10.2. POLYNOMIALS OF A SQUARE MATRIX 201

Example 70 For the matrix A= 3 −1 .

we found that −1 3

results in

Hence, A = T ΛT −1 and T= 11

1 −1

Λ = T −1AT = 20

04

p(A) = T p(Λ)T −1

=T p(2) 0 T −1

0 p(4)

= 1 p(2) + p(4) p(2) − p(4) .

2 p(2) − p(4) p(2) + p(4)

Suppose p is the characteristic polynomial of A. Since 2, 4 are the eigenvalues of A, we have

p(2) = p(4) = 0; hence p(A) = 0.

202 CHAPTER 10. EAT *

10.3 The matrix exponential: eA

Recall the exponential function:

f (λ) = eλ

For any complex number λ, eλ is given by the sum of a convergent series:

eλ = ∞ 1 λk = 1 + λ + 1 λ2 + 1 λ3 + . . .

k=0 k! 2 3!

The exponential of any matrix A is deﬁned by:

eA := ∞ 1 Ak = I + A + 1 A2 + . . .

k=0 k! 2!

The following properties can be readily deduced from the above deﬁnition.

(a) The matrices A and eA commute, that is,

AeA = eAA

(b) Suppose A is similar to another matrix B, that is, for some nonsingular T ,

A = T BT −1 .

Then

eA = T eBT −1 .

(c) Suppose A is diagonal, that is,

λ1 0 · · · 0

0 λ2 · · · 0

... ... ...

A = .

0 0 · · · λn

Then eλ1 0 · · · 0

0 eλ2 · · · 0

... ... ...

eA = .

0 0 · · · eλn

(d) Suppose A is diagonalizable, that is, for some nonsingular T ,

λ1 0 · · · 0

0 λ2 · · · 0

... ... ...

A = T T −1 .

0 0 · · · λn

10.3. THE MATRIX EXPONENTIAL: EA 203

Then eλ1 0 · · · 0

0 eλ2 · · · 0

... ... ...

eA = T T −1

0 0 · · · eλn

(e) If λ is an eigenvalue of A with eigenvector v, then eλ is an eigenvalue of eA with

eigenvector v.

204 CHAPTER 10. EAT *

10.4 The state transition matrix: eAt

Using the above deﬁnitions and abusing notation (eAt instead of etA),

eAt := ∞ 1 (tA)k = I + tA + 1 (tA)2 + 1 (tA)3 + . . .

k=0 k! 2! 3!

Letting Φ(t) := eAt

we have

Φ(0) = I

Φ˙ = AΦ

Hence, the solution of x˙ = Ax x(0) = x0

is given by x(t) = eAtx0

Some further properties of eAt. From the previous section, we obtain the following

properties:

(a) The matrices A and eAt commute, that is,

AeAt = eAtA

(b) Suppose A is similar to another matrix B, that is, for some nonsingular T ,

A = T BT −1 .

Then

eAt = T eBtT −1 .

(c) Suppose A is diagonal, that is,

λ1 0 · · · 0

0 λ2 · · · 0

... ... ...

A = .

0 0 · · · λn

Then eλ1t 0 · · · 0

0 eλ2t · · · 0

... ... ...

eAt = .

0 0 · · · eλnt

10.4. THE STATE TRANSITION MATRIX: EAT 205

(d) Suppose A is diagonalizable, that is, for some nonsingular T ,

λ1 0 · · · 0

0 λ2 · · · 0

... ... ...

A = T T −1 .

0 0 · · · λn

Then eλ1t 0 · · · 0

0 eλ2t · · · 0

... ... ...

eAt = T T −1

0 0 · · · eλnt

(e) If λ is an eigenvalue of A with eigenvector v, then eλt is an eigenvalue of eAt with

eigenvector v.

Properties (b)-(d) are useful for computing eAt.

206 CHAPTER 10. EAT *

10.5 Computation of eAt

10.5.1 MATLAB

>> help expm

EXPM Matrix exponential.

EXPM(X) is the matrix exponential of X. EXPM is computed using

a scaling and squaring algorithm with a Pade approximation.

Although it is not computed this way, if X has a full set

of eigenvectors V with corresponding eigenvalues D, then

[V,D] = EIG(X) and EXPM(X) = V*diag(exp(diag(D)))/V.

See EXPM1, EXPM2 and EXPM3 for alternative methods.

EXP(X) (that’s without the M) does it element-by-element.

10.5.2 Numerical simulation

Numerically solve

Φ˙ = AΦ, Φ(0) = I .

This can be carried out in in MATLAB as follows. For i = 1, 2, · · · , n let ei be the

n-vector corresponding to the i-th column of the n × n identiryt matrix, that is, the i-th

element of ei is one while all other elements are zero. Then use a diﬀerential solver to solve

φ˙1 = Aφ1 φ1(0) = e1

φ˙2 = Aφ2 φ2(0) = e2

... φn(0) = en

φ˙n = Aφn

Funally, let

Φ(t) = φ1(t) φ2(t) · · · φn(t)

10.5.3 Jordan form

eAt = T eΛtT −1

where Λ = T −1AT and Λ is the Jordan form of A. We have already seen this for diagonal-

izable systems.

10.5. COMPUTATION OF EAT 207

10.5.4 Laplace style

x˙ = Ax x(0) = x0

Suppose

X(s) = L(x)(s)

is the Laplace transform of x(·) evaluated at s ∈ C. Taking the Laplace transform of x˙ = Ax

yields:

sX(s) − x0 = AX(s)

Hence, except when s is an eigenvalue of A, sI − A is invertible and

L(x)(s) = X(s) = (sI − A)−1x0

Since x(t) = eAtx0 for all x0, we must have

L(eAt) = (sI − A)−1

L−1 ((sI − A)−1) = eAt

Example 71 Recall A= 3 −1

So, −1 3

Since s−3 −1

A)−1 (s−2)(s−4) (s−2)(s−4)

(sI − =

−1

s−3

(s−2)(s−4)

(s−2)(s−4)

s−3 = 1 s 1 2 + s 1 4

(s − 2)(s − 4) 2 − −

−1 = 1 s 1 2 − s 1 4

(s − 2)(s − 4) 2 − −

and L−1( s 1 2) = e2t L−1( s 1 4 ) = e4t

we have, − −

eAt = L−1 (sI − A)−1

= 1 e2t + e4t e2t − e4t

2 e2t − e4t e2t + e4t

Exercise 60 Compute eAt at t = ln(2) for

A= 01

10

using all methods of this section.

208 CHAPTER 10. EAT *

Chapter 11

Stability

11.1 Introduction

Consider a general system (linear or nonlinear) described by

x˙ = f (x) (11.1)

where the state x(t) is an n-vector and the real scalar variable t represents time. By a

solution of (11.1) we mean a continuous function x(·) which satisﬁes x˙ (t) = f (x(t)) for all

t ≥ 0.

In the following sections, we assume that xe is an equilibrium state of the above system

So, if the system starts at xe, it remains at xe for all time. What happens if the system does

not start at xe? If xe is a desirable equilibrium state, we would like the system to have the

property that if it is perturbed from xe, it returns to xe.

11.2 Stability

Global stability. An equilibrium state xe is globally asymptotically stable (GAS) if every solu-

tion converges to it, that is, every solution satisﬁes

lim x(t) = xe .

t→∞

If a system has a globally stable equilibrium state xe then, it cannot have any other

equilibrium states. In this case we say that the system is globally stable. Also, if a system has

a globally stable equilibrium state then, all solutions are bounded; the system cannot have

any unbounded solutions.

Example 72

x˙ = −x

The origin is GAS.

209

210 CHAPTER 11. STABILITY

Example 73

x˙ = −x3

The origin is GAS. Why?

