5 velocity graphical

AME 352 GRAPHICAL VELOCITY ANALYSIS

P.E. Nikravesh 5-1

5. GRAPHICAL VELOCITY ANALYSIS

Velocity analysis forms the heart of kinematics and dynamics of mechanical systems. Velocity analysis is usually performed following a position analysis; i.e., the position and orientation of all the links in a mechanism are assumed known. In this course we concentrate on one analytical and two graphical methods for planar mechanisms.

We start this chapter with some simple exercises to ensure that the fundamentals of velocity analysis using vector algebra are well understood. You may want to review these fundamentals in Chapter 2 of these notes.

Exercises

In these exercises take direct measurements from the figures for link lengths and the magnitudes of the velocity vectors. If it is stated that the angular velocity is known, assume ω = 1 rad/sec CCW unless it is stated otherwise. Write the position and velocity vector equations. Construct the velocity equation graphically in order to find the unknown(s). P.1

Known: VA and ω Determine: VB

A B

VA

P.2 Known: VA and VB Determine: ω

A B

VA

BV

P.3

Known: VA and VB Determine: ω What do you observe?

A B

VABV

P.4 Known: VA and ω are known.

Determine: VB , VC and VBC

A

B

C

VA

P.5

Known: VA and VB Determine: VC

A

B

C VA

BV

P.6 Known: VA and ω

Determine: VB , VC and VBC

A B C

VA

P.7 Known: VA and VB

Determine: VC

A B C

VA

P.8 Known: VA and ω

Determine: VB and VC What do you observe? Explain!

A B C VA


P.E. Nikravesh 5-2

P.9

Known: VA , ω and VBAs

Assume VBAs = 1 unit/sec positive

Determine: VB and VC

A B C VA

P.10

Known: VA , ω and VBAs

Assume VBAs = 1 unit/sec negative


A B C VA

P.11

Known: VA and VB

Determine: ω and VC

A B C VA

BV

P.12

Known: VA , ω i and ω j

Assume ω i = 1 rad/sec CCW and ω j = 1

rad/sec CW.


A B

C VA

(i)

(j)

P.13

Known: VA and VC

Determine: VB

A B

C

VACV

(i)

(j)

Polygon Method

Velocity polygon is a graphical pencil-and-paper approach for determining unknown velocities of a single degree-of-freedom mechanism. The method requires constructing a velocity loop equation (a polygon) graphically. A polygon may have three or more edges depending on the number of velocity vectors in the equation. For a vector loop equation, the polygon method is the graphical procedure of solving two algebraic equations in two unknowns. The velocity polygon method is demonstrated for several commonly used mechanisms. Four-bar

For a known four-bar mechanism, in a given configuration and for a known angular velocity of the crank, ω2 , we want to determine ω 3 and ω 4 . In this

example we assume ω2 is CCW.

For the position vector loop equation RAO2

+RBA − RBO4− RO4O2

= 0

the velocity equation is

VA +VBA − VB = 0 (a)

ABRBA

O2O4

RAO2

RBO4

RO4O2

Since vectors RAO2, RBA , and RBO2

have constant lengths, their corresponding velocity vectors


P.E. Nikravesh 5-3

are tangential; i.e.,

VAt +VBA

t − VBt = 0 (b)

or,

ω2RAO2

+ω 3RBA −ω 4RBO2= 0 (b)

The unknowns are ω 3 and ω 4 since all vector axes are known. Velocity polygon 1. Next to the diagram of the four-bar, select a point in a

convenient position as the reference for zero velocities.

Name this point OV (origin of velocities).

2. Compute the magnitude of VA as RAO2ω2 . From OV

construct vector VA perpendicular to RAO2 by rotating

RAO2 90o in the direction of ω2 .

3. From A draw a line perpendicular to RBA . VBA must

reside on this line.

4. From OV draw a line perpendicular to RBO4. VB must


5. Construct vectors VBA and VB .

6. Determine the magnitude of VBA from the polygon.

Compute ω 3 = VBA / LBA . Determine the direction of ω3 .

In this example it is CW since RBA must rotate 90o CW

to line up with VBA .

O V

VA

A

A

O2

RAO2

O V

VA

A

ABRBA

O V

VA

A

B

O4

RBO4

O V

VA

VBA

VB 7. Determine the magnitude of VB from the polygon. Compute ω 4 = VB / LBO4

. Determine the


P.E. Nikravesh 5-4

direction of ω4 . In this example it is CCW since RBO4 must rotate 90o CCW to line up with

VB .

Secondary equation(s) To determine the velocity of a secondary point, such

as a coupler point, we refer to the position expression and the corresponding velocity expression:

R PO2

= R AO2+ R PA

VP = VA + VPA =ω2R AO2

+ω3R PA

Since the angular velocities are already known, VA and

VPA are constructed. We add these two vectors

graphically to determine VP .

