Analytical Velocity Solutions

• With some intuitive feel for velocity

analysis, now turn to analytical methods:
– Based on relative velocity & closure eq.s
– Can find the unknown velocities in a
problem using:
• Time derivatives of polar-complex displacement
vectors
• Vector cross-products
• Time derivatives of displacement closure

MECH 335 Lecture Notes

© R.Podhorodeski, 2009
 Analytical Velocity Solutions
• Return to a familiar example
A

R2
ω2 R3

ω3 = ?
θ2

θ3
vB = ?
R0 R1
B

MECH 335 Lecture Notes

© R.Podhorodeski, 2009
 Analytical Velocity Solutions
• Solution 1: Use relative velocity eqns
A

R2
ω2 R3

ω3 = ?
θ2

θ3
vB = ?
R0 R1
B

MECH 335 Lecture Notes

© R.Podhorodeski, 2009
 Analytical Velocity Solutions
• Form the familiar relative velocity eqn:

– Now, express each term analytically:

Slide constrained to
horizontal motion

R2
ω2 R3

ω3 = ?
θ2

θ3
vB = ? =0
R0 R1
B MECH 335 Lecture Notes (Rigid link)
© R.Podhorodeski, 2009
 Analytical Velocity Solutions
• Form the familiar relative velocity eqn:

– Now, express each term analytically:

Since RB|A is defined from A to B, but
R3 is drawn from B to A

R2
ω2 R3

ω3 = ?
θ2

θ3
vB = ? =0
R0 R1
B MECH 335 Lecture Notes (Rigid link)
© R.Podhorodeski, 2009
 Analytical Velocity Solutions
• Form the familiar relative velocity eqn:

– Putting it all together:

A

R2
ω2 R3

ω3 = ?
θ2

θ3
vB = ?
R0 R1
B MECH 335 Lecture Notes
© R.Podhorodeski, 2009
 Analytical Velocity Solutions
• Substitute in for the polar-complex terms

A
»Simplify by noting:
R2
ω2 R3

ω3 = ?
θ2

θ3
vB = ?
R0 R1
B MECH 335 Lecture Notes
© R.Podhorodeski, 2009
 Analytical Velocity Solutions

• Split into real and imaginary equations:

– Re:
– Im:
• Now solve the Imag. equation for ω3:
A

R2
ω2
• Substitute into the real eq.
R3

ω3 = ?
θ2

θ3
vB = ?
R0 R1
B MECH 335 Lecture Notes
© R.Podhorodeski, 2009
 Analytical Velocity Solutions
• Solution 2: Use vector cross-products
A

R2
ω2 R3

ω3 = ?
θ2

θ3
vB = ?
R0 R1
B

MECH 335 Lecture Notes

© R.Podhorodeski, 2009
 Analytical Velocity Solutions
• Solution 2: Use vector cross-products:
– Quick review:
• Unit vectors directed along x y z
• Rotation in the plane of the mechanism
has only a component, with +
representing CCW rotation by the right
hand rule
A
•
R2 •
ω2 R3

ω3 = ?
θ2

θ3
vB = ?
R0 R1
B MECH 335 Lecture Notes
© R.Podhorodeski, 2009
 Analytical Velocity Solutions
• Solution 2: Use vector cross-products:
– Begin as before, with:
– By the slide constraint, we know:
– Compute VA using cross-products:

R2
ω2 R3

ω3 = ?
θ2

θ3
vB = ?
R0 R1
B MECH 335 Lecture Notes
© R.Podhorodeski, 2009
 Analytical Velocity Solutions
• Solution 2: Use vector cross-products:
– Similarly, compute VB|A:

R2
ω2 R3

ω3 = ?
θ2

θ3
vB = ?
R0 R1
B MECH 335 Lecture Notes
© R.Podhorodeski, 2009
 Analytical Velocity Solutions
• Solution 2: Use vector cross-products:
– Putting it all together:

– And splitting into & components:

R2 • Same equations as Solution 1, solved as:

ω2 R3

ω3 = ?
θ2

θ3
vB = ?
R0 R1
B MECH 335 Lecture Notes
© R.Podhorodeski, 2009
 Analytical Velocity Solutions
• Solution 3: Use d/dt of displ. closure
A

R2
ω2 R3

ω3 = ?
θ2

θ3
vB = ?
R0 R1
B

MECH 335 Lecture Notes

© R.Podhorodeski, 2009
 Analytical Velocity Solutions
• Solution 3: Use d/dt of displ. Closure
– Recall the displacement closure eqn:

– Substituting in the polar-complex forms:

A
»Now, differentiating w.r.t time:
R2
ω2 R3

ω3 = ?
θ2

θ3 =0
vB = ?
R0 R1
B MECH 335 Lecture Notes
© R.Podhorodeski, 2009
 Analytical Velocity Solutions
• Solution 3: Use d/dt of displ. Closure
– Removing zero terms:

– And substituting for polar-complex terms:

R2
ω2 R3

ω3 = ?
θ2

θ3
vB = ?
R0 R1
B MECH 335 Lecture Notes
© R.Podhorodeski, 2009
 Analytical Velocity Solutions
• Solution 3: Use d/dt of displ. Closure
– Split into real and imaginary components:

