CL to AoA

[Joshrk22]

I'm doing a physics (high school) project relating AoA and lift. Is there a formula to convert the coefficient of lift to angle of attack? I have found the Max CL for the airfoil but I want to convert that to angle of attack so I know the critical AoA.

I know for thin airfoils you can use AoA=2*pi*radians but I am using NACA 2412 (Cessna 172/182 airfoil) so the chord % is 12% with camber % being 2%. How can I adjust the equation to account for these factors?

Next, for the calculation of max CL I have:

CLmax=(2*L)/(S*rho*V^2)

L = 2,100 lbs/2.2=954.5*9.8 = 9354.5 N
S = (174 ft^2)/(3.28 ft)^2 = 16.17 m^2
rho = 1.225 kg/m^3
V = 48 KIAS/1.15 = 41.7 mi/h *(1609/3600) = 18.7 m/s

CLmax = (2*9354.5)/(16.17*1.225*18.7^2)
CLmax = 2.71

This seems kinda high for the CLmax on a 172. Is this right? Please help all you aeronautical engineers!

[Dugie8]

This probably won't help you solve the equation problem, but it may help you verify the answer one you get it. The critical angle of attack on most light airplanes (like your Cessna) is about 16-17 degrees.

[tgrayson]

Originally Posted by Joshrk22

I'm doing a physics (high school) project relating AoA and lift. Is there a formula to convert the coefficient of lift to angle of attack?

Well, sorta. It will vary slightly per airfoil and it will vary much with the aspect ratio of a particular wing. However, assuming an infinite aspect ratio, which is what you effectively have when you put a wing section in a wind tunnel, you get about .11 CL per degree AOA.

Originally Posted by Joshrk22

I have found the Max CL for the airfoil but I want to convert that to angle of attack so I know the critical AoA.

As I indicated above, you will only get an approximation using an approximate number.

Originally Posted by Joshrk22

I know for thin airfoils you can use AoA=2*pi*radians but I am using NACA 2412 (Cessna 172/182 airfoil) so the chord % is 12% with camber % being 2%. How can I adjust the equation to account for these factors?

By looking at actual wind tunnel data. By the way, 2*pi radians is equal to the .11 cl per degree that I mentioned. You'll find it works reasonable well for a 2412. Clmax is about 1.6; dividing by .11 gives a critial AOA of about 15, which looks about right glancing at the lift slope curve.

Next, for the calculation of max CL I have:

Originally Posted by Joshrk22

CLmax=(2*L)/(S*rho*V^2)

L = 2,100 lbs/2.2=954.5*9.8 = 9354.5 N
S = (174 ft^2)/(3.28 ft)^2 = 16.17 m^2
rho = 1.225 kg/m^3
V = 48 KIAS/1.15 = 41.7 mi/h *(1609/3600) = 18.7 m/s
CLmax = (2*9354.5)/(16.17*1.225*18.7^2)
CLmax = 2.71

This seems kinda high for the CLmax on a 172. Is this right? Please help all you aeronautical engineers!

Much too high for a C172. You're looking for around 1.6. One problem is you're using indicated airspeeds. You can't do that; you've got to use calibrated. I'll have to look over the rest of your conversion, since I'm not used to doing it in metric.

[tgrayson]

Originally Posted by Joshrk22

CLmax=(2*L)/(S*rho*V^2)

L = 2,100 lbs/2.2=954.5*9.8 = 9354.5 N
S = (174 ft^2)/(3.28 ft)^2 = 16.17 m^2
rho = 1.225 kg/m^3
V = 48 KIAS/1.15 = 41.7 mi/h *(1609/3600) = 18.7 m/s

CLmax = (2*9354.5)/(16.17*1.225*18.7^2)
CLmax = 2.71

This seems kinda high for the CLmax on a 172. Is this right? Please help all you aeronautical engineers!

Ok, redoing your conversions (using Google), I have:

L = W = 2300 lbs (not sure where you got 2100) = 10230.4 N
S = 174 sf = 16.16 m^2
rho = 1.225 kg/m^3
V = 50 KCAS = 25 m/s

clmax = (2 * 10230.4)/[(16.16 * 1.225 * 25^2) = 1.65

[Joshrk22]

tgrayson,

Thank you very much! You are of great help!

[Joshrk22]

Oh I forgot to post something...

