1 1

All figures taken from Design of Machinery, 3rd ed. Robert Norton 2003

MENG 372Chapter 6

Velocity Analysis

2 2

Velocity AnalysisDefinitions

RdtRdV

dtd

ipeipe

RVV

peR

iiPAPPA

iPA

Linear Velocity

Angular Velocity

Velocity of a point

Velocity is perpendicular to radius of rotation & tangent to path of motion

Multiplying by i rotates the vector by 90°

Link in pure rotation

3 3

cos sinire r i

sin cosiire r i

cosr sinr

r

Real

Imaginary

cosr

sinr

Vector r can be written as:

Multiplying by i gives:

Multiplying by i rotates a vector 90°

Velocity Analysis

4 4

If point A is moving

ipeV

VVVi

A

PAAP

Velocity Analysis

Graphical solution:

5 5

Graphical Velocity Analysis (3 & 4)• Given linkage configuration & 2. Find 3 and 4

• We know VA and direction of VB and VBA (perpendicular to AB)

• Draw vector triangle. V=r.

VBA

VB

VBA Direction

VBA

VB VB Direction

VA

3

4 4

BA

B

V AB

V O B

6 6

Graphical Velocity Analysis (VC)• After finding 3 and 4, find VC

• VC=VA+VCA

• Recall that 3 was in the opposite direction as 2

VCA

VC

VA

Double Scale

VCA

VC

7 7

Instant Center• A point common to two bodies in plane motion, which

has the same instantaneous velocity in each body.• In ENGR 214 we found the instant center between

links 1 and 3 (point on link 3 with no velocity)• Now we also have an instant center between links 2

and 4

8 8

Instant Centers• Kennedy’s rule: any three

links will have three instant centers and they will lie on a straight line

• The pins are instant centers• I13 is from links 1,2,3 and

1,3,4• I24 is from links 1,2,4 and

2,3,4

I13

I24

1 2 3I12

I23

I13

1 3 4I13

I34

I14

1 2 4I12

I24

I14

2 3 4I23

I34

I24

Links

IC’s

9 9

Instant Centers

I13

• I13 has zero velocity since link 1 is ground

• 3 is the same all over link 3 • Velocity relative to

ground=r, perpendicular to r• VA2=a2=VA3=p3

• From this, 3 must be in the opposite direction as 2, and smaller in magnitude since p>a A

a

VA2

VA3

p

3

3

10 10

Instant Centers

I24

• I24 has the same velocity on link 2 and link 4

• VI2=l22=VI4=l44

• From this, 4 is in the same direction as 2 and smaller in magnitude since l4>l2

4

VI4

l4

VI2l2

11 11

Instant Centers Practice Problems

O2

O4

A

B

O4O2

B

A

Power=Tinin=Toutout

12 12

Velocity Analysis of a 4-Bar LinkageGiven 2. Find 3 and 4

13 13

Velocity Analysis of a 4-Bar Linkage• Write the vector loop equation

• After solving the position analysis, take the derivative

or

where

01432 iiii decebeae

0432432 iceibeiae iii

0432432 iii ceibeiaei

0 BBAA VVV

4

3

2

4

3

2

iB

iBA

iA

ceiVbeiVaeiV

14 14

Velocity Analysis of a 4-Bar Linkage

• Take knowns to one side:

• Take conjugate to get 2nd equation:

• Put in matrix form:

• Invert matrix:

0432432 iii ceibeiaei

243243

iii aecebe

2

2

43

43

2

21

4

3

i

i

ii

ii

aeae

cebecebe

243243

iii aecebe

2

2

43

43

2

2

4

3

i

i

ii

ii

aeae

cebecebe

15 15

Inverted Crank Slider

Given 2. Find 3 andLink 3 is a slider link: its effective length, b, changes

b

16 16

Inverted Crank Slider• Given 2. Find 3 and• Write the vector loop equation:

• After solving the position analysis, take the derivative:

• To get another equation:or

so

01432 iiii decebeae

04332432 iiii ceibeiebaei

43 43

243323

iiii aeicebeieb

b

17 17

Inverted Crank Slider

• Take conjugate to get second equation:

• Put in matrix form:

243323

iiii aeicebeieb

243323

iiii aeicebeieb

2

2

433

433

2

2

3

i

i

iii

iii

aeiaeib

cebeiecebeie

2

2

433

433

2

21

3

i

i

iii

iii

aeiaei

cebeiecebeiebInvert:

18 18

Velocity of any Point on a Linkage• Write the vector for RP

• Take the derivative

• Similarly

332 iip peaeR

33232

iip peiaeiV

RP

22 iS seR

222

iS seiV

44 iU ueR

444

iU ueiV

19 19

Offset Crank SliderGiven 2. Find 3 and

a

b c

d