Chapter 4 The Equations of Motion

Chapter 4 The Equations of Motion Flight Mechanics and Control AEM 4303 Bérénice Mettler University of Minnesota Feb. 20-27, 2013 (v. 2/26/13) Bérénice Mettler (University of Minnesota) Chapter 4 The Equations of Motion Feb. 20-27, 2013 (v. 2/26/13) 1 / 26

Lecture Outline and Objectives Outline Newton-Euler equations The linearized equations of motion Aerodynamic derivatives, aerodynamic control derivatives, and power derivatives Equations of motion for small perturbations The decoupled longitudinal-vertical and lateral-directional equations of motion Dimensional vs. non-dimensional derivatives The state-space form Readings Chapter 4, Cook, The Equations of Motion Bérénice Mettler (University of Minnesota) Chapter 4 The Equations of Motion Feb. 20-27, 2013 (v. 2/26/13) 2 / 26

Newton-Euler Equations The Newton-Euler equations expressed in the inertial reference frame are derived from the principle of conservation of linear and angular momentum. For constant vehicle mass m and moment of inertia (inertial tensor I), they are: d I (mv) = F dt d I (Iω) = M dt where F = [X Y Z] T is the vector of external forces acting on the vehicle center of gravity and M = [L M N] T is the vector of external moments. The notation di d for the derivative means the derivative is taken w.r.t. the inertial frame. Bérénice Mettler (University of Minnesota) Chapter 4 The Equations of Motion Feb. 20-27, 2013 (v. 2/26/13) 3 / 26

Nomenclature Body-fixed frame axes and key states: lateral axis y, u, q Y, M aircraft cg longitudinal axis X, L x, v, p Z, N vertical axis Vector quantities: Body velocity v = [u v w] T Body angular rate ω = [p q r] T Resultant forces: F = [X Y Z] T Resultant moments: M = [L M N] T z, w, r Bérénice Mettler (University of Minnesota) Chapter 4 The Equations of Motion Feb. 20-27, 2013 (v. 2/26/13) 4 / 26

For the aircraft, it will be more convenient to express all components in the body-fixed coordinates. Since these coordinates are moving w.r.t. the inertial frame, we need to use the differentiation rule of relative motion. In a first step, let s define the differential with respect to the body-fixed reference frame (denoted by b ): m d I v dt d I (Iω) dt = m d b v dt = d b (Iω) dt + m(ω v) = F + (ω Iω) = M In this equation the vector components are still in inertial frame. In a second step, let s write all components in terms of the body coordinates: m d b v + m(ω v) = F dt I d b ω + (ω Iω) = M dt where v = [u v w] T and ω = [p q r] T are the fuselage velocities and angular rates. Notice that with the equation w.r.t. the body frame, the moment of inertia is constant and can be taken outside the derivative. Bérénice Mettler (University of Minnesota) Chapter 4 The Equations of Motion Feb. 20-27, 2013 (v. 2/26/13) 5 / 26

For a 6 DOF rigid-body system, the three ordinary differential equations describing the aircraft s translational motion about its three reference axes are 2 3 2 3 2 3 2 3 u p u X m 4 v 5 + m 4q5 4v 5 = 4Y 5 ẇ r w Z Written component-wise gives the three equations: m u + m(qw rv) = X m v + m(ru pw) = Y mẇ + m(pv qu) = Z Similarly, the three ordinary differential equations describing the aircraft s rotational motion: 2 3 2 3 2 3 2 3 2 3 2 3 I x I xy I xz ṗ p I x I xy I xz p L 4I yx I y I yz 5 4 q 5 + 4q5 4I yx I y I yz 5 4q5 = 4M5 I zx I zy I z ṙ r I zx I zy I z r N Bérénice Mettler (University of Minnesota) Chapter 4 The Equations of Motion Feb. 20-27, 2013 (v. 2/26/13) 6 / 26

