Velocities1

• Time derivatives of the loop-closure
expressions allow the analysis of velocities
& accelerations, i.e.:
Velocity & Acceleration Analysis
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
VELOCITY CLOSURE
ACCELERATION CLOSURE

• Review of time derivatives of displacement

• Example: Velocity analysis of the offset
slider-crank
2
3
1
0
Displacement closure:

• Taking the time derivative:
• Rearranging and substituting:

• Splitting into real and imaginary eqns
• The solution for is obtained by solving
the imaginary equation as:
REAL
IMAGINARY

• Substituting this solution back into the real
equation gives the other unknown:
REAL EQUATION

• The velocity of one point can be expressed
as the velocity of another point, plus the
relative velocity of the two points
Relative Velocity Analysis

• Example
RA
RB/A
RB
A
B
O4O2
θ2
θ3
θ4
ω2

• Example
RA
RB/A
RB
A
B
O4O2
θ2
θ3
θ4
ω2
VB
Absolute velocity of point B

• Example
RA
RB/A
RB
A
B
O4O2
θ2
θ3
θ4
ω2
VA
Absolute velocity of point A

• Example
RA
RB/A
RB
A
B
O4O2
θ2
θ3
θ4
ω2
VB/A
Relative velocity of point B w.r.t. point A

• Example
RA
RB/A
RB
A
B
O4O2
θ2
θ3
θ4
ω2
VA
VB/A
VB

• Note that:
– The direction and magnitude of VA is a known
function of the input angular velocity, ω2
– The mechanism’s joints define the direction of
many of the remaining relative and absolute
velocities
– This information can be manipulated to find the
velocity (direction and magnitude) of points not
on the input link

• There are 4 distinct cases where relative
velocity analysis is applied (though only 3
are non-trivial)
Same Point Different Points
Same
Link
Different
Links
Case 1
TRIVIAL CASE
Case 2
DIFFERENCE
MOTION
Case 3
RELATIVE MOTION
Case 4
DIFFERENCE & RELATIVE
MOTION

• Case 2: Different points on the same link
– Want to find velocity of point B w.r.t. point A (VB|A)
– Take the derivative of the rel. position vector (RB|A)
RA
RB/A
RB
A
B
O4O2
θ2
θ3
θ4
ω2
VA
VB/A
VB

– Examining this result gives simple method for
calculation:
RA
RB/A
RB
A
B
O4O2
θ2
θ3
θ4
ω2
VA
VB/A
VB
Always = 0 for a rigid link
(no length change)
Equivalent to rotation through 90° in the sense
(CW or CCW) of ωB|A (i.e. ω3)
So, for a rigid link :
VB|A = (rB|A)(ω3), ┴ RB|A

– Recalling that VB = VA + VB|A we can set up a system
of equations to solve for VB:
RA
RB/A
RB
A
B
O4O2
θ2
θ3
θ4
ω2
VA
VB/A
VB
Tricky to approach analytically, but graphical
methods can be used, and can be much more
intuitive

• Case 2 Example (Supp Ex V1)
C
B
A
O2 O4
3 4
2
ω2

• Solve using a graphical method
C
B
A
O2 O4
3 4
2
ω2

• Find VB by relative velocity analysis:
C
B
A
O2 O4
3 4
2
ω2
0V
1 mm = 5 mm/s

• VA: direction, magnitude both known
C
B
A
O2 O4
3 4
2
ω2
0V
1 mm = 5 mm/s

0V
1 mm = 5 mm/s
C
B
A
O2 O4
3 4
2
ω2

0V
VA
1 mm = 5 mm/s
A
C
B
A
O2 O4
3 4
2
ω2

0V
VA
1 mm = 5 mm/s
A
• VB|A: direction known, magnitude unknown
C
B
A
O2 O4
3 4
2
ω2

C
B
A
O2 O4
3 4
2
0V
VA
1 mm = 5 mm/s
A

C
B
A
O2 O4
3 4
2
0V
VA
dir(V
B|A)
1 mm = 5 mm/s
A

• VB: direction known, magnitude unknown
0V
VA
dir(V
B|A)
1 mm = 5 mm/s
A
C
B
A
O2 O4
3 4
2
ω2

