 '16 at 21:33
• Note that it is possible to maintain a banked turn with ailerons moving the opposite direction. This happens quite often in gliders where the pilot wants to tightly circle a thermal but does not want the aircraft to roll further so an opposite aileron input is made to maintain the turn. – slebetman Feb 14 '16 at 21:45

You need the ailerons only to bank the aircraft; once the bank angle is reached, the aileron deflection is mostly close to zero. I assume we are talking about stationary turns, so speed and aircraft mass will not change significantly over the duration of the turn.

While turning, several moments need to be balanced to keep the roll angle constant, and a good design does this without requiring aileron input:

• The rotation around the vertical axis causes more airspeed over the outer wing, increasing its lift. This causes a rolling moment into the turn.
• The rotation around the vertical axis causes a sideslip condition at the vertical tail which causes a yawing moment against the turn.
• The inertial forces try to pull the wings level. This causes a rolling moment against the turn.

However, the elevator needs to be held at a slightly more negative angle than in level flight to pull the required load factor $n_z$. In a turn, the load factor is proportional to the bank angle $\Phi$: $$n_z = \frac{1}{cos \,\Phi}$$ Radius $R$: $$R = \frac{v^2}{g\cdot tan\,\Phi}$$ Angular velocity $\Omega$ (rad/sec): $$\Omega = \frac{v}{R} = \frac{g\cdot tan\,\Phi}{v}$$ The amount of elevator deflection needed depends on the stability margin (expressed as $\frac{c_{m\alpha}}{c_{L\alpha}}$) of the airplane, its pitch damping (expressed as $c_{mq}$) and elevator effectiveness (expressed as $c_{m\eta_H}$).

• once the bank angle is reached, aileron deflection is mostly close to zero This will depend on the aircraft's stability about the longitudinal axis. If the OP is looking for precise measurements, he may need to take into account aileron inputs to maintain a bank. – J Walters Feb 14 '16 at 21:29
• The load factor formula is only true for level flight. A 60° bank may have a load factor of 1 in the right conditions. – J Walters Feb 14 '16 at 21:30
• @JonathanWalters … which would be a 60° dive. Yes, possible, but then acceleration in flight direction will be far from zero, so speed will change rather quickly. This is no stationary turn at all. – Peter Kämpf Feb 14 '16 at 22:19
• No, it needn't be a dive. As you point out, airspeed will certainly change, but all this could be accomplished in a climbing turn. – J Walters Feb 14 '16 at 22:29
• @JonathanWalters: yeah, right, a climbing turn with 60° bank. How realistic! If you have that power, why not climb straight up? If you don't, the turn will not last long. The key is "stationary" here! – Peter Kämpf Feb 14 '16 at 22:34

Aileron deflection affects your roll rate, but you can roll rapidly or slowly to whatever chosen angle of bank... and the angle of bank (along with true airspeed) is what determines your turn rate and radius. (Assuming a coorinated turn.)

Depending on various parameters, some small amount of rudder and elevator deflection may be necessary to maintain coordinated level flight once established in the turn, and depending on the aircraft, you may need some rudder input while the ailerons are deflected. That's all pretty well specific to whatever aircraft you're modeling. In a swept-wing jet, significant rudder deflection is not required (and what little deflection is needed is often provided by the yaw damper rather than the pilot), while in a long-winged glider with considerable adverse yaw, adding a fair bit of rudder along with aileron is required for coordinated flight.

• This answer brings to light some of the difficulties encountered in solving the question, but contains several statements that, in their current state of isolation, are inaccurate. 1) Aileron deflection does not give you roll rate on it's own; airspeed and other factors are also determinants. 2) Bank angle and true airspeed alone do not determine turn rate or radius; coordination/rudder input is also determining factors. 3) In a swept-wing jet rudder input is generally still needed in executing a coordinated roll and turn. – J Walters Feb 14 '16 at 18:12
• @JonathanWalters: Believe it or not, but if you fly a coordinated turn, bank angle and true air speed do give you enough information to accurately calculate turn rate. – Peter Kämpf Feb 14 '16 at 20:46
• @PeterKämpf Re-read my comment; I am pointing out exactly that: coordination is imperative to that calculation's accuracy. As you probably know, if you fly in a slip a bank angle of 20° may result in exactly zero turn rate regardless of true airspeed. – J Walters Feb 14 '16 at 21:24

Probably the relevant answer as a pilot: No

What you want to achieve is most probably more complex, flying a certain curve depends on so many things and is a dynamic process.

True answer as a physicist: Yes, of course

Whenever you have a given situation with the exact same parameters (as you described, all other parameters are put to constant, including wind, power, airspeed, density, etc.), you will end up with the same turn radius. It's deterministic...

Problem: Your premise of holding the other parameters constant is unrealistic

Combining the two: You never fly a curve, like the second answer suggests, where you have "a certain deflection" of the aileron, but change it during the turn to start and stop it as well as during the curve because you never have all the other parameters constant. You use, within other things, the rudder to compensate for other, changing parameters. So the question is, are you interested in making a whole turn (look at the first answer) or do you only care about the artificial steady state (second answer)?

Actually, I think your problem has been solved already by a lot of autopilots... I'd recommend to have a look at the state-of-the-art implementation of those.

Yes there is a relationship between control surface deflection and bank angle, but it is a complicated one, with a feedback loop, aerodynamic forces and inertia, double integrators, transformation from aircraft to earth axes etc.

Since it is a dynamic relationship, it is best characterised by the response to a step input. Start with control wheel at 0, set a sudden deflection at let's say wheel 30 deg for 5 seconds, then briskly return wheel to zero. Meanwhile, measure the aircraft bank angle response. This is the way that the flight dynamic model in flight simulators are tuned.

There is more information on this link (figure 5), although that explanation considers roll rate. In order to get to bank angle you need another integrator and axes transformation.

An autopilot will compare actual bank angle with a setpoint, and vary control wheel input until the required bank angle is reached. You may find source code examples in open source simulator programs such as FlightGear.