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3.3 Solid Object Motion

Linear and angular velocity vectors have derivatives which are known as linear and angular motion, that is:

      _.      

   ^BV_Q   =     ___  ^BV_Q   =  lim           ^BV_Q (t + Δt)  -  ^BV_{Q .} (t)           

                                     Δt        0   _________________________________             (3.20)

                                                                          Δt

and,

           .               

     ^BΩ_Q   =  ____   ^BΩ_Q  =   lim                   ^BΩ_Q (t + Δt)  -  ^BΩ_{Q .} (t)

                                        Δt        0    _________________________________            (3.21)

                                                                                Δt

With velocities, the reference frame of the differentiation become the universal frame, {u} .

                                        .

                ύ_A = ^u V _AORG                 (3.22)

and

                         .             .

                        ω_A =  ^u Ω _A              (3.23)

3.3.1  Linear motion  

Describe the velocity of a vector  ^B_Q  as:

                        ^AV_Q = ^A R ^BV_Q + ^AΩ_B  x  ^A R ^B_Q              (3.24)

                           ^{B                                B}

The left hand side of this equation show ^A_Q is changing in time. Rewrite (3.24) as:

            ___   ^A R ^B_Q      =  ^A R ^BV_Q  +  ^AΩ_B  x  ^A R ^B_Q            (3.25)

                     ^{B                     B                                     B}                                                                                                    

By differentiating (3.24), it can be deriving the expression for the acceleration of   ^B_Q     this become:

   .                                                      .                                            

^AV_Q     =  ___  ^A R ^BV_Q  + ^AΩ_B  x  ^A R ^B_Q  +  ^AΩ_B   x  ____  ^A R  ^B_Q         (3.26)                                                                                                          ^{B                                    B                                             B}

Now applying (3.25) twice, once to the first term and once to the last term. The right hand side of equation (3.26) becomes:

             .                                                    .

^A R ^BV_Q + ^AΩ_B  x  ^A R ^BV_Q + ^AΩ_B  x  ^A R ^BV_Q + ^AΩ_B  x  ^A R ^BV_Q  +   ^AΩ_B  x  ^A R ^B_Q       (3.27) 

^{B                                  B                                  B                                 B                                     B}   

By combining two terms this will become:

             .                                                     .

^A R ^BV_Q + 2^AΩ_B   x  ^A R ^BV_Q  +  ^AΩ_B   x  ^A R ^BV_Q  +  ^AΩ_B   x     ^AΩ_B  x  ^A R ^B_Q                (3.28) 

^{B                                     B                                     B                                                        B}

Finally, by adding one term this gives the linear acceleration resulting in the final general formula:

    .                           .                                          .

^A V_BORG  +  ^A R ^BV_Q  +  2^AΩ_B  x  ^A R ^BV_Q +  ^AΩ_B  x  ^A R ^B_Q  +  ^AΩ_B  x   ^AΩ_B  x  ^A R ^B_Q       (3.29)

                         ^{B                                       B                                    B                                    B}   

When  ^B_Q   is constant,

                                 .

            ^BV_Q =  ^BV_Q

By simplify (3.29) to,

                        .               .

                 ^AV_Q  =  ^A V_BORG  +  ^AΩ_B  x   ^AΩ_B  x  ^A R ^B_Q

                                                                            ^B

Use the result to calculate the linear motion of the link of a manipulator.

3.3.2  Angular motion

By rotating {B} relative to {A} with  ^AΩ_B and {C} rotating relative to{B} with ^BΩ_C . To calculate  ^AΩ_C the sum of vectors in frame {A} :

                           ^AΩ_B = ^AΩ_B  +  ^A R ^BΩ_C                           (3.31)

                                              ^B

By differentiating it can be obtain:

                          .             .

                      ^AΩ_C  =     ^AΩ_B  +  ___     ^A R ^BΩ_C                        (3.32)

                                                            ^B

Then by applying (3.25) to the last term of (3.32), this becomes:

                              .             .                                 

                        ^AΩ_C  = ^AΩ_B  +  ^A R ^BΩ_C  + ^AΩ_B  x  ^A R ^BΩ_C               (3.33)

                                                             ^{B                                   B}

Use the result to calculate the angular motion of a link of a manipulator.

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