mechanism velocity analysis

8/18/2019 mechanism velocity analysis

1/94

• Time derivatives of the loop-closure

expressions allow the analysis of velocities

& accelerations, i.e.:

Velocity & Acceleration Analysis

MECH 335 Lecture Notes© R.Podhorodeski, 2009

VELOCITY CLOSURE

ACCELERATION CLOSURE


2/94


3/94

• Example: Velocity analysis of the offset

slider-crank



2

3

1

0

Displacement closure:


4/94

• Taking the time derivative:

• Rearranging and substituting:




5/94






6/94






7/94






8/94

• Splitting into real and imaginary eqns

• The solution for is obtained by solving

the imaginary equation as:



REAL

IMAGINARY


9/94

• Substituting this solution back into the real

equation gives the other unknown:



REAL EQUATION


10/94

• The velocity of one point can be expressed

as the velocity of another point, plus the

relative velocity of the two points

Relative Velocity Analysis



11/94

• Example



R

R

B

R

B

A

B

O

42

θ 2

θ 3

θ 4

ω2


12/94

• Example



R

R

B

R

B

A

B

O

42

θ 2

θ 3

θ 4

ω2

V

B

Absolute velocity of point B


13/94

• Example



R

R

B

R

B

A

B

O

42

θ 2

θ 3

θ 4

ω2

V

Absolute velocity of point A


14/94

• Example



R

R

B

R

B

A

B

O

42

θ 2

θ 3

θ 4

ω2

V

B

Relative velocity of point B w.r.t. point A


15/94

• Example



R

R

B

R

B

A

B

O

42

θ 2

θ 3

θ 4

ω2

V

V

B

V

B


16/94

• Note that:

– The direction and magnitude of V A is a known

function of the input angular velocity, ω2

– The mechanism’s joints define the direction of

many of the remaining relative and absolute

velocities

– This information can be manipulated to find thevelocity (direction and magnitude) of points not

on the input link




17/94

• There are 4 distinct cases where relative

velocity analysis is applied (though only 3

are non-trivial)



Same Point Different Points

Same

Link

Different

Links

Case 1

TRIVIAL CASE

Case 2

DIFFERENCE

MOTION

Case 3

RELATIVE MOTION

Case 4

DIFFERENCE & RELATIVE

MOTION


18/94

• Case 2: Different points on the same link

– Want to find velocity of point B w.r.t. point A (V B|A)

– Take the derivative of the rel. position vector (RB|A)



R

R

B

R

B

A

B

O

42

θ 2

θ 3

θ 4

ω2

V

V

B

V

B


19/94


– Examining this result gives simple method for

calculation:



R

R

B

R

B

A

B

O

42

θ 2

θ 3

θ 4

ω2

V

V

B

V

B

Always = 0 for a rigid link

(no length change)

Equivalent to rotation through 90° in the sense

(CW or CCW) of ωB|A (i.e. ω3)

So, for a rigid link :V B|A = (r B|A)(ω3), ┴ R B|A


20/94


– Recalling that V B = V A + V B|A we can set up a system

of equations to solve for V B:



R

R

B

R

B

A

B

O

42

θ 2

θ 3

θ 4

ω2

V

V

B

V

B

Tricky to approach analytically, but graphical

methods can be used, and can be much more

intuitive


21/94

• Case 2 Example (Supp Ex V1)



C

B

AO2 O4

3 4

2

ω2


22/94

• Solve using a graphical method



C

B

AO2 O4

3 4

2

ω2


23/94

• Find V B by relative velocity analysis:



C

B

AO2 O4

3 4

2

ω2

0V

1 mm = 5 mm/s


24/94

• V A: direction, magnitude both known



C

B

AO2 O4

3 4

2

ω2

0V

1 mm = 5 mm/s


25/94




0V

1 mm = 5 mm/sC

B

AO2 O4

3 4

2

ω2


26/94

0V

V A

1 mm = 5 mm/s

A




C

B

AO2 O4

3 4

2

ω2


27/94

0V

V A

1 mm = 5 mm/s

A

• V B| A: direction known, magnitude unknown



C

B

AO2 O4

3 4

2

ω2


28/94




C

B

AO2 O4

3 4

2

0V

V A

1 mm = 5 mm/s

A


29/94




C

B

AO2 O4

3 4

2

0V

V A

d i r ( V B | A )

1 mm = 5 mm/s

A


30/94

• V B: direction known, magnitude unknown



0V

V A

d i r ( V B | A )

1 mm = 5 mm/s

A

C

B

AO2 O4

3 4

2

ω2


31/94




C

B

AO2 O4

3 4

2

0V

V A

d i r ( V B | A )

1 mm = 5 mm/s

A


32/94




C

B

AO2 O4

3 4

2

0V

V A

d i r ( V B )

d i r ( V B | A )

