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For a quad-copter, or a fixed-wing drone with quad-rotor formation, how much would the power of the motor decrease when a constant speed descent is needed to bring the drone back to the ground? Does the motor just need to ensure Lift equals the Weight so the drone can descend in constant speed?

Or, in other words, can the the motor power during descent be calculated using the climb power requirement equation, but just putting in negative speed to represent descent?

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    $\begingroup$ Off-topic here. Ask in the Drones and Model Aircraft StackExchange. $\endgroup$
    – MD88Fan
    Jan 28 at 18:45
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    $\begingroup$ On-topic here because the FAA considers all drones to be "Unmanned Aircraft Systems" -- see faa.gov/sites/faa.gov/files/2021-08/RemoteID_Final_Rule.pdf $\endgroup$ Jan 28 at 18:57
  • $\begingroup$ We still have an issue here about whether you really intend to ask whether Lift equals Weight minus the vertical component of Drag-- I don't want presume to edit the question, but you may wish to -- but this really this pertains to thrust rather than power -- $\endgroup$ Jan 28 at 20:42
  • $\begingroup$ @quietflyer, a good case could be made for either, but your quote equally bolsters an opposing view. Consider: BECAUSE the FAA considers all Unmanned Aircraft Systems to be drones, and BECAUSE Stack Exchange has a separate site for drones, THEREFORE the Drone SE site is the "best" place for this question... $\endgroup$ Jan 29 at 1:17
  • $\begingroup$ There may be multiple questions here --answer may be very different for quadcopter w/ no fixed wing vs quadcopter that also has a fixed wing. $\endgroup$ Jan 29 at 16:54

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can motor power during a descent be calculated from the climb power requirement?

Usually it's the other way around, but most importantly, one is quantifying rate of climb.

Power requirement for level flight is determined in fixed wing craft by the glide ratio. One you know your altitude loss over a certain distance, power (Thrust at glide speed) requirement for level flight is mass × glide ratio.

Climbing ability is determined by excess thrust available at that speed. As one angles up, lift requirement (perpendicular to the line of flight) becomes less than weight (by the cosine of climb angle × weight) and thrust must not only maintain airspeed but also contribute to vertical force to support the weight against gravity.

For a vertically descending helicopter, quad, landing Starship, whatever, the physics is a bit easier; it is just based on thrust, drag, and gravity.

What one must consider here is rate of descent. Aside from "lawn darts", most vertically descending objects will reach a "terminal" constant rate descent velocity from drag alone. Adding vertical thrust force will slow the rate of descent.

In a hover, vertical thrust force = weight. Ascending, vertical thrust = weight + drag (dependent on rate of ascent). Descending, vertical thrust = weight - drag (dependent on rate of descent).

Actual power requirement may also depend on technique of descent. Your quad may require less power (for a given rate of descent) if it is moving forward, as opposed to straight down, just like a helicopter. The forward motion actually helps it "glide" a bit.

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  • $\begingroup$ In conventional fixed-wing airplane, lift decreases (slightly) during climb or descent, and climb is due to excess thrust, descent due to lack of thrust. But in quadcopter, how do we define thrust? Wouldn't it be vertical in a vertical descent? And wouldn't it almost always point in quite a different direction than the direction of the flight path? I'm not sure that answers from fixed-wing airplanes readily translate over to this situation. Also answer may be very different for quadcopter w/ no fixed wing vs quadcopter that also has a fixed wing. $\endgroup$ Jan 29 at 16:51
  • $\begingroup$ @quietflyer well, one can reason a quad is simply 4 rotating wings that do not require forward motion to generate what we define as lift. "Quadracopter w/no fixed wing" still benefits from forward motion. $\endgroup$ Jan 29 at 19:33
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    $\begingroup$ For a helicopter the physics is actually much the same as for airplane. Note that auto-rotating helicopter – which is pretty much equivalent to fixed-wing gliding, it's just several wings gliding in circles – requires no power, but most of it's lift is still provided by the rotor, not drag on the body. $\endgroup$
    – Jan Hudec
    Jan 29 at 22:02
  • $\begingroup$ Absolutely, and, in the end, it takes a certain amount of power to fly, and more to climb. $\endgroup$ Jan 29 at 23:05
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Any decrease in power below what is required to maintain level flight will result in a descent.

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Lift will be less than weight in a descent. This results in a vertical speed component, that builds up until the vertical drag equals the force difference. Like any falling object reaching a terminal speed.

Or, in other words, can the the motor power during descent be calculated using the climb power requirement equation, but just putting in negative speed to represent descent?

Yes indeed, provided that vertical drag is a part of the climb power equation. Motor power is a funtion of generated thrust.

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  • $\begingroup$ Motor power is function of generated thrust and the velocity in the direction of thrust! After all, during autorotation the rotor still produces thrust to balance weight with no power input at all. $\endgroup$
    – Jan Hudec
    Jan 29 at 21:53
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The power required to maintain lift equal to weight¹ increases with climb rate and decreases with descent rate.

Thrust produced by a rotor (propeller) is equal to mass flow rate of air through the propeller disk times change in velocity of that air. This change in velocity increases momentum of the air (that balances the thrust), but it also increases its kinetic energy, and this energy has to be provided by the engine as induced power³.

But kinetic energy is equal to mass times square of velocity, which means the induced power is equal to mass flow rate times change in velocity times velocity² of the air. This velocity is some base value due to the rotor working, plus the vertical velocity of the craft.

So if the craft is climbing, the required power increases, and if the craft is descending, the power requirement decreases, even for the same lift.

If the angle of attack of the rotor blades is low enough so they don't stall when the flow direction changes to upward, the induced power will decrease below zero until it balances the parasite power³ and the rotor spins without any engine power at all. This is called autorotation, and allows helicopters to still make a safe landing in case of engine failure. However it requires the blades to have low enough angle of attack, so a quadcopter with fixed blade pitch won't autorotate, but the rotor will stall at higher rates of descent instead.


¹ Drag will increase the lift requirement in climb a bit, and decrease it in descent a bit, but with typical climb or descent rates – 100 ft/min ~ 1 knot – this is relatively small contribution.

² More precisely, $P = \dot m (v_0 \Delta v + \Delta v^2)$ where $P$ is the induced power, $\dot m$ is mass flow rate, $v_0$ is the air flow speed just above the rotor and $\Delta v$ is the increase in flow speed through the rotor. When the craft is descending, at some point $v_0$ will become negative, meaning the air flows from below.

³ Power required to turn a rotor (propeller) is induced power, which is the power required to increase kinetic energy of the air to produce thrust, plus parasite power, which is the power required to overcome parasite drag of the blades.

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