Why is the ratio maximum range speed/maximum endurance speed the same for any airplane?

Question

It seems counter-intuitive that the ratio of speeds for maximum range and maximum endurance is always the same, $3^{0.25} = 1.316...$ for any heavier-than-air aircraft, regardless of its mass, size or form. It's hard to believe that such ratio may be identical for an ultralight autogyro and for the F-35 fighter, but so it seems to be...

Why is it so?

Answer 1

The ratio is the same for all aeroplanes if you accept a number of assumptions:

• The propulsion efficiency is constant, regardless of speed or power setting
• Aerodynamic drag is the sum of parasite drag and induced drag
• Parasite drag is proportional to the square of airspeed:$ D_p = k_p \cdot V^2$
• Induced drag is inversely proportional to the square of airspeed: $D_i = \frac{k_i }{V^2}$
• There is no wind

Since we assume the propulsion efficiency is constant, the fuel consumption rate is directly proportional to the power required. The power required is drag times airspeed:

$P = D\cdot V = D_p\cdot V + D_i\cdot V = k_p\cdot V^3 + \frac{k_i}{V}$

For the maximum endurance we need to minimise the fuel consumption and thus we need to find the speed that minimises the power.

$\frac{dP}{dV} = 3 k_p V^2 - \frac{k_i}{V^2} = 0$

Solving for $V$ results in

$V_{endurance} = \sqrt[\uproot{1}4]{ \frac{k_i}{3k_p}} $

For the maximum range we need to find the speed that minimises the fuel consumption per distance travelled, which is found when the ratio of power to speed over ground is minimal. As we assume there is no wind, the ground speed and the airspeed are equal. Since the ratio of power to airspeed is drag, we have to find the speed for minimum drag:

$\frac{dD}{dV} = 2 k_p V - 2\frac{k_i}{V^3} = 0$

Solving for $V$ results in

$V_{range} = \sqrt[\uproot{1}4]{ \frac{k_i}{k_p}} $

We can now show that the ratio of maximum range speed to maximum endurance speed is:

$\frac{V_{range}}{V_{endurance}} = \left. \sqrt[\uproot{1}4]{ \frac{k_i}{k_p}} \middle/ \sqrt[\uproot{1}4]{\frac{k_i}{3k_p}} \right. = \left. 1 \middle/ \sqrt[\uproot{1}4]{\frac{1}{3}} \right. = \sqrt[\uproot{1}4]{ 3} = 1.316... $

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Now as long as the assumptions hold, for any aircraft the ratio of the speed will be approximately 1.3.

For the F-35 the ratio may be approximately right in the subsonic domain. Effects of compressibility will cause the drag to be higher when the aircraft gets into the transonic domain so depending on the max-range speed the last two assumptions may not hold.

For an autogyro at higher speeds, the relation of induced drag to airspeed is inversely proportional to the square of velocity, just like conventional fixed wing aircraft. This is because the fundamental way of creating lift by deflecting the incoming airflow downwards is the same for autogyros and aeroplanes. Therefore autogyros have also a factor pf approximately 1.3 between their maximum endurance and maximum range speeds.

Answer 2

Those ratios are only correct if Mach and engine efficiency aren't included in the calculations. If one considers that best range speed, from your calc, gets you into an unfavorable Mach number then it won't be correct. If best range speed forces engine operation at a inefficient power setting, (low power settings will be inefficient for turbines, high power settings are inefficient for gasoline engines) then that calculation will be incorrect.

I remember best rate of climb speeds in the T-38 were crazy high, like 540CAS until Mach 0.8 something, even though this was nowhere close to l/d max. The engines were straight jets, so performance was very good at speeds higher than what was ideal from an airframe perspective. This is largely why jets must fly so high to get reasonable efficiency, the engines don't lose much thrust with speed, but the airframe needs to be near l/d max CAS to be efficient.