Class 4 Velocity Analysis1 (1)

7/23/2019 Class 4 Velocity Analysis1 (1)

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Mechanics ofMachines

Dr. Mohammad Kilani

Class 4

Velocity Analysis



DERIVATIVE OF A ROTATINVECTOR



Deri!ati!e of a Rotatin" #nitVector

A unit vector in the θ direction, uθ , is a vector of unity

magnitude and an angle θ with the x- axis. It is

written as:

If uθ

rotates, it angle θ changes with time, the time

derivative of uθ is found by applying the standard

dierentiation rules on the expression of uθ

above

j i uθ

θ θ sincos +=

( ) j i u

j i u

θ

θ

θ θ θ

θ θ

θ θ

cossin

cossin

+−=

+−=

dt

d

dt

d

dt

d

dt

d

dt

d

θ

cosθ i

sinθ j

u θ




Noting that

the time derivative of the vector uθ is

( )

( ) θ π θ

θ π θ

sin2cos

cos2sin

−=+

=+

( ) j i u

j i u

θ

θ

θ θ θ

θ θ

cossin

sincos

+−=

+=

dt

d

dt

d

( )

( ) ( )[ ]

( )2

2sin2cos

cossin

π θ

θ

π θ π θ θ

θ θ θ

+=

+++=

+−=

uu

j i

u

j i u

θ

θ

θ

dt

d

dt

d

dt

d

dt

d

dt

d

dt

d θ

cosθ i

sinθ j

u θ




The derivative of a unit vector whose angle

with the x-axis is θ , θ changes in time!, is a

vector whose angle with the x-axis is θ+π/2

and whose magnitude is d θ/dt.

If we de"ne a vector $ as a vector in the %

direction of magnitude # $ d θ/dt , then

( )2

sincos

π θ

θ

θ θ

+=

+=

uu

j i u

θ

θ

dt

d

dt

d

θ

u θ

d u θ

/dt

θ + π/2

( ) θ θ uωu

u

k ω

×==

=

+ 2π θ

θ

θ

dt

d

dt

d

dt d



Deri!ati!e of a Rotatin" Vector

( )

( )

( )θ θ

θ

θ θ θ

θ θ

θ θ

θ

θ θ

r ωur

uωur

uur

uu

r ur

×+=

×+=

+=

+==

+

dt

dr

dt

d

r dt

dr

dt

d dt

d r

dt

dr

dt

d

dt

d r

dt

dr

dt

d

r

2π θ

θ θ

u θ

d u θ

/dt

θ + π/2

A vector r θ = r u

θ , is a general vector of magnitude r pointing in the θ

direction. The time derivative of r θ is found by applying the normal

dierentiation rules on the expression for r θ



Deri!ati!e of a Rotatin" Vector

θ

r θ

ω x r θ

θ + π/2

(dr/dt ) u θ

%iven a vector r θ = r u

θ which rotates relative to

the reference coordinates, the time derivative of

r θ

has two components& a component in the

direction of uθ and a component normal to u

θ in

the direction of u

θ+π/2

The magnitude of the component of d r θ /dt in the

direction of uθ is e'ual to dr/dt; that is the time

derivative of the length of r θ .

The magnitude of the component of d r θ /dt in the

direction of direction of uθ+π/2

is e'ual to ωr



VE&OCIT' ANA&'(I(

OF FO#R )ARMEC*ANI(M(



Deri!ati!e of the &oo+ Clos,reE-,ation for a Fo,r )ar KinematicChain The loop closure e'uation of a ()bar *inematic chain is written

as

+hen all the lin*s in the chain are of constant lengths, the

e'uation above reduces to

( )

( )

( ) ( ) ( )( ) ( )( ) 0

0

0

0

44113322

4132

4244121132332222

4132

4132

4132

4132

=+−+−+++

=−−+

=−−+

=−−+

+=+

++++ θ π θ θ π θ θ π θ θ π θ

θ θ θ θ

θ θ θ θ ur ur ur ur ur ur ur ur

ur ur ur ur dt

d

r r r r dt

d

r r r r

r r r r

( ) ( ) ( ) ( )

0

0

44113322

244211233222 4132

=×−×−×+×

=−−+++++

r ωr ωr ωr ω

π θ π θ π θ π θ θ θ θ θ ur ur ur ur



Deri!ati!e of the &oo+ Clos,reE-,ation for a Fo,r )arMechanism or a four bar mechanism with lin* - "xed we have

d θ 1 / dt = ω

1 = .

