SlideShare une entreprise Scribd logo
1  sur  3
CS 4733, Class Notes: Forward Kinematics II



1   Stanford Manipulator - First Three Joints




Figure 1: Stanford Robotic Arm. The frame diagram shows the first three joints, which are in a R-R-P
configuration (Revolute-Revolute-Prismatic.


                                               joint   θ    d        a     α
                                               nnn1    θ1   d1       0    -90
                                                2      θ2   d2       0     90
                                                3      0    d3       0     0
                                                4
                                                5
                                                6

                                                                                                 
                      C1     0 -S1     0           C2            0        S2   0           1 0 0 0
                     S1     0  C1     0  1  S2                0       -C2   0  2  0 1 0 0        
             T10   =                      T2 =                                  T3 =             
                     0      -1  0     d1        0             1        0    d2        0 0 1 d3   
                      0      0   0     1           0             0        0    1           0 0 0 1

                                                                                                       
                    C1 C2    -S1 C1 S2     -S1 d2           C1 C2 -S1 C1 S2 C1 S2 d3 - S1 d2
                   S 1 C2   C1  S1 S2     C1 d 2 0      1 C2
                                                          S      C1   S1 S2 S1 S 2 d3 + C 1 d 2           
           T20   =                                T3 =                                                 
                   -S2       0   C2         d1           -S2    0     C2       C2 d3 + d1              
                       0      0     0         1               0     0      0            1
                                                   
                                   1   0   0     0
                                  0   1   0     d2 
                              0
    if θ1 = θ2 = 0, d3 = 0: T3 =                    (Zero Position)
                                  0   0   1     d1 
                                   0   0   0     1
                                                            1
2    4-DOF Gantry Robot




Figure 2: Gantry Robot Arm. This arm is in a R-P-R-R configuration. θ1 , θ3 , θ4 are the revolute joint angle variables and
d2 is the prismatic joint variable. d4 is a constant.


                                                 joint   θ     d    a    α
                                                  1      θ1    0    0    0
                                                  2      0     d2   0   -90
                                                  3      θ3    0    0   90
                                                  4      θ4    d4   0    0


                                                                                            
                                       C1        -S1     0    0        1      0     0       0
                                      S1         C1     0    0  1   0      0     1       0 
                                A0 =                           A =                          
                                 1    0          0      1    0  2  0       -1    0       d2 
                                       0          0      0    1        0      0     0       1


                                                                                             
                                     C3      0     S3        0        C4      -S4       0    0
                                    S3
                                            0    -C3        0  3   S4      C4        0    0 
                               2
                              A3 =                             A =                            
                                    0       1     0         0  4  0         0        1    d4 
                                     0       0     0         1        0        0        0    1



                                                               2
                                                                                   
                     C1    -S1      0 0           1      0 0 0       C1                 0 -S1          0
                                        
                    S1    C1       0 0          0     0 1 0   S1                    0  C1          0 
              A0 = 
               2
                                                                 =                                      
                    0      0       1 0 
                                                 0     -1 0 d2   0                   -1  0          d2 
                      0     0       0 1          0     0 0 1        0                   0  0          1
                                        
                                                                                                            
             C3    0 S3         0       C4       -S4 0            0      C3 C4        -C3 S4     S3       S3 d4
                                    
            S3    0 -C3        0      S4       C4 0            0   S3 C4          -S3 S4    -C3      -C3 d4 
      A2 = 
       4
                                                                     =                                        
            0     1  0         0   
                                        0        0  1           d4   S4             C4        0          0   
              0    0  0         1       0        0  0           1         0            0        0          1
                                    
                                                                                                     
                              C1         0       -S1     0          C3 C 4   -C3 S4      S3      S3 d4
                                                             
                             S1        0        C1     0         S3 C4     -S3 S4     -C3     -C3 d4 
                                                                                                       
              A0 = A02A24 = 
               4
                             0         -1        0     d2   
                                                                   S4        C4         0         0   
                               0        0         0     1            0        0         0         1
                                                             
                                                                                                  
                      C1 C3 C4 - S1 S4 -C1 C3 S4 - C4 S1 C1 S3                         C1 S3 d4
                     C3 C4 S1 + C1 S4 -C3 S1 S4 + C1 C4 S1 S3                         S1 S3 d4    
                   =                                                                              
                         -S3 C4              S3 S4         C3                        C 3 d4 + d 2 
                              0                  0           0                              1

                                                         
                                    1        0   0     0
                                   0        1   0     0 
if θ1 = θ3 = θ4 = 0, d2 = 0: A04= 
                                   0
                                                           (Zero Position)
                                             0   1     d4 
                                    0        0   0     1
                                                             
                                     -1           0    0 0
                                    0            0    1 d4 
if θ1 = θ3 = θ4 = 90, d2 = D: A 4= 
                                0                             
                                    0            1    0 D 
                                      0           0    0 1




                                                             3

Contenu connexe

Similaire à Forward kinematics robotics m tech.

