class 4 velocity analysis1 (1)
TRANSCRIPT
7/23/2019 Class 4 Velocity Analysis1 (1)
http://slidepdf.com/reader/full/class-4-velocity-analysis1-1 1/43
Mechanics ofMachines
Dr. Mohammad Kilani
Class 4
Velocity Analysis
7/23/2019 Class 4 Velocity Analysis1 (1)
http://slidepdf.com/reader/full/class-4-velocity-analysis1-1 2/43
DERIVATIVE OF A ROTATINVECTOR
7/23/2019 Class 4 Velocity Analysis1 (1)
http://slidepdf.com/reader/full/class-4-velocity-analysis1-1 3/43
Deri!ati!e of a Rotatin" #nitVector
A unit vector in the θ direction, uθ , is a vector of unity
magnitude and an angle θ with the x- axis. It is
written as:
If uθ
rotates, it angle θ changes with time, the time
derivative of uθ is found by applying the standard
dierentiation rules on the expression of uθ
above
j i uθ
θ θ sincos +=
( ) j i u
j i u
θ
θ
θ θ θ
θ θ
θ θ
cossin
cossin
+−=
+−=
dt
d
dt
d
dt
d
dt
d
dt
d
θ
cosθ i
sinθ j
u θ
7/23/2019 Class 4 Velocity Analysis1 (1)
http://slidepdf.com/reader/full/class-4-velocity-analysis1-1 4/43
Deri!ati!e of a Rotatin" #nitVector
Noting that
the time derivative of the vector uθ is
( )
( ) θ π θ
θ π θ
sin2cos
cos2sin
−=+
=+
( ) j i u
j i u
θ
θ
θ θ θ
θ θ
cossin
sincos
+−=
+=
dt
d
dt
d
( )
( ) ( )[ ]
( )2
2sin2cos
cossin
π θ
θ
π θ π θ θ
θ θ θ
+=
+++=
+−=
uu
j i
u
j i u
θ
θ
θ
dt
d
dt
d
dt
d
dt
d
dt
d
dt
d θ
cosθ i
sinθ j
u θ
7/23/2019 Class 4 Velocity Analysis1 (1)
http://slidepdf.com/reader/full/class-4-velocity-analysis1-1 5/43
Deri!ati!e of a Rotatin" #nitVector
The derivative of a unit vector whose angle
with the x-axis is θ , θ changes in time!, is a
vector whose angle with the x-axis is θ+π/2
and whose magnitude is d θ/dt.
If we de"ne a vector $ as a vector in the %
direction of magnitude # $ d θ/dt , then
( )2
sincos
π θ
θ
θ θ
+=
+=
uu
j i u
θ
θ
dt
d
dt
d
θ
u θ
d u θ
/dt
θ + π/2
( ) θ θ uωu
u
k ω
×==
=
+ 2π θ
θ
θ
dt
d
dt
d
dt d
7/23/2019 Class 4 Velocity Analysis1 (1)
http://slidepdf.com/reader/full/class-4-velocity-analysis1-1 6/43
Deri!ati!e of a Rotatin" Vector
( )
( )
( )θ θ
θ
θ θ θ
θ θ
θ θ
θ
θ θ
r ωur
uωur
uur
uu
r ur
×+=
×+=
+=
+==
+
dt
dr
dt
d
r dt
dr
dt
d dt
d r
dt
dr
dt
d
dt
d r
dt
dr
dt
d
r
2π θ
θ θ
u θ
d u θ
/dt
θ + π/2
A vector r θ = r u
θ , is a general vector of magnitude r pointing in the θ
direction. The time derivative of r θ is found by applying the normal
dierentiation rules on the expression for r θ
7/23/2019 Class 4 Velocity Analysis1 (1)
http://slidepdf.com/reader/full/class-4-velocity-analysis1-1 7/43
Deri!ati!e of a Rotatin" Vector
θ
r θ
ω x r θ
θ + π/2
(dr/dt ) u θ
%iven a vector r θ = r u
θ which rotates relative to
the reference coordinates, the time derivative of
r θ
has two components& a component in the
direction of uθ and a component normal to u
θ in
the direction of u
θ+π/2
The magnitude of the component of d r θ /dt in the
direction of uθ is e'ual to dr/dt; that is the time
derivative of the length of r θ .
