CN107756400A - A kind of 6R Robotic inverse kinematics geometry solving methods based on spinor theory - Google Patents

A kind of 6R Robotic inverse kinematics geometry solving methods based on spinor theory Download PDF

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CN107756400A
CN107756400A CN201710953599.4A CN201710953599A CN107756400A CN 107756400 A CN107756400 A CN 107756400A CN 201710953599 A CN201710953599 A CN 201710953599A CN 107756400 A CN107756400 A CN 107756400A
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刘志峰
许静静
赵永胜
蔡力钢
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Beijing University of Technology
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    • B25—HAND TOOLS; PORTABLE POWER-DRIVEN TOOLS; MANIPULATORS
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Abstract

The invention discloses a kind of 6R Robotic inverse kinematics geometry solving methods based on spinor theory, belong to the inverse solution technique study field of robot kinematics.Basis coordinates system and tool coordinates system are established, 6R robot kinematics' parameters are determined by basis coordinates system and tool coordinates system, and establish positive kinematics model.The inverse solution Kinematic Decomposition in first three joint of 6R robots is described and establishes hexa-atomic quadratic equation group.Based on spinor positive kinematics model solution initial position qs1Corresponding target location qe1.This method is combined geometric description with spinor theory, and geometric meaning definitely, is solved for Algebraic Equation set by simplifying inverse arithmetic, effectively improves computational efficiency, can be that robot motion controls a kind of new inverse solution processing method of offer in real time.

Description

A kind of 6R Robotic inverse kinematics geometry solving methods based on spinor theory
Technical field
The invention belongs to the inverse solution technique study field of robot kinematics, more particularly to a kind of 6R machines based on spinor theory Device people's inverse kinematics geometry solving method.
Background technology
Kinematics analysis is to realize the basis of motion control, mainly establishes the mapping model of joint variable and end pose. Wherein against solution problem, i.e. known end pose solution joint variable the problem of, be one of robot field's study hotspot, it is solved Efficiency directly affects the real-time performance of motion planning and robot control.The inverse solution modeling of robot kinematics at present is based primarily upon D-H methods And spinor theory, researchers compared for two methods and find that the application of the latter has advantages below：It can avoid establishing part Coordinate system, simplified mathematical model simultaneously overcome singularity caused by local parameter；Its geometric meaning is clear and definite, can easily determine to produce The condition and number of raw more solutions.The Paden-Kahan subproblems for being currently based on spinor description are Converse solved by extensive robot In problem, but this method is not suitable for the six-DOF robot of arbitrary configuration, therefore most scholars are asked based on three kinds of basic sons Topic extend and has proposed some new subproblem models, as Tan subproblem two improve establish for " around The mathematical modeling of the rotary motion of two non-intersect axles " subproblem；Chen describe it is a kind of " around the motion of three axis rotations, Two of which diameter parallel and with the 3rd axis antarafacial " subproblem, carried out detailed spinor against solution for it and described.It is based on In six degree of freedom serial manipulator against being widely applied in solution problem, research shows to work as 6R robots these subproblem models Meet that there is its Inverse Kinematics Problem closing to solve during Piper criterions.Sariyildi etc., which is based on three subproblems, realizes 6R machines The inverse kinematics of people；Lv Shizeng etc. reduces dependence to subproblem by introducing Wu Ritt's method in spinor method, and will The inverse solution problem of 6R robots is converted into the Solve problems of hexa-atomic eight power journey group.In order to further simplify 6R robot inverse solution moulds The inverse solution problem of 6R robots is converted into a hexa-atomic quadratic equation group and one three by type, this method by Describing Method The solution of first quadratic equation group, so that its geometric meaning becomes apparent from, solution procedure is simpler.
The content of the invention
The purpose of the present invention aims to provide a kind of 6R Robotic inverse kinematics geometry solving methods based on spinor theory.Should Method is mainly characterized by by combining Describing Method and spinor theory, is proposed for 6R robot kinematics against solution problem A kind of geometric meaning is clear, the simple inverse kinematics model of solution procedure.
To achieve the above object, the technological means that the present invention uses is a kind of 6R robot inverses motion based on spinor theory Geometry solving method is learned, the implementation process of this method is as follows：
S1, basis coordinates system and tool coordinates system are established, 6R robot motions are determined by basis coordinates system and tool coordinates system Parameter is learned, and establishes positive kinematics model.
S2, such as Fig. 1, the inverse solution Kinematic Decomposition in first three joint of 6R robots is described：By point qeTo point c2=[x2 y2 z2] rotary motion around joint 1；Point c2To point c1=[x1 y1 z1] rotary motion around joint 2；Point c1To point qsAround joint 3 Rotary motion.Wherein qs, qeThe respectively initial and target location of robot end, the geometrical relationship description according to Fig. 1 are built Stand with x1,y1,z1,x2,y2And z2For the hexa-atomic quadratic equation group of variable.
S3, using in MATLAB " SOLVE " function solve above equation group, obtain c1And c2Coordinate vector.
S4, since then, the inverse solution in first three joint move whole story position, it is known that being built respectively based on Paden-Kahan subproblems 1 First three joint angle variable θ of vertical 6R robots1, θ2And θ3Explicit solution model, and solved.
S5, some initial position q as after in the inverse solution motion in three joints is taken on the axis in joint 6s1, and based on rotation Measure positive kinematics model solution initial position qs1Corresponding target location qe1。
S6, such as Fig. 2, the inverse solution Kinematic Decomposition in joint 4 and 5 is described：By point qe1To point c3=[x3y3z3] rotation fortune It is dynamic；By point c3To point qs1Rotary motion.Shown according to Fig. 2, established according to geometrical relationship with x3,y3And z3For the ternary two of variable Equation of n th order n group, and solved using " SOLVE " function, obtain c3Coordinate vector.
S7, since then, the inverse solution in joint 4 and joint 5 move whole story position, it is known that distinguishing based on Paden-Kahan subproblems 1 Establish the joint angle variable θ in joint 4 and joint 54And θ5Explicit solution model, and solved.
