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0Russian Journal of Nonlinear Dynamics, 2022, vol. 18, no. 4, pp. 661-680. Full-texts are available at http://nd.ics.org.ru DOI: 10.20537/nd221102
NONLINEAR ENGINEERING AND ROBOTICS
MSC 2010: 70E55, 70E60, 70Q05
Dynamics-Based Piecewise Constant Control
of an Omnivehicle
G. N. Moiseev, A. A. Zobova
We consider the dynamics of an omnidirectional vehicle moving on a perfectly rough horizontal plane. The vehicle has three omniwheels controlled by three direct current motors.
We study constant voltage dynamics for the symmetric model of the vehicle and get a general analytical solution for arbitrary initial conditions which is shown to be Lyapunov stable. Piecewise combination of the trajectories produces a solution to boundary-value problems for arbitrary initial and terminal mass center coordinates, course angles and their derivatives with one switch point. The proposed control combining translation and rotation of the vehicle is shown to be more energy-efficient than a control splitting these two types of motion.
For the nonsymmetrical vehicle configuration, we propose a numerical procedure of solving boundary-value problems that uses parametric continuation of the solution obtained for the symmetric vehicle. It shows that the proposed type of control can be used for an arbitrary vehicle configuration.
Keywords: omnidirectional vehicle, omniwheel, universal wheel, dynamics-based control, piecewise control, point-to-point path planning
1. Introduction
Wheeled mobile platforms serve in lots of areas as transporters or platforms for different tools and manipulators. Their precise positioning and stable control are of great importance for applications and are widely studied [1, 2]. Among the wheeled robots, there is a group of omnidirectional vehicles: their platforms can move instantly in an arbitrary direction without turning about a vertical [3]. A group of them have universal wheels, or omniwheels. Omniwheels
Received May 05, 2022 Accepted October 19, 2022
The work was supported by the Russian Foundation for Basic Research (project no. 19-01-0014).
Georgy N. Moiseev moiseev.georgii@gmail.com Alexandra A. Zobova alexandra.zobova@math.msu.ru
Lomonosov Moscow State University Leninsikie gory 1, Moscow, 119991 Russia
consist of a circular basis that carries a number of light rollers on the periphery. The axes of the rollers are tangent to the basis and lie in the basis plane [4, 5]. In each particular instant, the wheel moves with only one roller in contact with the supporting plane, and can move in transverse direction if the roller rotates. The supporting rollers change each other when the wheel rotates about the axis rigidly mounted in the platform. The platform with such wheels is omnidirectional [3, 6].
The majority of papers investigating omnivehicles can be divided into two vast groups: those concerning the assembly of real prototypes and enhancement of their maneuverability (for example, [7-10], see also [3]) and those studying the ways of control on the prescribed trajectories (for example, [11-16]). The analysis of their dynamics helps enhance the properties of the prototypes and the controllers.
The simplest dynamical model of the omnivehicle is based on the no-slip constraints of wheels: the omniwheel is modeled as a rigid disk which does not slip in the direction of the contact roller axis (so the dynamics of the rollers is neglected). Different methods are used to derive equations that govern the dynamics of the vehicle (for example, analysis of the reaction force [11, 17], Lagrange's multipliers method [6, 12, 14], Chaplygin's equations [18], Ferrers' equations [19], the Appel-Gibbs method [15, 20, 21], and Tatarinov's laconic equations [22, 23]). All of them give three ordinary differential equations that determine the change of the center of mass velocity and angular velocity due to controls. Martynenko et al. [21] studied symmetrical omnivehicle dynamics under DC motors controls with constant voltages. An analytical solution of dynamical equations was obtained there, and it was proven that this motion is globally stable with respect to pseudovelocities. For the inertial motion of the vehicle with an arbitrary assembling of the wheels, integrals have been found and phase portraits have been built [22, 23]. Besides, the stability and bifurcation of steady motions were studied for different parameters of the system under control. Borisov et al. [19] studied the motion of a four-wheeled vehicle with mecanum wheels on a plane and a sphere, describing the integrals of the system and bifurcations.
The extensions of the dynamical model consider the slipping of rollers, their masses and shape [3, 20, 24-27]. However, almost everywhere these modifications are used for the improvement of controllers like in [27]. Kosenko et al. [24] have built a sophisticated numerical model to study the dynamics of an omnivehicle considering the mass of rollers. Adamov et al. [20] numerically studied the dynamic model of a vehicle with four mecanum wheels. This study also considered various friction models for contact between a wheel and a plane.
For evaluating the controllers and for path-planning, the prescribed motions often consist of linear segments, circles or splines (for example, [15, 20, 28, 29]). These trajectories are not inspired by the dynamics of the vehicle itself, but are based solely on the kinematics and the common sense. Time-optimal trajectory planning for symmetric omnivehicles controlled by DC motors with the restriction on possible voltages was considered in [17]. Bang-bang solutions with constant controls turned out to be very close to the optimal ones. However, the wheel masses were neglected in dynamical equations, and other simplifications were made for decreasing the computational costs.