The magnitude of the component of d r θ /dt in the
direction of direction of uθ+π/2
is e'ual to ωr
7/23/2019 Class 4 Velocity Analysis1 (1)
http://slidepdf.com/reader/full/class-4-velocity-analysis1-1 8/43
VE&OCIT' ANA&'(I(
OF FO#R )ARMEC*ANI(M(
7/23/2019 Class 4 Velocity Analysis1 (1)
http://slidepdf.com/reader/full/class-4-velocity-analysis1-1 9/43
Deri!ati!e of the &oo+ Clos,reE-,ation for a Fo,r )ar KinematicChain The loop closure e'uation of a ()bar *inematic chain is written
as
+hen all the lin*s in the chain are of constant lengths, the
e'uation above reduces to
( )
( )
( ) ( ) ( )( ) ( )( ) 0
0
0
0
44113322
4132
4244121132332222
4132
4132
4132
4132
=+−+−+++
=−−+
=−−+
=−−+
+=+
++++ θ π θ θ π θ θ π θ θ π θ
θ θ θ θ
θ θ θ θ ur ur ur ur ur ur ur ur
ur ur ur ur dt
d
r r r r dt
d
r r r r
r r r r
( ) ( ) ( ) ( )
0
0
44113322
244211233222 4132
=×−×−×+×
=−−+++++
r ωr ωr ωr ω
π θ π θ π θ π θ θ θ θ θ ur ur ur ur
7/23/2019 Class 4 Velocity Analysis1 (1)
http://slidepdf.com/reader/full/class-4-velocity-analysis1-1 10/43
Deri!ati!e of the &oo+ Clos,reE-,ation for a Fo,r )arMechanism or a four bar mechanism with lin* - "xed we have
d θ 1 / dt = ω
1 = .
The vector e'uation above contains two scalar
e'uations and can be solved for two un*nowns.
/nowledge of d θ 2 / dt allows the calculation of d θ
3 / dt
and d θ 4 / dt .
( ) ( ) ( )244233222432
π θ π θ π θ θ θ θ +++
=+ ur ur ur
444333222
444333222
coscoscos
sinsinsin
θ θ θ θ θ θ
θ θ θ θ θ θ
r r r
r r r
=+
=+
7/23/2019 Class 4 Velocity Analysis1 (1)
http://slidepdf.com/reader/full/class-4-velocity-analysis1-1 11/43
Deri!ati!e of the &oo+ Clos,reE-,ation for a Fo,r )arMechanism
0liminate d θ 3 / dt by carrying out a dot product with u
θ3 on both sides
of the e'uation
Alternatively, eliminate d θ 4 / dt by carrying out a dot product with u
θ4
on both sides of the e'uation
( ) ( ) ( )244233222 432 π θ π θ π θ θ θ θ +++
=+ ur ur ur
( ) ( )
( ) ( )
( )
( )
( )
( ) 2
344
322
2
344
322
44
34443222
34443222
sin
sin
sin
sin
sinsin
2cos2cos
ω θ θ
θ θ θ
θ θ
θ θ θ ω
θ θ θ θ θ θ
θ π θ θ θ π θ θ
−
−=
−
−==
−=−
−+=−+
r
r
r
r
r r
r r
( ) ( )
( ) ( )
( )
( )
( )
( )
( )
( ) 2
343
422
2
433
422
2
433
422
33
43334222
43334222
sin
sin
sin
sin
sin
sin
sinsin
02cos2cos
ω θ θ
θ θ ω
θ θ
θ θ θ
θ θ
θ θ θ ω
θ θ θ θ θ θ
θ π θ θ θ π θ θ
−
−=
−
−−=
−
−−==
−−=−
=−++−+
r
r
r
r
r
r
r r
r r
7/23/2019 Class 4 Velocity Analysis1 (1)
http://slidepdf.com/reader/full/class-4-velocity-analysis1-1 12/43
An",lar Velocity Ratio andMechanical Ad!anta"e
The angular velocity ratio mV
is de"ned as the output angular velocity divided by the input angular velocity.
