Minimum-time lateral interception of a moving target by a Dubins car
MMinimum-time lateral interception of a moving target by aDubins car (cid:63)
Maksim Buzikov a , Andrey Galyaev a a V. A. Trapeznikov Institute of Control Sciences of Russian Academy of Sciences, Moscow, Russia
Abstract
This paper presents the problem of lateral interception by a Dubins car of a target that moves along an a priori knowntrajectory. This trajectory is given by two coordinates of a planar location and one angle of a heading orientation, every one ofthem is a continuous function of time. The optimal trajectory planning problem of constructing minimum-time trajectories fora Dubins car in the presence of an a priory known time-dependent wind vector field is a partial case of the considered problem.Using the properties of the three-dimensional reachable set of a Dubins car, it is proved that the optimal interception pointbelongs to a part of an analytically described surface in the three-dimensional space. The analytical description of the surfacemakes it possible to obtain 10 algebraic equations for calculating the parameters of the optimal control that implements theminimum-time lateral interception. These equations are generally transcendental and can be simplified for particular cases oftarget motion (e.g. resting target, straight-line uniform target motion). Finally, some particular cases of the optimal lateralinterception validate the developments of the paper and highlight its necessity to consider each of 10 algebraic equations inthe general case.
Key words:
Dubins vehicle; Moving target; Minimum-time interception; Path-planning; Optimal control.
Missions for autonomous ground, aerial and marine ve-hicles are extremely diverse. In most scenarios, the mis-sion involves a mobile vehicle path/trajectory planning(PP/TP) problem to be solved. Formalization of theproblem requires choosing a model of system dynamicsthat would satisfy several requirements at once. Firstof all, reference trajectories given by model dynamicsshould be close to feasible ones in practice. Secondly, thecomputing of these trajectories should be a time-rapidand memory-efficient process for an on-board computer.The analytical results obtained from such models tendto speed up the computing of trajectories and reducememory usage.Dubins car is a popular model of a non-holonomic vehiclethat moves only forward on the plane along trajectoriesof the bounded curvature. The optimal PP-problem offinding a minimum-length path with bounded curvature (cid:63)
The research was partially supported by Russian ScienceFoundation grant (No.21-19-00128). This paper was not pre-sented at any IFAC meeting.
Email addresses: [email protected] (MaksimBuzikov), [email protected] (Andrey Galyaev). was for the first time considered by Markov in the con-text of the railway track construction [22]. In [13], Du-bins solved a similar problem, when the orientation of theline is fixed at both ends, in contrast to Markov’s state-ment, when it is arbitrary at the final point and givenat the starting one. In [17], Isaacs considered pursuit-evasion games, and in one of the problems he used theequations of dynamics that corresponds to the motionwith a constant velocity and bounded maneuverability.Isaacs probably was the first who introduces the term”car” for such a model. In modern literature, such amodel is called ”Dubins car”. A complete analytical so-lution of the Dubins problem is obtained in [30,6]. In thecase when the distance between the initial and terminalpoints is relatively large it is possible to construct a log-ical classification scheme for time-rapid computation ofthe optimal control [33]. A large number of modificationsto the Dubins problem has been considered in the litera-ture. For example, recent studies devoted to the DubinsPP-problem deal with the case when the trajectory nec-essarily passes through a fixed intermediate point [10].Also recently, the problems of constructing the shortestpath to a circle [9] and the shortest path via a circularboundary [18] for Dubins car have been solved. Theseanalytical results make it affordable to implement effi-cient on-board algorithms.
Preprint submitted to Automatica 10 February 2021 a r X i v : . [ m a t h . O C ] F e b he Dubins’ model study is not limited to the PP/TPproblems. Some theoretically important works are de-voted to the study of the Dubins car reachable set. A de-tailed analysis of the planar evolution of the Dubins carreachable set is given in [11]. The evolution of a three-dimensional reachable set was studied in [29], and in [28],where a partial analytical description of the evolutionof the boundary of the set was obtained. From a practi-cal point of view, the reachable sets can be used, for ex-ample, to recover the vehicle trajectory from inaccuratemeasurements [5] and in the collision avoidance problem[31]. An analytical description of the planar reachableset made it possible to solve the problem of intercept-ing a moving target by a Dubins car at an unspecifiedinterception angle [8], as well as to clarify the necessaryconditions for interception along a geodetic line (relaxedDubins path) from [26].This paper solves the problem of minimum-time lateralinterception of a moving target by Dubins car when theplanar coordinates and the course angles of the targetand Dubins car must coincide at the interception mo-ment. The case of unlimited maneuverability, where thecriterion of optimality is the sum of the time and energyspent on maneuvers, is analyzed in [14]. A large num-ber of works deal with the problem of achieving the finalstate in the presence of wind instead of intercepting amoving target problem. The case of the presence of con-stant wind and the straight-line uniform motion of thetarget case are equivalent in the sense that there is a co-ordinate transformation in which one case passes into an-other [24]. Some properties of the constant wind case so-lution are investigated in [23]. The construction of a nu-merical scheme for computing the case of time-variablewind can be found in [25]. The first analytical resultsfor the constant wind case were obtained in the assump-tion of a sufficient relative distance between the targetand Dubins car when considering a selection of aerobi-ological samples mission [35]. Later, the same authorsobtained analytical results for the general case of the ini-tial target location [34]. The main result of these papersis that for some configurations there are explicit expres-sions for calculating the parameters of optimal control,and for other scenarios, in the worst case, it is neces-sary to solve a system of two transcendental algebraicequations with two unknowns. A numerical method forsolving a similar system of transcendental equations hasbeen considered in [2]. It is shown in [3] that the systemcan be simplified to a transcendental equation with oneunknown. Analytical results on the problem of achiev-ing by Dubins car the desired state in the presence ofconstant wind are summarized in [4]. In a recent paper[27], it was shown that by losing optimality, admissiblecontrol can be obtained without solving transcendentalequations for the constant wind case.Direct addressing of the problem of lateral interceptionof a target with straight-line uniform motion can befound in [15]. This article deals only with the case of intercepting a target at a right angle under the assump-tion of a large enough initial distance between the targetand Dubins car. The paper [16] generalizes this problemto the case of an arbitrary given angle of interception.The paper [21] provides a solution to a similar problem,but for the case of a target moving in a circle. It is as-sumed that the intercept velocity vectors should becomecollinear. A more complex case of a target moving alongthe racetrack path is discussed in [20].The statement of the problem considered in this papergeneralizes all of the above statements. No restrictionsare imposed on the target’s movement, except for thecontinuity of the trajectory on time. The area of the prac-tical application of such developments is significantlywide. It includes the tasks of constructing reference tra-jectories for the various controllable moving plants inthe presence of wind [32,12], the task of refueling an air-craft and other vehicles [7,21], the task of landing UAVson a moving platform [1]. The same task can be solved,including wind presence, in military applications [15],such as tasks of interception and destruction of targets,moving along program trajectories.This paper is organized as follows: Section 2 providesthe mathematical statement of the problem of lateralinterception of a moving target by a Dubins car andthe implied terminology. Next, Section 3 describes theposition of the optimal interception point and refinessome properties of the three-dimensional reachable set.Then Section 4 is devoted to obtaining equations forcalculating the optimal interception time moment anddetermining all parameters of the optimal control bythe time moment. Section 5 addresses considering someexamples of the optimal interception. Finally, Section 6gives conclusions and future work. In this section, we formulate the problem of lateral inter-ception of a moving target as an optimal control prob-lem. The subscript E used throughout this paper refersto the target’s configuration (endpoint for a Dubins car).We use the notation P ∈ R × S for a configuration,which is a planar position and orientation triplet. Thespace S consists of the real numbers, but the equality on S differs from the equality on R . For every ϕ, ψ ∈ S theequality ϕ = ψ holds on S when there exists a number k ∈ Z such that the equality ϕ = ψ + 2 πk holds on R .Also, we suppose that the absolute value of ϕ ∈ S calcu-lates according to the next rule | ϕ | = min(mod( ϕ, π ) , π − mod( ϕ, π )) . Modulo operation is supposed to be a remainder of divi-sion: mod( a, b ) = a − b (cid:98) a/b (cid:99) .