Following [17] and [21], we study the motions under piecewise constant voltages on DC motors, but consider the full dynamical problem for path-planning from one point in phase space into another. Thus, the obtained trajectories are inspired by the vehicle dynamics and depend on its mass-inertial characteristics; the concept could be of interest for heavy prototypes.
The structure of the paper is as follows. In Section 2 we state the problem in general and recall the equations of motion for an arbitrary geometry of the vehicle. In Section 3, we give a general analytical solution for constant control input for the symmetric configuration. The
path-planning task states a boundary-value problem for the equations of motion, which reduces to the solution of two linear algebraic systems on the control inputs. Section 4 presents the properties of piecewise constant control for the symmetric vehicle, including a detailed analysis of two basic types of motions, namely, rotation in place and translation along a straight line, and energy cost investigation. Finally, we compare the proposed method combining rotation and translation with one switch point to the control which splits these two types of motion [18]. Section 5 provides the numerical procedure which allows constructing the control for the vehicle with another geometry for which the analytical solution of the equations of motion is not available. The success of the procedure for all tested tasks implies that the proposed control is possible for other geometries of the omnivehicles. In Section 6, we summarize the results and discuss the prospects of the investigation. The appendices recall the procedure for deriving of the equations of motion in the case of the system with nonholonomic constraints and provide proofs for some propositions stated in the sections above.
2. Kinematics and dynamics of the vehicle
The omnivehicle consists of a horizontal base (platform) and three omniwheels that can rotate independently (Fig. 1a). The wheels can rotate only about their horizontal axes relatively to the base; their planes are vertical. Let (x, y) denote the coordinates of the vehicle's center of mass S (including the platform and the wheels) on a supporting plane in the inertial frame, and let Q be some fixed point on the base platform, and P\ the wheel centers. We will attach the frame Q(nz to the base platform: Qz is vertical, Qn is codirectional to qS, Q( completes the right-handed system (if Q coincides with S, Qn could be chosen arbitrarily); n1, n2 and n3 are perpendicular to the corresponding wheel plane. Let us introduce generalized coordinates: x1, X2 and x3 are the wheels' rotation angles (the positive direction is counterclockwise from the platform's point of view, so the wheels' relative angular velocity is equal to — X.n.); d is a course angle — the angle between Qx and Q( (the platform's angular velocity is dez). The vehicle geometry is defined by the following parameters: ai is the angle between Q£ and QSPp 3. is the angle between Q( and n.; A = \qS|, 5. = \QSi\. Without loss of generality we take the mass of the whole system to be m = 1, and the wheels' radii, R = 1. The problem does not have any time scale, so we will not specify a time dimension in what follows. We will also not specify the dimensions of voltage and voltage-related coefficients.
We model an omniwheel as a rigid disk, assuming that the velocity of its lowest point is perpendicular to the wheel plane [6]. Hence, the vehicle consists of 4 rigid bodies and the state of the system is described by 6 generalized coordinates: (x, y, d, x1, X2, X3). Its geometrical structure is defined by the constant values A, 5., 3. defined above. We consider a symmetric vehicle (Fig. 1b) in Section 3 and a nonsymmetric vehicle (Fig. 1c) with parallel wheels in Section 5.
Let the pseudovelocities be as follows:
v1 = Xcos d + ysin d, (2.1)
v2 = —XX sin d + ycos d, (2.2)
V3 = A0; (2.3)
where A2 is the inertia moment of the system about the vertical axis Sz. Thus, v1 and v2 are projections of the mass center velocity on the Q( and Qn axes, and v3 is proportional to the angular velocity of the platform.
model lel wheels model
Fig. 1. Omnivehicle notation
We assume that the velocity of the wheel contact point is perpendicular to the wheel plane. Expressing this constraint, for each wheel we get [23]
Xi = —v1 sin/3i + v2 cos /3i + (—A sin/i + Si cos( a^ — /))f)i, where i = 1, 2, 3. We will also use a matrix form of constraint equations:
a11 a12 1a 3
X = 2V, 2 = a21 a22 a23
a31 a32 a33
where
an = — sin ¡¡i, Ui2 =cos ¡¡i,
ai3 =
-Asin/j- +£-cos(aj-fa) A
The matrices 2 for the vehicles under consideration are nonsingular.
The wheels are controlled by the torques Mi = (c1 Ui — c2X(i)(—ni), c1 > 0, c2 > 0, where Ui are voltage inputs. The height of the mass center is constant, and so is the gravity force potential. We use the Tatarinov laconic method [30] for systems with constraints homogeneous in time and linear in pseudovelocities to derive differential equations governing the dynamics of the system. (The detailed algorithm is recalled in the Appendix A.) If we use the following notation: A = I + A22*2, a is a vector 3 x 1 with components (v2v3, —v1v3, 0), where A2 is the inertia moment of each wheel about its proper axis Pini and I is the 3 x 3 identity matrix, then the dynamical equations of an omnivehicle will be as follows:
Av = A_1a + c12*U — c2l
(2.4)
The sign * denotes the transpose of a real matrix.
