Basic Newton-Euler Mechanics
Mechanics is a vast and difficult subject. It is virtually impossible to provide a thorough introduction in a couple of sections. Here, the purpose instead is to overview some of the main concepts and to provide some models that may be used with the planning algorithms in Chapter 14. The presentation in this section and in Section 13.4 should hopefully stimulate some further studies in mechanics (see the suggested literature at the end of the chapter). On the other hand, if you are only interested in using the differential models, then you can safely skip their derivations. Just keep in mind that all differential models produced in this section end with the form x˙ = f(x, u), which is ready to use in planning algorithms.
There are two important points to keep in mind while studying mechanics:
- The models are based on maintaining consistency with experimental obser- vations about how bodies behave in the physical world. These observations depend on the kind of experiment. In a particular application, many effects may be insignificant or might not even be detectable by an experiment. For example, it is difficult to detect relativistic effects using a radar gun that measures automobile speed. It is therefore important to specify any simpli- fying assumptions regarding the world and the kind of experiments that will be performed in it.
- The approach is usually to express some laws that translate into constraints on the allowable velocities in the phase space. This means that implicit rep- resentations are usually obtained in mechanics, and they must be converted into parametric form. Furthermore, most treatments of mechanics do not ex- plicitly mention action variables; these arise from the intention of controlling the physical world. From the perspective of mechanics, the actions can be assumed to be already determined. Thus, constraints appear as g(x˙ , x) = 0, instead of g(x˙ , x, u) = 0.
Several formulations of mechanics arrive at the same differential constraints, but from different mathematical reasoning. The remainder of this chapter overviews three schools of thought, each of which is more elegant and modern than the one before. The easiest to understand is Newton-Euler mechanics, which follows from Newton’s famous laws of physics and is covered in this section. Lagrangian me- chanics is covered in Section 13.4.1 and arrives at the differential constraints using very general principles of optimization on a space of functions (i.e., calculus of variations). Hamiltonian mechanics, covered in Section 13.4.4, defines a higher dimensional state space on which the differential constraints can once again be obtained by optimization.
- The Newtonian Model
The most basic formulation of mechanics goes back to Newton and Euler, and parts of it are commonly studied in basic physics courses. Consider a world W
defined as in Section 3.1, except here a 1D world = R is allowed, in addition to 2D and 3D worlds. A notion of time is also needed. The space of motions that can be obtained in the space-time continuum can be formalized as a Galilean group [39]; however, the presentation here will utilize standard intuitive notions
W
of time and Euclidean space. It is also assumed that any relativistic effects due to curvature of the time-space continuum are nonexistent (Newton and Euler did not know about this, and it is insignificant for most small-scale mechanical systems on or near the earth).
Inertial coordinate frames Central to Newton-Euler mechanics is the idea that points in are expressed using an inertial coordinate frame. Imagine locating the origin and axes of somewhere in our universe. They need to be fixed in a way that does not interfere with our observations of the basic laws of motion. Imagine that we are playing racquetball in an indoor court and want to model the motion of the ball as it bounces from wall to wall. If the coordinate frame is rigidly attached to the ball, it will appear that the ball never moves; however, the walls, earth, and the rest of the universe will appear to spin wildly around the ball (imagine we have camera that points along some axis of the ball frame – you could quickly become ill trying to follow the movie). If the coordinate frame is fixed with respect to the court, then sensible measurements of the ball positions would result (the movie would also be easier to watch). For all practical purposes, we can consider this fixed coordinate frame to be inertial. Note, however, that the ball will dance around wildly if the coordinate frame is instead fixed with respect to the sun. The rotation and revolution of the earth would cause the ball to move at incredible speeds. In reality, inertial frames do not exist; nevertheless, it is a reasonable assumption for earth-based mechanical systems that an inertial frame may be fixed to the earth.
