### Position Analysis - UJ

```Mechanics of Machines
Class 3
Position Analysis
TYPES OF MECHANISM ANALYSESE
Position Analysis
 Position Analysis
 Given θ12 find θ23,
θ34 and θ14
 Find maximum and
minimum values of
θ14 .
 Find the location of
each point on the
mechanism for a
given value of θ12
and the curve traced
by the a point during
motion.
θ23
θ12
θ34
θ41
Velocity Analysis
 Position Analysis
 Given θ12 and ω12
find ω23, ω34 and ω14
 Find the speed of a
point in the
mechanism
θ 34, ω34
θ 23, ω23
θ 12, ω12
θ 41, ω41
Acceleration Analysis
 Position Analysis
 Given θ12 , ω12 and
α12 find α23, α34 and
α14
 Find the acceleration
of a point in the
mechanism
θ 34, ω34 , α34
θ 23, ω23 , α23
θ12, ω12, α12
θ 41, ω41 ,
α41
Question
 Why is acceleration
analysis important?
θ 34, ω34 , α34
θ 23, ω23 , α23
θ12, ω12, α12
θ 41, ω41 ,
α41
GRAPHICAL POSITION ANALYSIS
Graphical Position Analysis
 Use graphical methods,
(pen, ruler, compass
and protractors to
solve position analysis
problems.
may also be used as a
convenient alternative.
θ23
θ12
θ34
θ41
VECTOR POSITION ANALYSIS
Vector Position Analysis
 Use vector loop
closure equation to
solve position analysis
problems
Loop Closure Equation
 Use vector
loop closure
equation to
solve
position
analysis
problems

 
 
rP  r2  r3  r1  r4
Planar Four Bar Loop Closure Equation
 How many scalar
equations can be
written from the vector
loop-closure equation?
 How many unknowns
can be solved for

 
 
rP  r2  r3  r1  r4
Planar Four Bar Loop Closure Equation
 How many scalar
equations can be written
from the vector loopclosure equation?: Two
scalar equations could be
written from a vector
equation
 How many unknowns can
be solved for: Two
unknowns: In a standard
problem, θ1 and θ2 are
given. θ3 and θ4 are to be
solved for.
 We will focus on planar
mechanisms in this course

 
 
rP  r2  r3  r1  r4
Planar Four Bar Loop Closure Equation
 The closure condition
expresses the condition
that a loop of a linkage
closes on itself.
shown the closure
equation is


 
r2  r3  r1  r4




r2 u  2  r3 u  3  r1u  1  r4 u  4








r2 cos  2 i  sin  2 j  r3 cos  3 i  sin  3 j  r1 cos  1 i  sin  1 j  r4 cos  4 i  sin  4 j








Planar Four Bar Loop Closure Equation
 The closure condition
expresses the condition that a
loop of a linkage closes on
itself.
 For the four-bar linkage shown
the closure equation is


 
r2  r3  r1  r4




r2 u  2  r3 u  3  r1u  1  r4 u  4








r2 cos  2 i  sin  2 j  r3 cos  3 i  sin  3 j  r1 cos  1 i  sin  1 j  r4 cos  4 i  sin  4 j



 The following two scalar
equations are produced





r2 cos  2  r3 cos  3  r1 cos  1  r4 cos  4
r2 sin  2  r3 sin  3  r1 sin  1  r4 sin  4
Analytical Solution when θ2 is known
r2 cos  2  r3 cos  3  r1 cos  1  r4 cos  4
r2 sin  2  r3 sin  3  r1 sin  1  r4 sin  4
 The above two equations could be solved for two
unknowns. One approach is to eliminate one of
the unknown angles by isolating the trigonometric
function involving the angle on the left-hand
side of the equation.
