velocity & acceleration analysis - ariel.ac.il · •note that: –the direction and magnitude...
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• 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
Velocity & Acceleration Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
• Example: Velocity analysis of the offset slider-crank
Velocity & Acceleration Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
2
3
1
0
Displacement closure:
• Taking the time derivative:
• Rearranging and substituting:
Velocity & Acceleration Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
• Taking the time derivative:
• Rearranging and substituting:
Velocity & Acceleration Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
• Taking the time derivative:
• Rearranging and substituting:
Velocity & Acceleration Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
• Taking the time derivative:
• Rearranging and substituting:
Velocity & Acceleration Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
• Splitting into real and imaginary eqns
• The solution for is obtained by solving the imaginary equation as:
Velocity & Acceleration Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
REAL
IMAGINARY
• Substituting this solution back into the real equation gives the other unknown:
Velocity & Acceleration Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
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
MECH 335 Lecture Notes © R.Podhorodeski, 2009
• Example
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
RA
RB/A
RB
A
B
O4O2
θ2
θ3
θ4
ω2
• Example
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
RA
RB/A
RB
A
B
O4O2
θ2
θ3
θ4
ω2
VB
Absolute velocity of point B
• Example
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
RA
RB/A
RB
A
B
O4O2
θ2
θ3
θ4
ω2
VA
Absolute velocity of point A
• Example
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
RA
RB/A
RB
A
B
O4O2
θ2
θ3
θ4
ω2
VB/A
Relative velocity of point B w.r.t. point A
• Example
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
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
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
• There are 4 distinct cases where relative velocity analysis is applied (though only 3 are non-trivial)
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
Same Point Different Points
SameLink
Different Links
Case 1TRIVIAL CASE
Case 2DIFFERENCE
MOTION
Case 3RELATIVE MOTION
Case 4DIFFERENCE & 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)
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
RA
RB/A
RB
A
B
O4O2
θ2
θ3
θ4
ω2
VA
VB/A
VB
• Case 2: Different points on the same link
– Examining this result gives simple method for calculation:
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
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
• Case 2: Different points on the same link
– Recalling that VB = VA + VB|A we can set up a system of equations to solve for VB:
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
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)
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
C
B
AO2 O4
3 4
2
ω2
• Solve using a graphical method
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
C
B
AO2 O4
3 4
2
ω2
• Find VB by relative velocity analysis:
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
C
B
AO2 O4
3 4
2
ω2
0V
1 mm = 5 mm/s
• VA: direction, magnitude both known
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
C
B
AO2 O4
3 4
2
ω2
0V
1 mm = 5 mm/s
• VA: direction, magnitude both known
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
0V
1 mm = 5 mm/sC
B
AO2 O4
3 4
2
ω2
0V
VA
1 mm = 5 mm/s
A
• VA: direction, magnitude both known
