# Why would maximum fly-time (endurance) not coincide with max L/D operating point?

This question is motivated by another SE Aviation answer where @Peter Kampf writes the following two useful bits of information:

Aircraft like to fly near their optimum L/D ratio, where drag reaches its minimum.

Observation aircraft which want to optimize flying time will fly slower that what optimum L/D requires, especially if they use propellers.

Do aircraft use the majority of their fuel to overcome friction?

I'm trying to understand this more: Why would maximum fly-time not coincide with max L/D operating point?

Is this something to do with the jet engine fuel consumption characteristic? i.e. Does the minimum fuel-consumed-per-unit-time point not coincide with the optimum L/D point? But if your payload is fixed, so is your lift & hence your velocity choice that gives you that lift right?

i.e. For a given payload, lift is fixed. For a given lift, minimum drag leads to max fuel economy? And max economy means max dwell time? What gives. For simplicity I assume we can ignore the climb / descent / acceleration etc. for now. And only consider the level flight segment at fixed speed.

Maximum dwell time or maximum endurance occurs when the power required is minimum. Hence, in this case, the maximum endurance speed is one where the power required is minimum, while in case of maximum range speed, the thrust required is minimum.

For maximum endurance, we must minimize the fuel consumed per unit time i.e. the fuel flow. For maximum range, we must minimize the fuel used per unit distance traveled.

In case of propeller aircraft, the fuel flow rate is proportional to the power produced. Hence, the maximum endurance occurs at a point where the power is minimum. For (turbo)jets, the minimum fuel flow occurs when the thrust is minimum. Hence the maximum endurance occurs when the L/D is maximum. For turbofans, it is somewhere in-between.

Consider a propeller aircraft in a steady, level flight. For determining the condition where the energy expenditure is minimum, we have,

$P = W (\frac{C_{D}}{C_{L}})V$

is minimum. For steady flight, we have,

$V = \sqrt{\frac{W}{\frac{1}{2} \rho S C_{L}}}$

This gives,

$P = \sqrt{\frac{W}{\frac{1}{2} \rho S}}(\frac{C_{D}}{C_{L}^{\frac{3}{2}}})$

Thus, for propeller aircraft, the minimum power and maximum endurance occurs when $\frac{C_{L}^{\frac{3}{2}}}{C_{D}}$, rather than $\frac{C_{L}}{C_{D}}$ is maximum. Due to this, the minimum power (maximum endurance) condition occurs at a speed which is 76% of the minimum drag (maximum range) condition. Image from eaa1000.av.org

Also, see here and here

Thrust is a force that moves the aircraft. In steady, level flight, this is equal to the drag (if it is more/less, the aircraft will accelerate/decelerate). Power is the rate of doing work i.e. the energy consumed per unit time or rate of energy expenditure (by the a/c powerplant). This is why we are considering minimum power i.e. rate of energy expenditure for determining endurance.

Power is the product of force (thrust) and velocity. Think of it in this way- as speed increases, the drag decreases, reaches a minimum and then increases. However, as the power is product of drag (i.e. thrust) and velocity, it too follows the a similar path; however, the minimum is reached before the minimum drag. That speed gives the maximum endurance.

For jet engined aircraft, the speeds are different. In this case, the speed corresponding to minimum $\frac{C_{L}}{C_{D}}$ gives maximum endurance, while the speed corresponding to $\frac{C_{L}^{\frac{1}{2}}}{C_{D}}$ gives maximum range. Also, see here

• Can you expand on the difference between Power and Thrust? While it seems clear that max Range - where speed plays a role - happens at a faster speed than max Endurance (where speed does not), it isn't clear how overcoming MORE drag (i.e. by being left of L/D max) makes for better endurance. Is this relationship true for jets and props both? Thanks. – Ralph J Dec 26 '15 at 13:37
• @RalphJ I've added some points about thrust and power. Hope this helps. – aeroalias Dec 26 '15 at 15:40

The polar point for maximum flight time only coincides with the minimum drag point when engine thrust does not change over speed. This is approximately true for pure turbojets and rockets. If thrust creation involves accelerating a big mass flow of air, the declining thrust with increasing speed for a given engine power shifts the optimum to lower speeds.