r3 cos  3  r1 cos  1  r4 cos  4  r2 cos  2
r3 sin  3  r1 sin  1  r4 sin  4  r2 sin  2
 Squaring both sides of both equations and using
the identity sin 2   cos 2   1 we obtain
r3  r1  r2  r4  2 r1 r4 cos  1 cos  4  sin  1 sin  4 
2
2
2
2
 2 r1 r2 cos  1 cos  2  sin  1 sin  2   2 r2 r4 cos  2 cos  4  sin  2 sin  4 
Analytical Solution when θ2 is known
r3  r1  r2  r4  2 r1 r4 cos  1 cos  4  sin  1 sin  4 
2
2
2
2
 2 r1 r2 cos  1 cos  2  sin  1 sin  2   2 r2 r4 cos  2 cos  4  sin  2 sin  4 
 To obtain an explicit expression for θ4 in terms
of θ2 and the constant angle θ1 , we combine
the coefficients of cos θ4 and sin θ4 in the
equation above as follows:
A cos  4  B sin  4  C  0
 where
A  2 r1 r4 cos  1  2 r2 r4 cos  2
B  2 r1 r4 sin  1  2 r2 r4 sin  2
C  r1  r2  r4   r3  2 r1 r2 cos  1 cos  2  sin  1 sin  2 
2
2
2
2
Analytical Solution when θ2 is known
A cos  4  B sin  4  C  0
 The above equation can be solved for θ4 by
using the following half angle identities
sin  4 
cos  4 
2 tan  4 2 
 4
2
 4
2
tan  4
2
1  tan
1  tan
1
2
2
2
 After substitution and simplification, we get
C
 A t  2 Bt  ( A  C )  0
2
where
t  tan  4 2 
A  2 r1 r4 cos  1  2 r2 r4 cos  2
B  2 r1 r4 sin  1  2 r2 r4 sin  2
C  r1  r2  r4   r3  2 r1 r2 cos  1 cos  2  sin  1 sin  2 
2
2
2
2
Analytical Solution when θ2 is known
C
 A t  2 Bt  ( A  C )  0
2
A  2 r1 r4 cos  1  2 r2 r4 cos  2
B  2 r1 r4 sin  1  2 r2 r4 sin  2
C  r1  r2  r4   r3  2 r1 r2 cos  1 cos  2  sin  1 sin  2 
2
 Solving for t we obtain
t 
4 B  4 C  A C  A 
 2B 
2
2 C  A 
or
t 
B 
A  B C
2
AC
and
 4  2 tan
1
t 
2
2
2
2
2
Analytical Solution when θ2 is known
Given the values of θ1 and θ2 for a four
bar mechanism of known r1, r2, r3 and r4
the output angle θ4 is calculated as:
A  2 r1 r4 cos  1  2 r2 r4 cos  2
B  2 r1 r4 sin  1  2 r2 r4 sin  2
C  r1  r2  r4   r3
2
2
2
2
 2 r1 r2 cos  1 cos  2  sin  1 sin  2 
t 
B 
A  B C
2
2
2
AC
and
 4  2 tan
1
t 
The ± sign identifies the two possible
Analytical Solution when θ2 is known
Given the values of θ1 and θ2 for a four
bar mechanism of known r1, r2, r3 and r4
the output angle θ4 is calculated as:
A  2 r1 r4 cos  1  2 r2 r4 cos  2
B  2 r1 r4 sin  1  2 r2 r4 sin  2
C  r1  r2  r4   r3
2
2
2
2
 2 r1 r2 cos  1 cos  2  sin  1 sin  2 
t 
B 
A  B C
2
2
2
AC
and
 4  2 tan
1
t 
The ± sign identifies the two possible
Analytical Solution when θ2 is known
 Note that –π/2 ≤ tan-1 (t) ≤ –π/2. Therefore, θ4 will have the range –π ≤ θ4 ≤ –π.
 Unless the linkage is a Grashof type II linkage in one of the extreme positions of
its motion range, there are two valid solutions for θ4. These correspond to two
assembly modes or branches for the linkage.
A  2 r1 r4 cos  1  2 r2 r4 cos  2
B  2 r1 r4 sin  1  2 r2 r4 sin  2
C  r1  r2  r4   r3
2
2
2
2
 2 r1 r2 cos  1 cos  2  sin  1 sin  2 
t 
B 
A  B C
2
AC
and
 4  2 tan
1
t 
2
2
Analytical Solution when θ2 is known
 Because of the square root in the
expression for t, it can be complex if
(A2 + B2) < C2. In this case, the
mechanism is a Grashof type II
linkage in one of the extreme
positions of its motion range and it
cannot be assembled in specified
values of θ1 and θ2 .
