Given the following parameters for a small business aircraft:

  • Wing surface: $S=30m^2$
  • Lift Drag Polar in clean configuration: $C_D = 0.022 + 0.047 {C_L}^2$
  • Maximum Lift Coefficient $C_{L_{max}}=1.35$
  • Static sea-level Thrust: $T_0 = 12KN$
  • Thrust variation with altitude: $\frac{T}{T_0}=\frac{\rho}{\rho_0}$
  • Aircraft weight: $W = 50 KN$
  • Thrust specific fuel consumption (constant): $C_T = 0.7 N/Nh$

I need to calculate:

  • the optimum lift coefficient for maximum specific range
  • the airspeed for maximum specific range when operating at the specified weight and sea level conditions.

For $C_{L_{opt}}$ I have found a value of 0.395 using the lift drag polar and the fact that $C_{L_{opt}}= \sqrt{\frac{1}{3}C_{D_0}\pi A e}$.

I am then assuming that the airspeed for maximum range is given by $V=\sqrt{\frac{W}{S}\frac{2}{\rho}\frac{1}{C_{L_{opt}}}}$.

In my calculations, I am using $W=50000/9.80665=5098.581 KG$ and $\rho=1.225$, which gives me a value of $V = 26.504 m/s$

However this seems not to be the correct answer. What am I missing?

  • 2
    $\begingroup$ Without checking where your formulas come from: why are you using W in kg? W is a force and and the correct unit is N. You are not getting 26.504m/s but 26.504 sqrt(m). $\endgroup$
    – Gypaets
    Sep 3, 2016 at 20:25
  • $\begingroup$ @mns Yes you are absolutely correct! I don't know for some stupid reason I was persisting on trying to convert W to a mass whereas it is indeed a force. I am getting the right answer now: V=83 m/s. Thanks! If you put your comment as an answer I will accept it $\endgroup$ Sep 3, 2016 at 21:03

1 Answer 1


Check the units. Using $$W = 50000N$$ instead of $W=5099kg$ you get $$V=83m/s$$ (vs. $V=26.5\sqrt{m}$).


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