For a more general optimization goal we need to minimize not drag, but fuel flow. Since thrust varies with fuel flow for propeller and bypass engines (as it does for ramjets), this can be calculated when we model thrust $T$ over speed $v$ as $T \varpropto v^{n_v}$ with $n_v$ a negative number for propeller and turbofan engines, and positive for ramjets.

Starting from the equilibrium in steady flight $$T_0\cdot v^{n_v} = c_D\cdot\frac{\rho}{2}\cdot v^2\cdot S$$ we can express the speed $v$ in terms of the lift coefficient $c_L$ $$T_0 = c_D\cdot\left(\frac{\rho}{2}\cdot S\right)^{\frac{n_v}{2}}\cdot\left(\frac{m\cdot g}{c_L}\right)^{1-\frac{n_v}{2}}$$

$T_0$ is the reference thrust at a specific speed and depends only on fuel flow. You can see it equally as the thrust setting, and we want to minimize this. Therefore, we approximate the drag coefficient with the quadratic polar ($c_D = c_{D0} + \frac{c_L^2}{\pi\cdot AR\cdot\epsilon}$), differentiate the right part of the equation with respect to $c_L$ and look for the lift coefficient at which it is zero: $$0 = \frac{n_v-2}{2}\cdot c_{D0}\cdot c_L^{\frac{n_v-4}{2}} + \frac{n_v+2}{2\cdot\pi\cdot AR\cdot\epsilon}\cdot c_L^{\frac{n_v}{2}}$$ $$\Leftrightarrow c_L = \sqrt{\frac{2-n_v}{n_v+2}\cdot\pi\cdot AR\cdot\epsilon\cdot c_{D0}}$$ This by itself is not yet helpful, but if we look at the ratio of the drag components at specific values of $n_v$, the answer becomes clear: $$c_{Di} = \frac{c_L^2}{\pi\cdot AR\cdot\epsilon} = \frac{2-n_v}{n_v+2}\cdot c_{D0}$$ Propeller aircraft ($n_v$ = -1): $c_{Di} = 3\cdot c_{D0} \Rightarrow$ 76% of the speed for lowest drag

Turbofan aircraft ($n_v$ = -0.5): $c_{Di} = \frac{5}{3}\cdot c_{D0} \Rightarrow$ 88% of the speed for lowest drag

Turbojet aircraft ($n_v$ = 0): $c_{Di} = c_{D0} \Rightarrow$ 100% of the speed for lowest drag

For optimum loiter speed, the induced drag of a propeller aircraft needs to be three times as big as the zero-lift drag. As induced drag drops with increasing speed, only for turbojets will the optimum polar point for maximum flight duration be equal to that at the lowest drag.

Nomenclature:
$c_L \:\:\:$ lift coefficient
$n_v \:\:\:$ thrust exponent, as in $T = T_0\cdot v^{n_v}$
$\pi \:\:\:\:\:$ 3.14159$\dots$
$AR \:\:$ aspect ratio of the wing
$\epsilon \:\:\:\:\:$ the wing's Oswald factor
$c_{D0} \:$ zero-lift drag coefficient
$c_{Di} \:\:$ induced drag coefficient

• I'm still confused. Let's say we fly on a Cessna 172 at speed slower than its Vmd. We have now higher drag to overcome. Am I right in assuming, that by flying slower than Vmd, a propeller is actually a bit more efficient? As a consequence we can fly with lower RPM even if we are slower than our Vmd? Is this actually correct? Does lower RPM means lower fuel flow? Thanks and regards. – Electric Pilot Aug 9 '18 at 3:04
• @ElectricPilot: Propeller thrust is inversely related to speed. As long as the drag increase by slowing down is less than the thrust increase, the difference will be greater, hence less power is required to stay aloft. It is not about propeller efficiency but how much thrust can be gained from a given amount of power. Power is thrust times speed. – Peter Kämpf Aug 9 '18 at 20:10