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Page 104                              Wu. Intell Robot 2021;1(2):99-115  I http://dx.doi.org/10.20517/ir.2021.11


               with
                                             [                                      ]
                                    K    = diag    act,1     act,2  K        act,3  K        act,4     act,5  (15)
               where    act,   is the actuation stiffness and K    and K    are the upper and lower link stiffness matrices, respectively.

               To calculate the stiffness matrix of the loaded mode, a neighborhood in the loaded configuration in which the
               external loads and the joint location are supposed to be incremented by small values   F and     , which can
               still satisfy the equilibrium conditions, is considered, leading to

                                                                          0
                                      (J    +   J    ) G + (J    +   J    ) (F +   F) = K    (   −    +     )  (16)
               and the linearized kinematic constraint
                                                          t = J                                       (17)


               Based on Equation (14), expanding Equation (16) yields

                                                          
                                                                 
                                                
                                            H ⊗ G     + J   F + H ⊗ F     = K                         (18)
                                                
                                                          
                                                                 
               where the symbol ⊗ represents the Kronecker product between matrices and H    =   J    /    , H    =   J    /    .
               Combining Equations (17) and (18), the stiffness model of the robotic manipulator is reduced to
                                                 [           ] [  ]  [ ]
                                                  0      J       F      t                             (19)
                                                  J     K    − K           =  0
                                                     
               with
                                                                    
                                                           
                                                   K    = H ⊗ G + H ⊗ F                               (20)
                                                                    
                                                           
               From   F = K  t, the Cartesian stiffness matrix K of the robotic arm is calculated as
                                                     (               ) −1
                                                                 −1   
                                                 K = J    (K    − K    )  J                           (21)
               3.2.  Mass matrix
               The mass matrix can be derived from the expression of the system’s kinetic energy, consisting of energies of
               the revolute joints, links, and end-effector. The energy of the five active joints are
                                                    (  5        5         )
                                                   1  ∑     2  ∑        
                                                           ¤
                                                    =        ,      +       ,   v v   ,               (22)
                                                                        ,  
                                                   2
                                                       =1        =3
               with
                                                                   ¤
                                                      ¤
                                              v   ,3 = E 3   ;  v   ,   = E 45   ,    = 4, 5          (23)
               and
                                          [                        ]
                                     E 3 = z 0 × q 2  z 1 × (q 2 − q 1 )  0 3                         (24a)
                                           [                                       ]
                                     E 45 = z 0 × q 4  z 1 × (q 4 − q 1 )  z 2 × (q 4 − q 2 )  0 3×2  (24b)
               where      ,   is the moment of inertia of the   th joint,      ,   is the mass, and v   ,   is the velocity in the Cartesian
               space. Let I    = diag[     ,1 ,      ,2 , ...,      ,5 ]; then, Equation (22) can be written in a compact form, namely,

                                                           1  ¤   
                                                            =     M       ¤                           (25)
                                                           2
               with
                                                                                                      (26)
                                           M    = I    +      ,3 E E 3 + (     ,4 +      ,5 )E E 45
                                                        3                 45
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