S8, take any point q not on the axis of joint 6s2, based on spinor positive kinematics model solution qs2Corresponding mesh Cursor position qe2, then known to the inverse solution motion whole story position in joint 6.θ is similarly solved based on subproblem 16。
The method have the characteristics that the reverse kinematics formula of 6R robots is converted into one based on Describing Method The solution of hexa-atomic quadratic equation group and a ternary quadratic equation group, inverse solution model is simplified, and geometric meaning is definitely, is The real time kinematics control of 6R robots provides certain method support.This method is combined geometric description with spinor theory, geometry Meaning definitely, is solved for Algebraic Equation set by simplifying inverse arithmetic, effectively improves computational efficiency, can be robot Motion control in real time provides a kind of new inverse solution processing method.
Brief description of the drawings
The inverse solution kinematic geometry description in Fig. 1 joints 1,2 and 3；
The inverse solution kinematic geometry description in Fig. 2 joints 4 and 5；
Fig. 3 6R robots parameter coordinate system.
Embodiment
Below in conjunction with accompanying drawing 1-3, the present invention is described in detail.
S1 determines 6R robot kinematics parameter and establishes positive kinematics model
As shown in Figure 3, it is known that each joint position vector of original state and rotating vector of the 6R robots are as follows：
Wherein ri, 1≤i≤6 represent i joints in the position vector of basis coordinates system, ωiRepresent the rotating vector in i joints.
Based on spinor theory, the 6R robots positive kinematics model is expressed as,
Wherein gst(θ),gst(0) the initial pose and object pose of robot end is represented respectively,Represent that i is closed The exponent product form of rotary motion is saved,
θ in formulaiFor i-th of joint angular displacement；It is i joints rotating vector ωiAnother representation, by ωi= [ω1 ω2 ω3] be defined asThenνi It is the linear velocity of i joint motions, νi=-ωi×ri。
Given object pose,
S2 establishes hexa-atomic quadratic equation group for first three joint
Rotary motion based on first three joint of spinor theory is described as,
Q ' in formulasWith q 'eInitial position and target location of the 6R robot ends in the rotary motion are represented respectively.By S1 Understand,
q′s=[xs ys zs1]=[0 744 940 1]
q′e=[xe ye ze1]=[- 936.6611 631.7859 570.0752 1]
If move across point c1And c2, the geometric description according to Fig. 1 establishes relationship below group,
Wherein q1, q2And q3Any point respectively on joint 1, joint 2 and the rotation axis of joint 3, takes for simplified model q1=[0 0 0], q2=[0 150 250] and q3=[0 150 800], then formula (7) is expressed as equation group,
Equation group (8) is solved using " SOLVE " function in MATLAB, obtains x1,y1,z1,x2,y2And z2Solution.
S3 calculates joint angular displacement1, θ2And θ3
Obtain process point coordinates c1=[x1 y1 z1] and c2=[x2 y2 z2] after, by around joint 1, joint 2 and joint 3 Rotary motion is described as follows respectively,
Based on Paden-Kahan subproblems 1, it is as follows to obtain the display expression formula of joint angular displacement,
Step (4) establishes ternary quadratic equation group for joint 4 and joint 5
Known by formula (3),
WhereinOne is taken on the axis in joint 6 Point qs1=[xs1 ys1 zs1]=[0 744 0], its rotary motion around joint 4 and joint 5 is described as follows,
Wherein q 'e1=g1q′s1=[xe1 ye1 ze1 1]。
If it is c to move across point coordinates3=[x3 y3 z3], the geometric description according to Fig. 2 establishes relationship below group,
Wherein q4=[0 744 940].Then formula (14) is expressed as equation,
Equation group (15) is solved using " SOLVE " function in MATLAB, obtains x3,y3And z3Solution.
Step (5) calculates joint angular displacement4And θ5
Obtain process point coordinates c3=[x3 y3 z3] after, it will be described as follows respectively around the rotary motion in joint 4 and joint 5,
Based on Paden-Kahan subproblems 1, joint angular displacement is obtained4And θ5Display expression formula it is as follows,
Step (6) calculates joint angular displacement6
Take a not point q on the axis of joint 6s2=[0 750 940], its rotary motion around joint 6 are described as,
WhereinSimilarly it is based on Paden- Kahan subproblems 1, obtain joint angular displacement6Display expression formula it is as follows,
It is as shown in table 1 to obtain eight groups of Inverse Kinematics Solutions for solution more than.
