I have a question that I can't solve. Coefficient of drag= x, Coefficient of Lift= y, it is given that x= 0.015 + 0.05*(y^2), FInd Max Lift to drag ratio

and I know that Lift to drag ratio is L/D= Coefficient of lift/Coefficient of drag. Can anyone tell me how to solve this

Thanks in advance

  • 1
    $\begingroup$ This is just a calculus exercise, not an engineering question. $\endgroup$
    – alephzero
    Commented Feb 12, 2019 at 21:38
  • $\begingroup$ Hint: finding the maximum drag to lift ratio is easier than finding the minimum lift to drag ratio. $\endgroup$
    – alephzero
    Commented Feb 12, 2019 at 21:40
  • 2
    $\begingroup$ Take the first derivative and find where the slope is zero. That’ll be either the maximum or minimum. $\endgroup$
    – Eric S
    Commented Feb 12, 2019 at 21:41
  • $\begingroup$ @alephzero I think you may have got your ideas reversed. $\endgroup$
    – Phil Sweet
    Commented Feb 13, 2019 at 2:28
  • $\begingroup$ Translate L/D into an equation with y and x. The maximum of any function is defined by a mathematical analysis of the derivatives of the function. As noted, this is a calculus problem not an engineering one. $\endgroup$ Commented Feb 18, 2019 at 22:45

3 Answers 3


The lift(y) to drag(x) ratio can be expressed as: $LD_{ratio} = y/x = \frac{y}{0.015 + 0.05*(y^2)}$

You will get the maximum when the denominator is 0, so when: $0.015 + 0.05*(y^2)=0$

Hence, only when $$y = 0. - 0.547723i$$ or $$y = 0. + 0.547723i$$

...are you sure about the formula ?

  • $\begingroup$ Erm, it can't be zero, because both terms are positive for all y. $\endgroup$
    – Phil Sweet
    Commented Feb 13, 2019 at 2:26

As has been mentioned above by Eric Shain the correct answer is to take the partial with respect to y= CL and equate it to zero. Let's Call the max L/D, M.

$$M = \frac{C_L}{C_D} = \frac{C_L}{C_{D_0} +K*C_L^2}, \quad C_{D_0}=0.015\ and\ K=0.05 $$

$$ \frac{\partial M}{\partial C_L} =\ \frac{\partial}{\partial C_L} [ \frac{C_L}{0.015 +0.05*C_L^2}]\ =\quad \frac{[0.015 + 0.05*C_L^2 ]-C_L [0 + 2(0.05)C_L^2]}{[0.015 + 0.05*C_L^2 ]^2} =0$$

$$ \Rightarrow 0.015- 0.05C_L^2=0\\ C_L^2=0.015/0.05=0.3\\C_{L_M}=\sqrt{0.3} \quad and\ C_{D_M}= 2*C_{D_0}=0.03$$

And $$M=\frac{C_{L_M}}{C_{D_M}}= \frac{\sqrt{0.3}}{0.03}=18.257$$

Maximum lift to drag ratio is 18.257. please double check my arithmetic, but it looks correct.

Edit The following is a quote from Wikipedia on lift to drag ratio.

A House sparrow has a 4:1 L/D ratio, a Herring gull a 10:1 one, a Common tern 12:1 and an Albatross 20:1, to be compared to 8.3:1 for the Wright Flyer to 17.7:1 for a Boeing 747 in cruise.[4] A cruising Airbus A380 reaches 20:1.[5] The Concorde at takeoff and landing had a 4:1 L/D ratio, increasing to 12:1 at Mach 0.95 and 7.5:1 at Mach 2.[6] A Helicopter at 100 kn (190 km/h) has a 4.5:1 L/D ratio.[7] A Cessna 172 glides at a 10.9:1 ratio.[8] A cruising Lockheed U-2 has a 25.6 L/D ratio.[9] The Rutan Voyager had a 27:1 ratio and the Virgin Atlantic GlobalFlyer 37:1


As an alternative to solving with calculus, if you are provided with a graph of the L/D ratio, you can find the line which passes through the origin (0 lift and 0 drag), and is tangential to the L/D line on the graph.

The point at which this straight line touches the L/D graph is the maximum L/D.


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