Equations (2.4) are valid for mecanum wheels too, but the matrix 2 will be different. Yet the form of the matrix 2 is crucial for obtaining the analytical result we get in the subsequent chapters and may not work out for the case of mecanum wheels.
+
3. Control of the symmetric vehicle
3.1. Dynamic equations
For the most part we will consider a symmetric vehicle model (Fig. 1b), defined by
A = 0, ¿j = ¿2 = ¿3 = P, n n 7n
Based on system (2.4), the dynamical equations of the symmetric vehicle will be as follows:
\ (k 0 0\ (v\
( v A 1 v2v3
V 2 =L -v1v3
3) 0
0 k 0
+
(w
Wo
(3.1)
where
L
A(2 + 3A2 )
is the inertial characteristic of the vehicle,
Wl~Cl 2 + 3A2 ' 2 + 3A2
are linear combinations of control voltages on DC motors
, ^3 = Cr
pA(U + U2 + U3)
A2 + 3p2A2
3
K = Co
2 + 3A2'
K3 = c2
3 p2 A2 + 3p2A2
(3.2)
are constant coefficients which depend on c2 — the coefficient in the dissipative term of the motors' torque.
3.2. Constant control
We consider the dynamics under control with constant voltages U\ = const, i = 1, 2, 3 on some interval t G [t1, t2] (therefore, W\ = const). The system (3.1) could be integrated analytically. We will use a complex form for the first two equations of the system (3.1) with vL2 = = v1 + iv2, W1,2 = W1 + iW2:
VV1-2 = (-iL - K)V 1.2V3 + W1.2. The solution of system (3.1) will be
/ t \
v1-2(t) =
V
Wh2J eH(ti 'T) dr + Dh2
e
-Hh ,t)
/
V3 = D3e-K3t +
-*,t , W3
(3.3)
(3.4)
v
2
v
3
2
t
1
for integration constants D1,2 = D1 + iD2, D3, where
<p{ti, t) = x(t — t^) — i—----—.
K3
Next, we can integrate Eqs. (2.1)-(2.3) with respect to x, y, 6. We will use a complex form for the mass center coordinates: y = x + iy. Then the solution will be
tW* -D^e'***
e3A
0(t) = ^-(3-5)
K A
t j T
Y(t) = j 'T)+i&(T) Wh2 J 's) ds + D12
ti V ti J
dr + D4.5 (3.6)
for integration constants D45 = D4 + iD5, D6.
Thus, by consequent integration of differential equations, for any constant controls W.\ we had obtained the law of motion with some constant Di (i = 1, ..., 6), uniquely defined by the initial conditions of the mass center coordinates, the course angle x(t0), y(t0), 6(t0) and their derivatives X(t0), yy(t0), 6(t0) (or pseudovelocities v.(t0) since there is a bijective relation (Eqs. (2.1)-(2.3)) fo0 fixed 6(t0)).
3.3. The statement of the control problem
We are going to solve the following control problem: for given initial and terminal positions and velocities of the vehicle and for a given time interval T we find a control that ensures that the vehicle appears at the terminal position with the terminal velocity at time T.
Strictly speaking, for a given initial state x(0) = x0, y(0) = y0, 6(0) = 60, v.(0) = v.0, (i = = 1, 2, 3) we find the control functions W.(t) (i = 1, 2, 3) on some interval t G [0, T] to reach the given terminal state x(T) = xT, y(T) = yT, 6(T) = 6T, v.(T) = vf (i = 1, 2, 3).
The simplest type of control is constant voltages W.. However, for constant voltages during the whole interval of motion, the terminal position of the vehicle may not be possible in the general case. So, we divide the time interval [0, T] into two segments and switch the control voltages at t = tsw simultaneously. We show that it will be sufficient to build the solution for an arbitrary initial and terminal state of the vehicle.
3.4. Piecewise constant control
Based on the analytical solution derived in the previous chapter, here we construct the solution to the problem 3.3 by using piecewise constant control with one switch point tsw
(0 <tsw < T).
By substituting Wi on [0, tsw) and W+ on (tsw, T] in Eqs. (3.1) and then in Eqs. (2.1)-(2.3), we can obtain a trajectory in the phase space (x, y, 6, v1, v2, v3) that connects the initial and terminal states. Without loss of generality, we set x0 = 0, y0 = 0, 60 = 0.
We construct the solution in two steps: since the equations on 6, v3 separate from the other equations, we first solve the boundary problem in the subspace (6, v3) (step I); then we solve the boundary problem in the subspace (x, y, vl, v2) with known 6(t), v3(t) (step II).