or a four bar mechanism with lin* 1 as the input and lin* ( as the output this is expressed as
The e2ciency of a four bar lin*age is de"ned as the output power over the input power,
Assuming -3 e2ciency, which is normally approached by four bar mechanisms, we have
2
4
ω
ω
ω
ω
==in
out
V m
inin
out out
in
out
T
T
P
P
ω
ω η ==
V
T
out
in
inin
out
mm
T
T 1===
ω
ω
7/23/2019 Class 4 Velocity Analysis1 (1)
http://slidepdf.com/reader/full/class-4-velocity-analysis1-1 13/43
An",lar Velocity Ratio andMechanical Ad!anta"e
The mechanical advantage is de"ned as
the ratio between the output force to the
input force
out
in
T
inin
out out
in
out
Ar
r m
r T
r T
F
F m ===
7/23/2019 Class 4 Velocity Analysis1 (1)
http://slidepdf.com/reader/full/class-4-velocity-analysis1-1 14/43
VE&OCIT' ANA&'(I(
OF (&IDERCRANKMEC*ANI(M(
7/23/2019 Class 4 Velocity Analysis1 (1)
http://slidepdf.com/reader/full/class-4-velocity-analysis1-1 15/43
Velocity Analysis of a (liderCran%Mechanism
4esign parameters: r 2 , r
3, r
4, θ
1.
5osition analysis parameters:
r 1, θ
2 , θ
3
6elocity analysis parameter.
ind dr 17dt , dθ
2 7dt , dθ
37dt
To eliminate ω3 dot product both sides by u
θ 3
r 3
r 2
r1
r 4
r p
( ) ( ) 132 1233222
4132
θ π θ π θ ω ω ur ur ur
r r r r
=+
+=+
++
( ) ( )
( ) ( )
( )
( ) 2
13
232
1
3113222
3113222
cos
sin
cossin
cos2cos
ω θ θ
θ θ
θ θ θ θ ω
θ θ θ π θ ω
−−
=
−=−−
−=−+
r r
r r
r r
( ) ( )
( )
( ) 2
133
122
3
13331222
cos
cos
coscos
ω θ θ
θ θ ω
θ θ ω θ θ ω
−−
−=
−−=−
r
r
r r
To eliminate dr 17dt dot product both sides
by u/θ 1+π/2
0
7/23/2019 Class 4 Velocity Analysis1 (1)
http://slidepdf.com/reader/full/class-4-velocity-analysis1-1 16/43
Velocity Analysis of an In!erted(liderCran% Mechanism
%iven r 1
, r 2 , θ
1, θ
2 , ω
2
5osition analysis: ind r 3, θ
3
6elocity analysis: ind dr 37dt , dθ
37dt
To eliminate ω3 and "nd dr
17dt dot product both sides by u
θ 3
To eliminate dr 17dt and "nd ω
3 dot product both sides by u/
θ 3+π/20
( ) ( )2333222
312
312
332
312
π θ θ π θ
θ θ θ
ω ω ++ +=+=
+=
ur ur ur
ur ur ur
r r r
( )
( ) 32322
33222
sin
2cos
r r
r r
=−
=−+
θ θ ω
θ π θ ω
r 2
r1
θ 1
θ 2
r 3
θ 3
( )
( )2
3
232
3
333222
cos
cos
ω θ θ
ω
ω θ θ ω
r
r
r r
−=
=−
7/23/2019 Class 4 Velocity Analysis1 (1)
http://slidepdf.com/reader/full/class-4-velocity-analysis1-1 17/43
Velocity Analysis of an In!erted(liderCran% Mechanism
%iven r 1
, r 2 , r
4, θ
1, θ
2 , ω
2
5osition analysis: ind r 3, θ
3 , θ
4
6elocity analysis: ind dr 37dt , dθ
37dt , dθ
47dt
r 3
r 2 r
1
r 4
( ) ( ) ( )
( ) ( ) 34423333222
34
24423333222
41332
4132
32
432
412
2
θ π θ θ π θ
π θ π θ θ π θ
θ θ θ θ
ω ω ω π θ θ
ω ω ω
ur ur ur ur
ur ur ur ur
ur ur ur ur
r r r r
−=+++=
=++
+=+
+=+
++
+++
with
7/23/2019 Class 4 Velocity Analysis1 (1)