W
W
The properties that inertial frames should technically possess are 1) the laws of motions appear the same in any inertial frame, and 2) any frame that moves at constant speed without rotation with respect to an inertial frame is itself inertial. As an example of the second condition, suppose that the racquetball experiment is performed inside of a big truck that is driving along a highway. Ignoring vibrations, if the truck moves at constant speed on a straight stretch of road, then an inertial coordinate frame can be fixed to the truck itself, and the ball will appear to bounce as if the court was not moving. If, however, the road curves or the truck changes its speed, the ball will not bounce the right way. If we still believe that the frame attached to the truck is inertial, then the laws of motion will appear strange. The inertial frame must be attached to the earth in this case to correctly model the behavior of the truck and ball together.
Closed system Another important aspect of the Newton-Euler model is that the system of bodies for which motions are modeled is closed, which means that no bodies other than those that are explicitly modeled can have any affect on the motions (imagine, for example, the effect if we forget to account for a black hole
that is a few hundred meters away from the racquetball court).
Newton’s laws The motions of bodies are based on three laws that were exper- imentally verified by Newton and should hold in any inertial frame:
- An object at rest tends to stay at rest, and an object in motion tends to stay in motion with fixed speed, unless a nonzero resultant8 force acts upon it.
- The relationship between a body mass m, its acceleration a, and an applied force f is f = ma.
- The interaction forces between two bodies are of equal magnitude and in opposite directions.
Based on these laws, the differential constraints on a system of moving bodies can be modeled.
- Motions of Particles
The Newton-Euler model is described in terms of particles. Each particle is con- sidered as a point that has an associated mass m. Forces may act on any particle. The motion of a rigid body, covered in Section 13.3.3, is actually determined by modeling the body as a collection of particles that are stuck together. Therefore, it is helpful to first understand how particles behave.
- Motion of a single particle
Consider the case of a single particle of mass m that moves in = R. The force becomes a scalar, f R. Let q(t) denote the position of the particle in at time t. Using this notation, acceleration is q¨, and Newton’s second law becomes
∈ W
W
f = mq¨. This can be solved for q¨ to yield
q¨ = f/m. (13.50)
If f is interpreted as an action variable u, and if m = 1, then (13.50) is precisely the double integrator q¨ = u from Example 13.3. Phase variables x1 = q and x2 = q˙ can be introduced to obtain a state vector x = (q, q˙). This means that for a fixed u, the motion of the particle from any initial state can be captured by a vector field on R2. The state transition equation is
x˙ 1 = x2
u x˙ 2 = m,
(13.51)
in which x1 = q, x2 = q˙, and u = f. Let U = [ fmax, fmax], in which fmax repre- sents the maximum magnitude of force that can be applied to the particle. Forces
−
8This is the sum of all forces acting on the point.
of arbitrarily high magnitude are not allowed because this would be physically unrealistic.
Now generalize the particle motion to W = R2 and W = R3. Let n denote the dimension of W, which may be n = 2 or n = 3. Let q denote the position of the particle in W. Once again, Newton’s second law yields f = mq¨, but in
this case there are n independent equations of the form fi = mq¨i. Each of these may be considered as an independent example of the double integrator, scaled by
- Each component fi of the force can be considered as an action variable ui. A 2n-dimensional state space can be defined as x = (q, q˙). The state transition equation for n = 2 becomes
x˙ 1 = x3 x˙ 2 = x4
x˙ 3 = u1/m (13.52)
x˙ 4 = u2/m,
and for n = 3 it becomes
x˙ 1 = x4 x˙ 2 = x5 x˙ 3 = x6
x˙ 4 = u1/m
x˙ 5 = u2/m (13.53)
x˙ 6 = u3/m.
For a fixed action, these equations define vector fields on R4 and R6, respectively. The action set should also be bounded, as in the one-dimensional case. Suppose that
U = {u ∈ R | IuI ≤ fmax}. (13.54)
n
Now suppose that multiple forces act on the same particle. In this case, the vector sum
-
F = f (13.55)
yields the resultant force over all f taken from a collection of forces. The resultant force F represents a single force that is equivalent, in terms of its effect on the particle, to the combined forces in the collection. This enables Newton’s second law to be formulated as F = mq¨. The next two examples illustrate state transi- tion equations that arise from a collection of forces, some of which correspond to actions.