A

RPA

P

O2

RAO2 RPO2

VP

VA

VPA

x

y

O V

Example FB-VP-1 A four-bar mechanism has the following constant

data: LAO2

= 1.0 , LBA = 4.0 , LBO4

= 3.0 , LO4O2

= 3.0 ,

LPA = 1.8 , β3 = 75o . The crank angle is at θ2 = 170o

with an angular velocity of ω2 = 1.0 rad/sec CCW.

The velocity polygon is constructed and the following velocities are determined from the polygon:

ω3 =

VBA

LBA

=0.964.0

= 0.24 rad/sec, CCW

ω4 =

VB

LBO4

=0.843.0

= 0.28 rad/sec, CCW

A second polygon provides the velocity of point P

as VP = 1.3 in the direction shown.

A

P

O2

B

O4

VA

VBA

VBO

VP

VPA

VA

O

Slider-crank (inversion 1)

This slider-crank mechanism in the given configuration has a known angular velocity of the crank, ω2 . We want to determine ω 3 and the velocity of the slider block. In this example we

assume ω2 is CCW. The position vector loop equation is:

RAO2+RBA − RBO2

= 0

The velocity (loop) equation is expressed as VA +VBA − VB = 0

We note that VA and VBA are tangential and VB is of

slip type (along the axis of RBO2). Therefore the

velocity equation can be expressed as

ω2RAO2

+ω 3RBA − VBs = 0

A

B

RBA

O2

RAO2


P.E. Nikravesh 5-5

Velocity polygon 1. Next to the diagram of the mechanism, select a point for

the origin of velocities.

2. Compute the magnitude of VA as RAO2ω2 . From OV

construct vector VA perpendicular to RAO2 by rotating

RAO2 90o in the direction of ω2 .

3. From A draw a line perpendicular to RBA . VBA must

reside on this line

4. From OV draw a line parallel to the axis of the slider; i.e.,

parallel to RBO2. VB must reside on this line.

5. Construct vectors VBA and VB , considering their signs in

the velocity equation. 6. Determine the magnitude of VBA from the polygon.

Compute ω 3 = VBA / RBA . Determine the direction of ω3 ,

which is CW in this example. 7. Determine the magnitude of VB from the polygon. The

direction of this vector indicates that the slider block is moving to the left.

O V

VA

A

A

O2

RAO2

O V

VA

A

A

RBA

B

VBA

VBO V

VA

Example SC-VP-1

For a slider-crank mechanism, the following lengths are given:

LAO2

= 1.5, LBA = 3.0

The crank angle is θ2 = 120o , and ω2 = 1 rad/sec CW.

The mechanism is drawn for θ2 = 120o . For the given

angular velocity, the velocity polygon is constructed. The following velocities are determined from the polygon:

ω3 = 0.28 rad/sec, CW; VB = 0.94 to the right.

VBA

VB

VA

O

A

O2

B


For this slider-crank mechanism (inversion 2), in the given configuration and for a known angular velocity of the crank, ω2 , construct the velocity polygon. Then determine ω 4 and the

velocity of the slider block. Assume ω2 is CW.

The position vector loop equation is RAO2− RO4O2

− RAO4= 0 . Since RAO2

is a rotating fixed-

length vector, VAO2 is tangential. However, RAO4

is a variable-length, variable-angle vector, and

therefore VAO4 contains both tangential and slip components.


P.E. Nikravesh 5-6

Hence the velocity (loop) equation is written as

VAO2

t − VAO4

s − VAO4

t =ω2RAO2− VAO4

s −ω 4RAO4= 0

Note: In this velocity equation we do not drop the index of the non-moving points, O2 and O4 . Dropping them

may cause confusion.

A

O2 O4

(3) (4)

RO4O2

RAO4

RAO2

Velocity polygon 1. Select a point for the origin of velocities.

2. Construct vector VAO2.

3. From the end of VAO2 draw a line parallel to RAO4

.

VAO4

s should reside on this line

4. From OV draw a line perpendicular to the axis of the

slider; i.e., perpendicular to RBO2. VAO4

t should


5. Construct vectors VAO4

s and VAO4

t , considering their

signs in the velocity equation.

6. Determine the magnitude of VAO4

t . Compute

ω 4 = VAO4

t / RAO4. Determine the direction of ω4 . In

this example ω4 is CW.

7. Determine the magnitude of VAO4

s from the polygon.

O2

A

RAO2

O V O V VAO2

O4

A

RAO4

O V VAO2

VAO4

s

VAO4

t

O V VAO2

Secondary equation(s)

Determine the velocity of a secondary point, such P on link 4, where L4 = RPO4

is a known constant.

