# Helicopter total power required: zero wind speed while climbing --vs-- down wind speed while hovering

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?

• 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. Feb 11 at 21:49

• @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. Feb 12 at 8:46