all figures taken from design of machinery , 3 rd ed. robert norton 2003
DESCRIPTION
MENG 372 Chapter 6 Velocity Analysis. All figures taken from Design of Machinery , 3 rd ed. Robert Norton 2003. Velocity Analysis. Definitions. Multiplying by i rotates the vector by 90°. Velocity is perpendicular to radius of rotation & tangent to path of motion. Linear Velocity. - PowerPoint PPT PresentationTRANSCRIPT
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All figures taken from Design of Machinery, 3rd ed. Robert Norton 2003
MENG 372Chapter 6
Velocity Analysis
2 2
Velocity AnalysisDefinitions
RdtRdV
dtd
ipeipe
RVV
peR
iiPAPPA
iPA
Linear Velocity
Angular Velocity
Velocity of a point
Velocity is perpendicular to radius of rotation & tangent to path of motion
Multiplying by i rotates the vector by 90°
Link in pure rotation
3 3
cos sinire r i
sin cosiire r i
cosr sinr
r
Real
Imaginary
cosr
sinr
Vector r can be written as:
Multiplying by i gives:
Multiplying by i rotates a vector 90°
Velocity Analysis
4 4
If point A is moving
ipeV
VVVi
A
PAAP
Velocity Analysis
Graphical solution:
5 5
Graphical Velocity Analysis (3 & 4)• Given linkage configuration & 2. Find 3 and 4
• We know VA and direction of VB and VBA (perpendicular to AB)
• Draw vector triangle. V=r.
VBA
VB
VBA Direction
VBA
VB VB Direction
VA
3
4 4
BA
B
V AB
V O B
6 6
Graphical Velocity Analysis (VC)• After finding 3 and 4, find VC
• VC=VA+VCA
• Recall that 3 was in the opposite direction as 2
VCA
VC
VA
Double Scale
VCA
VC
7 7
Instant Center• A point common to two bodies in plane motion, which
has the same instantaneous velocity in each body.• In ENGR 214 we found the instant center between
links 1 and 3 (point on link 3 with no velocity)• Now we also have an instant center between links 2
and 4
8 8
Instant Centers• Kennedy’s rule: any three
links will have three instant centers and they will lie on a straight line
• The pins are instant centers• I13 is from links 1,2,3 and
1,3,4• I24 is from links 1,2,4 and
2,3,4
I13
I24
1 2 3I12
I23
I13
1 3 4I13
I34
I14
1 2 4I12
I24
I14
2 3 4I23
I34
I24
Links
IC’s
9 9
Instant Centers
I13
• I13 has zero velocity since link 1 is ground
• 3 is the same all over link 3 • Velocity relative to
ground=r, perpendicular to r• VA2=a2=VA3=p3
• From this, 3 must be in the opposite direction as 2, and smaller in magnitude since p>a A
a
VA2
VA3
p
3
3
10 10
Instant Centers
I24
• I24 has the same velocity on link 2 and link 4
• VI2=l22=VI4=l44
• From this, 4 is in the same direction as 2 and smaller in magnitude since l4>l2
4
VI4
l4
VI2l2
11 11
Instant Centers Practice Problems
O2
O4
A
B
O4O2
B
A
Power=Tinin=Toutout
12 12
Velocity Analysis of a 4-Bar LinkageGiven 2. Find 3 and 4
13 13
Velocity Analysis of a 4-Bar Linkage• Write the vector loop equation
• After solving the position analysis, take the derivative
or
where
01432 iiii decebeae
0432432 iceibeiae iii
0432432 iii ceibeiaei
0 BBAA VVV
4
3
2
4
3
2
iB
iBA
iA
ceiVbeiVaeiV
14 14
Velocity Analysis of a 4-Bar Linkage
• Take knowns to one side:
• Take conjugate to get 2nd equation:
• Put in matrix form:
• Invert matrix:
0432432 iii ceibeiaei
243243
iii aecebe
2
2
43
43
2
21
4
3
i
i
ii
ii
aeae
cebecebe
243243
iii aecebe
2
2
43
43
2
2
4
3
i
i
ii
ii
aeae
cebecebe
15 15
Inverted Crank Slider
Given 2. Find 3 andLink 3 is a slider link: its effective length, b, changes
b
16 16
Inverted Crank Slider• Given 2. Find 3 and• Write the vector loop equation:
• After solving the position analysis, take the derivative:
• To get another equation:or
so
01432 iiii decebeae
04332432 iiii ceibeiebaei
43 43
243323
iiii aeicebeieb
b
17 17
Inverted Crank Slider
• Take conjugate to get second equation:
• Put in matrix form:
243323
iiii aeicebeieb
243323
iiii aeicebeieb
2
2
433
433
2
2
3
i
i
iii
iii
aeiaeib
cebeiecebeie
2
2
433
433
2
21
3
i
i
iii
iii
aeiaei
cebeiecebeiebInvert:
18 18
Velocity of any Point on a Linkage• Write the vector for RP
• Take the derivative
• Similarly
332 iip peaeR
33232
iip peiaeiV
RP
22 iS seR
222
iS seiV
44 iU ueR
444
iU ueiV
19 19
Offset Crank SliderGiven 2. Find 3 and
a
b c
d