For nonlinear systems, we also need the following deﬁnition of stability.

Asymptotic stability. An equilibrium state xe is asymptotically stable (AS) if it has a neigh-

borhood with the property that whenever the initial state is in this neighborhood then, the

resulting solution converges to xe.

The above deﬁnition of stability does not require that every solution converge to xe but,

only those solutions which originate in some neighborhood of xe. Only nonlinear systems

can have equilibrium states which are asymptotically stable but not globally asymptotically

stable. For linear systems, asymptotically stability and global asymptotically stability are

equivalent concepts. So, if the origin of a linear system is asymptotically stable then the

origin is globally asymptotically stable and all solutions are bounded. However, a nonlinear

system can have a asymptotically stable equilibrium state but also have some unbounded

solutions. This is illustrated in the next example.

Example 74

x˙ = −x(1 − x2)

The zero state is AS but not GAS. This system has some unbounded solutions.

Example 75 Damped simple pendulum

x˙ 1 = x2

x˙ 2 = − sin x1 − x2

The origin is AS but not GAS.

11.2. STABILITY 211

Figure 11.1: Region of attraction

Region of attraction. When an equilibrium state is asymptotically stable, we refer to

its region of attraction as the set of states with the following property: if the system starts

at one of these states, the resulting solution converges to the equilibrium state. Hence,

an equilibrium state is globally asymptotically stable if it is asymptotically stable with the

whole state space as its region of attraction.

Example 76

x˙ = −x(ǫ2 − x2)

Linear systems. Consider now a linear system described by

x˙ = Ax (11.2)

where A is a constant n × n matrix. If A has n distinct eigenvalues then, every solution of

the above system can be expressed as

m

x(t) = eλitvi

i=1

where λ1, . . . , λm are the eigenvalues of A and vi is an eigenvector for λi. From this general

form of the solution we have the following conclusion.

The zero state of x˙ = Ax is globally asymptotically stable if and only if all the eigenvalues

of A have negative real parts.

It can be shown that the above statement also holds when A does not have n distinct

eigenvalues; hence it holds for any square matrix A.

Suppose A is a 2 × 2 matrix and

p(s) = s2 + a1s + a0

is the characteristic polynomial of A. We have shown before that both roots of such a second

order polynomial have negative real part if and only if both a0 and a1 are positive. Hence, if

212 CHAPTER 11. STABILITY

A is a 2 × 2 matrix then, x˙ = Ax is stable if and only if the coeﬃcients of the characteristic

polynomial of A are positive.

For a general linear system x˙ = Ax, one can determine whether or not all the eigen-

values of matrix A have negative real part by applying the Routh test to the characteristic

polynomial of A.

Example 77 The system

x˙ 1 = −x1 + x2

x˙ 2 = −x2

is GS.

Example 78 The system

x˙ 1 = −x1 + x2

x˙ 2 = −x1 − x2

is GS.

Example 79 The system

x˙ 1 = x2

x˙ 2 = −x1

is not GS.

Example 80 The system

x˙ 1 = x2

x˙ 2 = x1

is not GS.

11.3 Neutral stability and instability

Consider a linear undamped spring mass system described by q¨+q = 0. All nonzero solutions

oscillate with a constant amplitude which depends on the initial state. Also, the amplitude

is small if the magnitude of the initial state is small. The origin is not stable in the sense

just deﬁned. However it is neutrally stable.

We say that an equilibrium state xe of a system is neutrally stable (NS) or marginally stable

if it is not asymptotically stable as deﬁned previously, but has the following property. If the

perturbation x(0)−xe of the initial state from the equilibrium state is suﬃciently small, then

the perturbation x(t)−xe of the resulting solution from the equilibrium state is small.

11.3. NEUTRAL STABILITY AND INSTABILITY 213

Figure 11.2: Neutral stability

Instability. An equilibrium state is unstable (US) if it is neither stable nor neutrally stable.

For a linear system, it can be shown that neutral stability of the origin implies all solutions

are bounded; also, instability of the origin implies the system has some unbounded solutions.

However, for a nonlinear system, instability of an equilibrium state does not imply that the

system has some unbounded solutions. Also, the existence of a stable or neutrally stable

equilibrium state does not imply all solutions are bounded.

Example 81

x˙ = 0

Every state is a neutrally stable equilibrium state. All solutions are bounded.

Example 82

x˙ = x

The origin is unstable. Also the system has unbounded solutions.

Example 83 Undamped oscillator

x˙ 1 = x2

x˙ 2 = −x1

The origin is neutrally stable. All solutions are bounded.

Example 84 Simple pendulum

x˙ 1 = x2

x˙ 2 = − sin x1

(0, 0) is neutrally stable; (π, 0) is unstable.

Example 85

x˙ = x(1 − x2)

The origin here is unstable. However, all solutions of this system are bounded. This cannot

occur in a linear system.

214 CHAPTER 11. STABILITY

Example 86 Van der Pol oscillator This nonlinear system is interesting in that is has a

single equilibrium at the origin and this equilibrium is unstable. However all solutions of the

system are bounded,

x˙ 1 = x2

x˙ 2 = (1 − x21)x2 − x1

Figure 11.3: Van der Pol oscillator

Linear systems. Consider now a linear system described by x˙ = Ax where A is a constant

n × n matrix. If A has n distinct eigenvalues, we have seen that every solution of the above

system has the form m

x(t) = eλitvi

i=1

where λ1, . . . , λm are eigenvalues of A. From the general form of the solution we have the

following conclusion:

If the matrix A has at least one eigenvalue with a positive real part then, the system

x˙ = Ax is unstable about zero and has some unbounded solutions.

The above conclusion also holds when A does not have n distinct eigenvalues; so, it holds

for any square A.

Example 87 The system

x˙ 1 = x2

x˙ 2 = x1

is US about zero.

We also have the following conclusion from the above general form of the solution:

Suppose that none of the eigenvalues of the matrix A have positive real part and there is

at least one eigenvalue with zero real part. Then, the system x˙ = Ax is neutrally stable about

zero and all solutions are bounded.

Example 88 The system

x˙ 1 = x2

x˙ 2 = −x1

is NS about 0.

11.3. NEUTRAL STABILITY AND INSTABILITY 215

One has to be a little careful in applying the last conclusion above. It may not hold if

A has an eigenvalue λ with a zero real part and this eigenvalue is a repeated root of the

characteristic polynomial of A. This is illustrated in the next example.

Example 89 Unattached mass

x˙ 1 = x2

x˙ 2 = 0

Here 01

00

A=

and the characteristic polynomial of A is p(s) = s2. Hence, A has 0 a single eigenvalue 0

and this eigenvalue is a repeated root of the characteristic polynomial of A. This system is

unstable about zero and has unbounded solutions. To see this consider the solution

x(t) = t

1

11.3.1 Exercises

Exercise 61 Determine the stability properties of the zero equilibrium state of the following

systems.

(a)

x˙ = 2x

(b)

x˙ 1 = 2x2

x˙ 2 = −2x1

(c)

x˙ 1 = 2x2

x˙ 2 = 2x1

Exercise 62 Determine the stability properties of the following systems.

(♠)

x˙ 1 = −2x1 + x2

x˙ 2 = x1 − x2

(♣)

x˙ 1 = −2x1 + x2

x˙ 2 = x1 + x2

216 CHAPTER 11. STABILITY

(♦)

x˙ 1 = −x1 + x2

x˙ 2 = −4x1 − x2

(♥)

x˙ 1 = 2x2

x˙ 2 = −4x1

Exercise 63 Determine whether or not the following system is stable, neutrally stable, or

unstable about the origin.

x˙ 1 = x2

x˙ 2 = x3

x˙ 3 = x4

x˙ 4 = x1

Exercise 64 Determine the stability properties of the following systems about the zero

solution.

