REFERENCE STEADY STATE
If all the disturbance quantities are set equal to zero in the foregoing equations, then they apply to the reference flight condition. When this is done, we get the following relations, which may be used to eliminate all the reference forces and moments from the equations:
X0 — mg sin 90 = 0 Fo = 0
Z0 + mg cos в0 = 0 (4.9,6)
L0 = M0 = N0 = 0
xEo = n0 cos 60, >>,, = 0, zEo = – U0 sin e{)
We further postulate that in the reference steady state the aileron and rudder angles are zero. When (4.9,6) are substituted into (4.9,2 to 4.9,5), (4.9,3) are solved for p and r, and the equations are rearranged, the result is (4.9,7 to 4.9,10).
AX An =——– gA0cos в0 m 
(a) 

A Y v =—— 1 g<p cos в0 — u0r m 
(b) 
(4.9,7) 
AZ w =——– gAO sin 60 + uQq m 
(c) 

p = (/,/, – fir’ (f. AL + IXZAN) 
(a) 

AM ^ /, 
(b) 
(4.9,8) 
r = (/,/. – /:,)‘(/,tAL + I „AN) 
(c) 

A e=q 
(a) 

= p + r tan 0O; p = ф — ф sin в0 
(b) 
(4.9,9) 
ф = r sec в0 
(c) 

г = An cos в0 — u0AQ sin в0 + w sin в0 
(a) 

Aye = и0фсos в0 + v 
(b) 
(4.9,10) 
= —An sin в0 — noA0cos в0 + w cos 60 
(c) 
Leave a reply