Suppose an unpowered, rigid, mechanical, lightweight aircraft (e.g. sport glider) is descending at a steady glide in a constant trim (hands off) configuration in a standard atmosphere. The pilot applies a quick elevator input and then goes hands off again, causing the nose to pitch up and the glider to gain altitude momentarily before eventually returning to a steady glide.

Suppose we measure the maximum angle of attack reached, (or max pitch angle, if that's easier), the maximum altitude gain, and the time it took to reach those maximums after the control input.

How would those values (max altitude gain, max angle, and times to get there) differ if the same control input occurred at 5,000 feet vs 10,000 feet above sea level? Support your answer using physics/ aviation formulas or reputable references.

My thoughts so far

Since the "effective" air speed of the glider in a particular trim configuration is expected to remain the same across altitudes, then I would assume the lift and drag force changes generated by the control input would be the same at any altitude.

However, the "true" airspeed of the glider is from what I could find about 10% higher at 10k, which results in greater momentum to resist changes in motion and greater kinetic energy that can translate into height.

However, I'm not sure if those two (momentum vs kinetic energy) would cancel each other out.

If kinetic energy alone is considered, I'd expect the altitude gain to be greater at higher altitudes, and I'm not sure about max angles or the times to get there.

  • $\begingroup$ Control effectiveness as in rate of change in direction will be greater in denser air. At higher altitudes, momentum is greater as you pointed out, for the same directional change force. Much easier to stall. What is interesting is altitude gain after control input if it is sustained. $\endgroup$ Commented Aug 13, 2023 at 6:17
  • $\begingroup$ @RobertDiGiovanni Do you know of any proofs of "easier to stall"/ easier to change angle, or even a research paper/ controlled experimental data that verifies this for fixed wing aircraft? $\endgroup$ Commented Aug 15, 2023 at 1:10
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    $\begingroup$ My thoughts, not really very consolidated yet: inertial effects are much larger, in relation to aerodynamic effects, at high altitude. So for the same control input (elevator moves x degrees and stays there for y seconds and then is allowed to go back to trim), the change in angle-of-attack is bigger at low altitude. Or is it? Because once the glider begins pitching up, there is less aerodynamic damping at high altitude, so the motion will continue for longer after the pilot releases all pressure on the stick. ??? It's complicated. $\endgroup$ Commented Aug 15, 2023 at 1:52
  • $\begingroup$ @quietflyer Indeed it is complicated. So far my search has led me to flight dynamics textbooks, which are full of very long and complicated equations modeling this in 6-9 axes. I'm just hoping this simple aspect of it has been figured out already, in a way that doesn't require me to take a year of physics and math courses. $\endgroup$ Commented Aug 15, 2023 at 2:01
  • $\begingroup$ @Oleg very hard to stall a paper airplane as it's pitch change can keep up with change in AoA. As you go faster at higher altitude, aerodynamic force from elevator is the same but momentum is greater. The plane rotates at the same rate but pitches (changes direction) slower. $\endgroup$ Commented Aug 15, 2023 at 21:41

1 Answer 1


What you describe is called a doublet. You fly at a trimmed state, then move the stick momentarily to a different position and back to the original position again. Then you watch what happens.

Traditionally, differential equations were used to describe the results and I guess that is what you still find in flight dynamics textbooks. There, the second derivative term describes inertial effects and the first derivative term describes damping effects. So far, you only looked at the inertial effects, which do indeed get bigger with the inverse of air density.

However, damping is at least as important, and this term also is affected by air density: Damping goes down with the square root of density. Not only will the excitations from a doublet grow from inertial effects, but the countering forces from stability will become weaker, so it will take longer for them to stop the resulting motion.

I'm not sure about max angles or the times to get there

Lowered damping due to a higher altitude will make the angle deviations larger and reduce the eigenfrequency of the resulting motion by the inverse of the square root of the density ratio between your reference and actual altitude. The excursions will become slower and larger by the increase in true air speed. Things will take longer to die down, so it will take more oscillations until the motion dies down.

If the trimmed state does not change with altitude, the true air speed goes up with the inverse of the square root of the change in density. The speed of the pitch response is proportional to the frequency of the short period mode which again depends on the stability margin, pitch damping and density as explained in the linked answer.

  • $\begingroup$ Thank you, you seem to know a lot about the topic. Is there any chance you can go into more detail, mathematically, about that INITIAL pitch change? By what percent will the angle/altitude differ between 5k and 10k? By what percent faster will they occur? How much of that is due to inertial vs dampening effects? Are there simpler equations I could use to calculate this myself or with a math-only tutor? $\endgroup$ Commented Aug 15, 2023 at 21:50
  • $\begingroup$ I looked into flight dynamics with the help of some tutors. I understand now that the math is too complicated for a "simple" answer to my post. $\endgroup$ Commented Apr 23 at 22:03
  • $\begingroup$ You mentioned "traditionally". What is the modern alternative? Computer simulation? $\endgroup$ Commented Apr 23 at 22:04
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    $\begingroup$ @Oleg Of course. Specifically, the finite difference method. For the equation of the time dependency of the pitch response see the newly linked answer, which contains an approximation for the short period mode frequency. $\endgroup$ Commented Apr 23 at 23:33

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