Proceedings of the 2011 IEEE International Conference on Mechatronics and Automation August 7 - 10, Beijing, China
Optimum Inverse Kinematic Method for a 12 DOF Manipulator Ali Pyambri Paramani BSc Student Sharif University of Technology ,International Campus ,Iran
[email protected] I. INTRODUCTION The area of inverse kinematics of robots, mainly manipulators, has been widely researched, and several solutions exist. The solutions provided by analytical methods are specific to a particular robot configuration and are not applicable to other robots. Most researchers resort to iterative methods for inverse kinematics using the Jacobian matrix to avoid the difficulty of finding a closed-form joint solution. Since a closed-form joint solution, if available, has many advantages over iterative methods [1]. A humanoid robot is a multi-jointed mechanism that mechanically emulates a human’s functions, movements and activities. It can be considered as a biped robot with an upper main body, linking two arms, a neck and a head, or as a combination of multiple manipulators, which are themselves linked together through waist and neck joints to emulate a human’s functions Most researchers often use iterative methods for controlling humanoid robots [2],[3],[4]. One of the iterative methods makes use of the Jacobian matrix [5]. Singularity, redundancy, and computational complexity are the main drawbacks of using the inverse-Jacobian-matrix approach . Another commonly adopted method for inverse kinematics is the geometric method [6], [7]. However, the geometric method requires geometric intuition in solving the joint solution of a manipulator, and it may become more difficult to obtain the joint solution when more than four or five joints are involved. Another method, called the inversetransform technique, was presented by Paul et al [8] to obtain the inverse kinematic joint solution of a 6-DOF robot manipulator. Cui et al [9] derived a closed-form joint solution for a 6-DOF humanoid robot arm but only the solution of joint angles within a certain range was considered and the singularities were not discussed.Solving kinematic equations is carried out by nonlinear methods in which efforts are made to calculate the values of the joint angles with the numerical values of inverse matrix. For instance, will have twelve equations and six indeterminate equations for the manipulator with 6 degree of freedom. However, only three equations out of nine equations are independent [10] .These equations are nonlinear and non-algebraic and their solutions may be difficult. It is clear that kinematic equations would be much more complex for a general mechanism with six degree of freedom in which all interface parameters are nonzero. Each manipulator is soluble when solver can determine all the sets of joint variables related to the supposed place and orientation by the algorithmic pursuit. The present paper analyzes inverse kinematic of Longshank robot with 12 DOF. “Fig 1”
Abstract - In General , there are two methods to analyse the inverse kinematic of manipulators, one of which can be selected with respect to the conditions and the type of the manipulator. One of the methods is the closed solution which is based on the analytical expressions or forth degree or less polynomial solution in which the calculations are non-repetitive. The other method is the numerical solution. In the numerical solutions , the numbers are repeated and generally it is much slower than the closed solutions. The slowness of this method is so noticeable in such a way that principally there is no interest to use the numerical solutions to solve kinematic equations . The purpose of the present paper is to present a compound method that is made up of the numerical and the closed solutions which does not have the problems of the current methods and can be generalized to similar robots with different degrees of freedom . The method which will be presented is based on the connecting lines between the pairs of links and the conditions under which the size of the connecting lines are created . The inverse matrix equations are not applied in this analysis. However, the positions of all links are expressed based on the geometrical position of the interfaces and the use of the mathematical function. Index –algorithm, condition, interval, angular displacement
Fig. 1 Schematic of LongShank Robot with 12 DOF
978-1-4244-8114-9/11/$26.00 ©2011 IEEE
2020
Table I
II. ANALYSES METHOD A. Basic rules: Consider link OA and AB, with magnitude |a| , and they are connected together with joint A. distance between point B and O equals to L .And L can change between 0 and 2a , Know consider the case that has position of point O and A in reference coordinate system, and then get the value of all possible positions for point B for different length of L . Solving this equation system:
Longshank Characteristics Degree of freedom joint Goal Rigid Body Coordinate system Link Length | OO′| Triangle EGE ′
(1)
12 Revolute (12R) G Triangle EGE ′ with end effectors G Based on point O a |EE ′|
|EG|
d
g .
|E ′G|
g
connecting lines
If initialized the value of L between 0 and 2a with required accuracy then find all possible position for point B. Outcome: managing position of link AB and OA with magnitude of line L. In other system with more than 3 links ,and replace all two sets of link with one line L, the magnitude and direction of L is equals two vector sum of 2 link.
