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In the very interesting answer from Jan Hudec to a question about the various flight control computers on Airbus planes, two different means to trim the aircraft are mentioned: The stabiliser and the elevator.

How does this work in practice? How is the aircraft trimmed in flight, both as seen from the flight deck (i.e., the user interface to the pilots) and as actual input to the control surfaces?

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  • $\begingroup$ Are you asking about trim in normal cruise phase under normal law with all systems operating normally or some other flight phase or condition or flight law? $\endgroup$ – RedGrittyBrick Jun 1 '15 at 11:12
  • $\begingroup$ @RedGrittyBrick Let's stick to normal law, which after all, er, normal. It would be nice to know if there are differences in how it works for different phases (or conditions) of flight, though. $\endgroup$ – Monolo Jun 1 '15 at 12:01
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The faster an aircraft is, the wider range of elevator deflections it needs. And so do aircraft with large change of weight (amount of fuel used) during flight. However larger elevator deflection causes higher drag. It is more efficient to move the whole horizontal stabilizer instead. Therefore jet aircraft generally have the forward part of the horizontal stabilizer movable.

Now it would be possible to simply have the whole stabilizer move. And supersonic aircraft generally have just that. It is called stabilators. However having the stabilizer split lends itself to elegant solution for trim.

This takes advantage of the fact, that the aerodynamic forces tend to put the elevator in the position of lowest drag almost straight (slightly bent in direction of less lift) behind the forward part. So when the forward part is moved, the lift changes without changing the force on the elevator. So the elevator is used for the small adjustments where the elevator is intended to return to the neutral position, which the aerodynamic forces help with, and the stabilizer is used to trim for current airspeed.

In aircraft with mechanical controls, the elevator force (or a fraction of it via power drive) acts directly back on the control column and the pilot can feel it, while the horizontal stabilizer is actuated via mechanism that holds selected position (may be electrical with jack-screw like on DC-9 or hydraulic, but with no feedback) and connected to the trim, which is usually pair of large wheels on the sides of the centre console.

Now in Airbus A320 and all newer models the side-stick does not have mechanical link and the pilot does not feel the force on the elevator anyway. In normal law, the ELACs automaically adjust the elevator and stabilizer so that with side-stick in neutral position, the aircraft maintains 1 G vertical acceleration and therefore straight flight (at any flight path angle). Since the trim wheel does have direct mechanical link, it turns as the ELACs (or SECs) adjust trim.

However the distinction between elevator and trim is still useful in case of failure. When the system degrades into direct law, the side-stick position corresponds to the elevator deflection and the trim wheel position corresponds to the stabilizer position. Since the side-stick still returns to neutral position, the distinction of using the elevator for momentary adjustments and trim to balance the aircraft to fly (approximately) straight at current speed holds (just because the side-stick is spring-loaded, the force needed to deflect it does not increase with speed as it does for mechanical controls).

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  • $\begingroup$ Great answer, thanks! I now also see why the trim wheels are implemented as they are. $\endgroup$ – Monolo Jun 1 '15 at 13:46
  • $\begingroup$ Why the faster the wider the range? I thought the range of movables was sized by low speed,where due to small dynamic pressure we have small efficiencies... $\endgroup$ – GHB Mar 8 '16 at 17:53
  • $\begingroup$ @GHB, the size is determined by need for sufficient authority at low speed. But on faster plane it needs to adapt to larger range of angles of attack. $\endgroup$ – Jan Hudec Mar 9 '16 at 10:00
  • $\begingroup$ @JanHudec sorry I still don't get it! The Aero pitching moment that the elev can generate will be dictated by both the size of it and the deflection of it: $M_a = q S b Cm_{\delta_{elev}} \delta_{elev} $, p dynamic pressure, S wing area, b wing span, $ \delta_{elev} $ elev deflection. The magnitude of the control derivative $Cm_{\delta_{elev}}$ will be given by the size of the elevator itself ($Cm_{\delta_{elev}} \leq Cm_{\delta_{stab}} $ ). Do you mean: on faster aircraft, being the low speed constraining, the elev shall provide a wider range of deflect than considering high speed only? $\endgroup$ – GHB Mar 9 '16 at 23:45
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    $\begingroup$ @RustyCore, Airbus has trim wheels and they are to be adjusted manually in direct and mechanical laws (mechanical law was deleted from A380 onward). Unlike 737 they are hydraulic though, so they are easier to turn and need fewer revolutions. $\endgroup$ – Jan Hudec Jun 4 at 8:13
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It is fully hydraulic, the active power is the hydraulic even when you act directly on the wheels, that’s why there is no handles to the Wheels, the wheels will control a valve block to drive the screw.

There is no thumb switch to operate the trim, when you move the stick the elevators are operated for short term action to get the desired load factor, followed by the trim which moves automatically to maintain that load factor, and replaces at long run the elevators effect, thus at long run the THS and the elevators are aligned which reduces the drag.

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  • $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review $\endgroup$ – Sean Jun 4 at 1:56
  • $\begingroup$ @Sean, ok, I added a useful information just now to complete the answer. $\endgroup$ – user40476 Jun 4 at 7:43

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