http://slidepdf.com/reader/full/class-4-velocity-analysis1-1 18/43
Velocity Analysis of an In!erted(liderCran% Mechanism
r 3
r 2 r
1
r 4
( ) ( )
( )
( )
( )
( )
( )
( )
( ) ( )
( )( )
2322
3
322
43
2
3
322
433222
2
3
322
34
4433222
4433222
3
2
3
322
3
333222
2
34423333222
sincos
cossin
cos
sin
2cos
cos
0cos
3
32
ω θ θ θ θ
ω θ θ
θ θ ω
ω θ θ
ω ω
ω θ θ ω
ω θ π θ ω
ω θ θ
ω
ω θ θ ω
ω ω ω
θ
π θ
θ π θ θ π θ
−+
−=
−=+−−
−−==
−=+−−
−=+−+
−−=
=+−
−=++
+
++
r r
r r r
r
r r r r
r
r
r r r
r r r
u
r
r
r r
u
ur ur ur ur
8ut
bysidesboth4ot
bysidesboth4ot
7/23/2019 Class 4 Velocity Analysis1 (1)
http://slidepdf.com/reader/full/class-4-velocity-analysis1-1 19/43
E1am+le
( )
( )
( ) ( ) ( ) ( )
( ) ( )
( ) ( ) ( ) ( )
( )
( )
( )
−+−+
=
−+=+
−+=+
=−
+−±
=⇒=+++−
=++
+−−+++=+
−−−−−+++=+
−+=+
+=++
+=++
−
−
224411
2244111
2,13
2244113
2244113
2,1
1
2,14
222
2,1
2
44
424242241441
2
4
2
2
2
1
2
424212411441
2
4
2
2
2
1
2
241
152
412
coscoscos
coscossintan
coscossinsin
coscoscoscos
tan2,02
0sincos
sinsincoscos2cos2cos2
cos2cos2cos2
2413
61532
4132
θ θ θ
θ θ θ θ
θ θ θ θ
θ θ θ θ
θ
θ θ
θ θ θ θ θ θ
θ θ θ θ θ θ
θ θ θ θ
θ θ θ θ θ
θ θ θ θ
r r r
r r r
r r r ba
r r r ba
t C A
AC B Bt AC Bt t AC
C B A
r r r r r r r r r ba
r r r r r r r r r ba
ur ur ur uba
u sur ur uaur
ur ur ubaur
1!r'uationclosureloopfromAlso
itself bye'uationtheof sidesboth4ot
-!0'uation9losure:oop
II:oop
I:oop
a
r 2
r1
b
r 4
r 5
s
a
r 2
r1
b
r 4
r 5
s
( )
( )2tan
cos2
sin2
cos22
4
2
221
2
4
2
2
2
1
142
24241
θ
θ
θ
θ
=
+−−++=
−=
−=
t
bar r r r r C
r r B
r r r r A
7/23/2019 Class 4 Velocity Analysis1 (1)
http://slidepdf.com/reader/full/class-4-velocity-analysis1-1 20/43
E1am+le
( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( )
( ) ( ) ( ) ( )
( )
( )
( )
−+−+
=
−+=+
−+=+
=−
+−±=⇒=+++−
−+++=+
−−−+−−−−
−−+++===
−−−+−−−−
−−−−−++++=
−−+=
−
−
224411
2244111
2,13
2244113
2244113
2,1
1
2,14
222
2,1
2
241441
2
4
2
2
2
1
2
32226213262
31221
22
2
22
1
2
5
61
32226213262
1311221161
22
2
22
1
2
5
215
coscoscos
coscossintan
coscossinsin
coscoscoscos
tan2,02
cos2cos2
cos2cos2cos2cos2
cos2cos2
2,0
cos2cos2cos2cos2
cos2cos2cos2
32615
θ θ θ
θ θ θ θ
θ θ θ θ
θ θ θ θ
θ
θ θ
θ θ θ θ θ θ θ θ
θ θ
π θ θ
θ θ θ θ θ θ θ θ
θ θ θ θ θ θ
θ θ θ θ θ
r r r
r r r
r r r ba
r r r ba
t C A
AC B Bt AC Bt t AC
r r r r r r r ba
ar sr as sr
ar r r ar sr r
ar sr as sr
ar r r sr ar sr r
uaur u sur ur
1!r'uationclosureloopfromAlso
:et
itself bye'uationtheof sidesboth4ot
1!0'uation9losure:oop
a
r 2
r1
b
r 4
r 5
s
a
r 2
r1
b
r 4
r 5
s
( )
( )2tan
cos2
sin2
cos22
4
2
221
2
4
2
2
2
1
142
24241
θ
θ
θ
θ
=
+−−++=
−=
−=
t
bar r r r r C
r r B
r r r r A
7/23/2019 Class 4 Velocity Analysis1 (1)
http://slidepdf.com/reader/full/class-4-velocity-analysis1-1 21/43
*23
5678 569 /:;c08 5<6/:08 544.