Dynamic Response Analysis of Automatic Generation Control in a 2-Area (Reheat...
Dynamic Response Analysis of Automatic Generation Control in a 2-Area (Reheat...Dynamic Response Analysis of Automatic Generation Control in a 2-Area (Reheat...
Dynamic Response Analysis of Automatic Generation Control in a 2-Area (Reheat...IJMER
 
ว เคราะห ข_อสอบล_คก__
ว เคราะห ข_อสอบล_คก__ว เคราะห ข_อสอบล_คก__
ว เคราะห ข_อสอบล_คก__4821010054
 
Rotation3
Rotation3Rotation3
Rotation3hprmup
 
2009-2 기말고사문제(초고주파공학)
2009-2 기말고사문제(초고주파공학)2009-2 기말고사문제(초고주파공학)
2009-2 기말고사문제(초고주파공학)Yong Heui Cho
 
Chapter 9(laplace transform)
Chapter 9(laplace transform)Chapter 9(laplace transform)
Chapter 9(laplace transform)Eko Wijayanto
 
TU4.L09 - FOUR-COMPONENT SCATTERING POWER DECOMPOSITION WITH ROTATION OF COHE...
TU4.L09 - FOUR-COMPONENT SCATTERING POWER DECOMPOSITION WITH ROTATION OF COHE...TU4.L09 - FOUR-COMPONENT SCATTERING POWER DECOMPOSITION WITH ROTATION OF COHE...
TU4.L09 - FOUR-COMPONENT SCATTERING POWER DECOMPOSITION WITH ROTATION OF COHE...grssieee
 
TU4.L09 - FOUR-COMPONENT SCATTERING POWER DECOMPOSITION WITH ROTATION OF COHE...
TU4.L09 - FOUR-COMPONENT SCATTERING POWER DECOMPOSITION WITH ROTATION OF COHE...TU4.L09 - FOUR-COMPONENT SCATTERING POWER DECOMPOSITION WITH ROTATION OF COHE...
TU4.L09 - FOUR-COMPONENT SCATTERING POWER DECOMPOSITION WITH ROTATION OF COHE...grssieee
 
Manuel m. altamirano
Manuel m. altamiranoManuel m. altamirano
Manuel m. altamiranoEPMMA
 
Resultados de evaluacion de sector 2010-2011
Resultados de evaluacion de sector 2010-2011Resultados de evaluacion de sector 2010-2011
Resultados de evaluacion de sector 2010-2011EPMMA
 

Similaire à Forward kinematics robotics m tech. (12)

Dynamic Response Analysis of Automatic Generation Control in a 2-Area (Reheat...
Dynamic Response Analysis of Automatic Generation Control in a 2-Area (Reheat...Dynamic Response Analysis of Automatic Generation Control in a 2-Area (Reheat...
Dynamic Response Analysis of Automatic Generation Control in a 2-Area (Reheat...
 
ว เคราะห ข_อสอบล_คก__
ว เคราะห ข_อสอบล_คก__ว เคราะห ข_อสอบล_คก__
ว เคราะห ข_อสอบล_คก__
 
Rotation3
Rotation3Rotation3
Rotation3
 
2009-2 기말고사문제(초고주파공학)
2009-2 기말고사문제(초고주파공학)2009-2 기말고사문제(초고주파공학)
2009-2 기말고사문제(초고주파공학)
 
Chapter 9(laplace transform)
Chapter 9(laplace transform)Chapter 9(laplace transform)
Chapter 9(laplace transform)
 
16%20 lecture
16%20 lecture16%20 lecture
16%20 lecture
 
8. terapan hk1 n
8. terapan hk1 n8. terapan hk1 n
8. terapan hk1 n
 
TU4.L09 - FOUR-COMPONENT SCATTERING POWER DECOMPOSITION WITH ROTATION OF COHE...
TU4.L09 - FOUR-COMPONENT SCATTERING POWER DECOMPOSITION WITH ROTATION OF COHE...TU4.L09 - FOUR-COMPONENT SCATTERING POWER DECOMPOSITION WITH ROTATION OF COHE...
TU4.L09 - FOUR-COMPONENT SCATTERING POWER DECOMPOSITION WITH ROTATION OF COHE...
 
