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

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.

• Perhaps you cannot find those values as the military don’t want you to... – Solar Mike Sep 15 '19 at 21:26