R2
ω2 R3

ω3 = ?
θ2

θ3
vB = ?
R0 R1
B MECH 335 Lecture Notes
© R.Podhorodeski, 2009
 Analytical Velocity Solutions
• Solution 3: Use d/dt of displ. Closure
– By now, we should recognize these two
equations! The solution is again found:

R2
ω2 R3

ω3 = ?
θ2

θ3
vB = ?
R0 R1
B MECH 335 Lecture Notes
© R.Podhorodeski, 2009
 Instantaneous Center Analysis
• Turn now to another tool for velocity
analysis: the Instantaneous Center (IC)
– The IC of two links is a point which has the
same velocity when considered part of the
first link as when it is considered part of the
second link
– The IC is thus a point at which the relative
motion of the two links is purely rotational
(i.e. there is no relative translation at the IC)

MECH 335 Lecture Notes

© R.Podhorodeski, 2009
 Instantaneous Center Analysis
• Consider a simple example:
– Where is the IC between links 1 and 2?
– In general, the IC of two links pinned
together will be at the pin (Why?)

(1,2)

MECH 335 Lecture Notes

© R.Podhorodeski, 2009
 Instantaneous Center Analysis
• Since ICs only exist between pairs of
links:
# of ICs in an # of unique pairs
=
n-link mechanism among the n links

• From statistics, the number, p, of unique

pairs from a population of size n (not
allowing self-pairing) is:

MECH 335 Lecture Notes

© R.Podhorodeski, 2009
 Instantaneous Center Analysis
• A handy bookkeeping method for
counting off ICs (6-bar example):
– For a 6-bar, we expect ICs
– Count them off graphically:
1 2
(1,2) (1,3) (1,4) (1,5) (1,6)
(2,3) (2,4) (2,5) (2,6)
6 3
(3,4) (3,5) (3,6)
(4,5) (4,6)
5 4 (5,6)

MECH 335 Lecture Notes

© R.Podhorodeski, 2009
 Instantaneous Center Analysis
• Now, let’s try to locate the ICs of a 4-bar
3 (3,4)

(2,3)
• We already know
that all the pins
2 4 are ICs
(1,2)
1
• Do the graphical
bookkeeping
1
1
(1,4)
• How can we find
(1,3) and (2,4)?
4 2

MECH 335 Lecture Notes

3 © R.Podhorodeski, 2009
 Instantaneous Center Analysis
• Let’s look at (1,3) first:
• Imagine standing at the
3
IC (1,3) and looking at 3
• Since we’re at the IC, 3
4
appears to rotate around
2
us at (1,3)
1
• So, we’d expect the
1
velocity of any point P on
3 to be ┴ to a vector
from (1,3) to P (Why?)
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
 Instantaneous Center Analysis
• Let’s look at (1,3) first:
• Of course, we can’t draw
3
this vector (1,3) -> P
since we don’t know
4
where (1,3) is yet
2
• But, if we know the
1
velocity of any point P on
1
3, we know the vector
from the IC through P is
perp. to this velocity
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
 Instantaneous Center Analysis
• Let’s look at (1,3) first:
• Fortunately, we DO know
3
the direction of the
absolute velocity for at
4
least 2 points on 3
2
• But, these velocities are
1
necessarily perp. to link
1
2 and link 4!
• The IC (1,3) will thus lie
on the projected
intersection of 2 & 4
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
 Instantaneous Center Analysis
• So, all that remains is to locate (2,4)
3 (3,4)

(2,3)
• Need to step back
to a simple case to
2 4 figure this out
(1,2)
1

1
(1,4)
1

4 2
(1,3)

MECH 335 Lecture Notes

3 © R.Podhorodeski, 2009
 Instantaneous Center Analysis
• Consider 2 links as shown below

• To find (2,3), we
P
3 need a point that
would have the
same velocity
2
regardless of which
link we attach it to.
1
• Take the point P
• Is P the IC (2,3)?
1
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
 Instantaneous Center Analysis
• Consider 2 links as shown below

• To test if P is the IC,

P VP,2
VP,3
3 determine VP ,
considering P first
as a member of 2,
2
then as a member
of 3
1
• Clearly, P is not
the IC, since
1
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
 Instantaneous Center Analysis
• Consider 2 links as shown below

• We know that VP is
P VP,2
VP,3
3 always perp. to the
vector from P to
the link’s center of
2
rotation
• So, for P to be an
1
IC, the vectors to P
from each link’s
1
MECH 335 Lecture Notes
base joint must be
© R.Podhorodeski, 2009 parallel
 Instantaneous Center Analysis
• Consider 2 links as shown below

• But, this condition

3 can only be met on
the line between
the ICs (1,2) & (1,3)
2
• The location of P
(1,3)
must be such that
P = (2,3)
(1,2) , where:
1
VP,2
1 VP,3
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
 Instantaneous Center Analysis
• We still don’t seem much closer to
finding (2,4) from the 4-bar…
• Fortunately, the same argument just
made about the location of P can be
proven generally, giving Kennedy’s
Theorem:

