For this aircraft, you are given the following parameters:

l(wing)=34.9m, l(tail)-67.8m, S=437 m^2, S_H=93 m^2, C_=8.25m, a=4.8/rad , a_t=5.2/rad.
Furthermore we will assume that the downwash angle is equal to 10% of the angle of attack of the wing. Given that the centre of gravity of this A340-600 is situated 39.2 metres from the nose, find the static margin (in metres) of this A340.

1 answer

The static margin of an aircraft is the distance between the center of gravity (CG) and the neutral point (NP), expressed as a percentage of the mean aerodynamic chord (MAC) length.

To calculate the static margin, we need to find the MAC length first.

MAC = (2/3) * C [where C is the mean aerodynamic chord]

Given:
l(wing) = 34.9m
C_ = 8.25m

The mean aerodynamic chord can be calculated as:
C = (2/3) * C_

C = (2/3) * 8.25m = 5.5m

Now, we need to find the neutral point (NP).

Neutral point (NP) = l(tail) - (S_H/S) * MAC

Given:
l(tail) = 67.8m
S_H = 93m^2
S = 437m^2
MAC = 5.5m

NP = 67.8m - (93m^2 / 437m^2) * 5.5m
NP = 67.8m - (0.2128) * 5.5m
NP = 67.8m - 1.1692m
NP = 66.6308m

Finally, we can calculate the static margin (SM) as:

SM = CG - NP

Given:
CG = 39.2m
NP = 66.6308m

SM = 39.2m - 66.6308m
SM = -27.4308m

Therefore, the static margin of this A340-600 aircraft is -27.4308 meters.