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8/3/2019 6 IC General

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AME 352 INSTANT CENTERS

P.E. Nikravesh 1

Instant Center of Velocities

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. In

this lesson we review the fundamentals of instant centers.

A Single Link and The Ground

Assume that the link shown is pinned to the ground at O. The link has a known angular

velocity (assume CCW). The magnitude of the velocity of any point on the link, such asA,B,

or C, can be computed as

V =R

The direction of the velocity vector is determined by rotating the position vector R 90

in the

direction of . Using the rotated vector notation, we have

V =

R (v.1)

This equation represents both the magnitude and the direction of

the velocity vector. Note that the position vector has a constant

length.

This problem can also be stated as: For a link that is pinnedto the ground at O, the velocity of a point such asA is given.

Determine the angular velocity of the link and the velocity of

pointB (or C).

The magnitude of the angular velocity is obtained by

measuring the magnitude of vector RAO

and then the above

equation is used to obtain the angular velocity. After finding ,

the velocity ofB is determined.

1, iI

(i)

A

B

C

O

RC,O

VC

B,OR

BV

A,OR

AV

The pin joint that connects linki to the ground can be viewed as two coinciding points: point

Oi

on the link and point O1

on the ground (the ground is always given index 1). Point Oihas the

same velocity as point O1

. Obviously, since the velocity ofO1

is zero, the velocity ofOi

is zero

as well. These two points that are on two different bodies but coincide have identicalvelocitiesthey form an instant centerbetween the two bodies. This instant center is denoted as

I1,i

(or Ii,1

).

In this example, the instant center between linki and the ground is an actual pin joint. As we

will see next, an instant center may be an imaginary pin joint.

A side note: In the example shown

below, the given velocity is incorrect! The

velocity ofA must be perpendicular to RAO

.

(i)

A

O

A,OR

AV

Another incorrect problem statement is

shown below. Velocity ofA suggests a CW

rotation but velocity ofB suggests the opposite.

O

(i)

A

C

RC,O

VC

A,OR

AV


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P.E. Nikravesh 2

In the example shown here, linki is not pinned to the ground.

Assume that the velocities of two or more points on the link are

given. The following observations can be made:

a) The normal axes to the velocity vectors, each passing through

its corresponding point, intersect at one point, denoted as I1,i

.

b) The distances of these points from I1,i yield

VA

RA,I1,i

=

VB

RB,I1,i

=

VC

RC,I1,i

=

c) All velocity vectors give the same sense of direction for the

angular velocity (CCW in this illustration).

The point I1,i

is the instant centerbetween this link and the

ground. This center can be viewed as an imaginary pin joint

between the link and the ground. We may imagine that the area of

the link is extended to cover the point where the instant center is,

and then place an imaginary pin joint there. Note that the velocity

of this point is zero.

1, iI

(i)

A

B

CVC

BV

AV

A, IR 1, i

B, IR1, i

RC, I1, i

To find the instant center between a link and

the ground, we may have the velocity of one

point on the link and the angular velocity of the

link (magnitude and direction). First we

compute R =V / . Then, based on the

direction of we locate the instant center.

(i)

A

AV

(i)

A

AV

A, I

R1, i

1, iI

In case we have the velocity of one point

on the link and the axis of the velocity of a

second point, the intersection of the normal to

the two velocity axes is the instant center as

shown.

(i)

A

B

BV

AV

1, iI

(i)

A

B

BV

AV

Two Moving Links

We consider two links

connected by a pin joint at P.

Assume that linki has an angular

velocity i, CCW, and linkj has

an angular velocity j , CW. The

velocities of several typical points

on the two links relative to P are

shown. Note that as a point gets

closer to P, its relative velocity

gets smaller until it becomes zero

when the point reaches the pin

joint. The center of the pin joint

(i)

(j)

A

B

C

D

E

P

APR

BPR

CPR

DPR

EPR

APV

BPV

CPV

DPV

EPV

i, jI

may be viewed as two coinciding points: Pi

on linki and Pj on linkj. These two points have

identical absolute velocities. The center of the pin joint is the instant center of velocities between


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P.E. Nikravesh 3

these two links and it is denoted as Ii, j (or Ij,i ).

If the absolute velocity of one point on each link (for exampleA andE) and the angular

velocities of the links are known, the velocity ofP can be computed in two ways:

VP=V

AV

AP

= VEV

EP

Therefore the velocity ofE, for example, can be determined as VE = VAVAP +VEP . However,

the instant center method provides a simpler process for such analyses by taking advantage of

other instant centers as we will see later in this lesson.

In this example we consider two links that are

not connected to each other by a pin joint. Assume

that the velocity ofA on linki, the velocity ofB on

linkj, and the angular velocities for both links are

given. There are two points, Pi

on linki and Pj on

linkj, that have identical absolute velocities and

they coincide at P as shown. These points may be

viewed on the imaginary extensions of these links.