There are two possible ways to determine VP :

(a) We can position P with respect to the ground point

O2 as RPO2= RAO2

+ RPA . The corresponding

velocity expression is VP = VA + VPA , where VPA

contains both tangential and slip components. Computing the tangential component requires the magnitude of RPA , which must be determined based

on L4 − RAO4. The slip component of VPA must be

based on the slip component of RAO4. It should be

obvious that we have made a simple problem unnecessarily difficult!

(b) We can position P with respect to the ground point

O2 as RPO2= RO4O2

+ RPO4. The corresponding

velocity expression is VP = VPO4

t =ω 4RPO4 as shown

on the figure based on a CW direction of ω4 .

A

O2

P

RAO2

RPA

RPO2

(a)

O2

O4

P

RPO4

RO4O2

RPO2

VP

(b)


P.E. Nikravesh 5-7

Example SC-VP-2

Consider the following lengths for a slider-crank,

inversion 2: LAO2

= 1.0, LO4O2= 2.0, LPO4

= 4.0 . The crank

angle is in θ2 = 125o orientation. The angular velocity of the

crank is ω2 = 1 rad/sec CW.

The mechanism is drawn for the given crank angle. The velocity polygon is constructed and the following values are

obtained: ω3 =ω4 = 0.3 rad/sec CW; VAO4

s = 0.61 in the

direction shown. The velocity of point P is computed as

VP = (4.0)(0.3) = 0.12 , in the direction shown.

Q: Is the slider moving away from O4 ?

VP

O2 O4

P A

VAO4

s

VAO4

t

VAO2O


For this slider-crank mechanism (inversion 3), in the given configuration and for a known angular velocity of the crank, ω2 , construct the velocity polygon. Then determine ω 4 and the

velocity of the slider block. Assume ω2 is CW.

The position vector loop equation is written as:

RAO2+ RO4A − RO4O2

= 0

The corresponding velocity equation is

VAO2

t + VO4As + VO4A

t =ω2RAO2+ VO4A

s +ω 3RO4A = 0

Velocity polygon 1. Select a point for the origin of velocities.

2. Construct vector VAO2.

3. From the end of VAO2 draw a line parallel to RO4A .

VO4As should reside on this line

4. From OV draw a line perpendicular to the axis of the

slider; i.e., perpendicular to RO4A . VO4At must reside

on this line.

5. Construct vectors VAO4

s and VAO4

t , considering their

signs in the velocity equation.

O2 O4

(3)

(4)

A

RAO2

RO4O2

RO4A

O V

VAO2

A

RAO2

O2

O V

VAO2

A

O4

RO4A

O V

VAO2

VO4At

VO4As


P.E. Nikravesh 5-8

6. Determine the magnitude of VO4At . Compute ω 3 =ω 4 = VO4A

t / RO4A . Determine the direction of

ω3 . In this example, it is CW.

4. Determine the magnitude of VO4As .

Secondary point Determine the velocity of point P on link 3, where

RPA = L3 is a known constant.

Point P can be positioned with respect to the ground

point O2 as RPO2= RAO2

+ RPA . The corresponding

velocity expression is

VP = VA + VPA =ω2R AO2

+ω3R PA

Since both angular velocities are known, VP can be

constructed graphically.

RAO2

RPO4

A

O2 O4

(3)

P

RPA

VPAt

P

O V

VP

VAO2

Example SC-VP-3 The following lengths are provided for a slider-crank

(inversion 3) mechanism: LAO2

= 1.0, LO4O2= 1.5, LPA = 0.5 .

The crank angle in the shown configuration is θ2 = 30o .

The angular velocity of the crank is ω2 = 1 rad/sec CCW.

The velocity polygon is constructed and the following

velocities are determined from the polygon: ω3 =ω4 = 0.46

rad/sec CW, VO4As = 0.9 in the direction shown.

A second polygon provides the velocity of P as VP = 1.1

in the direction shown.

O2 O4

P A

VO4At

VO4As

O

VO4At

VPAt

VP

VO4At

O

Exercises

In these exercises take direct measurements from the figures for link lengths and the magnitudes of velocity vectors. Construct velocity polygons to determine the unknowns.

Exercises P.1 – P.4 are examples of four-bar mechanism. Assume known value and direction

for ω2 . Determine ω3 , ω4 , and VP .

P.1

(2)

(3)(4)

P

P.2

(2)(3)

(4)

P


P.E. Nikravesh 5-9

P.3 (2)

(3)

(4)P

P.4

(2)(3)

(4)

P

Exercises P.5 – P.8 are examples of slider-crank mechanism. Assume known value and direction for ω2 . For P.5 and P.6 determine ω3 , ω4 , and the velocity of the slider block. For

P.7 and P.8 determine ω3 , ω4 , and VP . P.5

(2)(3)

(4)

P.6

(2)(3)

(4)

P.7

(2)(3)

(4)

P

P.8

(2)

(3)

(4)

P

P.9

For this six-bar mechanism ω2

is given. Determine ω5 , velocity of P, and the velocity of the slider block.