C
B
A
O2 O4
3 4
2
0V
VA
dir(V
B|A)
1 mm = 5 mm/s
A

C
B
A
O2 O4
3 4
2
0V
VA
dir(VB)
dir(V
B|A)
1 mm = 5 mm/s
A

• Solution is obtained by intersection &
measurement
0V
VA
dir(VB)
dir(V
B|A)
1 mm = 5 mm/s
A
C
B
A
O2 O4
3 4
2
ω2

measurement
0V
VA
dir(VB)
dir(V
B|A)
1 mm = 5 mm/s
A
B
VB
C
B
A
O2 O4
3 4
2
ω2

measurement
0V
VA
dir(VB)
dir(V
B|A)
1 mm = 5 mm/s
A
B
VB
X
C
B
A
O2 O4
3 4
2
ω2

• Noticing that , we measure:
C
B
A
O2 O4
3 4
2
ω2
X
0V
VA
dir(VB)
dir(V
B|A)
1 mm = 5 mm/s
A
B
VB

• Noticing that , we measure:
C
B
A
O2 O4
3 4
2
ω2
0V
VA
dir(VB)
dir(V
B|A)
1 mm = 5 mm/s
V
B|A
A
B
VB
X

• And compute:
• Where the direction of rotation is inferred
from the direction of
C
B
A
O2 O4
3 4
2
ω2
0V
VA
dir(VB)
dir(V
B|A)
1 mm = 5 mm/s
V
B|A
A
B
VB
X X

• Similarly:
C
B
A
O2 O4
3 4
2
ω2
0V
VA
dir(VB)
dir(V
B|A)
1 mm = 5 mm/s
V
B|A
A
B
VB
X X X

0V
VA
dir(VB)
dir(V
B|A)
1 mm = 5 mm/s
V
B|A
A
B
VB
• Now, VC can be found in a variety of ways:
– Intersect relative velocity directions w.r.t. A & B
– Compute directly, e.g. VC = VA+(ω3 X AC)
C
B
A
O2 O4
3 4
2
ω2
X X X

• Using the first method, note that:
X X X
C
B
A
O2 O4
3 4
2
ω2
0V
VA
dir(VB)
dir(V
B|A)
1 mm = 5 mm/s
V
B|A
A
B
VB

C
B
A
O2 O4
3 4
2
ω2
X X
and ,
X
0V
VA
dir(VB)
dir(V
B|A)
1 mm = 5 mm/s
V
B|A
A
B
VB

C
B
A
O2 O4
3 4
2
ω2
X X
and ,
0V
VA
dir(VB)
dir(V
B|A)
1 mm = 5 mm/s
dir(VC|A)
V
B|A
A
B
VB
X

C
B
A
O2 O4
3 4
2
ω2
X X
and ,
X
0V
VA
dir(VB)
dir(V
B|A)
1 mm = 5 mm/s
dir(VC|A)
V
B|A
A
B
VB

C
B
A
O2 O4
3 4
2
ω2
X X
and ,
0V
VA
dir(VB)
dir(V
B|A)
1 mm = 5 mm/s
dir(VC|A)
V
B|A
dir(VC|B
)
A
B
VB
X

• Intersection gives the solution for VC
C
B
A
O2 O4
3 4
2
ω2
X X X
0V
VA
dir(VB)
dir(V
B|A)
1 mm = 5 mm/s
dir(VC|A)
V
B|A
dir(VC|B
)
A
B
VB

C
B
A
O2 O4
3 4
2
ω2
X X
0V
VA
dir(VB)
dir(V
B|A)
1 mm = 5 mm/s
dir(VC|A)
V
B|A
VC
dir(VC|B
)
A C
B
VB
X

C
B
A
O2 O4
3 4
2
ω2
X X
0V
VA
dir(VB)
dir(V
B|A)
1 mm = 5 mm/s
dir(VC|A)
V
B|A
VC
dir(VC|B
)
A C
B
VB
X X