Written component-wise gives the three equations: I xṗ (I y I z)qr + I xy (pr q) I xz(pq + ṙ) + I yz(r 2 q 2 ) = L I y q + (I x I z)pr + I yz(pq ṙ) + I xz(p 2 r 2 ) I xy (qr + ṗ) = M I zṙ (I x I y )pq I yz(pr + q) + I xz(qr ṗ) + I xy (q 2 p 2 ) = N The moment equation can be simplified considering that AC are symmetric about the (oxz) plane. Therefore, I xy = I yz = 0. I xṗ (I y I z)qr I xz(pq + ṙ) = L I y q + (I x I z)pr + I xz(p 2 r 2 ) = M I zṙ (I x I y )pq + I xz(qr ṗ) = N Furthermore, if the body axes are aligned with the principal inertia axes, I xz = 0. Even if this is not the case, in general I xz << I x, I y or I z. 2 I x 0 3 (I xz) I = 4 0 I y 0 5 (I zx) 0 I z Bérénice Mettler (University of Minnesota) Chapter 4 The Equations of Motion Feb. 20-27, 2013 (v. 2/26/13) 7 / 26

4.1.4 Disturbance forces and moments The force and moment components can be decomposed into contributions due to the aerodynamics, gravity, controls, power and atmospheric disturbances. and m u + m(qw rv) = X a + X g + X c + X p + X d m v + m(ru pw) = Y a + Y g + Y c + Y p + Y d mẇ + m(pv qu) = Z a + Z g + Z c + Z p + Z d I xṗ (I y I z)qr I xz(pq + ṙ) = L a + L g + L c + L p + L d I y q + (I x I z)pr + I xz(p 2 r 2 ) = M a + M g + M c + M p + M d I zṙ (I x I y )pq + I xz(qr ṗ) = N a + N g + N c + N p + N d The next step in modeling the AC dynamics would involve deriving the detailed expressions for this different contributions. This task is complex since these effects are dependent on geometry and the AC motion itself. For analytical purpose, the equations are linearized about an equilibrium (trim) operating point of interest. The force contributions can then be simplified by the product of derivatives and perturbations in the states about trim conditions. Bérénice Mettler (University of Minnesota) Chapter 4 The Equations of Motion Feb. 20-27, 2013 (v. 2/26/13) 8 / 26

4.2 The Linearized Equations of Motion Let s assume a rectilinear, steady trim condition given by airspeed V 0 and the AC body velocities (U e, V e, W e). In the absence of side slip, V e = 0. Consider small perturbations about the trim conditions given by (u, v, w) and (p, q, r): U = U e + u V = V e + v = v W = W e + w Since the perturbations are small, taking their square or product is much smaller and can be ignored. Substituting the perturbation terms in the equations of motion results in m( u + W eq) = X a + X g + X c + X p m( v + W ep + U er) = Y a + Y g + Y c + Y p m(ẇ U eq) = Z a + Z g + Z c + Z p and I xṗ I xzṙ = L a + L g + L c + L p I y q = M a + M g + M c + M p I zṙ I xzṗ = N a + N g + N c + N p Bérénice Mettler (University of Minnesota) Chapter 4 The Equations of Motion Feb. 20-27, 2013 (v. 2/26/13) 9 / 26

Next, let s express the force and moment components associated with small perturbation about the trim conditions and deflection of the control surfaces. Gravitation terms For simplicity consider the symmetric trim condition (wings level) as shown here (φ e = ψ e = 0). For this trim condition, the components of weight in the body axes only appear in the xz plane of symmetry. Using the direction cosine matrix D: 2 3 X ge 2 3 mg sin θ e 4Y ge 5 = 4 Z ge 0 mg cos θ e 5 (1) The perturbations in weight resulting from the perturbations in attitude are determined from the small-angle approximation of the cosine direction matrix. 2 3 2 3 2 3 2 3 2 3 X g 1 ψ θ X ge 1 ψ θ mg sin θ e 4Y g 5 = 4 ψ 1 φ 5 4Z ge 5 = 4 ψ 1 φ 5 4 0 5 (2) Z g θ φ 1 Z ge θ φ 1 mg cos θ e Bérénice Mettler (University of Minnesota) Chapter 4 The Equations of Motion Feb. 20-27, 2013 (v. 2/26/13) 10 / 26