Step I. On each time interval [0, tsw) and (tsw, T], we have the solutions (3.4) and (3.5) with the constant controls W± and integration constants D±, D± where the superscripts " —" and " +" correspond to the left and right time intervals, respectively.
Let us substitute the boundary conditions
W-
"3(0) = ^3" + —
Xo
j(T ) = D+e-K3T +
Ko
D-
- D+e-K3T
0(T ) =
K3A
+ D+.
The solution is continuous on the whole interval [0, T] if
W- , W+
K
K
=
t W-- — D-e-K3tsw uswrr 3 q
K3A
+ D- =
t W+- D+ e X3tsw sw 3 -
K3A
+ D+.
(3.7)
(3.8)
(3.9)
(3.10)
(3.11)
(3.12)
Equations (3.7)-(3.12) constitute the system of linear equations for D3 , D+, W3 , W3+, D-, D+:
P>
D \ fv0)
D+ 0
W3- T v3
W3+ 0
D- 0
VTJ
(3.13)
where
P.
1
e-K3 tjs
0 4-
—e-K3tsw e x3T
0
]_ J_
0 0
e~J<3tsw e~"3taw ts
x:iA
x3A
\
0
x3A 0
J_
x3
J_
x3 0
^sw
x3A T
0 0 \
0 0
0 0
1 0
1 -1
0 lJ
Proposition 1. The .system of linear algebraic equations (3.13) has a unique solution if 0 <tsw < T.
Proof. The determinant of Pg is
e-{T+t
A(tsw, T) =
x6A3
^^, where f(t) = (t- teTx3 + Tetx3 - T).
The function f(t) is differentiable, and f(0) = 0, f(T) = 0. The function derivative f'(t) is strongly monotonic on the interval and f'(0) < 0. Thus, the function f (t) does not have any other zeros than t = 0, t = T. Therefore,
A(tsw, T) = 0 for tsw e (0,T). RUSSIAN JOURNAL OF NONLINEAR DYNAMICS, 2022, 18(4), 661 680
□
3
v
0
x. A
x3 A
K3A
Step II. In the subspace (x, y, v1, v2) (using the analytic solution (3.3), (3.6)) the boundary conditions v°,2 = v° + iv°, vf2 = vf + ivf, y° =0, yT = xT + iyT read
v1.2(0) = Dh2 = v0.2,
T
12(T) = W+V^w'T) J e^sw'T) dr + D+2'T) = vT2,
6 S ^ "" tsw
Y (0) = D-5 = 0,
T ( t
Y(T) = J e~^(tsw,T)+id(T)
T
W+2J e^sw ,s) ds + D+2 dr + D+5 = yT
tsw
and the continuity conditions for the instant tsw read
vh2(tsw) = W-^-^'^) J e^(°'T) dr + D-^-^^™) = D+2,
°
V / T \
Y(tsw) = J e-^(°'T)+i0(T) i W- J e^(°'s) ds + D-2 1 dT + D-5 = D+5. Eliminating the constants D- (i = 1, 2, 4, 5), we obtain a linear system on the complex con-
stants W±2 and D±2:
P
fW-2\ (- -v(0,2e~^(o'tsw A
W+2 T
D+2 -V°.2^(0, tsw)
\D4+5/ \ T YT
(3.14)
where
P
($(0, tsw) 0 -1 0 \
0 $(tsw, T) e~^(tsw ,T) 0
r(0, tsW) 0 0 -1
V 0 r(tsw, T) ^(tsw, T) 1 j
„2 t
r(t 1} t2) = J e~^(ti >T)+ie(T) J e^(ti ,s) dsdr,
h h t2
1} t2) = J e~^(ti'T)+id(T) dr, h
h
$(t1, t2) = e~^(ti't2) J e^(t'i'T) dr.
t
i
The linear system (3.14) of 4 complex-valued equations can be rewritten as a linear system of 8 real-valued equations with variables (W—, W2-, W+, W+, D+, D+, D+, D+).
The analysis of the system (3.14) is rather challenging because of its size and sophisticated components r(-, •), •), $(•, •). Nonetheless, all the numerical attempts to solve the system for 0 < tsw < T and arbitrary dT, v0 and v3T have succeeded. It seems that the analogue of Proposition 1 takes place.
Therefore, the problem of reaching an arbitrary terminal configuration from an arbitrary initial state using piecewise constant control with one switch point was reduced to the problem of consequent solution of two systems of linear equations for defined time parameters tsw, T
(0 <tsw < T).