(a)

x˙ 1 = x3

x˙ 2 = x2 + x3

x˙ 3 = −x2 + x3

(b)

x˙ 1 = 2x2

x˙ 2 = 2x1

(c)

x˙ 1 = −x1 + 2x2

x˙ 2 = −2x1 − x2

11.4. LINEARIZATION AND STABILITY 217

11.4 Linearization and stability

We know that a linear system is stable is stable if all its associated eigenvalues have negative

real part. How do we determine the stability properties of an equilibrium state of a nonlinear

system? On approach is to linearize the nonlinear system about the equilibrium state and

look at the stability properties of the linearized system, The big question then is: what do

the stability properties of the linearized system tell us about the nonlinear system? We will

see that, except in a special case, the local stability behavior of the nonlinear system about

the equilibrium state is predicted by the stability property of the linearized system.

Consider a nonlinear time-invariant system described by

x˙ = f (x)

where x(t) is an n-vector at each time t. Suppose xe is an equilibrium state for this system

and let

δx˙ = Aδx

be the linearization of this system about xe. Then we have the following fundamental results.

If all the eigenvalues of the matrix A have negative real part, then the nonlinear system is

stable about xe.

If the matrix A matrix has at least one eigenvalue with positive real part, then the nonlinear

system is unstable about xe.

If none of the eigenvalues of the matrix A have positive real part and there is at least one

eigenvalue with zero real part, then, one cannot make any conclusions on the stability of

the equilibrium state based on linearization.

Note that we cannot conclude global stability of a nonlinear system based on linearization.

We can only conclude local results.

Example 90

x˙ = −x(1 − x2)

Example 91 Undamped pendulum.

x˙ 1 = x2

x˙ 2 = − sin x1

Example 92 Pendulum with linear damping.

x˙ 1 = x2

x˙ 2 = −x2 − sin x1

Example 93 Van der Pol oscillator

x˙ 1 = x2

x˙ 2 = (1 − x12)x2 − x1

Example 94

x˙ = ax3

218 CHAPTER 11. STABILITY

11.4.1 Exercises

Exercise 65 For each of the following systems, determine (if possible) the stability proper-

ties of the zero equilibrium state. If not possible, state the reason. Justify your results.

(a)

x˙ = −x3

(b)

x˙ = x − 10x3

(c)

x˙ 1 = x2

x˙ 2 = − sin x1 − (1 − x12)x2

Exercise 66 For each of following set(s) of diﬀerential equations,

(a)

(1 + y2)y¨ + (|y˙| − 1)y˙ + cos(y˙) sin(y) = 0

(b)

x˙ 1 = x31 + x2

x˙ 2 = −x1 + x23

answer the following questions.

(i) Linearize the diﬀerential equation(s) about the zero solution.

(ii) What are the stability properties of the linearized diﬀerential equation(s)?

(iii) Based on part (ii), what can you say about the stability properties of the original set

of nonlinear equation(s)?

Exercise 67 Consider the system described by

y¨1 + (sin y2)y¨2 + (cos y1)y˙22 − sin y2 = 0

(1−ey1)y¨1 + y¨2 + y2y˙12 + e−y1 − 1 = 0

where y1 and y2 are real scalar variables. Determine whether or not this system is stable

about the zero solution (y1(t) ≡ y2(t) ≡ 0.)

Exercise 68 For each of the following second order systems, determine whether they are

stable, unstable, or neutrally stable about the zero solution.

11.4. LINEARIZATION AND STABILITY 219

(a)

y¨ − y˙ + y = 0

(b)

−y¨ − y˙ − y = 0

(c)

y¨ + y˙ = 0

(d)

y¨ − y˙ = 0

(e)

y¨ + y˙ − sin y = 0

(f )

y¨ + y˙3 + y = 0

Exercise 69 If possible, use linearization to determine whether or not the following system

is stable, unstable, or marginally stable about the equilibrium solution ω1e = 1, ω2e = ω3e = 0.

I1ω˙ 1 + (I3 − I2)ω2ω3 = 0

I2ω˙ 2 + (I1 − I3)ω1ω3 = 0

I3ω˙ 3 + (I2 − I1)ω1ω2 = 0

If not possible, state why. Consider three diﬀerent cases:

(a)

I1 = 1, I2 = 2, I3 = 3

(b)

I1 = 2, I2 = 3, I3 = 1

(c)

I1 = 3, I2 = 1, I3 = 2

Exercise 70 (Simple pendulum in drag)

Recall the simple pendulum in drag whose motion is described by

mlθ¨ + κV (lθ˙ − w sin θ) + mg sin θ = 0

where

V = l2θ˙2 + w2 − 2lw sin(θ)θ˙

(a) Obtain (by hand) the equilibrium values θe of the angle θ.

(a) For each equilibrium condition, determine the range of values of the wind speed w for

which that condition is stable.

220 CHAPTER 11. STABILITY

Figure 11.4: Pendulum in drag

11.5 Root loci

Here, we look at how the roots of a polynomial or the eigenvalues of a matrix vary with a

scalar parameter.

11.5.1 Polynomials

Suppose n and d are two polynomials and k is a real scalar parameter. Consider now the

parameter dependent polynomial given by

p(s; k) = d(s) + kn(s) .

This new polynomial depends on k in an aﬃne linear manner. A root locus consists of a

plot in the complex plane of the roots of p for values of k between 0 and ∞. This can be

accomplished with the MATLAB command

rlocus

We will not go into the usual root locus rules here. It is assumed that they are covered

elsewhere.

Example 95 Consider the polynomials n(s) = s2 + 2s + 2 .

d(s) = s2 − 2s + 2 and

Then

d(s) + kn(s) = (1 + k)s2 + 2(k − 1)s + 2(1 + k) .

We now illustrate the use of the MATLAB command rlocus. The output of that command

is shown in Figure 11.5.

d=[1 -2 2];

n=[1 2 2];

roots(d)

ans =

1.0000 + 1.0000i

1.0000 - 1.0000i

roots(n)

11.5. ROOT LOCI 221

ans =

-1.0000 + 1.0000i

-1.0000 - 1.0000i

rlocus(n,d)

1.5

1

0.5

Imag Axis 0

−0.5

−1

−1.5

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2

Real Axis

Figure 11.5: A root locus

Example 96 The MATLAB output for rlocus(d,n) corresponding to

d(s) = (s − 1)(s + 2)(s + 4) = s3 + 5s2 + 2s − 8 and n(s) = s + 3

is shown in Figure 11.6.

11.5.2 Matrices

Here we look at the manner in which the eigenvalues of a matrix vary with a scalar parameter.

To do this we need the following result on determinants.

Fact 3 Suppose M is an n × m matrix and N is m × n matrix. Then

det(In + MN ) = det(Im + N M) . (11.3)

In particular, if m = 1, then

det(In + MN ) = 1 + N M . (11.4)

Consider now a parameter dependent matrix described by

A − bkc

222 CHAPTER 11. STABILITY

Imag Axis 5

4

3 3

2

1

0

−1

−2

−3

−4

−5

−5 −4 −3 −2 −1 0 1 2

Real Axis

Figure 11.6: Another root locus

where k is a scalar parameter and the matrices A, b and c are n × n, n × 1, and 1 × n,

respectively. Here a root locus consists of a plot in the complex plane of the eigenvalues of

A − bkc for values of k between 0 and ∞.