Expressed in components the gravitational terms are: X g = mg sin θ e mgθ cos θ e Y g = mgψ sin θ e + mgφ cos θ e Z g = mg cos θ e mgθ sin θ e Note that φ, θ and ψ are all perturbations from trim conditions φ e = 0, θ e and ψ e = 0. a a If the trim is a non-symmetric configuration (φ e 0) the subsequent derivations can easily be modified to account for the lateral components Aerodynamic terms For an AC operating about trim condition, a disturbance or control deflection will result in a perturbation in the equilibrium aerodynamic forces (X, Y, Z) and moments (L, M, N). The force and moment perturbations and their relationship to the perturbation in the AC state (speeds u, v, w and angular rates p, q, r) can be described using a Taylor series. For example, for the axial force components X a: X X a = X ae + u u + 2 X u 2 u 2 2! + 3 X u 3 u 3 3! + 4 X u 4 u 4 X + u u + 2 X u 2 u 2 2! + 3 X u 3 u 3 3! + 4 X u 4 u 4 4! + «4! + «+ + higher order derivatives Bérénice Mettler (University of Minnesota) Chapter 4 The Equations of Motion Feb. 20-27, 2013 (v. 2/26/13) 11 / 26

Aerodynamic derivatives Since the perturbations in motion are small, the higher-order terms and higher-order derivatives are usually negligible except for z-component acceleration ẇ, which is typically included. X a = X ae + X u u + X v v + X w w + X p p + X q q + X r r + X ẇ ẇ (3) The derivatives are usually written as X = X : X a = X ae + X uu + X v v + X w w + X pp + X qq + X r r + Xẇ ẇ (4) The derivatives X u, X v... Xẇ are called the aerodynamic derivatives. The remaining force and moment components are written similarly, for example for the pitching moment M a = M ae + M uu + M v v + M w w + M pp + M qq + M r r + Mẇ ẇ (5) Concretely, the aerodynamic derivatives describe the force or moment perturbation that results from perturbation in states from the equilibrium. For example, M v is the roll moment resulting from the perturbation in lateral velocity v. Bérénice Mettler (University of Minnesota) Chapter 4 The Equations of Motion Feb. 20-27, 2013 (v. 2/26/13) 12 / 26

Aerodynamic control terms: control derivatives The primary AC control surfaces are the elevator η, the aileron ξ and the rudder ζ. Their force and moment contributions are due to aerodynamic effects, therefore they are described by aerodynamic control derivatives or simply control derivatives. For example, the pitching moment due to the control surfaces is described by: M c = M η η + M ξ ξ + M ζ ζ (6) Note that the elevator η is the primary source of pitch control moment. The presence of control derivatives for the other surfaces (η,ξ,ζ) accounts for secondary effects. Remember that the derivatives in the linearized equations describe the perturbation in forces and moments with respect to the trim conditions. Therefore η, ξ and ζ are perturbations about the trim control settings η e, ξ e and ζ e, respectively. M c = M ηη + M ξ ξ + M ζ ζ (7) The remaining derivatives are expressed in a similar manner. For the rolling moments, for example, L c = L ηη + L ξ ξ + L ζ ζ (8) The same general form applies to other control surfaces such as flaps or spoilers. Bérénice Mettler (University of Minnesota) Chapter 4 The Equations of Motion Feb. 20-27, 2013 (v. 2/26/13) 13 / 26

Power terms A powered AC is equipped with a propulsion system such as a turbofan, jet or propeller driven by an internal combustion, turbo-shaft or electric motor. The thrust is controlled throttle lever angle ɛ, however, in the linearized equation of motion, the normal force due to the thrust is described as a perturbation in thrust force from the trim force necessary to maintain the equilibrium condition. For example, for the normal force, the thrust perturbation τ relative to the equilibrium thrust τ e is given as: Z p = Z τ τ (9) The thrust contributions to the other force and moment components are described in a similar way. Bérénice Mettler (University of Minnesota) Chapter 4 The Equations of Motion Feb. 20-27, 2013 (v. 2/26/13) 14 / 26

Equations of motion for small perturbations Now that all force and moment contributions are accounted for we can substitute these in the linearized Newton-Euler equations to obtain the complete equations of motion for small perturbations. and m( u + W eq) =X ae + X uu + X v v + X w w + X pp + X qq + X r r + Xẇ ẇ mg sin θ e mgθ cos θ e + X ηη + X ξ ξ + X ζ ζ + X τ τ m( v + W ep + U er) =Y ae + Y uu + Y v v + Y w w + Y pp + Y qq + Y r r + Yẇ ẇ + mgψ sin θ e + mgφ cos θ e + Y ηη + Y ξ ξ + Y ζ ζ + Y τ τ m(ẇ U eq) =Z ae + Z uu + Z v v + Z w w + Z pp + Z qq + Z r r + Zẇ ẇ + mg cos θ e mgθ sin θ e + Z ηη + Z ξ ξ + Z ζ ζ + Z τ τ I xṗ I xzṙ =L ae + L uu + L v v + L w w + L pp + L qq + L r r + Lẇ ẇ + L ηη + L ξ ξ + L ζ ζ + L τ τ I y q =M ae + M uu + M v v + M w w + M pp + M qq + M r r + Mẇ ẇ + M ηη + M ξ ξ + M ζ ζ + M τ τ I zṙ I xzṗ =N ae + N uu + N v v + N w w + N pp + N qq + N r r + Nẇ ẇ + N ηη + N ξ ξ + N ζ ζ + N τ τ Bérénice Mettler (University of Minnesota) Chapter 4 The Equations of Motion Feb. 20-27, 2013 (v. 2/26/13) 15 / 26

