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The Iterative Newton-Euler Dynamics Algorithm

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The Iterative Newton-Euler Dynamics Algorithm Manipulator Dynamics (2) Read Chapter 6 Dynamic equations for each link • For each link there are two equations to describe the effects of force and torque to the motion: • Newton equation = • Euler’s equation ̇ τ2 = + × τ xc1 ̇ 1 {Ci} has its origin at the center of mass of the x link and has the same orientation as the link 1 frame {i} Dynamic equations for all the links • Use the iterative Newton-Euler algorithm i F zi zi+1 xi+1 xi i +1n i +1 i n f i f i N • Between links there are action and reaction forces (torques) • For link n, we have n+1 f = 0 and n +1n = 0; therefore the equations can start from the last link and go inwards Compute forces and torques • Look the top figure in the previous slide, one has: = +1 = × × − +1 +1 +1 +1 − +1 − − +1 • From the above, one can obtain the following iterative equations: = + +1 +1 = + + × + × +1 +1 +1 +1 +1 • Note that n+1 f = 0 and n +1n = 0; so we can start the iteration from link n. • Since every joint has only one torque applied, we have = ( ) Question: how the above equation should̂ be changed for a prismatic joint? The iterative Newton-Euler dynamics algorithm (1) • The iterative algorithm takes two steps - First step: perform outward iterations to compute velocities and accelerations - Second step: perform inward iterations to computer forces and torques • First step: - Angular acceleration = + = + ( ) +1 +1 +1 = + ( ) ̇+1 ̂ +1 +1 +1 ̇+1 ̂= + × + - Linear acceleration+1 +1 +1 +1 +1 ̇ ̇ ̇+1 ̂ ̈+1 ̂ = ( + × ) +1 +1 = +1 [ × + ×( × ) + ] - For the+1 center+ of1 the link: +1 +1 ̇ ̇ +1 +1 ̇ = × + ×( × )+ +1 +1 +1 +1 +1 +1 +1 ̇+1 ̇ +1 +1 ̇ The iterative Newton-Euler dynamics algorithm (2) • Second step: - Draw the Newton-Euler equations = +1 +1 +1 ̇+1 = + × +1 +1 +1 +1 +1 +1 - Inward iteration from ̇link n to link 1: = + +1 +1 = + + × + × +1 +1 +1 +1 +1 = ( ) ̂ Iterative algorithm example • Use the OSU hexapod as an example which is similar to the manipulator shown below – to simplify the problem, let m1 and m2 be point mass and located at the tip of the link τ1 m2 Ze Xe D1 Z1 m1 X1 X2 D2 Z2 D 0 Z0 X0 Use the two iterations • Determine the values needed in the algorithm - The position vectors of the center of mass = 1 1 = 1�1 2 2 - The inertia tensor at2 the�2 center of mass = 0 1 1 = 0 2 - Rotational matrices2 0 0 = 0 = 0 1 1 2 2 1 0 0 1 2 0 1 0 0 −1 1 1 −2 2 − • Outward iteration - For link 1 0 0 = = 0 = = 0 1 1 1 1 ω ̇ ̂ ̇ ̈1̂1 1 ̇ 1̈ 0 0 0 0 ( ) ( ) = 0 0 = 0 = + 0 2 + = 2 1 1 1 1 1 ̇ 1 1 0 0 1 1 0 − 0̇ 1 − 1 ̇ −1 1 ̇ 1̈1 ̇ 1̈1 ( ) ( ) 0 = 2 = 2 1 ̇ 1 1 1 ̇ 1 = 0 1 − − 1 1 0 1̈1 11̈1 1 - For link 2 0 + = + = + 0 = 2 ̇1 2̇1 = + 2 2 0 2 1 2 2 1 1 1 1 2 2 2 1 ̇ ̇ ̈ ω ̇ ̂ ̇ ̂ 2̇1 ̇ 2 ̇2 ̇ −2̇1̇2 2̈1 ̇2 0 ( ) ( ) ̈2 = 0 2 = ( ) 2 2 2 1 ̇ 1 −21 ̇ 1 2 0 1 0 − 2 ̇ −2 2 1̈1 21 ̇ 1 − 1 ̈1 − = × + × × + 2 2 2 2 2 2 2 ̇2 ̇ 2 2 ̇ + ( ) = + × 0 + × × 0 + ( ) 2 + 2̇1̇2 2̈1 2 2̇1 2̇1 2 −21 ̇ 1 2 2 0 0 2 0 2 ̇ −2̇1̇2 2̈1 2̇1 2̇1 21 ̇ 1 2 ̈2 ̇2 ̇2 −1̈1 0 + + 2 2 2 2 2 = + 2 + ( ) 2+ = 2 −2 2 ̇1 − 2 ̇2 2 1 ̇1 2 −2 2+̇1 − 2 ̇2 +− 21 ̇1 + 2 2 2 − 2 2 2 2 ̈2 2 ̇ 222 ̇1 21 ̇1 2 2̈2 222 ̇1 21 ̇1 2 2 2 1 2 2 2 1 ̇ ̇ − ̈ 1 1 22̇1̇2 − ̈ 222̇1̇2 −2 2̈1 −1 ̈1 + 0 = 2 2 2 2 = 0 −2 2+̇1 − 2 ̇2 +− 21 ̇1 + 2 2 2 2 2 2 0 2̈2 222 ̇1 21 ̇1 2 222̇1̇2 −2 2̈1 −1 ̈1 • Inward iteration - For link 2 = 2 2 0 = 0 × = ( ) 2 2 0 2 ( + + + ) 2 2 2 2 ̇1 ̇2 2 2 ̈1 1 ̈1 − 2 −2 − 2 2 2̈2 222 ̇1 21 ̇1 2 - For link 1 0 + ( ) 2 2 2 = + = 0 0 1 2 + + + + 2 2 2 −2 2 ̇1 − 2 ̇2 − 21 ̇1 2 1 1 ̇ 1 1 1 2 1 − 0 2 2 − 2 2 − 2̈2 222 ̇1 21 ̇1 2 11̈1 2 2 1 222̇1̇2 −2 2̈1 −1 ̈1 = + × + × Figure out the details? 1 1 2 1 1 1 1 2 2 1 2 2 Final dynamic equation • Since every joint has only one degree of freedom, one has: = ( ) 1 1 1 = ( ) ̂ 2 2 Centrifugal force • Ultimately one2 can obtain: ̂ Coriolis force = M( ) + V( , )+G( ) • where Θ Θ̈ Θ Θ̇ Θ Gravity force = = 1 1 Θ 2 2 Lagrangian formulation • Another approach is using the Lagrangian formulation - Step 1 obtain the Lagrangian of a manipulator ( , )= ( , ) - ( ) where Θ Θ(̇ , ) =Θ Θ̇ = Θ ( ) + ( ) 1 1 is the kinetic energy of the manipulator, and Θ Θ̇ ∑ ∑ 2 2 ( )= = [ ( ) ] 0 0 is the potential energy of the manipulator Θ ∑ ∑ − • The dynamic equation can be expressed as - = - + = Becomes the same as the one ̇ Θ Θ Θ̇ Θ Θ shown earlier = M( ) + V( , )+G( ) Θ Θ̈ Θ Θ̇ Θ.
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