2e consider throughout this paper the following metricon the space R × S ρ ( P, P ) = (cid:112) ( x − x ) + ( y − y ) + | ϕ − ϕ | , where P = ( x, y, ϕ ) ∈ R × S , P = ( x , y , ϕ ) ∈ R × S . Dubins’ model describes a vehicle’s planar motion withthe constant-speed and the limited maneuverability [6].We suppose that time and length scales are chosen insuch a way that the velocity and the minimum turn-radius of the Dubins car are units. That is, the Dubinscar motion is described by the system of equations ˙ x = cos ϕ, ˙ y = sin ϕ, ˙ ϕ = u, (1)where ( x ( t ) , y ( t )) ∈ R are car’s Cartesian coordinates(see Fig. 1), ϕ ( t ) ∈ S is the heading of the car’s forwardvelocity (we use counterclockwise system and supposethat x-axis is a reference direction), u ( t ) ∈ [ − ,
1] isthe control input at time t . Without loss of generality,we assume that every considered scenario starts at thetime moment t = 0 with the initial configuration of theDubins car being( x (0) , y (0) , ϕ (0)) = (0 , , π/ . (2) The configuration of the target is declared by a contin-uous on time vector-function E = ( x E , y E , ϕ E ), where( x E ( t ) , y E ( t )) ∈ R are the target’s Cartesian coordi-nates (see Fig. 1), ϕ E ( t ) ∈ S is the target’s orientation.Generally speaking, the heading of the target’s veloc-ity should not be the same as the target’s orientation.Moreover, the target’s velocity value may not exist when( x E , y E ) is not differentiable. The lateral interceptionoccurs when car and target configurations are coincidewith each other( x ( T ) , y ( T ) , ϕ ( T )) = ( x E ( T ) , y E ( T ) , ϕ E ( T )) , (3)where T ∈ R +0 is an interception time.We examine the optimal-control problem, where the goalof the Dubins car control is to minimize the interceptiontime T , as it is pointed below J [ u ] def = T (cid:90) dt → min u . -2 -1 0 1 2 3 x -10123 y ϕ ( t ) ( x (0) , y (0))( x ( t ) , y ( t )) ϕ E ( t )( x E (0) , y E (0))( x E ( t ) , y E ( t )) x ( T ) = x E ( T ) y ( T ) = y E ( T ) ϕ ( T ) = ϕ E ( T ) Fig. 1. The Dubins car planar path is a solid line. It startsfrom the configuration (0 , , π/
2) and ends on (3 , , π/ ϕ E ( t ) = π/ According to Theorem 1 in [11], the piecewise-constantcontrol functions are enough to produce interception tra-jectories in the considered problem.
Throughout this paper, we use the common notation andsome known results relating to the Dubins’ model. Simi-lar to [6] we suppose that C denotes an arc of a unit circleand S denotes a straight line segment. Sequences CSC,CCC are assigned to the trajectory consisting only of C,S consecutive parts. For definiteness, the controls raisingthese trajectories are supposed to be right-continuouspiecewise-constant functions [29]. If C corresponds to aclockwise or counterclockwise turn, it will be replacedby R or L respectively. So L corresponds to the control u ( t ) = 1 and R corresponds to u ( t ) = −
1, but S matchesthe control action u ( t ) = 0.Further analysis significantly uses the reachable setproperties. Let R ( t ) be the three-dimensional reachableset of the Dubins car at time t ∈ R +0 . It means that forany point P ∈ R ( t ) there exists an admissible control,the usage of which leads to the point P at time t . Wenote that R : R +0 → R × S is a multi-valued mapping.According to general results of optimal control theory[19] the set R ( t ) is closed and bounded. B ( t ) denotesa boundary of the set R ( t ). The main result of [29] isthat CSC- and CCC- trajectories are enough to reachany point of B ( t ) at time t , but not all trajectories ofthese types lead to B ( t ), some of them lead to the in-ternal points of R ( t ). Let E ( t ) be a set of all possible3erminations of CSC- and CCC- trajectories at time t .It follows from above that B ( t ) ⊂ E ( t ) ⊂ R ( t )at any time moment t ∈ R +0 . Further, we use thepartition E ( t ) = E CSC ( t ) ∪ E CCC ( t ), where E CSC ( t )corresponds to the CSC-trajectories terminal pointsand E CCC ( t ) consists of the CCC-trajectories terminalpoints. This section addresses new properties of mappings B , E , R and contains the analytical description of the map-ping E . These properties allow to determine the optimalinterception point configuration and to reduce the spaceof optimal trajectory candidates. First of all, we have to succinctly describe controls thatallow reaching points of E ( t ). Let B = {− , } be a bi-nary set. Lemma 1
Let s, σ ∈ B , τ ∈ [0 , π ) , τ ∈ [ τ , + ∞ ) .Then any point of E CSC ( t ) is attainable using the control u s,σCSC ( t ; τ , τ ) = s, t ∈ [0 , τ ) , , t ∈ [ τ , τ ) ,σ, t ∈ [ τ , + ∞ ) , (4) and any point of E CCC ( t ) is attainable using the control u sCCC ( t ; τ , τ ) = s, t ∈ [0 , τ ) , − s, t ∈ [ τ , τ ) ,s, t ∈ [ τ , + ∞ ) , (5) where τ − τ ∈ [0 , π ) for the CCC-trajectory case. PROOF.
