Mechanics of Machines

Dr. Mohammad Kilani

Class 4Velocity Analysis

DERIVATIVE OF A ROTATING VECTOR

Derivative of a Rotating Unit Vector

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 differentiation rules on the

expression of uθ

above

jiuθ sincos

jiu

jiu

θ

θ

cossin

cossin

dt

d

dt

ddt

d

dt

d

dt

d

θ

cosθ i

sinθ j

uθ

Derivative of a Rotating Unit Vector

Noting that

the time derivative of the vector uθ

is

sin2cos

cos2sin

jiu

jiu

θ

θ

cossin

sincos

dt

d

dt

d

2

2sin2cos

cossin

uu

jiu

jiu

θ

θ

θ

dt

d

dt

ddt

d

dt

ddt

d

dt

dθ

cosθ i

sinθ j

uθ

Derivative of a Rotating Unit Vector

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 dθ/dt.

If we define a vector ω as a vector in the k direction of

magnitude ω = dθ/dt, then

2

sincos

uu

jiu

θ

θ

dt

d

dt

d

θ

uθ

duθ

/dt

θ + π/2

θθ uωu

u

kω

2

dt

d

dt

ddt

d

Derivative of a Rotating Vector

θθθ

θθθ

θθ

θθ

θ

θθ

rωur

uωur

uur

uu

r

ur

dt

dr

dt

d

rdt

dr

dt

ddt

dr

dt

dr

dt

ddt

dr

dt

dr

dt

d

r

2

θ

uθ

duθ

/dt

θ + π/2

A vector rθ

= r uθ

, is a general vector of magnitude r pointing in the θ direction. The time

derivative of rθ

is found by applying the normal differentiation rules on the expression for rθ

Derivative of a Rotating Vector

θ

rθ

ω x rθ

θ + π/2

(dr/dt) uθ

Given a vector rθ

= ruθ

which rotates relative to the

reference coordinates, the time derivative of rθ

has two

components; a component in the direction of uθ

and a

component normal to uθ

in the direction of uθ+π/2

The magnitude of the component of drθ

/dt in the direction

of uθ

is equal to dr/dt; that is the time derivative of the length

of rθ

.

The magnitude of the component of drθ

/dt in the direction

of direction of uθ+π/2

is equal to ωr

VELOCITY ANALYSIS OF FOUR BAR MECHANISMS

Derivative of the Loop Closure Equation for a Four Bar Kinematic Chain

The loop closure equation of a 4-bar kinematic chain is written as

When all the links in the chain are of constant lengths, the equation above

reduces to

0

0

0

0

44113322

4132

4244121132332222

4132

4132

4132

4132

urururururururur

ururururdt

d

rrrrdt

d

rrrr

rrrr

0

0

44113322

244211233222 4132

rωrωrωrω

urururur

Derivative of the Loop Closure Equation for a Four Bar Mechanism

For a four bar mechanism with link 1 fixed we have dθ1/dt = ω

1 =

0.

The vector equation above contains two scalar equations and can

be solved for two unknowns. Knowledge of dθ2/dt allows the

calculation of dθ3/dt and dθ

4/dt .

244233222 432 ururur

444333222

444333222

coscoscos

sinsinsin

rrr

rrr

Derivative of the Loop Closure Equation for a Four Bar Mechanism

Eliminate dθ3/dt by carrying out a dot product with u

θ3 on both sides of the equation

Alternatively, eliminate dθ4/dt by carrying out a dot product with u

θ4 on both sides of

the equation

244233222 432 ururur

2

344

3222

344

32244

34443222

34443222

sin

sin

sin

sin

sinsin

2cos2cos

r

r

r

r

rr

rr

2

343

4222

433

4222

433

42233

43334222

43334222

sin

sin

sin

sin

sin

sin

sinsin

02cos2cos

r

r

r

r

r

r

rr

rr

Angular Velocity Ratio and Mechanical Advantage

The angular velocity ratio mV

is defined as the output angular velocity divided by the input angular velocity. For a four bar mechanism

with link 2 as the input and link 4 as the output this is expressed as

The efficiency of a four bar linkage is defined as the output power over the input power,

Assuming 100% efficiency, which is normally approached by four bar mechanisms, we have

2

4

in

outVm

inin

outout

in

out

T

T

P

P

VT

out

in

inin

out

mm

T

T 1

Angular Velocity Ratio and Mechanical Advantage

The mechanical advantage is defined as the ratio

between the output force to the input force

out

inT

inin

outout

in

outA r

rm

rT

rT

F

Fm

VELOCITY ANALYSIS OF SLIDER-CRANK MECHANISMS

Velocity Analysis of a Slider-Crank Mechanism

Design parameters: r2

, r3

, r4

, θ1

.