Let

d(s) = det(sI − A) and g(s) = c(sI − A)−1b . (11.5)

Then, it can be shown that g can be uniquely expressed as

g(s) = n(s) (11.6)

d(s)

where n is a polynomial of degree strictly less than d. We now show that

det(sI − A + bkc) = d(s) + kn(s) (11.7)

This yields the following conclusion: The eigenvalues of the matrix A − bkc are the roots of

the polynomial d(s) + kn(s). Also, the coeﬃcients of the characteristic polynomial of A − bkc

depend on k in an aﬃne linear fashion.

To demonstrate the above result, consider any complex number s which is not an eigen-

value of A. Then sI − A is an invertible matrix and

sI − A + bkc = (sI − A)(I + (sI − A)−1bkc) .

Using Fact 3 and other properties of determinants, we obtain that

det(sI − A + bkc) = det (sI − A)(I + (sI − A−1)bkc)

= det(sI − A) det(I + (sI − A)−1bkc)

= det(sI − A)(1 + kc(sI − A)−1b)

= d(s)(1 + kg(s))

= d(s) + kn(s) .

11.5. ROOT LOCI 223

That is, det(sI − A + bkc) = d(s) + kn(s). Since we have shown that this relationship holds

for every s except the eigenvalues of A, and both sides of this equation have no poles at the

eigenvalues of A, the relationship also holds at the eigenvalues of A.

To use the MATLAB command rlocus, we do not actually have to compute n and d.

We can simply type

rlocus(A,b,c,0)

Notice the “0”.

Example 97 With

0 1 0 0

A = 0 0 1 b = 0 c= 3 1 0 ,

8 −2 −5 1

Figure 11.7 illustrates the output of the MATLAB command rlocus(A,b,c,0). Why are

Imag Axis 5

4

3 3

2

1

0

−1

−2

−3

−4

−5

−5 −4 −3 −2 −1 0 1 2

Real Axis

Figure 11.7: Yet another root locus

Figures 11.6 and 11.7 the same?

224 CHAPTER 11. STABILITY

Chapter 12

Equations of motion for general ﬂight

In this chapter we derive the equations of motion for the general three-dimensional motion

of an aircraft. We can regard an aircraft as a dynamical system with the following control

inputs:

th: throttle setting

el: elevator deﬂection

δa: deﬂection of the ailerons

δr: deﬂection of the rudder

If we model an aircraft as a single rigid body, we can completely describe its motion relative

to the earth by specifying the position of its mass center relative to some earth ﬁxed point

and by specifying the orientation of the aircraft relative to the earth. Hence the complete

state of an aircraft can be described by the position and velocity of aircraft mass center, and

the orientation and angular velocity of the aircraft.

We have already looked at the longitudinal dynamics of an aircraft. Before proceeding

with the general aircraft dynamics, we look at a simpler scenario to obtain some familiarity

with some of the variables involved in the lateral dynamics of an aircraft.

12.1 Ice-racer

Our Ice-racer is basically an aircraft moving on a lake of ice; we assume that the vehicle has

a plane of symmetry which is vertical and we ignore friction between the vehicle and the ice;

see Figure 12.1. The inputs to the vehicle are the throttle setting th and the rudder deﬂection

δr. The throttle setting aﬀects the thrust T ; we assume that the thrust vector lies in the

plane of symmetry of the vehicle. The rudder deﬂection produces a yawing moment on the

vehicle. The purpose of studying this system is to gain an initial appreciation of the lateral

dynamics of an aircraft.

Kinematics. Introducing reference frame e ﬁxed in the ice, the location of the mass center

of the vehicle can be described by the range coordinates p1 and p2 which are the cartesian

coordinates of the mass center of the vehicle relative to e; see Figure 12.2. The orientation

of the vehicle can be described by the yaw angle ψ which is the angle between a horizontal

225

226 CHAPTER 12. EQUATIONS OF MOTION FOR GENERAL FLIGHT

Figure 12.1: Ice-racer

Figure 12.2: Ice-racer kinematics

reference line ﬁxed in the plane of symmetry of the vehicle and the eˆ1 axis. It is considered

positive as shown, that is, when it corresponds to a positive rotation about the eˆ3 axis. We

let r be the yaw rate, that is, r = ψ˙ .

(12.1)

The the velocity V¯ of the vehicle mass center is given by

V¯ = p˙1eˆ1 + p˙2eˆ2 (12.2)

We can also describe the velocity V¯ by its magnitude V (speed) and the sideslip angle β; this

is the angle between V¯ and the reference line ﬁxed in the vehicle. Then

V¯ = V cos(ψ + β)eˆ1 + V sin(ψ + β)eˆ2

and recalling (12.2) we obtain the following navigation equations:

p˙1 = V cos(ψ + β) (12.3a)

p˙2 = V sin(ψ + β) (12.3b)

Forces and moments. Since the vehicle moves in a horizontal plane, we only consider

the forces in that plane and the moments perpendicular to that plane. The corresponding

free-body-diagram of the vehicle is contained in Figure 12.3.

Here F¯ is the total aerodynamic force on the vehicle and N is the yawing moment N. We

can decompose F¯ in two ways:

12.1. ICE-RACER 227

Figure 12.3: Relevant ice-racer force and moments

Drag D and Crosswind C. The drag force is always in the direction opposite the velocity

vector V¯ . The crosswind force is perpendicular to the drag and considered positive as shown.

Figure 12.4: Drag and crosswind

If we introduce the wind reference frame which is determined by V¯ and shown in Figure

12.5, then

F¯ = −Dwˆ1 − Cwˆ2

Figure 12.5: w-frame

Force component X and sideforce Y . These are the components of F¯ relative to a reference

frame b ﬁxed in the vehicle.

F¯ = Xˆb1 + Y ˆb2

228 CHAPTER 12. EQUATIONS OF MOTION FOR GENERAL FLIGHT

Figure 12.6: X and sideforce Y

The sideforce force Y¯ is perpendicular to the vehicle and Y is considered positive as shown,

that is to the right of the vehicle.

From Figure 12.7 we see that Y = −C cos β − D sin β; hence the crosswind C can be

expressed in terms of D and Y :

C cos β = −Y − D sin β

Figure 12.7: Resolving forces

Dimensionless coeﬃcients. As usual, we let q¯ = 1 ρV 2 be the dynamic pressure where

ρ is the air density. 2

We express the drag D as

D = q¯SCD

where CD is the drag coeﬃcient and S is a reference area associated with the vehicle. We

will consider CD to be constant here.

We can express the sideforce Y as

Y = q¯SCY

where CY is called the sideforce coeﬃcient. The static value of CY depends mainly on β and

δr. These dependencies can be described by the derivatives

CYβ = ∂Y and CYδr = ∂Y

∂β ∂δr

12.1. ICE-RACER 229

We will consider CYβ and CYδr constant. Including rate dependent terms, we approximate

CY by

CY = CYβ β + b CYr r + CYδr δr (12.4)

2V

where b is a reference length like the wing span. Note that a lot of the sideforce is due to the

vertical tail and rudder. Usually

CYβ < 0 , CYr > 0 , CYδr > 0

We can express the yawing moment as

N = q¯SbCn

where Cn is the yawing moment coeﬃcient and is usually expressed as

Cn = Cnβ β + b Cnr r + Cnδr δr (12.5)

2V

Usually,

Cnβ > 0 , Cnr < 0 , Cnδr < 0 ,

Equations of motion. Looking at the free-body-diagram in Figure 12.8 and applying

Newtons’ Second Law we have

D¯ + C¯ + T¯ = ma¯

where a¯ is the inertial acceleration of the vehicle mass center.