Point P that is viewed as an imaginary pin joint isthe instant center between the two links.

i, jI

(i)

(j)

A

B

AV

BV

P

The three instant centers between two moving links and the ground can be used together to

perform velocity analysis as will be seen shortly.

Kennedys Rule

Kennedys rule states that there are three

instant centers among three planar links, and the

three centers lie on a straight line. This rule does

not tell us where the line is or where the centers are

on that line. But the rule can be used to find the

instant centers when we consider a complete

mechanism.

When using Kennedys rule, ground can be

considered as one of the three bodies. Inclusion of

the ground as one of the bodies yields the most

useful configuration in velocity analysis. Note that

the ground is always given the index 1.

i, jI

k, iI

j, kI

(i)

(j)

(k)

i, jI

1, iI

1, jI

(i)

(j)

Using Three Centers

This exercise illustrates how the three instant

centers between two moving bodies and the ground

can be used to determine unknown velocities. The

problem statement in this exercise is very typicalother slight variations of this problem statement

will be encountered in the velocity analysis of

mechanisms.

Assume that the location of the three instant

centers between links i,j, and the ground (link1)

are given. The velocity of pointA on linki is

known. What is the velocity of pointB on linkj?

i, jI

1, iI

1, jI

(i) (j)

A BAV


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P.E. Nikravesh 4

We note that the three centers are on a straight

line. To determine the velocity ofB, we consider

the following four steps:

1. We start with the link that has a known

velocity. Linki 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 linki as i=V

A/R

A,I1,i. The direction

of this angular velocity is CW.

2. The instant center Ii, j is an imaginary point

on linki, and therefore we can determine its

velocity. We measure the length of vector RIi , j ,I1,i ,

then we compute VIi , j =iRIi , j ,I1,i . The direction of

VIi,

j

is established based on the direction ofi.

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

point on linkj, and we already know VIi , j . Linkj

rotates about the imaginary pin joint to the ground

at I1, j . We measure the length of vector RIi , j ,I1, j ,

then j =VIi , j /RIi , j ,I1, j is computed. The direction

ofj is established to be CCW.

4. Point B is attached to linkj that rotatesabout an imaginary pin joint at I

1, j . We construct

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

compute VB =j RB,I1,j . The direction ofVB is

established based on the direction of the angular

velocity of linkj.

1, iI

(i)

A AV

A, IR

1, i

i

ii, j

I

1, iI

(i)

IV

i, j

I , IR 1, ii, j

i, jI

1, jI

(j)

IV i, j

I , IR 1, ji, j

j

1, jI

(j)

B

B, IR 1, j

BV

j

Number of Instant Centers

In a mechanism with n links (count the ground as one of the links), there are the followingnumber of instant centers:

C=n(n 1)

2

As an example, in a four-bar mechanism there are 6 ICs ( n = 4 ). The same number of

centers exists in every slider-crank mechanism. For any six-bar mechanism C=15 .


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P.E. Nikravesh 5

Finding Four-bar Instant Centers

A four-bar mechanism has six instant centers

regardless of the dimensions or orientation of the

links. For bookkeeping purposes in locating the

ICs, 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, butbecomes very useful when mechanisms with

greater number of links are considered.

(1)

(2)

(3)

(4)

1

2

3

4

Pin joints are instant centers. For a four-bar

with four pin joints, four ICs are immediately

identified. Each found IC is marked on the circle

as a line drawn between two indices. These four

ICs are actual (not imaginary) pin joints. In order

to find the other ICs, we must apply Kennedys

rule over and over. (1)

(2)

(3)

(4)

1,2I

2,3I

3,4I

1,4I

1

2

3

4

The ICs 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 ICs 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 tell us what center to find next. The red line between links 2

and 4 indicates that this is the center we want to find. This line is shared between two triangles

with known ICs. 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 .

1,2I

2,3I

3,4I

1,4I

2,4I

(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

.

1,2I

2,3I

3,4I

1,4I

1,3I

2,4I

(1)

(2)

(3)

(4)

1

2

3

4


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P.E. Nikravesh 6

Sliding Joint

Between two bodies

that form a sliding joint

there is an instant center

located in infinity along any

axis perpendicular to thesliding joint axis.

(i)

(j)

i, jI

When one of the links of

a sliding joint is fixed to the

ground, neither link can

rotate. In this case, all the

points on the moving linkhave identical velocities. In

the illustration shown,

VA= V

B.

1, iI

(i)

BV

AVA

B

The three ICs between links

i, j, and the ground lie on a

straight line. Note that Ii, j is

located in infinity on the linegoing through I1,i and I1, j .