4. Properties of symmetric vehicle motion
Here we study the properties of the obtained solution including basic motions such as rotation about a fixed vertical going through the center of mass and the translational motion along a straight line. Besides, we study the possibility of the consumed energy optimization by varying the switch point; the strategy in the case of bounded controls and the advantage of the dynamics-based control over the control that splits the rotation and translation. While some results are obtained analytically, for the most of them we use numerical experiments with the following values of the parameters:
3 5 1
p=~, A = A = Ci = 0.01, c2 = 0.00025. (4.1)
The inertial characteristics are based on our choice of units. The motor constants are based on the experimental data of several mobile vehicle's direct current drives (Maxon RE-10, RE-13 and RE-30, transmission included). Qt [31] and QCustomPlot [32] were used for app UI and plots. The systems of linear algebraic equations are solved by LU-decomposition implemented in GNU Scientific Library [33]. The Cauchy problems are solved by the adaptive Dorman - Prince method [34]. The problems of integral computation are reduced to the Cauchy problems. The energy expense functional for DC motors is
T
E = J (U)2 + (U2)2 + (U3)2) dt. (4.2)
0
For our type of control the functional (4.2) can be written as
E = tsw ((U-)2 + (U—)2 + (U3-)2) + (T - tsw) ((U+)2 + (U+)2 + (U+ )2).
4.1. Rotation about a vertical
Let us consider a simple motion: the vehicle rotates about a vertical axis through a fixed angle eT from the state of rest to the state of rest, while the center of mass stays in the same place. The corresponding boundary conditions are
v0 = vT = 0, i = 1, 2, 3, xT = 0, yT = 0, eT = 0. (4.3)
The solution of system (3.13) with these boundary conditions is
eK3 tsw — eK3T „
W-7 =-r-— v-^,
3 t —t eK3T - T + eK3tsw 3 '
vsw vsw^ ^ 1 ^
eK3tsw — 1
Wt =-rp-;—x3A0T.
3 tmn - tmne*zT -T + ex3 s™
sw
T±
Controls Wh2 = 0 satisfy the system (3.14) for the boundary condition (4.3); for this control, we get v1 = 0, v2 = 0, x = 0, y = 0, i. e., the center of mass is fixed.
Proposition 2. If the vehicle moves from the state v0 = 0 to the state vj = 0, then the course angle d(t) is either identically equal to zero (if dT = 0), or a strongly monotonic function
(if eT = 0).
Proof. The function v3(t) is a strongly monotonic or a constant function on the constant control intervals [0, tsw) and (tsw, T] (see (3.4)). Considering v3(0) = 0 and v3(T) = 0, v3(t) is identically equal to zero on [0, T] or strongly increases (strongly decreases) on [0, tsw) and strongly decreases (strongly increases) on (tsw, T]. In the second case the sign of v3(t) is constant on [0, T] and it is not equal to zero on (0, T). Thus, from (2.3) we find that d(t) is identically equal to zero (if dT = 0) or is a strongly monotonic function: increasing if dT > 0 and decreasing otherwise. □
From this proof we obtain the following proposition.
Proposition 3. Motion that satisfies the boundary conditions v0'T = 0, d0,T = 0 is the translation motion of a base platform.
Let us consider the optimization problem in terms of minimizing energy expense functional (4.2) by varying the time of switch tsw. For rotation the voltages on the wheels' axes are identical on the wheels
, 2 (A2 + 3p2A2) ,
Ut= V 3 JWt, i = 1,2,3.
Considering the definition (3.2), the expense will be
E(tSun T) = P^ 2 (9T) gx {tsu„ T),
Ci
-1
(eK3tsw - eK3T)2 tsW + (eK3tsw - ^ 2 (T - tsw)
where g {taw, T) = ±---7 "" Tv -^-(4.4)
3 (tsw - tsweK3 - T + eK3^w)2
We analyze the function (tsw, T) numerically for T =1 as a function of tsw G (0, T). It is positive and has one global minimum at tsw = 0.500061. Thus, the optimal time of switch is close to the middle of the time interval, but is slightly moved to the right due to the counter-electromotive force.
4.2. Motion along a straight line
Let us consider another case: a vehicle moves from one point to another without rotation (dT = 0) from the state of rest to a state of rest:
v0 = vT = 0 (i = 1, 2, 3), eT = 0, x(T) = xT, y(T) = yT.
We proved (Proposition 3) that in this case the platform does not rotate. Here we state that the mass center moves along a straight line:
Proposition 4. For the motion from the rest v0 = 0 (i = 1, 2, 3) to the rest vf = 0 (i = 1, 2, 3) without rotation (dT = 0), the mass center trajectory is a straight-line segment or a point (if xT = yT = 0).
The voltages on the wheels surely depend on the direction of the terminal position. However, due to the symmetry of the system, the energy expense depends only on the distance from the initial point:
Proposition 5. For fixed tsw, T the value of energy expense functional depends only on (xT)2 + (yT)2.
See Appendix B for the proofs and illustrations.