Figure 12.8: Free body diagram

If we express V¯ as

V¯ = V wˆ1

then, noting that eω¯ w = (r + β˙)wˆ3, we use the BKE an expression for a¯:

a¯ = edV¯ = wdV¯ + eω¯ w × V¯ = V˙ wˆ1 + (r + β˙)V wˆ2 .

dt dt

230 CHAPTER 12. EQUATIONS OF MOTION FOR GENERAL FLIGHT

Newtons second law now results in (12.6)

mV˙ = −D + T cos β

mV β˙ = −C − T sin β − mV r

where the crosswind C is given by

C cos β = −Y − D sin β

Moment considerations about the yaw axis through the mass center results in

J33r˙ = N (12.7)

and J33 is the moment of inertia of the vehicle about the yaw axis at the CM.

Thus we have a state-space description with two inputs T and δr and three state variables:

V , β and r.

Linearization. Consider a trim condition corresponding to V = V e and β = r = 0. The

linearized ice-racer is described by

δV˙ = −(DV − TV )δV (12.8)

V eδβ˙ = Yβδβ + (Yr − V e)δr + Yδr δr (12.9)

(12.10)

δr˙ = Nβδβ + Nrδr + Nδr δr

where

Yβ = 1 ∂Y = q¯S CYβ <0

m ∂β m

Yr = 1 ∂Y = q¯SbCYr >0

m ∂r 2mV e

Nβ = 1 ∂N = q¯SbCnβ >0

J33 ∂β J33

Nr = 1 ∂N = q¯S b2 Cnr <0

J33 ∂r 2J33V e

and

Yδr = 1 ∂Y = q¯SCYδr > 0

m ∂δr m

Nδr = 1 ∂Y = q¯SbCnδr < 0

J33 ∂δr J33

Note that the dynamics of δV are decoupled from the dynamics of δβ, δr. The A matrix

for the latter dynamics is given by

Yβ/V e Yr/V e − 1

Nβ Nr

12.1. ICE-RACER 231

Example 98 (Cessna 182) Some data:

W = 2650 lbs

J33 = 1967 slug ft2

b = 36 ft

S = 174 ft2

CYβ = -0.3930

CYr = 0.2140

Cnβ = 0.0587

Cnr = -0.0937

CYδr = 0.1870

Cnδr = -0.0645

V e = 150 ft sec−2

Consider the A matrix corresponding to the δβ, δr dynamics, we have

A= −0.1872 −0.9917

9.2719 −1.2104

whose eigenvalues are

−0.6988 ± 2.9888

232 CHAPTER 12. EQUATIONS OF MOTION FOR GENERAL FLIGHT

12.2 Translational kinematics

12.2.1 Cartesian and spherical components of a vector

Suppose (ˆb1, ˆb2, ˆb3) is a orthonormal set of basis vectors. Consider any vector V¯ and let V

be the magnitude of V¯ , that is, V = |V¯ |. If we let V¯I be the projection of V¯ onto the ˆb1 −ˆb3

plane then,

V¯ = V¯I + V2ˆb2 ;

see Figure 12.9.

Figure 12.9: Components

Let β be the angle between V¯ and its projection V¯I where β is considered positive if V2 is

positive. Note that β lies between −90o and 90o and can be regarded as the angle between

V¯ and the ˆb1 −ˆb3 plane. Also,

V2 = V sin β and VI = V cos β

where VI denotes the magnitude of V¯I. Let α be the angle between the vector V¯I and ˆb1

where α is considered positive if it is clockwise as seen by ˆb2; see Figure 12.10. Resolve V¯I

Figure 12.10: Components

into components along ˆb1 and ˆb3 to obtain

V¯I = VI cos α ˆb1 + VI sin α ˆb3

It now follows that V¯ = V1ˆb1 + V2ˆb2 + V3ˆb3

12.2. TRANSLATIONAL KINEMATICS 233

where V1 = V cos β cos α

V2 = V sin β

V3 = V cos β sin α

Inverting these relationships, we obtain

V = V12 + V22 + V32

α = arctan(V3/V1)

β = arcsin(V2/V )

The three scalars, V1, V2, V3, are called the cartesian components of V¯ relative to (ˆb1, ˆb2, ˆb3).

The three scalars V, α, β, are called the spherical components of V¯ relative to (ˆb1, ˆb2, ˆb3).

12.2.2 Aircraft translational kinematics

The conﬁguration of a rigid aircraft relative to the earth can be completely speciﬁed by

specifying the position of the mass center of the aircraft and the orientation of the aircraft.

The translational kinematics of the aircraft describe the motion of the aircraft mass center.

Earth-ﬁxed reference frame. As an inertial reference frame, we will choose a reference

frame

e = (eˆ1, eˆ2, eˆ3)

ﬁxed in the earth where eˆ3 points vertically downward and eˆ1 and eˆ3 lie in the surface of the

(assumed) ﬂat earth; see Figure 12.11.

Figure 12.11: Position coordinates

The position vector r¯. Choosing an arbitrary point ﬁxed in the earth as the origin of

the inertial reference frame, the position of the aircraft mass center relative to this point can

be expressed as

r¯ = p1eˆ1 + p2eˆ2 − heˆ3 (12.11)

234 CHAPTER 12. EQUATIONS OF MOTION FOR GENERAL FLIGHT

where h is the altitude or height of the aircraft. So, the scalars p1, p2, −h are the cartesian

components of r¯ relative to e; see Figure 12.11.

The velocity vector V¯ . If we let V¯ be the velocity of the mass center of the aircraft

relative to earth, then V¯ = p˙1eˆ1 + p˙2eˆ2 − h˙ eˆ3 .

(12.12)

So, we can regard the scalars p˙1, p˙2, −h˙ as the cartesian components of V¯ relative to e.

Aircraft ﬁxed reference frame. We choose a reference frame

b = (ˆb1, ˆb2, ˆb3)

ﬁxed in the aircraft with

ˆb1 (aircraft X-axis) in the plane of symmetry of the aircraft and pointing forward

ˆb2 (aircraft Y-axis) perpendicular to the plane of symmetry of the aircraft and pointing starboard

ˆb3 (aircraft Z-axis) in the plane of symmetry of the aircraft and pointing downward

Thus, the ˆb1 − ˆb3 plane is the aircraft plane of symmetry; see Figure 12.12.

Figure 12.12: Aircraft-ﬁxed frame

It is common to describe the vector V¯ in components taken relative to the aircraft frame.

Relative to the aircraft frame, the cartesian components of V¯ are denoted by u, v, w ; hence,

V¯ = uˆb1 + vˆb2 + wˆb3

The spherical components of V¯ relative to the aircraft frame are denoted by V, α, β and

described as follows; see Figure 12.13

airspeed: V = |V¯ |.

sideslip angle, β is the angle between the velocity vector and the plane of symmetry of the

aircraft; it is considered positive if it points out the right of the aircraft. Thus, β is the angle

between V¯ and the the aircraft-ﬁxed ˆb1 − ˆb3 plane (aircraft X-Z plane). It is considered

positive if v, the ˆb2 (aircraft Y) component of V¯ , is positive.

angle of attack, α is the angle between the projection of the velocity vector onto the the plane

of symmetry of the aircraft and the reference line in the plane of symmetry; it is considered

12.2. TRANSLATIONAL KINEMATICS 235

Figure 12.13: Velocity description

positive if it is clockwise as seen from right of aircraft. Thus α is the angle between the

projection of the velocity V¯ onto the ˆb1−ˆb3 plane and the ˆb1 (aircraft X-axis). It is considered

positive if it is clockwise as seen by ˆb2 (aircraft Y-axis.)