1 mm = 5 mm/s

A


33/94

• Solution is obtained by intersection &measurement



0V

V A

d i r ( V B )

d i r ( V B | A )

1 mm = 5 mm/s

A

C

B

AO2 O4

3 4

2

ω2


34/94




0V

V A

d i r ( V B )

d i r ( V B | A )

1 mm = 5 mm/s

A

B V B

C

B

AO2 O4

3 4

2

ω2


35/94




0V

V A

d i r ( V B )

d i r ( V B | A )

1 mm = 5 mm/s

A

B V B

X

C

B

AO2 O4

3 4

2

ω2


36/94

• Noticing that , we measure:



C

B

AO2 O4

3 4

2

ω2

X

0V

V A

d i r ( V B )

d i r ( V B | A )

1 mm = 5 mm/s

A

B V B


37/94




C

B

AO2 O4

3 4

2

ω2

0V

V A

d i r ( V B )

d i r ( V B | A )

1 mm = 5 mm/s

V B | A

A

B V B

X


38/94




C

B

AO2 O4

3 4

2

ω2

0V

V A

d i r ( V B )

d i r ( V B | A )

1 mm = 5 mm/s

V B | A

A

B V B

X


39/94

• And compute:

• Where the direction of rotation is inferred

from the direction of



C

B

AO2 O4

3 4

2

ω2

0V

V A

d i r ( V B )

d i r ( V B | A )

1 mm = 5 mm/s

V B | A

A

B V B

X X


40/94

• Similarly:



C

B

AO2 O4

3 4

2

ω2

0V

V A

d i r ( V B )

d i r ( V B | A )

1 mm = 5 mm/s

V B | A

A

B V B

X X X


41/94

0V

V A

d i r ( V B )

d i r ( V B | A )

1 mm = 5 mm/s

V B | A

A

B V B

• Now, V C can be found in a variety of ways:

– Intersect relative velocity directions w.r.t. A & B

– Compute directly, e.g. V C = V A+(ω3 X AC)



C

B

AO2 O4

3 4

2

ω2

X X X


42/94

• Using the first method, note that:



X X X

C

B

AO2 O4

3 4

2

ω2

0V

V A

d i r ( V B )

d i r ( V B | A )

1 mm = 5 mm/s

V B | A

A

B V B


43/94




C

B

AO2 O4

3 4

2

ω2

X X

and ,

X

0V

V A

d i r ( V B )

d i r ( V B | A )

1 mm = 5 mm/s

V B | A

A

B V B


44/94




C

B

AO2 O4

3 4

2

ω2

X X

and ,

0V

V A

d i r ( V B )

d i r ( V B | A )

1 mm = 5 mm/s

dir( VC|A)

V B | A

A

B V B

X


45/94




C

B

AO2 O4

3 4

2

ω2

X X

and ,

X

0V

V A

d i r ( V B )

d i r ( V B | A )

1 mm = 5 mm/s

dir( VC|A)

V B | A

A

B V B


46/94




C

B

AO2 O4

3 4

2

ω2

X X

and ,

0V

V A

d i r ( V B )

d i r ( V B | A )

1 mm = 5 mm/s

dir( VC|A)

V B | A

d i r ( V

C | B )

A

B V B

X


47/94

• Intersection gives the solution for V C



C

B

AO2 O4

3 4

2

ω2

X X X

0V

V A

d i r ( V B )

d i r ( V B | A )

1 mm = 5 mm/s

dir( VC|A)

V B | A

d i r ( V C

| B )

A

B V B


48/94




C

B

AO2 O4

3 4

2

ω2

X X

0V

V A

d i r ( V B )

d i r ( V B | A )

1 mm = 5 mm/s

dir( VC|A)

V B | A

V C

d i r ( V C

| B )

A C

B V B

X


49/94




C

B

AO2 O4

3 4

2

ω2

X X

0V

V A

d i r ( V B )

d i r ( V B | A )

1 mm = 5 mm/s

dir( VC|A)

V B | A

V C

d i r

( V C | B )

A C

B V B

X X


50/94

• Another Case 2 Example (Supp Ex V2)



C

B

A

O2 3

2

ω2

4


51/94

• Solve using the same graphical method:



C

B

A

O2 3

2

ω2

40V

1 mm = 10 mm/s


52/94

• Find V B by case 2 analysis:



