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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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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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AME 352 INSTANT CENTERS
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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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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AME 352 INSTANT CENTERS
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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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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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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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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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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 .
1
2
3
4
5
6
1
2
3
4
5
6
1
2
3
4
5
6