How do you determine the L/D ratio of F-117 Nighthawk?

Question

I was trying to look for an L/D ratio published everywhere on the web and I can't seem to find it (for a project). Since I couldn't find it, I wanted to try calculating myself. Can't seem to find $C_l$ or $C_d$ though without some sort of paywall on research papers.

The equations are:

L = $\frac{1}{2} C_{l} \cdot \rho v^2 A$

D = $\frac{1}{2} C_{d} \cdot \rho v^2 A$

So L/D would just be:

$\frac{L}{D} = \frac{C_l}{C_d}$

or would the effective area be different for L and D, where L is the area of the top/bottom view of the plane, and D is the area of the front/back view?

The only quantities I have are:

v = Mach 0.92 = 315.62 $\frac{m}{s}$..............................top speed

$\rho$ = 0.003996 $\frac{kg}{m^3}$....................density of air at 40,000 ft (assuming it's flying at its maximum altitude)

I couldn't really find any area of the plane; only wingspan and vertical length, but that would give an inaccurate area of the plane. How would I determine this? If $\frac{L}{D}$ is just $\frac{C_l}{C_d}$ does anyone happen to know $C_l$ and $C_d$? Thank you in advance.

Answer 1

Glide ratio, or glide angle meaning when you trim the plane to glide level with the engines off, is numerically equal to L/D ratio.

As a start guess I would say with a wing load of 330kg/m2 and very small wing aspect ratio it shouldn't be more than 12- 14.

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