Angle of Attack for F-117A at 30,000ft and 400kts: 0.648

[jd3005]

You are flying an F-117A fully equipped, which means that your aircraft weighs 52,500 pounds. You want to maintain equilibrium in straight and level flight at an altitude of 30,000 feet, cruising at 400 knots to conserve fuel. The aircraft's wing area is 1,140 square feet. At what angle of attack should the F-117A be set to maintain equilibrium?
The air density at FL300 is .000889


This was an assignment (independent study) assigned in math. I am working on it however I felt I completed the problem properly, however the answer seems to disagree with me.


L = (1/2) d v2 s CL

* L = Lift, which must equal the airplane's weight in pounds
* d = density of the air. This will change due to altitude. These values can be found in a I.C.A.O. Standard Atmosphere Table.
* v = velocity of an aircraft expressed in feet per second
* s = the wing area of an aircraft in square feet
* CL = Coefficient of lift , which is determined by the type of airfoil and angle of attack.




52,500 = (1/2) (.000889) (400^2 ) (1140) CL

52,500 = (81076.8) CL

0.648 = CL

I'm not concerning significant digits.

 

[Nate810]

Therefore the angle of attack must be 0.648 to maintain equilibrium in straight and level flight at an altitude of 30,000 feet, cruising at 400 knots.

 

[tomcue]

Based on the given information and using the lift equation, the angle of attack for the F-117A at 30,000ft and 400kts should be approximately 0.648 degrees to maintain equilibrium in straight and level flight. This means that the wings of the aircraft should be angled at 0.648 degrees relative to the oncoming airflow to generate enough lift to balance out the weight of the aircraft. However, it is important to note that this is an ideal calculation and the actual angle of attack may vary due to factors such as air turbulence and pilot control.