Derivatives in trim conditions In steady-state, equilibrium (trim) condition, all derivatives are by definition equal to zero. The steady-state equation of motion are therefore: X ae Y ae = 0 Z ae L ae = 0 M ae = 0 N ae = 0 = mg sin θ e = mg cos θ e These equations define the constant trim terms X ae,..., N ae, which can be substituted back into the equations of motion. m u =X uu + X v v + X w w + X pp + (X q mw e)q + X r r + Xẇ ẇ (10) mgθ cos θ e + X ηη + X ξ ξ + X ζ ζ + X τ τ (11) m v =Y uu + Y v v + Y w w + (Y p mw e)p + Y qq + (Y r mu e)r + Yẇ ẇ (12) + mgψ sin θ e + mgφ cos θ e + Y ηη + Y ξ ξ + Y ζ ζ + Y τ τ (13) mẇ =Z uu + Z v v + Z w w + Z pp + (Z q + U e)q + Z r r + Zẇ ẇ (14) mgθ sin θ e + Z ηη + Z ξ ξ + Z ζ ζ + Z τ τ (15) Bérénice Mettler (University of Minnesota) Chapter 4 The Equations of Motion Feb. 20-27, 2013 (v. 2/26/13) 16 / 26

I xṗ I xzṙ =L uu + L v v + L w w + L pp + L qq + L r r + Lẇ ẇ (16) + L ηη + L ξ ξ + L ζ ζ + L τ τ (17) I y q =M uu + M v v + M w w + M pp + M qq + M r r + Mẇ ẇ (18) + M ηη + M ξ ξ + M ζ ζ + M τ τ (19) I zṙ I xzṗ =N uu + N v v + N w w + N pp + N qq + N r r + Nẇ ẇ (20) + N ηη + N ξ ξ + N ζ ζ + N τ τ (21) The two sets of differential equations describe the response of the aircraft s linear and angular motion to small perturbations about the equilibrium conditions given by airspeed components (U e, V e, W e) and attitude angles (φ e, θ e, ψ e). For a statically stable AC, the responses to small perturbations are transient. For these conditions, the coupling between the longitudinal and lateral responses is typically small enough to be neglected. This makes it possible to use decoupled sets of equations. These, being simpler, give more intuitive understanding of the AC dynamic behavior. Bérénice Mettler (University of Minnesota) Chapter 4 The Equations of Motion Feb. 20-27, 2013 (v. 2/26/13) 17 / 26

The Decoupled Equations of Motion Longitudinal-vertical equations of motion The longitudinal-vertical EOM are confined to the (oxz) plane. The aircraft motion in this plane is described by the x and z component linear motion and the pitching motion. The other components are ignored, are simply assumed to be zero. Therefore all the force and moment derivatives and control derivatives associated with these motions are taken as zero. Furthermore, for the longitudinal-vertical motion, only the primary control derivatives are retained. Therefore, the aileron, rudder control derivatives are also taken as zero. This results in a set of three equations for the longitudinal x, vertical z and pitching motion θ: m u = X uu + X w w + (X q mw e)q + Xẇ ẇ mgθ cos θ e + X ηη + X τ τ (m Zẇ )ẇ = Z uu + Z w w + (Z q + U e)q mgθ sin θ e + Z ηη + Z τ τ I y q = M uu + M w w + M qq + Mẇ ẇ + M ηη + M τ τ Bérénice Mettler (University of Minnesota) Chapter 4 The Equations of Motion Feb. 20-27, 2013 (v. 2/26/13) 18 / 26