The vector e'uation above contains two scalar

e'uations and can be solved for two un*nowns.

/nowledge of d θ 2 / dt allows the calculation of d θ

3 / dt

and d θ 4 / dt .

( ) ( ) ( )244233222432

π θ π θ π θ θ θ θ +++

=+ ur ur ur

444333222

444333222

coscoscos

sinsinsin

θ θ θ θ θ θ

θ θ θ θ θ θ

r r r

r r r

=+

=+



Deri!ati!e of the &oo+ Clos,reE-,ation for a Fo,r )arMechanism

0liminate d θ 3 / dt by carrying out a dot product with u

θ3 on both sides

of the e'uation

Alternatively, eliminate d θ 4 / dt by carrying out a dot product with u

θ4

on both sides of the e'uation

( ) ( ) ( )244233222 432 π θ π θ π θ θ θ θ +++

=+ ur ur ur

( ) ( )

( ) ( )

( )

( )

( )

( ) 2

344

322

2

344

322

44

34443222

34443222

sin

sin

sin

sin

sinsin

2cos2cos

ω θ θ

θ θ θ

θ θ

θ θ θ ω

θ θ θ θ θ θ

θ π θ θ θ π θ θ

−

−=

−

−==

−=−

−+=−+

r

r

r

r

r r

r r

( ) ( )

( ) ( )

( )

( )

( )

( )

( )

( ) 2

343

422

2

433

422

2

433

422

33

43334222

43334222

sin

sin

sin

sin

sin

sin

sinsin

02cos2cos

ω θ θ

θ θ ω

θ θ

θ θ θ

θ θ

θ θ θ ω

θ θ θ θ θ θ

θ π θ θ θ π θ θ

−

−=

−

−−=

−

−−==

−−=−

=−++−+

r

r

r

r

r

r

r r

r r



An",lar Velocity Ratio andMechanical Ad!anta"e

The angular velocity ratio mV

is de"ned as the output angular velocity divided by the input angular velocity.

or a four bar mechanism with lin* 1 as the input and lin* ( as the output this is expressed as

The e2ciency of a four bar lin*age is de"ned as the output power over the input power,

Assuming -3 e2ciency, which is normally approached by four bar mechanisms, we have

2

4

ω

ω

ω

ω

==in

out

V m

inin

out out

in

out

T

T

P

P

ω

ω η ==

V

T

out

in

inin

out

mm

T

T 1===

ω

ω



An",lar Velocity Ratio andMechanical Ad!anta"e

The mechanical advantage is de"ned as

the ratio between the output force to the

input force

out

in

T

inin

out out

in

out

Ar

r m

r T

r T

F

F m ===



VE&OCIT' ANA&'(I(

OF (&IDERCRANKMEC*ANI(M(



Velocity Analysis of a (liderCran%Mechanism

4esign parameters: r 2 , r

3, r

4, θ

1.

5osition analysis parameters:

r 1, θ

2 , θ

3

6elocity analysis parameter.

ind dr 17dt , dθ

2 7dt , dθ

37dt

To eliminate ω3 dot product both sides by u

θ 3

r 3

r 2

r1

r 4

r p

( ) ( ) 132 1233222

4132

θ π θ π θ ω ω ur ur ur

r r r r

=+

+=+

++

( ) ( )

( ) ( )

( )

( ) 2

13

232

1

3113222

3113222

cos

sin

cossin

cos2cos

ω θ θ

θ θ

θ θ θ θ ω

θ θ θ π θ ω

−−

=

−=−−

−=−+

r r

r r

r r

( ) ( )

( )

( ) 2

133

122

3

13331222

cos

cos

coscos

ω θ θ

θ θ ω

θ θ ω θ θ ω

−−

−=

−−=−

r

r

r r

To eliminate dr 17dt dot product both sides

by u/θ 1+π/2

0



Velocity Analysis of an In!erted(liderCran% Mechanism

%iven r 1

, r 2 , θ

1, θ

2 , ω

2

5osition analysis: ind r 3, θ

3

6elocity analysis: ind dr 37dt , dθ

37dt

To eliminate ω3 and "nd dr

17dt dot product both sides by u

θ 3

To eliminate dr 17dt and "nd ω

3 dot product both sides by u/

θ 3+π/20

( ) ( )2333222

312

312

332

312

π θ θ π θ

θ θ θ

ω ω ++ +=+=

+=

ur ur ur

ur ur ur

r r r

( )