 The assembly configurations would
then appear as shown below
A  2 r1 r4 cos  1  2 r2 r4 cos  2
B  2 r1 r4 sin  1  2 r2 r4 sin  2
C  r1  r2  r4   r3
2
2
2
2
 2 r1 r2 cos  1 cos  2  sin  1 sin  2 
t 
B 
A  B C
2
AC
and
 4  2 tan
1
t 
2
2
Analytical Solution when θ2 is known
 After θ4 is known, an expression for
θ3 can be obtained by solving the
loop closure equation to obtain:
 3  tan
1
 r1 sin  1  r4 sin  4  r2 sin  2

 r1 cos  1  r4 cos  4  r2 cos  2



 Note that it is essential that the sign
of the numerator and denominator
be maintained to determine the
quadrant in which the angle θ3 lies.
This can be done by using the ATAN2
function. The form of this function
is:
ATAN2 (sin θ3, cos θ3)
r3 cos  3  r1 cos  1  r4 cos  4  r2 cos  2
r3 sin  3  r1 sin  1  r4 sin  4  r2 sin  2
A  2 r1 r4 cos  1  2 r2 r4 cos  2
B  2 r1 r4 sin  1  2 r2 r4 sin  2
C  r1  r2  r4   r3
2
2
2
2
 2 r1 r2 cos  1 cos  2  sin  1 sin  2 
t 
B 
A  B C
2
AC
and
 4  2 tan
1
t 
2
2
Analytical Solution when θ2 is known
 Once all of the angular
quantities are known, it is
relatively straightforward to
compute the coordinates of any
point on the mechanism.
 In particular, the coordinates of
Q,P, and R are given by




rQ  r2  r2 cos  2 i  sin  2 j









rp  r2  r3  r2 cos  2 i  sin  2 j  r3 cos  3 i  sin  3 j





 
rp  r1  r4  r1 cos  1i  sin  1 j  r4 cos  4 i  sin  4 j






rp  r1  r1 cos  1i  sin  1 j








Analytical Solution when θ3 is known
 If the coupler angle θ3 is given, and θ2
and θ3 are to be determined, graphical
solution will require an iterative trial and
error solution.
 The analytical procedure, in contrast,
follows exactly the same procedure as
when θ2 is given. It starts by writing the
loop closure equations with θ2
replacing θ3 as the variable to be
eliminated.
 The equations set are of exactly the same
form except that the indices 2 and 3 are
interchanged. Therefore, we can use
directly the position solution derived for
the case of known θ2 while
interchanging the indices 2 and 3.
r3 cos  3  r1 cos  1  r4 cos  4  r2 cos  2
r3 sin  3  r1 sin  1  r4 sin  4  r2 sin  2
r2 cos  2  r1 cos  1  r4 cos  4  r3 cos  3
r2 sin  2  r1 sin  1  r4 sin  4  r3 sin  3
Analytical Solution when θ3 is known
 When the coupler angle θ3 is given,
there is an assembly-mode ambiguity
similar to that occurine when θ2 is
given.
 It is necessary to know the
before the analysis is begun;. The
mode is determined by the + or – sign
used for the square root term when
calculating t.
 Once the assembly mode is
determined, it remains the same for
any position of the input link unless
passes through a singular
(indeterminate) position .
r2 cos  2  r1 cos  1  r4 cos  4  r3 cos  3
r2 sin  2  r1 sin  1  r4 sin  4  r3 sin  3
Example:
Position Analysis of a Four Bar Linkage
 A four bar linkage with r1 = 1, r2 = 2, r3 = 3.5, r4 = 4, and θ1 = 0, find θ3 and
θ4 for each of the solution branches when the driving crank is in the
positions θ2 = 0, π/2, π,and - -π/2.
A  2 r1 r4 cos  1  2 r2 r4 cos  2
A  2 (1 )( 4 )  2 ( 2 )( 4 )   8
B  2 r1 r4 sin  1  2 r2 r4 sin  2  0
C  r1  r2  r4   r3
2
2
2
2
 2 r1 r2 cos  1 cos  2  sin  1 sin  2 
C  1  2  4  3 . 5  2 (1 )( 2 )  4 . 75
2
t 
t 
2
B 
2
2
A  B C
2
2
2
AC
0
  8 2
 4  2 tan
 0   4 . 75 
2
 8  4 . 75
1
  0 . 5049  
2
  0 . 5049
 53 . 58

Example:
Position Analysis of a Four Bar Linkage
 A four bar linkage with r1 = 1, r2 = 2, r3 = 3.5, r4 = 4, and θ1 = 0, find θ3 and θ4
for each of the solution branches when the driving crank is in the positions
θ2 = 0, π/2, π,and - -π/2.
