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In the picture below you see a King Air 200, an aircraft with a so-called T-tail. One of the primary reasons of placing the tail higher up is to keep the tail out of the wake and downwash of the wing. Here, you may assume that .

Image courtesy of Mark Jones Jr., CC - BY

For this aircraft, we will in this exercise investigate the required position of the centre of gravity to guarantee static, longitudinal stability. The first step in this analysis is to set up the moment equation around the centre of gravity of this aircraft.

1. Derive the moment equation around the centre of gravity of this King Air 200, as a function of the total lift , the tail lift , the centre of gravity distance , the tail arm and the aerodynamic centre moment .

Tips:

1. Explicitly indicate multiplication with a * symbol, indicate division with a / symbol, and use an underscore to indicate subscripts (so is written as "l_cg").

2. Use the sign convention "Pitch up = positive".

3. Take a positive moment in the aerodynamic centre.

4. Use a capital 'H' for all subscripts denoting the horizontal tail.

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2. Non-dimensionalise the equation just derived and give the result below. You are allowed to use the quantities , , , , , , and .

Tips:

1. This problem segment does not support a double subscript, so write as "C_mac" and as "C_LH".

2. Don't introduce the tail volume as the quantity yet.

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3. Derive this equation with respect to the angle of attack and give your result for . You are allowed to use the quantities , , (the tail volume), (the wing lift coefficient gradient) and (the tail lift coefficient gradient).

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4. What is the requirement on the position of the centre of gravity? You are allowed to use the quantities , , (the tail volume), (the wing lift coefficient gradient) and (the tail lift coefficient gradient).

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1 answer

To derive the required equations for the static longitudinal stability of the King Air 200 with the given parameters, we will follow through each given section sequentially.

1. Moment Equation around the Centre of Gravity

The moment equation around the center of gravity for the King Air 200 can be expressed considering the lift from the main wing and the lift from the horizontal tail, as well as their distances from the center of gravity.

Let:

  • \( L \) = total lift of the aircraft
  • \( L_H \) = lift generated by the horizontal tail
  • \( l_{cg} \) = distance from the aerodynamic center (AC) to the center of gravity (cg)
  • \( l_H \) = distance from the aerodynamic center (AC) to the horizontal tail
  • \( M_{AC} \) = moment about the aerodynamic center

The moment due to the total lift about the center of gravity is negative since it will pitch the aircraft down:

  • Moment from the wing lift: \( -L * l_{cg} \)

The moment due to the tail lift about the center of gravity is positive since it will pitch the aircraft up:

  • Moment from the tail lift: \( +L_H * l_H \)

Thus, we can express the overall moment about the center of gravity as: \[ M_{cg} = -L * l_{cg} + L_H * l_H + M_{AC} \]

2. Non-dimensionalised Moment Equation

To non-dimensionalise the moment equation, let's define the following parameters:

  • \( W \) = weight of the aircraft
  • \( S \) = wing area
  • \( C_{mac} \) = mean aerodynamic chord
  • \( H = \frac{L_H}{W} \) = non-dimensional tail lift
  • \( C_L \) = lift coefficient of the wing
  • \( C_{LH} \) = lift coefficient of the tail

The non-dimensional moment equation can be given in terms of these coefficients. The non-dimensional form takes into account \( \frac{L}{W} \) and leads to a form similar to the following:

\[ \frac{M_{cg}}{W * C_{mac}} = -\left(\frac{L}{W} * \frac{l_{cg}}{C_{mac}}\right) + H * \left(\frac{l_H}{C_{mac}}\right) + \frac{M_{AC}}{W * C_{mac}} \]

This represents the non-dimensional moment about the center of gravity.

3. Derivation with Respect to Angle of Attack

To derive the equation with respect to the angle of attack \( \alpha \), we consider the relationship of the lift coefficients with respect to angle of attack:

  • \( C_L = C_{L_0} + C_{L_\alpha} * \alpha \)
  • \( C_{LH} = C_{LH_0} + C_{LH_\alpha} * \alpha \)

Thus, taking the derivative of the non-dimensionalised moment with respect to \( \alpha \):

\[ \frac{dM_{cg}}{d\alpha} \propto -\left(C_{L_\alpha}\right) * \frac{l_{cg}}{C_{mac}} + \left(\frac{H}{W}\right) * \left(C_{LH_\alpha}\right) * \frac{l_H}{C_{mac}} \]

This will yield the final expression required for this part.

4. Requirement on the Position of the Centre of Gravity

For static longitudinal stability, we require the derivative to be negative: \[ -\left(C_{L_\alpha}\right) * \frac{l_{cg}}{C_{mac}} + \left(\frac{H}{W}\right) * \left(C_{LH_\alpha}\right) * \frac{l_H}{C_{mac}} < 0 \]

Reorganizing terms provides the requirement on the position of the center of gravity \( (l_{cg}) \):

\[ l_{cg} < \left(\frac{H}{C_{L_\alpha}}\right) * \left(C_{LH_\alpha}\right) * l_H \]

This final result can be interpreted as a condition for ensuring that the center of gravity's position will lead to a statically stable longitudinal behavior of the aircraft.

These equations and conditions will collectively allow you to investigate the static longitudinal stability of the King Air 200 based on its aerodynamic characteristics and geometry.