LEFT SHANK L1 BD L3 OB
OE OD
L4 L5
O`E` O`D`
RIGHT SHANK L6 O`B` L7 B`D'
L8 L9
the Position of point G (Goal Point) from point E is fixed and is equal to g (it moves on circle with radius g and center of E). It should be satisfy the conditions and terms of motion which is defined in next part. Now, with draw lines L and L the workspace of joint E is as follows. With respect to “Fig 2” it is clear that Point E is located on the circles with radius of g and center of G and it moves on the circle
B. analytical analysis : Generally, analytical methods can be used in position analysis to yield results with high degree of accuracy. This accuracy comes with a price in that the methods often become numerically intensive. Complex method involving higher order math, have been developed for position analysis. Using geometrical conceptions, complex methods can be expressed in a simpler way and algorithms can be reduced. Another problem that may rise in solving kinematic equations is the existence of multiple solutions [11].Consider planar manipulator with two Interfaces in which the final operator is located in a specified place and orientation as point A: (AX, AY). For the inverse kinematic, instead of calculation at Point A: (AX, AY), we can perform the calculations at Point B :(0 , BY). In case BX=0, BY=|OA|, it then moved all the obtained solutions to Point B. Application of this method has some advantages. For instance, if a configuration is found on Point B, it is just required to acquire another configuration by symmetrizing the solutions on the X-axis and then the configuration is transferred to the first point. If there is a large number of links, this strategy can increase the speed of calculations and can reduce the number of calculations.
B. Preliminary Analysis joint D Point D is located on a circle with center of E and radius "a" .The workspace of Point D in terms of Point E is expressed based on the following relations. In addition, the limit of Point YD L and YD 0 can be expressed as: D, for XD (2) The Goal is finding exact interval of L3 from Eq.”(2)” . , , (3) (4) (5) .
(6)
Put “(3)”,”(4)”,”(5)”,”(6)” in “(2)” :
II. CONDITION In this section, the analysis basis will be discussed. The inverse matrix equations will not be used in this analysis; however, positions of all the links are expressed based on the geometrical position of interfaces and through mathematical function. The connecting lines, based on which the analysis is carried out, are introduced in Table I. A. Preliminary Analysis joint E The analysis is based on joint E and joint D which is indicating position and workspace of system. It obvious that
Fig. 2 Schematics of LongShank Robot with connecting line
2021
.
Table II
(7) ,
0.4
(8)(9)
i, j
, 1 4 1 2 Now, in case of selecting L3 value from the interval and putting it in the equation “(7)” the workspace of joint D will be calculated correctly. In other words, all the positions that joint D can be placed are calculable based on different values of L3. , 10 , 11 2
1
1 4 2 1
2
,
,
(13)
By solving the following system of equations, a specific and unique coordinates can be obtained for joint E and for each of the obtained coordinates of Point D: /2
,
2
|
j
k
2
2 ;}
|;
(20)
;}
(21)
(16)
′ ′
|
|
{ .
n, 1
;
{
(14)(15)
′
i
The workspace of Points E, B and C, can be expressed through the following formula in case the value of L3 is specified. The specified functions stated the positions of Points E, B and C in terms of each line of the following Table II with respect to the constants values of L3 and L4 and different values of L5. As stated in analytical analysis section, all calculations can be done according to the Point: β βX , βY 0, L ,and it then transfers the obtained solutions to the obtained Points D in which: 19
12
,
N, 1
i L4 L3 L5 min L5 max 1 L3 2 L3 … … … … … n-1 L3 L3 n Subtitles i and j specified accuracy of L4 and L5 interval
4
17
′
.
22
0,
(23)
′ ′
18
′
C. Preliminary Analysis joint A.B,C
.