Position analysis parameters:

r1

, θ2

, θ3

Velocity analysis parameter.

Find dr1

/dt, dθ2

/dt , dθ3

/dt

To eliminate ω3

dot product both sides by uθ3

r3

r2

r1

r4

rp

132 1233222

4132

ururur

rrrr

2

13

2321

3113222

3113222

cos

sin

cossin

cos2cos

rr

rr

rr

2

133

1223

13331222

cos

cos

coscos

r

r

rr

To eliminate dr1

/dt dot product both sides by

u(θ1+π/2

)

Velocity Analysis of an Inverted Slider-Crank Mechanism

Given r1

, r2

, θ1

, θ2

, ω2

Position analysis: Find r3

, θ3

Velocity analysis: Find dr3

/dt, dθ3

/dt

To eliminate ω3

and find dr1

/dt dot product both sides by uθ3

To eliminate dr1

/dt and find ω3

dot product both sides by u(θ3+π/2

)

2333222

312

312

332

312

ururur

ururur

rrr

32322

33222

sin

2cos

rr

rr

r2

r1

θ1

θ2

r3

θ3

23

2323

333222

cos

cos

r

r

rr

Velocity Analysis of an Inverted Slider-Crank Mechanism

Given r1

, r2

, r4

, θ1

, θ2

, ω2

Position analysis: Find r3

, θ3

, θ4

Velocity analysis: Find dr3

/dt, dθ3

/dt , dθ4

/dt

r3

r2 r

1

r4

34423333222

34

24423333222

41332

4132

32

432

412

2

urururur

urururur

urururur

rrrr

with

Velocity Analysis of an Inverted Slider-Crank Mechanism

r3

r2 r

1

r4

23223

32243

23

322433222

23

32234

4433222

4433222

3

23

3223

333222

2

34423333222

sincos

cossin

cos

sin

2cos

cos

0cos3

32

rr

rrr

r

rrrr

r

r

rrr

rrr

u

r

r

rr

u

urururur

But

by sides both Dot

by sides both Dot

Example

224411

22441112,13

2244113

2244113

2,11

2,14

222

2,12

44

42424224144124

22

21

2

42421241144124

22

21

2

241

152

412

coscoscos

coscossintan

coscossinsin

coscoscoscos

tan2,02

0sincos

sinsincoscos2cos2cos2

cos2cos2cos2

2413

61532

4132

rrr

rrr

rrrba

rrrba

tCA

ACBBtACBttAC

CBA

rrrrrrrrrba

rrrrrrrrrba

urururuba

usururuaur

ururubaur

(2) rquation closure loop from Also

itself by equation the of sides both Dot

(1) Equation Closure Loop

II Loop

I Loop

a

r2

r1

b

r4

r5

s

a

r2

r1

b

r4

r5

s

2tan

cos2

sin2

cos22

4

2221

24

22

21

142

24241

t

barrrrrC

rrB

rrrrA

Example

224411

22441112,13

2244113

2244113

2,11

2,14

222

2,12

24144124

22

21

2

32226213262

3122122

222

125

61

32226213262

131122116122

222

125

215

coscoscos

coscossintan

coscossinsin

coscoscoscos

tan2,02

cos2cos2

cos2cos2cos2cos2

cos2cos2

2,0

cos2cos2cos2cos2

cos2cos2cos2

32615

rrr

rrr

rrrba

rrrba

tCA

ACBBtACBttAC

rrrrrrrba

arsrassr

arrrarsrr

arsrassr

arrrsrarsrr

uaurusurur

(2) rquation closure loop from Also

Let

itself by equation the of sides both Dot

(2) Equation Closure Loop

a

r2

r1

b

r4

r5

s

a

r2

r1

b

r4

r5

s

2tan

cos2

sin2

cos22

4

2221

24

22

21

142

24241

t

barrrrrC

rrB

rrrrA

HW#3

6-17, 6-18 (b+c), 6-21 (b), 6-44.

METHOD OF INSTANT CENTERS

Relative Velocities Between Two Points on a Rigid Body

Given any two points A and B that lie on a

rigid body, let the line AB be a line passing

through A and B, then the components of

the velocity of A and the velocity of B on

the line AB must be equal.

The velocity of point B relative to A must

be normal to the line AB

A

B

A

B

ABBA vvv

Instant Center Relative to the Ground

If line AA’ is drawn normal to the direction of vA

, 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 BB’ is drawn perpendicular to the direction

of vB

, 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 zero.

A

BI

A’

B’

IBB

IAAIAA

IAA

IAIA

rv

rvrv

rkv

rkvv

Instant Center Relative to the Ground

Point I determined as before is called the instant center of

zero 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 link is

proportional to its distance from the point I.