( ) 32322

33222

sin

2cos

r r

r r

=−

=−+

θ θ ω

θ π θ ω

r 2

r1

θ 1

θ 2

r 3

θ 3

( )

( )2

3

232

3

333222

cos

cos

ω θ θ

ω

ω θ θ ω

r

r

r r

−=

=−




%iven r 1

, r 2 , r

4, θ

1, θ

2 , ω

2

5osition analysis: ind r 3, θ

3 , θ

4

6elocity analysis: ind dr 37dt , dθ

37dt , dθ

47dt

r 3

r 2 r

1

r 4

( ) ( ) ( )

( ) ( ) 34423333222

34

24423333222

41332

4132

32

432

412

2

θ π θ θ π θ

π θ π θ θ π θ

θ θ θ θ

ω ω ω π θ θ

ω ω ω

ur ur ur ur

ur ur ur ur

ur ur ur ur

r r r r

−=+++=

=++

+=+

+=+

++

+++

with




r 3

r 2 r

1

r 4

( ) ( )

( )

( )

( )

( )

( )

( )

( ) ( )

( )( )

2322

3

322

43

2

3

322

433222

2

3

322

34

4433222

4433222

3

2

3

322

3

333222

2

34423333222

sincos

cossin

cos

sin

2cos

cos

0cos

3

32

ω θ θ θ θ

ω θ θ

θ θ ω

ω θ θ

ω ω

ω θ θ ω

ω θ π θ ω

ω θ θ

ω

ω θ θ ω

ω ω ω

θ

π θ

θ π θ θ π θ

−+

−=

−=+−−

−−==

−=+−−

−=+−+

−−=

=+−

−=++

+

++

r r

r r r

r

r r r r

r

r

r r r

r r r

u

r

r

r r

u

ur ur ur ur

8ut

bysidesboth4ot

bysidesboth4ot



E1am+le

( )

( )

( ) ( ) ( ) ( )

( ) ( )

( ) ( ) ( ) ( )

( )

( )

( )

−+−+

=

−+=+

−+=+

=−

+−±

=⇒=+++−

=++

+−−+++=+

−−−−−+++=+

−+=+

+=++

+=++

−

−

224411

2244111

2,13

2244113

2244113

2,1

1

2,14

222

2,1

2

44

424242241441

2

4

2

2

2

1

2

424212411441

2

4

2

2

2

1

2

241

152

412

coscoscos

coscossintan

coscossinsin

coscoscoscos

tan2,02

0sincos

sinsincoscos2cos2cos2

cos2cos2cos2

2413

61532

4132

θ θ θ

θ θ θ θ

θ θ θ θ

θ θ θ θ

θ

θ θ

θ θ θ θ θ θ

θ θ θ θ θ θ

θ θ θ θ

θ θ θ θ θ

θ θ θ θ

r r r

r r r

r r r ba

r r r ba

t C A

AC B Bt AC Bt t AC

C B A

r r r r r r r r r ba

r r r r r r r r r ba

ur ur ur uba

u sur ur uaur

ur ur ubaur

1!r'uationclosureloopfromAlso

itself bye'uationtheof sidesboth4ot

-!0'uation9losure:oop

II:oop

I:oop

a

r 2

r1

b

r 4

r 5

s

a

r 2

r1

b

r 4

r 5

s

( )

( )2tan

cos2

sin2

cos22

4

2

221

2

4

2

2

2

1

142

24241

θ

θ

θ

θ

=

+−−++=

−=

−=

t

bar r r r r C

r r B

r r r r A



E1am+le

( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( )

( ) ( ) ( ) ( )

( )

( )

( )

−+−+

=

−+=+

−+=+

=−

+−±=⇒=+++−

−+++=+

−−−+−−−−

−−+++===

−−−+−−−−

−−−−−++++=

−−+=

−

−

224411

2244111

2,13

2244113

2244113

2,1

1

2,14

222

2,1

2

241441

2

4

2

2

2

1

2

32226213262

31221

22

2

22

1

2

5

61

32226213262

1311221161

22

2

22

1

2

5

215

coscoscos

coscossintan

coscossinsin

coscoscoscos

tan2,02

cos2cos2

cos2cos2cos2cos2

cos2cos2

2,0

cos2cos2cos2cos2

cos2cos2cos2

32615

θ θ θ

θ θ θ θ

θ θ θ θ

θ θ θ θ

θ

θ θ

θ θ θ θ θ θ θ θ

θ θ

π θ θ

θ θ θ θ θ θ θ θ

θ θ θ θ θ θ

θ θ θ θ θ

r r r

r r r

r r r ba

r r r ba

t C A

AC B Bt AC Bt t AC

r r r r r r r ba

ar sr as sr

ar r r ar sr r

ar sr as sr

ar r r sr ar sr r

uaur u sur ur

1!r'uationclosureloopfromAlso

:et

itself bye'uationtheof sidesboth4ot

1!0'uation9losure:oop

a

r 2

r1

b

r 4

r 5

s

a

r 2

r1

b

r 4

r 5

s

( )