For the special case of level flight (W e = 0) and when the body axes coincide with the wind axes, i.e., θ e = 0 the equations of motion further simplify to State-space form m u = X uu + X w w + X qq + Xẇ ẇ + X ηη + X τ τ (m Zẇ )ẇ = Z uu + Z w w + (Z q + U e)q + Z ηη + Z τ τ I y q = M uu + M w w + M qq + Mẇ ẇ + M ηη + M τ τ The equations of motion are a set of coupled, linear differential equations. It is convenient to represent them using matrix algebra. This description involves collecting the state variables and control inputs in a state vector x R n and input vector u R m, and grouping the aerodynamic and control derivatives in matrices leading to the following form: ẋ(t) = Ax(t) + Bu(t) (22) where A is a n n stability derivatives matrix and B is a m m control derivatives matrix. The state-space form also includes a set of output equations that describes which state are accessible as measurements. y(t) = Cx(t) + Du(t) (23) where y R r is the output vector, C is (n n) output matrix and D is (n m) direct matrix. Bérénice Mettler (University of Minnesota) Chapter 4 The Equations of Motion Feb. 20-27, 2013 (v. 2/26/13) 19 / 26

Mathematical definition for linear time invariant (LTI) model The linear model makes it possible to analyze the aircraft s stability and control properties using powerful linear systems mathematical tools (see next Chapter). For each trim condition, the A and B matrices have constant coefficient, therefore, the system described by the linear matrix-differential equation is a linear, time-invariant (LTI) system. For the nonlinear EOM, such as the ones obtained from the Newton-Euler equations and other first principles lead to a set of nonlinear differential equations. These equations can be described by ẋ(t) = f (x(t), u(t)) (24) The linear state-space equations (LTI model) can be derived using a linearization about the trim condition given by the state x and control input u : ẋ(t) = Ax(t) + Bu(t) (25) ««f (x, u) f (x, u) = x(t) + u(t) (26) x x,u u x,u Therefore the A and B matrices can be determine by taking the derivative of the nonlinear model f w.r.t. the state and input vectors and evaluating the derivatives at the trim condition. Bérénice Mettler (University of Minnesota) Chapter 4 The Equations of Motion Feb. 20-27, 2013 (v. 2/26/13) 20 / 26

To convert the set of differential equations in matrix form, the terms involving the time derivatives are collected on the left of the equality and the remaining terms to the right. m u Xẇ ẇ = X uu + X w w + (X q mw e)q mgθ cos θ e + X ηη + X τ τ mẇ Zẇ ẇ = Z uu + Z w w + (Z q + U e)q mgθ sin θ e + Z ηη + Z τ τ I y q Mẇ ẇ = M uu + M w w + M qq + M ηη + M τ τ θ = q Note the last equation was added to the original set, giving us four differential equations needed for the four states of the state vector x = [u, w, q, θ] T and two inputs u = [η, τ] T in the longitudinal-vertical system. This system can be written as 2 3 2 3 m Xẇ 0 0 X u X w (X q mw e) mg cos θ e 6 0 (m Zẇ ) 0 0 7 4 0 Mẇ I y 05 ẋ(t) = 6 Z u Z w (Z q + mu e) mg sin θ e 7 4M u M w M q 0 5 x(t) (27) 0 0 0 I 0 0 1 0 2 3 X η X τ + 6 Z η Z τ 7 4M η M τ 5 u(t) (28) 0 0 Bérénice Mettler (University of Minnesota) Chapter 4 The Equations of Motion Feb. 20-27, 2013 (v. 2/26/13) 21 / 26

Notice that these equations are in a form: Mẋ(t) = A x(t) + B u(t) (29) which can be converted in the standard form my multiplying the equation by M 1, i.e., ẋ(t) = M 1 A x(t) + M 1 B u(t) (30) Therefore, state and control matrix are A = M 1 A and B = M 1 B, respectively. The standard LTI form is usually written as 2 3 2 3 2 3 2 3 u x u x w x q x θ u x η x τ» 6 ẇ 7 4 q 5 = 6 z u z w z q z θ 7 6w 7 4m u m w m q m θ 5 4q 5 + 6 y η z τ 7 η 4m η m τ 5 τ θ 0 0 1 0 θ 0 0 (31) Bérénice Mettler (University of Minnesota) Chapter 4 The Equations of Motion Feb. 20-27, 2013 (v. 2/26/13) 22 / 26