Assume initially that a CSC-trajectory ter-minating at some point P ∈ E CSC ( t ) has cycles at thefirst circle arc. Consider a new trajectory obtained fromthe previous trajectory by shifting these cycles to thelast circle arc. The new trajectory corresponds to thecontrol (4) and terminates at the point P . The same rea-soning is applicable when we consider the case of CCC-trajectories with possible cycles. These cycles have to beshifted from the first and second circle arcs to the last.Such a new trajectory corresponds to the control (5) andterminates at the same point as the previous one does.Using this lemma, now we may obtain the analytical de-scription of the set E ( t ). According to this lemma, for any point P s,σCSC ( t ; τ , τ ) ∈ E CSC ( t ) there exists a con-trol (4) such that corresponding trajectory terminatesat this point at time t . Now, lets integrate equations (1)using the initial conditions (2) and the control (4). Itgives the following values x s,σCSC ( t ; τ ,τ ) = s (cos τ − − ( τ − τ ) sin τ )+ σ (cos( sτ + σ ( t − τ )) − cos τ ) ,y s,σCSC ( t ; τ ,τ ) = sin τ + ( τ − τ ) cos τ − σ ( s sin τ − sin( sτ + σ ( t − τ ))) ,ϕ s,σCSC ( t ; τ ,τ ) = π/ sτ + σ ( t − τ ) , (6)where we consider that P s,σCSC ( t ; τ , τ ) def = ( x s,σCSC ( t ; τ , τ ) , y s,σCSC ( t ; τ , τ ) ,ϕ s,σCSC ( t ; τ , τ )) ∈ R × S . Therefore, introducing domains of parameters s , σ , τ , τ , we obtain that E CSC ( t ) = { P s,σCSC ( t ; τ , τ ) : τ ∈ [0 , π ) , τ ∈ [ τ , t ] , ( s, σ ) ∈ B } . This reasoning is valid also for the analytical descriptionof E CCC ( t ). Let P sCCC ( t ; τ , τ ) ∈ E CCC ( t ). Then, theintegrating of the system (1) using the initial conditions(2) and the control (5) gives x sCCC ( t ; τ ,τ ) = s (2 cos τ − − τ − τ )+ cos(2 τ − τ + t )) ,y sCCC ( t ; τ ,τ ) = 2 sin τ − τ − τ )+ sin(2 τ − τ + t ) ,ϕ sCCC ( t ; τ ,τ ) = π/ s (2 τ − τ + t ) , (7)where we also suppose that P sCCC ( t ; τ , τ ) def = ( x sCCC ( t ; τ , τ ) , y sCCC ( t ; τ , τ ) ,ϕ sCCC ( t ; τ , τ )) ∈ R × S . Finally, we obtain that E CCC ( t ) = { P sCCC ( t ; τ ,τ ) : τ ∈ [0 , π ) , τ ∈ [ τ , t ] ,τ − τ ∈ [0 , π ) , s ∈ B } . E ( t )The analytical description of E ( t ) allows to state thefollowing lemma. Lemma 2 E ( t ) is a closed set at any moment of time t ∈ R +0 . ROOF.
Since P s,σCSC ( t ; τ , τ ), P sCCC ( t ; τ , τ ) dependcontinuously on τ , τ , only strict inequalities in the an-alytical descriptions of E CSC ( t ), E CCC ( t ) may break thecloseness of E ( t ). Lets prove that E CSC ( t ) is closed. Astrict inequality is used only for τ ∈ [0 , π ) in the ex-plicit expression of E CSC ( t ). Let τ = 2 π , τ ∈ [ τ , t ], t ≥ π . Then, calculations give P s,σCSC ( t ; 2 π, τ ) = P s,σCSC ( t ; 0 , τ − π ) ∈ E CSC ( t ) . Now, lets show that E CCC ( t ) is closed. Strict inequalitiesare used only for τ ∈ [0 , π ) and τ − τ ∈ [0 , π ) in theanalytical description of E CCC ( t ). If τ = 2 π , τ ∈ [ τ , t ], τ − τ ∈ [0 , π ), t ≥ π , then P sCCC ( t ; 2 π, τ ) = P sCCC ( t ; 0 , τ − π ) ∈ E CCC ( t ) . If τ = 2 π , τ ∈ [ τ , t ], τ − τ = 2 π , t ≥ π , then P sCCC ( t ; 2 π, π ) = P sCCC ( t ; 0 , ∈ E CCC ( t ) . If τ ∈ [0 , π ), τ ∈ [ τ , t ], τ − τ = 2 π , t ≥ τ + 2 π , then P sCCC ( t ; τ , τ + 2 π ) = P sCCC ( t ; τ , τ ) ∈ E CCC ( t ) . For the further purposes we formulate some key defini-tions related to multi-valued mappings.
Definition 3
A multi-valued mapping M is said to belower semicontinuous at the time moment t when forany sufficiently small ε > and any point P ∈ M ( t ) there exists δ > such that for any t : | t − t | < δ thereexists a point P ∈ M ( t ) such that ρ ( P, P ) < ε . Definition 4
A multi-valued mapping M is said to beupper semicontinuous at the time moment t when forany sufficiently small ε > there exists δ > such thatfor any t : | t − t | < δ and any point P ∈ M ( t ) thereexists a point P ∈ M ( t ) such that ρ ( P, P ) < ε . Definition 5
If a multi-valued mapping M is lower andupper semicontinuous at the time moment t , then it iscontinuous at time t . The continuity of the multi-valued mapping R is an es-tablished property [19], but a some multi-valued map-ping M may not be continuous even if M ( t ) ⊂ R ( t ) re-mains for any time t ∈ R +0 . For example, the evolutionof the boundary of the Dubins car planar reachable setis not continuous at the some time moment [8]. Lemma 6
The multi-valued mapping E is continuous atany time moment t > . PROOF.
Lets prove that E is lower semicontinuous.For any point P = P s,σCSC ( t ; τ , τ ) ∈ E CSC ( t ) we sup-pose that if τ < t , then P = P s,σCSC ( t ; τ , τ ), if τ = t and τ < τ , then P = P s,σCSC ( t ; τ , t ), if τ = τ = t ,then P = P s,σCSC ( t ; t, t ). Lets suppose δ = ε/ √
2. Now, if τ < t or τ = τ = t , then ρ ( P, P ) ≤ (cid:112) − t − t ) + ( t − t ) ≤ √ | t − t | < δ √ ε, if τ = t and τ < τ , then ρ ( P, P ) ≤ | t − t | < δ < ε. For any P = P sCCC ( t ; τ , τ ) ∈ E CCC ( t ) we supposethat if τ < t , then P = P sCCC ( t ; τ , τ ), if τ = t and τ < τ , then P = P sCCC ( t ; τ , t ), if τ = τ = t , then P = P sCCC ( t ; t, t ). Now, we have ρ ( P, P ) ≤ (cid:112) − t − t ) + ( t − t ) ≤ √ | t − t | < δ √ ε. Now, lets show that E is upper semicontinuous. Letschoose δ = ε/ √
2. For any time t : | t − t | < δ andany point P = P s,σCSC ( t ; τ , τ ) ∈ E CSC ( t ) we sup-pose that if τ < t , then P = P s,σCSC ( t ; τ , τ ), if τ = t and τ < τ , then P = P s,σCSC ( t ; τ , t ), if τ = τ = t , then P = P s,σCSC ( t ; t , t ). For anypoint P = P sCCC ( t ; τ , τ ) ∈ E CCC ( t ) we suppose thatif τ < t , then P = P sCCC ( t ; τ , τ ), if τ = t and τ < τ , then P = P sCCC ( t ; τ , t ), if τ = τ = t , then P = P sCCC ( t ; t , t ). By the same way we concludethat ρ ( P, P ) < ε . The listed above statements allow to prove the main the-orem of this section. This theorem describes the config-uration of the optimal interception point lying on thetarget’s trajectory.
