kom

1. P. Nikravesh, AME, U of A
Instant Centers of Velocities
Introduction
Instant Centers of Velocities
A simple method to perform velocity analysis in planar systems is the instant
center (IC) method. This is a graphical method that requires drawing lines,
finding intersections, finding ratios, etc.
In this presentation we will review the followings:
1. Definition of instant centers
2. Number of IC’s for a given mechanism
3. Kennedy’s rule
4. Finding IC’s
5. Finding velocities

2. P. Nikravesh, AME, U of A
Instant Centers of Velocities
Definition of an instant center
Between every two bodies in plane
motion there is a common point that has
the same instantaneous velocity in each
body. This common point is called the
instant center (IC) of velocity.
An instant center may be an actual
common point between two bodies,
such as the pin joint shown in (a), or it
may be a virtual common point as
shown in (b).
An instant center between links (bodies)
i and j is labeled as Ii,j or Ij,i .
Note: For convenience we agree to
reserve “1” as a label for the non-
moving body; i.e., the ground.
Definition
I i,j
(a)
(i)
(j)
I i,j
(b)
(j)
(i)

3. P. Nikravesh, AME, U of A
Instant Centers of Velocities
Obvious instant centers
Obvious instant centers
Some instant centers are very easy to
identify:
1. Two bodies connected by a pin
joint: The center of the pin joint is
the IC between the two bodies as
shown in (a). Since the pin joint
remains between the bodies as
they move, this IC is a permanent
center as well.
2. Two bodies connected by a sliding
joint: Since the two bodies rotate
together; i.e., there is no relative
rotation between them, the instant
center is in infinity on an axis
perpendicular to the axis of sliding
as shown in (b). Note: ωi = ωj
I i,j
(a)
(i)
(j)
(i)
(j)
I i,j
(b)

4. P. Nikravesh, AME, U of A
Instant Centers of Velocities
Number of instant centers
For a mechanism containing n links, there are
n (n − 1) ∕ 2
instant centers. For example for a fourbar mechanism,
there are
4 (4 − 1) ∕ 2 = 6
instant centers. The same is true for or a slider-crank
mechanism.
Book keeping: A small circle will help us keep track of
locating the instant centers. On the circumference of the
circle we put as many marks as the number of links. For
example if a system contains 4 links and they are
numbered 1, 2, 3 and 4, on the circumference we put four
marks. Each time we find a center between two
links, we draw a line between the corresponding marks on
the circle.
Number of instant centers
1 2
4 3
►

5. P. Nikravesh, AME, U of A
Instant Centers of Velocities
Kennedy’s rule
Kennedy’s rule
Between every three links (bodies), there are 3 (3 − 1) ∕ 2 = 3 instant centers.
These three IC’s lie on the same straight line.
For the three links that are labeled in the figure as i, j and k, the three centers
are shown schematically to be on a straight line. The three centers may
appear in different orders when the orientation of mechanism changes.
We will use this rule extensively to find non-obvious IC’s in mechanisms as
will be seen in other presentations.
I i,j
(i)
(j)
I k,i
(k)
I j,k

6. P. Nikravesh, AME, U of A
Instant Centers of Velocities
O2 = I1,2
= I4,1
A = I2,3
B = I3,4
Finding instant centers
In other presentations on instant
centers, we will see how to find all the
IC’s for a given mechanism. We first
locate all the obvious IC’s. We then
apply Kennedy’s rule to find the non-
obvious IC’s.
As an example, for the slider-crank
mechanism shown, the obvious IC’s
are the three pin joints and the sliding
joint.
The non-obvious IC’s are found by
applying Kennedy’s rule.
Finding instant centers
1 2
4 3
I1,3
I2,4
►
►

7. P. Nikravesh, AME, U of A
Instant Centers of Velocities
Finding velocities
Finding velocities
In the following slides we review several cases for determining velocities.
These cases are presented in a general form. In all these cases always two
moving bodies are involved.
Known: It is assumed that for one body (body i) we know either its angular
velocity or the velocity (absolute) of a point on that body.
Unknown: We want to find either the angular velocity of the second body
(body j) or the velocity (absolute) of a point on that body.
Third body: In any analysis we need to get a third body involved. We use
the ground (body 1) as the third body.
Instant centers: Between bodies, 1, i and j, we have three IC’s: I1,i, I1,j and
Ii,j. These are the three centers to use!
I i,j
(i)
(j)
I 1,i
I 1,j

8. P. Nikravesh, AME, U of A
Instant Centers of Velocities
Problem 1
Problem 1
Assume ωi is known and we need to find ωj.
Solution: We solve this problem in two steps.
(i)
(j)
ωi
Ii,j I1,i
I1,j

9. P. Nikravesh, AME, U of A
Instant Centers of Velocities
Step 1: The instant center Ii,j is a point on link i. Link i is pinned to the
ground at I1,i (real or imaginary). Since we know ωi we can determine VIi,j
(magnitude and direction):
VIi,j = ωi ∙ RIi,jI1,i
Its direction is obtained by rotating R Ii,jI1,i 90° in the direction of ωi.
For the next step we no longer need vector R Ii,jI1,i and the IC I1,i .
Problem 1; step 1
►
(i)
(j)
ωi
VIi,j
R Ii,jI1,i
Ii,j I1,i

10. P. Nikravesh, AME, U of A
Instant Centers of Velocities
Step 2: The instant center Ii,j is also a point on link j. Link j is pinned to the
ground at I1,j (real or imaginary). Since we know the velocity of Ii,j, we can
determine ωj:
ωj = VIi,j ∕ RIi,jI1,j
The direction of the angular velocity is found by rotating R Ii,jI1,j 90° to match
the direction of VIi,j .
Problem 1; step 2
►
(i)
(j)
R Ii,jI1,j
ωj
VIi,j
Ii,j
I1,j

11. P. Nikravesh, AME, U of A
Instant Centers of Velocities
Problem 2
Problem 2
Assume that instead of ωi, we have been given the velocity of point P on link i
and we need to find ωj.
(i)
(j)
Ii,j I1,i
I1,j
P
VP

12. P. Nikravesh, AME, U of A
Instant Centers of Velocities
Problem 2; solution
Solution: We first note that the axis of VP is not arbitrary. This axis must be
perpendicular to RPI1,i .
We measure the length RPI1,i then we compute ωi as:
ωi = VP ∕ RPI1,i
Now that we have determined ωi, we can continue as in Problem 1.
(i)
(j)
ωi
Ii,j I1,i
I1,j
P
VP
RPI1,i
►
►

13. P. Nikravesh, AME, U of A
Instant Centers of Velocities
Problem 3
Problem 3
In this problem we add one more unknown to Problem 1 (or Problem 2).
Determine the velocity of point B on link j.
(i)
(j)
ωi
Ii,j I1,i
I1,j
B

14. P. Nikravesh, AME, U of A
Instant Centers of Velocities
Problem 3; solution
Solution: We start solving this problem as in Problem 1 (or Problem 2) until
we have ωj.
After that we can easily find the velocity of point B:
VB = ωj ∙ RBI1,j
(i)
(j)
ωi
Ii,j I1,i
I1,j
B
►
ωj
►
RBI1,j
VB

15. P. Nikravesh, AME, U of A
Instant Centers of Velocities
General comments
The procedures we have seen
in this presentation for finding
instant centers and then using
the centers to find velocities
can be applied to any planar
mechanism.
We will see the process for
four-bar and slider-crank
mechanisms in detail in the
coming presentations.
General comments