(6)(2)

(3)(4)

P

(5)

Q

P.10 For this six-bar mechanism ω2

is given. Determine ω5 and the velocity of the slider block (6).

(2)(3)

(4)

(5)

(6)

Instant Center Method


P.E. Nikravesh 5-10

Instant center of velocities is a simple graphical method for performing velocity analysis on

mechanisms. The method provides visual understanding on how velocity vectors are related. What is An Instant Center? Instant center of velocities between two links is the location at which two coinciding points, one

on each link, have identical velocities. The most obvious instant center of velocities, or simply the

instant center (IC), between two links that are pinned to each other is the point at the center of the pin joint. For example, the center of the pin joint between links i and j can be viewed as two coinciding points, Pi on link i and Pj on link j, that have the same velocities.

The instant center between these two links is denoted as Ii, j or I j ,i .

The instant center of velocities may not be located within the physical boundaries of a link. As shown in the second figure, the IC between links k and h, Ik ,h , is located on imaginary extensions of both links.

(j)

Ii, jVPi= VPj

(i) Pi Pj

(k) I1, j

(h) Ik ,h

VPk= VPh

Ph

Pk

Instant Center Between A Link and The Ground

Consider link i that is pinned to the ground at O. Point O is the instant center between links i and the ground, and it is denoted I1,i or

Ii,1 (the ground is always given the index 1).

If the link has a non-zero angular velocity ω , every point on the link has a non-zero velocity except for point O. The velocity of any

point on the link is determined as V =ωR , where R is the position vector of that point with respect to O. Note that all the velocity

vectors are tangent to circles with a common center at I1,i .

Now consider link j that is not connected to the ground directly. If the link has a non-zero angular velocity, the velocity vectors of all the points on the link must be tangent to circles with a common center. This common center is the instant center between link j and the ground; i.e., this point has a zero velocity. This center acts as an imaginary pin joint between the link and the ground. It should be

obvious that VA / RA,I1, j= VB / RB,I1, j

= VC / RC ,I1, j=ω .

Note: If we know the velocity (absolute) of a point on a link, the

instant center between that link and the ground must be located on an axis perpendicular to the velocity vector passing through the point.

1, i I

(i)

A

B

C

RC,O

V C

B,ORBV

A,ORAV O

(j)

A

BCV C

BV

AV

I1, j

RC ,I1, j

RB ,I1, j

RA ,I1, j

(i) 1, i

I

AAV

Two Links Connected by A Sliding Joint

The instant center between two links that are connected by a sliding joint is located in infinity on any axis perpendicular to the sliding axis. The reason for this instant center being in infinity will be discussed later.

(i)

(j)

∞ i, j I

1, i I

(i)

∞BV

AV A

B


P.E. Nikravesh 5-11

Number of Instant Centers In a mechanism with n links (count the ground as one of the links), the number of instant

centers is determined as:

C =n(n −1)

2

As an example, in a four-bar mechanism or a slider-crank, there are six IC’s ( n = 4 ). For any six-bar mechanism, C = 15 . Kennedy’s Rule

The three instant centers between three planar links must lie on a straight line.

This rule does not tell us where the line is or where the centers are on that line. However, the rule can be used to find the instant centers when we consider a mechanism.

i, j I

k, i I

j, k I

(i)

(j)

(k)

i, j

I 1, i

I 1, j I

(i)

(j)

Instant Centers of A Four-bar

A four-bar mechanism has six instant centers regardless of the dimensions or orientation of the links. For bookkeeping purposes in locating the IC’s, we draw a circle and place link indices on the circle in any desired order. This bookkeeping procedure may not be necessary for a four-bar, but becomes very useful when mechanisms with greater number of links are considered.

(1)

(2)

(3)

(4)

1

2

3

4

Since pin joints are instant centers, for a four-bar with four pin joints, four IC’s are immediately identified. Each found IC is marked on the circle as a line drawn between the two corresponding link indices. These four IC’s are actual (not imaginary) pin joints. In order to find the other IC’s, we apply Kennedy’s rule. (1)

(2)

(3)

(4)

1,2 I

2,3 I

3,4 I

1,4 I

1

2

3

4 The IC’s between links 2, 3 and 4 must lie on a straight line. These are I2,3 , I3,4 , and I2,4 .

Since we already have I2,3 and I3,4 , we draw a line through them; I2,4 must also be on this line.

The IC’s between links 1, 2 and 4 must lie on a straight line. These are I1,2 , I1,4 , and I2,4 . Since

we already have I1,2 and I1,4 , we draw a line through them; I2,4 must also be on this line. The

intersection of these two lines is I2,4 .