1 eight groups of Inverse Kinematics Solutions of table

Claims (2)

  1. A kind of 1. 6R Robotic inverse kinematics geometry solving methods based on spinor theory, it is characterised in that：The realization of this method Process is as follows：
    S1, basis coordinates system and tool coordinates system are established, determine that 6R robot kinematics join by basis coordinates system and tool coordinates system Number, and establish positive kinematics model；
    S2, the inverse solution Kinematic Decomposition description by first three joint of 6R robots：By point qeTo point c2=[x2 y2 z2] around joint 1 Rotary motion；Point c2To point c1=[x1 y1 z1] rotary motion around joint 2；Point c1To point qsRotary motion around joint 3； Wherein qs, qeThe respectively initial and target location of robot end, established according to geometrical relationship description with x1,y1,z1,x2,y2 And z2For the hexa-atomic quadratic equation group of variable；
    S3, using in MATLAB " SOLVE " function solve above equation group, obtain c1And c2Coordinate vector；
    S4, since then, the inverse solution in first three joint move whole story position, it is known that establishing 6R respectively based on Paden-Kahan subproblems 1 First three joint angle variable θ of robot1, θ2And θ3Explicit solution model, and solved；
    S5, some initial position q as after in the inverse solution motion in three joints is taken on the axis in joint 6s1, and based on spinor just Kinematics model solves initial position qs1Corresponding target location qe1；
    S6, the inverse solution Kinematic Decomposition description by joint 4 and 5：By point qe1To point c3=[x3 y3 z3] rotary motion；By point c3Arrive Point qs1Rotary motion；Established according to geometrical relationship with x3,y3And z3For the ternary quadratic equation group of variable, and use " SOLVE " Function solves, and obtains c3Coordinate vector；
    S7, since then, the inverse solution in joint 4 and joint 5 move whole story position, it is known that being established respectively based on Paden-Kahan subproblems 1 Joint 4 and the joint angle variable θ in joint 54And θ5Explicit solution model, and solved；
    S8, take any point q not on the axis of joint 6s2, based on spinor positive kinematics model solution qs2Corresponding target location qe2, then known to the inverse solution motion whole story position in joint 6；θ is similarly solved based on subproblem 16。
  2. 2. a kind of 6R Robotic inverse kinematics geometry solving methods based on spinor theory according to claim 1, it is special Sign is：S1 determines 6R robot kinematics parameter and establishes positive kinematics model
    Each joint position vector of original state and rotating vector of the known 6R robots are as follows：
    <mrow> <mtable> <mtr> <mtd> <mtable> <mtr> <mtd> <mrow> <msub> <mi>r</mi> <mn>1</mn> </msub> <mo>=</mo> <mfenced open = "[" close = "]"> <mtable> <mtr> <mtd> <mn>0</mn> </mtd> <mtd> <mn>0</mn> </mtd> <mtd> <mn>0</mn> </mtd> </mtr> </mtable> </mfenced> </mrow> </mtd> <mtd> <mrow> <msub> <mi>r</mi> <mn>2</mn> </msub> <mo>=</mo> <mfenced open = "[" close = "]"> <mtable> <mtr> <mtd> <mn>0</mn> </mtd> <mtd> <mn>150</mn> </mtd> <mtd> <mn>250</mn> </mtd> </mtr> </mtable> </mfenced> </mrow> </mtd> <mtd> <mrow> <msub> <mi>r</mi> <mn>3</mn> </msub> <mo>=</mo> <mfenced open = "[" close = "]"> <mtable> <mtr> <mtd> <mn>0</mn> </mtd> <mtd> <mn>150</mn> </mtd> <mtd> <mn>800</mn> </mtd> </mtr> </mtable> </mfenced> </mrow> </mtd> </mtr> </mtable> </mtd> </mtr> <mtr> <mtd> <mrow> <msub> <mi>r</mi> <mn>4</mn> </msub> <mo>=</mo> <msub> <mi>r</mi> <mn>5</mn> </msub> <mo>=</mo> <msub> <mi>r</mi> <mn>6</mn> </msub> <mo>=</mo> <mfenced open = "[" close = "]"> <mtable> <mtr> <mtd> <mn>0</mn> </mtd> <mtd> <mn>744</mn> </mtd> <mtd> <mn>940</mn> </mtd> </mtr> </mtable> </mfenced> </mrow> </mtd> </mtr> </mtable> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>1</mn> <mo>)</mo> </mrow> </mrow>