Using the results of the previous section, we obtain the following relationships between

the cartesian components and the spherical components.

u = V cos α cos β

v = V sin β

w = V sin α cos β

and √

V = u2 + v2 + w2

α = arctan(w/u)

β = arcsin(v/V )

236 CHAPTER 12. EQUATIONS OF MOTION FOR GENERAL FLIGHT

12.3 Rotational kinematics

To discuss the rotational kinematics of a rigid body, we need only discuss the rotational

kinematics of a reference frame ﬁxed in the body. So, to discuss the rotational kinematics

of a rigid body, we will consider the rotation of one reference frame relative to another.

12.3.1 Simple rotations

Consider any two reference frames

e = (eˆ1, eˆ2, eˆ3) and b = (ˆb1, ˆb2, ˆb3) .

Suppose reference frame b is obtained from reference frame e by a simple counter-clockwise

rotation through angle θ about eˆ2; see Figure 12.14. Then

Figure 12.14: A simple rotation

ˆb1 = cos θ eˆ1 − sin θ eˆ3 (12.13a)

ˆb2 = eˆ2 (12.13b)

ˆb3 = sin θ eˆ1 + cos θ eˆ3 . (12.13c)

Rotation matrix. The rotation matrix R associated with the above rotation of b relative

to e is deﬁned by

cos θ 0 sin θ

R = 01 0 . (12.14)

− sin θ 0 cos θ

Note that this matrix is obtained by letting its ﬁrst, second, and third columns consist of

the cartesian components of ˆb1, ˆb2, ˆb3, respectively, relative to e. Note also that

RT R = I (12.15)

where I is the 3 × 3 identity matrix. From this it follows that R is invertible and its inverse

is its transpose, that is, R−1 = RT . Also, RRT = I.

Exercise 71 Verify (12.15).

12.3. ROTATIONAL KINEMATICS 237

Coordinate transformations. Consider any vector V¯ (it does not have to be a velocity

vector) and suppose we express it in components relative to e and b, that is,

V¯ = eV1eˆ1 + eV2eˆ2 + eV3eˆ3 and V¯ = bV1ˆb1 + bV2ˆb2 + bV3ˆb3 ,

respectively. Introducing the coordinate vectors,

eV eV1 bV bV1

= eV2 and = bV2 , (12.16)

eV3 bV3

one may readily obtain the following transformation relationship from (12.13):

eV = R bV (12.17)

Premultiplying both sides of the last equation by RT and using RT R = I, we also have

bV = RT eV (12.18)

For example, if V¯ = ˆb1, then

1 cos θ

bV = 0 and eV = 0 .

0 − sin θ

Clearly eV = R bV holds for this vector.

Exercise 72 Verify (12.17).

Angular velocity vector. Recalling the above rotation of b relative to e, we deﬁne the

angular velocity of b relative to e as follows:

eω¯b = θ˙eˆ2 . (12.19)

We can obtain the angular velocity by diﬀerentiating the rotation matrix R with respect

to time. To see this, deﬁne Ω := RT R˙

(12.20)

then, one may readily obtain that

0 0 θ˙

Ω = 0 0 0 . (12.21)

−θ˙ 0 0

Thus, the matrix Ω is skew symmetric (ΩT = −Ω) and its 13 element equals θ˙.

Exercise 73 Using (12.20) as the deﬁnition of Ω where R is given by (12.14), prove that Ω

is given by (12.21).

238 CHAPTER 12. EQUATIONS OF MOTION FOR GENERAL FLIGHT

Basic Kinematic Equation (BKE). Consider any vector V¯ and let

edV¯ and bdV¯

dt dt

denote the time rate of change of V¯ relative to reference frames e and b, respectively, that

is,

edV¯ = eV˙1eˆ1 + eV˙2eˆ2 + eV˙3eˆ3

dt

bdV¯

dt = bV˙1ˆb1 + bV˙2ˆb2 + bV˙3ˆb3 .

Since the cartesian components of the vector e dV¯ relative to e are given by

dt

eV˙ eV˙1

eV˙2

= ,

eV˙3

it follows from relationship (12.18), that the cartesian components of e dV¯ relative to b are

dt

by RT eV˙ . bdV¯

given The cartesian components of the vector dt relative to b are given by

bV˙ bV˙1

bV˙2

= .

bV˙3

If we time-diﬀerentiate the relationship eV = R bV , we obtain

eV˙ = R bV˙ + R˙ bV .

Pre-multiplying both sides of this relationship by RT yields

RT eV˙ = bV˙ + Ω bV (12.22)

where Ω = RT R˙ . Recalling the expression (12.21) for Ω, we see that

Ω bV θ˙ bV3

= 0 .

−θ˙ bV1

Noting that

eω¯b × V¯ = (θ˙ˆb2) × ( bV1ˆb1 + bV2ˆb2 + bV3ˆb3) = (θ˙ bV3) ˆb1 − (θ˙ bV1) ˆb3 ,

we see that Ω bV is the component vector of eω¯b ×V¯ relative to frame b. Hence, relationship

(12.22) yields the basic kinematic equation (BKE):

edV¯ b dV¯

dt dt

= + eω¯b × V¯ (12.23)

12.3. ROTATIONAL KINEMATICS 239

12.3.2 General rotations

Here we look at various ways of representing the rotation of any reference frame b =

(ˆb1, ˆb2, ˆb3) relative to any other reference frame e = (eˆ1, eˆ2, eˆ3).

Rotation matrix (R): Suppose we express each of the b-vectors in terms of the e-vectors,

that is, ˆb1 = R11eˆ1 + R21eˆ2 + R31eˆ3

ˆb2 = R12eˆ1 + R22eˆ2 + R32eˆ3

ˆb3 = R13eˆ1 + R23eˆ2 + R33eˆ3 . (12.24)

We deﬁne the rotation matrix R associated with the rotation of reference frame b relative to

reference frame e by R11 R12 R13

R = R21 R22 R23 .

R31 R32 R33

Note that this matrix is obtained by letting its ﬁrst, second, and third columns consist of

the scalar components of ˆb1, ˆb2 ˆb3, respectively, relative to e. Note also that

RT R = I (12.25)

where I is the 3 × 3 identity matrix. From this it follows that R is invertible and its inverse

is its transpose, that is, R−1 = RT . Also, RRT = I.

Exercise 74 Verify (12.25).

Coordinate transformations. Consider any vector V¯ and suppose we express it in com-

ponents relative to e and b, that is,

V¯ = eV1eˆ1 + eV2eˆ2 + eV3eˆ3 and V¯ = bV1ˆb1 + bV2ˆb2 + bV3ˆb3,

respectively. Now introduce the coordinate vectors,

eV eV1 bV bV1

= eV2 and = bV2 .

eV3 bV3

Using relationships (12.24) one may readily obtain the following transformation relationship:

eV = R bV (12.26)

Premultiplying both sides by RT and using RT R = I, we also have

bV = RT eV (12.27)

For example, if V¯ = ˆb1, then

1 R11

bV = 0 and eV = R21 .

0 R31

Clearly eV = R bV holds here.

Exercise 75 Verify (12.26).

240 CHAPTER 12. EQUATIONS OF MOTION FOR GENERAL FLIGHT

Composition of rotations. If f and g are any two reference frames, we let fRg denote

the rotation matrix associated with the rotation of frame g relative to frame f . We now

demonstrate the following fact: If a, b and c are any three reference frames, then

aRc = aRb bRc . (12.28)

To see this, consider any vector V¯ and let aV , bV and cV be the coordinate vectors of V¯

relative to reference frames a, b and c, respectively. Utilizing relationship (12.26) between

a and b and between b and c, we obtain aV = aRb bV and bV = bRc cV , respectively. It

now follows that

aV = aRb bRc cV .

Now utilizing relationship (12.26) between a and c, we also have aV = aRc cV ; hence

aRc cV = aRb bRc cV .