I got the 2,100 lbs because isn't some of the lift created from the tail section? So the wing is effectively creating less lift than the weight of the aircraft...

[tgrayson]

Originally Posted by Joshrk22

Oh I forgot to post something...

I got the 2,100 lbs because isn't some of the lift created from the tail section? So the wing is effectively creating less lift than the weight of the aircraft...

Actually, most of the time the tail is generating down lift, so the main wing must support the aircraft weight as well as the downwards lift produced by the tail.

At some rearward CG positions at low airspeeds, the tail may be generating positive lift.

[aerospacepilot]

In theory, an infinite wing will have a lift slope of 2*pi. That translates to .1096 per degree.

Fortunately, there is a magic formula that converts lift slope for an infinite wing to lift slope for a finite wing.
a = lift slope of finite wing
a_0 = lift slope of infinite wing
AR = aspect ratio
e = oswald efficiency number. Technically, this changes for an infinite wing versus a finite wing, but we can approximate it as being the same.


a = a_0 / (1 + (180*a_0 / (pi^2 *e*AR)) )

e = 0.8 for NACA 2412
AR = 7.32
To find a_0, we must find the lift slope of the NACA 2412 airfoil. It is not ideal (like the .1096 per degree stated above). My book has a graph of the NACA 2412 airfoil data. Here is what I picked off of it.
At -8 degrees AoA, Cl is -0.6.
At 8 degrees AoA, Cl is 1.08
Change in coefficient of lift over change in Aoa = (1.08 - -0.6) / (8 - -8)
That gives an a_0 = 0.105.

Using the formula above, we get a = 0.0791 per degree. So for every degree increase in angle of attack, you get a .0791 increase in Cl.

If you want to find the critical angle of attack, you need to find the zero lift angle of attack. For any cambered airfoil, this will be a negative angle of attack. From my chart of the NACA 2412 airfoil, it looks like the zero lift angle of attack is -2 degrees.

A fully loaded Cessna-172 stalls with no flaps at 44KIAS. However, at this AoA, there is a big difference between indicated and calibrated airpseed, so lets use the fact that it stalls at 51KCAS. To convert from knots to m/s, multiply by .514444. Stall speed is 26.236m/s.
Using your equation to find Cl,max, we get 1.462.
Same thing Tgrayson and you did, but I am using 2,300lbs (fully loaded) and 51KCAS. That translates to 26.236m/s, and this problem is very sensitive to a small change in speed.

There you have it. Cl,max for a Cessna 172 is 1.462 (fully loaded, no flaps).

To find the critical angle of attack, we will use the fact that the lift slope is 0.0791 per degree. Take 1.462/0.0791, and we get 18.48 degrees. That is the number of degrees between the zero lift angle of attack and the stalling angle of attack. Since we know that the zero lift angle of attack is at -2 degrees, we take 18.48 minus 2 degrees and we get 16.48.

There you have it. The critical AoA is 16.48 degrees.

This agrees with the poster above that said the C-172 stalls between 16 and 17 degrees AoA with flaps up.
Hope this helps.
Feel free to ask more questions.
Ross

[tgrayson]

Originally Posted by aerospacepilot

Hope this helps.

Good analysis. And you made the correction from absolute AOA to "regular" AOA that I neglected to make.

[Joshrk22]

Awesome, this helps me out tremendously. I've searched up and down for how to do this and came up with nothing. Thank you both! I will let you know if I have anymore questions.

[Joshrk22]

Okay just remembered something... If I want to graph the data CL vs. AoA wouldn't it just give me a linear line? How would I graph it after the wing has stalled and you get a decrease in the CL as the AoA continues to increase?

[aerospacepilot]

Originally Posted by Joshrk22

Okay just remembered something... If I want to graph the data CL vs. AoA wouldn't it just give me a linear line? How would I graph it after the wing has stalled and you get a decrease in the CL as the AoA continues to increase?

Yes, the lift slope is a linear line. However, it is only valid up to the critical AoA. After that, it drops off drastically. There is no way to solve for this. You just have to measure it experimentally. I have a graph of the lift slope for a NACA 2412. If you'd like, I could scan it in and give it to you.
Just let me know if you'd like it.

[Joshrk22]

aerospace pilot,

Thank you! I don't think I need you to scan it though. I found one online from GA Tech but I will defintely PM you if I need that. You have been of great help!