Example 13.7 (Lunar Lander) Using the Newton-Euler model of a particle, an example will be constructed for which X = R4. A lunar lander is modeled as a particle with mass m in a 2D world shown in Figure 13.8. It is not allowed to rotate, implying that = R2. There are three thrusters on the lander, which are on the left, right, and bottom of the lander. The forces acting on the lander are
C
shown in Figure 13.8. The activation of each thruster is considered as a binary switch. Each has its own associated binary action variable, in which the value 1 means that the thruster is firing and 0 means the thruster is dormant. The left and right lateral thrusters provide forces of magnitude fl and fr, respectively, when activated (note that the left thruster provides a force to the right, and vice versa).
Figure 13.8: There are three thrusters on the lunar lander, and it is under the influence of lunar gravity. It is treated as a particle; therefore, no rotations are possible. Four orthogonal forces may act on the lander: Three arise from thrusters that can be switched on or off, and the remaining arises from the acceleration of gravity.
The upward thruster, mounted to the bottom of the lander, provides a force of magnitude fu when activated. Let g denote the scalar acceleration constant for gravity (this is approximately 1.622 m/s2 for the moon).
From (13.55) and Newton’s second law, F = mq¨. In the horizontal direction, this becomes
and in the vertical direction,
mq¨1 = ul_f_l − ur_f_r, (13.56)
mq¨2 = uu_f_u − mg. (13.57)
Opposing forces are subtracted because only the magnitudes are given by fl, fr, fu, and g. If they were instead expressed as vectors in R2, then they would be added.
The lunar lander model can be transformed into a four-dimensional phase space in which x = (q1, q2, q˙1, q˙2). By replacing q¨1 and q¨2 with x˙ 3 and x˙ 4, respectively, (13.56) and (13.57) can be written as
and
1
x˙ 3 = m(ul_f_l − ur_f_r) (13.58)
x˙ 4
uu_f_u
= m − g. (13.59)
Figure 13.9: The pendulum is a simple and important example of a nonlinear system.
Using x˙ 1 = x3 and x˙ 2 = x4, the state transition equation becomes
x˙ = x
1
3
fs
x˙ = (u f
3
m
l
l
− u f )
x˙ 2
r
r
= x4
x˙ 4
uu_f_u
= m − g, (13.60)
which is in the desired form, x˙ = f(x, u). The action space U consists of eight
elements, which indicate whether each of the three thrusters is turned on or off. Each action vector is of the form (ul, ur, uu), in which each component is 0 or 1.
．
The next example illustrates the importance of Newton’s third law.
Example 13.8 (Pendulum) A simple and very important model is the pendu- lum shown in Figure 13.9. Let m denote the mass of the attached particle (the string is assumed to have no mass). Let g denote the acceleration constant due to gravity. Let L denote the length of the pendulum string. Let θ denote the angular displacement of the pendulum, which characterizes the pendulum configuration. Using Newton’s second law and assuming the pendulum moves in a vacuum (no wind resistance), the constraint
mLθ¨ = −mg sin θ (13.61)
is obtained. A 2D state space can be formulated in which x1 = θ and x2 = θ˙. This leads to
x˙ 1 = x2
g
(13.62)
x˙ 2 = −L sin x1,
which has no actions (the form of (13.62) is x˙ = f(x)).
A linear drag term kLθ˙ can be added to the model to account for wind resis- tance. This yields
which becomes
mLθ¨ = −mg sin θ − kLθ˙, (13.63)
x˙ 1 = x2
g k (13.64)
in the state space form.
x˙ 2 = −L sin x1 − mx2
Now consider applying a force uf on the particle, in a direction perpendicular to the string. This action can be imagined as having a thruster attached to the side of the particle. This adds the term uf to (13.63). Its sign depends on the choice of the perpendicular vector (thrust to the left or to the right). The state transition equation x˙ = f(x, u) then becomes
x˙ 1 = x2
g k 1
x˙ 2 = −L sin x1 − mx2 + mLuf .
