Towards Trajectory Planning for a 6-Degree-of-Freedom Robot Manipulator Considering the Orientation of the End-efector Using Computer Algebra

Abstract

We consider the trajectory planning of a 6-Degree-of-Freedom (DOF) robot manipulator using computer algebra, with controlling the orientation of the end-efector. As a first step towards the objective, we present a solution to the inverse kinematics problem of the manipulator such that the orientation of the end-efector remains constant using computer algebra. Trajectory planning, Inverse kinematic problem, Robot manipulator, Gröbner basis (M. Mikawa) This paper discusses the trajectory planning of a 6-Degree-of-Freedom (DOF) robot manipulator. A manipulator is a robot resembling a human hand and comprises links that function as a human arm and joints that function as human joints. Each link is connected for movement relative to each other by a joint. The first link is connected to the ground, and the last link called end-efector, contains the hand, which can be moved freely. In this paper, we consider a manipulator called “myCobot 280" [1] (hereafter called “myCobot”) that has six joints connected in series that can only rotate around a certain axis. Note that each joint has one degree of freedom. Therefore, myCobot has at most six degrees of freedom.

1. Introduction

LGOBE ∗Corresponding author.

(CC BY 4.0). CEUR Workshop Proceedings

ceur-ws.org ISSN1613-0073 4 3 1 6 4 5 3 2 1 5 Systems (CGS) and certifying the existence of a solution to the inverse kinematic problem using the CGS-QE, or the quantifier elimination (QE) based on the CGS computation.

In this paper, we consider the trajectory planning of a 6-DOF robot manipulator while controlling the orientation of the end-efector using computer algebra. As a first step towards the objective, we present a solution to the inverse kinematics problem such that the end-efector’s orientation remains constant. More precisely, we present a solution to the inverse kinematics problem under the condition that the end-efector’s local coordinate system overlaps the global coordinate system by translation.

The paper is organized as follows. In Section 2, the coordinate system and the notation of the manipulator are introduced. In Section 3, the analytical solution of general inverse kinematics problems is first described, followed by the solution in myCobot. In Section

4, future initiatives are described.

2. Preliminaries

2.1. Coordinate systems

For each joint in myCobot, the coordinate system is defined as follows (see Figure 1). Let each joint be numbered as joint from the base to the end-efector in increasing order, and the end-efector be numbered as joint 7. Let be the coordinate system of joint . Here, 1 is the reference (global) coordinate system, while the other coordinate systems are local. The axes of each coordinate system are defined as follows: the axis in the direction of the joint axis (rotational axis in the case of a revolute joint), axis in the direction of the common normal of the and +1 axes, and axis to be a right-handed system.

2.2. Notation

For the index , the matrices are denoted by , and vectors are denoted by +1 , in which subscripts are used to distinguish points, and superscripts denote the local coordinate system to which they refer (vectors with no superscripts are referenced in 1). If the vector is referenced with respect to a diferent coordinate system, it is enclosed within brackets, and a separate superscript is added outside the brackets. (e.g. [+1 ]). Scalars are expressed in lowercase variables, and subscripts, such as , are used where necessary. In , the origin is denoted by and the unit vectors are denoted by

,

and . In vector , the -th component is denoted by [] .

2.3. The Denavit-Hartenberg convention

normal and the axis.

The ‘Denavit-Hartenberg parameters’ [11] are used as a transformation method between coordinate systems. The following parameters are used in the transformations: as the length of the common normal between the and +1 axes, as the angle from the -axis towards the +1 -axis around the +1 -axis in the clockwise direction, as the length between the common normal and the origin of the coordinate system , and as the angle between the common

Let be the transformation matrix from +1 and . Then, as the product of matrices representing rotation and translation, is expressed by using the above parameters as = ⎜ ⎛ cos sin − sin cos cos cos sin 0 cos

0 sin sin

cos − cos sin sin ⎟ .

1 − 1 ⎞ ⎟ ⎠ − sin ⎟ . − cos ⎟ ⎞ ⎠ ( 1 ) ( 2 ) +1 = −1 , where

For expressing the transformation of the coordinate system, afine coordinates are used as follows: if the coordinates of a point is expressed as ( , , ) and also as +1 (+1 , +1 , +1 ) , they are denoted by = ( , , , 1) and +1 = (+1 , +1 , +1 , 1) , respectively, in which the last coordinates represent translation. Then, by the definition of , we have = +1 and −1 = ⎜ ⎛ ⎝

cos − sin cos ⎜ sin sin 0

sin cos cos − cos sin 0 sin cos 0 0

If + 1 coordinate systems are given, there exist coordinate transformations between neighboring coordinate systems. Therefore, if the coordinates of a point are given as 7 with respect to 7 (the end-efector), the coordinates of the same point with respect to 1 (the global 1 coordinate system) is obtained by multiplying in sequence, such that 1 = Aeq7 ,

Aeq = 1 2 3 4 5 6.

