Chapter 4 The Equations of Motion

Chapter 4 The Equations of MotionFlight Mechanics and Control – AEM 4303

Berenice Mettler

University of Minnesota

Feb. 20 - 27, 2013 (v. 2/26/13)

Berenice 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


Newton-Euler Equations

The Newton-Euler equations expressed in the inertial reference frame are derived fromthe principle of conservation of linear and angular momentum. For constant vehicle massm and moment of inertia (inertial tensor I), they are:

d I (mv)

dt= F

d I (Iω)

dt= M

where F = [X Y Z ]T is the vector of external forces acting on the vehicle center ofgravity and M = [L M N]T is the vector of external moments.

The notation d I ·d· for the derivative means the derivative is taken w.r.t. the inertial frame.


Nomenclature

Body-fixed frame axes and key states:

y, u, q

x, v, p

z, w, r

Z, N

X, L

Y, M

lateral axis

longitudinal axis

vertical axis

aircraft cg

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


For the aircraft, it will be more convenient to express all components in the body-fixedcoordinates. Since these coordinates are moving w.r.t. the inertial frame, we need to usethe differentiation rule of relative motion. In a first step, let’s define the differential withrespect to the body-fixed reference frame (denoted by b):

md Iv

dt= m

dbv

dt+ m(ω × v) = F

d I (Iω)

dt=

db(Iω)

dt+ (ω × Iω) = M

In this equation the vector components are still in inertial frame. In a second step, let’swrite all components in terms of the body coordinates:

mdbv

dt+ m(ω × v) = F

Idbω

dt+ (ω × Iω) = M

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 constantand can be taken outside the derivative.


For a 6 DOF rigid-body system, the three ordinary differential equations describing theaircraft’s translational motion about its three reference axes are

m

24 uvw

35+ m

24pqr

35×24u

vw

35 =

24XYZ

35Written component-wise gives the three equations:

mu + m(qw − rv) = X

mv + m(ru − pw) = Y

mw + m(pv − qu) = Z

Similarly, the three ordinary differential equations describing the aircraft’s rotationalmotion: 24 Ix Ixy Ixz

Iyx Iy IyzIzx Izy Iz

3524pqr

35+

24pqr

35×24 Ix Ixy Ixz

Iyx Iy IyzIzx Izy Iz

3524pqr

35 =

24 LMN

35


Written component-wise gives the three equations:

Ix p − (Iy − Iz)qr + Ixy (pr − q)− Ixz(pq + r) + Iyz(r 2 − q2) = L

Iy q + (Ix − Iz)pr + Iyz(pq − r) + Ixz(p2 − r 2)− Ixy (qr + p) = M

Iz r − (Ix − Iy )pq − Iyz(pr + q) + Ixz(qr − p) + Ixy (q2 − p2) = N

The moment equation can be simplified considering that AC are symmetric about the(oxz) plane. Therefore, Ixy = Iyz = 0.

Ix p − (Iy − Iz)qr − Ixz(pq + r) = L

Iy q + (Ix − Iz)pr + Ixz(p2 − r 2) = M

Iz r − (Ix − Iy )pq + Ixz(qr − p) = N

Furthermore, if the body axes are aligned with the principal inertia axes, Ixz = 0. Even ifthis is not the case, in general Ixz << Ix , Iy or Iz .

I =

24 Ix 0 (Ixz)0 Iy 0

(Izx) 0 Iz

35


4.1.4 Disturbance forces and moments

The force and moment components can be decomposed into contributions due to theaerodynamics, gravity, controls, power and atmospheric disturbances.

mu + m(qw − rv) = Xa + Xg + Xc + Xp + Xd

mv + m(ru − pw) = Ya + Yg + Yc + Yp + Yd

mw + m(pv − qu) = Za + Zg + Zc + Zp + Zd

and

Ix p − (Iy − Iz)qr − Ixz(pq + r) = La + Lg + Lc + Lp + Ld

Iy q + (Ix − Iz)pr + Ixz(p2 − r 2) = Ma + Mg + Mc + Mp + Md

Iz r − (Ix − Iy )pq + Ixz(qr − p) = Na + Ng + Nc + Np + Nd

The next step in modeling the AC dynamics would involve deriving the detailedexpressions for this different contributions. This task is complex since these effects aredependent on geometry and the AC motion itself.