( )2tan

cos2

sin2

cos22

4

2

221

2

4

2

2

2

1

142

24241

θ

θ

θ

θ

=

+−−++=

−=

−=

t

bar r r r r C

r r B

r r r r A



*23

5678 569 /:;c08 5<6/:08 544.



MET*OD OF IN(TANT CENTER(



Relati!e Velocities )et=een T=o>oints on a Ri"id )ody

%iven any two points A and B that

lie on a rigid body, let the line AB

be a line passing through A and 8,

then the components of the

velocity of A and the velocity of B

on the line AB must be e'ual.

The velocity of point B relative to

A must be normal to the line AB

A

B

A

B

A B B A vvv +=



Instant Center Relati!e to thero,nd

If line AA; is drawn normal to the direction of v A

, then the

velocity of any point on the rigid body that falls on line AC

must be perpendicular to the line AA;

In a similar manner, if line 88; is drawn perpendicular to

the direction of v B

, then the velocity of any point on the

rigid body that falls on this line must be perpendicular to

the line BB;

If point I is the intersection of lines AA; and BB;, then the

velocity of point I must be perpendicular to both AA; and

BB;. This can only happen when the velocity of point I is

<ero.

A

B I

A’

B’

I B B

I A A I A A

I A A

I A I A

r v

r vr v

r k v

r k vv

ω

ω ω

ω

ω

=

=⇒=

×=

×+=



Instant Center Relati!e to thero,nd

5oint I determined as before is called the instant

center of <ero velocity of the rigid body with

respect to the ground. The direction of the

velocity of point A is normal to the line IA. The

direction of the velocity of line B is normal to the

line IB.

The direction of the velocity of any other point C

on the same body must be normal to the line IC.

The magnitude of the velocity of any point on

the lin* is proportional to its distance from the

point I.

A

B I

A’

B’

C

IC

v

IB

v

IA

v A B A

==



Velocity of a >oint on a &in% :yInstant Center Method

9onsider two points A and B on a rigid lin*. et v A

and v B

be

the velocities of points A and 8 at a given instant. If v A

is

*nown in magnitude and direction and v B

in direction only,

then the magnitude of v B

may be determined by the

instantaneous centre method.

4raw AI and BI perpendiculars to the directions v A

and v B

respectively. et these lines intersect at I, which is *nown as

instantaneous centre or virtual centre of the lin*. The

complete rigid lin* rotates about the centre I at the given

instant. The relations shown may be used to determine the

magnitude of the velocity of point B.

I B B

I A A I A A

I A A

I A I A

r v

r vr v

r k v

r k vv

ω

ω ω

ω

ω

=

=⇒=

×=

×+=



Instant Center )et=een T=o Ri"id)odies

In the foregoing discussion, the velocities of

points A and B were assumed to be reference to

a coordinate system attached to the ground.

The resulting center is called an instant center

relative to the ground.

The velocities of points A and B could as well be

ta*en relative to coordinate system attached to

another body. The resulting point in this case

will have a <ero velocity with respect to that

body. In other words, point I will have the same

velocity for both bodies.

A

B I

A’

B’




A instant center between two bodies is a:

A point on both bodies

A point at which the two bodies have

no relative velocity.

A point about which one body may

be considered to rotate around the

other body at a given instant.

A

B I

A’

B’




+hen two lin*s are connected to

one another by a revolute =oint, the

center of the connecting =oint is an

instant center for the two lin*s.

+hen two lin*s are not connected,

an instant center between the two

lin*s will also exist and can be

determined if the velocities of both

lin*s is *nown.

A

A’




The number of instant centers in a constrained *inematic chain is

e'ual to the number of possible combinations of two lin*s.

The number of pairs of lin*s or the number of instantaneous

centers is the number of combinations of L lin*s ta*en two at a

time. >athematically, number of instant centers is,

( )

2

1−=

L L N



Ty+es of Instantaneo,s Centers

The instant centers for a mechanism are of the

following three types :

-. 5rimary 5ermanent! instant centers. They

can be "xed or moving

1. ?econdary Instant centers Not permanent!