Theorem 7
If the minimum-time lateral interceptionof the moving target by a Dubins car is possible, so theoptimal interception time T is finite, then the optimalinterception point E ( T ) ∈ ˜ B ( T ) , where ˜ B ( T ) def = { P (cid:48) ∈ E ( T ) : lim t → T − min P ∈B ( t ) ρ ( P, P (cid:48) ) = 0 } . PROOF.
Let M be an arbitrary continuous multi-valued mapping, such that M ( t ) is a closed set. Let ρ M ( t ) def = min P ∈M ( t ) ρ ( P, E ( t )) . Since the vector-function E is continuous then ρ M is acontinuous function. Therefore, ρ R , ρ E are continuous,5ut ρ B is not necessarily continuous. For any time t > ρ R ( t ) ≤ ρ E ( t ) ≤ ρ B ( t )since B ( t ) ⊂ E ( t ) ⊂ R ( t ). According to the definitionof the minimum-time interception it holds at any time t ∈ [0 , T ), that E ( t ) / ∈ R ( t ) and E ( T ) ∈ R ( T ). It impliesthat ρ R ( t ) > t ∈ [0 , T ) and ρ R ( T ) = 0.Since R ( t ) is closed, E ( t ) / ∈ R ( t ), and B ( t ) is a boundaryof R ( t ) we deduce that ρ R ( t ) = ρ B ( t ) at any time t ∈ [0 , T ). Now we conclude thatlim t → T − ρ E ( t ) ≤ lim t → T − ρ B ( t ) = lim t → T − ρ R ( t ) = 0 . From the continuity of ρ E we get ρ E ( T ) = 0, so E ( T ) ∈E ( T ). According to the ˜ B ( T ) definition we obtain that E ( T ) ∈ ˜ B ( T ) since ρ B ( T −
0) = 0.Thus, the analytical description of E ( T ) ⊃ ˜ B ( T ) givesthe desired algebraic equations which solution allows todetermine the optimal control parameters. But it is stillpossible to reduce the space of optimal control candi-dates. Consider the next mapping for the following pur-poses: S τ ( x, y, ϕ ) = ( x + τ cos ϕ, y + τ sin ϕ, ϕ ) . Note that if the Dubins car is located at a point P =( x , y , ϕ ) ∈ R ( t ) at the time moment t and its con-trol u ( t ) = 0 for any t ≥ t , then the Dubins car is lo-cated at the point S t − t ( P ) ∈ R ( t ) at time t . Lemma 8
If a point P = ( x , y , ϕ ) is an internalpoint of R ( t ) , then for any time moment t > t thereexists δ > such that for any δ ∈ [0 , δ ) the point S t − t ( P ) is an internal point of R ( t − δ ) . PROOF.
Since P is an internal point there exists aball B δ ( P ) = { P ∈ R × S : ρ ( P, P ) < δ } which isa subset of R ( t ). Lets demand that δ < t − t andconsider an arbitrary δ ∈ [0 , δ ). According to propertiesof the mapping S t − t − δ : S t − t − δ ( B δ ( P )) ⊂ R ( t − δ ) . It is obvious that S δ ( P ) ∈ B δ ( P ) since ρ ( S δ ( P ) , P ) = δ < δ . Moreover, S δ ( P ) is an internal point of B δ ( P ). Themapping S t − t − δ is a continuous mapping. Hence, animage S t − t − δ ( B δ ( P )) ⊂ R ( t − δ ) is an open set andthe point S t − t − δ ( S δ ( P )) = S t − t ( P ) is an internalpoint of the image. Lemma 9
Let t − τ ≥ π and τ − τ > . If one ofthe inequalities τ > or t − τ > π is fulfilled, then P s,σCSC ( t ; τ , τ ) / ∈ ˜ B ( t ) . PROOF.
According to ˜ B ( t ) definition it is suffice toprove that there exists δ such that for any δ ∈ [0 , δ ) thepoint P s,σCSC ( t ; τ , τ ) holds an internal point for R ( t − δ ).Suppose that τ = ( τ − τ ) / P s,σCSC ( t − τ ; τ , τ − τ ) ∈ R ( t − τ ) . This point can be attained using the trajectory that con-tains a cycle. This cycle could be shifted to any partof the trajectory and according to Lemma 4 from [29]the point P s,σCSC ( t − τ ; τ , τ − τ ) is an internal point of R ( t − τ ). Hence, according to Lemma 8 we conclude that P s,σCSC ( t ; τ , τ ) holds an internal point for R ( t − δ ) since S τ ( P s,σCSC ( t − τ ; τ , τ − τ )) = P s,σCSC ( t ; τ , τ ) . The following lemma allows to avoid a case with twocycles when we consider an interception problem.
Lemma 10
The configuration (0 , , π/ / ∈ ˜ B (4 π ) . PROOF.
According to the definition of ˜ B it is enoughto prove that there exists δ > δ ∈ [0 , δ ) the point (0 , , π/
2) is an internal point of R (4 π − δ ). Lets consider the control given by u ( t ) = , t ∈ [0 , π/ χ ) , / ξ, t ∈ [ π/ χ, π + χ ) , , t ∈ [2 π + χ, π/ χ ) , / η, t ∈ [5 π/ χ, + ∞ ) . Integration of the system (1) with that control functionand the initial conditions given by (2) leads to x = ( η − ξ ) cos( π (1 + ξ ))( + ξ )( + η ) − π (cid:18) π (cid:18) ξ (cid:19)(cid:19) + cos( π (1 + ξ ) + ( + η )( t − π − χ )) + η − + ξ ,y = ( η − ξ ) sin( π (1 + ξ ))( + ξ )( + η ) + π (cid:18) π (cid:18) ξ (cid:19)(cid:19) + sin( π (1 + ξ ) + (cid:0) + η (cid:1) ( t − π − χ )) + η + π χ,ϕ = 3 π ξ ) + (cid:18)
23 + η (cid:19) (cid:18) t − π − χ (cid:19)
6t time t ≥ π/ χ . If t = 4 π , ξ = 0, η = 0, χ = 0, thenthe point ( x, y, ϕ ) = (0 , , π/
2) = (0 , , π/ ∈ R (4 π ).The calculation of the Jacobi matrix determinant leadsto the following value (cid:12)(cid:12)(cid:12)(cid:12) ∂ ( x, y, ϕ ) ∂ ( ξ, η, χ ) (cid:12)(cid:12)(cid:12)(cid:12) ξ =0 η =0 χ =0 t =4 π = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) π − π π π π − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = − π − π (cid:54) = 0 . Since this value does not vanish and the correspondingmapping of ( ξ, η, χ ) to ( x, y, ϕ ) is continuous for smallenough δ > t ∈ (4 π − δ , π + δ ) thismapping transforms any small enough neighborhood of( ξ, η, χ ) = (0 , ,
0) to the neighborhood of ( x, y, ϕ ) =(0 , , π/ ξ, η, χ ). It implies that theneighborhood of ( x, y, ϕ ) = (0 , , π/
2) is a subset of R ( t ) at any time t ∈ (4 π − δ , π + δ ). Therefore, thepoint (0 , , π/
2) is an internal point of R (4 π − δ ) for any δ ∈ [0 , δ ).Now, it is possible to prove a statement which is a gen-eralization of [34, Lemma 2]. According to Lemma 10and Theorem 7 the point (0 , , π/
2) is not an optimalinterception point at time t = 4 π . Moreover, Lemma 9excludes some trajectories with cycles. We use it for theoptimal control candidates reduction in the followingtheorem. Theorem 11
For attaining any point P ∈ ˜ B ( t ) attime t ∈ R +0 it suffices to use the controls (4) , (5) with t − τ ∈ [0 , π ) or the controls (4) , (5) with τ = 0 , t − τ = 2 π . PROOF.