For analytical purpose, the equations are linearized about an equilibrium (trim) operatingpoint of interest. The force contributions can then be simplified by the product ofderivatives and perturbations in the states about trim conditions.


4.2 The Linearized Equations of Motion

Let’s assume a rectilinear, steady trim condition given by airspeed V0 and the AC bodyvelocities (Ue ,Ve ,We). In the absence of side slip, Ve = 0. Consider small perturbationsabout the trim conditions given by (u, v ,w) and (p, q, r):

U = Ue + u

V = Ve + v = v

W = We + w

Since the perturbations are small, taking their square or product is much smaller and canbe ignored. Substituting the perturbation terms in the equations of motion results in

m(u + Weq) = Xa + Xg + Xc + Xp

m(v + Wep + Uer) = Ya + Yg + Yc + Yp

m(w − Ueq) = Za + Zg + Zc + Zp

and

Ix p − Ixz r = La + Lg + Lc + Lp

Iy q = Ma + Mg + Mc + Mp

Iz r − Ixz p = Na + Ng + Nc + Np


Next, let’s express the force and moment components associated with small perturbationabout the trim conditions and deflection of the control surfaces.

Gravitation terms

For simplicity consider the symmetrictrim condition (wings level) as shownhere (φe = ψe = 0).

For this trim condition, the componentsof weight in the body axes only appearin the xz plane of symmetry. Using thedirection cosine matrix D:24Xge

Yge

Zge

35 =

24−mg sin θe

0mg cos θe

35 (1)

The perturbations in weight resultingfrom the perturbations in attitude aredetermined from the small-angleapproximation of the cosine directionmatrix.24Xg

Yg

Zg

35 =

24 1 ψ −θ−ψ 1 φθ −φ 1

3524Xge

Zge

Zge

35 =

24 1 ψ −θ−ψ 1 φθ −φ 1

3524−mg sin θe

0mg cos θe

35 (2)


Expressed in components the gravitational terms are:

Xg = −mg sin θe −mgθ cos θe

Yg = mgψ sin θe + mgφ cos θe

Zg = mg cos θe −mgθ sin θe

Note that φ, θ and ψ are all perturbations from trim conditions φe = 0, θe and ψe = 0. a

aIf the trim is a non-symmetric configuration (φe 6= 0) the subsequent derivations can easily be modified toaccount for the lateral components

Aerodynamic terms

For an AC operating about trim condition, a disturbance or control deflection will result ina 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 ACstate (speeds u, v ,w and angular rates p, q, r) can be described using a Taylor series. Forexample, for the axial force components Xa:

Xa = Xae +

„∂X

∂uu +

∂2X

∂u2

u2

2!+∂3X

∂u3

u3

3!+∂4X

∂u4

u4

4!+ · · ·

«· · ·+

„∂X

∂uu +

∂2X

∂u2

u2

2!+∂3X

∂u3

u3

3!+∂4X

∂u4

u4

4!+ · · ·

«+ · · ·+ higher order derivatives


Aerodynamic derivatives

Since the perturbations in motion are small, the higher-order terms and higher-orderderivatives are usually negligible except for z-component acceleration w , which istypically included.

Xa = Xae +∂X

∂uu +

∂X

∂vv +

∂X

∂ww +

∂X

∂pp +

∂X

∂qq +

∂X

∂rr +

∂X

∂ww (3)

The derivatives are usually written as X• = ∂X∂• :

Xa = Xae + Xuu + Xvv + Xww + Xpp + Xqq + Xr r + Xw w (4)

The derivatives Xu,Xv . . .Xw are called the aerodynamic derivatives.

The remaining force and moment components are written similarly, for example for thepitching moment

Ma = Mae + Muu + Mvv + Mww + Mpp + Mqq + Mr r + Mw w (5)

Concretely, the aerodynamic derivatives describe the force or moment perturbation thatresults from perturbation in states from the equilibrium. For example, Mv is the rollmoment resulting from the perturbation in lateral velocity v .