C

B

A

O2 3

2

ω2

40V

1 mm = 10 mm/s


53/94

• V A : magnitude, direction both known



C

B

A

O2 3

2

ω2

40V

1 mm = 10 mm/s


54/94




C

B

A

O2 3

2

ω2

40V

1 mm = 10 mm/s


55/94



MECH 335 Lecture Notes

© R.Podhorodeski, 2009

C

B

A

O2 3

2

ω2

40V

AV A

1 mm = 10 mm/s


56/94

• V B| A : direction known, magnitude unknown




C

B

A

O2 3

2

ω2

40V

AV A

1 mm = 10 mm/s


57/94





C

B

A

O2 3

2

ω2

40V

AV A

1 mm = 10 mm/s


58/94





C

B

A

O2 3

2

ω2

40V

d i r ( V

B | A ) A

V A

1 mm = 10 mm/s


59/94

• V B : direction known, magnitude unknown




0V

d i r ( V

B | A ) A

V A

1 mm = 10 mm/s

C

B

A

O2 3

2

ω2

4


60/94

• V B : direction known, magnitude unknown




0V

d i r ( V

B | A ) A

V A

dir(VB)

1 mm = 10 mm/s

C

B

A

O2 3

2

ω2

4


61/94

• Intersection & measurement give:




C

B

A

O2 3

2

ω2

40V

d i r ( V

B | A ) A

V A

dir(VB) VB

V B | A

B

1 mm = 10 mm/s


62/94

• Intersection & measurement give:




C

B

A

O2 3

2

ω2

40V

d i r ( V

B | A ) A

V A

dir(VB) VB

V B | A

B

1 mm = 10 mm/s

X


63/94

• And ω3 is found from:




C

B

A

O2 3

2

ω2

4

X0V

d i r ( V

B | A ) A

V A

dir(VB) VB

V B | A

B

1 mm = 10 mm/s

X


64/94

• Solve for V C by intersection:




C

B

A

O2 3

2

ω2

40V

d i r ( V

B | A ) A

V A

dir(VB) VB

V B | A

B

1 mm = 10 mm/s

X X


65/94





C

B

A

O2 3

2

ω2

40V

d i r ( V

B | A ) A

V A

dir(VB) VB

V B | A

B

d i r ( V

C | A

)

1 mm = 10 mm/s

X X


66/94





C

B

A

O2 3

2

ω2

40V

d i r ( V

B | A ) A

V A

dir(VB) VB

V B | A

d i r ( V

C | B )

B

d i r ( V

C | A

)

1 mm = 10 mm/s

X X


67/94

• Measuring then gives:




C

B

A

O2 3

2

ω2

40V

d i r ( V

B | A ) A

V A

dir(VB) VB

V B | A

d i r ( V

C | B )

B

C V C

d i r ( V

C | A

)

1 mm = 10 mm/s

X XX


68/94

• Case 3: Coincident points on different links

– Occurs for slides & pistons, cams & followers:

• Two points on different links momentarily occupy

the same point in the plane• Each has a different absolute velocity, therefore a

relative velocity exists





69/94


– Calculate the slide (relative) velocity, V B3|B4




B

2

B

3

B

4

4

2

3

• So far, we have taken the derivative of

the relative position vector, RB3|B4• But how can we express this vector for

two coincident points?

• Intuitively, we can imagine displacing the

slide by some small distance along the

slide, then drawing RB3|B4• Taking the limit as the displacement

approaches zero, we can see that RB3|B4has zero length, and is directed along the

tangent to the slide (link 4) at point B


70/94






θ slide

B

2

B

3

B

4

4

2

3

• So:

• Taking the derivative:


71/94






θ slide

B

2

B

3

B

4

4

2

3

• Simplifying gives the final expression:

• Note that this deceptively simple

expression hides the potentially difficult

task of finding the slide tangent angle

• In the following examples, straight slidesare used to avoid this hassle (tangent

angle = link angle for a straight slide)


72/94

• Case 3 Example (Supp Ex V3)




C

B

O2

2

ω2

4

O4

3


73/94

C

B

O2

2

ω2

4

O4

3

• Solve graphically




0V

1 mm = 10 mm/s


74/94

C

B

O2

2

ω2

4

O4

3

• Use case 3 analysis to find V B4:




0V

1 mm = 10 mm/s


75/94

C

B

O2

2

ω2

4

O4

3

0V

1 mm = 10 mm/sC

B

O2

2

ω2

4

O4

3

• V B2: direction, magnitude both known:




0V

B2

V B 2

1 mm = 10 mm/s


76/94

0V

B2

V B 2

1 mm = 10 mm/s

0V

B2

V B 2

d i r ( V B 2 | B

4 )

1 mm = 10 mm/sC

B

O2

2ω2

4

O4

3

• V B2|B4: only direction is known




C

B

O2

2ω2

4

O4

3


77/94

C

B

O2

2ω2

4

O4

3

C

B

O2

2ω2

4

O4

3

0V

B2

V B 2

d i r ( V B 2 | B

4 )

1 mm = 10 mm/s

0V

d i r ( V B 4

)