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
C
B
AO2 O4
3 4
2
ω2
0V
VA
1 mm = 5 mm/s
A
• VB|A: direction known, magnitude unknown
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
C
B
AO2 O4
3 4
2
ω2
• VB|A: direction known, magnitude unknown
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
C
B
AO2 O4
3 4
2
0V
VA
1 mm = 5 mm/s
A
• VB|A: direction known, magnitude unknown
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
C
B
AO2 O4
3 4
2
0V
VA
dir(VB|A )
1 mm = 5 mm/s
A
• VB: direction known, magnitude unknown
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
0V
VA
dir(VB|A )
1 mm = 5 mm/s
A
C
B
AO2 O4
3 4
2
ω2
• VB: direction known, magnitude unknown
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
C
B
AO2 O4
3 4
2
0V
VA
dir(VB|A )
1 mm = 5 mm/s
A
• VB: direction known, magnitude unknown
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
C
B
AO2 O4
3 4
2
0V
VA
dir(VB)
dir(VB|A )
1 mm = 5 mm/s
A
• Solution is obtained by intersection & measurement
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
0V
VA
dir(VB)
dir(VB|A )
1 mm = 5 mm/s
A
C
B
AO2 O4
3 4
2
ω2
• Solution is obtained by intersection & measurement
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
0V
VA
dir(VB)
dir(VB|A )
1 mm = 5 mm/s
A
BVB
C
B
AO2 O4
3 4
2
ω2
• Solution is obtained by intersection & measurement
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
0V
VA
dir(VB)
dir(VB|A )
1 mm = 5 mm/s
A
BVB
X
C
B
AO2 O4
3 4
2
ω2
• Noticing that , we measure:
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
C
B
AO2 O4
3 4
2
ω2
X
0V
VA
dir(VB)
dir(VB|A )
1 mm = 5 mm/s
A
BVB
• Noticing that , we measure:
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
C
B
AO2 O4
3 4
2
ω2
0V
VA
dir(VB)
dir(VB|A )
1 mm = 5 mm/s
VB|A
A
BVB
X
• Noticing that , we measure:
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
C
B
AO2 O4
3 4
2
ω2
0V
VA
dir(VB)
dir(VB|A )
1 mm = 5 mm/s
VB|A
A
BVB
X
• And compute:
• Where the direction of rotation is inferred from the direction of
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
C
B
AO2 O4
3 4
2
ω2
0V
VA
dir(VB)
dir(VB|A )
1 mm = 5 mm/s
VB|A
A
BVB
X X
• Similarly:
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
C
B
AO2 O4
3 4
2
ω2
0V
VA
dir(VB)
dir(VB|A )
1 mm = 5 mm/s
VB|A
A
BVB
X X X
0V
VA
dir(VB)
dir(VB|A )
1 mm = 5 mm/s
VB|A
A
BVB
• 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)
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
C
B
AO2 O4
3 4
2
ω2
X X X
• Using the first method, note that:
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
X X X
C
B
AO2 O4
3 4
2
ω2
0V
VA
dir(VB)
dir(VB|A )
1 mm = 5 mm/s
VB|A
A
BVB
• Using the first method, note that:
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
C
B
AO2 O4
3 4
2
ω2
X X
and ,
X
0V
VA
dir(VB)
dir(VB|A )
1 mm = 5 mm/s
VB|A
A
BVB
• Using the first method, note that:
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
C
B
AO2 O4
3 4
2
ω2
X X
and ,
0V
VA
dir(VB)
dir(VB|A )
1 mm = 5 mm/s
dir(VC|A)
VB|A
A
BVB
X
• Using the first method, note that:
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
C
B
AO2 O4
3 4
2
ω2
X X
and ,
X
0V
VA
dir(VB)
dir(VB|A )
1 mm = 5 mm/s
dir(VC|A)
VB|A
A
BVB
• Using the first method, note that:
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
C
B
AO2 O4
3 4
2
ω2
X X
and ,
0V
VA
dir(VB)
dir(VB|A )
1 mm = 5 mm/s
dir(VC|A)
VB|A
dir(V
C|B )
A
BVB
X
• Intersection gives the solution for VC
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
C
B
AO2 O4
3 4
2
ω2
X X X
0V
VA
dir(VB)
dir(VB|A )
1 mm = 5 mm/s
dir(VC|A)
VB|A
dir(V
C|B )
A
BVB
• Intersection gives the solution for VC
