# How to pick a good airfoil based on Cl and Cd

Let's assume I have 10 different airfoils graphs of Cl x Aoa, Cd x Aoa and Cl/Cd x Aoa. How do I pick the best one that will give me the most efficiency (the longest flight time)?

I understand the information that those graphs provide and IMHO I think I should always pick an airfoil that has the best Cl/Cd ratio at Aoa (alpha) of 0 since at 0 the drag is usually the least. This is my "common sense" but nobody says this on the internet so I am pretty sure I am wrong!

So based on the three graphs I have of each airfoil, what math should I do in order to get the most efficient (more flight time) airfoil?

PS: the plane will be a floater/slow flier/glider/sailplane. Of course it will be a slow plane cause my focus is on efficiency and fast planes usually are not efficient.

• Note that the longest flight time is only ONE measure of efficiency and it happens to be the one that most designers consider not important. This is because another measure of efficiency is the furthest distance which is usually more important because most airplanes are designed to go somewhere. Interestingly, the most efficient airfoil for furthest distance is never the same as longest flight time. That's why most gliders have flaps so that you can manipulate the camber of the airfoil as a compromise Apr 22 '18 at 3:28

The $\alpha$=0 point on the polar would be a poor metric - there is nothing special about it, and your aircraft will not fly at this polar point when you strive for longest endurance. But your common sense is correct in predicting that this will be a slow airplane.

The precise answer normally depends on your means of propulsion. Since this is for gliders, the answer will be quite simple.

Your definition of efficiency is minimum energy loss. Energy means potential energy in this case $E_{pot} = m\cdot g\cdot h$, and the loss of potential $\frac{dh}{dt}$ energy over time is expressed as the sink speed $v_s$. So we need to find the polar point at which the glider will have the lowest possible sink speed.

Let's start with the parabolic drag equation which splits the total drag coefficient $c_D$ in one component which is constant over the lift coefficient range and one which changes with the square of the lift coefficient. $$c_D = c_{D0}+\frac{c_L^2}{\pi\cdot AR\cdot\epsilon}$$ The symbols are:
$\kern{5mm} m\:\:\:\:$ aircraft mass
$\kern{5mm} g\:\:\:\:\;$ gravitational acceleration
$\kern{5mm} \rho\:\:\:\:\:$ air density
$\kern{5mm} v\:\:\:\:\:$ velocity
$\kern{5mm} v_z\:\:\;$ sink speed
$\kern{5mm} c_{D0} \:$ zero-lift drag coefficient
$\kern{5mm} c_L \:\:\:$ lift coefficient
$\kern{5mm} \pi \:\:\:\:\:$ 3.14159$\dots$
$\kern{5mm} AR \:\:$ aspect ratio of the wing
$\kern{5mm} \epsilon \:\:\:\:\:\:$ the wing's Oswald factor
$\kern{5mm} S \:\:\:\:\:$ wing area

Next, we need an expression for the sink speed. This should be easy: It is the forward speed times the tangent of the glide path angle $\gamma$. If you allow to approximate the tangent function for small angles by the radian of the angle, you can write: $$v_z = v\cdot tan\gamma ≈ v\cdot\gamma = v\cdot\frac{c_D}{c_L}$$

Now insert the drag coefficient $$v_z = v\cdot\left(\frac{c_{D0}}{c_L} + \frac{c_L}{\pi\cdot AR\cdot\epsilon}\right)$$

and make the speed dependency of the lift coefficient obvious: $$v_z = \frac{c_{D0}\cdot\frac{\rho\cdot v^3}{2}\cdot S}{m\cdot g} + \frac{m\cdot g}{\pi\cdot AR\cdot\epsilon\cdot\frac{\rho\cdot v}{2}\cdot S} = v^3\cdot\frac{c_{D0}\cdot\rho\cdot S}{2\cdot m\cdot g} + \frac{1}{v}\cdot\frac{2\cdot m\cdot g}{\pi\cdot AR\cdot\epsilon\cdot\rho\cdot S}$$

Now we are ready to differentiate with respect to $v$ and find the condition when the derivation will become zero. $$\frac{dv_z}{dv} = 3\cdot v^2\cdot\frac{c_{D0}\cdot\rho\cdot S}{2\cdot m\cdot g} - \frac{1}{2\cdot v^2}\cdot\frac{2\cdot m\cdot g}{\pi\cdot AR\cdot\epsilon\cdot\rho\cdot S}$$