A

BI

A’

B’

C

IC

v

IB

v

IA

v ABA

Velocity of a Point on a Link byInstant Center Method

Consider two points A and B on a rigid link. Let vA

and vB

be the velocities

of points A and B at a given instant. If vA

is known in magnitude and

direction and vB

in direction only, then the magnitude of vB

may be

determined by the instantaneous centre method.

Draw AI and BI perpendiculars to the directions vA

and vB

respectively. Let

these lines intersect at I, which is known as instantaneous centre or virtual

centre of the link. The complete rigid link 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.

IBB

IAAIAA

IAA

IAIA

rv

rvrv

rkv

rkvv

Instant Center Between Two Rigid Bodies

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 taken

relative to coordinate system attached to another body. The

resulting point in this case will have a zero velocity with

respect to that body. In other words, point I will have the

same velocity for both bodies.

A

BI

A’

B’

Instant Center Between Two Rigid Bodies

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

BI

A’

B’

Instant Center Between Two Rigid Bodies

When two links are connected to one

another by a revolute joint, the center of

the connecting joint is an instant center for

the two links.

When two links are not connected, an

instant center between the two links will

also exist and can be determined if the

velocities of both links is known.

A

A’

Instant Center Between Two Rigid Bodies

The number of instant centers in a constrained kinematic chain is equal to the

number of possible combinations of two links.

The number of pairs of links or the number of instantaneous centers is the number

of combinations of L links taken two at a time. Mathematically, number of instant

centers is,

2

1

LLN

Types of Instantaneous Centers

The instant centers for a mechanism are of the following three

types :

1. Primary (Permanent) instant centers. They can be fixed

or moving

2. Secondary Instant centers (Not permanent)

2

1

nnC

Instant Centers of a Four Bar Mechanism

Consider 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 fixed instant centers as they remain in the

same place for all configurations of the mechanism. The instant centers I23

and

I34

are permanent instant centers as they move when the mechanism moves,

but the joints are of permanent nature. The instantaneous centres I13

and I24

are neither fixed nor permanent as they vary with the configuration of the

mechanism.

2

1

nnC

6

2

12

2

144

N A

B

C

D

1 (Ground)

23

4

Location of Instant Centers

When the two links are connected by a pin joint (or pivot joint), the instant

center lies on the center of the pin as. Such an instant center is of

permanent nature. If one of the links is fixed, the instant center will be of

fixed type.

When the two links 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 links.

When the two links 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 Line) Theorem

Kennedy’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 require that the three bodies be

connected in any way. We can use this rule, in conjunction

with the linear graph, to find the remaining lCs which are not

obvious from inspection.

Example: Instant Centers of a Four Bar Mechanism

Draw a circle with all links numbered around the circumference

Locate as many ICs as possible by inspection. All pin joints will be permanent

ICs . Connect the links numbered on the circle to create a linear graph and

record those found.

Identify a link combination for which the IC has not been found, and draw a

dotted line connecting those two link 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. Use Kennedy’s theorem to locate

the needed IC.

Example: Instant Centers of a Four Bar Mechanism

Example: Instant Centers of a Slider-Crank Mechanism

Example: Instant Centers of a Slider-Crank Mechanism

Example: Instant Centers of a Cam and Follower: Common Normal Method

Example: Instant Centers of a Cam and Follower: Effective Link Method

Velocity Analysis with Instant Centers

Once the instant centers (ICs) of a linkage with

respect to the ground link have been found, they

can be used for a very rapid graphic analysis for

that link.

Note that some of the ICs may be very far removed

from the links. For example, if links 2 and 4 are

nearly parallel, their extended lines will intersect at

a point far away and not be practically available for

velocity analysis.

Velocity Analysis with Instant Centers

From the definition of the instant center, both links sharing the instant center will

have identical velocity at that point.

Instant center I13

involves the coupler (link 3) which is in general plane motion,

and the ground link which is stationary. All points on the ground link have zero

velocity in the global coordinate system, which is embedded in link 1. Therefore,

I13

must have zero velocity at this instant, and it can be considered to be an

instantaneous "fixed pivot" about which link 3 is in pure rotation with respect to

link 1.

A moment later, I13

will move to a new location and link 3 will be "pivoting" about

a new instant center.

Velocity Analysis with Instant Centers

If ω2

is known for the mechanism shown, the magnitude of the velocity of

point A can be computed as vA

= 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 link 2 and as part of link 3. Since link 3 is effectively pivoting about I13

at

this instant, the angular velocity ω3

can be found by ω3

= vA

/AI13

. Once

ω3

is known, the magnitude of vB

can also be found from vB

= ω3

AI13

Once vB

is known, ω4

can also be found from ω4

= vB

/BO4

= vB

/BO4

.

Finally, vC

(or the velocity of any other point on the coupler) can be found

from vC

= ω3

CI13

133 AI

vA

133BIvB