Note that the inverse transformation is given as Aeq−1 = 6−1 5−1 4−1 3−1 2−1 1−1.

3. Solving the inverse kinematic problem

3.1. The forward kinematics

We first consider the forward kinematics problem of myCobot. Let = ⃖⃖⃖⃖⃖⃖⃖⃗7 be the vector 1 from the origin of 1 (position of the root of the manipulator) to the origin of 7 (position of the end-efector), and let = (1, 2, 3) = 1[7 ],

= ( 1, 2, 3) = 1[7 ], = ( 1, 2, 3) = 1[7 ] be the vectors that are parallel to the unit vectors on the , , and axes of 7, respectively. Then, , , , is represented with respect to 1 as ⎛ 1 ⎞ ⎜ 3⎟ ⎝0⎠

1 ⎛ ⎞ ⎜0⎟ ⎝0⎠ ⎛ 1

⎞ ⎜ 3⎟ ⎝ 0 ⎠ ⎜ 2⎟ = Aeq ⎜0⎟ , ⎜ 2⎟ = Aeq ⎜1⎟ , ⎜ 2⎟ = Aeq ⎜0⎟ ,

⎜ 2⎟ = Aeq ⎜0⎟ , 0 ⎛ ⎞ ⎜1⎟ ⎝0⎠ ⎛ 1

⎞ ⎜ 3⎟ ⎝ 1 ⎠

0 ⎛ ⎞ ⎜0⎟ ⎝1⎠ 0 ⎛ ⎞ ⎜0⎟ ⎝0⎠ ⎛ 1

⎞ ⎜ 3⎟ ⎝ 0 ⎠ therefore, each component of Aeq and Aeq−1 is obtained as ⎛ ⎝0 1 1 1 1 Aeq = ⎜ 2 2 2 2⎟ ,

2 0 2 3 3 0

−(1 1 + 2 2 + 3 3) −( 1 1 + 2 2 + 3 3)⎟ . −( 1 1 + 2 2 + 3 3) ⎟ 1 ⎞ ⎠ ( 3 ) As seen in eq. ( 3 ), by putting the angles of the joints 1, … , 6 into the transformation matrix , the position and orientation of the end-efector are obtained.

3.2. The inverse kinematic problem

To make use of eq. ( 3 ) in myCobot, we substitute some of the joint parameters , , , of myCobot into the transformation matrix . As shown in the schematic diagram of myCobot in taking a positive counterclockwise direction with respect to the axis, the joint parameters are given as { 1, 2, 3, 4, 5, 6} = { /2, 0, 0, /2, /2, 0}, 1 = 2 = 3 = 4 = 5 = 6 = 0.

( 4 )

By substituting the joint parameters in eq. ( 4 ) into the transformation matrix in eq. ( 1 ), Aeq is calculated. Then, by comparing the components of Aeq with the components of Aeq in eq. ( 3 ), a system of 12 polynomial equations in the variables and ( = 1, … , 6 ) is obtained, where = sin and = cos . Each can be obtained by selecting the appropriate equation(s) from the above system and finding the solution. However, in the computation of Gröbner basis, the coeficients of the equations may expand, which makes it dificult to find the solution. Therefore, we focus on the structure of myCobot and solve its inverse kinematic problem from a diferent perspective as follows. 3.3. Solving the inverse kinematic problem with a fixed orientation Inverse kinematics problems for robot manipulators of a certain structure can be solved analytically [12, 13]. Pieper [13] has pointed out that when the end efector of a 6-DOF manipulator has a spherical joint, the inverse kinematics problem can be separated into position and orientation problems of the end-efector. He has further noted that when the rotational axes of three consecutive rotational joints of a 6-DOF manipulator intersect at a single point, the inverse kinematics problem can also be separated into position and orientation problems of the end-efector.

Pieper’s argument suggests that if there exists a combination of three consecutive rotational joints whose rotational axes intersect at a single point, it becomes possible to solve the inverse kinematics problem analytically. However, unfortunately, in the case of myCobot, there are no combinations of three consecutive rotational joints whose rotational axes intersect at a single point, although there are combinations of two consecutive joints whose rotational axes intersect at a single point. Therefore, following Pieper’s approach, we solve the inverse kinematics problem by imposing constraints on the orientation of the end-efector.