B2

V B 2

d i r ( V B 2 | B

4 )

1 mm = 10 mm/s

• V B4: direction known, magnitude unknown





78/94

• Obtain V B4 by intersection




C

B

O2

2ω2

4

O4

3

0V

d i r ( V B 4

)

B2

V B 2

d i r ( V B 2 | B

4 )

V B 4

B4

V B 2 | B

4

1 mm = 10 mm/s

X

l l l


79/94

• ω4 follows immediately from V B4:




C

B

O2

2ω2

4

O4

3

0V

d i r ( V B 4

)

B2

V B 2

d i r ( V B 2 | B

4 )

V B 4

B4

V B 2 | B

4

1 mm = 10 mm/s

X X

l l l


80/94

• V C follows immediately from ω4, or by:

•




C

B

O2

2ω2

4

O4

3

0V

d i r ( V B 4

)

B2

V B 2

d i r ( V B 2 | B

4 )

V B 4

B4

V B 2 | B

4

1 mm = 10 mm/s

X X

l i l i l i


81/94


• ,




C

B

O2

2ω2

4

O4

3

X X0V

d i r ( V B 4

)

B2

V B 2

d i r ( V B 2 | B

4 )

V B 4

B4

V B 2 | B

4

d i r ( V C | B 4 )

1 mm = 10 mm/s

R l i V l i A l i


82/94


• , ,




C

B

O2

2ω2

4

O4

3

X X0V

d i r ( V B 4

)

B2

V B 2

d i r ( V B 2 | B

4 )

V B 4

B4

V B 2 | B

4

d i r ( V

C )

d i r ( V C | B 4 )

1 mm = 10 mm/s

R l i V l i A l i


83/94

C

B

O2

2ω2

4

O4

3

• Measuring:




0V

d i r ( V B 4

)

B2

V B 2

d i r ( V B 2 | B

4 )

V B 4

B4

V B 2 | B

4

d i r ( V

C )

d i r ( V C | B 4 )

C

V C

1 mm = 10 mm/s

X X X

R l i V l i A l i


84/94

• Another Case 3 Example (Supp Ex V4)




C

B

AO2 O4

3 4

2

ω2

D5, D6

5

6

R l ti V l it A l i


85/94

• Note: The 4-Bar was solved in Ex V1




C

B

AO2 O4

3 4

2

ω2

D5, D6

5

6

R l ti V l it A l i


86/94

C

B

AO2 O4

3 4

2

ω2

D5, D6

5

6

• V C is found by scaling arguments:




V A

V B

V B | A

A

B

0V

1 mm = 5 mm/s

R l ti V l it A l i


87/94

C

B

AO2 O4

3 4

2

ω2

D5, D6

5

6

• V C is found by scaling arguments:




V A

V B

V B | A

A

B

V C | A

0V

1 mm = 5 mm/s

R l ti V l it A l i


88/94

C

B

AO2 O4

3 4

2

ω2

D5, D6

5

6

• Measuring:




V A

V B

V B | A

A

B

V C | A

0V

C

VC

1 mm = 5 mm/s

X

R l ti V l it A l i


89/94

C

B

AO2 O4

3 4

2

ω2

D5, D6

5

6

• V D5 is found by Case 3 analysis:




X

V A

V B

V B | A

A

B

V C | A

0V

C

VC

1 mm = 5 mm/s

R l ti V l it A l i


90/94

C

B

AO2 O4

3 4

2

ω2

D5, D6

5

6

• V D5 : only the direction is known




X

V

A

V B

V B | A

A

B

V C | A

d i r ( V

D 5 )

0V

C

VC

1 mm = 5 mm/s

R l ti V l it A l i


91/94

C

B

AO2 O4

3 4

2

ω2

D5, D6

5

6

• V D5|C : only the direction is known




X

V

A

V B

V B | A

A

B

V C | A

d i r ( V

D 5 )

0V

d i r ( V

D 5 | C )

C

VC

1 mm = 5 mm/s



92/94

C

B

AO2 O4

3 4

2

ω2

D5, D6

5

6

• Measuring:




X

V

A

V B

V B | A

A

BV D 5

V C | A

d i r ( V

D 5 )

0V

d i r ( V

D 5 | C )

V D 5 | C

C

D5VC

1 mm = 5 mm/s

X



93/94

C

B

AO2 O4

3 4

2

ω2

D5, D6

5

6

• ω5 is found immediately from:




V

A

V B

V B | A

A

BV D 5

V C | A

d i r ( V

D 5 )

0V

d i r ( V

D 5 | C )

V D 5 | C

C

D5VC

1 mm = 5 mm/s

X X X


94/94

END OF LECTURE PACK 3

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mechanism velocity analysis