    <mrow> <mtable> <mtr> <mtd> <mrow> <msub> <mi>&amp;omega;</mi> <mn>1</mn> </msub> <mo>=</mo> <mfenced open = "[" close = "]"> <mtable> <mtr> <mtd> <mn>0</mn> </mtd> <mtd> <mn>0</mn> </mtd> <mtd> <mn>1</mn> </mtd> </mtr> </mtable> </mfenced> </mrow> </mtd> <mtd> <mrow> <msub> <mi>&amp;omega;</mi> <mn>2</mn> </msub> <mo>=</mo> <mfenced open = "[" close = "]"> <mtable> <mtr> <mtd> <mn>1</mn> </mtd> <mtd> <mn>0</mn> </mtd> <mtd> <mn>0</mn> </mtd> </mtr> </mtable> </mfenced> </mrow> </mtd> <mtd> <mrow> <msub> <mi>&amp;omega;</mi> <mn>3</mn> </msub> <mo>=</mo> <mfenced open = "[" close = "]"> <mtable> <mtr> <mtd> <mn>1</mn> </mtd> <mtd> <mn>0</mn> </mtd> <mtd> <mn>0</mn> </mtd> </mtr> </mtable> </mfenced> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <msub> <mi>&amp;omega;</mi> <mn>4</mn> </msub> <mo>=</mo> <mfenced open = "[" close = "]"> <mtable> <mtr> <mtd> <mn>0</mn> </mtd> <mtd> <mn>1</mn> </mtd> <mtd> <mn>0</mn> </mtd> </mtr> </mtable> </mfenced> </mrow> </mtd> <mtd> <mrow> <msub> <mi>&amp;omega;</mi> <mn>5</mn> </msub> <mo>=</mo> <mfenced open = "[" close = "]"> <mtable> <mtr> <mtd> <mn>1</mn> </mtd> <mtd> <mn>0</mn> </mtd> <mtd> <mn>0</mn> </mtd> </mtr> </mtable> </mfenced> </mrow> </mtd> <mtd> <mrow> <msub> <mi>&amp;omega;</mi> <mn>6</mn> </msub> <mo>=</mo> <mfenced open = "[" close = "]"> <mtable> <mtr> <mtd> <mn>0</mn> </mtd> <mtd> <mn>0</mn> </mtd> <mtd> <mn>1</mn> </mtd> </mtr> </mtable> </mfenced> </mrow> </mtd> </mtr> </mtable> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>2</mn> <mo>)</mo> </mrow> </mrow>
    Wherein ri, 1≤i≤6 represent i joints in the position vector of basis coordinates system, ωiRepresent the rotating vector in i joints；
    Based on spinor theory, the 6R robots positive kinematics model is expressed as,
    <mrow> <msub> <mi>g</mi> <mrow> <mi>s</mi> <mi>t</mi> </mrow> </msub> <mrow> <mo>(</mo> <mi>&amp;theta;</mi> <mo>)</mo> </mrow> <mo>=</mo> <mi>exp</mi> <mrow> <mo>(</mo> <msub> <mover> <mi>&amp;xi;</mi> <mo>^</mo> </mover> <mn>1</mn> </msub> <msub> <mi>&amp;theta;</mi> <mn>1</mn> </msub> <mo>)</mo> </mrow> <mi>exp</mi> <mrow> <mo>(</mo> <msub> <mover> <mi>&amp;xi;</mi> <mo>^</mo> </mover> <mn>2</mn> </msub> <msub> <mi>&amp;theta;</mi> <mn>2</mn> </msub> <mo>)</mo> </mrow> <mo>...</mo> <mi>exp</mi> <mrow> <mo>(</mo> <msub> <mover> <mi>&amp;xi;</mi> <mo>^</mo> </mover> <mn>6</mn> </msub> <msub> <mi>&amp;theta;</mi> <mn>6</mn> </msub> <mo>)</mo> </mrow> <msub> <mi>g</mi> <mrow> <mi>s</mi> <mi>t</mi> </mrow> </msub> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>3</mn> <mo>)</mo> </mrow> </mrow>
    Wherein gst(θ),gst(0) the initial pose and object pose of robot end is represented respectively,Represent the rotation of i joints The dynamic exponent product form of transhipment,
    <mrow> <mi>exp</mi> <mrow> <mo>(</mo> <mrow> <msub> <mover> <mi>&amp;xi;</mi> <mo>^</mo> </mover> <mi>i</mi> </msub> <msub> <mi>&amp;theta;</mi> <mi>i</mi> </msub> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <mfenced open = "[" close = "]"> <mtable> <mtr> <mtd> <mrow> <mi>exp</mi> <mrow> <mo>(</mo> <mrow> <msub> <mover> <mi>&amp;omega;</mi> <mo>^</mo> </mover> <mi>i</mi> </msub> <msub> <mi>&amp;theta;</mi> <mi>i</mi> </msub> </mrow> <mo>)</mo> </mrow> </mrow> </mtd> <mtd> <mrow> <mrow> <mo>(</mo> <mrow> <mi>I</mi> <mo>-</mo> <mi>exp</mi> <mrow> <mo>(</mo> <mrow> <msub> <mover> <mi>&amp;omega;</mi> <mo>^</mo> </mover> <mi>i</mi> </msub> <msub> <mi>&amp;theta;</mi> <mi>i</mi> </msub> </mrow> <mo>)</mo> </mrow> </mrow> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <mrow> <msub> <mi>&amp;omega;</mi> <mi>i</mi> </msub> <mo>&amp;times;</mo> <msub> <mi>v</mi> <mi>i</mi> </msub> </mrow> <mo>)</mo> </mrow> <mo>+</mo> <msub> <mi>&amp;theta;</mi> <mi>i</mi> </msub> <msub> <mi>&amp;omega;</mi> <mi>i</mi> </msub> <msup> <msub> <mi>&amp;omega;</mi> <mi>i</mi> </msub> <mi>T</mi> </msup> <msub> <mi>v</mi> <mi>i</mi> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mn>0</mn> </mtd> <mtd> <mn>1</mn> </mtd> </mtr> </mtable> </mfenced> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>4</mn> <mo>)</mo> </mrow> </mrow>
    θ in formulaiFor i-th of joint angular displacement；It is i joints rotating vector ωiAnother representation, by ωi=[ω1 ω2 ω3] be defined asThenνiIt is that i is closed Save the linear velocity of motion, νi=-ωi×ri；
    Given object pose,
    <mrow> <msub> <mi>g</mi> <mrow> <mi>s</mi> <mi>t</mi> </mrow> </msub> <mrow> <mo>(</mo> <mi>&amp;theta;</mi> <mo>)</mo> </mrow> <mo>=</mo> <mfenced open = "[" close = "]"> <mtable> <mtr> <mtd> <mn>0.5234</mn> </mtd> <mtd> <mn>0.4033</mn> </mtd> <mtd> <mrow> <mo>-</mo> <mn>0.7506</mn> </mrow> </mtd> <mtd> <mrow> <mo>-</mo> <mn>936.6611</mn> </mrow> </mtd> </mtr> <mtr> <mtd> <mn>0.5979</mn> </mtd> <mtd> <mn>0.4537</mn> </mtd> <mtd> <mn>0.6608</mn> </mtd> <mtd> <mn>631.7859</mn> </mtd> </mtr> <mtr> <mtd> <mn>0.6071</mn> </mtd> <mtd> <mrow> <mo>-</mo> <mn>0.7946</mn> </mrow> </mtd> <mtd> <mrow> <mo>-</mo> <mn>0.0036</mn> </mrow> </mtd> <mtd> <mn>570.0752</mn> </mtd> </mtr> <mtr> <mtd> <mn>0</mn> </mtd> <mtd> <mn>0</mn> </mtd> <mtd> <mn>0</mn> </mtd> <mtd> <mn>1</mn> </mtd> </mtr> </mtable> </mfenced> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>5</mn> <mo>)</mo> </mrow> </mrow>