Aerodynamic control terms: control derivatives

The primary AC control surfaces are the elevator η, the aileron ξ and the rudder ζ. Theirforce and moment contributions are due to aerodynamic effects, therefore they aredescribed by aerodynamic control derivatives or simply control derivatives.For example, the pitching moment due to the control surfaces is described by:

Mc =∂M

∂ηη +

∂M

∂ξξ +

∂M

∂ζζ (6)

Note that the elevator η is the primary source of pitch control moment. The presence ofcontrol derivatives for the other surfaces (η,ξ,ζ) accounts for secondary effects.

Remember that the derivatives in the linearized equations describe the perturbation inforces and moments with respect to the trim conditions. Therefore η, ξ and ζ areperturbations about the trim control settings ηe , ξe and ζe , respectively.

Mc = Mηη + Mξξ + Mζζ (7)

The remaining derivatives are expressed in a similar manner. For the rolling moments, forexample,

Lc = Lηη + Lξξ + Lζζ (8)

The same general form applies to other control surfaces such as flaps or spoilers.


Power terms

A powered AC is equipped with a propulsion system such as a turbofan, jet orpropeller driven by an internal combustion, turbo-shaft or electric motor.

The thrust is controlled throttle lever angle ε, however, in the linearized equation ofmotion, the normal force due to the thrust is described as a perturbation in thrustforce from the trim force necessary to maintain the equilibrium condition.

For example, for the normal force, the thrust perturbation τ relative to theequilibrium thrust τe is given as:

Zp = Zττ (9)

The thrust contributions to the other force and moment components are describedin a similar way.


Equations of motion for small perturbations

Now that all force and moment contributions are accounted for we can substitute thesein the linearized Newton-Euler equations to obtain the complete equations of motion forsmall perturbations.

m(u + Weq) =Xae + Xuu + Xvv + Xww + Xpp + Xqq + Xr r + Xw w

−mg sin θe −mgθ cos θe + Xηη + Xξξ + Xζζ + Xττ

m(v + Wep + Uer) =Yae + Yuu + Yvv + Yww + Ypp + Yqq + Yr r + Yw w

+ mgψ sin θe + mgφ cos θe + Yηη + Yξξ + Yζζ + Yττ

m(w − Ueq) =Zae + Zuu + Zvv + Zww + Zpp + Zqq + Zr r + Zw w

+ mg cos θe −mgθ sin θe + Zηη + Zξξ + Zζζ + Zττ

and

Ix p − Ixz r =Lae + Luu + Lvv + Lww + Lpp + Lqq + Lr r + Lw w

+ Lηη + Lξξ + Lζζ + Lττ

Iy q =Mae + Muu + Mvv + Mww + Mpp + Mqq + Mr r + Mw w

+ Mηη + Mξξ + Mζζ + Mττ

Iz r − Ixz p =Nae + Nuu + Nvv + Nww + Npp + Nqq + Nr r + Nw w

+ Nηη + Nξξ + Nζζ + Nττ


Derivatives in trim conditions

In steady-state, equilibrium (trim) condition, all derivatives are by definition equal tozero. The steady-state equation of motion are therefore:

Xae = mg sin θe

Yae = 0

Zae = −mg cos θe

Lae = 0

Mae = 0

Nae = 0

These equations define the constant trim terms Xae , . . . ,Nae , which can be substitutedback into the equations of motion.

mu =Xuu + Xvv + Xww + Xpp + (Xq −mWe)q + Xr r + Xw w (10)

−mgθ cos θe + Xηη + Xξξ + Xζζ + Xττ (11)

mv =Yuu + Yvv + Yww + (Yp −mWe)p + Yqq + (Yr −mUe)r + Yw w (12)

+ mgψ sin θe + mgφ cos θe + Yηη + Yξξ + Yζζ + Yττ (13)

mw =Zuu + Zvv + Zww + Zpp + (Zq + Ue)q + Zr r + Zw w (14)