• Another Case 2 Example (Supp Ex V2)
C
B
A
O2
3
2
ω2
4

• Solve using the same graphical method:
C
B
A
O2
3
2
ω2
4
0V
1 mm = 10 mm/s

• Find VB by case 2 analysis:
C
B
A
O2
3
2
ω2
4
0V
1 mm = 10 mm/s

• VA : magnitude, direction both known
C
B
A
O2
3
2
ω2
4
0V
1 mm = 10 mm/s

• VA : magnitude, direction both known
C
B
A
O2
3
2
ω2
4
0V
A
V
A
1 mm = 10 mm/s

• VB|A : direction known, magnitude unknown
C
B
A
O2
3
2
ω2
4
0V
A
V
A
1 mm = 10 mm/s

• VB|A : direction known, magnitude unknown
C
B
A
O2
3
2
ω2
4
0V
dir(VB|A)
A
V
A
1 mm = 10 mm/s

• VB : direction known, magnitude unknown
0V
dir(VB|A)
A
V
A
1 mm = 10 mm/s
C
B
A
O2
3
2
ω2
4

• VB : direction known, magnitude unknown
0V
dir(VB|A)
A
V
A
dir(VB)
1 mm = 10 mm/s
C
B
A
O2
3
2
ω2
4

• Intersection & measurement give:
C
B
A
O2
3
2
ω2
4
0V
dir(VB|A)
A
V
A
dir(VB) VBVB|A
B
1 mm = 10 mm/s

• Intersection & measurement give:
C
B
A
O2
3
2
ω2
4
0V
dir(VB|A)
A
V
A
dir(VB) VBVB|A
B
1 mm = 10 mm/s
X

• And ω3 is found from:
C
B
A
O2
3
2
ω2
4
X
0V
dir(VB|A)
A
V
A
dir(VB) VBVB|A
B
1 mm = 10 mm/s
X

• Solve for VC by intersection:
C
B
A
O2
3
2
ω2
4
0V
dir(VB|A)
A
V
A
dir(VB) VBVB|A
B
1 mm = 10 mm/s
X X

C
B
A
O2
3
2
ω2
4
0V
dir(VB|A)
A
V
A
dir(VB) VBVB|A
B
dir(VC|A)
1 mm = 10 mm/s
X X

C
B
A
O2
3
2
ω2
4
0V
dir(VB|A)
A
V
A
dir(VB) VBVB|A
dir(VC|B
)
B
dir(VC|A)
1 mm = 10 mm/s
X X

• Measuring then gives:
C
B
A
O2
3
2
ω2
4
0V
dir(VB|A)
A
V
A
dir(VB) VBVB|A
dir(VC|B
)
B
C VC
dir(VC|A)
1 mm = 10 mm/s
X XX

• Case 3: Coincident points on different links
– Occurs for slides & pistons, cams & followers:
• Two points on different links momentarily occupy
the same point in the plane
• Each has a different absolute velocity, therefore a
relative velocity exists

– Calculate the slide (relative) velocity, VB3|B4
B2, B3, B4
4
2
3
• So far, we have taken the derivative of
the relative position vector, RB3|B4
• But how can we express this vector for
two coincident points?
• Intuitively, we can imagine displacing the
slide by some small distance along the
slide, then drawing RB3|B4
• Taking the limit as the displacement
approaches zero, we can see that RB3|B4
has zero length, and is directed along the
tangent to the slide (link 4) at point B

θslide
B2, B3, B4
4
2
3
• So:
• Taking the derivative:

θslide
B2, B3, B4
4
2
3
• Simplifying gives the final expression:
• Note that this deceptively simple
expression hides the potentially difficult
task of finding the slide tangent angle
• In the following examples, straight slides
are used to avoid this hassle (tangent
angle = link angle for a straight slide)

• Case 3 Example (Supp Ex V3)
C
B
O2
2
ω2
4
O4
3

C
B
O2
2
ω2
4
O4
3
• Solve graphically
0V
1 mm = 10 mm/s

C
B
O2
2
ω2
4
O4
3
• Use case 3 analysis to find VB4:
0V
1 mm = 10 mm/s

C
B
O2
2
ω2
4
O4
3
0V
1 mm = 10 mm/sC
B
O2
2
ω2
4
O4
3
• VB2: direction, magnitude both known:
0V
B2
VB2
1 mm = 10 mm/s

0V
B2
VB2
1 mm = 10 mm/s
0V
B2
VB2
dir(VB2|B4)
1 mm = 10 mm/sC
B
O2
2
ω2
4
O4
3
• VB2|B4: only direction is known
C
B
O2
2
ω2
4
O4
3

