Position Analysis
• Useful indices for Position Analysis:
a) Types of Mechanism:
•
•
•
•
Grashof vs. Non-Grashof:
Grashof: at least 1 pair of adjacent links is capable of
full rotation
Test for Grashof mech: “Grashof’s Law”
“The sum of the shortest and longest links’ lengths is
less than the sum of the lengths of the remaining two
links”
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
Position Analysis
• Useful indices for Position Analysis:
a) Types of Mechanism:
•
•
4-bar mechanisms can be classified based on Grashof
criterion
For those that meet the criterion, the identity of the
shortest link defines the mechanism type
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
Position Analysis
• Example: Types of 4-bar
– Grashof 4-bar with rshort = r2  Crank-Rocker
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
Position Analysis
• Example: Types of 4-bar
– Grashof 4-bar with rshort = r1  Drag-Link
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
Position Analysis
• Example: Types of 4-bar
– Grashof 4-bar with rshort = r3 (or 4)  Rocker-Rocker
Note that the
extreme angular
positions of the crank
and follower do not
necessarily coincide
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
Position Analysis
• Example: Types of 4-bar
– Any non-Grashof 4-bar is a Triple-Rocker
Again, extreme
angular positions of
the crank and
follower do not
necessarily coincide
Also, there are points
where the motion of
the mechanism
diverges, i.e. one or
more outputs are
possible for a given
crank input
Which way will link 3
rotate when the
input is rotated CW
from this position?
How about link 4?
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
Position Analysis
• Useful indices for Position Analysis:
b) Inversions of a mechanism
•
•
Inversions change which of the mechanism’s links is
fixed
An n-link mechanism has n inversions
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
Position Analysis
• Example: Inversions of the slider-crank
– Standard slider-crank
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
Position Analysis
• Example: Inversions of the slider-crank
– Freeing the slide, and fixing link 2
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
Position Analysis
• Example: Inversions of the slider-crank
– Or, fixing link 3 (note: 4 is still free to rotate)
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
Position Analysis
• Example: Inversions of the slider-crank
– Or, fixing 4 (1 slides, but does not rotate, w.r.t. 4)
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
Position Analysis
• Useful indices for Position Analysis:
c) Transmission (γ) and Deviation (δ) angles
•
•
Transmission Angle (γ): The acute angle between the
directions of the velocity-difference vector of the
driving point , and the absolute velocity vector of the
driven point
Deviation Angle (δ): The angle between the absolute
direction of travel of the driven point and the direction
of the force exerted by the driving link
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
Position Analysis
• Example:
– Transmission Angle (γ):
The acute angle between
the directions of:
the absolute velocity vector
of the driven point
2
A1,2
VB
3
INPUT
1
4
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
4
VB2/A2
the velocity-difference
vector of the driving point
and
B2,3
γ
3
Position Analysis
• Example:
– Deviation Angle (δ):
B2,3
The angle between:
The absolute direction of
travel of the driven point
and
the direction of the force
exerted by the driving link
FB2→B3
δ
2
A1,2
VB
3
INPUT
1
4
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
4
3
Displacement Analysis
• Any single-loop, planar mechanism will have
two unknowns (dependent variables)
• Need a method to solve for these variables in
terms of known mechanism parameters
• 2 main methods presented:
– Graphical
– Analytical
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
Displacement Analysis
• Example: 4-bar mech.
– Vectors Ri used to
represent links
– All link lengths known
– Input (independent)
variable is crank angle θ2
– What are the dependent
variables?
B
R3
A
R4
R2
θ2
A0
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
R1
B0
Displacement Analysis
• Example: 4-bar mech.
– Vectors Ri used to
represent links
– All link lengths known
– Input (independent)
variable is crank angle θ2
– What are the dependent
variables?
B
R3
θ3
A
R4
R2
• Need a method to solve for
A0
these variables
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
θ4
θ2
R1
B0
Displacement Analysis
• Graphical (approximate) method
• Solution lies at the
intersection of the possible
positions of links 3 & 4
A
R2
θ2
A0
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
R1
B0
Displacement Analysis
• Graphical (approximate) method
• Solution lies at the
intersection of the possible
positions of links 3 & 4
• Note presence of 2 possible
solutions!
A
θ2
A0
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
B0
Displacement Analysis
• Graphical (approximate) method
• Solution lies at the
intersection of the possible
positions of links 3 & 4
• Note presence of 2 possible
solutions!