    S2 establishes hexa-atomic quadratic equation group for first three joint
    Rotary motion based on first three joint of spinor theory is described as,
    <mrow> <mi>exp</mi> <mrow> <mo>(</mo> <msub> <mover> <mi>&amp;xi;</mi> <mo>^</mo> </mover> <mn>1</mn> </msub> <msub> <mi>&amp;theta;</mi> <mn>1</mn> </msub> <mo>)</mo> </mrow> <mi>exp</mi> <mrow> <mo>(</mo> <msub> <mover> <mi>&amp;xi;</mi> <mo>^</mo> </mover> <mn>2</mn> </msub> <msub> <mi>&amp;theta;</mi> <mn>2</mn> </msub> <mo>)</mo> </mrow> <mi>exp</mi> <mrow> <mo>(</mo> <msub> <mover> <mi>&amp;xi;</mi> <mo>^</mo> </mover> <mn>3</mn> </msub> <msub> <mi>&amp;theta;</mi> <mn>3</mn> </msub> <mo>)</mo> </mrow> <msubsup> <mi>q</mi> <mi>s</mi> <mo>&amp;prime;</mo> </msubsup> <mo>=</mo> <msubsup> <mi>q</mi> <mi>e</mi> <mo>&amp;prime;</mo> </msubsup> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>6</mn> <mo>)</mo> </mrow> </mrow>
    Q ' in formulasWith q 'eInitial position and target location of the 6R robot ends in the rotary motion are represented respectively；From S1,
    q′s=[xs ys zs1]=[0 744 940 1]
    q′e=[xe ye ze1]=[- 936.6611 631.7859 570.0752 1]
    If move across point c1And c2, relationship below group is established according to geometric description,
    <mrow> <mfenced open = "{" close = ""> <mtable> <mtr> <mtd> <mrow> <mo>|</mo> <mo>|</mo> <msub> <mi>q</mi> <mn>1</mn> </msub> <mo>-</mo> <msub> <mi>q</mi> <mi>e</mi> </msub> <mo>|</mo> <mo>|</mo> <mo>=</mo> <mo>|</mo> <mo>|</mo> <msub> <mi>q</mi> <mn>1</mn> </msub> <mo>-</mo> <msub> <mi>c</mi> <mn>2</mn> </msub> <mo>|</mo> <mo>|</mo> <mo>,</mo> </mrow> </mtd> <mtd> <mrow> <mo>(</mo> <msub> <mi>q</mi> <mi>e</mi> </msub> <mo>-</mo> <msub> <mi>c</mi> <mn>2</mn> </msub> <mo>)</mo> <msup> <msub> <mi>&amp;omega;</mi> <mn>1</mn> </msub> <mi>T</mi> </msup> <mo>=</mo> <mn>0</mn> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>|</mo> <mo>|</mo> <msub> <mi>q</mi> <mn>2</mn> </msub> <mo>-</mo> <msub> <mi>c</mi> <mn>2</mn> </msub> <mo>|</mo> <mo>|</mo> <mo>=</mo> <mo>|</mo> <mo>|</mo> <msub> <mi>q</mi> <mn>2</mn> </msub> <mo>-</mo> <msub> <mi>c</mi> <mn>1</mn> </msub> <mo>|</mo> <mo>|</mo> <mo>,</mo> </mrow> </mtd> <mtd> <mrow> <mo>(</mo> <msub> <mi>c</mi> <mn>2</mn> </msub> <mo>-</mo> <msub> <mi>c</mi> <mn>1</mn> </msub> <mo>)</mo> <msup> <msub> <mi>&amp;omega;</mi> <mn>2</mn> </msub> <mi>T</mi> </msup> <mo>=</mo> <mn>0</mn> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>|</mo> <mo>|</mo> <msub> <mi>q</mi> <mn>3</mn> </msub> <mo>-</mo> <msub> <mi>c</mi> <mn>1</mn> </msub> <mo>|</mo> <mo>|</mo> <mo>=</mo> <mo>|</mo> <mo>|</mo> <msub> <mi>q</mi> <mn>3</mn> </msub> <mo>-</mo> <msub> <mi>q</mi> <mi>s</mi> </msub> <mo>|</mo> <mo>|</mo> <mo>,</mo> </mrow> </mtd> <mtd> <mrow> <mo>(</mo> <msub> <mi>c</mi> <mn>1</mn> </msub> <mo>-</mo> <msub> <mi>q</mi> <mi>s</mi> </msub> <mo>)</mo> <msup> <msub> <mi>&amp;omega;</mi> <mn>3</mn> </msub> <mi>T</mi> </msup> <mo>=</mo> <mn>0</mn> </mrow> </mtd> </mtr> </mtable> </mfenced> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>7</mn> <mo>)</mo> </mrow> </mrow>
    Wherein q1, q2And q3Any point respectively on joint 1, joint 2 and the rotation axis of joint 3, q is taken for simplified model1= [0 0 0], q2=[0 150 250] and q3=[0 150 800], then formula (7) is expressed as equation group,
    <mrow> <mfenced open = "{" close = ""> <mtable> <mtr> <mtd> <mrow> <msup> <msub> <mi>x</mi> <mn>2</mn> </msub> <mn>2</mn> </msup> <mo>+</mo> <msup> <msub> <mi>y</mi> <mn>2</mn> </msub> <mn>2</mn> </msup> <mo>=</mo> <msup> <msub> <mi>x</mi> <mi>e</mi> </msub> <mn>2</mn> </msup> <mo>+</mo> <msup> <msub> <mi>y</mi> <mi>e</mi> </msub> <mn>2</mn> </msup> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <msup> <msub> <mi>y</mi> <mn>1</mn> </msub> <mn>2</mn> </msup> <mo>-</mo> <mn>300</mn> <msub> <mi>y</mi> <mn>1</mn> </msub> <mo>+</mo> <msup> <msub> <mi>z</mi> <mn>1</mn> </msub> <mn>2</mn> </msup> <mo>-</mo> <mn>500</mn> <msub> <mi>z</mi> <mn>1</mn> </msub> <mo>=</mo> <msup> <msub> <mi>y</mi> <mn>2</mn> </msub> <mn>3</mn> </msup> <mo>-</mo> <mn>300</mn> <msub> <mi>y</mi> <mn>2</mn> </msub> <mo>+</mo> <msup> <msub> <mi>z</mi> <mn>2</mn> </msub> <mn>2</mn> </msup> <mo>-</mo> <mn>500</mn> <msub> <mi>z</mi> <mn>2</mn> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <msup> <msub> <mi>y</mi> <mn>1</mn> </msub> <mn>2</mn> </msup> <mo>-</mo> <mn>300</mn> <msub> <mi>y</mi> <mn>1</mn> </msub> <mo>+</mo> <msup> <msub> <mi>z</mi> <mn>1</mn> </msub> <mn>2</mn> </msup> <mo>-</mo> <mn>1600</mn> <msub> <mi>z</mi> <mn>1</mn> </msub> <mo>=</mo> <msup> <msub> <mi>y</mi> <mi>s</mi> </msub> <mn>2</mn> </msup> <mo>-</mo> <mn>300</mn> <msub> <mi>y</mi> <mi>s</mi> </msub> <mo>+</mo> <msup> <msub> <mi>z</mi> <mi>s</mi> </msub> <mn>2</mn> </msup> <mo>-</mo> <mn>1600</mn> <msub> <mi>z</mi> <mi>s</mi> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <msub> <mi>z</mi> <mn>2</mn> </msub> <mo>=</mo> <msub> <mi>z</mi> <mi>e</mi> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <msub> <mi>x</mi> <mn>1</mn> </msub> <mo>=</mo> <mn>0</mn> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <msub> <mi>x</mi> <mn>2</mn> </msub> <mo>=</mo> <mn>0</mn> </mrow> </mtd> </mtr> </mtable> </mfenced> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>8</mn> <mo>)</mo> </mrow> </mrow>
