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While I understand the derivation of Power required, I fail to understand why minimum power required corresponds to a velocity that is lower than velocity corresponding to minimum thrust required.

My understanding is that Minimum Thrust is the most efficient speed an airplane should fly at and how can flying lower than that consume even lesser power.

Please correct my understanding if wrong. Introduction to flight 9th Edition

Edit: While the given explanations make sense thinking about Piston driven engines, where thrust decreases with increase in velocity. I am still having trouble understanding the concept with JET engines where thrust is fairly consistent with speed. Would really appreciate if somebody could clarify

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  • $\begingroup$ Do you mean by a mathematical or by a physical point of view? Or both? $\endgroup$
    – sophit
    Mar 14, 2023 at 11:09

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There are two ways to look at efficiency of flight, and it depends on the objective of the flight which type off efficiency is applicable.

If the objective is to stay airborne as long as possible, we want to minimise the energy consumption per unit of time. In this case, the speed to fly at is the speed which has the minimum power required.

If the object is to fly as far as possible, we want to minimise the energy consumption per unit of distance traveled. Since energy consumed equals force (thrust) times distance, the minimum energy per unit of distance is found when the thrust and thus drag is lowest.

The minimum drag speed (most efficient for travel) is found at a speed that is higher that the minimum power speed (most efficient for staying airborne as long as possible) .

Flying at the minimum drag velocity requires more power than flying at the minimum power velocity.

For transport flights, cruising speeds are selected close to the minimum thrust speed, but holding speeds are selected close to minimum power speed.

Drag and power graph

Own work using Desmos graphing calculator

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  • $\begingroup$ Re "cruising speeds are set close to the minimum thrust speed" - that's not true. Minimum thrust speed occurs at the same speed as $V_{L/D \ max}$. What is special about $V_{L/D \ max}$? - it is also the best glide speed; and the best glide speed is generally nowhere near the cruise speed. $\endgroup$ Mar 14, 2023 at 11:20
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    $\begingroup$ @AdityaSharma best glide and max range speed are typically very close to each other. Differences are due to the fact that propulsive efficiency is not constant over the speed envelope. For best glide, only L/D matters, because there is no propulsive efficiency to take into account. For maximum range cruise, the variation with speed of the efficiency of the propulsive system needs to be taken into account as well, but the offset from best-glide is typically low. For operational reasons (time is money) a slightly higher speed is chosen, accepting extra fuel costs because it saves time. $\endgroup$
    – DeltaLima
    Mar 14, 2023 at 12:13
  • $\begingroup$ @AdityaSharma see also aviation.stackexchange.com/q/81664/19 $\endgroup$
    – DeltaLima
    Mar 14, 2023 at 12:20
  • $\begingroup$ That's why airliner Vbg is close to their cruising speed (with flaps up). $\endgroup$ Mar 14, 2023 at 16:56
  • $\begingroup$ But Aditya Sharna is still correct, because cruise speeds are nowhere near maximum range cruise speed. They are much faster. Other considerations than fuel consumption matter as well. $\endgroup$ Mar 20, 2023 at 14:07
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While the given explanations make sense thinking about Piston driven engines... I am still having trouble understanding the concept with jet engines.

The kind of propulsion is actually not part of the problem. And therefore I think that the easiest way to understand it is just using a... sailplane 🙂

A sailplane basically loses altitude in order to overcome aerodynamic drag: it continuously trades potential energy (thanks to the loss in altitude) with kinetic energy (which is being dissipated due to drag). When is this exchange maximised? This exchange is obviously maximised when the sailplane loses as little kinetic energy as possible because of the drag i.e. when the drag is at its minimum (which happens to be also the point for best L/D). The relevant speed is termed "speed for best glide ratio".

Now, what happens when the sailplane reaches a thermal? Do we still need to minimise drag also in this case? No, the goal now is not only to gain as much altitude as possible but, since we don't know for how long the thermal is going to exist, the goal is also to do it in the quickest way as possible: now not only the travelled space (aka altitude aka potential energy) is important but the needed time as well; what has to be optimised now is therefore their ratio i.e. power by definition: the power extracted from the thermal.

So, if you want to:

  • fly as much space as possible, you must fly at the speed for minimum drag (best L/D).
  • fly as much time as possible, you must fly at the speed for minimum power required.

As said, note that this is a very general result not related to the particular propulsive system or aerodynamic characteristics of the airplane and therefore it applies equally well to an airplane with engine(s): obviously minimising the needed power minimises in this case the fuel consumption since that is the source of the power.

