Anyone who has transitioned from light GA aircraft to large turbofan airliners will notice the difference: the control column on a large aircraft is much heavier. While still within limits of a healthy adult, it takes a considerably larger force to move the controls.

Why is this? Is this intentional, or just a by-product of the early days mechanical linkages? Airbus has shown that even on airliners, the controls can be made as light as simple joystick. So why can't the controls of a Boeing 747 feel the same as a Cessna 172?


The control forces are there by design. All airliners have artificial force feel from springs, dampers etc. A B747 flies through the skies a lot faster than a Cessna 172 and bad things could happen if it were easy to deflect the flying controls at M 0.85.

The B737 elevator is a good case in point. It has q-feel (artificial feel) provided by a spring device, and the spring gradients stiffen up as a function of airspeed. At higher speeds it is quite hard to deflect the column by pushing/pulling and much easier to use the trim button to control the pitch. For Level D simulator control forces we measured the forces necessary to deflect the column fully when trimmed in mid position, which turned out to be a serious body-building exercise:

  • About 200 N at zero airspeed
  • About 500 N at 400 knots 20,000 ft. (Measurement taken on ground with a pressure hose attached to the pitot tube, and aircraft on jacks with the landing gear retracted).

By contrast, helicopter flying controls are light because helicopters are inherently unstable and a lot of stick action is necessary, particularly in the hover. In most helicopters the feel forces can be switched off completely with the trim button, resulting in a very light stick, perfectly OK with the relatively docile response of the helicopter.

But a 747 is a stable airframe. A fast aircraft with large control deflections results in high accelerations, which can result in Pilot Induced Oscillations or structural damage to the airframe.

The two measurements of B737 control column, digitised from the original plots and on the same scale: column position on the x-axis, force on the y-axis. By design, force required to deflect the column fully from about mid position is about 40 lbf at zero knots, 100 lbf at 400 knots. This is from an artificial q-feel system and the feel is implemented by design. The vertical distance between the sides of the curve is caused by friction.

enter image description here

At 0 knots:

  • Notice that around mid the spring gradient is very steep: the breakout, to indicate neutral position.
  • Then between 0 and 5 inch a spring gradient of 4 lbf/inch.
  • Beyond 5 inch a relatively flat gradient, to make it not too hard to deflect the column fully.

enter image description here

At 400 knots: it is made hard to fully deflect the column. 100 lbf pull from a seated position is nothing to sneeze at.

  • 4
    $\begingroup$ Unit conversions: 200 N is about 45 lbs, 500 N is about 110 lbs. $\endgroup$ – Tanner Swett Aug 20 '17 at 16:23
  • 1
    $\begingroup$ I'm curious is that 400 knots indicated or true airspeed? At 400 KIAS a B737 would have disintegrated... $\endgroup$ – kevin Aug 20 '17 at 19:58
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    $\begingroup$ @kevin The measurements were taken on ground, with a measurement set with a position and force transducer attached to the column. A pressure hose was attached to the pitot tube: 400 KIAS. And then an untinterrupted, continuous, slow sweep of the control, starting from neutral, push slowly until in the fwd stop, release slowly then pull into aft stop, then relese slowly again. The plot is of control force vs position and shows the spring gradients, friction, travel limits, cable stiffness. $\endgroup$ – Koyovis Aug 20 '17 at 22:04
  • $\begingroup$ Adding to what @TannerSwett wrote above, for those of us more used to SI units but who don't necessarily have an intuitive grasp of newtons, it's often close enough to just say that 10 N = 1 kgf, so divide forces in newtons by 10 and you get the equivalent force in kgf. I'll be the first to admit that this is not a perfect equivalence, but unless you are writing a scientific or detailed engineering paper, it's good enough for most purposes. The exact value is 9.80665 newton per kgf. $\endgroup$ – a CVn Aug 21 '17 at 11:37
  • $\begingroup$ @TannerSwett correction: pounds of force (lbf), not pounds (lb). The former is force, the latter is mass. $\endgroup$ – John Dvorak Aug 21 '17 at 18:23

Adding to Koyovis' answer. In an Airbus the stick movement does not correspond to control surface deflection.

The stick commands a load-factor / roll-rate for the pitch/roll. So the faster the plane, the smaller the deflections will be for the same stick movement.

The Airbus A320 stick stop upper-limits are +2.5/-1 g (pitch) and 15°/s (roll) when in clean configuration.


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