(13.65)
．
Although sufficient information has been given to specify differential models for a particle, several other concepts are useful to introduce, especially in the extension to multiple particles and rigid bodies. The main idea is that conservation laws can be derived from Newton’s laws. The linear momentum (or just momentum) d of the particle is defined as
d = mq˙. (13.66)
This is obtained by integrating f = mq¨ with respect to time.
It will be convenient when rigid-body rotations are covered to work with the moment of momentum (or angular momentum). A version of momentum that is based on moments can be obtained by first defining the moment of force (or
torque) for a force f acting at a point q ∈ W as
n = q × f, (13.67)
in which denotes the vector cross product in R3. For a particle that has linear momentum d, the moment of momentum e is defined as
×
e = q × d. (13.68)
It can be shown that
de
= n, (13.69)
dt
which is equivalent to Newton’s second law but is expressed in terms of momentum. For the motion of a particle in a closed system, the linear momentum and moment of momentum are conserved if there are no external forces acting on it. This is essentially a restatement of Newton’s first law.
This idea can alternatively be expressed in terms of energy, which depends on the same variables as linear momentum. The kinetic energy of a particle is
1
·
T = mq˙ q˙, (13.70)
2
in which is the familiar inner product (or dot product). The total kinetic energy of a system of particles is obtained by summing the kinetic energies of the individual particles.
·
- Motion of a set of particles
The concepts expressed so far naturally extend to a set of particles that move in a closed system. This provides a smooth transition to rigid bodies, which are modeled as a collection of infinitesimal particles that are “stuck together,” causing forces between neighboring particles to cancel. In the present model, the particles are independently moving. If a pair of particles collides, then, by Newton’s third law, they receive forces of equal magnitude and opposite directions at the instant of impact.
It can be shown that all momentum expressions extend to sums over the par- ticles [681]. For a set of particles, the linear momentum of each can be summed to yield the linear momentum of the system as
D = - d. (13.71)
The total external force can be determined as
F = - fi, (13.72)
which is a kind of resultant force for the whole system. The relationship dD/dt = F holds, which extends the case of a single particle. The total mass can be summed to yield
-
M = m, (13.73)
and the center of mass of the system is
1
-
p = mq, (13.74)
M
in which m and q are the mass and position of each particle, respectively. The
expressions D = M p˙
and F = M p¨ hold, which are the analogs of d = mq˙
and
f = mq¨ for a single particle.
So far the translational part of the motion has been captured; however, rotation of the system is also important. This was the motivation for introducing the moment concepts. Let the total moment of force (or total torque) be
N = - q × f, (13.75)
and let the moment of momentum of the system be
E = - q × d. (13.76)
It can be shown that dE/dt = N, which behaves in the same way as in the single- particle case.
The ideas given so far make a system of particles appear very much as a single particle. It is important, however, when conducting a simulation of their behavior to consider the collisions between the particles. Detecting these collisions and calculating the resulting impact forces ensures that correct motions are obtained. As the number of particles tends to infinity, consider the limiting case of a rigid body. In this case, the particles are “sewn” together, which cancels their internal forces. It will be sufficient only to handle the forces that act on the boundary of the rigid body. The expressions for the motion of a rigid body are given in Section
13.3.3. The expressions can alternatively be obtained using other concepts, such as those in Section 13.4.
- Motion of a Rigid Body
For a free-floating 3D rigid body, recall from Section 4.2.2 that its C-space has six dimensions. Suppose that actions are applied to the body as external forces. These directly cause accelerations that result in second-order differential equations. By defining a state to be (q, q˙), first-order differential equations can be obtained in a twelve-dimensional phase space X.
