3
$\begingroup$

I've been stumped on this for a while. Let's say you have a helicopter in climb mode (i.e., 100 ft/min) with zero wind speed present. Would the total power requirement be the same as if the helicopter was hovering, but with 100 ft/min wind speed in the downward direction? Note, I'm assuming steady (constant velocity) flight.

Blade element theory (BET) would say that the power requirements would be the same due to equivalent torque on the rotor (i.e., the blade element for both cases will experience the same angle of attack and magnitude of the relative wind). However, in the former case isn't the helicopter gaining gravitational potential energy, so there would be an additional climb power term equivalent to the weight times climb speed? Or does this term get excluded with BET analysis? Can I think of both scenarios as if the helicopter is gaining "air-referenced" potential energy, where it's climbing relative to the air, meaning that the calculated power requirements would be the same regardless?

$\endgroup$
1
  • $\begingroup$ I suppose that in the first case you have energy associated with a) compensating for the weight, b) winning the aerodynamic drag and c) gaining height. In the second case you have energy associated with a) compensating for the weight, b) winning the aerodynamic drag and c) moving air downward. But I'm not a physicist. $\endgroup$
    – sophit
    Feb 11, 2023 at 21:49

1 Answer 1

2
$\begingroup$

Yes, the power required to hover in a 100 ft/min downdraft is the same as the power required to make a 100 ft/min climb in calm air, all other things being equal. And yes, the helicopter that's climbing is gaining potential energy.

What you're missing is that the downdraft is removing potential energy from the helicopter. If we replace the helicopter with a neutrally-buoyant balloon, it's easy to see that a 100 ft/min downdraft would cause the balloon to descend at 100 ft/min, removing potential energy from it.

To answer your last question, yes, blade element theory completely ignores the effect of gravity. BET is solely concerned with the performance of propeller blades, so, in fact, it ignores the effects of everything besides the blade and things that directly affect the blade.

$\endgroup$
2
  • $\begingroup$ Thank you for your answer. It makes perfect sense to look at it this way. So with wind present, climb power is more accurately determined by weight times vertical airspeed (not just vertical ground speed), just as parasite power is determined by drag times horizontal airspeed in steady, level flight, correct? $\endgroup$ Feb 12, 2023 at 0:13
  • 1
    $\begingroup$ @AdamYassine: Aircrafts move in respect to air, not ground, and therefore everything should be calculated in respect to air. Longitudinally, vertically and laterally. For example in the usual $L=½ \rho V² S C_l$, speed $V$ is measured in respect to wind, not ground. $\endgroup$
    – sophit
    Feb 12, 2023 at 8:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .