I recently read in the manual for The Beechcraft T-6 Texan II (a turboprop aircraft) about engine failure during flight at different altitudes and the gains from pitching up after the engine failure occurs to trade airspeed for altitude.

They show graphs that depict the altitude gain from pitching 20° nose up attitude bleeding airspeed from 200 KIAS to 145 KIAS at different Pressure Altitudes. The not so intuitive part is that shows how the heavier the aircraft gets (by having more fuel) the more altitude gain by pitching the same amount of nose up (20°) until airspeed bleeds off from 200 to 145 KIAS in case of engine failure. could someone please explain how could this be possible? I doubt there is a mistake in their manuals. There are a couple of these graphs from 250 to 145 and 200 to 145 at different Pressure altitudes.

At a first glance, it may seem intuitive to think that, if the heavier aircraft encounters an engine failure and it pitches up (same amount nose up attitude as the lighter aircraft) it will produce the same amount of lift as the lighter one (at the same pressure altitude and same airspeed) but with more weight the less lift to bring the aircraft to a higher altitude compared to a lighter aircraft. I guess that is not right?

If there is a vector analysis/equations or any math proving this I would appreciate it a lot.

In other words: two identical turboprop aircraft, except A weighs less than B, fly at same KIAS, engine fails, both pitch up 20° nose up, which one gains more Alt?

The graph of weight vs altitude gain:

graph from POH


2 Answers 2


To gain altitude (by any means) you need to gain potential energy. Since the engine has stopped, the only thing you can do is trade kinetic energy for potential energy (slow down).

Since both kinetic energy ($E_k = \frac12 m v^2$) and potential energy ($E_p = m h g$) are proportional to mass, for a spherical cow trading the same speed would result in the same gain of altitude independent of its mass.

But an airplane isn't a spherical cow. There is one more force we need to account for: the drag. Drag has two components:

  • Induced drag (the cost of producing lift) is proportional to lift which is to sufficient approximation equal to weight (at 20° the approximation that $\sin x = x$ is still good enough).
  • Parasitic drag (the cost of moving through the air), however, is not proportional to lift; it depends only on the speed, air density and shape of the aircraft.

Since all the variables except parasitic drag are proportional to weight, the parasitic drag will take relatively less energy for a heavier aircraft (more energy to begin with), indeed allowing the heavier one to climb higher.

For a similar case that might be a bit more intuitive: consider having two balls of similar size, but significantly different weight. A ping-pong ball and a golf ball may be a good example, because they are light enough that accelerating your hand becomes the limiting factor, so you can throw the lighter ping-pong ball as fast as the golf one, but not much faster. But the golf one will then fly far, while the ping-pong one will slow down and drop much closer. This is because the air drag is relatively much bigger for the lighter ball.

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    $\begingroup$ Your ping pong and golf ball comparison really nailed it for me! $\endgroup$ Commented May 1, 2022 at 14:20
  • $\begingroup$ @Jan Hudec great thoughts TY. When you mentioned that "Induced drag (the cost of producing lift) is proportional to lift which is to sufficient approximation equal to weight..." I do see how induced drag is L*sinx and 20 being small (pi/9 ~ 0.3) might be a small amount to consider but what does weight has to do with them. are you saying the heavier one will create proportionally more lift? not sure. I thought both examples produce the same lift because of the same pitch attitude. $\endgroup$ Commented May 1, 2022 at 17:31
  • $\begingroup$ @Jan Hudec What about conservation of momentum? do you think that is why the parasite drag affects the lighter one more than the heavier one? therefore taking energy from the system at a less rate or less compared to the lighter aircraft. Lastly, extra bonus follow-up question. once both aircraft reach their max altitude gained, do you think the heavier one will have a greater descent rate? $\endgroup$ Commented May 1, 2022 at 17:44
  • $\begingroup$ No. The reason parasite drag affects the lighter aircraft relatively more is that drag is taking away energy – drag is force and force times path is work, that is energy – at the same rate, but there is less energy to begin with. Conservation of momentum is a reason why induced drag exists, and why it is proportional to the lift (and decreases with wing span). $\endgroup$
    – Jan Hudec
    Commented May 1, 2022 at 20:18
  • $\begingroup$ @YamchaAviator It's basically just that a heavier aircraft has more energy for the same speed than a lighter aircraft (the heavy one will have had to have already burned more fuel to gain that energy/speed). So an effect that reduces the aircraft's energy by a magnitude that does not depend upon weight will take less speed from the heavier aircraft than from the lighter one. $\endgroup$
    – Ben
    Commented May 2, 2022 at 8:59

A few issues with the concept unless hard test flight data is shown.

  • Kinetic Energy vs drag

Glider data shows us that two gliders of different weight will glide the same distance. The heavier one must glide faster but has more potential energy. The two perfectly cancel each other out.

So Vbg for the fuel laden craft will be higher, therefor it must break off its ascent sooner.

  • both craft must turn to ascend

Obviously, the heavier craft must either pitch harder or fly a larger radius to reach 20 degrees nose up

Once at 20 degrees nose up, both parasite and induced drag will be greater because the entire plane has a higher AoA.

Procedurally, yes, pitch up for Vbg, but be aware of what it is for your weight. 145 knots may be too slow for a plane with a large fuel load.

  • $\begingroup$ Do you think conservation of moment has to do with the energy that is stored in the system at the beginning of the climb? that is why drag has a larger effect on the lighter aircraft (less mass, less momentum) ..maybe? $\endgroup$ Commented May 1, 2022 at 17:49
  • $\begingroup$ @YamchaAviator Yes, Jan Hudec answer is correct. I wouldn't go down to 145 if heavy. The real key to Vbg is optimum AoA. The heavier plane should break at a faster airspeed. Really wanted to see curves. They are linear. Thanks! $\endgroup$ Commented May 1, 2022 at 18:14
  • $\begingroup$ TY for the input! just as the last thought once they have reached max altitude and start descending, which one do you think will have a greater descent rate ? $\endgroup$ Commented May 1, 2022 at 18:39
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    $\begingroup$ @YamchaAviator theoretically, if both planes can hold optimal AOA, both planes descend at the same angle (going just as far) but (by geometry) the heavier plane has the faster descent rate. But, for Vbg, it's the angle that counts. $\endgroup$ Commented May 1, 2022 at 19:17
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    $\begingroup$ @YamchaAviator yes, and thanks for an excellent question. $\endgroup$ Commented May 1, 2022 at 21:29

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