Since the above relationship holds for any vector cV we must have the desired relationship:

aRc = aRb bRc.

Angular velocity vector. Consider any two reference frames e and b and let R be the

rotation matrix associated with the rotation of b relative to e. Deﬁne the following matrix

Ω := RT R˙ (12.29)

Using the relationship RT R = I, one can readily show that Ω is skew-symmetric, that is,

ΩT = −Ω .

Hence, Ω must have the following structure:

0 Ω12 Ω13

0

Ω = −Ω12 Ω23 .

−Ω23

−Ω13 0

Suppose we let

ω1 := −Ω23 , ω2 := −Ω31 , ω3 := −Ω12 .

Then 0 −ω3 ω2

0

Ω = ω3 −ω1

ω1

−ω2 0

We deﬁne the angular velocity of b relative to e as :

eω¯b = ω1ˆb1 + ω2ˆb2 + ω3ˆb3

Composition of angular velocities. One can readily demonstrate the following fact: If

a, b and c are any three reference frames, then

aω¯c = aω¯b + bω¯c . (12.30)

12.3. ROTATIONAL KINEMATICS 241

Example 99 Consider the machine illustrated in Figure 12.15.

Figure 12.15: Example

242 CHAPTER 12. EQUATIONS OF MOTION FOR GENERAL FLIGHT

Basic Kinematic Equation (BKE). We let

edV¯ and bdV¯

dt dt

denote the time rate of change of V¯ relative to reference frames e and b, respectively; thus

edV¯ = eV˙1eˆ1 + eV˙2eˆ2 + eV˙3eˆ3

dt

bdV¯

dt = bV˙1ˆb1 + bV˙2ˆb2 + bV˙3ˆb3

The scalar components of the vector e dV¯ relative to e are given by eV˙ . Recalling relationship

dt

(12.27), the scalar components of this vector relative to b are given by RT eV˙ . The scalar

bdV¯

components of the vector dt relative relative to b are given by bV˙ . If we time-diﬀerentiate

relationship (12.26) we obtain V˙ e = R bV˙ + R˙ bV

Pre-multiplying both sides of this relationship by RT and using RT R = I yields

RT eV˙ = bV˙ + Ω bV

where Ω = RT R˙ . Recalling the above expression for Ω, we see that

Ω bV ω2 bV3 − ω3 bV2

= ω3 bV1 − ω1 bV3

ω1 bV2 − ω2 bV1

Noting that

eω¯b × V¯ = (ω2 bV3 − ω3 bV2) eˆ1 + (ω3 bV1 − ω1 bV3) eˆ2 + (ω1 bV2 − ω2 bV1) eˆ3

we see that Ω bV is the component vector of eω¯b × V¯ relative to frame b. Hence, we obtain

the basic kinematic equation (BKE):

edV¯ = bdV¯ + eω¯ b × V¯

dt dt

12.3.3 Euler angles

We have seen how to represent rotations with rotations with rotation matrices. Here we look

at a more intuitive way of representing rotations. Consider a rotation of a reference frame

b relative to another reference frame e obtained by the following sequence of three simple

rotations:

Rotation of b about ˆb3 through angle ψ.

Rotation of b about (new) ˆb2 through angle θ.

Rotation of b about (new) ˆb1 through angle φ.

12.3. ROTATIONAL KINEMATICS 243

This is called a 3-2-1 Euler angle sequence.

Letting Rψ denote the rotation matrix for the ﬁrst rotation, we have

cos ψ − sin ψ 0

Rψ = sin ψ cos ψ 0 .

0 01

Letting Rθ denote the rotation matrix for the second rotation, we have

cos θ 0 sin θ

Rθ = 0 1 0

− sin θ 0 cos θ

and the rotation matrix for the composition of the ﬁrst two rotations is

cos ψ cos θ − sin ψ cos ψ sin θ

Rψ Rθ = sin ψ cos θ cos ψ sin ψ sin θ

− sin θ 0 cos θ

Letting Rφ denote the rotation matrix for the third rotation, we have

1 0 0

Rφ = 0 cos φ − sin φ .

0 sin φ cos φ

The rotation matrix R for the overall rotation which is a composition of the three simple

rotations is given by

R = RψRθRφ .

Carrying our the matrix multiplications yields

cos ψ cos θ − sin ψ cos φ + cos ψ sin θ sin φ sin ψ sin φ + cos ψ sin θ cos φ

cos ψ cos φ + sin ψ sin θ sin φ

R= sin ψ cos θ − cos ψ sin φ + sin ψ sin θ cos φ .

cos θ sin φ

− sin θ cos θ cos φ

(12.31)

The angular velocity of b in e is now given by

eω¯b = ω1ˆb1 + ω2ˆb2 + ω3ˆb3 .

where 0 −ω3 ω2

ω3 0 −ω1 = Ω = RT R˙ .

−ω2 ω1 0

Diﬀerentiating R and pre-multiplying it by RT yields

ω1 = φ˙ − ψ˙ sin θ (12.32)

ω2 = θ˙ cos φ + ψ˙ cos θ sin φ

ω3 = −θ˙ sin φ + ψ˙ cos θ cos φ

244 CHAPTER 12. EQUATIONS OF MOTION FOR GENERAL FLIGHT

These relationships may also be written as

ω1 1 φ˙ 0 θ˙ − sin θ ψ˙

ω2 = 0 + cos φ + cos θ sin φ

ω3 0 − sin φ cos θ cos φ

Note that these equations do not depend on the angle ψ. Solving for φ˙, θ˙, ψ˙ in terms of

ω1, ω2, ω3 yields

φ˙ = ω1 + tan θ sin φ ω2 + tan θ cos φ ω3 (12.33a)

θ˙ = cos φ ω2 − sin φ ω3 (12.33b)

ψ˙ = (sin φ/ cos θ)ω2 + (cos φ/ cos θ)ω3 (12.33c)

12.3.4 Aircraft rotational kinematics

Recall the earth-ﬁxed frame e and the aircraft-ﬁxed b frame introduced earlier. We will

describe the orientation of b relative to e with a 3-2-1 Euler angle sequence.

In going from earth frame e to aircraft frame b carry out the following simple rotations

in sequence..

Rotate about ˆb3 (aircraft Z-axis) yaw angle ψ Positive when nose right

Rotate about new ˆb2 (aircraft Y-axis) pitch angle θ Positive when nose up

Rotate about new ˆb1 (aircraft X-axis) roll angle φ Positive when right wing down

Figure 12.16: Yaw

Rotation matrices.

cos ψ − sin ψ 0 cos θ 0 sin θ 1 0 0

Rψ = sin ψ cos ψ 0 Rθ = 0 1 0 Rφ = 0 cos φ − sin φ

00 1 − sin θ 0 cos θ 0 sin φ cos φ

12.3. ROTATIONAL KINEMATICS 245

Figure 12.17: Pitch

Figure 12.18: Roll

cos ψ cos θ − sin ψ cos φ + cos ψ sin θ sin φ sin ψ sin φ + cos ψ sin θ cos φ

cos ψ cos φ + sin ψ sin θ sin φ

R = Rψ Rθ Rφ = sin ψ cos θ − cos ψ sin φ + sin ψ sin θ cos φ

cos θ sin φ

− sin θ cos θ cos φ

Angular velocity. The notation for the angular velocity of the aircraft relative to the earth

is

eω¯b = pˆb1 + qˆb2 + rˆb3

Sometimes the following nomenclature is used:

p roll rate (12.34)

q pitch rate

r yaw rate

From the previous section, we have that

p = φ˙ − ψ˙ sin θ

q = θ˙ cos φ + ψ˙ cos θ sin φ

r = −θ˙ sin φ + ψ˙ cos θ cos φ

246 CHAPTER 12. EQUATIONS OF MOTION FOR GENERAL FLIGHT

Utilizing (12.33) we can obtain the following diﬀerential equations for the rotational kine-

matics:

φ˙ = p + tan θ sin φ q + tan θ cos φ r

θ˙ = cos φ q − sin φ r (12.35)

ψ˙ = (sin φ/ cos θ) q + (cos φ/ cos θ) r

Also, using

eV = R bV

we obtain the following diﬀerential equations for the translational kinematics:

p˙1 = cos ψ cos θ u +(− sin ψ cos φ + cos ψ sin θ sin φ) v +(sin ψ sin φ + cos ψ sin θ cos φ) w

p˙2 = sin ψ cos θ u +(cos ψ cos φ + sin ψ sin θ sin φ) v +(− cos ψ sin φ + sin ψ sin θ cos φ) w

h˙ = sin θ u

− cos θ sin φ v − cos θ cos φ w

(12.36)

Two other reference frames in common use in describing the dynamics of an aircraft are

the stability reference frame and the wind reference frame.