“For any set of 3 links in relative motion, all three

instantaneous centers lie on a straight line”
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
 Instantaneous Center Analysis
• Now, back to the 4-bar
3 (3,4)

(2,3)
• Can see that K.T.
holds for the ICs
2 4 found so far
(1,2)
1
• To find (2,4), we
look first at the ICs
1
1
(1,4)
of links 2,3,4
• Have (2,3), (3,4)
2
4
(1,3)
• So (2,4) is on the
MECH 335 Lecture Notes
line passing
3 © R.Podhorodeski, 2009 through (2,3)&(3,4)
 Instantaneous Center Analysis
• Now, back to the 4-bar
3 (3,4)

(2,3)
• Now, by looking at
(2,4)
links 1,2,4, we can
2 4 draw another line
(1,2)
1
to locate (2,4)
• (2,4) must lie on
1
1
(1,4)
the same line as
(1,2) & (1,4)
4 2
(1,3)

MECH 335 Lecture Notes

3 © R.Podhorodeski, 2009
 Instantaneous Center Analysis
• Note: if two lines that should intersect at
an IC end up being parallel (happens for
slides), the intersection is said to occur
at infinity, in the direction of the parallel
lines

MECH 335 Lecture Notes

© R.Podhorodeski, 2009
 Instantaneous Center Analysis
• So, how can ICs help our analyses?
• Recall the earlier example:
• Here, (2,3) lies between (1,2)
and (1,3)
3 • This alone would tell us that 2
and 3 are rotating in opposite
2

(1,3)
directions!
(1,2)
P = (2,3) • Similarly, if (2,3) was NOT
1

1
VP,2
VP,3
between (1,2)&(1,3), it would
indicate that 2 and 3 are
rotating in the SAME direction
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
 Instantaneous Center Analysis
• So, how can ICs help our analyses?
• Recall the earlier example:
• So, we get some qualitative
information just from the
3 relative location of ICs
• But also get quantitative info,
2

(1,3)
since:
P = (2,3)
(1,2) 1
VP,2
1 VP,3

MECH 335 Lecture Notes

© R.Podhorodeski, 2009
 Instantaneous Center Analysis
• So, how can ICs help our analyses?
• Recall the earlier example:
• This is a powerful tool, since
there could be any number of
3 links in the mechanism chain
between 2 and 3, and we could
2 still compute ω3 from ω2 and
(1,3)
P = (2,3)
the locations of the 3 ICs
• Next example shows the use
(1,2) 1
VP,2
1 VP,3
of ICs in graphical velocity
analysis
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
 Instantaneous Center Analysis
• 4-Bar from Ex. V1
B • Let’s find VB using
(3,4)
ICs
• First, locate all the
1 C ICs
4 2
3 4
• Joints first
(1,3)
• Apply K.T. with
3

links 1, 4 & 3
O2 (1,2) 2
(2,3)
O4
• Apply K.T. with
A (1,4)

ω2
links 1, 2 & 3
• Gives (1,3)
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
 Instantaneous Center Analysis
• 4-Bar from Ex. V1
B • Apply K.T. to links
(3,4)
1, 2 & 4
• Apply K.T. to links
1 C 2, 3 & 4
4 2
3 4
• Gives (2,4)
(1,3)
• All ICs have been
3

found
(2,3)
O2 (1,2) 2 (1,4) O4
A
(2,4)

MECH 335 Lecture Notes

© R.Podhorodeski, 2009
 Instantaneous Center Analysis
• 4-Bar from Ex. V1
B • To find VB , first
(3,4)
draw VA (to any
scale)
1 C • Now rotate VA
4 2
3 4
about (1,2), down
(1,3)
to the 1,2,4 IC line
3

• Find the equivalent

O2 (1,2) 2
(2,3)
O4
velocity at (2,4),
A (1,4)

(2,4)
with respect to
(1,2) by scaling
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
 Instantaneous Center Analysis
• 4-Bar from Ex. V1
B • Now, rotate the
(3,4)
scaled velocity
about (1,4) up to
1 C the 1,3,4 IC line
4 2
3 4
• Now, draw VB by
(1,3)
scaling from (1,4)
3

• Compute the ratio

O2 (1,2) 2
(2,3)
O4
of this vector’s
A (1,4)

(2,4)
length to the
original vector VA :
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
 Instantaneous Center Analysis
• 4-Bar from Ex. V1
B •
(3,4)

• Computing the
1 C same ratio from Ex.
4 2
3 4
V1, we get 0.61!
(1,3)
• ~ 30% relative
3

error
O2 (1,2) 2
(2,3)
O4
• A reminder of the
A (1,4)

(2,4)
precision benefits
of analytical
MECH 335 Lecture Notes
© R.Podhorodeski, 2009 methods…
 Instantaneous Center Analysis
• This method may seem complicated
compared to the earlier graphical method
• That’s because it is!
• But, this method pays off when the
mechanism is more complex, since you
don’t have to compute a large number of
relative velocities along the way
• Also, we will see this material again when
we talk about mechanical advantage…
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
END OF LECTURE PACK 4