Due to the particular shape of the power required (as given for example in the plot in your question) the speed for minimum power is lower than the speed for minimum drag ($1/\sqrt{3}=0.76$ smaller to be precise) and the relevant drag is some 15.5% bigger.

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  • $\begingroup$ Neat little trick that fixed pitch props have: highest $\eta$ under partial power for V min power, then ... because forward speed and RPM are both higher ... higher $\eta$ at Vy! $\endgroup$ Mar 15, 2023 at 22:19
  • $\begingroup$ Note that the general relationships described in this answer are true regardless of the shape of the thrust-required curve or drag-required curve-- the differences between the two speeds are just more pronounced with some shapes. $\endgroup$ Mar 16, 2023 at 16:54
  • $\begingroup$ @quietflyer: yes correct, shall I update my answer or do you want do it? $\endgroup$
    – sophit
    Mar 16, 2023 at 17:11
  • $\begingroup$ Alas, the "power extracted from the thermal" can only be applied as gravitational force, "thrust" = mg × sin glide angle $\theta$. $\endgroup$ Mar 16, 2023 at 17:22
  • $\begingroup$ @sophit -- it was just a comment of general interest. I won't update it and you already implied it with phrase "or aerodynamic characteristics of the airplane" but I think it would make a nice addition to answer to point it out explicitly-- I'll leave it up to you. Feel free to borrow/adapt any language you wish from the comment. Maybe simply a final sentence pointing out that with curves other shapes, the exact difference between speeds will be different but the min-power-req speed will still always be lower.... $\endgroup$ Mar 16, 2023 at 17:48
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This is because as speed decreases, it requires less power to produce a given amount of thrust. The following is a thrust-speed curve for a 1 watt engine:

enter image description here

The engine is operating at a constant propulsive efficiency ($η$) of 1 - this means that the shaft-power (RPM × Torque) gets converted into thrust-power (thrust × speed) with 100% efficiency¹. We can see that at 1 m/s, we obtain 1 newton of thrust from the 1 watt engine. The important thing to remember is that thrust is inversely proportional to speed.

If the aircraft slows down from the minimum thrust speed, it would certainly need a greater amount of thrust to oppose the drag. However, since the speed has reduced, the power required to produce that thrust has also reduced. As a result, the minimum power speed is slightly less than the minimum thrust speed - at this speed, the power required (thrust × speed) is minimum, even though thrust alone is not.


¹of course in practice, $η$ is not constant - it itself varies with speed, but the reasoning for why the minimum power speed is less than the minimum thrust speed remains the same.

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    $\begingroup$ Thanks for your reply. While it makes sense for propeller driven aircraft. How would you explain Jet aircraft where thrust is contents across range of velocities. Again assuming constant air density $\endgroup$ Mar 15, 2023 at 9:14
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all aircraft are moved with thrust

THRUST ACCELERATES AIRCRAFT UNTIL DRAG EQUALS THRUST

"Potential energy" of fuel is turned into thrust, no matter if it's a propeller, a rotor, a jet, or a rocket.

Minimum "power" is the lowest aircraft fuel consumption rate, in liters per second, the keep the aircraft in level flight.

Minimum drag is the airspeed at which the lowest amount of drag is produced. This is expressed as best L/D ratio

Why are not minimum power and minimum drag speeds the same?

Because the efficiency, $\eta$, of propellers is affected by the freestream velocity or airspeed and the propeller angle of attack.

For a given RPM, propellers can produce more thrust at a lower airspeed because force, or thrust, is derived as air m(V2 - V1). The slower the freestream, V1, the greater the thrust per unit of fuel burned.

Even though aerodynamic drag is a bit more at V min power, for the lowest amount of fuel burned, thrust = drag at V min power.

So, if you have to glide, do it at best L/D, because $\eta$ of gravity does not change with speed!!

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  • $\begingroup$ Oh, won't somebody PLEASE make a graph of Drag vs Velocity. $\endgroup$ Mar 14, 2023 at 13:59
  • $\begingroup$ You mean "thrust required" curve? That's in the question. $\endgroup$ Mar 14, 2023 at 22:43
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    $\begingroup$ Yes, absolutely, thrust = drag. That is why, at Vbg, the angle of glide (relative to horizon) is least. "Thrust" = g × sin $\theta$. What could be plotted on the Drag/Thrust graph is fuel consumption, which for props, will be < Vbg at its minimum due to $\eta$ factor. $\endgroup$ Mar 14, 2023 at 23:27

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