    Equation group (8) is solved using " SOLVE " function in MATLAB, obtains x1,y1,z1,x2,y2And z2Solution；
    S3 calculates joint angular displacement1, θ2And θ3
    Obtain process point coordinates c1=[x1 y1 z1] and c2=[x2 y2 z2] after, by around the rotation in joint 1, joint 2 and joint 3 Motion is described as follows respectively,
    <mrow> <mfenced open = "{" close = ""> <mtable> <mtr> <mtd> <mrow> <mi>exp</mi> <mrow> <mo>(</mo> <msub> <mover> <mi>&amp;xi;</mi> <mo>^</mo> </mover> <mn>1</mn> </msub> <msub> <mi>&amp;theta;</mi> <mn>1</mn> </msub> <mo>)</mo> </mrow> <msubsup> <mi>c</mi> <mn>2</mn> <mo>&amp;prime;</mo> </msubsup> <mo>=</mo> <msubsup> <mi>q</mi> <mi>e</mi> <mo>&amp;prime;</mo> </msubsup> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mi>exp</mi> <mrow> <mo>(</mo> <msub> <mover> <mi>&amp;xi;</mi> <mo>^</mo> </mover> <mn>2</mn> </msub> <msub> <mi>&amp;theta;</mi> <mn>2</mn> </msub> <mo>)</mo> </mrow> <msubsup> <mi>c</mi> <mn>1</mn> <mo>&amp;prime;</mo> </msubsup> <mo>=</mo> <msubsup> <mi>c</mi> <mn>2</mn> <mo>&amp;prime;</mo> </msubsup> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mi>exp</mi> <mrow> <mo>(</mo> <msub> <mover> <mi>&amp;xi;</mi> <mo>^</mo> </mover> <mn>3</mn> </msub> <msub> <mi>&amp;theta;</mi> <mn>3</mn> </msub> <mo>)</mo> </mrow> <msubsup> <mi>q</mi> <mi>s</mi> <mo>&amp;prime;</mo> </msubsup> <mo>=</mo> <msubsup> <mi>c</mi> <mn>1</mn> <mo>&amp;prime;</mo> </msubsup> </mrow> </mtd> </mtr> </mtable> </mfenced> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>9</mn> <mo>)</mo> </mrow> </mrow>
    Based on Paden-Kahan subproblems 1, it is as follows to obtain the display expression formula of joint angular displacement,
    <mrow> <msub> <mi>&amp;theta;</mi> <mn>1</mn> </msub> <mo>=</mo> <mi>a</mi> <mi> </mi> <mi>t</mi> <mi>a</mi> <mi>n</mi> <mrow> <mo>(</mo> <mfrac> <mrow> <msub> <mi>x</mi> <mi>e</mi> </msub> <msub> <mi>y</mi> <mn>2</mn> </msub> <mo>-</mo> <msub> <mi>y</mi> <mi>e</mi> </msub> <msub> <mi>x</mi> <mn>2</mn> </msub> </mrow> <mrow> <msub> <mi>x</mi> <mn>2</mn> </msub> <msub> <mi>x</mi> <mi>e</mi> </msub> <mo>+</mo> <msub> <mi>y</mi> <mn>2</mn> </msub> <msub> <mi>y</mi> <mi>e</mi> </msub> </mrow> </mfrac> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>10</mn> <mo>)</mo> </mrow> </mrow>
    <mrow> <msub> <mi>&amp;theta;</mi> <mn>2</mn> </msub> <mo>=</mo> <mi>a</mi> <mi> </mi> <mi>t</mi> <mi>a</mi> <mi>n</mi> <mrow> <mo>(</mo> <mfrac> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mn>1</mn> </msub> <mo>-</mo> <mn>150</mn> <mo>)</mo> <mo>(</mo> <msub> <mi>z</mi> <mn>2</mn> </msub> <mo>-</mo> <mn>250</mn> <mo>)</mo> <mo>-</mo> <mo>(</mo> <msub> <mi>y</mi> <mn>2</mn> </msub> <mo>-</mo> <mn>150</mn> <mo>)</mo> <mo>(</mo> <msub> <mi>z</mi> <mn>1</mn> </msub> <mo>-</mo> <mn>250</mn> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mn>1</mn> </msub> <mo>-</mo> <mn>150</mn> <mo>)</mo> <mo>(</mo> <msub> <mi>y</mi> <mn>2</mn> </msub> <mo>-</mo> <mn>150</mn> <mo>)</mo> <mo>+</mo> <mo>(</mo> <msub> <mi>z</mi> <mn>1</mn> </msub> <mo>-</mo> <mn>250</mn> <mo>)</mo> <mo>(</mo> <msub> <mi>z</mi> <mn>2</mn> </msub> <mo>-</mo> <mn>250</mn> <mo>)</mo> </mrow> </mfrac> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>11</mn> <mo>)</mo> </mrow> </mrow>
    <mrow> <msub> <mi>&amp;theta;</mi> <mn>3</mn> </msub> <mo>=</mo> <mi>a</mi> <mi> </mi> <mi>t</mi> <mi>a</mi> <mi>n</mi> <mrow> <mo>(</mo> <mfrac> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mi>s</mi> </msub> <mo>-</mo> <mn>150</mn> <mo>)</mo> <mo>(</mo> <msub> <mi>z</mi> <mn>1</mn> </msub> <mo>-</mo> <mn>800</mn> <mo>)</mo> <mo>-</mo> <mo>(</mo> <msub> <mi>y</mi> <mn>1</mn> </msub> <mo>-</mo> <mn>150</mn> <mo>)</mo> <mo>(</mo> <msub> <mi>z</mi> <mi>s</mi> </msub> <mo>-</mo> <mn>800</mn> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mi>s</mi> </msub> <mo>-</mo> <mn>150</mn> <mo>)</mo> <mo>(</mo> <msub> <mi>y</mi> <mn>1</mn> </msub> <mo>-</mo> <mn>150</mn> <mo>)</mo> <mo>+</mo> <mo>(</mo> <msub> <mi>z</mi> <mi>s</mi> </msub> <mo>-</mo> <mn>800</mn> <mo>)</mo> <mo>(</mo> <msub> <mi>z</mi> <mn>1</mn> </msub> <mo>-</mo> <mn>800</mn> <mo>)</mo> </mrow> </mfrac> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>12</mn> <mo>)</mo> </mrow> </mrow>
