PPRuNe Forums - View Single Post - Total drag questions
View Single Post
Old 17th Oct 2008, 23:11
  #23 (permalink)  
ft
 
Join Date: Oct 2000
Location: N. Europe
Posts: 436
Likes: 0
Received 0 Likes on 0 Posts
Allright, let's do the full math in your preferred choice of coordinate system then.

Earth-fixed steady state.

ca = climb angle

Vertical force balance gives

W + D*sin(ca) = T*sin(ca) + L*cos(ca) (1)

Horizontal force balance gives

T*cos(ca) = D*cos(ca) + L*sin(ca) (2)

(2) gives

(T-D) = L*sin(ca)/cos(ca) (3)

(1) gives

(T-D) = (W-L*cos(ca)/sin(ca) (4)

(3) + (4)

L*sin(ca)/cos(ca) = (W-L*cos(ca))/sin(ca)

L*sin2(ca)/cos(ca) = W-L*cos(ca)

L*(sin2(ca)/cos(ca)+cos(ca)) = W

L = W/(sin2(ca)/cos(ca)+cos(ca)) = W/(1/cos(ca)*(sin2(ca)+cos2(ca))) =
= W/(1/cos(ca)*1) = W*cos(ca)

L = W*cos(ca)

Now, where did I hear that before... ?

The point where you go wrong is when you mistakenly assume that the thrust can safely be ignored in the vertical force balance. Take a look at the relative rates of change of the vertical component of thrust and the vertical component of lift with increasing climb angle. The longitudinal component of the weight of the aircraft increases with the sine of the climb angle and thrust has to increase with it to keep the aircraft from decelerating.

The vertical component of thrust also increases with the sine of the climb angle. The vertical component of lift, however, will only decrease with the cosine of the climb angle. In order to maintain equilibrium, the magnitude of the lift force has to decrease.
ft is offline