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When the pilot sets the takeoff engines thrust on the Flight Management thrust limit page that displays on the CDU (or MCDU) interface, the selection will finally show as a percentage.

The direct consequence of this is that, « at first sight », the pilot does not know the exact thrust that the engines will develop in pounds-force.

Considering that all operators do not necessarily choose the same rating for a similar combination of aircraft-engine, does it mean the same percentage of takeoff thrust might provide different thrusts according to the operator?

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Does it mean the same percentage of takeoff thrust might provide different thrusts according to the operator?

Not just per operator, it is different per engine installation in one fleet. That's why calculations are done according to each airframe's AFM. And as the engine ages, its thrust decreases* (and fuel consumption increases), and this is reflected in the calculations. (The fuel flow (and drag) offsets are based on data from previous flights, and are presented to the pilots as corrections/biases to input.)

The selection will finally show as a percentage.

The final selection will be percent N1, EPR, or for the latest planes (A350 and 787 with R-R engines) as an actual available thrust percentage. So it depends on the plane/engine/technology.

As to why, because lbf or kN mean nothing at first glance.

  1. The calculations are very complex to do by hand.
  2. Density altitude would make a correct number feel incorrect. In other words, 95% N1 (for example) is a different thrust force at SL at 15°C from at 5,000' MSL at 35°C.

The number simply shows without guesswork or confusion** how much of the available thrust will be used. And it makes the AFM (or EFB) easier to present/use.

* As an example, a heavily used fan will not be in a good condition, so two similar fans (but different ages) running at the same rotation (N1 percentage) won't deliver the same force. (The same for the conditions of the rest of the stages and combustor.)

** Plus, the engine controllers have no way of measuring force, this is another reason why the selected level during flight prep must match the engine instrumentations, otherwise the crew can't verify the takeoff thrust that's been preset, has been achieved.

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  • $\begingroup$ Thank you for answering. I understand well your answer, however for the same computed thrust %, the engines(and more particularly for a twin engines aircraft) will produce at takeoff the same thrust even though of different age while their fuel consumption will be different. So what I mean is that the computed thrust% will depend very little on the corrections biaises, while the predictive fuel consumption is effectively directly related. $\endgroup$
    – user40476
    Jun 6, 2019 at 20:40
  • $\begingroup$ I am using two comments because I lack space. So thank you for answering. The second part of my question is more general, for instance considering a specific terrain, a specific aircraft, a given takeoff weight, why not entering in the thrust limit page a thrust in thousands pounds-force since finally what is needed is a FORCE. How this force for a given altitude is translated in % would be the FMS job to calculate. $\endgroup$
    – user40476
    Jun 6, 2019 at 20:51
  • $\begingroup$ You're welcome @user40476, I've responded to both comments by adding footnotes, let me know if something is still unclear. $\endgroup$
    – user14897
    Jun 6, 2019 at 21:07
  • $\begingroup$ Percentage gives a good indication of the performance capability and margin the aircraft has on that specific takeoff; provided the forces have been calculated in the background according to regulations, the margin available to the crew might be more important to them in the takeoff preparation (I know I’d be more interested in knowing whether on firewalling I could expect “much more oomph” or “no more oomph” rather than what that oomph is in force units, because converting the latter to an expected additional climb gradient is outside of what I can safely do during takeoff preparation). $\endgroup$
    – Cpt Reynolds
    Jun 7, 2019 at 6:49

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