    Step (4) establishes ternary quadratic equation group for joint 4 and joint 5
    Known by formula (3),
    <mrow> <mi>exp</mi> <mrow> <mo>(</mo> <msub> <mover> <mi>&amp;xi;</mi> <mo>^</mo> </mover> <mn>4</mn> </msub> <msub> <mi>&amp;theta;</mi> <mn>4</mn> </msub> <mo>)</mo> </mrow> <mi>exp</mi> <mrow> <mo>(</mo> <msub> <mover> <mi>&amp;xi;</mi> <mo>^</mo> </mover> <mn>5</mn> </msub> <msub> <mi>&amp;theta;</mi> <mn>5</mn> </msub> <mo>)</mo> </mrow> <mi>exp</mi> <mrow> <mo>(</mo> <msub> <mover> <mi>&amp;xi;</mi> <mo>^</mo> </mover> <mn>6</mn> </msub> <msub> <mi>&amp;theta;</mi> <mn>6</mn> </msub> <mo>)</mo> </mrow> <mo>=</mo> <msub> <mi>g</mi> <mn>1</mn> </msub> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>12</mn> <mo>)</mo> </mrow> </mrow>
    WhereinA point q is taken on the axis in joint 6s1 =[xs1 ys1 zs1]=[0 744 0], its rotary motion around joint 4 and joint 5 is described as follows,
    <mrow> <mi>exp</mi> <mrow> <mo>(</mo> <msub> <mover> <mi>&amp;xi;</mi> <mo>^</mo> </mover> <mn>4</mn> </msub> <msub> <mi>&amp;theta;</mi> <mn>4</mn> </msub> <mo>)</mo> </mrow> <mi>exp</mi> <mrow> <mo>(</mo> <msub> <mover> <mi>&amp;xi;</mi> <mo>^</mo> </mover> <mn>5</mn> </msub> <msub> <mi>&amp;theta;</mi> <mn>5</mn> </msub> <mo>)</mo> </mrow> <msubsup> <mi>q</mi> <mrow> <mi>s</mi> <mn>1</mn> </mrow> <mo>&amp;prime;</mo> </msubsup> <mo>=</mo> <msubsup> <mi>q</mi> <mrow> <mi>e</mi> <mn>1</mn> </mrow> <mo>&amp;prime;</mo> </msubsup> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>13</mn> <mo>)</mo> </mrow> </mrow>
    Wherein
    If it is c to move across point coordinates3=[x3 y3 z3], relationship below group is established according to geometric description,
    <mrow> <mfenced open = "{" close = ""> <mtable> <mtr> <mtd> <mrow> <mo>|</mo> <mo>|</mo> <msub> <mi>c</mi> <mn>3</mn> </msub> <mo>-</mo> <msub> <mi>q</mi> <mn>4</mn> </msub> <mo>|</mo> <mo>|</mo> <mo>=</mo> <mo>|</mo> <mo>|</mo> <msub> <mi>q</mi> <mrow> <mi>e</mi> <mn>1</mn> </mrow> </msub> <mo>-</mo> <msub> <mi>q</mi> <mn>4</mn> </msub> <mo>|</mo> <mo>|</mo> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>(</mo> <msub> <mi>c</mi> <mn>3</mn> </msub> <mo>-</mo> <msub> <mi>q</mi> <mrow> <mi>e</mi> <mn>1</mn> </mrow> </msub> <mo>)</mo> <msup> <msub> <mi>&amp;omega;</mi> <mn>4</mn> </msub> <mi>T</mi> </msup> <mo>=</mo> <mn>0</mn> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>(</mo> <msub> <mi>c</mi> <mn>3</mn> </msub> <mo>-</mo> <msub> <mi>q</mi> <mrow> <mi>s</mi> <mn>1</mn> </mrow> </msub> <mo>)</mo> <msup> <msub> <mi>&amp;omega;</mi> <mn>5</mn> </msub> <mi>T</mi> </msup> <mo>=</mo> <mn>0</mn> </mrow> </mtd> </mtr> </mtable> </mfenced> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>14</mn> <mo>)</mo> </mrow> </mrow>
    Wherein q4=[0 744 940]；Then formula (14) is expressed as equation,
    <mrow> <mfenced open = "{" close = ""> <mtable> <mtr> <mtd> <mrow> <msup> <msub> <mi>x</mi> <mn>3</mn> </msub> <mn>2</mn> </msup> <mo>+</mo> <msup> <msub> <mi>z</mi> <mn>3</mn> </msub> <mn>2</mn> </msup> <mo>-</mo> <mn>1880</mn> <msub> <mi>z</mi> <mn>3</mn> </msub> <mo>=</mo> <msup> <msub> <mi>x</mi> <mrow> <mi>e</mi> <mn>1</mn> </mrow> </msub> <mn>2</mn> </msup> <mo>+</mo> <msup> <msub> <mi>z</mi> <mrow> <mi>e</mi> <mn>1</mn> </mrow> </msub> <mn>2</mn> </msup> <mo>-</mo> <mn>1880</mn> <msub> <mi>z</mi> <mrow> <mi>e</mi> <mn>1</mn> </mrow> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <msub> <mi>x</mi> <mn>3</mn> </msub> <mo>=</mo> <msub> <mi>x</mi> <mrow> <mi>s</mi> <mn>1</mn> </mrow> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <msub> <mi>y</mi> <mn>3</mn> </msub> <mo>=</mo> <msub> <mi>y</mi> <mrow> <mi>e</mi> <mn>1</mn> </mrow> </msub> </mrow> </mtd> </mtr> </mtable> </mfenced> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>15</mn> <mo>)</mo> </mrow> </mrow>
    Equation group (15) is solved using " SOLVE " function in MATLAB, obtains x3,y3And z3Solution；
    Step (5) calculates joint angular displacement4And θ5
    Obtain process point coordinates c3=[x3 y3 z3] after, it will be described as follows respectively around the rotary motion in joint 4 and joint 5,