Note how the circle is used to decide which center to find next. The red line between links 2 and 4 indicates the center we are after. This line is shared between two triangles with known IC’s. The triangles tell us to draw a line between I1,2 and I1,4 , then draw another line between

I2,3 and I3,4 . The intersection is I2,4 .


P.E. Nikravesh 5-12

1,2 I

2,3 I

3,4 I

1,4 I

2,4 I

(1)

(2)

(3)

(4)

1

2

3

4 According to the circle, the last center to find is between links 1 and 3. The two triangles that

share this new red line tell us to draw a line between I1,2 and I2,3 , and a second line between I1,4

and I3,4 . The intersection of these two lines is I1,3 .

1,2 I

2,3 I

3,4 I

1,4 I

1,3 I

2,4 I

(1)

(2)

(3)

(4)

1

2

3

4 Now we have found all six centers.

Instant Centers of A Slider-crank A slider-crank mechanism has six

instant centers regardless of which inversion it is. Again, for bookkeeping purposes, we draw a circle with link indices. (1)

(2) (3)

(4)

1

2

3

4 Pin joints provide three of the

instant centers, I1,2 , I2,3 , and I3,4 . The

center between the slider block and the ground, I1,4 , is in infinity on an axis

perpendicular to the sliding axis. 1,2 I

2,3 I

1,4 I

(1)

(2) (3)

(4)

3,4 I 1

2

3

4 I2,4 must lie on the axis of I2,3

and I3,4 , and on the axis of I1,2 and

I1,4 . The intersection of these two axes

is I2,4 .

2,4 I

1,2 I

2,3 I

1,4 I

(1)

(2) (3)

(4)

3,4 I 1

2

3

4


P.E. Nikravesh 5-13

I1,3 must lie on the axis of I2,3 and

I1,2 , and on the axis of I3,4 and I1,4 .

Note that I1,4 is in infinity on an axis

perpendicular to the slider. The intersection of these two axes is I1,3 .

Now we have all six centers.

1,3 I

2,4 I

1,2 I

2,3 I

1,4 I

(1)

(2) (3)

(4)

3,4 I 1

2

3

4 Instant Centers of A Six-bar

In this example we consider a six-bar mechanism containing a four-bar and an inverted slider-crank that share one link and one pin joint. A circle is constructed with link indices 1 – 6.

(1)

(2)

(3)

(4) (5)

(6)

1

2

3

4

5

6

We first find the six IC’s that belong to the four-bar.

1,2 I

2,3 I

3,4 I

1,4 I

1,3 I

2,4 I

(1)

(2)

(3)

(4) (5)

(6)

1

2

3

4

5

6

Next we find the IC’s for the slider-crank. Note that I1,4 is shared between the two sub-

mechanisms.

1,2 I

2,3 I

3,4 I

1,4 I

1,3 I

2,4 I

(1)

(2)

(3)

(4) (5)

(6)

1,6 I

1,5 I

4,5 I

5,6 I

4,6 I

1

2

3

4

5

6

Next, we use the circle to guide us in finding the next IC. I2,6 must be on the intersection of

lines I2,4 - I4,6 and I1,2 - I1,6 (blue lines). I3,5 is found at the intersection of lines I1,3 – I1,5 and

I3,4 – I4,5 (red lines).


P.E. Nikravesh 5-14

1,2 I

2,3 I

3,4 I

1,4 I

1,3 I

2,4 I

(1)

(2)

(3)

(4) (5)

(6)

1,6 I

4,5 I

5,6 I

4,6 I

2,6 I

1

2

3

4

5

6

3,6 I

2,5 I

1,5 I

3,5 I

The next IC to find is I3,6 . This center is at the intersection of lines I3,4 – I4,6 and I1,3 – I1,6

(green lines). The last center, I2,5 , is found at the intersection of I2,4 – I4,5 and I1,2 – I1,5 (purple

lines). Now we have all the centers.

1,2 I

2,3 I

3,4 I

1,4 I

1,3 I

2,4 I

(1)

(2)

(3)

(4) (5)

(6)

1,6 I

1,5 I

4,5 I

5,6 I

4,6 I

2,6 I

3,5 I

1

2

3

4

5

6

Strategy

The instant center method is a graphical process to perform velocity analysis. A graphical process is a pencil-and-paper approach that requires locating points, drawing lines, finding intersections, and finally taking direct length measurements from the drawing. All of these steps have graphical and measurement errors. Therefore, the accuracy of the analysis depends on the accuracy of our drawings and measurements.

For four-bars and slider-cranks, since four links are involved, there are only six centers to locate. For mechanisms with more links that four, there are many more centers to find. Locating some of the centers requires using some of the other centers that have already been found. The following strategy can reduce the graphical error in locating some of the centers.