Suppose that the controls (4), (5) with t − τ ≥ π leads to the point P ∈ ˜ B ( t ). According to theBellman’s principle these controls remain the optimalfrom the time t − π to t . Considering the new time-optimal interception problem from the time t − π raisesa contradiction with Lemma 10 and Theorem 7 since thenew time-optimal problem interception point is locatedat (0 , , π/
2) at time t = 4 π . So, the controls (4), (5)with t − τ ≥ π are redundant.Now, lets suppose that the control (4) with t − τ ≥ π leads to the point P = P s,σCSC ( t ; τ , τ ) ∈ ˜ B ( t ).We associate the case when τ − τ = 0 with CCC-trajectories and consider it further. If Lemma 9 caseshold, then P / ∈ ˜ B ( t ). Otherwise, τ = 0 and t − τ =2 π .Next, lets suppose that the control (5) with t − τ ≥ π leads to the point P = P sCCC ( t ; τ , τ ) ∈ ˜ B ( t ). Consider the case when τ >
0. According to [29, Lemma2] if τ + ( t − τ ) > τ − τ , then P is an internal point of R ( t ). Consider thislemma for a terminal point at time t ∈ ( t − τ / , t ]. Itfollows that the terminal point holds internal for R ( t )since τ + t − τ > τ + t − τ − τ = τ t − τ > π > τ − τ . Therefore, P / ∈ ˜ B ( t ). Now, suppose that τ = 0 and τ − τ >
0. Lets consider a new control u − sCCC ( · ; τ (cid:48) , τ (cid:48) )with τ (cid:48) = τ , τ (cid:48) − τ (cid:48) = t − τ − π which leads to thesame point P . For the arbitrary t ∈ ( t − τ (cid:48) / , t ] thecorresponding terminal point also holds internal for R ( t )since τ (cid:48) + t − τ (cid:48) > τ (cid:48) t − τ (cid:48) = τ (cid:48) π > π > τ (cid:48) − τ (cid:48) . If τ = 0 and τ − τ = 0, then the control u − sCCC ( · ; τ (cid:48) , τ (cid:48) )with τ (cid:48) = 0, t − τ (cid:48) = 2 π leads to the point P . In this section, we consider some scenarios of the optimalinterception. These scenarios are enough to completelysolve the problem. Finally, an algorithm of the optimalcontrol selection is described.
According to Theorem 7 the optimal interception point E ( T ) ∈ ˜ B ( T ) ⊂ E ( T ). Suppose that the using of theCSC-trajectory is optimal. It follows from (3) that E ( T ) = P s,σCSC ( T ; τ , τ ) . Next, we use the equations given by (6) and infer that ξ s,σCSC ( T ) def = sx E ( T ) + 1 − sσ sin ϕ E ( T )= (1 − sσ ) cos τ − ( τ − τ ) sin τ ,η σCSC ( T ) def = y E ( T ) + σ cos ϕ E ( T )= (1 − sσ ) sin τ + ( τ − τ ) cos τ ,ϕ E ( T ) = π sτ + σ ( T − τ ) + 2 πk, k ∈ Z . (8)Consider a new function given by ρ ,s,σCSC ( T ) def = ( ξ s,σCSC ( T )) + ( η σCSC ( T )) = (1 − sσ ) + ( τ − τ ) . τ ≥ τ . Itgives the following equation τ − τ = (cid:113) ρ ,s,σCSC ( T ) − (1 − sσ ) . Now we substitute the value τ − τ to the first and secondequations of (8) and after some algebra it provides thefollowing C s,σCSC ( T ) def = η σCSC ( T ) ρ ,s,σCSC ( T ) (cid:113) ρ ,s,σCSC ( T ) − (1 − sσ ) + (1 − sσ ) ξ s,σCSC ( T ) ρ ,s,σCSC ( T ) = cos τ ,S s,σCSC ( T ) def = − ξ s,σCSC ( T ) ρ ,s,σCSC ( T ) (cid:113) ρ ,s,σCSC ( T ) − (1 − sσ ) + (1 − sσ ) η σCSC ( T ) ρ ,s,σCSC ( T ) = sin τ . For the further purposes we use the next function:arctan ( y, x ) def = arccos x (cid:112) x + y , y ≥ , π − arccos x (cid:112) x + y , y < . The previous equations and τ ∈ [0 , π ) provide the val-ues τ = θ ,s,σCSC ( T ) def = arctan ( S s,σCSC ( T ) , C s,σCSC ( T )) ,τ = θ ,s,σCSC ( T ) def = θ ,s,σCSC ( T ) + (cid:113) ρ ,s,σCSC ( T ) − (1 − sσ ) . The third equation of (8) and T ≥ τ give the followingproblem: ϕ E ( T ) − π − sθ ,s,σCSC ( T ) − σ ( T − θ ,s,σCSC ( T )) = 2 πk,T ≥ θ ,s,σCSC ( T ) , k ∈ Z . The addictive restriction from the Theorem 11 con-straints the value of T − τ < π in the cycle-free case.Therefore, the optimal interception time T is subject to F s,σCSC ( T ) def = − T + θ ,s,σCSC ( T )+ mod (cid:16) σ (cid:16) ϕ E ( T ) − π (cid:17) − sσθ ,s,σCSC ( T ) , π (cid:17) = 0 . (9)Hence, we have four equations (9) since the trajec-tory’s configuration ( s, σ ) ∈ B is not established. Foreach configuration ( s, σ ) we should find a minimal non-negative solution T s,σCSC of the algebraic equation (9)which is transcendental in the general case. If this equa-tion does not have a solution for some configuration ( s, σ ) ∈ B , then we suppose that T s,σCSC = + ∞ . Theoptimal control u s,σCSC ( · ; τ , τ ) is defined by (4) with thefollowing choice of parameters:( s, σ ) = arg min ( s,σ ) ∈ B T s,σCSC ,τ = θ ,s,σCSC ( T s,σCSC ) , τ = θ ,s,σCSC ( T s,σCSC ) . (10)For the optimal configuration ( s, σ ) we suppose that theoptimal interception time is T CSC = T s,σCSC in the con-sidered case of the cycle-free CSC-trajectories.If there is a cycle in the optimal CSC-trajectory anda non-vanished S-part, then according to Theorem 11 τ = 0, T − τ = 2 π . In this case, the optimal interceptiontime T is subject to F SC ( T ) def = ρ ( E ( T ) , (0 , T − π, π/ . (11)We should find a minimal solution T SC ≥ π of the alge-braic equation (11) which is transcendental in the gen-eral case. If this equation does not have a solution T SC ≥ π , then we suppose that the optimal interception time T SC = + ∞ . The optimal control u s,σCSC ( · ; τ , τ ) is de-fined by (4) with the arbitrary choice of parameters( s, σ ) ∈ B and τ = 0, τ = T SC − π .If the S-part in the optimal CSC-trajectory vanishes,then we relate this case to the CCC-trajectory case whichis considered further. Lets consider that the using of the CCC-trajectory isoptimal. The line of reasoning is similar to the CSC-trajectory case. It follows from (3) that E ( T ) = P sCCC ( T ; τ , τ ) . Next, we use the equations given by (7) and infer that ξ sCCC ( T ) def = ( sx E ( T ) + 1 − sin ϕ E ( T )) /
2= (1 − cos( τ − τ )) cos τ − sin( τ − τ ) sin τ ,η sCCC ( T ) def = ( y E ( T ) + s cos ϕ E ( T )) /
2= (1 − cos( τ − τ )) sin τ + sin( τ − τ ) cos τ ,ϕ E ( T ) = π/ s (2 τ − τ + T ) + 2 πk, k ∈ Z . (12)Consider a new function given by ρ ,sCCC ( T ) def = ( ξ sCCC ( T )) + ( η sCCC ( T )) = 2 − τ − τ ) .