(i)

(j)

1, jI

1, iI

i, jI

Using Three Centers

Here is an example of using three centers to find unknown

velocities. Assume the velocity ofA on linki is given and we

are asked to find the velocity ofB on linkj. The centers

between the two links and the ground are also given.

We follow similar steps as we saw with the pin joints.

However, a slider makes some of the steps simpler. Linki is

pinned (imaginary) to the ground at I1,i

. The angular velocity

of linki is computed as i=V

A/R

A,I1,i, CCW. Since the two

links are connected by a sliding joint, j =i . Linkj rotates

about I1, j . Velocity ofB is computed as VB =jRB,I1,j . The

direction is established based on the direction of the angular

velocity.

(i)

(j)

1, jI

1, iI

i, jI

B

A

AV

BV

A, IR 1, i

B, IR1, j

Another example is provided here where one of the links, linki, is connected to the ground

by a sliding joint. The velocity ofA on linki is given and we are asked to find the velocity ofB

on linkj. The centers between the two links and the ground are also given.

Since linki slides relative to the ground, all

points on linki have identical velocities. The

center Ii, j is a point on linki (as well as a point on

linkj), therefore VIi , j = VA . Linkj rotates with

respect to the ground about I1, j . The angular

velocity of linkj is computed as j =VIi , j /RIi , j ,I1, j ,

CCW. The velocity ofB is computed as

VB =

jR

B,I1,jin the direction shown.

1, iI

(i)

AV A

B (j)

1, jI

i, jI

BV

B

B, IR1, j

IV i, j

I , IR

1, ji, j


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P.E. Nikravesh 7

Finding Slider-Crank Instant Centers

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. The center between the

slider block and the ground is in

infinity on an axis perpendicular to the

sliding axis.1,2

I

2,3I

1,4I

(1)

(2) (3)

(4)

3,4I 1

2

3

4 I

2,4must lie on the axis of I

2,3and

I3,4

, and on the axis of I1,2

and I1,4

.

The intersection of these two axes is

I2,4 .

2,4I

1,2I

2,3I

1,4I

(1)

(2) (3)

(4)

3,4I 1

2

3

4 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

.

1,3I

2,4I

1,2I

2,3I

1,4I

(1)

(2) (3)

(4)

3,4I 1

2

3

4

Six-bar MechanismIn 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 ICs that belong to the four-bar. Next we find the ICs for the slider-

crank. Note that I1,4

is shared between the two sub-mechanisms.

1,2I

2,3I

3,4I

1,4I

1,3I

2,4I

(1)

(2)

(3)

(4)

(5)

(6)

1

2

3

4

5

6


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P.E. Nikravesh 8

1,2I

2,3I

3,4I

1,4I

1,3I

2,4I

(1)

(2)

(3)

(4)

(5)

(6)

1,6I

1,5I

4,5I

5,6I

4,6I

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

the line I2,4

- I4,6

and the line 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).

1,2I

2,3I

3,4I

1,4I

1,3I

2,4I

(1)

(2)

(3)

(4)

(5)

(6)

1,6I

1,5I

4,5I

5,6I

4,6I

2,6I

3,5I

1

2

3

4

5

6

The next IC to find is I3,6 . This center is at the intersection of the 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).

1,2I

2,3I

3,4I

1,4I

1,3I

2,4I

(1)

(2)

(3)

(4)

(5)

(6)

1,6I

4,5I

5,6I

4,6I

2,6I

1

2

3

4

5

6

3,6I

2,5I

1,5I

3,5I


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P.E. Nikravesh 9

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 only four links are involved, there are only six centers

to locate. For mechanisms with more links that four, for example a six-bar, there are many more

centers to find. Locating some of the centers requires using some other centers that have already

been found. The following strategy can reduce the graphical error in locating some of the centers.

Lets use the above six-bar mechanism as

an example. The first eleven centers that we

found belonged to the four-bar and the inverted

slider-crank sub-mechanisms. The accuracy of

the positions of these eleven centers depends

on how accurately we marked the centers of the

pin joints and the intersections of the lines.

Next we located I2,6 and I3,5 using some

of the first eleven centers. Therefore the

positions of these two new centers contain their

own graphical error on top of the errors from

the original centers.

We located I3,6 using the centers I3,4 - I4,6

and I1,3

- I3,6

. We could have used one of these

lines and the line going through I2,3 - I2,6 (or

I3,5

- I5,6

), but we did not! Why? The reason is

that I2,6 (or I3,5 ) has twice the error than the

original eleven centers, therefore we try not touse it if possible.

We followed the same strategy in finding

I2,5

. We used the centers I2,4

- I4,5

and I1,2

-

I1,5

we did not use I2,6 or I3,5 .

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