Let us use the previous six-bar mechanism as an example. The first seven centers that we locate are at the center of the pin joints. Marking these centers by hand on a diagram contain certain amount or error that we call Order-1 level:

O −1: I1,2 , I2,3, I3,4 , I1,4 , I4,5 , I5,6 , I1,6

Next we locate I1,3 , I2,4 , I4,6 and I1,5 using the first seven centers.

1

2

3

4

5

6


P.E. Nikravesh 5-15

These centers add more errors on top of the errors from the original seven. We consider these new centers to contain errors at Order-2 level:

O − 2 : I1,3, I2,4 , I4,6 , I1,5

Next we locate I2,6 and I3,5 using centers with O-1 and O-2 level

errors. Therefore these two centers contain their own graphical error on top of the errors from the other centers:

O − 3 : I2,6 , I3,5

Up to this point we did not have any other choices in how to locate the centers, but for the remaining centers we may have more than one choice. For example, to locate I3,6 we can use the intersection between

any two of these four axes: I3,4 - I4,6 , I1,3 - I1,6 , I2,3 - I2,6 , and I3,5 - I5,6 .

Considering the error level in I2,6 and I3,5 , we should not use I2,3 - I2,6 ,

and I3,5 - I5,6 axes. Instead, we should use the intersection of I3,4 - I4,6

and I1,3 - I1,6 to locate I3,6 :

O − 3 : I3,6

Note: When locating a new center, use existing centers with the lowest amount of error.

1

2

3

4

5

6

1

2

3

4

5

6

1

2

3

4

5

6

Determining Unknown Velocities Instant centers are used to determine unknown velocities in a mechanism. Typically the process requires finding the velocity of a point on one link, based on known velocity of a point on another link. The process, in its most efficient form, requires using three instant centers—the centers between the two links and the ground.

Assume that the location of the three instant centers between links i, j, and the ground (link 1), and the velocity of point A on link i are given. The objective is to determine the velocity of point B on link j.

To determine the velocity of B, we consider the following four steps:

1. We start with the link that has a known velocity.

Link i rotates about an imaginary pin joint at I1,i .

We construct vector RA,I1,i and determine its

magnitude. We compute the angular velocity of link

i as ω i = VA / RA,I1,i. The direction of this angular

velocity is CW.

2. The instant center Ii, j is an imaginary point on link i, and therefore we can determine its velocity. We measure the length of vector R Ii , j ,I1,i

and compute

VIi , j=ω iRIi , j ,I1,i

. The direction of VIi , j is established

based on the direction of ω i .

i, j I

1, i I

1, j I

(i) (j)

A BAV

1, i I

(i)

A AV

A, I R1, i

ω i

ω ii, j

I

1, i I I V

i, j

I , I R1, i i, j


P.E. Nikravesh 5-16

3. The instant center Ii, j is also an imaginary point on

link j, and we already know VIi , j. Link j rotates

about the imaginary pin joint to the ground at I1, j .

We measure the length of vector R Ii , j ,I1, j, then

ω j = VIi , j/ RIi , j ,I1, j

is computed. The direction of ω j

is established to be CCW. 4. Point B is attached to link j that rotates about an

imaginary pin joint at I1, j . We construct vector

RB,I1, j and determine its magnitude. We then

compute VB =ω j RB,I1, j. The direction of VB is

established based on the direction of the angular velocity of link j.

i, j I

1, j I

I V i, j

I , I R1, j i, j

ω j

1, j I

(j)

B

B, I R1, j

BV

ω j

These four steps, either as they are presented or with slight variations, can be applied to find any unknown velocities in mechanisms. Here is another example of applying these four steps when an instant center is in infinity.

Assume that the centers between two links connected by a sliding joint and the ground, and the velocity of A on link i are given. The objective is to find the velocity of B on link j. Here are the four steps, slightly revised: 1. Link i is pinned (imaginary) to the ground at I1,i .

The angular velocity of link i is computed as

ω i = VA / RA,I1,i, CCW.

2-3. The two links are connected by a sliding joint,

therefore ω j =ω i .

4. Link j rotates about I1, j . Velocity of B is computed

as VB =ω j RB,I1, j. The direction is established based

on the direction of the angular velocity.

∞

(i)

(j)

1, j I

1, i I

i, j I

B

A

AV

BV

A, I R1, i

B, I R1, j

In a third example link i is connected to the ground by a sliding joint. The velocity of A on

link i is given and the objective is to find the velocity of B on link j. 1. Since link i slides relative to the ground, ω i = 0 .

2. Ii, j is a point on link i, therefore VIi , j= VA .

3. Link j rotates with respect to the ground about I1, j .

The angular velocity of link j is computed as

ω j = VIi , j/ RIi , j ,I1, j

, CCW.

