Max Lift to drag ratio

Question

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

This is just a calculus exercise, not an engineering question. — alephzero, Feb 12 '19 at 21:38
Hint: finding the maximum drag to lift ratio is easier than finding the minimum lift to drag ratio. — alephzero, Feb 12 '19 at 21:40
Take the first derivative and find where the slope is zero. That’ll be either the maximum or minimum. — Eric S, Feb 12 '19 at 21:41
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. — Jeffrey J Weimer, Feb 18 '19 at 22:45

kamran · Answer 1 · 2019-02-14T00:58:58.087

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

score 1 · Answer 2 · answered Feb 12 '19 at 23:03

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.

james · Accepted Answer · 2019-02-13T06:51:07.613

-2

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 ?

edited Feb 13 '19 at 06:51

answered Feb 12 '19 at 21:52

james

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Erm, it can't be zero, because both terms are positive for all y. – Phil Sweet Feb 13 '19 at 02:26

Max Lift to drag ratio

3 Answers3