BRIAN L. STEVENS
FRANK L. LEWIS
ERIC N. JOHNSON
AIRCRAFT CONTROL
AND SIMULATION
DYNAMICS, CONTROLS DESIGN,
AND AUTONOMOUS SYSTEMS
THIRD EDITION
Instructor’s Solution Manual to Accompany
Aircraft Control and Simulation
Third Edition,
John Wiley & Sons, 2015
by
Brian L. Stevens
Frank L. Lewis
Eric N. Johnson
CHAPTER 1
Problem 1.2-1: Show that u.(vw)= signed volume of the
parallelepiped defined by u,v,w.
If the cross-product follows a right-handed convention, and u,v,w
form a right-handed set, the formula will yield a positive volume.
—————–
Problem 1.2-2: Show that: u(vw) + v(wu) + w(uv) = 0
Expanding each term using the vector triple product formula
shows that the terms sum to zero:
Problem 1.2-3a: Shortest distance between two trajectories.
Let d be a vector representing the shortest distance between
the trajectories, and defined by:
where t1 and t2 are the times of closest approach of the respective
particles to the other trajectory. The trajectories are
Also, from (A),
Take the dot product of wv and this equation, and make use of
(B),
Therefore, the shortest between the trajectories is given by,
(b) Shortest distance between particles
At any time, t, the vector distance between the particles is:
This solution could also have been derived very simply by keeping
p0 stationary, applying the relative velocity at s0, and solving
the right-triangle for the closest approach to p0.
—————–
Problem 1.2-4: Derive the vector expressions in Fig. 1.2-1.
Problem 1.3-1: Derive the cross-product matrix used in Eqn. 1.3-3.
z
y
xy
xz
v
uu
0
——————-
Problem 1.3-2: Self explanatory
——————-
Problem 1.3-3:
Find the DCM for a NED to FRD rotation with a ZXY rotation
sequence.
),(2atan
2221
cc
We now have a 2-quadrant-only roll angle, and must consider the
desirability of restricting it to ±900.
——————-
Problem 1.3-4: For the rotation Cfrd/ned with all Euler Angles= -900,
find:
(a) Eigenvalues.
1
0
202
000
)(,1
vectorEAIadjfor
T
(c) In the NED System, the E-vector is pointing N and 450 down.
(d) A 1800 rotation around this axis will cause a vehicle that is
level, and heading North, to head vertically downward, with
underside facing North.
—————–
Problem 1.3-5: Rotation matrix from known vectors:
Let the two coordinate systems “a” and “b” share a common
origin, and the known position vectors of the two different
Also, the cross-product of U and V exists, and is linearly
independent of the vectors themselves. Therefore, the relation:
u
provides a third linearly independent matrix equation. These
three equations can be combined as the single matrix equation:
since linear independence guarantees the existence of the
indicated inverse.
(b) Vehicle attitude determination.
A star almanac gives the polar coordinates of stars in an
“inertial” coordinate system with its origin at the Earth’s center
of mass (i.e. an Earth-centered Inertial, or ECI, system). With
——————
SECTION 1.4 Rotational Kinematics.
Problem 1.4-1: Prove that the derivative of the angular velocity
vector is the same in either of the frames that it describes.
From the text, wωww ab
ba
/
Problem 1.4-2: Show that the Centripetal Acceleration vector is
orthogonal to the angular velocity vector.
Proof: Take the dot-product of centripetal acceleration with a
unit vector in the direction of the angular velocity vector:
0||)()(||
ωrωrωω
Therefore, the Centripetal Acceleration is Orthogonal to the
Angular Velocity vector.
——————
Problem 1.4-3: Find the Euler-angle rates, as in Equation (1.4-5),
Following the same steps as the textbook,
0
0
which leads to,
ccs
s
csc
R
Q
P
0
10
0
and the inverse is:
Q
P
tcts
sc
1
0
SECTION 1.5: TRANSLATIONAL KINEMATICS
Problem 1.5-1: Coriolis deflection of a falling mass.
ieePeP m/// 2ωvg
Neglect aerodynamic forces (F≡0); place the NED coordinate origin
at some point, O, at latitude φ, on the surface of the Earth,
E
E
D
v
g
cos
sin
Therefore, if [x y z]T are the NED components of the position
vector from O, the scalar differential equations of motion are:
.. = –ωEy
.sinφ
z
which are nonlinear and coupled. For a particle falling
so we shall neglect the North motion. In addition, the y
.
contribution to the vertical acceleration is negligible. The
equations then have closed-form solutions:
y= ωE cos(φ)(gDt3/3 + t2z
Case (b), dropped from height h: z(0)= -h, z
.(0)=y
.(0)=y(0)=0.
The time of fall is obtained by solving the z equation for z=0:
Substituting for t in the y equation gives the Coriolis deflection
Case (a), thrown vertically: z
.(0)= -u, z(0)=y(0)=y
.(0)=0.
Find the total time of flight by setting z=0 and solving for t:
To compare this with case (b), relate h to u by solving the z
equation with t=u/gD (the time to max height, obtained from
Thus, the deflection for case (b) can be written as:
Therefore, when the particle is dropped from the max. height
reached in (a), the Coriolis deflection on reaching the ground is
only a quarter of the magnitude, and is in the opposite direction.
——————
SECTION 1.6, GEODESY
Problem 1.6-1: Derive the Meridian radius of curvature formula.
By definition, radius of curvature, ρ, is given by:
prime meridian ), and find x in terms of φ.
Differentiate the ellipse equation with respect to x:
Differentiate to find an expression for dx/dφ
Substitute for dx/dφ and sinφ in the radius of curvature formula,
———————
Prob. 1.6-2: Derive the Geocentric Radius formula (1.6-15) and use
this to derive the formulae (1.6-13) for Geocentric Latitude.
(a) Refer to Figure 1.6-3. Geocentric Radius, r, in terms of ECEF
position coordinates of the point, P, above the Spheroid:
(b) From Figure 1.6-3, Geocentric Latitude is,
r
r
Substituting for r, from part (a), and removing the common factor
(N+h), gives,
212 ]sin)2(1[
nn
and, for cosψ,
———————
Problem 1.6-3: Derivation of the ECEF Gravitation model.
Equation 1.6-24 is
2
r
The gravitation components in geocentric coordinates are found
from the gradient of this function in the positive directions of
the respective coordinate axes:
sGG
G
sc
r
r
r
)cot(
0
Since Gecf is independent of longitude, , we will write it in terms
of the geocentric position vector ECEF components
——————–
Problem 1.6-4: Program to evaluate G and g.
An m file to run the gravity and gravitation calculations and
make plots is as follows:
% Program to run the Gravity function 02 April 2004
clear all
rtod=180/pi;
%rphi=input(‘Enter Geodetic Latitude : ‘)/rtod;
The calculations of the gravity and gravitation ECEF and NED
components are performed in the following function:
function [Gx,Gz,gN,gD]= gravity(phi,h)
sspsi5=5*pz*pz/rsq;
Gx= G0*(1+T*(1-sspsi5))*px/r; % Gravitation ECEF components
Gz= G0*(1+T*(3-sspsi5))*pz/r;