POSITION ANALYSIS FOR A RIGID BODY WHEN
TWO POINTS ARE KNOWN
Position Analysis for a Rigid Body When
Two Points are Knows
 Given the kinematic properties of one
point on a rigid body and the angular
position, angular velocity, and angular
acceleration of the body, we can
compute the position, velocity, and
acceleration of any defined point on
the rigid body.
 For the rigid body shown. Assume that
A and B are two points attached to an
point is defined relative to the line
between points A and B by the angle β
and the distance rC/A, which is
represented as r6. Then the position of
point C can be computed directly if rA
and θ5 are known.
Position Analysis for a Rigid Body When
Two Points are Knows
 The position of point C is given as:






rC  rA  r6  rA  r6 cos  6 i  sin  6 j 
where
6    5
 If θ5 If is known, the equation above can be
used to calculate the location of point C directly.
 We often know the position vectors of two
points A and B on the rigid body. The value of θ5
can be calculated from the from the x and y
components of the position vectors for A and B
using
 5  tan
1
 rB y  rA y

r r
Ax
 Bx




POSITION ANALYSIS FOR A SLIDER-CRANK
MECHANISM
HW#2
(Prob. 4-6 and 4-7, with data in row (a)
Table P4-1).
(4-10, 4-12, 4-18(f))
Position Analysis for a Slider-Crank
Mechanism
 Next to the fourbar linkage, the
slider-crank is
probably the
most commonly
used mechanism.
 It appears in all
internal
combustion
engines and in
numerous
industrial and
household
devices.
Position Analysis for a Slider-Crank
Mechanism
 To develop the closure
equations, locate vectors r2
and r3 as was done in the
 One of the other two
vectors is taken in the
direction of the slider
velocity and the other is
taken perpendicular to the
velocity direction. The loop
closure equation is



 
rp  r2  r3  r1  r4
r3
r4
r2
rp
r1
Position Analysis for a Slider-Crank
Mechanism
 Writing the loop closure
equation in terms of the vector
angles, we obtain
r3


 
r2  r3  r1  r4




r2 u  2  r3 u  3  r1u  1  r4 u  4




r2 cos  2 i  sin  2 j  r3 cos  3 i  sin  3 j




 r1 cos  1i  sin  1 j  r4 cos  4 i  sin  4 j





r4
r2


rp
r1

where
 4  1   2
 Two scalar equations are
produced. The equations can
be solved for two unknowns.
r2 cos  2  r3 cos  3  r1 cos  1  r4 cos  4
r2 sin  2  r3 sin  3  r1 sin  1  r4 sin  4
Position Analysis for a Slider-Crank
Mechanism
 Unlike the four-bar linkage loop
lengths are known, the piston
displacement r1 is an unknown in the
slider-crank equation. The constraint
resulting from a known r1 is replaced
by the constraint θ4 = θ1 + π/2.
r2 cos  2  r3 cos  3  r1 cos  1  r4 cos  4
r2 sin  2  r3 sin  3  r1 sin  1  r4 sin  4
r3
 The following problem statements
are possible
 Crank angle θ2 given , find θ3
and r1
 Piston displacement r1 given,
find θ2 and θ3
 Coupler angle θ3 given, find θ2
and r1.