TU4.L09 - FOUR-COMPONENT SCATTERING POWER DECOMPOSITION WITH ROTATION OF COHE...
TU4.L09 - FOUR-COMPONENT SCATTERING POWER DECOMPOSITION WITH ROTATION OF COHE...TU4.L09 - FOUR-COMPONENT SCATTERING POWER DECOMPOSITION WITH ROTATION OF COHE...
TU4.L09 - FOUR-COMPONENT SCATTERING POWER DECOMPOSITION WITH ROTATION OF COHE...
 
Lab1 verilog
Lab1 verilogLab1 verilog
Lab1 verilog
 
Manuel m. altamirano
Manuel m. altamiranoManuel m. altamirano
Manuel m. altamirano
 
Resultados de evaluacion de sector 2010-2011
Resultados de evaluacion de sector 2010-2011Resultados de evaluacion de sector 2010-2011
Resultados de evaluacion de sector 2010-2011
 

Forward kinematics robotics m tech.

  • 1. CS 4733, Class Notes: Forward Kinematics II 1 Stanford Manipulator - First Three Joints Figure 1: Stanford Robotic Arm. The frame diagram shows the first three joints, which are in a R-R-P configuration (Revolute-Revolute-Prismatic. joint θ d a α nnn1 θ1 d1 0 -90 2 θ2 d2 0 90 3 0 d3 0 0 4 5 6       C1 0 -S1 0 C2 0 S2 0 1 0 0 0  S1 0 C1 0  1  S2 0 -C2 0  2  0 1 0 0  T10 =  T2 =   T3 =    0 -1 0 d1   0 1 0 d2   0 0 1 d3  0 0 0 1 0 0 0 1 0 0 0 1     C1 C2 -S1 C1 S2 -S1 d2 C1 C2 -S1 C1 S2 C1 S2 d3 - S1 d2  S 1 C2 C1 S1 S2 C1 d 2 0  1 C2 S C1 S1 S2 S1 S 2 d3 + C 1 d 2  T20 =  T3 =    -S2 0 C2 d1   -S2 0 C2 C2 d3 + d1  0 0 0 1 0 0 0 1   1 0 0 0  0 1 0 d2  0 if θ1 = θ2 = 0, d3 = 0: T3 =   (Zero Position)  0 0 1 d1  0 0 0 1 1
  • 2. 2 4-DOF Gantry Robot Figure 2: Gantry Robot Arm. This arm is in a R-P-R-R configuration. θ1 , θ3 , θ4 are the revolute joint angle variables and d2 is the prismatic joint variable. d4 is a constant. joint θ d a α 1 θ1 0 0 0 2 0 d2 0 -90 3 θ3 0 0 90 4 θ4 d4 0 0     C1 -S1 0 0 1 0 0 0  S1 C1 0 0  1  0 0 1 0  A0 =  A =   1  0 0 1 0  2  0 -1 0 d2  0 0 0 1 0 0 0 1     C3 0 S3 0 C4 -S4 0 0  S3  0 -C3 0  3  S4 C4 0 0  2 A3 = A =    0 1 0 0  4  0 0 1 d4  0 0 0 1 0 0 0 1 2
  • 3.     C1 -S1 0 0 1 0 0 0 C1 0 -S1 0   S1 C1 0 0  0 0 1 0   S1 0 C1 0  A0 =  2  =   0 0 1 0   0 -1 0 d2   0 -1 0 d2  0 0 0 1  0 0 0 1 0 0 0 1       C3 0 S3 0 C4 -S4 0 0 C3 C4 -C3 S4 S3 S3 d4   S3 0 -C3 0  S4 C4 0 0   S3 C4 -S3 S4 -C3 -C3 d4  A2 =  4  =   0 1 0 0   0 0 1 d4   S4 C4 0 0  0 0 0 1  0 0 0 1 0 0 0 1     C1 0 -S1 0 C3 C 4 -C3 S4 S3 S3 d4   S1 0 C1 0  S3 C4 -S3 S4 -C3 -C3 d4   A0 = A02A24 =  4  0 -1 0 d2   S4 C4 0 0  0 0 0 1  0 0 0 1    C1 C3 C4 - S1 S4 -C1 C3 S4 - C4 S1 C1 S3 C1 S3 d4  C3 C4 S1 + C1 S4 -C3 S1 S4 + C1 C4 S1 S3 S1 S3 d4  =   -S3 C4 S3 S4 C3 C 3 d4 + d 2  0 0 0 1   1 0 0 0  0 1 0 0  if θ1 = θ3 = θ4 = 0, d2 = 0: A04=   0  (Zero Position) 0 1 d4  0 0 0 1   -1 0 0 0  0 0 1 d4  if θ1 = θ3 = θ4 = 90, d2 = D: A 4=  0   0 1 0 D  0 0 0 1 3