C
B
O2
2
ω2
4
O4
3
C
B
O2
2
ω2
4
O4
3
0V
B2
VB2
dir(VB2|B4)
1 mm = 10 mm/s
0V
dir(VB4)
B2
VB2
dir(VB2|B4)
1 mm = 10 mm/s
• VB4: direction known, magnitude unknown

• Obtain VB4 by intersection
C
B
O2
2
ω2
4
O4
3
0V
dir(VB4)
B2
VB2
dir(VB2|B4)
VB4
B4
VB2|B4
1 mm = 10 mm/s
X

• ω4 follows immediately from VB4:
C
B
O2
2
ω2
4
O4
3
0V
dir(VB4)
B2
VB2
dir(VB2|B4)
VB4
B4
VB2|B4
1 mm = 10 mm/s
X X

• VC follows immediately from ω4, or by:
•
C
B
O2
2
ω2
4
O4
3
0V
dir(VB4)
B2
VB2
dir(VB2|B4)
VB4
B4
VB2|B4
1 mm = 10 mm/s
X X

• ,
C
B
O2
2
ω2
4
O4
3
X X
0V
dir(VB4)
B2
VB2
dir(VB2|B4)
VB4
B4
VB2|B4
dir(VC|B4)
1 mm = 10 mm/s

• , ,
C
B
O2
2
ω2
4
O4
3
X X
0V
dir(VB4)
B2
VB2
dir(VB2|B4)
VB4
B4
VB2|B4
dir(VC)
dir(VC|B4)
1 mm = 10 mm/s

C
B
O2
2
ω2
4
O4
3
• Measuring:
0V
dir(VB4)
B2
VB2
dir(VB2|B4)
VB4
B4
VB2|B4
dir(VC)
dir(VC|B4)
C
VC
1 mm = 10 mm/s
X X X

• Another Case 3 Example (Supp Ex V4)
C
B
A
O2 O4
3 4
2
ω2
D5, D6
5
6

• Note: The 4-Bar was solved in Ex V1
C
B
A
O2 O4
3 4
2
ω2
D5, D6
5
6

C
B
A
O2 O4
3 4
2
ω2
D5, D6
5
6
• VC is found by scaling arguments:
VA
VB
V
B|A
A
B
0V
1 mm = 5 mm/s

C
B
A
O2 O4
3 4
2
ω2
D5, D6
5
6
• VC is found by scaling arguments:
VA
VB
V
B|A
A
B
V
C|A
0V
1 mm = 5 mm/s

C
B
A
O2 O4
3 4
2
ω2
D5, D6
5
6
• Measuring:
VA
VB
V
B|A
A
B
V
C|A
0V
C
VC
1 mm = 5 mm/s
X

C
B
A
O2 O4
3 4
2
ω2
D5, D6
5
6
• VD5 is found by Case 3 analysis:
X
VA
VB
V
B|A
A
B
V
C|A
0V
C
VC
1 mm = 5 mm/s

C
B
A
O2 O4
3 4
2
ω2
D5, D6
5
6
• VD5 : only the direction is known
X
VA
VB
V
B|A
A
B
V
C|A
dir(VD5
)
0V
C
VC
1 mm = 5 mm/s

C
B
A
O2 O4
3 4
2
ω2
D5, D6
5
6
• VD5|C : only the direction is known
X
VA
VB
V
B|A
A
B
V
C|A
dir(VD5
)
0V
dir(VD5|C
)
C
VC
1 mm = 5 mm/s

C
B
A
O2 O4
3 4
2
ω2
D5, D6
5
6
• Measuring:
X
VA
VB
V
B|A
A
B
VD5
V
C|A
dir(VD5
)
0V
dir(VD5|C
)
VD5|C
C
D5
VC
1 mm = 5 mm/s
X

C
B
A
O2 O4
3 4
2
ω2
D5, D6
5
6
• ω5 is found immediately from:
VA
VB
V
B|A
A
B
VD5
V
C|A
dir(VD5
)
0V
dir(VD5|C
)
VD5|C
C
D5
VC
1 mm = 5 mm/s
X X X

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