C
Let A ⊆ R3 denote a free-floating rigid body. Let σ(r) denote the body density
at r ∈ A. Let m denote the total mass of A, which is defined using the density as
r
m =
A
σ(r)dr, (13.77)
in which dr = dr1dr2dr3 represents a volume element in R3. Let p ∈ R3 denote the center of mass of A, which is defined for p = (p1, p2, p3) as
r
1
pi =
m A
riσ(r)dr. (13.78)
Suppose that a collection of external forces acts on (it is assumed that all internal forces in cancel each other out). Each force f acts at a point on the boundary, as shown in Figure 13.10 (note that any point along the line of force may alternatively be used). The set of forces can be combined into a single force
A
A
f
Figure 13.10: A force f acting on A at r produces a moment about p of r × f.
and moment that both act about the center of mass p. Let F denote the total external force acting on A. Let N denote the total external moment about the center of mass of A. These are given by
F = - f (13.79)
and
N = - r × f (13.80)
for the collection of external forces. The terms F and N are often called the resultant force and resultant moment of a collection of forces. It was shown by Poinsot that every system of forces is equivalent to a single force and a moment parallel to the line of action of the force. The result is called a wrench, which is the force-based analog of a screw; see [681] for a nice discussion.
Actions of the form u U can be expressed as external forces and/or moments that act on the rigid body. For example, a thruster may exert a force on the body when activated. For a given u, the total force and moment can be resolved to obtain F(u) and N(u).
∈
Important frames Three different coordinate frames will become important during the presentation:
- Inertial frame: The global coordinate frame that is fixed with respect to all motions of interest.
- Translating frame: A moving frame that has its origin at the center of mass of A and its axes aligned with the inertial frame.
- Body frame: A frame that again has its origin at the center of mass of , but its axes are rigidly attached to . This is the same frame that was used to define bodies in Chapter 3.
A
A
The translational part The state transition equation involves 12 scalar equa- tions. Six of these are straightforward to obtain by characterizing the linear veloc- ity. For this case, it can be imagined that the body does not rotate with respect to the inertial frame. The linear momentum is D = mp˙, and Newton’s second law implies that
dD
F(u) =
dt
= mp¨. (13.81)
This immediately yields half of the state transition equation by solving for p¨. This yields a 3D version of the double integrator in Example 13.3, scaled by m. Let (p1, p2, p3) denote the coordinates of p. Let (v1, v2, v3) denote the linear velocity the center of mass. Three scalar equations of the state transition equation are p˙i = vi for i = 1, 2, 3. Three more are obtained as v˙i = Fi(u)/m for i = 1, 2, 3. If there are no moments and the body is not rotating with respect to the inertial frame, then these six equations are sufficient to describe its motion. This may occur for a spacecraft that is initially at rest, and its thrusters apply a total force only through the center of mass.
The rotational part The six equations derived so far are valid even if rotates with respect to the inertial frame. They are just the translational part of the motion. The rotational part can be decoupled from the translational part by using the translating frame. All translational aspects of the motion have already been considered. Imagine that is only rotating while its center of mass remains fixed. Once the rotational part of the motion has been determined, it can be combined with the translational part by simply viewing things from the inertial frame. Therefore, the motion of is now considered with respect to the translating frame, which makes it appear to be pure rotation.
A
A
A
Unfortunately, characterizing the rotational part of the motion is substantially more complicated than the translation case and the 2D rotation case. This should not be surprising in light of the difficulties associated with 3D rotations in Chapters 3 and 4.
Following from Newton’s second law, the change in the moment of momentum
is
N(u) =
dE
. (13.82)
dt
The remaining challenge is to express the right-hand side of (13.82) in a form that can be inserted into the state transition equation.
Differential rotations To express the change in the moment of momentum in detail, the concept of a differential rotation is needed. In the plane, it is straight- forward to define ω = θ˙; however, for SO(3), it is more complicated. One choice is to define derivatives with respect to yaw-pitch-roll variables, but this leads to distortions and singularities, which are problematic for the Newton-Euler formu- lation. Instead, a differential rotation is defined as shown in Figure 13.11. Let v
x
Figure 13.11: The angular velocity is defined as a rotation rate of the coordinate frame about an axis.