−mgθ sin θe + Zηη + Zξξ + Zζζ + Zττ (15)


Ix p − Ixz r =Luu + Lvv + Lww + Lpp + Lqq + Lr r + Lw w (16)

+ Lηη + Lξξ + Lζζ + Lττ (17)

Iy q =Muu + Mvv + Mww + Mpp + Mqq + Mr r + Mw w (18)

+ Mηη + Mξξ + Mζζ + Mττ (19)

Iz r − Ixz p =Nuu + Nvv + Nww + Npp + Nqq + Nr r + Nw w (20)

+ Nηη + Nξξ + Nζζ + Nττ (21)

The two sets of differential equations describe the response of the aircraft’s linear andangular motion to small perturbations about the equilibrium conditions given by airspeedcomponents (Ue ,Ve ,We) and attitude angles (φe , θe , ψe).

For a statically stable AC, the responses to small perturbations are transient. For theseconditions, the coupling between the longitudinal and lateral responses is typically smallenough 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.


The Decoupled Equations of Motion

Longitudinal-vertical equations of motion

The longitudinal-vertical EOM are confined to the (oxz) plane. The aircraft motionin this plane is described by the x and z component linear motion and the pitchingmotion.

The other components are ignored, are simply assumed to be zero. Therefore all theforce and moment derivatives and control derivatives associated with these motionsare taken as zero.

Furthermore, for the longitudinal-vertical motion, only the primary control derivativesare 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 pitchingmotion θ:

mu = Xuu + Xww + (Xq −mWe)q + Xw w −mgθ cos θe + Xηη + Xττ

(m − Zw )w = Zuu + Zww + (Zq + Ue)q −mgθ sin θe + Zηη + Zττ

Iy q = Muu + Mww + Mqq + Mw w + Mηη + Mττ


For the special case of level flight (We = 0) and when the body axes coincide with thewind axes, i.e., θe = 0 the equations of motion further simplify to

mu = Xuu + Xww + Xqq + Xw w + Xηη + Xττ

(m − Zw )w = Zuu + Zww + (Zq + Ue)q + Zηη + Zττ

Iy q = Muu + Mww + Mqq + Mw w + Mηη + Mττ

State-space form

The equations of motion are a set of coupled, linear differential equations. It isconvenient to represent them using matrix algebra. This description involves collectingthe state variables and control inputs in a state vector x ∈ Rn and input vector u ∈ Rm,and grouping the aerodynamic and control derivatives in matrices leading to the followingform:

x(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 stateare 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) directmatrix.


Mathematical definition for linear time invariant (LTI) model

The linear model makes it possible to analyze the aircraft’s stability and controlproperties using powerful linear systems mathematical tools (see next Chapter).

For each trim condition, the A and B matrices have constant coefficient, therefore, thesystem 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 andother first principles lead to a set of nonlinear differential equations. These equations canbe described by

x(t) = f (x(t), u(t)) (24)

The linear state-space equations (LTI model) can be derived using a linearization aboutthe trim condition given by the state x∗ and control input u∗:

x(t) = Ax(t) + Bu(t) (25)

=

„∂f (x, u)

∂x

«x∗,u∗

x(t) +

„∂f (x, u)

∂u

«x∗,u∗

u(t) (26)

Therefore the A and B matrices can be determine by taking the derivative of thenonlinear model f w.r.t. the state and input vectors and evaluating the derivatives at thetrim condition.


To convert the set of differential equations in matrix form, the terms involving the timederivatives are collected on the left of the equality and the remaining terms to the right.

mu − Xw w = Xuu + Xww + (Xq −mWe)q −mgθ cos θe + Xηη + Xττ

mw − Zw w = Zuu + Zww + (Zq + Ue)q −mgθ sin θe + Zηη + Zττ

Iy q −Mw w = Muu + Mww + Mqq + Mηη + Mττ

θ = q

Note the last equation was added to the original set, giving us four differential equationsneeded 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 as2664m −Xw 0 00 (m − Zw ) 0 00 −Mw Iy 00 0 0 I

