To find the static margin of the A340-600, we need to calculate the distance between the center of gravity and the wing aerodynamic center.
Firstly, let's determine the aerodynamic center of the wing. The wing aerodynamic center is typically located around 25% to 30% of the mean aerodynamic chord (MAC) behind the leading edge. For the A340-600, we will assume it is at 27.5% of the MAC.
Given that the wing has a sweep angle of 30 degrees, we can use simple trigonometry to calculate the length of the MAC. Assuming a reference chord length of 1 meter, the MAC is given by:
MAC = cos(30 degrees) = 1 * cos(30 degrees) = 1 * 0.866 = 0.866 meters.
To find the distance between the center of gravity and the aerodynamic center of the wing, we need to calculate the moment arm.
Moment arm = nose to CG distance + MAC * static margin.
Let's assume a static margin of "x" meters.
Moment arm = 39.2 meters + 0.866 meters * x.
Since we know the downwash angle is equal to 10% of the angle of attack, we can assume the downwash angle to be equivalent to 0.1 * angle of attack. In this case, the downwash angle is assumed to be 0.1 * x.
Using trigonometry again, we have:
Tangent of downwash angle = downwash height / moment arm.
For small angles, the tangent is close to the angle itself, so we can say:
Downwash angle (radians) = downwash height / moment arm.
Therefore,
0.1 * x = tan(downwash angle) = downwash height / moment arm.
Rearranging the equation gives:
Downwash height = 0.1 * x * moment arm.
We know that the downwash height is equal to the vertical distance between the center of gravity and the aerodynamic center of the wing. Thus,
Downwash height = x * 0.866 meters.
Setting the two equations for downwash height equal to each other gives:
0.1 * x * moment arm = x * 0.866 meters.
Substituting the expression for the moment arm, we have:
0.1 * x * (39.2 meters + 0.866 meters * x) = x * 0.866 meters.
Expanding and simplifying:
3.92x + 0.0866x^2 = 0.866x.
Rearranging the equation:
0.0866x^2 - 0.866x + 3.92x = 0.
0.0866x^2 + 3.054x = 0.
Dividing throughout by x, we get:
0.0866x + 3.054 = 0.
0.0866x = -3.054.
x = -3.054 / 0.0866.
x ≈ -35.27 meters.
Since the static margin cannot be negative, we discard this solution.
Therefore, the static margin of this A340-600 is approximately 35.27 meters.
For this aircraft, you are given the following parameters:
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.
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