denote a unit vector in R3, and let θ denote a rotation that is analogous to the 2D case. Let ω denote the angular velocity vector,
ω = v
dθ
. (13.83)
dt
This provides a natural expression for angular velocity.9 The change in a rotation matrix R with respect to time is
R˙ = ω × R. (13.84)
This relationship can be used to derive expressions that relate ω to yaw-pitch-roll angles or quaternions. For example, using the yaw-pitch-roll matrix (3.42) the conversion from ω to the change yaw, pitch, and roll angles is
γ˙
˙
1
cos α sin α 0
ω1
β = − sin α cos β cos α cos β 0
cos β
α˙
ω2 . (13.85)
ω3
α˙
cos α sin β sin α sin β − cos β
ω3
Inertia matrix An inertia matrix (also called an inertia tensor or inertia oper- ator) will be derived by considering as a collection of particles that are rigidly attached together (all contact forces between them cancel due to Newton’s third
A
9One important issue to be aware of is that the integral of ω is not path-invariant (see Example 2.15 of [994]).
law). The expression σ(r)dr in (13.77) represents the mass of an infinitesimal par- ticle of A. The moment of momentum of the infinitesimal particle is r × r˙σ(r)dr. This means that the total moment of momentum of A is
E = r
A(q)
(r × r˙) σ(r)dr. (13.86)
By using the fact that r˙ = ω × r, the expression becomes
E = r × (ω × r) σ(r)dr. (13.87)
r
A(q)
Observe that r now appears twice in the integrand. By doing some algebraic ma- nipulations, ω can be removed from the integrand, and a function that is quadratic in the r variables is obtained (since r is a vector, the function is technically a quadratic form). The first step is to apply the identity a (b c) = (a c)b (a b)c to obtain
× × · − ·
E = r
A(q)
（(r · r)ω − (r · ω)r）σ(r)dr. (13.88)
The angular velocity can be moved to the right to obtain
E = （r
A(q)
（(r · r)I3 − rrT ）σ(r)dr＼ ω, (13.89)
in which the integral now occurs over a 3 3 matrix and I3 is the 3 3 identity matrix.
× ×
Let I be called the inertia matrix and be defined as
Using the definition,
I(q) = （r
A(q)
（(r · r)I3 − rrT ）σ(r)dr＼ . (13.90)
E = Iω. (13.91)
This simplification enables a concise expression of (13.82) as
dE
N(u) =
dt
d(Iω)
=
dt
dω
= I +
dt
dI
ω, (13.92)
dt
which makes use of the chain rule.
Simplifying the inertia matrix Now the inertia matrix will be considered more carefully. It is a symmetric 3 × 3 matrix, which can be expressed as
I11(q) I12(q) I13(q)
I
I(q) =
12
(q) I22
(q) I23
(q)
. (13.93)
I13(q) I23(q) I33(q)
For each i 1, 2, 3 , the entry Iii(q) is called a moment of inertia. The three cases are
∈ { }
I11(q) = r (r2 + r2)σ(r)dr, (13.94)
A(q)
2
3
I22(q) = r (r2 + r2)σ(r)dr, (13.95)
and
1 3
A(q)
I33(q) = r (r2 + r2)σ(r)dr. (13.96)
A(q)
A(q)
1
2
The remaining entries are defined as follows. For each i, j ∈ {1, 2, 3} such that
i /= j, the product of inertia is
and Iij (q) = −Hij (q).
