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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.) enter image description here

The cruise Cl values are as under: enter image description here

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

NitinG
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Nooradila Noordin
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    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 03 '15 at 16:21
  • possible duplicate of How to calculate the Lift Coefficient for the A320? – fooot Jun 03 '15 at 17:09
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    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 03 '15 at 18:43
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    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 04 '15 at 08:03
  • The ratio of $c_L$ over $c_D$ should be somewhere between 12 and 18. Your $c_L$ values are too small by more than an order of magnitude. – Peter Kämpf Jun 04 '15 at 15:01
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    @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 05 '15 at 03:56
  • @GürkanÇetin we could also add that L is most probably slightly bigger than W, since in a nose-down attitude (usually the case in a descent) the thrust has a component that helps the weight (a mirror situation of this case) – Federico Jun 05 '15 at 08:10

1 Answers1

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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.

DeltaLima
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