4.3. Restrictions on voltages
Previously, we did not impose any restrictions on controls Wi (linear combinations of control voltages Ui). But these restrictions always exist in applications. Let us numerically show that we can satisfy the restrictions by varying "time of acceleration" tsw and "time of deceleration" T -twn.
1618
141.6
27.66
9.528
-4.445
Fig. 2. Maximum voltage value heatmap for v0 = vT = 0, xT = 10, yT = 10, 0T = 2n motion
We plot the heat map (Fig. 2) with intensity
U = max \m\
i =1, 2, 3; 11 s = -, +
to prove our assumption on the problem v0 = vT = 0, xT = 10, yT = 10, dT = 6.28 (the vehicle moves from rest to rest to the point (10, 10) and rotates on a full circle while moving dT ~ 2n). Every point (tsw, T) on a map corresponds to a solution of a problem with fixed times of acceleration and deceleration, the scale for U is shown on the right.
The map shows that for bigger T we have smaller voltages U on the trajectories; for fixed T the minimal U is achieved for tsw ~ j. The maps for other boundary problems have similar properties.
4.4. Comparison of one-switch and three-switches strategies
Using only one switch of constant controls, the vehicle could approach an arbitrary position on the plane. If 9T (xT)2 + (yT) = 0, the vehicle rotates and translates simultaneously, and
the trajectory of the center is, in general, quite complicated (see Fig. 3, lines with 1' and 2' tags).
Splitting the rotation about the center of mass and translational motion is widely used (see, e. g., [18]). To reach an arbitrary position, the vehicle first rotates about a vertical (d goes from 0 to dT, Section 4.1) and then moves along a straight line to xT, yT, Section 4.2. The trajectory of the mass center is a straight line. On both intervals, the piecewise constant control can be used, so the full control has three switch points.
Let us compare the energy expense (4.2) for one-switch and three-switch strategies. For both cases, the instant of the switch is chosen such that the energy expense on the interval is minimal.
For numerical computation we use the parameters (4.1) and T = 10, the results are presented in Table 1. The last column provides the ratio of the energy expenses for the one-switch strategy to the three-switch strategies. We see that combining the motion (for dT ^ 2n) is on average four times more efficient that splitting it into rotation and translation.
Table 1. Comparison of one-switch and three-switches strategies
Terminal state Expenses' Ratio
T T y eT
1 0 0.785 13.05%
1 0 1.57 13.40%
1 0 2.355 15.68%
1 0 3 .14 18.16%
1 0 4 . 71 22.75%
1 0 6.28 26.73%
5 5 0.785 33.48%
5 5 1.57 24.40%
5 5 2.355 20.69%
5 5 3.14 19.15%
5 5 4.71 19.27%
5 5 6.28 19.99%
10 -10 0.785 44.65%
10 -10 1.57 34.78%
10 -10 2.355 30.39%
10 -10 3.14 28.30%
10 -10 4.71 28.46%
10 -10 6.28 26.19%
5. Nonsymmetric vehicle
Let us define the three-wheeled omnivehicle model by its geometrical and inertial characteristics: variables (a1, a2, a3), (P1, (32, (33), (¿1, 52, 53), A and A. They define the 2 matrix, kinetic energy and dynamical equations. The symmetric vehicle is defined by the following values
of these variables: A = 0, 51 = 52 = 53 = p, a1 = f31 = a2 = f32 = a3 = f33 = Let us study another vehicle model (Fig. 1c):
52 > A > 0, 51 = 53 = 5,
a1 = ft = 0,
n
«2=^2=2' a3=Aj=7r-
The inertia moment A is the same as in the symmetric model. We call this model a "vehicle with parallel wheels".
The constraint matrix and the kinetic energy matrix are as follows:
/0
1
1 0
01
I ^
( 1 + A2
A =
A £
A J 0
1 + 2A2
A2
8n -A
Direct computations show that (in contrast to the symmetric case) no equation separates from the others in Eqs. (2.4). The problem of analytical integration of this system for constant voltages is unsolved.
We solve a similar boundary problem of motion with piecewise constant voltages with one switch point for the vehicle with parallel wheels numerically. This problem is reduced to the problem of solving a nonlinear system of six differential equations with variables Ui (three variables in acceleration stage t G (0, tsw) and three variables in slowdown stage t G (tsw, T)). We use the values of voltages for the symmetric vehicle problem with the same boundary conditions as the initial approximation.
The problem is solved by a numerical algorithm which computes voltages of motors for an arbitrary vehicle (although in this paper we demonstrate only examples of solution for vehicles with parallel wheels) using step-by-step increments of variables (a1, a2, a3), (P1, (32, (33), (51, 52, 53), A, A in the described order, starting from the symmetric vehicle model. The step size is dynamic and auto-decreases if Newton's method diverges. We also use some empiric additions to boost computations. It helps in the case of vehicles with parallel wheels, but may result in the contrary for some other models. These empiric additions are as follows:
• increment of ai also increases /3i by the same value for i = 1, 2, 3;
• increment of a1 also decreases a3 by the same value;
• increment of /31 also decreases (33 by the same value;
• increment of 51 also increases 53 by the same value;
• increment of 52 also increases A by the same value.