Hij (q) = r
A(q)
ri_r_j σ(r)dr, (13.97)
One problem with the formulation so far is that the inertia matrix changes
as the body rotates because all entries depend on the orientation q. Recall that it was derived by considering as a collection of infinitesimal particles in the translating frame. It is possible, however, to express the inertia matrix in the body frame of . In this case, the inertia matrix can be denoted as I because it does not depend on the orientation of with respect to the translational frame. The original inertia matrix is then recovered by applying a rotation that relates the body frame to the translational frame: I(q) = RI, in which R is a rotation matrix. It can be shown (see Equation (2.91) and Section 3.2 of [994]) that after performing this substitution, (13.92) simplifies to
A
A
A
dω
×
N(u) = I + ω (Iω). (13.98)
dt
The body frame of must have its origin at the center of mass p; however, its orientation has not been constrained. For different orientations, different inertia matrices will be obtained. Since I captures the physical characteristics of , any two inertia matrices differ only by a rotation. This means for a given , all inertia matrices that can be defined by different body frame orientations have the same eigenvalues and eigenvectors. Consider the positive definite quadratic form xT Ix = 1, which represents the equation of an ellipsoid. A standard technique in linear algebra is to compute the principle axes of an ellipsoid, which turn out to be the eigenvectors of I. The lengths of the ellipsoid axes are given by the eigenvalues. An axis-aligned expression of the ellipsoid can be obtained by defining x′ = Rx, in which R is the matrix formed by columns of eigenvectors. Therefore, there exists an orientation of the body frame in which the inertia matrix simplifies to
A
A
A
I11 0 0
0
I = 0 I22 (13.99)
0 0 I33
and the diagonal elements are the eigenvalues. If the body happens to be an ellipsoid, the principle axes correspond to the ellipsoid axes. Moment of inertia tables are given in many texts [690]; in these cases, the principle axes are usually chosen as the axis of the body frame because they result in the simplest expression of I.
Completing the state transition equation Assume that the body frame of aligns with the principle axes. The remaining six equations of motion can finally
A
be given in a nice form. Using (13.99), the expression (13.98) reduces to [681]
N1(u) I11 0 0 ω˙ 1
N (u)
=
0 I
0
ω˙
+
2 22 2
N3(u)
0 0 I33
0 −ω3 ω2
ω˙ 3
I11 0 0 ω1
(13.100)
ω3 0 −ω1
0 I22 0
ω2 .
−ω2 ω1 0
Multiplying out (13.100) yields
0 0 I33 ω3
N1(u) = I11ω˙ 1 + (I33 − I22)ω2ω3
N2(u) = I22ω˙ 2 + (I11 − I33)ω3ω1 N3(u) = I33ω˙ 3 + (I22 − I11)ω1ω2.
To prepare for the state transition equation form, solving for ω˙
ω˙ 1 = N1(u) + (I22 − I33)ω2ω3 /I11 ω˙ 2 = N2(u) + (I33 − I11)ω3ω1 /I22
（ ）
（ ）
ω˙ 3 = （N3(u) + (I11 − I22)ω1ω2）/I33.
yields
(13.101)
(13.102)
One final complication is that ω needs to be related to angles that are used to express an element of SO(3). The mapping between these depends on the particu- lar parameterization of SO(3). Suppose that quaternions of the form (a, b, c, d) are used to express rotation. Recall that a can be recovered once b, c, and d are given using a2 + b2 + c2 + d2 = 1. The relationship between ω and the time derivatives of the quaternion components is obtained by using (13.84) (see [690], p. 433):
b˙ = ω3c − ω2d
c˙ = ω1d − ω3b
d˙ = ω2b − ω1c.
This finally completes the specification of x˙ = f(x, u), in which
(13.103)
x = (p1, p2, p3, v1, v2, v3, b, c, d, ω1, ω2, ω3) (13.104)
is a twelve-dimensional phase vector. For convenience, the full specification of the state transition equation is
p˙1 = v1 p˙2 = v2 p˙3 = v3
v˙1 = F1(u)/m v˙2 = F2(u)/m v˙3 = F3(u)/m
b˙ = ω3c − ω2d c˙ = ω1d − ω3b
d˙ = ω2b − ω1c (13.105)
ω˙ 1 = N1(u) + (I22 − I33)ω2ω3 /I11 ω˙ 2 = N2(u) + (I33 − I11)ω3ω1 /I22
（ ）
（ ）
ω˙ 3 = （N3(u) + (I11 − I22)ω1ω2）/I33.
The relationship between inertia matrices and ellipsoids is actually much deeper than presented here. The kinetic energy due to rotation only is elegantly expressed as
T = 1 ωT Iω. (13.106)
2
A fascinating interpretation of rotational motion in the absence of external forces was given by Poinsot [39, 681]. As the body rotates, its motion is equivalent to that of the inertia ellipsoid, given by (13.106), rolling (without sliding) down a plane with normal vector Iω in R3.