3775 x(t) =

2664Xu Xw (Xq −mWe) −mg cos θe

Zu Zw (Zq + mUe) −mg sin θe

Mu Mw Mq 00 0 1 0

3775 x(t) (27)

+

2664Xη XτZη ZτMη Mτ

0 0

3775 u(t) (28)


Notice that these equations are in a form:

M x(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.,

x(t) = M−1A′x(t) + M−1B ′u(t) (30)

Therefore, state and control matrix are A = M−1A′ and B = M−1B ′, respectively.The standard LTI form is usually written as2664

uwq

θ

3775 =

2664xu xw xq xθzu zw zq zθmu mw mq mθ

0 0 1 0

37752664

uwqθ

3775+

2664xη xτyη zτmη mτ

0 0

3775»ητ–

(31)


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 fullycoupled EOM (10-21) we have:

mv =Yvv + (Yp −mWe)p + (Yr −mUe)r + Yw w

+ mgψ sin θe + mgφ cos θe + Yξξ + Yζζ

Ix p − Ixz r =Lvv + Lpp + Lr r

Iz r + Ixz p =Nvv + Npp + Nr r + Nξξ + Nζζ

For level flight and body axes coinciding with wind axes, θe = We = 0

mv =Yvv + Ypp + (Yr −mUe)r + mgφ+ Yξξ + Yζζ

Ix p − Ixz r =Lvv + Lpp + Lr r

Iz r + Ixz p =Nvv + Npp + Nr r + Nξξ + Nζζ

Next, these sets of coupled differential equations, similar to the longitudinal-vertical EOMare presented in matrix form, with state vector x = [v , p, r , φ, ψ]T and input u = [ξ, ζ]T .


266664m 0 0 0 00 Ix −Ixz 0 00 −Ixz Ix 0 00 0 0 1 00 0 0 0 1

377775266664

vpr

φ

ψ

377775 =

266664Yv Yp + We Yr − Ue mg cos θe mg sin θe

Lv Lp Lr 0 0Nv Np Nr 0 00 1 0 0 00 0 1 0 0

377775266664

vprφψ

377775

+

266664Yξ YζLξ LζNξ Nζ0 00 0

377775»ξζ

–


Dimensional vs. non-dimensional derivatives

Dimensionless forms are common in aerodynamics field. It’s main purpose is that itallows, for example, comparing the derivatives of different AC across a broad range ofscale.

For the longitudinal equations

mu = Xuu + Xww + Xqq + Xw w + Xηη + Xττ

the dimension of the derivatives is force. Therefore they can be non-dimensionalized bydividing them by 1

2ρV 2

0 S , where S is the wing surface area, which is used as a referencevalue.

The dimensionless form of other variables and parameters are:

Time: t = tσ

, where σ = m12ρV0S

Longitudinal and lateral relative density factor: µ1 = m12ρS ¯c

and µ1 = m12ρSb

Velocities: u = uV0

Angular rate q = qσ

In level flight lift and weight are equal: mg = 12ρV 2

0 SCL

Moment of inertias ix = Ixmb2 , iy =

Iym¯c2 , iz = Iz

mb2 , ixz = Ixzmb2


scales

the aerodynamics

When implemented for the longitudinal EOM gives:

u

V0σ =

Xu

12ρV 2

0 S

!u

V0+

Xw

12ρV 2

0 S

!w

V0+

Xq

12ρV 2

0 S ¯c

!qσ

µ1+

mg12ρV 2

0 Sθ

+

Xw

12ρS ¯c

!wσ

V0µ1+

Xη

12ρV 2

0 S

!η + Xτ

τ

12ρV 2

0 S

!

The dimensionless longitudinal equations would be given as:

˙u = Xu u + Xw w + Xqq

µ1+ CLθ + Xw

˙w

µ1+ Xηη + Xτ τ

where the non-dimensional derivatives and the non-dimensional variables and parametersare obtained from above.

Cook uses the notation◦X u,

◦X v . . .

◦X w for the dimensional form of the derivatives and the

plain form for the dimensionless form. Appendix 2 in Cook provides the dimensionlessaerodynamic and control derivatives.


Chapter 4 The Equations of Motion

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