r4
r2
rp
r1
Analytical Solution when θ2 is known
 The analytical solution procedure
follows the same major steps as in the
four-bar linkage case. To eliminate θ3 ,
first isolate it in the loop closure
equations as follows:
r2 cos  2  r3 cos  3  r1 cos  1  r4 cos  4
r2 sin  2  r3 sin  3  r1 sin  1  r4 sin  4
r3
r3 cos  3  r1 cos  1  r4 cos  4  r2 cos  2
r3 sin  3  r1 sin  1  r4 sin  4  r2 sin  2
 Squaring both sides of both equations
and using the identity sin   cos   1
we obtain
2
r4
r2
rp
r1
2
r3  r1  r2  r4  2 r1 r4 cos  1 cos  4  sin  1 sin  4 
2
2
2
2
 2 r1 r2 cos  1 cos  2  sin  1 sin  2   2 r2 r4 cos  2 cos  4  sin  2 sin  4 
Analytical Solution when θ2 is known
r3  r1  r2  r4  2 r1 r4 cos  1 cos  4  sin  1 sin  4 
2
2
2
2
 2 r1 r2 cos  1 cos  2  sin  1 sin  2   2 r2 r4 cos  2 cos  4  sin  2 sin  4 
 The expression gives r1 in a
involving θ2 and the other
known variables. To obtain
a solution, collect together
the coefficients of the
different powers of r1 as
follows
r3
r4
r2
rp
r1
r1  Ar1  B  0
2
where
A  2 r4 cos  1 cos  4  sin  1 sin  4   2 r2 cos  1 cos  2  sin  1 sin  2 
B  r2  r4  r3  2 r2 r4 cos  2 cos  4  sin  2 sin  4 
2
2
2
Analytical Solution when θ2 is known
r1  Ar1  B  0
2
where
A  2 r4 cos  1 cos  4  sin  1 sin  4   r2 cos  1 cos  2  sin  1 sin  2 
B  r2  r4  r3  2 r2 r4 cos  2 cos  4  sin  2 sin  4 
2
2
2
 The expression for A could
be simplified by noting that
cos  1 cos  4
r3
 sin  1 sin  4 
r4
 cos(  1   4 )  cos(   2 )  0
giving
A   r2 cos  1 cos  2  sin  1 sin  2 
r2
rp
r1
Analytical Solution when θ2 is known
r1  Ar1  B  0
2
where
A   r2 cos  1 cos  2  sin  1 sin  2 
B  r2  r4  r3
2
2
2
 2 r2 r4 cos  2 cos  4  sin  2 sin  4 
 Solving for r1 gives
r1 
 A
A  4B
2
2
possible assembly modes
for the same θ2.
Analytical Solution when θ2 is known
 Because of the square
root in the expression for
r1, it becomes complex
when A2 < 4B.
 If this happens, the
mechanism cannot be
assembled in the
position specified.
A   r2 cos  1 cos  2  sin  1 sin  2 
B  r2  r4  r3
2
2
2
 2 r2 r4 cos  2 cos  4  sin  2 sin  4 
r1 
 A
A  4B
2
2
Analytical Solution when θ2 is known
A   r2 cos  1 cos  2  sin  1 sin  2 
 Once a value for r1 is determined, the
closure equations can be solved for
θ3 to give
B  r2  r4  r3
2
2
2
 2 r2 r4 cos  2 cos  4  sin  2 sin  4 
r3 cos  3  r1 cos  1  r4 cos  4  r2 cos  2
r3 sin  3  r1 sin  1  r4 sin  4  r2 sin  2
 3  tan
1
 r1 sin  1  r4 sin  4  r2 sin  2

 r1 cos  1  r4 cos  4  r2 cos  2
 As in the case of the four-bar linkage,
it is essential that the signs of the
numerator and denominator in the
above expression be maintained to
determine the quadrant in which the
angle θ3 lies.



r1 
 A
A  4B
2
2
r3
r4
r2
rp
r1
Analytical Solution when θ2 is known
A   r2 cos  1 cos  2  sin  1 sin  2 
 Once all of the angular quantities are
known, it is relatively straightforward
to compute the coordinates of any
point on the vector loops used in the
closure equations.
B  r2  r4  r3
2

r1 
 A
A  4B
2
2

r3







rp  r2  r3  r2 cos  2 i  sin  2 j  r3 cos  3 i  sin  3 j

2
 2 r2 r4 cos  2 cos  4  sin  2 sin  4 
 In particular, the coordinates of Q
and P are given by




rQ  r2  r2 cos  2 i  sin  2 j
2



r4
r2
rp
r1
Analytical Solution when r1 is known
 The analytical solution procedure
follows the same major steps as in the
previous case.