The 2D case The dynamics of a 2D rigid body that moves in the plane can be handled as a special case of a 3D body. Let R2 be a 2D body, expressed in its body frame. The total external forces acting on can be expressed in terms of
A
A ⊂
a two-dimensional total force through the center of mass and a moment through the center of mass. The phase space for this model has six dimensions. Three come from the degrees of freedom of SE(2), two come from linear velocity, and one comes from angular velocity.
The translational part is once again expressed as
dD
F(u) =
= mp¨. (13.107)
dt
This provides four components of the state transition equation.
All rotations must occur with respect to the z-axis in the 2D formulation. This means that the angular velocity ω is a scalar value. Let θ denote the orientation of . The relationship between ω and θ is given by θ˙ = ω, which yields one more component of the state transition equation.
A
At this point, only one component remains. Recall (13.92). By inspecting (13.101) it can be seen that the inertia-based terms vanish. In that formulation, ω3 is equivalent to the scalar ω for the 2D case. The final terms of all three equations vanish because ω1 = ω2 = 0. The first terms of the first two equations also vanish because ω˙ 1 = ω˙ 2 = 0. This leaves N3(u) = I33ω˙ 3. In the 2D case, this can be notationally simplified to
dE
N(u) =
dt
d(Iω) dω
= = I
dt dt
= Iω˙ , (13.108)
in which I is now a scalar. Note that for the 3D case, the angular velocity can change, even when N(u) = 0. In the 2D case, however, this is not possible. In both cases, the moment of momentum is conserved; in the 2D case, this happens to imply that ω is fixed. The sixth component of the state transition equation is obtained by solving (13.108) for ω˙ .
The state transition equation for a 2D rigid body in the plane is therefore
p˙1 = v1 p˙2 = v2
v˙1 = F1(u)/m
v˙2 = F2(u)/m (13.109)
θ˙ = ω
ω˙ = N(u)/I.
A car with tire skidding This section concludes by introducing a car model that considers it as a skidding rigid body in the plane. This model was suggested by Jim Bernard. The C-space is = R2 S1, in which q = (x, y, θ). Suppose that
C ×
as the car moves at high speeds, the tires are able to skid laterally in a direction
perpendicular to the main axis of the car (i.e., parallel to the rear axle). Let ω denote the angular velocity of the car. Let v denote the lateral skidding velocity, which is another state variable. This results in a five-dimensional state space in which each state is a vector of the form (x, y, θ, ω, v).
The position of the rear axle center can be expressed as
x˙ = s cos θ − v sin θ
y˙ = s sin θ + v cos θ,
(13.110)
which yields two components of the state transition equation. Let ω = θ˙ denote the angular velocity, which yields one more component of the state transition equation. This leaves only two equations, which are derived from 2D rigid body mechanics (which will be covered in Section 13.3.3). The state transition is
x˙ = s cos θ − v sin θ y˙ = s sin θ + v cos θ
θ˙ = ω
(13.111)
ω˙ = (aff − bfr)/I
v˙ = −sω + (ff + fr)/m,
in which ff and fr are the front and rear tire forces, m is the mass, I is the moment of inertia, and a and b are the distances from the center of mass to the front and rear axles, respectively. The first force is
ff = cf (v + aω)/s + φ , (13.112)
（ ）
in which cf is the front cornering stiffness, and φ is the steering angle. The second force is
fr = cr(v − bω)/s, (13.113)
in which cr is the rear cornering stiffness. The steering angle can be designated as an action variable: uφ = φ. An integrator can be placed in front of the speed to allow accelerations. This increases the state space dimension by one.
Reasonable values for the parameters for an automotive application are: m = 1460 kg, cf = 17000, cr = 20000, a = 1.2 m, b = 1.5 m, I = 2170 kg/m2,
and s = 27 m/sec. This state transition equation involves a linear tire skidding model, which is a poor approximation in many applications. Nonlinear tire models provide better approximations to the actual behavior of cars [91]. For a thorough introduction to the dynamics of cars, see [822].