For simplicity, we solve the inverse kinematic problem with the condition that the orientation of the end-efector remains constant, i.e. the axis in 7 is parallel to the axis in 1 preserving the same direction. Let = ( 1, 0, 0 ), = ( 0, 1, 0 ), = ( 0, 0, 1 ), then Aeq and Aeq−1 in eq. ( 3 ) become as

Aeq = ⎜ ⎛ Let be the intersection point of axes 4 and 5 , expressed as 1 = ⃖⃖⃖⃖⃖⃗= 1

(, , , 1). By equating two vectors 7 above, we have 7 = 6−1 5−1 ⎜0⎟ = ⎜

7 = Aeq−1 ⎜ ⎟ = ⎜ 0 ⎛ ⎞ ⎜0⎟ ⎝1⎠ ⎛ ⎜ ⎝ − 5 sin 6 − 5 cos 6⎟ , ⎞ ⎟ ⎠ − 6

1 1 − sin 6 = , cos 6 = 2 − Note that is the position of Joint 5. We first express sin and cos ( = 1, … , 6) with the coordinate of the intersection .

Let 7 = ⃖⃖⃖⃖⃖⃗, then 7 is expressed in two ways as 7 ⎛ ⎜ ⎜ ⎝

sin 1 ⎞ − cos 1⎟ , 0 0 ⎟ ⎠ Let 4 be the unit vector in the direction of 4 -axis. Then, 4 and 7 [ 4] are expressed as 4 = 1 2 3

7 [ 4] = 6−1 5−1 4−1 ⎜− sin 6 sin 5⎟ .

By using 7 [ 4] above and Aeq, we obtain another expression for 4 as

4 = Aeq7 [ 4] = (cos 6 sin 5, − sin 6 sin 5, − cos 5, 0).

By comparing each component of 4 in eqs. ( 7 ) and ( 8 ), respectively, we have cos 5 = 0, sin 5 = ±1, cos 1 = ± sin 6 = ± 1 − , sin 1 = ± cos 6 = ± 2 −

. ⎛ ⎞ ⎜ ⎟ ⎝1⎠ . ⎛ ⎜ ⎝ ⎛ ⎝ − 1 − 2⎟ . ⎜ − 3⎟ ⎞ ⎠ 1 cos 6 sin 5 ⎞ − cos 5 0 ⎟ ⎠ 5 ⎟ ⎠ ( 6 ) ( 7 ) ( 8 ) (9) (10) (11) (12) (13) 7[ 5] = 6−1 5−1 ⎜0⎟ = ⎜

0 ⎛ ⎞ ⎜1⎟ ⎝0⎠ ⎛cos 6⎟ ,

⎞ ⎜ 0 ⎟ ⎝ 0 ⎠ Next, 3 = ⃖⃖⃖⃖⃖⃗ is expressed as 1

⎜ 3 = 1 2 3 ⎜ 4⎟

⎟ = ⎜ 0 0 ⎛ ⎞ ⎝ 1 ⎠ ⎛ ⎜ ⎝

Therefore,

5 1 4 sin 1 + 2 cos 1 cos 2 + 3 cos 1(cos 2 cos 3 − sin 2 sin 3) ⎞ − 4 cos 1 + 2 sin 1 cos 2 + 3 sin 1(cos 2 cos 3 − sin 2 sin 3)⎟ .

1 + 2 sin 2 + 3(sin 2 cos 3 + cos 2 sin 3) From the third component of 3 and the constraints,we have = 2 + 2 + ( − 1)2 = 22 + 32 + 42 + 2 2 3 cos 3. cos 3 = 2 + 2 + ( − 1)2 − 22 − 32 − 42 2 2 3

, sin 3 = ±√1 − (cos 3)2 1 + 2 sin 2 + 3(sin 2 cos 3 + cos 2 sin 3) = , (sin 2)2 + (cos 2)2 = 1.

On the other hand, 3 is also expressed as 1 = (, , , 1) 1 [1]2 + 1 [2]2 + (1 [3] − 1)2 = 3[1]2 + 3[2]2 + ( 3[3] − 1)2, then we have as shown in eq. ( 5 ). Let = using 7 [ 5] expressed as If cos 3, sin 3 is obtained, cos 2, sin 2 is easily found from eq. (12).

Finally to find

cos 4 and sin 4, let 5 be the unit vector in the direction of 5 -axis. Then, by we have the expression for 5 in two ways as

Next, from eq. (6) and the trigonometric identity, we have

− 6 = − 3. ( 1 − 2 ) = 1.

) + ( 2 − 2 5 5

Finally, by equating the first and the third components in the vector in the right-most-hand of eq. (10) with and , respectively, we have 4 sin 1 + 2 cos 1 cos 2 + 3 cos 1(cos 2 cos 3 − sin 2 sin 3) = ,

1 + 2 sin 2 + 3(sin 2 cos 3 + cos 2 sin 3) = .