Stability reference frame. Obtained via a clockwise rotation of aircraft frame about ˆb2

(aircraft Y-axis) through α. The corresponding rotation matrix is:

cos α 0 − sin α

Rα = 0 1 0

sin α 0 cos α

Wind reference frame. Rotate stability frame counterclockwise about sˆ3 (stability Z-

axis) through β. Note that wˆ1 (the wind X-axis) is aligned with V¯ . The corresponding

rotation matrix is: cos β − sin β 0

Rβ = sin β cos β 0

0 01

The rotation matrix corresponding to the rotation of the wind frame relative to the

aircraft frame is:

cos α cos β − cos α sin β − sin α

Rαβ = RαRβ = sin β cos β 0

sin α cos β − sin α sin β cos α

The angular velocity of the wind frame relative to the aircraft frame is

bω¯w = −α˙ sin β wˆ1 − α˙ cos β wˆ2 + β˙ wˆ3

12.4. FORCES AND MOMENTS ON AIRCRAFT 247

12.4 Forces and moments on aircraft

12.4.1 Aerodynamic and thrust

Forces. A common way to model the total aerodynamic force F¯A is (12.37)

F¯A = −Dwˆ1 − Cwˆ2 − Lwˆ3

where D is drag, C is called the crosswind and L is lift. In terms of the body-ﬁxed frame b

we have

F¯A = XAˆb1 + YAˆb2 + ZAˆb3

where XA −D

YA = Rαβ −C

ZA −L

Let F¯T be the resultant thrust; usually,

F¯T = T Cǫˆb1 + T Sǫˆb3 .

We will let the resultant aerodynamic and thrust force be given by

F¯A + F¯T = X ˆb1 + Y ˆb2 + Z ˆb3 (12.38)

Note that we are using Y to mean two diﬀerent things. The meaning of Y will be clear

from the context

Moments. We will let the resultant moment about mass center due to aerodynamic and

thrust forces be given by

M¯ A + M¯ T = L♣ ˆb1 + M ˆb2 + N ˆb3 (12.39)

12.4.2 Gravity

W¯ = m g¯ , g¯ = geˆ3

Aircraft-ﬁxed components. Using

0

bg = RT eg and eg = 0

g

we obtain

g¯ = −g sin θ ˆb1 + g cos θ sin φ ˆb2 + g cos θ cos φ ˆb3 (12.40)

248 CHAPTER 12. EQUATIONS OF MOTION FOR GENERAL FLIGHT

12.5 Equations of motion

12.5.1 Translational dynamics

Apply

ΣF¯ = ma¯

to mass center of aircraft to yield

e dV¯

dt

m = W¯ + F¯A + F¯T . (12.41)

Body ﬁxed components. We shall consider components of (12.41) relative to the aircraft

ﬁxed frame. Since

V¯ = u ˆb1 + v ˆb2 + w ˆb3

and

eω¯b = p ˆb1 + q ˆb2 + r ˆb3 ,

application of the BKE yields

edV¯ = bdV¯ + eω¯b × V¯

dt dt

= (u˙ + qw − rv) ˆb1 + (v˙ + ru − pw) ˆb2 + (w˙ + pv − qu) ˆb3 .

Recalling (12.41), (12.38), and (12.40), we obtain

u˙ = rv − qw − g sin θ + X/m

v˙ = −ru + pw + g cos θ sin φ + Y /m (12.42)

w˙ = qu − pv + g cos θ cos φ + Z/m

12.5.2 Rotational dynamics

The rotational dynamics of an aircraft are governed by

ΣM¯ = edH¯ (12.43)

dt

where H¯ is the angular momentum of the aircraft about its mass center and ΣM¯ is the

resultant moment about the mass center of all forces acting on the aircraft. Hence

edH¯ = M¯ A + M¯ T . (12.44)

dt

If we use the aircraft-ﬁxed frame, we have

H¯ = H1ˆb1 + H2ˆb2 + H3ˆb3

12.5. EQUATIONS OF MOTION 249

where H1 p

H2 = Jω and ω = q

H3 r

while J11 0 J13

J = 0 J22 0

J13 0 J33

is the inertia matrix of the aircraft about its mass center and relative to the aircraft-ﬁxed

reference frame. Since the aircraft is symmetric wrt to the ˆb1–ˆb3 plane; J21 and J23 are both

zero. Thus,

H1 = J11p + J13r

H2 = J22q (12.45)

H3 = J13p + J33r

Applying the BKE with eω¯b = pˆb1 + q ˆb2 + r ˆb3 yields

edH¯ = bdH¯ + eω¯ b × H¯

dt dt

= (H˙1 + qH3 − rH2) ˆb1 + (H˙2 + rH1 − pH3) ˆb2 + (H˙3 + pH2 − qH1) ˆb3(12.46)

Substituting (12.46), (12.45) and

M¯ A + M¯ T = L♣ˆb1 + Mˆb2 + Nˆb3 (12.47)

into (12.44) we obtain

J11p˙ + J13r˙ = (J22 − J33)qr − J13pq + L♣

J22q˙ = (J33 − J11)pr + J13(p2 − r2) + M (12.48)

J13p˙ + J33r˙ = (J11 − J22)pq + J13qr +N

250 CHAPTER 12. EQUATIONS OF MOTION FOR GENERAL FLIGHT

12.6 The whole story

The general three dimensional motion of an aircraft can be completely described by the

following 12 diﬀerential equations:

p˙1 = cos ψ cos θ u + (− sin ψ cos φ + cos ψ sin θ sin φ) v + (sin ψ sin φ + cos ψ sin θ cos φ) w

p˙2 = sin ψ cos θ u + (cos ψ cos φ + sin ψ sin θ sin φ) v + (− cos ψ sin φ + sin ψ sin θ cos φ) w

h˙ = sin θ u − cos θ sin φ v − cos θ cos φ w

φ˙ = p + tan θ sin φ q + tan θ cos φ r

θ˙ = cos φ q − sin φ r

ψ˙ = (sin φ/ cos θ) q + (cos φ/ cos θ) r

u˙ = rv − qw − g sin θ + X/m

v˙ = −ru + pw + g cos θ sin φ + Y /m

w˙ = qu − pv + g cos θ cos φ + Z/m

J11p˙ + J13r˙ = (J22 − J33)qr − J13pq + L♣

J22q˙ = (J33 − J11)pr + J13(p2 − r2) + M

+J13p˙ + J33r˙ = (J11 − J22)pq + J13qr + N

This is a state space description with state and control input

p1

p2

h

φ

θ th

el

ψ δa

u

x = and u = ,

v δr

w

p

q

r

respectively.