The lateral-directional equations of motion The lateral-directional EOM are obtained similarly to the longitudinal-vertical EOM. Isolating the side force (Y ), rolling (L) and yawing moment (N) terms from the fully coupled EOM (10-21) we have: m v =Y v v + (Y p mw e)p + (Y r mu e)r + Yẇ ẇ + mgψ sin θ e + mgφ cos θ e + Y ξ ξ + Y ζ ζ I xṗ I xzṙ =L v v + L pp + L r r I zṙ + I xzṗ =N v v + N pp + N r r + N ξ ξ + N ζ ζ For level flight and body axes coinciding with wind axes, θ e = W e = 0 m v =Y v v + Y pp + (Y r mu e)r + mgφ + Y ξ ξ + Y ζ ζ I xṗ I xzṙ =L v v + L pp + L r r I zṙ + I xzṗ =N v v + N pp + N r r + N ξ ξ + N ζ ζ Next, these sets of coupled differential equations, similar to the longitudinal-vertical EOM are presented in matrix form, with state vector x = [v, p, r, φ, ψ] T and input u = [ξ, ζ] T. Bérénice Mettler (University of Minnesota) Chapter 4 The Equations of Motion Feb. 20-27, 2013 (v. 2/26/13) 23 / 26

2 3 2 3 2 3 2 3 m 0 0 0 0 v Y v Y p + W e Y r U e mg cos θ e mg sin θ e v 0 I x I xz 0 0 ṗ 6 0 I xz I x 0 0 7 6 ṙ 7 4 0 0 0 1 05 4 φ 5 = L v L p L r 0 0 p 6N v N p N r 0 0 7 6r 7 4 0 1 0 0 0 5 4φ5 0 0 0 0 1 ψ 0 0 1 0 0 ψ 2 3 + 6 4 Y ξ L ξ N ξ Y ζ L ζ N ζ 0 0 0 0 7 5» ξ ζ Bérénice Mettler (University of Minnesota) Chapter 4 The Equations of Motion Feb. 20-27, 2013 (v. 2/26/13) 24 / 26

Dimensional vs. non-dimensional derivatives Dimensionless forms are common in aerodynamics field. It s main purpose is that it allows, for example, comparing the derivatives of different AC across a broad range of scale. For the longitudinal equations m u = X uu + X w w + X qq + Xẇ ẇ + X ηη + X τ τ the dimension of the derivatives is force. Therefore they can be non-dimensionalized by dividing them by 1 2 ρv 2 0 S, where S is the wing surface area, which is used as a reference value. The dimensionless form of other variables and parameters are: Time: ˆt = t, where σ = σ 1 2 ρv 0S Longitudinal and lateral relative density factor: µ 1 = Velocities: û = u V 0 Angular rate ˆq = qσ In level flight lift and weight are equal: mg = 1 2 ρv 2 0 SC L Moment of inertias i x = m Ix, i mb 2 y = Iy, i m c 2 z = 1 2 Iz, i mb 2 xz = Ixz mb 2 m m and µ1 = ρs c 1 2 ρsb Bérénice Mettler (University of Minnesota) Chapter 4 The Equations of Motion Feb. 20-27, 2013 (v. 2/26/13) 25 / 26

When implemented for the longitudinal EOM gives:!!! u X u u X w w X q qσ σ = 1 V 0 ρv 2 2 0 S + 1 V 0 ρv 2 2 0 S + 1 V 0 ρv 2 2 0 S c + mg 1 µ 1 ρv 2 2 0 S θ!!! ẇσ X η τ + + η + X τ V 0µ 1 Xẇ 1 2 ρs c 1 2 ρv 2 0 S 1 2 ρv 2 0 S The dimensionless longitudinal equations would be given as: ˆq ŵ û = X uû + X w ŵ + X q + C L θ + Xẇ + X ηη + X τ ˆτ µ 1 µ 1 where the non-dimensional derivatives and the non-dimensional variables and parameters are obtained from above. Cook uses the notation X u, X v... X ẇ for the dimensional form of the derivatives and the plain form for the dimensionless form. Appendix 2 in Cook provides the dimensionless aerodynamic and control derivatives. Bérénice Mettler (University of Minnesota) Chapter 4 The Equations of Motion Feb. 20-27, 2013 (v. 2/26/13) 26 / 26