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
C
B
AO2 O4
3 4
2
ω2
X X
0V
VA
dir(VB)
dir(VB|A )
1 mm = 5 mm/s
dir(VC|A)
VB|A
VC
dir(V
C|B )
A C
BVB
X
• Intersection gives the solution for VC
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
C
B
AO2 O4
3 4
2
ω2
X X
0V
VA
dir(VB)
dir(VB|A )
1 mm = 5 mm/s
dir(VC|A)
VB|A
VC
dir(V
C|B )
A C
BVB
X X
• Another Case 2 Example (Supp Ex V2)
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
C
B
A
O23
2
ω2
4
• Solve using the same graphical method:
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
C
B
A
O23
2
ω2
40V
1 mm = 10 mm/s
• Find VB by case 2 analysis:
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
C
B
A
O23
2
ω2
40V
1 mm = 10 mm/s
• VA : magnitude, direction both known
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
C
B
A
O23
2
ω2
40V
1 mm = 10 mm/s
• VA : magnitude, direction both known
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
C
B
A
O23
2
ω2
40V
1 mm = 10 mm/s
• VA : magnitude, direction both known
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
C
B
A
O23
2
ω2
40V
AV
A
1 mm = 10 mm/s
• VB|A : direction known, magnitude unknown
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
C
B
A
O23
2
ω2
40V
AV
A
1 mm = 10 mm/s
• VB|A : direction known, magnitude unknown
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
C
B
A
O23
2
ω2
40V
AV
A
1 mm = 10 mm/s
• VB|A : direction known, magnitude unknown
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
C
B
A
O23
2
ω2
40V
dir
(VB
|A)
AV
A
1 mm = 10 mm/s
• VB : direction known, magnitude unknown
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
0Vdir
(VB
|A)
AV
A
1 mm = 10 mm/s
C
B
A
O23
2
ω2
4
• VB : direction known, magnitude unknown
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
0Vdir
(VB
|A)
AV
A
dir(VB)
1 mm = 10 mm/s
C
B
A
O23
2
ω2
4
• Intersection & measurement give:
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
C
B
A
O23
2
ω2
40V
dir
(VB
|A)
AV
A
dir(VB) VBV
B|A
B
1 mm = 10 mm/s
• Intersection & measurement give:
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
C
B
A
O23
2
ω2
40V
dir
(VB
|A)
AV
A
dir(VB) VBV
B|A
B
1 mm = 10 mm/s
X
• And ω3 is found from:
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
C
B
A
O23
2
ω2
4
X0V
dir
(VB
|A)
AV
A
dir(VB) VBV
B|A
B
1 mm = 10 mm/s
X
• Solve for VC by intersection:
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
C
B
A
O23
2
ω2
40V
dir
(VB
|A)
AV
A
dir(VB) VBV
B|A
B
1 mm = 10 mm/s
X X
• Solve for VC by intersection:
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
C
B
A
O23
2
ω2
40V
dir
(VB
|A)
AV
A
dir(VB) VBV
B|A
B
dir(V
C|A )
1 mm = 10 mm/s
X X
• Solve for VC by intersection:
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
C
B
A
O23
2
ω2
40V
dir
(VB
|A)
AV
A
dir(VB) VBV
B|A
dir(VC|B
)
B
dir(V
C|A )
1 mm = 10 mm/s
X X
• Measuring then gives:
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
C
B
A
O23
2
ω2
40V
dir
(VB
|A)
AV
A
dir(VB) VBV
B|A
dir(VC|B
)
B
C VC
dir(V
C|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
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
• Case 3: Coincident points on different links
– Calculate the slide (relative) velocity, VB3|B4
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
B2, B3, B4
42
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
• Case 3: Coincident points on different links
– Calculate the slide (relative) velocity, VB3|B4
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
θslide
B2, B3, B4
42
3
• So:
• Taking the derivative:
• Case 3: Coincident points on different links
– Calculate the slide (relative) velocity, VB3|B4
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
θslide
B2, B3, B4
42