0
4 6
x
(a) xT = 10, yT = 10, 0T = 0 (trajectories are the same)
S>
-20
10 20
x
(c) xT = 40, yT = -20, 6T = 9.5
(e) xT = 10, yT = 10, 9T = 100
5s
0.0002 0.0001 0
-0.0001 -0.0002
/1 \ I I I
/ \ I I I I
1 1 1 — \ I > 1
. 1 . 1 . 1 • 1
■ ... i ..... .
4 6
x
(b) xT = 10, yT = 0, eT = 0
10 0
56-10 -20 -30
10
20
30
(d) xT = 10, yT = 10, 6T = 30
8 10
1
\
k ______-©V^-
_ VTA : 7
.... ¡I .... i ... .
(f) xT = 10, yT = 10, dT = 200
Fig. 3. Comparison of the center of mass trajectories: the vehicle with parallel wheels (continuous, 1 and 2) vs. the symmetric vehicle (dashed, 1' and 2')
The trajectories in Fig. 3 were obtained by using this algorithm for the following boundary values: x(0) = 0, y(0) = 0, 0(0) = 0, = 0, i = 1, 2, 3, tsw = 5, T = 10, with terminal
coordinates and angles depicted in the figures and their subscripts. The trajectory of a vehicle with parallel wheels is shown as a continuous line, where the line tagged with 1 is the motion before switch and the line tagged with 2 is the motion after switch; the corresponding trajectory of the symmetric vehicle is shown as a dashed line, where the line tagged with 1' is the motion before switch and the line tagged with 2' is the motion after switch.
6. Conclusion
We consider the motions of omnivehicles controlled by direct current motors. For a symmetric vehicle, the problem of reaching an arbitrary terminal state from an initial state using piecewise constant control with one switch point is reduced to the problem of solving two systems of linear algebraic equations. We investigated analytically two basic solutions — translations without rotation and rotations in place. To minimize the energy expenses, one should switch the controls shortly after the middle of the time interval.
Restrictions (if any) on the maximum values of control voltages could be satisfied by the motion time's increase. The motions with one switch point are more energy-efficient for moderate terminal course angles than a sequential realization of splitted rotation and translation with three switch points.
The idea of a numerical algorithm for solving the same boundary problem for an arbitrary construction of a three-wheeled omnivehicle is proposed. It converges for the vehicle with parallel wheels.
The proposed method could be used for the omnivehicle's path planning and enlarges the collection of trajectories with energy- and time-efficient ones. For applications, it could be combined with feedback control to adjust the vehicle trajectory. We plan to further expand this theme in future works.
Appendix A: Derivation of dynamics equations
We use the Tatarinov laconic method to derive differential equations [30] that govern the dynamics of the system. Let L(qi, qi, t) be a Lagrange function of a system with generalized coordinates qi (i = 1, ..., n) and differential constraints iji = vi(vj, q), where Vj (j = 1, ..., m)
are pseudovelocities with independent variations, the rank of is maximal. Then the equations
for a system with constraints homogeneous in time and linear in pseudovelocities can be written
as
^ + iP,,L>} = U±„]Pj\+±Qj^, , = (A.1)
The index • denotes substitution of constraints, Qj is a generalized force corresponding to the
coordinate q^ and {/, g} = [m~'§f~ ~ a Poisson bracket on the variables (qpj,
i_1 1 1 11/
where pi is a formal impulse. We can obtain functions Pi from the equations
Y.P Vj = £ Pivi(vJ, j_i i_i
i.e., Pi is a linear combination of formal impulses pi with coefficients depending on generalized
' dL '
coordinates. It is needed to substitute pi = j in Eqs. (A.l) after computing the Poisson brackets.
The kinetic energy of a three-wheeled omnivehicle will be as follows:
2T = x2 + y2 + A262 + A2 (x2 + x2 + XX3), (A.2)
where A2 is the inertia moment of each wheel about its proper axis and A2 is the inertia moment of a vehicle about the vertical axis that goes through the mass center. By substituting the constraints into the kinetic energy, we get
2T* = v* (E + 2*2) v = v* Av,
where 2 is a constraint matrix from XX = 2v. The sign * is used to denote real matrix transposition.
The height of the mass center is constant due to model specification, and so is the gravity force potential. Thus, the Lagrangian function L coincides with the kinetic energy T (L* will be the Lagrangian function for T*).
By substituting the constraints, we get
3
i=1
3
y cosT
i=1
3
p* = j; + J2'Pi(Ji 3'
i=1
where aij is 2 matrix components.
