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Velocity Analysis

Velocity Analysis

Chapter 6

Definition

• Rate of change of position with

respect to time

– Angular

– Linear

V

d

dt

dR

dt

– Position Vector

R

PA

pe

j

Velocity

V

pa

R

dt

PA

pje

j

d

dt

pe

j

Definition

Velocity (absolute)

PA j

d

VPA

R pje

dt dt

pe

j

• The velocity is always in a

direction perpendicular to

the radius of rotation and is

tangent to the path of

motion

V

PA

pj

cos

jsin

p

sin

j cos

VPA V P

"Absolute"

Definition

Velocity (difference)

V

PA

V

P

V

A

V

P

V

A

V

PA

"on thesame body "

– Relative Velocity

V

PA

V

P

V

A

Graphical Analysis

• Graphical Velocity Analysis

V

P

V

V

A

V

v r

PA

– Solve for

angular velocities

; 3

, 4

linear velocities; A , B,

C

Graphical Analysis

• Graphical Velocity Analysis

V

P

V

A

V

PA

V

v r

Graphical Analysis

• Example 6-1

– Given θ 2 , θ 3 , θ 4 , ω 2 find ω 3 ,

ω 4 , V A , V B and V C

– Position analysis already

performed

– 1. Start at the end of the

linkage about which you

have the most information.

Calculate the magnitude of

the velocity of point A,

v A

AO 2

2

Graphical Analysis

• Example 6-1

– 2. Draw the velocity V A

– 3. Move next to a point

which you have some

information, point B. Draw

the construction line pp

through B perpendicular to

BO 4

– 4. Write the velocity

difference equation for point

B vs. A

V V V

B

A

BA

Graphical Analysis

• Example 6-1

– 5. Draw construction line qq

through point B and

perpendicular to BA to

represent the direction of V BA

– 6. The vector equation can

be solve graphically by

drawing the following vector

diagram

V V V

B

A

BA

Graphical Analysis

• Example 6-1

– 7. The angular velocities of

link 3 and 4 can be

calculated,

4

v B

BO

4

3

v BA

BA

– 8. Solve for V C

V

C

V

A

V

CA

v CA

c 3

Instant Center of

Velocity

• An instant center of velocity is a point, common

to two bodies in plane motion, which point has

the same instantaneous velocity in each body

• The numbers of IC is calculated with;

C

n

n 1

2

• Linear graph is a useful way to keep track of

which IC have been found

• Kennedy’s Rule

Instant Center of

Velocity

– Any three bodies in plane motion will have

exactly three instant centers, and they will lie

an the same straight line

Instant Center of

Velocity

Instant Center of

Velocity

• Slider-Crank Linkage

• Slider-Crank Linkage

Instant Center of

Velocity

• Slider-Crank Linkage

• Check Example 6-4:

IC for a Cam-

Follower Mechanism

Instant Center of

Velocity

Velocity Analysis

with IC

• Once the ICs have

been found, they can

be used to do a very

rapid graphical velocity

analysis of the linkage

v A

AO 2

2

v A

v B

3

AI

4

1,3 BO

v B

BI 1,3

3

v C

,3 3

4

CI 1

Velocity Analysis

with IC

• A rapid graphical

solution for the

magnitudes at B and C

are found from vectors

drawn perpendicular to

that line at the

intersection of the arcs

and line AI 1,3 (V B’ , V C’ )