7/23/2019 Class 4 Velocity Analysis1 (1)
http://slidepdf.com/reader/full/class-4-velocity-analysis1-1 22/43
MET*OD OF IN(TANT CENTER(
7/23/2019 Class 4 Velocity Analysis1 (1)
http://slidepdf.com/reader/full/class-4-velocity-analysis1-1 23/43
Relati!e Velocities )et=een T=o>oints on a Ri"id )ody
%iven any two points A and B that
lie on a rigid body, let the line AB
be a line passing through A and 8,
then the components of the
velocity of A and the velocity of B
on the line AB must be e'ual.
The velocity of point B relative to
A must be normal to the line AB
A
B
A
B
A B B A vvv +=
7/23/2019 Class 4 Velocity Analysis1 (1)
http://slidepdf.com/reader/full/class-4-velocity-analysis1-1 24/43
Instant Center Relati!e to thero,nd
If line AA; is drawn normal to the direction of v A
, then the
velocity of any point on the rigid body that falls on line AC
must be perpendicular to the line AA;
In a similar manner, if line 88; is drawn perpendicular to
the direction of v B
, then the velocity of any point on the
rigid body that falls on this line must be perpendicular to
the line BB;
If point I is the intersection of lines AA; and BB;, then the
velocity of point I must be perpendicular to both AA; and
BB;. This can only happen when the velocity of point I is
<ero.
A
B I
A’
B’
I B B
I A A I A A
I A A
I A I A
r v
r vr v
r k v
r k vv
ω
ω ω
ω
ω
=
=⇒=
×=
×+=
7/23/2019 Class 4 Velocity Analysis1 (1)
http://slidepdf.com/reader/full/class-4-velocity-analysis1-1 25/43
Instant Center Relati!e to thero,nd
5oint I determined as before is called the instant
center of <ero velocity of the rigid body with
respect to the ground. The direction of the
velocity of point A is normal to the line IA. The
direction of the velocity of line B is normal to the
line IB.
The direction of the velocity of any other point C
on the same body must be normal to the line IC.
The magnitude of the velocity of any point on
the lin* is proportional to its distance from the
point I.
A
B I
A’
B’
C
IC
v
IB
v
IA
v A B A
==
7/23/2019 Class 4 Velocity Analysis1 (1)
http://slidepdf.com/reader/full/class-4-velocity-analysis1-1 26/43
Velocity of a >oint on a &in% :yInstant Center Method
9onsider two points A and B on a rigid lin*. et v A
and v B
be
the velocities of points A and 8 at a given instant. If v A
is
*nown in magnitude and direction and v B
in direction only,
then the magnitude of v B
may be determined by the
instantaneous centre method.
4raw AI and BI perpendiculars to the directions v A
and v B
respectively. et these lines intersect at I, which is *nown as
instantaneous centre or virtual centre of the lin*. The
complete rigid lin* rotates about the centre I at the given
instant. The relations shown may be used to determine the
magnitude of the velocity of point B.
I B B
I A A I A A
I A A
I A I A
r v
r vr v
r k v
r k vv
ω
ω ω
ω
ω
=
=⇒=
×=
×+=
7/23/2019 Class 4 Velocity Analysis1 (1)
http://slidepdf.com/reader/full/class-4-velocity-analysis1-1 27/43
Instant Center )et=een T=o Ri"id)odies
In the foregoing discussion, the velocities of
points A and B were assumed to be reference to
a coordinate system attached to the ground.