    <mrow> <mfenced open = "{" close = ""> <mtable> <mtr> <mtd> <mrow> <mi>exp</mi> <mrow> <mo>(</mo> <msub> <mover> <mi>&amp;xi;</mi> <mo>^</mo> </mover> <mn>4</mn> </msub> <msub> <mi>&amp;theta;</mi> <mn>4</mn> </msub> <mo>)</mo> </mrow> <msubsup> <mi>c</mi> <mn>3</mn> <mo>&amp;prime;</mo> </msubsup> <mo>=</mo> <msubsup> <mi>q</mi> <mrow> <mi>e</mi> <mn>1</mn> </mrow> <mo>&amp;prime;</mo> </msubsup> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mi>exp</mi> <mrow> <mo>(</mo> <msub> <mover> <mi>&amp;xi;</mi> <mo>^</mo> </mover> <mn>5</mn> </msub> <msub> <mi>&amp;theta;</mi> <mn>5</mn> </msub> <mo>)</mo> </mrow> <msubsup> <mi>q</mi> <mrow> <mi>s</mi> <mn>1</mn> </mrow> <mo>&amp;prime;</mo> </msubsup> <mo>=</mo> <msubsup> <mi>c</mi> <mn>3</mn> <mo>&amp;prime;</mo> </msubsup> </mrow> </mtd> </mtr> </mtable> </mfenced> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>16</mn> <mo>)</mo> </mrow> </mrow>
    Based on Paden-Kahan subproblems 1, joint angular displacement is obtained4And θ5Display expression formula it is as follows,
    <mrow> <msub> <mi>&amp;theta;</mi> <mn>4</mn> </msub> <mo>=</mo> <mi>a</mi> <mi> </mi> <mi>t</mi> <mi>a</mi> <mi>n</mi> <mrow> <mo>(</mo> <mfrac> <mrow> <mo>(</mo> <msub> <mi>z</mi> <mn>3</mn> </msub> <mo>-</mo> <mn>940</mn> <mo>)</mo> <msub> <mi>x</mi> <mrow> <mi>e</mi> <mn>1</mn> </mrow> </msub> <mo>-</mo> <msub> <mi>x</mi> <mn>3</mn> </msub> <mo>(</mo> <msub> <mi>z</mi> <mrow> <mi>e</mi> <mn>1</mn> </mrow> </msub> <mo>-</mo> <mn>940</mn> <mo>)</mo> </mrow> <mrow> <msub> <mi>x</mi> <mn>3</mn> </msub> <msub> <mi>x</mi> <mrow> <mi>e</mi> <mn>1</mn> </mrow> </msub> <mo>+</mo> <mrow> <mo>(</mo> <msub> <mi>z</mi> <mn>3</mn> </msub> <mo>-</mo> <mn>940</mn> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <msub> <mi>z</mi> <mrow> <mi>e</mi> <mn>1</mn> </mrow> </msub> <mo>-</mo> <mn>940</mn> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>17</mn> <mo>)</mo> </mrow> </mrow>
    <mrow> <msub> <mi>&amp;theta;</mi> <mn>5</mn> </msub> <mo>=</mo> <mi>a</mi> <mi> </mi> <mi>t</mi> <mi>a</mi> <mi>n</mi> <mrow> <mo>(</mo> <mfrac> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>s</mi> <mn>1</mn> </mrow> </msub> <mo>-</mo> <mn>744</mn> <mo>)</mo> <mo>(</mo> <msub> <mi>z</mi> <mn>3</mn> </msub> <mo>-</mo> <mn>940</mn> <mo>)</mo> <mo>-</mo> <mo>(</mo> <msub> <mi>z</mi> <mrow> <mi>s</mi> <mn>1</mn> </mrow> </msub> <mo>-</mo> <mn>940</mn> <mo>)</mo> <mo>(</mo> <msub> <mi>y</mi> <mn>3</mn> </msub> <mo>-</mo> <mn>744</mn> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>s</mi> <mn>1</mn> </mrow> </msub> <mo>-</mo> <mn>744</mn> <mo>)</mo> <mo>(</mo> <msub> <mi>y</mi> <mn>3</mn> </msub> <mo>-</mo> <mn>744</mn> <mo>)</mo> <mo>+</mo> <mo>(</mo> <msub> <mi>z</mi> <mn>3</mn> </msub> <mo>-</mo> <mn>940</mn> <mo>)</mo> <mo>(</mo> <msub> <mi>z</mi> <mrow> <mi>s</mi> <mn>1</mn> </mrow> </msub> <mo>-</mo> <mn>940</mn> <mo>)</mo> </mrow> </mfrac> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>18</mn> <mo>)</mo> </mrow> </mrow>
    Step (6) calculates joint angular displacement6
    Take a not point q on the axis of joint 6s2=[0 750 940], its rotary motion around joint 6 are described as,
    <mrow> <mi>exp</mi> <mrow> <mo>(</mo> <msub> <mover> <mi>&amp;xi;</mi> <mo>^</mo> </mover> <mn>6</mn> </msub> <msub> <mi>&amp;theta;</mi> <mn>6</mn> </msub> <mo>)</mo> </mrow> <msubsup> <mi>q</mi> <mrow> <mi>s</mi> <mn>2</mn> </mrow> <mo>&amp;prime;</mo> </msubsup> <mo>=</mo> <msubsup> <mi>q</mi> <mrow> <mi>e</mi> <mn>2</mn> </mrow> <mo>&amp;prime;</mo> </msubsup> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>19</mn> <mo>)</mo> </mrow> </mrow>
    WhereinSimilarly it is based on Paden-Kahan Subproblem 1, obtain joint angular displacement6Display expression formula it is as follows,
    <mrow> <msub> <mi>&amp;theta;</mi> <mn>5</mn> </msub> <mo>=</mo> <mi>a</mi> <mi> </mi> <mi>t</mi> <mi>a</mi> <mi>n</mi> <mrow> <mo>(</mo> <mfrac> <mrow> <msub> <mi>x</mi> <mrow> <mi>s</mi> <mn>2</mn> </mrow> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>e</mi> <mn>2</mn> </mrow> </msub> <mo>-</mo> <mn>744</mn> <mo>)</mo> </mrow> <mo>-</mo> <msub> <mi>x</mi> <mrow> <mi>e</mi> <mn>2</mn> </mrow> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>s</mi> <mn>2</mn> </mrow> </msub> <mo>-</mo> <mn>744</mn> <mo>)</mo> </mrow> </mrow> <mrow> <msub> <mi>x</mi> <mrow> <mi>s</mi> <mn>2</mn> </mrow> </msub> <msub> <mi>x</mi> <mrow> <mi>e</mi> <mn>2</mn> </mrow> </msub> <mo>+</mo> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>s</mi> <mn>2</mn> </mrow> </msub> <mo>-</mo> <mn>744</mn> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>e</mi> <mn>2</mn> </mrow> </msub> <mo>-</mo> <mn>744</mn> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>20</mn> <mo>)</mo> </mrow> </mrow>
    Solved more than and obtain eight groups of Inverse Kinematics Solutions.
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