r1  r2  r4  r3  2 r1 r4 cos  1 cos  4  sin  1 sin  4 
2
2
2
2
 2 r1 r2 cos  1 cos  2  sin  1 sin  2   2 r2 r4 cos  2 cos  4  sin  2 sin  4   0
 After eliminating θ3 from the loop
closure equation, we simplify the
resulting equation as follows
r3
A cos  2  B sin  2  C  0
r2
 where
A   2 r1 r2 cos  1  2 r2 r4 cos  4
B   2 r1 r2 sin  1  2 r2 r4 sin  4
C  r1  r2  r4  r3  2 r1 r4 cos  1 cos  4  sin  1 sin  4 
2
2
2
2
r4
rp
r1
Analytical Solution when r1 is known
A cos  2  B sin  2  C  0
 The trigonometric half-angle identities
can be used to solve the equation above
for θ2 . Using these identities and
simplifying gives
C
 A t  2 Bt  ( A  C )  0
A   2 r1 r2 cos  1  2 r2 r4 cos  4
B   2 r1 r2 sin  1  2 r2 r4 sin  4
C  r1  r2  r4  r3
2
2
2
 2 r1 r4 cos  1 cos  4  sin  1 sin  4 
sin  2 
2
cos  2 
where
2
t  tan  2 2 
2 tan  2 2 
 2
2
 2
2
tan  2
2
1  tan
1  tan
1
2
2
2
 Solving for t gives
t 
t 
r3
4 B  4 C  A C  A 
 2B 
2
2 C  A 
B 
A  B C
2
2
AC
and  2  2 tan
1
t 
2
r2
r4
rp
r1
Analytical Solution when r1 is known
A   2 r1 r2 cos  1  2 r2 r4 cos  4
t 
t 
4 B  4 C  A C  A 
 2B 
2
2 C  A 
B 
A  B C
2
2
2
AC
and  2  2 tan
1
t 
assembly modes. Typically, there
are two valid solutions for θ2.
 Because tan-1 has a valid range
of π/2 ≤ tan-1 (t) ≤ –π/2, θ2 will
have the range –π ≤ θ2 ≤ –π.
B   2 r1 r2 sin  1  2 r2 r4 sin  4
C  r1  r2  r4  r3
2
2
2
2
 2 r1 r4 cos  1 cos  4  sin  1 sin  4 
Analytical Solution when r1 is known
A   2 r1 r2 cos  1  2 r2 r4 cos  4
t 
t 
4 B  4 C  A C  A 
 2B 
2
2 C  A 
B 
A  B C
2
2
2
AC
and  2  2 tan
1
t 
 Because of the square root
in the equation for the
variable t , it becomes
complex when (A2 + B2) < C2
 If this happens, the
mechanism cannot be
assembled for the specified
value of r1 .
B   2 r1 r2 sin  1  2 r2 r4 sin  4
C  r1  r2  r4  r3
2
2
2
2
 2 r1 r4 cos  1 cos  4  sin  1 sin  4 
Analytical Solution when r1 is known
A   2 r1 r2 cos  1  2 r2 r4 cos  4
t 
t 
4 B  4 C  A C  A 
 2B 
2
2 C  A 
B 
A  B C
2
2
2
AC
and  2  2 tan
1
t 
 Knowing θ2 , the closure
equations can be solved
for θ3. As in the previous
cases, it is essential that
the signs of the numerator
and denominator be
maintained to determine
angle θ3 lies.
B   2 r1 r2 sin  1  2 r2 r4 sin  4
C  r1  r2  r4  r3
2
2
2
2
 2 r1 r4 cos  1 cos  4  sin  1 sin  4 
Analytical Solution when θ3 is known
 When the coupler angle θ3 is the input link,
the analytical procedure for solving the
position equations follows the same major
steps as when θ2 is the input.
 we can assume that θ1 , θ3, θ3 are known and
that θ2 and r1 are to be found. The link
lengths r2 and r2 and the angles θ1, and θ4 are
constants.
 For the position analysis, again begin with the
loop closure equations and isolate the terms
with θ2 The resulting equations of the same
form obtained when θ2 is the input except
r2 cos  3  r1 cos  1  r4 cos  4  r3 cos  2
that the indices 2 and 3 are interchanged.
r2 sin  3  r1 sin  1  r4 sin  4  r3 sin  2
Therefore, we can use directly the position
solution derived for the case when θ2 is the
input and interchange the indices 2 and 3.
POSITION ANALYSIS FOR AN INVERTED
SLIDER-CRANK MECHANISM
Position Analysis for an Inverted SliderCrank Mechanism
 An inversion of the fourbar
which the sliding joint is
between links 3 and 4 is
shown. The slider block
point O4.