4. The velocity of B is computed as VB =ω j RB,I1, j in the

direction shown. 1, i I

(i)

∞AV

A

B (j)

1, j I

i, j I

BV

B

B, I R1, j

I V i, j

I , I R1, j i, j

Angular Velocity Ratio A formula, known as the angular velocity ratio, can be derived between the angular velocities

of any two links of a mechanism regardless of the type of joints used in that mechanism. This formula, in general, can be established for links i and j. Between these two links and the ground

link 1, there are three instant centers I1,i , Ii, j , and I1, j . The velocity of the common center Ii, j


P.E. Nikravesh 5-17

can be determined as VIi , j= RIi , j I1,i

ω i = RIi , j I1, jω j . Therefore, the angular velocity ratio between the

two moving links is expressed as RIi , j I1,i

RIi , j I1, j

=ω j

ω i

i, j I

1, i I

1, j I

ω j

ωi

R Ii , j I1, j

R Ii , j I1,i

i, j I

1, i I

1, j I

ω j

ωiR Ii , j I1, j

R Ii , j I1, j

(a) (b)

Note: If Ii, j is between I1,i and I1, j , as in (a), ω i and ω j are in opposite directions ( R Ii , j I1,i and

R Ii , j I1, j are in opposite directions). But if Ii, j is not located between I1,i and I1, j , as in (b),

ω i and ω j are in the same direction ( R Ii , j I1,i and R Ii , j I1, j

are in the same direction).

In the following examples we use instant centers to perform velocity analysis for several

mechanisms. It is always assumed that either the angular velocity of one link or the linear velocity of one point is given. Four-bar Mechanism

For this four-bar mechanism, we have already found the instant centers. Assume the angular velocity of link 2 is given, CCW. Find: (a) the velocity of point P

on link 3, and (b) the angular velocity of

link 4.

P

1,2 I

2,3 I

3,4 I

1,4 I

1,3 I

2,4 I

(1)

(2)

(3)

(4)

(a) The angular velocity of link 2 is known, and we want to

find the velocity of point P on link 3. We need to pick a third link, and that link is always the ground, link 1. So,

we pick the three centers between these three links: I1,2 ,

I2,3 , and I1,3 . (We can ignore the other centers and

links at this point.) Link 2 rotates about I1,2 . The

magnitude of the velocity of I2,3 is computed as

VI2,3=ω2RI2,3I1,2

and its direction is shown. I2,3 is also a

point on link 3, and link 3 rotates about I1,3 . The

angular velocity of link 3 is computed as

ω 3 = VI2,3/ RI2,3I1,3

, CW. (We could have also used the

P

1,2 I

2,3 I

1,3 I

(1)

(2)

(3)

ω 2

ω 3

R I2,3I1,2

R I2,3I1,3

RPI1,3

VP

VI2,3

angular velocity ratio formula to find ω 3 .) Link 3 rotates about I1,3 , therefore velocity of P is

computed as VP =ω 3RP,I1,3. The direction is shown on the diagram.


P.E. Nikravesh 5-18

b) In order to move from link 2 to link 4, we only need I1,2 , I2,4 , and I1,4 . We can ignore the

other centers. Link 2 rotates about I1,2 . The magnitude of the velocity of I2,4 is

VI2,4=ω2RI2,4I1,2

and its

direction is as shown. I2,4 is

also a point on link 4, and link 4 rotates about I1,4 . The

angular velocity of link 4 is

ω 4 = VI2,4/ RI2,4I1,4

, CCW.

(We could have also used the angular velocity ratio formula.)

1,2 I

1,4 I

2,4 I

(1)

(2) (4)

ω 4

VI2,4

R I2,4 I1,2

R I2,4 I1,4

ω 2


The instant centers of this slider-crank have already been located. Assume the angular velocity of link 2 is given, CW. The objective is to find the velocity of link 4.

Since we have the angular velocity of link 2 and we are interested in the velocity of link 4, we pick the instant centers I1,2 , I2,4 , and

I1,4 .

The center I2,4 is a point on link 2. The

magnitude of its velocity is computed as

VI2,4=ω2RI2,4I1,2

, and its direction is as shown.

I2,4 is also a point on link 4. Since link 4

does not rotate, all the points on this link have

the same velocity. Therefore, VB = VI2,4.

1,3 I

2,4 I

1,2 I

2,3 I

1,4 I

(1)

(2) (3)

(4)

3,4 I

VI2,4

R I2,4 I1,2

2,4 I

1,2 I

1,4 I

(1)

(2)

(4)

ω2 B VB


For the third inversion of the slider-crank mechanism, the angular velocity of link 2 is given, CCW. We are asked to find the angular velocity of link 4.

The six instant centers are found as shown. We can determine the angular velocity of link 4 two different sets of instant centers.

1,3 I

2,4 I 1,2

I

2,3 I

3,4 I

(2)

(3)

(4)

1,4 I

(1) We use the instant centers I1,2 , I2,4 , and

I1,4 . The angular velocity formula yields

ω 4 =ω2RI2,4I1,2/ RI2,4I1,4

, CCW.