( )21−=

nnC



Instant Centers of a Fo,r )arMechanism

9onsider a four bar mechanism ABCD as shown. The number of

instant centers N! in a four bar mechanism is given by

The instant centers I12

and I14

are "xed instant centers as they

remain in the same place for all con"gurations of the

mechanism. The instant centers I23

and I34

are permanent

instant centers as they move when the mechanism moves, but

the =oints are of permanent nature. The instantaneous centres

I13

and I24

are neither "xed nor permanent as they vary with

the con"guration of the mechanism.

( )

2

1−=

nnC

( ) 62

12

2

144==

−

= N A

B

C

D

1 (Ground)

2

3

4



&ocation of Instant Centers

+hen the two lin*s are connected by a pin =oint or pivot

=oint!, the instant center lies on the center of the pin as.

?uch an instant center is of permanent nature. If one of the

lin*s is "xed, the instant center will be of "xed type.

+hen the two lin*s have a pure rolling contact without

slipping!, the instantaneous centre lies on their point of

contact, as this point will have the same velocity on both

lin*s.

+hen the two lin*s have a sliding contact, the instant center

lies at the center of curvature of the path of contact. This

points lies at the common normal at the point of contact.



Kennedy /or Three Centers in&ine0 Theorem

/ennedy;s theorem states that any three bodies

in plane motion will have exactly three instant

centers, and the three centers will lie on the

same straight line.

Note that this rule does not re'uire that the

three bodies be connected in any way. +e can

use this rule, in con=unction with the linear

graph, to "nd the remaining l9s which are not

obvious from inspection.



E1am+le? Instant Centers of a Fo,r)ar Mechanism

4raw a circle with all lin*s numbered around the circumference

ocate as many ICs as possible by inspection. All pin =oints will

be permanent ICs . 9onnect the lin*s numbered on the circle

to create a linear graph and record those found.

Identify a lin* combination for which the IC has not been found,

and draw a dotted line connecting those two lin* numbers.

Identify two triangles on the graph which each contains the

dotted line and whose two other sides are solid lines

representing the ICs the already found. @se /ennedy;s

theorem to locate the needed I9.



E1am+le? Instant Centers of a Fo,r)ar Mechanism



E1am+le? Instant Centers of a(liderCran% Mechanism



E1am+le? Instant Centers of a(liderCran% Mechanism

E1am+le? Instant Centers of a Cam



E1am+le? Instant Centers of a Camand Follo=er? Common NormalMethod

E1am+le? Instant Centers of a Cam



E1am+le? Instant Centers of a Camand Follo=er? E@ecti!e &in%Method



Velocity Analysis =ith InstantCenters

nce the instant centers I9s! of a lin*age

with respect to the ground lin* have been

found, they can be used for a very rapid

graphic analysis for that lin*.

Note that some of the I9s may be very far

removed from the lin*s. or example, if

lin*s 1 and ( are nearly parallel, their

extended lines will intersect at a point far

away and not be practically available for

velocity analysis.




rom the de"nition of the instant center, both lin*s sharing the

instant center will have identical velocity at that point.

Instant center I13

involves the coupler lin* B! which is in general

plane motion, and the ground lin* which is stationary. All points on

the ground lin* have <ero velocity in the global coordinate

system, which is embedded in lin* -. Therefore, I13

must have

<ero velocity at this instant, and it can be considered to be an

instantaneous C"xed pivotC about which lin* B is in pure rotation

with respect to lin* -.

A moment later, I

13

will move to a new location and lin* B will be

CpivotingC about a new instant center.




If ω2 is *nown for the mechanism shown, the magnitude of

the velocity of point A can be computed as v A

$ ω2 O2 A Its

direction and sense can be determined by inspection.

Note that point A is also instant center and it has the same

velocity as part of lin* 1 and as part of lin* B. ?ince lin* B is

eectively pivoting about I13 at this instant, the angular

velocity ω3 can be found by ω

3 $ v

A 7 AI

13 . nce ω

3 is

*nown, the magnitude of v B

can also be found from v B

$ ω3

AI13

nce v B

is *nown, ω4

can also be found from ω4 $ v

B 7BO

4

$ v B

7BO4 . inally, v

C or the velocity of any other point on

the coupler! can be found from v C

$ ω3 CI

13

13

3

AI

v A

=ω

133 BI v

B ω =

Documents

Class 4 Velocity Analysis1 (1)