 Note that the angle
fixed and is equal to π/2,
and that link 1 is assumed
to be aligned with the xaxis of the reference
coordinate, θ1 = 0.
r3
r2
r1
r4
Position Analysis for an Inverted SliderCrank Mechanism
 To develop the closure equations,
locate vectors r2 and r3 as was
done in the original slider-crank
linkage. The magnitude of r3 is
variable depending on the
location of the slider.
 Vectors r1 is taken in the direction
of the ground link, and vector r4 is
taken perpendicular to r3. The
loop closure equation is
 
 
r2  r3  r1  r4
r3
r2
r1
r4
Position Analysis for an Inverted SliderCrank Mechanism
 Writing the loop closure
equation in terms of the vector
angles, we obtain


 
r2  r3  r1  r4




r2 u  2  r3 u  3  r1u  1  r4 u  4




r2 cos  2 i  sin  2 j  r3 cos  3 i  sin  3 j




 r1 cos  1i  sin  1 j  r4 cos  4 i  sin  4 j





r3



r2
r1
r4
where
4  3   2
 Two scalar equations are
produced. The equations can
be solved for two unknowns.
r2 cos  2  r3 cos  3  r1 cos  1  r4 cos  4
r2 sin  2  r3 sin  3  r1 sin  1  r4 sin  4
Position Analysis for an Inverted SliderCrank Mechanism
 The piston displacement r3 is an
unknown in the equation. The
constraint resulting from a known r3
is replaced by the constraint θ4 = θ3
+ π/2.
cos  4   sin  3
cos  1  1
sin  4  cos  3
sin  1  0
 The following problem statements
are possible
 Crank angle θ2 given , find θ4
and r3
 Piston displacement r3 given,
find θ2 and θ4
 Output angle θ4 given, find θ2
and r3.
r3
r2
r4
r1
r2 cos  2  r3 sin  4  r1  r4 cos  4
r2 sin  2  r3 cos  4  r4 sin  4
Analytical Solution when θ2 is known
 The analytical solution procedure
follows the same major steps as in the
non-inverted slider crank. To eliminate
θ4 , first isolate it in the loop closure
equations as follows:
r2 cos  2  r3 sin  4  r1  r4 cos  4
r2 sin  2  r3 cos  4  r4 sin  4
r4 cos  4  r3 sin  4  r2 cos  2  r1
r4 sin  4  r3 cos  4  r2 sin  2
r3
 Squaring both sides of both equations
and using the identity sin   cos   1
we obtain
2
r4  r3  r1  r2  2 r1 r2 cos  2
2
2
2
2
2
r2
r1
r4
Analytical Solution when θ2 is known
 The equation can be solved for
r3, which is the one unknown.
 Substituting the resulting r3
value in the first closure
equation, we obtain an
expression with only θ3
unknown
r2 cos  2  r3 sin  4  r1  r4 cos  4
r2 sin  2  r3 cos  4  r4 sin  4
r4  r3  r1  r2  2 r1 r2 cos  2
2
r2 cos  2  r3 sin  4  r1  r4 cos  4
 The above expression can be
solve using the previous half
angle identities
2
2
2
r3
r2
r1
r4
Trigonometric Solution when θ2 is known
 The problem could be
solved using
trignonometry by
solving first for the
the triangle O2AO4, of
which two sides and an
angle are known.
O4A and the triangle
ABO4 the length r3 and
θ4 could be found.
r3
r2
r4
r1
O 4 A 2
 r1  r2  2 r1 r2 cos  2
O 4 A 2
 r1  r3
2
2
2
2
Analytical Solution when θ2 is known
r1  r2  r3  r4  2 r2 r3 cos  2 cos  3  sin  2 sin  3 
2
2
2
2
 2 r1 r2 cos  1 cos  2  sin  1 sin  2   2 r1 r3 cos  1 cos  3  sin  1 sin  3   0
 The expression can be seen to be the
same as that obtained for a noninverted slider-crank when r1 is known
with interchanging the indices 1 and 2
and the indices 3 and 4.
r3
r2
r1
r4
TRANSMISSION ANGLES
Analytical Solution when θ3 is known
 The transmission angle μ is defined
as the angle between the output
link and the coupler of a four-bar
mechanism. It is usually taken as
the absolute value of the acute
angle of the pair of angles at the
intersection of the two links and