The resulting center is called an instant center
relative to the ground.
The velocities of points A and B could as well be
ta*en relative to coordinate system attached to
another body. The resulting point in this case
will have a <ero velocity with respect to that
body. In other words, point I will have the same
velocity for both bodies.
A
B I
A’
B’
7/23/2019 Class 4 Velocity Analysis1 (1)
http://slidepdf.com/reader/full/class-4-velocity-analysis1-1 28/43
Instant Center )et=een T=o Ri"id)odies
A instant center between two bodies is a:
A point on both bodies
A point at which the two bodies have
no relative velocity.
A point about which one body may
be considered to rotate around the
other body at a given instant.
A
B I
A’
B’
7/23/2019 Class 4 Velocity Analysis1 (1)
http://slidepdf.com/reader/full/class-4-velocity-analysis1-1 29/43
Instant Center )et=een T=o Ri"id)odies
+hen two lin*s are connected to
one another by a revolute =oint, the
center of the connecting =oint is an
instant center for the two lin*s.
+hen two lin*s are not connected,
an instant center between the two
lin*s will also exist and can be
determined if the velocities of both
lin*s is *nown.
A
A’
7/23/2019 Class 4 Velocity Analysis1 (1)
http://slidepdf.com/reader/full/class-4-velocity-analysis1-1 30/43
Instant Center )et=een T=o Ri"id)odies
The number of instant centers in a constrained *inematic chain is
e'ual to the number of possible combinations of two lin*s.
The number of pairs of lin*s or the number of instantaneous
centers is the number of combinations of L lin*s ta*en two at a
time. >athematically, number of instant centers is,
( )
2
1−=
L L N
7/23/2019 Class 4 Velocity Analysis1 (1)
http://slidepdf.com/reader/full/class-4-velocity-analysis1-1 31/43
Ty+es of Instantaneo,s Centers
The instant centers for a mechanism are of the
following three types :
-. 5rimary 5ermanent! instant centers. They
can be "xed or moving
1. ?econdary Instant centers Not permanent!
( )21−=
nnC
7/23/2019 Class 4 Velocity Analysis1 (1)
http://slidepdf.com/reader/full/class-4-velocity-analysis1-1 32/43
Instant Centers of a Fo,r )arMechanism
9onsider a four bar mechanism ABCD as shown. The number of
instant centers N! in a four bar mechanism is given by
The instant centers I12
and I14
are "xed instant centers as they
remain in the same place for all con"gurations of the
mechanism. The instant centers I23
and I34
are permanent
instant centers as they move when the mechanism moves, but
the =oints are of permanent nature. The instantaneous centres
I13
and I24
are neither "xed nor permanent as they vary with
the con"guration of the mechanism.
( )
2
1−=
nnC
( ) 62
12
2
144==
−
= N A
B
C
D
1 (Ground)
2
3
4
7/23/2019 Class 4 Velocity Analysis1 (1)
http://slidepdf.com/reader/full/class-4-velocity-analysis1-1 33/43
&ocation of Instant Centers
+hen the two lin*s are connected by a pin =oint or pivot
=oint!, the instant center lies on the center of the pin as.
?uch an instant center is of permanent nature. If one of the
lin*s is "xed, the instant center will be of "xed type.
+hen the two lin*s have a pure rolling contact without
slipping!, the instantaneous centre lies on their point of
contact, as this point will have the same velocity on both
lin*s.
+hen the two lin*s have a sliding contact, the instant center
lies at the center of curvature of the path of contact. This
points lies at the common normal at the point of contact.
7/23/2019 Class 4 Velocity Analysis1 (1)
http://slidepdf.com/reader/full/class-4-velocity-analysis1-1 34/43
Kennedy /or Three Centers in&ine0 Theorem
/ennedy;s theorem states that any three bodies
in plane motion will have exactly three instant
centers, and the three centers will lie on the
same straight line.
Note that this rule does not re'uire that the
three bodies be connected in any way. +e can
use this rule, in con=unction with the linear
graph, to "nd the remaining l9s which are not
obvious from inspection.