For design situation, these complex methods can be difficult to understand and impalement. The following section describes how to find the intersection points between circles on a plane with radius L4 and L5 and with center O and D, the following notation is used. The aim is to find points: ,
,
,
,
,
2
• •
4
25 .
26
(19)(20)(21) 4
2
If they exist .First calculate the distance d between the centers of the circles. d = || L3 ||. •
24
2
2
If L3 > L4 +L5 then there are no solutions, the circles are separate. If L3 < |L4 – L5| then there are no solutions because one circle is contained within the other. If L3 = 0 and L4 = L5 then the circles are coincident and there are an infinite number of solutions
2
4
.
27
.
28
. 29 4 2 With respect to the explanations presented in analytical analysis Section, it is clear that equations “(23)” to “(29)” only result in the configurations on the one side of the “vertical” axis. Therefore, to find the remaining configurations, it is
2022
Table III
required to symmetries all the obtained coordinates in proportion to the axis of ordinates. ;
;
,
,
,
,
,
_
,
,
}
, ,
, ,
(34) ,
,
,
,
A′ , B ′ , C ′, D′ A′, B ′ , C ′
,
′
′
For each coordinate E, there are two possible coordinates for joint ’ which it is calculated through solving in the O coordinate system. (42) To analyze the right shank, it would be enough to change the names of some parameters in the equations of the left shank (shown in Table III).
Now, in case of choosing the value of L7 from the interval “(45)” and putting it in equation “(46)”, the position of ’ will be obtained:
(36)(37)(38)
.
′
B. Analysis joint
XD , YD , point “O” in fixed: ,
A , B , C, D A, B, C
A. Preliminary Analysis joint
(35)
The magnitude of the angular displacement vector is the angle between the initial and final of a link during an interval. This magnitude will be in rotational unit (degree, radians), and denoting either clockwise or counter clockwise specifies the direction. The aim is to find points A, B, C, for transfer rigid body ( O, ijA, ijB , ijC, β) With Angular displacement of θD for 0, L3 to point D
L L L LC ′
(33)
D. Angular displacement
Point β
L L L LC
(30)(31)(32)
(39)
′
.
′
| |
,
,| |
0,5
0,4
43 44 45
(40)(41) ′′
2
′′
,
2
′′
′′ ′
′
′
′
46 47
′′ ′
′
′′
48
′′ ′
C. Analysis joint
49
′
′
′
′
The workplace of joints ’, ’ and ’ can be expressed by knowing the value of L7 and by the following formula. The specified functions expressed the positions of Points ’, ’ and ’, for each line of the following table for the values of L7, L9 and L8. As mentioned in analytical analysis section, all the calculations can be made in terms of the Point (50) , 0, And then we displace the obtained configuration as a solid object. Point ′ is displaced to Point ’ obtained from the previous equations and Point O is displaced to Point ’.
III. RAIGHT SHANK ANALYSES For the known coordinates of Points G and E, there are two coordinates for Point ′ . By careful attention to the structure of the robot, it is clear that the right shank of Points: O′ E′ Is the left Shank that has moved up or down by a certain amount and it has moved from Point O to O′ by . Due to this, to calculate the inverse kinematic of the right shank, one can first calculate its inverse kinematic on the O coordinate system and then transfer the obtained configuration as a solid object to Point O′ by “ ”.
2 {
2023
;
2 ;}
(51)
|
|
2 When
|;
|
{
;
, 1
, 1
,
(52)
3.5394,11.0402 ,
,
,
′
,
′
′
, ′,
′
′
′
′
′
,
′
′
′
′
′
′
′
,
,
75 76
In term of required accuracy select interval for i , j Then based on Eq” (19)”,” (35)” calculate all configuration of Left shank for XD = 0, YD =21.14 (all step shown in TABLE VI , V) and “Fig 3”and ,”Fig 4 “)
,
,
(63)(64) (65)(66)
′
′
′
′
′
′
′
′
′
′
′
′
′
′
18.032 ,20.8457
B. STEP II : Tables
(59)(60) (61)(62)
′
0,
′
′
(74)
Select, D = (10, 18.62) from interval “(75)”and” (76)” and then calculate E = (10.55, 30.6)
(56)(57)(58)
D. Angular displacement ′
22.17
(53)(54)(55)
Equations of this section are quite similar to the inverse kinematic equations of the left shank and avoid repeating them here.