8t follows thatcos( τ − τ ) = 1 − ρ ,sCCC ( T )2 , sin( τ − τ ) = µ (cid:118)(cid:117)(cid:117)(cid:116) − (cid:32) − ρ ,sCCC ( T )2 (cid:33) , where µ ∈ B . According to the Lemma 1 we supposethat τ − τ ∈ [0 , π ). If µ = 1, then the length of thesecond circle arc of the CCC-trajectory is τ − τ ∈ [0 , π ].Otherwise, if µ = −
1, then τ − τ ∈ ( π, π ). Therefore, τ = τ − π ( µ −
1) + µ arccos (cid:32) − ρ ,sCCC ( T )2 (cid:33) . Now we substitute the values cos( τ − τ ) and sin( τ − τ )to the first and second equations of (12) and after somealgebra it provides the following C s,µCCC ( T ) def = µ η sCCC ( T ) ρ ,sCCC ( T ) (cid:118)(cid:117)(cid:117)(cid:116) − (cid:32) − ρ ,sCCC ( T )2 (cid:33) + ξ sCCC ( T )2 = cos τ ,S s,µCCC ( T ) def = − µ ξ sCCC ( T ) ρ ,sCCC ( T ) (cid:118)(cid:117)(cid:117)(cid:116) − (cid:32) − ρ ,sCCC ( T )2 (cid:33) + η sCCC ( T )2 = sin τ . The previous equations and τ ∈ [0 , π ) provide the val-ues τ = θ ,s,µCCC ( T ) def = arctan ( S s,µCCC ( T ) , C s,µCCC ( T )) ,τ = θ ,s,µCCC ( T ) def = θ ,s,µCCC ( T ) − π ( µ − µ arccos (cid:32) − ρ ,sCCC ( T )2 (cid:33) . The third equation of (12) and T ≥ τ give the followingproblem: ϕ E ( T ) − π − sθ ,s,µCCC ( T ) + 2 sθ ,s,µCCC ( T ) − sT = 2 πk,T ≥ θ ,s,µCCC ( T ) , k ∈ Z . The addictive restriction from the Theorem 11 con-straints the value of T − τ < π in the cycle-free case.Therefore, the optimal interception time T is subject to F s,µCCC ( T ) def = − T + θ ,s,µCCC ( T ) +mod( s ( ϕ E ( T ) − π − θ ,s,µCCC ( T ) + θ ,s,µCCC ( T ) , π ) = 0 . (13) Hence, we have four equations (13) since the trajec-tory’s configuration ( s, µ ) ∈ B is not established. Foreach configuration ( s, µ ) we should find a minimal non-negative solution T s,µCCC of the algebraic equation (13)which is transcendental in the general case. If this equa-tion does not have a solution for some configuration( s, µ ) ∈ B , then we suppose that T s,µCSC = + ∞ . The op-timal control u sCCC ( · ; τ , τ ) is defined by (5) with thefollowing choice of parameters:( s, µ ) = arg min ( s,µ ) ∈ B T s,µCCC ,τ = θ ,s,µCCC ( T s,µCCC ) , τ = θ ,s,µCCC ( T s,µCCC ) . (14)For the optimal configuration ( s, µ ) we suppose that theoptimal interception time is T CCC = T s,µCCC in the con-sidered case of the cycle-free CCC-trajectories.If there is a cycle in the optimal CCC-trajectory, thenaccording to Theorem 11 the first circle arc is vanished τ = 0 and T − τ = 2 π . In this case, the optimal inter-ception time T is subject to F CC ( T ) def = ρ ( E ( T ) , ((1 − cos T ) sgn x E ( T ) , sin T,π/ − T sgn x E ( T ))) = 0 . (15)We should find a minimal solution T CC ≥ π of the alge-braic equation (15) which is transcendental in the gen-eral case. If this equation does not have a solution T CC ≥ π , then we suppose that the optimal interception time T CC = + ∞ . The optimal control u sCCC ( · ; τ , τ ) is de-fined by (5) with the parameters s = sgn x E ( T ), τ = 0, τ = T CC − π . The definition of ˜ B ( T ) allows to consider that anypoint of ˜ B ( T ) is attainable using either CSC- or CCC-trajectories. We consider the brute force algorithm ofsearching of the optimal control. This algorithm calcu-lates minimal roots of 10 equations and compares them.It supposes that the non-negative minimal root T s,σCSC of the equation (9) is defined by some root-finding algo-rithm for each ( s, σ ) ∈ B . The corresponding optimalcycle-free CSC-trajectory is defined by the parameters(10) and the interception time is T CSC . If T CSC = + ∞ ,then the equation (9) does not have non-negative rootsfor each ( s, σ ) ∈ B . It means that the minimum timeinterception along any cycle-free CSC-trajectory is im-possible. The CSC-trajectory with a cycle and non-zeroS-part is examined by solution T SC ≥ π of the alge-braic equation (11).The algorithm also supposes that the non-negative min-imal root T s,µCCC of the equation (13) is defined by some9oot-finding algorithm for each ( s, µ ) ∈ B . The corre-sponding optimal cycle-free CCC-trajectory is definedby the parameters (14) and the interception time is T CCC . If T CCC = + ∞ , then the equation (13) does nothave non-negative roots for each ( s, µ ) ∈ B . The cycledCCC-trajectory is examined by solution T CC ≥ π ofthe algebraic equation (15). The final step of the algo-rithm is a comparison of T CSC , T SC , T CCC and T CC values. The following theorem determines the bruteforce algorithm. Theorem 12