7/23/2019 Class 4 Velocity Analysis1 (1)
http://slidepdf.com/reader/full/class-4-velocity-analysis1-1 35/43
E1am+le? Instant Centers of a Fo,r)ar Mechanism
4raw a circle with all lin*s numbered around the circumference
ocate as many ICs as possible by inspection. All pin =oints will
be permanent ICs . 9onnect the lin*s numbered on the circle
to create a linear graph and record those found.
Identify a lin* combination for which the IC has not been found,
and draw a dotted line connecting those two lin* numbers.
Identify two triangles on the graph which each contains the
dotted line and whose two other sides are solid lines
representing the ICs the already found. @se /ennedy;s
theorem to locate the needed I9.
7/23/2019 Class 4 Velocity Analysis1 (1)
http://slidepdf.com/reader/full/class-4-velocity-analysis1-1 36/43
E1am+le? Instant Centers of a Fo,r)ar Mechanism
7/23/2019 Class 4 Velocity Analysis1 (1)
http://slidepdf.com/reader/full/class-4-velocity-analysis1-1 37/43
E1am+le? Instant Centers of a(liderCran% Mechanism
7/23/2019 Class 4 Velocity Analysis1 (1)
http://slidepdf.com/reader/full/class-4-velocity-analysis1-1 38/43
E1am+le? Instant Centers of a(liderCran% Mechanism
E1am+le? Instant Centers of a Cam
7/23/2019 Class 4 Velocity Analysis1 (1)
http://slidepdf.com/reader/full/class-4-velocity-analysis1-1 39/43
E1am+le? Instant Centers of a Camand Follo=er? Common NormalMethod
E1am+le? Instant Centers of a Cam
7/23/2019 Class 4 Velocity Analysis1 (1)
http://slidepdf.com/reader/full/class-4-velocity-analysis1-1 40/43
E1am+le? Instant Centers of a Camand Follo=er? E@ecti!e &in%Method
7/23/2019 Class 4 Velocity Analysis1 (1)
http://slidepdf.com/reader/full/class-4-velocity-analysis1-1 41/43
Velocity Analysis =ith InstantCenters
nce the instant centers I9s! of a lin*age
with respect to the ground lin* have been
found, they can be used for a very rapid
graphic analysis for that lin*.
Note that some of the I9s may be very far
removed from the lin*s. or example, if
lin*s 1 and ( are nearly parallel, their
extended lines will intersect at a point far
away and not be practically available for
velocity analysis.
7/23/2019 Class 4 Velocity Analysis1 (1)
http://slidepdf.com/reader/full/class-4-velocity-analysis1-1 42/43
Velocity Analysis =ith InstantCenters
rom the de"nition of the instant center, both lin*s sharing the
instant center will have identical velocity at that point.
Instant center I13
involves the coupler lin* B! which is in general
plane motion, and the ground lin* which is stationary. All points on
the ground lin* have <ero velocity in the global coordinate
system, which is embedded in lin* -. Therefore, I13
must have
<ero velocity at this instant, and it can be considered to be an
instantaneous C"xed pivotC about which lin* B is in pure rotation
with respect to lin* -.
A moment later, I
13
will move to a new location and lin* B will be
CpivotingC about a new instant center.
7/23/2019 Class 4 Velocity Analysis1 (1)
http://slidepdf.com/reader/full/class-4-velocity-analysis1-1 43/43
Velocity Analysis =ith InstantCenters
If ω2 is *nown for the mechanism shown, the magnitude of
the velocity of point A can be computed as v A
$ ω2 O2 A Its
direction and sense can be determined by inspection.
Note that point A is also instant center and it has the same
velocity as part of lin* 1 and as part of lin* B. ?ince lin* B is
eectively pivoting about I13 at this instant, the angular
velocity ω3 can be found by ω
3 $ v
A 7 AI
13 . nce ω
3 is
*nown, the magnitude of v B
can also be found from v B
$ ω3
AI13
nce v B
is *nown, ω4
can also be found from ω4 $ v
B 7BO
4
$ v B
7BO4 . inally, v
C or the velocity of any other point on
the coupler! can be found from v C
$ ω3 CI
13
13
3
AI
v A
=ω
133 BI v
B ω =