′
0.375 .
21.1441
′ ′
(67) (68)
′
12,0
8,
{
24 21.1141 21.1141
{
| |21.1141 |21.1441
;
(77) (78) (78) 24 ;} (79)
24 |;
;} (80)
In terms of Eq” (36)”to“(41)” For i=3, j=2 with L3=21.14, L4=5, L5=19.
′
6.960, 9.774 ,
′ ′
21.1141
C. STEP III : Angular Displacement
′
2
,1
′
2.693, 16.901 ,
(81)(82)
′
8.540 , 12.684
(83)
′ ′
′
′
In terms of Eq “(2)”to”(41)” find one of the specific configuration of left shank of Robot:
′ ′
′
′
′
′
′
28.26° 97.19° 52.68° 95.71°
IV. INITIALIZED LEFT SHANKS A. STEP I : Basic Calculation ,
, ,
, ,
12,32,12,4,2,2
1.501 , 11.905
(69)
10 , 18.62
Calculate basic interval From Eq” (2-13)”: 12 0.375
32 .
14 1168
10.37 , 13.61 10.55 , 30.6
(70) 14
Solving Equation System “(71)”: 0.5 ∆ 4 1168 1 0.375 32 20.54,48.021 0,48 ∆ 0
72
20.54,48
(73)
1.520 15.20 12 32
(84)
85 86 87 88 89 90
Table IV L4
(71)
14
10 18.62 1.501 11.905 10.37 13.61 1.520 15.20
i=1 i=2 i=3 … i=10 i=11 i=12
2024
1 3 5 … 19 21 23
L3 21.1141 21.1141 21.1141 21.1141 21.1141 21.1141 21.1141
L5 min 20.1141 18.1141 16.1141 … 2.1141 0.1141 1.8859
L5 max 22.1141 24 24 … 24 24 24
Table V
i=1 i=2 i=3 i=4 i=5 i=6
L 1 3 5 7 9 11
j=1 L min 20.11 18.11 16.11 14.11 12.11 10.11
j=2
j=3
j=4
j=5
21.11 21.11 19.11 15.11 13.11 11.11
22.11 20.11 17.11 15.11 13.11
22.11 18.11 16.11 14.11
20.11 18.11 16.11
V. REDUCTION OF CALCULATIONS OVERLAPPING AND TRANSFER
j=6 L max
23.11 21.11 19.11
BY
j=7
22.11 20.11
DATA
Avoiding making repetitive calculations is one of the methods for reducing calculations in the numerical methods. In LongShank robot, some possible solutions for the configuration of the right Shank can be calculated by calculating inverse kinematic of the left Shank and only by using previous data. Number of these solutions, with respect to the specifications of the robot: YG , XG , d , a , a Based on the required accuracy of the problem, a certain number of configurations were calculated for the left Shank. If all the configurations are regarded as a solid object, to calculate the configurations of the right shank, after specifying Points ′ and ′ , some of the configurations of the left Shank in which : L , Can be displaced as a solid object to the right Shank. Consider, the value of L3 is within the interval of: 20.45 48 (91) And regarding the Points ′ and ′ , the values are : 19.22 43.22 (92) If define ∆ as the intersection of the two interval of L3and L7: ∆ 20.45 , 43.22 (93) it means that eighty percent of the interval L3 and L7 has intersection. This point confirms that all the configurations of the left Shank in which: 20.45 , 43.22 (94) Can be displaced to the right Shank ∆ 95 if L is selected, all of the inverse kinematic calculations of the right shank will be similar to the ones of the left side. Only some changes are made in the equations of Angular displacement and translation section.
Fig. 3 Work Space of joint A,B,C for L3=21.14 and XD=0 , YD=21.14 (GREEN: Joint A,RED: Joint B , BLUE : Joint C)
Fig. 4 all configuration of Left shank for XD=0 , YD=21.14 0
