# What are the opimum values of Cl during descent?

Statistical Analysis of A320 aircraft shows the Cl values to be as under: cl_value. L is assumed to be 0.5*MaximumLandingWeight of A320 as the Aircraft is about to land.

The Units used are: Flight Level(feet), rho(lb/feet^3, V(feet/s), for the sixth column, 1/2(rho.V^2.S.)

The cruise Cl values are as under:

When compared to cruise the CL during descent is very small. Is this an acceptable result?

• If I'm understanding your question correctly, you should be assuming very close to 1.0 * MaximumLandingWeight. An airplane needs to lift its full weight in a stabilized descent. – Fred Larson Jun 3 '15 at 16:21
• possible duplicate of How to calculate the Lift Coefficient for the A320? – fooot Jun 3 '15 at 17:09
• I assume you want to work in SI unit. If so, your third column should be in kg/m^3, currently the values are wrong. The fourth column I expect to be in knots, the fifth should be in m/s and is wrong. Consequently the sixth column isn't correct either and that explains why the C_L doesn't match your expectations. – DeltaLima Jun 3 '15 at 18:43
• Yes, CL is smaller during descent, and No, the values in your tables are not correct. You have a formula/unit conversion problem somewhere. Also 0.5 MLW is a bit too much a correction for descent. You are basically assuming 60deg downslope with is a lot even for a dive bomber. 3deg is more likely, and 0.998 is a better slope correction factor. – Radu094 Jun 4 '15 at 8:03
• @FredLarson and Radu094 has said it well. It can drop like a rock, if it's landing? if there's a substantial downwards gravitational force (if L is not equal to W), the aircraft will accelerate downwards (assuming small angles for a decent landing). The question needs to be corrected of these errors, to avoid unnecessary explanations further. – Gürkan Çetin Jun 5 '15 at 3:56

I am going to work in SI units for all engineering work, anything else would cause instant insanity on my side. I'll leave the altitude in feet (isn't used) and convert speed from knots to m/s so you can follow the flow easily. Since $C_L$ is unitless it should not matter what units you use in the calculation as long as you are consistent.

• Altitude: 34000 [ft] (not used)
• $V$ (true air speed): 442 [knots] = 227 [m/s]
• S = 112.6 [m2]
• $\rho_0$ = 1.225 [kg/m3]
• $\frac{\rho}{\rho_0}$=0.322 [-] (from your table)
• $\rho$ = $1.225 \cdot 0.322 = 0.394$ [kg/m3]
• $\frac{1}{2}\rho V^2 S$ = $\frac{1}{2} 0.394\cdot 227^2 \cdot112.6 = 125\cdot10^4$[N]
• $m$ = MLM = 64500 [kg]
• W = $m\cdot g$ = $64500\cdot 9.81$ = $63\cdot10^4$ [N]
• $C_L$ = $\frac{W}{\frac{1}{2}\rho V^2 S}$=$\frac{63\cdot 10^4}{125\cdot 10^4}$ = 0.5

I did take full maximum landing mass for this calculation. If you want to apply corrections then note that the $C_L$ value scales linearly with the mass.

One further note:

I assume your data is derived from ADS-B. ADS-B transmits groundspeed, whilst you should work with true airspeed in these formula's. The difference between them is the tailwind (or headwind) component.

• The difference between them is the tailwind (or headwind) component. It's a vector sum, so also the sidewind has an influence. – Federico Jun 4 '15 at 12:02
• @Federico correct, but generally the effect is negligible for this kind of analysis. I didn't want to overcomplicate the matter. – DeltaLima Jun 4 '15 at 12:41
• For the analysis I might agree, for the TAS vs GS difference, a little less. – Federico Jun 4 '15 at 13:22
• It should be noted that speed is "true airspeed". and a small correction: there's a "not" missing in the "it should matter" part.. – Gürkan Çetin Jun 5 '15 at 4:00
• @GürkanÇetin thank you, I've corrected it – DeltaLima Jun 5 '15 at 6:55