• Unknown angles are
measured from drawing
B
θ3
A
θ4
θ2
A0
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
B0
Displacement Analysis
• Analytical methods
– Graphical methods inexact, inconvenient for
repeated analysis
– 2 Analytical methods presented:
• Geometric analysis
• Loop-closure
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
Displacement Analysis
• Example 1: Geometric analysis
B
R3
A
γ
θ3
Δ
R2
arg(D)
D
R4
φ
θ4
θ2
A0
B0
R1
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
4
Displacement Analysis
• Loop Closure – Based Displacement Analysis
– Represent each link by a vector,
– Sum vectors to close the loop
– Split the real and imaginary components of this
equation to yield two equations by applying:
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
Displacement Analysis
• Example 2: Loop Closure Analysis
B
R3
A
θ3
R4
R2
θ4
θ2
A0
B0
R1
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
4
Displacement Analysis
• Example 2: Loop Closure Analysis
B
EQUATION FROM REAL PARTS
4
R3
A
θ3
R4
R2
EQUATION FROM IMAGINARY PARTS
θ4
θ2
A0
B0
R1
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
Displacement Analysis
• Example 3: Loop Closure, offset slider-crank
4
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
Displacement Analysis
• Example 3: Loop Closure, offset slider-crank
2
3
4
0
1
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
Displacement Analysis
• Example 3: Loop Closure, offset slider-crank
4
2
3
0
1
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
Displacement Analysis
• Example 3: Loop Closure, offset slider-crank
4
2
3
0
1
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
Displacement Analysis
• Example 3: Loop Closure, offset slider-crank
4
2
3
0
1
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
Displacement Analysis
• Example 3: Loop Closure, offset slider-crank
4
2
3
0
1
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
Displacement Analysis
• Example 3: Loop Closure, offset slider-crank
4
2
3
0
1
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
Displacement Analysis
• Example 3: Loop Closure, offset slider-crank
• Solution is correct, but “weak”, since
4
multiple
angles have the same sine
2
3
0
1
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
Displacement Analysis
• Example 3: Loop Closure, offset slider-crank
• A stronger solution can be found by writing:
4
2
3
0
1
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
Displacement Analysis
• Example 3: Loop Closure, offset slider-crank
• A stronger solution can be found by writing:
4
2
3
0
1
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
Displacement Analysis
• Example 3: Loop Closure, offset slider-crank
• A stronger solution can be found by writing:
4
2
3
0
1
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
Displacement Analysis
• Example 3: Loop Closure, offset slider-crank
• A stronger solution can be found by writing:
4
2
3
0
1
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
Displacement Analysis
• Example 3: Loop Closure, offset slider-crank
• A stronger solution can be found by writing:
4
2
3
0
1
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
Displacement Analysis
• Example 3: Loop Closure, offset slider-crank
• A stronger solution can be found by writing:
4
2
3
0
1
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
Displacement Analysis
• Example 3: Loop Closure, offset slider-crank
• A stronger solution can be found by writing:
4
2
3
0
1
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
Displacement Analysis
• Example 3: Loop Closure, offset slider-crank
• A stronger solution can be found by writing:
4
2
3
0
1
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
Displacement Analysis
• Example 3: Loop Closure, offset slider-crank
• A stronger solution can be found by writing:
4
2
3
0
1
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
Displacement Analysis
• Example 3: Loop Closure, offset slider-crank
• A stronger solution can be found by writing:
4
2
3
0
1
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
Displacement Analysis
• Example 3: Loop Closure, offset slider-crank
• A stronger solution can be found by writing:
4
2
3
0
1
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
Displacement Analysis
• Example 3: Loop Closure, offset slider-crank
• A stronger solution can be found by writing:
4
2
3
0
1
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
Displacement Analysis
• Example 3: Loop Closure, offset slider-crank
• A stronger solution can be found by writing:
4
2
3
0
1
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
Displacement Analysis
• Example 3: Loop Closure, offset slider-crank
• A stronger solution can be found by writing:
4
2
3
0
1
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
Displacement Analysis
• Iterative Loop-Closure based displacement
solution
– If the dependent variables are inaccurate
– An iterative algorithm for closure can then be
based on:
and
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
Displacement Analysis
• Example: Newton-Raphson iterative closure
Loop closure error:
2
3
0
1
Input:
Unknowns:
,
i.e.
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
Displacement Analysis
• Example: Newton-Raphson iterative closure
2
3
If is correct,
Estimate of can be improved
by iteration, i.e.
0
1
Input:
Unknowns:
,
i.e.
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
Displacement Analysis
• Example: Newton-Raphson iterative closure
2
3
If is correct,
Estimate of can be improved
by iteration, i.e.
0
1
Input:
Unknowns:
,
i.e.
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
Displacement Analysis
• Example: Newton-Raphson iterative closure
For this example:
2
3
0
1
Input:
Unknowns:
Then, from (3):
,
i.e.
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
Displacement Analysis
• Example: Newton-Raphson iterative closure
Inverting (4), substituting into
(5), and then into (2) gives:
2
3
0
1
Input:
Unknowns:
i.e.
,
Where
is given in (1).
Equation (6) would be used to
iteratively improve the
estimates of , until:
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
Displacement Analysis
• Numerical Example:
2
3
0
1
» Suppose we start by guessing:
» Analysis tolerance:
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
Displacement Analysis
• Numerical Example:
2
3
• Now we iterate from that starting point
0
1
MECH 335 Lecture Notes
© R.Podhorodeski, 2009