Motion Analysis Theory

7/21/2019 Motion Analysis Theory

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1 Motion Analysis - Theory

We identify a kinematic analysis, as a study which only considers the geometries andthe constraints of a machine, with no regard of the forces, torques or masses of the system.A dynamic analysis studies the system, while considering the driving forces, torques,

and masses of the system. Before discussing the methods of performing such studies, letsclarify some main definitions.

1.1 Inputs and outputs of a system

The systems we study in motion analysis are real life machines which are driven by ac-tuators, and which produces desired motion. The inputs of the system are the degrees offreedom (DOF) related to actuation. Consider a motor driving a gear pair. Lets take thegear pair as the mechanism. The inputs to the system for the purposes of a kinematicstudy are the motor acceleration, angular speed, and motor shaft angle. For a dynamicstudy the input is the torque produced by the motor. The outputs of the system wouldbe the output gear speed, rotation angle and acceleration.

1.2 Four problems in motion studies

In general, motion analysis of mechanisms are classified in to four problems.

Forward kinematics This analysis is concerned with finding the output kinematic variablesby using the known quantities of input kinematic variables to the system.

Inverse kinematics The inverse kinematic problem attempts to find the input kinematicvariables by using the known quantities of the output kinematic variables.

Forward dynamics The forward dynamics problem attempts to find the output kinematic

variables of the system by using the known quantities of the input forces to thesystem.

Inverse dynamics The inverse dynamics problem attempts to find the input forces/torquesof the mechanism for given known quantities of input kinematic variables.

The solution strategy for these four problems are introduced using the following example.

1.3 Example - The RR manipulator

1.3.1 Forward kinematics

Position analysis

For the forward position analysis the knowns are the input DOF 1, 2. The unknownsare the output DOF x, y. The final end effector position is marked as x, y. Using vectorloop equations for the loop OAB we can find:

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2

x,y

1

O

A

B

111

222

{WA}

{WB}{1A}

{1C}

{2A}

{2C}

m1I1

m2I2

l1

l2

Figure 1: The RR manipulator

rOA+ rAB rOB =0rOB =rOA+ rAB

x

y

=

l1cos(1) + l2cos(1+ 2)l1sin(1) + l2sin(1+ 2)

(1)

Define the input DOF as q and output DOF as p. Then a general expression for theforward position analysis equations can be written as:

p= f(q) (2)

Velocity analysis: For the forward velocity analysis the knowns are the input DOF speedsand positions1, 2, 1, 2. The unknowns are the output DOF speeds x, y. The velocityequations can be found by differentiating the position equations.

x

y

=

l1sin(1)1 l2sin(1+ 2)(1+ 2)l1cos(1)1+ l2cos(1+ 2)(1+ 2)

(3)

This can be arranged in a convenient matrix form. Notice that a compact notation is usedfor sines and cosines. ex: sin(1) =S1, sin(1+ 2) =S12

x

y

=

l1S1 l2S12 l2S12l1C1+ l2C12 l2C12

12

(4)

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This can be expressed in a compact form by defining the matrix as Jv.

p= Jvq (5)

Acceleration analysis: For the forward acceleration analysis the knowns are the input

DOF accelerations, speeds and positions 1, 2, 1,

2,

1,

2. The unknowns are theoutput DOF accelerations x, y. The acceleration equations can be found by differentiating

the velocity equations. This can be arranged in a matrix form as:

x

y

=

l1S1 l2S12 l2S12l1C1+ l2C12 l2C12

12

+l1C11 l2C121 l2C122 l2C121 l2C122

l1S11 l2S121 l2S122 l2S121 l2S122

12

(6)

p= Jvq + Jaq (7)

1.3.2 Inverse kinematics

Position analysis: For the inverse position analysis the knowns are the output DOFpositions x, y. The unknowns are the input DOF accelerations 1, 2. Apply cosine ruleto angle BOA, to find 1. The solution for 2 is found by identifying an expression fortan(1+ 2), using the equation set (1).

1 =aTan2(y, x) cos1

l21+x2+y2l2

2

2l1

x2+y2

2 =aTan2(y l1sin 1, x l1cos 1) 1(8)

There are two solutions for the angles 1, 2, for a given position x, y.

2

x,y

1

O

A

B

{WA}

{1A}

-2

Figure 2: The two solutions for the inverse position analysis

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1

-m1a1x

l1

-m1a1y

-I11

-f12x

-f12y

f01y

f01x

01

-12

Input (external) forces

internal forces

Inertial forces

1

-m2a2x

l2

-m2a2y

-I22

f12y

f12x

122

Figure 3: FBD- link1 and 2

Velocity analysis: Rearrange the forward velocity analysis equations to find qwhen pisknown.

q= J1v p (9)

Acceleration analysis: Rearrange the forward acceleration analysis equations to find qwhen p is known.

q= J1v (p Jaq) (10)

1.3.3 Force analysis - Inverse dynamics

To perform a dynamic analysis first the free body diagrams for the two links are drawn

indicating all the forces.

The force balance equations applied to link 1:

f01x f12x m1a1x= 0f01y f12y m1a1y = 001 12 I11+ f01xl1S1 f01yl1C1= 0

(11)

The force balance equations applied to link 2:

f12x

m2a2x= 0f12y m2a2y= 012 I22+ f12xl2S12 f12yl2C12= 0

(12)

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q=[1, 2]q=[1,2]q=[1,2]q =[1, 2] p=[x, y]...

Initial condition set

1,2

Initial condition set

1, 2

Figure 4: Dynamic simulation block diagram

Now the set of unknowns for the forward dynamics study are the accelerations of the inputDOF and the internal forces. The vector of unknowns is identified as:

f01x f01y f12x f12y 1 2T=

fi qT (18)

There are 6 equations and 6 unknowns which is solvable. Arrange the equation in thematrix form:

1 0 1 0 m1l1S1 00 1 0 1 m1l1C1 0

l1S1 l1C1 0 0 I1 00 0 1 0 m2(l1S1 l2S12) m2(l2S12)0 0 0 1 m2(l1C1+ l2C12) m2(l2C12)0 0 l2S12

l2C12 0

I2

f01xf01yf12xf12y12

=

m1l1C121m1l1S1

21

01+ 12m2((l1C11 l2C121 l2C122)1+ (l2C121 l2C122)2)m2((l1S11 l2S121 l2S122)1+ (l2S121 l2S122)2)

12

(19)

The solution can be found by matrix inversion:

C

fiq

= d

fi

q

= C1d

(20)

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Motion Analysis Theory