# What is the average aerodynamic load on a control surface of a commuter-sized airplane?

I'm doing a study about aircraft hydraulic pump sizing. In order to do that, I need to know the size of a flight control actuator, then determine the maximum flow and pressure that the actuator needs.

Therefore, I need to know how much force that the flight control surface needs to be moved.

What is the average aerodynamic load on a flight control surface of a commuter-sized airplane (10-20 passengers)?

• If you’re not too much into in aerodynamic calculations and researching associated literature (I’m not really sure where best to start there...), another approach might be to find which actuators were used in real aircraft and make conclusions from that. – Cpt Reynolds Jul 17 '18 at 10:45

I think you don't actually want to know the load but rather the hinge moment of control surfaces. The actuator load is the hinge moment divided by the length of the control horn. Below is a rather poor sketch for a typical aileron linkage, but the principle is correct (source):

It already shows one popular way of reducing control forces: The tab, a little auxiliary control surface which moves against the "real" control surface. This reduces effectivity a bit, but forces by a lot. Here, the amount of deflection of the tab is controlled by a spring in its linkage, which is a clever way to adjust its deflection such that the actuation forces become more constant over speed.

Another way of reducing control forces is a horn: An extension of the surface forward of its hinge line, so the aerodynamic loads here balance those on the surface behind the hinge line. The picture below shows the left aileron of the ATR-72 which is moved by mechanical linkage (source)

This way, the lift loads on the control surface are mostly carried by the hinge and only the actuation loads need to be carried by the control rod or actuator. If you think you don't need all those nifty tricks, your actuator and hydraulic system will become much heavier than needed.

Why are two different methods used? The tab reduces the loads from deflection changes while the horn reduces those from angle of attack changes, too. When sized properly, both together will drive the hinge moment close to zero.

Why do I explain all this? It shows that your question does not have a simple answer. Rather, you need to specify exactly how your control surface looks and is moved, and only then can you start to calculate the actuator loads. I also want to show that a subsonic airplane for 10 - 20 passengers will be perfectly flyable with manual controls. The ATR-72 needs hydraulics only for the flaps, the spoilers, the brakes and the landing gear. Avoiding hydraulics for primary flight controls also lets it get away with single redundancy in its hydraulics system.

As you noted, there are lots of complexities to be considered when sizing control actuators, including the size of the control surface, the desired deflection angle, the actual deflection angle, hinge moments, boundary layer effects etc. You can obtain a rough approximation of force on the surface though starting with the simple definition $$P = \frac{F}{A}$$

Where P is the dynamic pressure, q, and A is the exposed control surface area. Solving for force gives: $$F = PA$$ Then substituting dynamic pressure and exposed area results in: $$F = \frac{1}{2} \rho V^2A sin\delta$$

Where A is the control surface area, and multiplying it by the sine of the deflection angle yields exposed area.

For a light transport aircraft, say a Beech 1900 (19 pax), the elevator has an area of 19.3 sqft. Using the logic above, this surface when deflected 5 degrees at cruise speed would feel a force of approximately 421 lbf. Your actuator sizing will ultimately need to account for the parameters above (and more), but hopefully this is an informative starting point.

• There are two things missing: The control surface is part of an empennage which carries a load even without control surface deflections. Next, the question should be more about the hinge moment on that surface - that, divided by control horn length, is what determines actuator loads. – Peter Kämpf Jul 17 '18 at 17:46
• @peterkampf I see. When designing the control mechanisms, wouldn't the the hinge moment be a design target? In other words, wouldn't you tweak the control horn and hinge locations to obtain an adequate hinge moment, where "adequate" is driven by the amount of force the control surface must oppose when deflected? – Geoff Jul 17 '18 at 18:13
• Yes, if you add area ahead of the hinge, loads go down. The best mechanism I have personally seen was on a Canberra - its ailerons could be moved with very low forces at flight speeds up to Mach 0.8. The downside here is an increased flutter risk - such mechanisms were the result of desperation. – Peter Kämpf Jul 17 '18 at 20:22
• Sin5°=-0.9589,sin 25=-0.1324,As angle of deflection increases,exposed area(Asin sigma)should increase not decrease......do you think this invalidates the above formula(the 3rd one)? – David Teahay Nov 15 '18 at 10:38