R I2,4 I1,2

2,4 I

1,2 I

(2)

(4)

1,4 I R I2,4 I1,4

ω 2


P.E. Nikravesh 5-19

(2) We use the instant centers I1,2 , I2,3 , and

I1,3 . The angular velocity formula yields

ω 3 =ω2RI2,3I1,2/ RI2,3I1,3

, CCW. Since links

3 and 4 form a sliding joint, they have the same angular velocities. Therefore,

ω 4 =ω 3 , CCW.

1,3 I

1,2 I

2,3 I

(2)(3)

R I2,3I1,2

R I2,3I1,3

Six-bar Mechanism

Assume that for this six-bar mechanism the angular velocity of link 6 is given in the CCW direction. We are asked to find (a) the angular velocity of link 3 and (b) the velocity of point A. We already know where the IC’s are from an earlier exercise.

1,2 I

2,3 I

3,4 I

1,4 I

1,3 I

2,4 I

(1)

(2)

(3)

(4) (5)

(6)

1,6 I

4,5 I

5,6 I

4,6 I

2,6 I

3,6 I

2,5 I

1,5 I

3,5 I

A

(a) The three IC’s between links 6, 3,

and 1 are: I1,6 , I3,6 , and I1,3 . The

angular velocity ratio formula yields

ω 3 =ω6RI3,6I1,6/ RI3,6I1,3

, CCW.

(b) We can determine the velocity of point A using two different ways: (1) Point A is a point on link 3

which rotates about I1,3 . The

magnitude of the velocity of A is VA =ω 3RAI1,3

and its direction is

perpendicular to RAI1,3 as shown.

(2) Since A is also a point on link 2,

we can use I1,6 , I2,6 , and I1,2 .

We use the angular velocity ratio formula to determine ω2 , then

we determine the velocity of A.

1,3 I

(3) (6)

1,6 I

3,6 I

A ω 6

R I3,6I1,6

R I3,6I1,3

RAI1,3

VA

ω 3

1,2 I

(2)

(6)

1,6 I

2,6 I

A

R I2,6I1,2

R I2,6I1,6

RAI1,2

VA ω 6

ω 2


P.E. Nikravesh 5-20

Exercises

In these exercises take direct measurements from the figures for link lengths and the magnitudes of velocity vectors. If it is stated that the angular velocity is known, assume ω = 1 rad/sec CCW, unless it is stated otherwise. P.1

VA and ω are known.

Determine VB .

A B

VA

P.2

VA and VB are known.

Determine ω .

A B

VA

BV

P.3


Determine ω . What do you observe?

A B

VABV

P.4


Determine VB , VC and VBC .

A

B

C

VA

P.5


Determine VC .

A

B

C VA

BV

P.6


Determine VB , VC and VBC .

A B C

VA

P.7


Determine VC .

A B C VA

BV

The following exercises (P.8 – P.10) are not typical velocity analysis problems for using the

instant center method. They are provided to make you think, to apply the fundamentals of the IC method back and forth, and to better understand the concept and the meaning of the instant centers. The solution to some of the following four exercises can be tricky! P.8


Determine ω and VC .

P.9

VA , ω i and ω j are known. Assume ω i = 1

rad/sec CCW and ω j = 1 rad/sec CW.

Determine VB and VC .


P.E. Nikravesh 5-21

A B C VA

BV

A B

C VA

(i)

(j)

P.10

VA and VC are known.

Determine VB .

A B

C VA

CV

(i)

(j)

The following exercises are typical problems using the instant center method. Each exercise

is a complete mechanism. Exercises P.11 – P.14 are examples of four-bar mechanism. In each problem, find the instant

centers. Assume ω2 is given, then determine ω3 , ω4 , and VP .

P.11

(2)

(3)(4)

P

P.12

(2)(3)

(4)

P

P.13

(2)

(3)

(4)P

P.14

(2)(3)

(4)

P

Exercises P.15 – P.18 are examples of slider-crank mechanism. In each problem, find the

instant centers. Assume ω2 is given, then:

For P.15 and P.16 determine ω3 , ω4 , and the velocity of the slider block;

For P.17 and P.18 determine ω3 , ω4 , and VP .

P.15

(2)(3)

(4)

P.16

(2)(3)

(4)


P.E. Nikravesh 5-22

P.17

(2)(3)

(4)

P

P.18

(2)

(3)

(4)

P

For these six-bar mechanisms ω2 is given. Find the instant centers. Determine ω5 , and the velocity of the slider block 6. In P.19, also find the velocity of P.

P.19

(6)(2)

(3)(4)

P

(5)

P

P.20

(2)(3)

(4)

(5)

(6)

5 velocity graphical

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