Now re-insert the lift coefficient, which simplifies the equation mightily: $$\frac{dv_z}{dv} = 3\cdot\frac{c_{D0}}{c_L} - \frac{c_L}{\pi\cdot AR\cdot\epsilon}\,\overset{!}{=}\, 0$$ $$\Rightarrow 3\cdot c_{D0} = \frac{c^2_L}{\pi\cdot AR\cdot\epsilon}$$

So your energy loss will be minimized when the induced drag is three times bigger than the zero-lift drag. The polar point is then: $$c_L = \sqrt{3\cdot c_{D0}\cdot \pi\cdot AR\cdot\epsilon}\;\;\text{and}\;\;c_D = 4\cdot c_{D0}$$ With wings of a high aspect ratio this gives a rather high lift coefficient, so you might need to pick a high-lift airfoil in order to actually trim this polar point. The figure of merit when selecting the airfoil should be the ratio $\frac{c^{³⁄₂}_L}{c_D}$; this should reach its maximum at the lift coefficient of minimum energy loss. Enter the lift and drag coefficients into a spreadsheet and create a new column for $\frac{c^{³⁄₂}_L}{c_D}$ (or $\frac{c^3_L}{c_D^2}$; doesn't matter). Select an airfoil that maximises this value at the calculated optimum lift coefficient.

• I think I followed what you wrote right up to the figure of merit formula, for which I don't understand the source. Where did it come from? Thanks for taking the time to explain so much to so many. Apr 20 '18 at 16:21
• The cl3/cd2 is the endurance parameter, a good explanation (including range related functions etc) is here : dept.aoe.vt.edu/~lutze/AOE3104/range&endurance.pdf Apr 20 '18 at 16:57
• Page 16, (as it’s a rather long document.) Apr 20 '18 at 16:58
• @GürkanÇetin: Yes, thank you, I could have produced the same here, but given that nobody gets the explanation for the polar point (just see what people vote on …), I found it not worth the effort. Apr 21 '18 at 10:33
• Funny but the chosen one is a WRONG answer. The endurance is not proportional to lift-to-drag ratio, but to cl3/cd2. Apr 21 '18 at 11:53

If you only care about efficiency, you want the one that has the highest Cl3/Cd2 for endurance and highest Cl/Cd for range at the design Cl, which may not be zero AoA. Determine desired Cl using required lift (weight plus tail download) in the following:

L = Cl * A * .5 * r * V^2

where L is lift, Cl is coefficient of lift, A is wing area, r is density and V is Velocity. Explanation is at:

https://www.grc.nasa.gov/www/k-12/airplane/lifteq.html

Look through your Cl/AoA charts at that Cl then get Cd using the associated AoA. Pick the airfoil with the highest value of Cl cubed / Cd squared for maximum endurance. Pick the highest CL/Cd using the associated AoA for maximum range.

• Your answer makes sense but I didnt understand why do I have to find the L (lift). Should I first get the L or the Cl of my plane? Then using one or other I find the other right? Then I should go to the Cl/AoA (according to what you said), why? Why shouldnt I go directly to the Cl/Cd x AoA diagram? Apr 19 '18 at 19:22
• L is the lift you want your wing to produce, the aircraft loaded weight plus tail download if that is your design. The formula gives you Cl at your design point, which you use to get AoA, which gives you Cl/Cd. You can't go directly to Cl/Cd because you don't know Cd or AoA before using the charts. Apr 19 '18 at 19:40
• thanks for commenting. So correct me if I am wrong: FIRST I get the total weight of my plane (I will not take into account the tail download cause I have no idea how to do that and I assume the tail download will be small compared to the weight) and make L = weight. SECOND I calculate Cl (great!). THIRD: I try to find (on each airfoil) the AoA that matches the required Cl. FORTH: I go to the Cl/Cd x AoA and pick the airfoil that has the HIGHEST Cl/Cd at the chosen AoA. Is it right? Apr 19 '18 at 19:48
• Yes. Use 10% for tail download absent anything better. Apr 19 '18 at 19:50
• thanks! I think, using your analogy of this answer it would be possible to get an answer to the other question I did. Since with your answer I can pick the right airfoil (which I will do right now to 3 of my planes!) maybe it can be used as an iterative process to find the right chord (it will have to be an iterative process cause when I find a new chord it will change wing area and the math should be redone). Anyway my friend, you answer is by far the best one in the internet regarding this subject! Apr 19 '18 at 19:56