• Angular Velocity Ratio

– Output angular velocity

divided by the input

angular velocity

m V

4

– Can be derived by

constructing a pair of

effective links

2

– Effective link pairs is

two lines, mutually

parallel, drawn through

the fixed pivot and

intersecting the coupler

extended

O

O

sin

2A

2A

O

V A

4B

2B

O

sin

O A 2

2

V

A V B

m V

4

2

O2

A

O B

4

O2

Asin

O Bsin

4

– Now the effective links

are colinear and

intersect the coupler at

the same point, I 2,4

m V

4

2

O I

2

O I

4

2,4

2,4

• Mechanical Advantage

– Power in a mechanical system,

P

– For rotating system,

– Mechanical efficiency,

F V

F V

– Mechanical Advantage,

m

A

F

F

out

in

T

T

out

in

r

r

in

out

in

x

x

P T

out

P

P

out

in

r

r

in

out

F

y

V

T

T

out

in

y

in

out

O

4Bsin

O2

Asin

r

r

in

out

Centrodes

• The path, or locus, created by a IC at

successive positions

Centrodes

Velocity of Slip

• Used when there is a sliding joint between

two links and neither one is the ground

– Example 6-5, 6-6

V slip

VA

V

42 4slip

A 2 slip

4

3

V A

4

AO

4

Velocity of Slip

4

3

V A

3

AO

3

Analysis Solution

• Position Analysis (revisited)

0

1

4

3

2

R

R

R

R

0

1

4

3

2

j

j

j

j

de

ce

be

ae

A

AC

B

B

2

4

2arctan

2

4 2

1,

D

DF

E

E

2

4

2arctan

2

3 2

1,

Analysis Solution

Velocity Analysis

0

1

4

3

2

j

j

j

j

de

ce

be

ae

0

1

4

3

2

j

j

j

j

de

ce

be

ae

dt

d

0

4

3

2 4

3

2

dt

d

jce

dt

d

jbe

dt

d

jae

j

j

j

0

4

3

2

4

3

2

j

j

j

e

jc

e

jb

e

ja

Analysis Solution

Velocity Analysis

V

A

V

V

V

V

A

BA

B

BA

V

ja

e

2

4

3

jc

e

B

jb

e

j

j

2

j

4

0

3

Euler identity

real part

imaginary part

Analysis Solution

Velocity Analysis

4

3

2

4

2

3

sin

sin

b

a

3

4

3

2

2

4

sin

sin

c

a

4

4

4

4

4

4

3

3

3

3

3

3

2

2

2

2

2

2

cos

sin

sin

cos

cos

sin

sin

cos

cos

sin

sin

cos

j

c

j

jc

j

b

j

jb

j

a

j

ja

B

BA

A

V

V

V

Analysis Solution

– Slider-Crank

0

1

4

3

2

j

j

j

j

de

ce

be

ae

0

3

2

3

2

d

e

jb

e

ja

j

j

0

1

4

3

2

R

R

R

R

BA

A

B

V

V

V

0

B

AB

A

V

V

V

BA

AB

V

V

Analysis Solution

– Slider - Crank

3

a

b

sin

2

2

sin

3

d

a b

2

sin

2

3

sin

3

V

B

V

A

V

BA

V

V

V

A

AB

BA

a

2

V

sin

j cos

b

sin

j cos

3

AB

2

3

2

3

Review - InvertedSlider

-Crank

Geared Fivebar

0

1

5

4

3

2

j

j

j

j

j

fe

de

ce

be

ae

0

5

4

3

2

5

4

3

2

j

j

j

j

e

jd

e

jc

e

jb

e

ja

0

1

5

4

3

2

R

R

R

R

R

2

5

5 2

Geared Fivebar

3

4

3

5

4

5

4

2

2

4

3

cos

2

cos

sin

sin

2sin

b

d

a

4

5

5

3

3

2

2

4

sin

sin

sin

sin

c

d

b

a

BA

A

B

C

BA

A

j

d

j

b

j

a

V

V

V

V

V

V

5

5

5

3

3

3

2

2

2

cos

sin

cos

sin

cos

sin

Velocity of Any Point

• Once the angular velocities of all the links are

found it is easy to define and calculate the

velocity of any point on any link for any input

position of the linkage

Velocity of Any Point

• To find the velocity of points S & U

2 2

scos

j

R S 0

R S

sin

R U

se j

2 2 2

2

2

V S

jse j

2

sin

2

s

j

2 2 2 2

cos

4 4

ucos

j

ue j

4 4 4

4

4

0

sin

V U

jse j

4

sin

4

u

j

4 4 4 4

cos

2

4

2

4

Velocity of Any Point

R

R

PA

P

V

V

P

• To find the velocity of point P

pe

R

PA

A

A

j

3

pcos

j

3

R

jpe

V

j

PA

3 3

sin

3

p

sin

j

3

V

PA

3 3

3 3

cos

3

3

3

3

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