Let the minimum-time lateral intercep-tion of the moving target by the Dubins car be possibleand T ∗ = min( T CSC , T SC , T CCC , T CC ) . If T ∗ = T CSC ,then u = u s,σCSC ( · ; τ , τ ) is an optimal control, where s , σ , τ , τ are defined by (10) . If T ∗ = T SC , then u = u s,σCSC ( · ; τ , τ ) is an optimal control, where ( s, σ ) ∈ B , τ = 0 , τ = T SC − π . If T ∗ = T CCC , then u = u sCCC ( · ; τ , τ ) is an optimal control, where s , τ , τ aredefined by (14) . If T ∗ = T CC , then u = u sCCC ( · ; τ , τ ) is an optimal control, where s = sgn x E ( T CC ) , τ = 0 , τ = T CC − π . This section provides a set of optimal interception ex-amples. The main goal of these examples is to highlightthe necessity of considering all 10 algebraic equations inthe general case. Also, we demonstrate an application ofthe proposed brute force algorithm.The Dubins car reachable set has a plane of symmetry x = 0 in R × S . For the CSC-optimal trajectory case itmeans that if there exists an example when the optimalinterception of the target E ( t ) = ( x E ( t ) , y E ( t ) , ϕ E ( t ))proceeds using the CSC-trajectory and the control u s,σCSC ( · ; τ , τ ) is an optimal, then the other target E ∗ ( t ) = ( − x E ( t ) , y E ( t ) , π − ϕ E ( t )) (16)could be optimally intercepted using the mirror control u − s, − σCSC ( · ; τ , τ ). The same situation occurs when thecontrol u sCCC ( · ; τ , τ ) is optimal. The target E ∗ could beoptimally intercepted using the control u − sCCC ( · ; τ , τ ).This symmetry allows considering only 6 cases for thedemonstration that it is necessary to solve each of 10equations in the general case.Here and further in figures, the target’s orientation onits Cartesian plane path is presented by the light trian-gles and Dubins car orientation by filled black ones. Atthe interception moment, the black triangular coincideswith the light one. The continuous line demonstrates anoptimal interception path on the plane. / / / − − x (0) , y (0))( x ( τ ) , y ( τ )) ( x ( τ ) , y ( τ ))( x E (0) , y E (0))( x ( T ) , y ( T ))( x E ( T ) , y E ( T )) x E ( t ) = 3 −
32 cos (cid:18) πt π + 1) (cid:19) y E ( t ) = −
32 + 32 sin (cid:18) πt π + 1) (cid:19) ϕ E ( t ) = − πt π + 1) τ = π τ = π T = π + 1 Fig. 2. RSR-optimal interception case ( s = − σ = − t = 0 t = ( π + 1) / t = 2( π + 1) / t = π + 10 π π π π F − , +1 CCC ( t ) F − , − CSC ( t ) F +1 , − CCC ( t ) F +1 , +1 CSC ( t )00.01 Fig. 3. Functions F +1 , +1 CSC , F − , − CSC , F − , +1 CCC , F +1 , − CCC graphs:RSR-optimal interception case.
Consider the case of the circular movement of the tar-get as shown in Fig. 2 by a dashed line. The analysis ofthe equations (9), (13) is presented in Figs. 3, 4. In thesefigures, one can see that t = π + 1 is a minimal time mo-ment when the function F − , − CSC becomes equal to zero.The other function graphs in the figures do not cross thehorizontal axis zero level on the section [0 , π + 1]. Since T = π + 1 < π we should not analyse the cases of thepresence of a cycle. Hence, the RSR-trajectory ( s = − σ = −
1) is an optimal trajectory. The mirror symmetry10 = 0 t = ( π + 1) / t = 2( π + 1) / t = π + 10 π π π π F +1 , − CSC ( t ) F − , − CCC ( t ) F − , +1 CSC ( t ) F +1 , +1 CCC ( t ) Fig. 4. Functions F +1 , − CSC , F − , +1 CSC , F +1 , +1 CCC , F − , − CCC graphs:RSR-optimal interception case. allows to state that the LSL-trajectory ( s = +1, σ = +1)is able to be optimal for the E ∗ target’s trajectory. − / − / / x (0) , y (0))( x ( τ ) , y ( τ ))( x ( τ ) , y ( τ ))( x E (0) , y E (0))( x ( T ) , y ( T )) = ( x E ( T ) , y E ( T )) x E ( t ) = − y E ( t ) = 2 tπ + 1 ϕ E ( t ) = π τ = π τ = π T = π + 1 Fig. 5. LSR-optimal interception case ( s = +1, σ = − Next, consider the case of the uniform rectilinear move-ment of the target (see Fig. 5). The analysis of the equa-tions (9), (13) in Figs. 6, 7 demonstrates that the func-tion F +1 , − CSC becomes equal to zero at t = π + 1 andthe other functions in the figures do not. Similarly, since T = π + 1 < π we should not analyse the cases of thepresence of a cycle. Hence, the LSR-trajectory ( s = +1, σ = −
1) is an optimal trajectory. The mirror symme-try allows to state that the RSL-trajectory ( s = − σ = +1) is able to be optimal for the E ∗ target’s trajec-tory. t = 0 t = ( π + 1) / t = 2( π + 1) / t = π + 10 π π π F − , − CSC ( t ) F +1 , − CSC ( t ) F − , +1 CCC ( t ) F +1 , − CCC ( t ) Fig. 6. Functions F − , +1 CCC , F − , − CSC , F +1 , − CCC , F +1 , − CSC graphs:LSR-optimal interception case. t = 0 t = ( π + 1) / t = 2( π + 1) / t = π + 10 π π π π π F − , +1 CSC ( t ) F +1 , − CSC ( t ) F − , − CCC ( t ) F +1 , +1 CCC ( t ) Fig. 7. Functions F − , +1 CSC , F +1 , +1 CCC , F +1 , − CSC , F − , − CCC graphs:LSR-optimal interception case.
Finally, consider the case of the circular movement ofthe target (see Fig. 8). The analysis of the equations (9),(11), (13), (15) in Figs. 9, 10 reveals that the function F SC becomes equal to zero at t = 2 π + 1 and the otherfunctions do not. In this case, all 10 candidates have tobe analysed since T = 2 π + 1 > π . As it is seen inFig. 8, the optimal interception trajectory of the Dubinscar contains a cycle and as it is mentioned above theparameters ( s, σ ) may be arbitrary chosen.11 − / / / − x (0) , y (0)) = ( x ( τ ) , y ( τ ))( x ( T ) , y ( T )) = ( x ( τ ) , y ( τ ))= ( x E ( T ) , y E ( T ))( x E (0) , y E (0)) x E ( t ) = cos ϕ E ( t ), y E ( t ) = sin ϕ E ( t ), ϕ E ( t ) = 3 π − πt π + 1 τ = 0, τ = 1, T = 2 π + 1 Fig. 8. SC-optimal interception case ( σ = − t = 0 t = 2 π/ / t = 4 π/ / t = 2 π + 1 − π − π π π π F +1 , +1 CSC ( t ) F − , +1 CCC ( t ) F +1 , − CCC ( t ) F +1 , − CSC ( t ) F CC ( t ) Fig. 9. Functions F +1 , − CCC , F +1 , +1 CSC , F CC , F − , +1 CCC , F +1 , − CSC graphs: SC-optimal interception case.