Then, by substituting sin 1 = 5 1− in eq. (9) into the above equations, assuming 2− and cos 1 = 5 ifrst that sin 5 = 1, we have the following system of equations in sin 2, cos 2, sin 3, cos 3: 2 − 4 5

1 − + 2 5 cos 2 + 3 1 −5 (cos 2 cos 3 − sin 2 sin 3) = , 1 + 2 sin 2 + 3(sin 2 cos 3 + cos 2 sin 3) = , (sin 2)2 + (cos 2)2 = 1, (sin 3)2 + (cos 3)2 = 1, in which the last two equations are added as trigonometric identities.

Then, the solution of the system in eq. (16) gives the value of cos 3 as cos 3 = 2 2 3(11 − ) 2 (− 22 12 − 32 12 + 12 12 + 22 42 + 2 22 1 + 2 32 1 − 2 1 12 − 2 2 4 5 − 22 2 − 32 2 + 12 2 + 52 2 − 2 2 42 + 2 4 5 + 42 2 − 2 12 1 + 4 1 1 − 2 1 2 (14) (15) (16) + 12 2 − 2 1 2 + 2 2). (17) By putting cos 3 in eq. (17) into the first equation in eq. ( 11) and multiplying both sides by ( 1 − ) 2, we have 42 12 + 42 22 − 2( 42 1 + 4 5 2) + ( 42 + 52 − 12) 2 + 2 1 3 − 4

− 2 42 2 + 2 4 5 + ( 42 − 12) 2 + 2 1 2 − 2 2 = 0. (18) In the case sin 5 = −1, perform the same calculation with sin 1 = 5 1− . In 2− and cos 1 = 5 this case, the sign in eqs. (17) and (18) changes in part, which gives the following equations. − 22 2 − 32 2 + 12 2 + 52 2 − 2 2 42 − 2 4 5 + 42 2 − 2 12 1 + 4 1 1 − 2 1 2 + 12 2 − 2 1 2 + 2 2), (19) 42 12 + 42 22 − 2( 42 1 − 4 5 2) + ( 42 + 52 − 12) 2 + 2 1 3 − 4

Furthermore, eq. (18) or eq. (20) together with eqs. (14) and (15), we obtain a system of polynomial equations in , , . Solving the system (by using Gröbner basis computation, etc.) gives the position of the intersection point .

4. Concluding remarks

In this paper, we have proposed a solution for the inverse kinematic problem of a 6-DOF manipulator under the condition that the orientation of the end-efector remains constant.

Our first task from here includes the verification of the solution with the CGS-QE method and eficiently solving the system of polynomial equations by using CGS, as we have proposed in the previous work. In addition, the orientation was specified this time for simplicity. However, orientation is not always constant in real-world manipulators. It is therefore necessary to develop the problem into an inverse kinematics problem for arbitrary orientations. There are several conditions on the geometry of the 6-DOF manipulator to be analytically solvable [13], and a method with computer algebra has been proposed for solving the inverse kinematic problem of the 6-DOF manipulator [3]. We will look for better solutions with reference to these methods. [9] S. Otaki, A. Terui, M. Mikawa, A Design and an Implementation of an Inverse Kinematics Computation in Robotics Using Real Quantifier Elimination based on Comprehensive Gröbner Systems, Preprint, 2021. doi:10.48550/arXiv.2111.00384, arXiv:2111.00384. [10] M. Yoshizawa, A. Terui, M. Mikawa, Inverse Kinematics and Path Planning of Manipulator Using Real Quantifier Elimination Based on Comprehensive Gröbner Systems, in: Computer Algebra in Scientific Computing: CASC 2023, volume 14139 of Lecture Notes in Computer Science, Springer, 2023, pp. 393–419. doi:10.1007/978-3-031-41724-5_21. [11] B. Siciliano, L. Sciavicco, L. Villani, G. Oriolo, Robotics: Modelling, Planning and Control,

Springer, 2008. doi:10.1007/978-1-84628-642-1. [12] A. A. Brandstötter, Mathias, M. Hofbaur, An analytical solution of the inverse kinematics problem of industrial serial manipulators with an ortho-parallel basis and a spherical wrist, in: Proceedings of the Austrian Robotics Workshop 2014, 2014, pp. 7–11. [13] D. L. Pieper, The kinematics of manipulators under computer control, Stanford University, 1969. URL: https://apps.dtic.mil/sti/citations/AD0680036, accessed 2024-08-05, Ph.D. Thesis.

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