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)
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
C
B
O2
2
ω2
4
O4
3
C
B
O2
2
ω2
4
O4
3
• Solve graphically
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
0V
1 mm = 10 mm/s
C
B
O2
2
ω2
4
O4
3
• Use case 3 analysis to find VB4:
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
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:
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
0V
B2
VB2
1 mm = 10 mm/s
0V
B2
VB2
1 mm = 10 mm/s
0V
B2
VB2
dir(V
B2|B
4 )
1 mm = 10 mm/sC
B
O2
2
ω2
4
O4
3
• VB2|B4: only direction is known
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
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(V
B2|B
4 )
1 mm = 10 mm/s
0V
dir(VB4)
B2
VB2
dir(V
B2|B
4 )
1 mm = 10 mm/s
• VB4: direction known, magnitude unknown
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
• Obtain VB4 by intersection
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
C
B
O2
2
ω2
4
O4
3
0V
dir(VB4)
B2
VB2
dir(V
B2|B
4 )
VB4
B4
VB
2|B
4
1 mm = 10 mm/s
X
• ω4 follows immediately from VB4:
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
C
B
O2
2
ω2
4
O4
3
0V
dir(VB4)
B2
VB2
dir(V
B2|B
4 )
VB4
B4
VB
2|B
4
1 mm = 10 mm/s
X X
• VC follows immediately from ω4, or by:
•
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
C
B
O2
2
ω2
4
O4
3
0V
dir(VB4)
B2
VB2
dir(V
B2|B
4 )
VB4
B4
VB
2|B
4
1 mm = 10 mm/s
X X
• VC follows immediately from ω4, or by:
• ,
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
C
B
O2
2
ω2
4
O4
3
X X0V
dir(VB4)
B2
VB2
dir(V
B2|B
4 )
VB4
B4
VB
2|B
4
dir(VC|B4)
1 mm = 10 mm/s
• VC follows immediately from ω4, or by:
• , ,
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
C
B
O2
2
ω2
4
O4
3
X X0V
dir(VB4)
B2
VB2
dir(V
B2|B
4 )
VB4
B4
VB
2|B
4
dir
(VC)
dir(VC|B4)
1 mm = 10 mm/s
C
B
O2
2
ω2
4
O4
3
• Measuring:
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
0V
dir(VB4)
B2
VB2
dir(V
B2|B
4 )
VB4
B4
VB
2|B
4
dir
(VC)
dir(VC|B4)
C
VC
1 mm = 10 mm/s
X X X
• Another Case 3 Example (Supp Ex V4)
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
C
B
AO2 O4
3 4
2
ω2
D5, D6
5
6
• Note: The 4-Bar was solved in Ex V1
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
C
B
AO2 O4
3 4
2
ω2
D5, D6
5
6
C
B
AO2 O4
3 4
2
ω2
D5, D6
5
6
• VC is found by scaling arguments:
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
VA
VB
VB|A
A
B
0V
1 mm = 5 mm/s
C
B
AO2 O4
3 4
2
ω2
D5, D6
5
6
• VC is found by scaling arguments:
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
VA
VB
VB|A
A
B
VC|A
0V
1 mm = 5 mm/s
C
B
AO2 O4
3 4
2
ω2
D5, D6
5
6
• Measuring:
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
VA
VB
VB|A
A
B
VC|A
0V
C
VC
1 mm = 5 mm/s
X
C
B
AO2 O4
3 4
2
ω2
D5, D6
5
6
• VD5 is found by Case 3 analysis:
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
X
VA
VB
VB|A
A
B
VC|A
0V
C
VC
1 mm = 5 mm/s
C
B
AO2 O4
3 4
2
ω2
D5, D6
5
6
• VD5 : only the direction is known
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
X
VA
VB
VB|A
A
B
VC|A
dir(VD5 )
0V
C
VC
1 mm = 5 mm/s
C
B
AO2 O4
3 4
2
ω2
D5, D6
5
6
• VD5|C : only the direction is known
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
X
VA
VB
VB|A
A
B
VC|A
dir(VD5 )
0V
dir(VD5|C
)
C
VC
1 mm = 5 mm/s
C
B
AO2 O4
3 4
2
ω2
D5, D6
5
6
• Measuring:
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
X
VA
VB
VB|A
A
B
VD5
VC|A
dir(VD5 )
0V
dir(VD5|C
)
VD5|C
C
D5
VC
1 mm = 5 mm/s
X
C
B
AO2 O4
3 4
2
ω2
D5, D6
5
6
• ω5 is found immediately from:
Relative Velocity Analysis
MECH 335 Lecture Notes © R.Podhorodeski, 2009
VA
VB
VB|A
A
B
VD5
VC|A
dir(VD5 )
0V
dir(VD5|C
)
VD5|C
C
D5
VC
1 mm = 5 mm/s
X X X
END OF LECTURE PACK 3