The Lagrangian function L* does not depend on generalized coordinates and impulses, thus {Pi, L•} = 0. For {Pi, Pj} we have
= + + = 0, 17 2 dpx dx dx dpx
P1 = px cos 6 + py sin 6 + piai
i=1
3
P2 = -Px sin 6 + Py cos 6 + Y^ Pi°i2,
{Pl, P3} =
dP1 dP3 -px sin 6 + py cos 6
d6 dps A
{P2, P3} =
dP2 dP3 -px cos 6 — py sin 6
d6 dpd A
Let us obtain the expressions of px, py in terms of px = = x, Py = %j = y, where L is the Lagrangian function in the form of (A.2). Thus,
{P1, P2} = 0,
{P2, P3} =
The wheels are controlled by torques Mi = (c1 Ui — c2Xci)(—ei), c1 > 0, c2 > 0, where ei =
sP.
= Elementary work can be written as ¿u^), where = 50ez — ?>Xiei- Thus,
Qx. = ci Ui — c2 Xi-Finally, we get the following equations:
Av = A-1a + c1 — c2S*Sv, where a is a 3 x 1 vector with components (v2v3, —v1v3, 0).
Appendix B: Motion along a straight line
Proposition. For the motion from the rest v0 = 0 (i = 1, 2, 3) to the rest vf = 0 (i = = 1, 2, 3) without rotation (dT = 0), the mass center trajectory is a straight-line segment or a point (if xT = yT = 0).
Proof. Let us recall that, according to Proposition 2, the conditions v0 = vT = 0, dT = 0 lead to v3(t) = 0, d(t) = 0 on t e [0, T].
Then we can rewrite Eqs. (2.1), (2.2) and (3.1) as
v1 = —kv1 + W1, (B.1)
v2 = —k v2 + W2, (B.2)
X = v1, (B.3)
y = V2. (B.4)
After integrating Eqs. (B.1)-(B.2) on the constant control intervals considering the boundary conditions, the solution will be
—^ (l — e~xt), te[0,t8W],
V1.2(t) = ' K
W+
K
The continuity conditions will be as follows:
f (l-e-^), te[tsw,T}.
W'2 (1 — e'^) = W+2 (1 — e~K(tsw~T^. (B.5)
After integrating Eqs. (B.3) and (B.4) for fixed xT and yT, we obtain the following solution:
x(t) = TK+ (B.6)
Wr ^Lit), K t e [0, tsw],
K t e [tsw, T])
t e [0, tsw],
W+, T K t e [tsw, T] ,
y(t) = (B.7)
/ e-xt _ 1
f+(t) = t - T +
K
e-K(t-T) _ 1 '
K
The continuity conditions for the trajectory will be as follows:
W-f-(tsw ) = W+f+(tsw) + kxt , (B.8)
W-f-(tsw ) = W2+f+(tsw) + KyT. (B.9)
By solving Eqs. (B.5), (B.8)-(B.9), we finally obtain the control values:
W- = k xt , W+ = k+xT,
1 — T ' + + T ' (B.10)
W2- = k-yT, W2+ = k+y ,
where
k=
k (eKtsw - eKT)
TeKtsw - T - tsweKT + ts
k (eKtsw - 1)
^ =_
+ Te^w-T-tswe-T + tsw
The denominator is not equal to zero, as follows from the proof of Proposition 1.
By substituting the controls W±2 into the trajectory (B.6)-(B.7) and eliminating time t, we obtain a straight-line segment formula (if (xT)2 + (yT)2 = 0):
yTx(t) - xTy(t) = 0,
which reduces to a point if xT = yT = 0. This completes the proof. □
Proposition. For fixed tsw, T the value of the energy expense functional is equal for any problem with boundary conditions v0 = vf = 0 (i = 1, 2, 3), dT = 0, (xT)2 + (yT)2 = const, i. e, the amount of spent energy on straight-line motion does not depend on the direction.
Proof. Controls (B.10), W± = 0 carry out the desired motion. Let us write down the value of a functional on this motion with the results of the proof of Proposition 4. Let us write down the U± using its definition W = cA- 12T U:
2 + 3A
U± U±
2q
-i^i \ 3 /
\U*J
At the same time, we have the following equation:
(Utf + + (Utf = ¿j (l + ^A2)2 ((^f)2 + (^)2).
Then the expense functional considering the definition (3.2) can be written as
3c2
E(tsw, T) = ({xTf + (yT)2) gx(tswlT),
where
, (, T-. _ (e**™ ~ exT?tsw + (eMsw - 1 )2(T - J
j - (t№ - - T + e^)2 ' ( j
which proves the statement. □
Notice that gK(tsw, T) function (B.11) is similar to gKg(tsw, T) function (4.4) from the previous subsection. It reaches its minimum at tsw = 0.50003. Again, as in the previous case, the optimal time switch is slightly shifted to the right from the middle of the time interval.
Conflict of interest
The authors declare that they have no conflict of interest.
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