Consider the other case of the uniform rectilinear move-ment of the target (see Fig. 11). The analysis of theequations (9), (13) in Figs. 12, 13 demonstrates that thefunction F − , − CCC becomes to equal zero at time t = 2 π and the other functions in the figures do not. Since thetarget is not located at (0 , , π/
2) at time t = 2 π weshould not analyse the cases of the presence of a cycle.Therefore, the RLR-trajectory ( s = − µ = −
1) is anoptimal trajectory. The mirror symmetry allows to statethat the LRL-trajectory ( s = +1, µ = −
1) is able to be t = 0 t = 2 π/ / t = 4 π/ / t = 2 π + 1 − π − π π π π F − , − CSC ( t ) F SC ( t ) F − , +1 CSC ( t ) F +1 , +1 CCC ( t ) F − , − CCC ( t ) Fig. 10. Functions F − , − CSC , F − , − CCC , F SC , F − , +1 CSC , F +1 , +1 CCC graphs: SC-optimal interception case. − / / / x (0) , y (0)) ( x ( τ ) , y ( τ ))( x ( τ ) , y ( τ ))( x ( T ) , y ( T )) = ( x E ( T ) , y E ( T )) ( x E (0) , y E (0)) x E ( t ) = 2 − (1 + √ π t , y E ( t ) = 1 + √ ϕ E ( t ) = π τ = π/ τ = 7 π/ T = 2 π Fig. 11. RLR-optimal interception case ( s = − µ = − optimal for the E ∗ target’s trajectory.Next, consider the case of the uniform rectilinear move-ment of the target (see Fig. 14). The target’s orientation ϕ E ( t ) changes clockwise and uniformly and starts from ϕ E (0) = π/
2. The analysis of the equations (9), (13)in Figs. 15, 16 demonstrates that the function F − , +1 CCC becomes to equal zero at time t = 5 π/ T = 5 π/ < π we may analyse only the cycle-free cases. Hence, theRLR-trajectory ( s = − µ = +1) is an optimal tra-jectory. The mirror symmetry allows to state that theLRL-trajectory ( s = +1, µ = +1) is able to be optimal12 = 0 t = 2 π/ t = 4 π/ t = 2 π − π π π π π F +1 , − CSC ( t ) F +1 , +1 CSC ( t ) F − , +1 CCC ( t ) F +1 , − CCC ( t ) Fig. 12. Functions F +1 , − CSC , F +1 , − CCC , F − , +1 CCC , F +1 , +1 CSC graphs:RLR-optimal interception case ( µ = − t = 0 t = 2 π/ t = 4 π/ t = 2 π − π π π π π F − , − CSC ( t ) F +1 , +1 CCC ( t ) F − , − CCC ( t ) F − , +1 CSC ( t ) Fig. 13. Functions F − , − CSC , F +1 , +1 CCC , F − , − CCC , F − , +1 CSC graphs:RLR-optimal interception case ( µ = − for the E ∗ target’s trajectory.Finally, consider the case of the uniform rectilinearmovement of the target in shown Fig. 17. The analy-sis of the equations (9), (11), (13), (15) in Figs. 18, 19demonstrates that the function F CC becomes to equalzero at time t = 9 π/ -1 − / / x (0) , y (0))( x ( τ ) , y ( τ ))( x ( τ ) , y ( τ ))( x ( T ) , y ( T )) = ( x E ( T ) , y E ( T ))( x E (0) , y E (0)) x E ( t ) = 1 − √ y E ( t ) = (8 + 2 √ t π , ϕ E ( t ) = π − t τ = π/ τ = π , T = 5 π/ Fig. 14. RLR-optimal interception case ( s = − µ = +1). t = 0 t = 5 π/ t = 5 π/ t = 5 π/ π π π F +1 , − CSC ( t ) F − , +1 CCC ( t ) F +1 , − CCC ( t ) F +1 , +1 CSC ( t ) Fig. 15. Functions F +1 , − CCC , F +1 , +1 CSC , F +1 , − CSC , F − , +1 CCC graphs:RLR-optimal interception case ( µ = +1). For the problem of the time-minimal lateral intercep-tion of a predefined moving target by the Dubins car,we obtained algebraic equations, the solution of whichallows one to calculate all the parameters of the op-timal control. Since the target movement is specifiedin the form of an arbitrary continuous vector-function E = ( x E , y E , ϕ E ), the obtained analytical results aresuitable for a wide range of cases of the target’s move-ment. Moreover, the obtained results are applicable tothe problem of the time-minimal achievement to the13 = 0 t = 5 π/ t = 5 π/ t = 5 π/ π π π π F − , − CSC ( t ) F − , +1 CSC ( t ) F +1 , +1 CCC ( t ) F − , − CCC ( t ) Fig. 16. Functions F − , +1 CSC , F − , − CSC , F − , − CCC , F +1 , +1 CCC graphs:RLR-optimal interception case ( µ = +1). − / − − / / / − − x E (0) , y E (0))( x (0) , y (0)) = ( x ( τ ) , y ( τ ))( x ( T ) , y ( T )) = ( x E ( T ) , y E ( T ))= ( x ( τ ) , y ( τ )) x E ( t ) = 1 − √ y E ( t ) = (4 + 2 √ t π − ϕ E ( t ) = π τ = 0, τ = π/ T = 9 π/ Fig. 17. CC-optimal interception case ( s = +1). desired state for the Dubins car moving under time-depended wind-conditions.Instead of the maximum principle application, we usedanother methodological approach which deals with geo-metric properties of the reachable set R ( t ) and the cor-responding multi-valued mapping. This technique can t = 0 t = 3 π/ t = 3 π/ t = 9 π/ − π − π π π π F +1 , − CSC ( t ) F +1 , +1 CSC ( t ) F − , +1 CCC ( t ) F +1 , − CCC ( t ) F CC ( t ) Fig. 18. Functions F +1 , − CSC , F +1 , +1 CSC , F +1 , − CCC , F CC , F − , +1 CCC graphs: CC-optimal interception case. t = 0 t = 3 π/ t = 3 π/ t = 9 π/ − π − π π π π F − , − CSC ( t ) F +1 , +1 CCC ( t ) F − , − CCC ( t ) F SC ( t ) Fig. 19. Functions F − , − CCC , F SC , F − , − CSC , F +1 , +1 CCC graphs:CC-optimal interception case. be useful for considering interception problems by mo-bile vehicles whose dynamics differ from the dynamics ofthe Dubins car. Theorem 7 determines the optimal inter-ception location on the reachable set, and the analyticaldescription of the multi-valued mapping E makes it pos-sible to reduce the initial problem to solving algebraicequations. Theorem 11 gives the additional constraintson the parameters of the optimal control without losingthe optimality of the solution and reduces the number ofalgebraic equations required for consideration to 10. Thedescribed examples approve the necessity to consider all140 equations in the general case.Topics to be addressed in future work include an inves-tigation of effective numerical methods for solving theobtained algebraic transcendental equations. These al-gebraic equations are discontinuous and not defined onthe entire number line, which raises a sophisticated chal-lenge for the standard numerical methods. Acknowledgements
The research was partially supported by Russian ScienceFoundation grant (No.21-19- 00128).
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