Linkages
FUNDAMENTALS Topic 4
Gerald Rothenhofer
9/21/2009
What is a Linkage?
• A mechanical linkage is a series of rigid links connected with joints
to form a closed chain, or a series of closed chains. Each link has
two or more joints, and the joints have various degrees of freedom
to allow motion between the links. A linkage is called a mechanism if
two or more links are movable with respect to a fixed link.
Mechanical linkages are usually designed to take an input and
produce a different output, altering the motion, velocity, acceleration,
and applying mechanical advantage.
• A linkage designed to be stationary is called a structure.
en.wikipedia.org/wiki/Linkage_(mechanical)
History
• Leonardo da Vinci (1452, 1519), Codex Madrid I.
• Industrial Revolution was the boom age of linkages: cloth
making, power conversion, speed regulation, mechanical
computation, typewriting and machining
Linkages Today
• In many applications (typewriting) linkages have been replaced by
electronics.
• Still linkages can have a cost advantage over electronic solutions:
Couple different outputs by a mechanism rather than using one
motor per output and electronics to achieve the coupling.
• Current applications: Sports Equipment, Automotive (HVAC
modules), Precision Machinery (Compliant Mechanisms), Medical
Devices
Linkage Categorization
• Planar Linkages
– Three bar
– Four bar
• Slider Crank
– Five bar
– Six bar
– …be creative…
• Spatial Linkages
Degrees of Freedom
• Planar Linkages:
– F=3*(N-1)-2*J1-Jh
•
•
•
•
F – total degrees of freedom
N – number of links
J1 – constraints by 1DOF joints
Jh – constraints by 2DOF joints
Four Bar
• Grashof: The sum of the shortest (S) and longest (L) links of a
planar four-bar linkage must be smaller than the sum of the
remaining two links (P, Q). In this case the shortest link can rotate
360degree relative to the longest link.
• L + S < P + Q: crank-rocker, double-crank, rocker-crank, doublerocker
• L + S = P + Q: crank-rocker, double-crank, rocker-crank, doublerocker,  note: linkage can change its closure in singularity
positions (all links aligned)
• If L + S > P + Q, double-rocker, no continuous rotation of any link
Transmission Angles – Four Bar
C
B
A
γ
D
•
•
•
•
A – ground link
B – input link
C – coupler
D – output link
• Angle between coupler and output link
should be 40º≤γ≤140º
 Zero torque at output link if γ=0º or
γ=180º
Transmission Angles – Slider Crank
y
Θ2
b
a
α
T
Θ1
x
• Especially important in critical position such as within the
main working range or high load positions
• Minimize α
• Minimize |θ2-90°|
• No stick condition: 1/tan(α)<μ
Four Bar Synthesis
C
D
B
A
•
•
•
•
A – ground link
B – input link
C – coupler
D – output link
• Function Generation (input/output relation)
• Line Path Generation (line on coupler)
• Point Path Generation (coupler point)
Four Bar Two Position Synthesis
Four Bar Three Position Synthesis
Cognate Mechanisms
• Provide identical
motion of a point
or link
• Here: coupler
point cognate
Four Bar Function Generation
• Two angular displacements
• Only one initial position;
either primary or secondary
side can be chosen freely
(here 60°)
• E.g.:
– Primary side moves by
2x 20 °
– Secondary side moves
by 35°+30°
Crank Rocker Design
• Design in extreme positions
• Typically design for crank
movement >180º
depending on required
transmission ratio i.e. rocker
should move slowly when
load is heavy, the return fast
• In this example rocker
moves through 60º while the
crank moves through
180º+10º=190º
Slider Crank Synthesis I
• Two point synthesis
Slider Crank Synthesis II
• Three point
synthesis by
geometrical
inversion
Other Basic Four Bar Design
Methods
• Approximate function generation
• Approximate coupler point path generation
– Uncorrelated with input
– Correlated with input
• Slider crank synthesis by approximation
What is possible with advanced
design methods?
• Four coupler position synthesis
• In some cases five coupler position synthesis is
achievable
• Straight line motion
• Complex linkages (more than four bars)
• Spatial linkages
Five Bar
• How many degrees of freedom?
• Why does it work?
Six Bar
• Watt & Stephenson Linkages
• approximate dwells or better MTB suspension
Four Bar Analysis I
lawof cosine:
rp 2 =a 2 +b 2 -2abcos(180-1+ )
b
ψ
rp 2 =c 2 +d 2 -2cdcos(180- 2 )
a 2 +b 2 +2abcos( 1 - )=d 2 +c 2 +2cdcos(  2 )
c sin( 2 )  a sin(1 )
sin( ) 
b
d  c cos( 2 )  a sin(1 )
cos ( ) 
b
aftersome math :
ey
rp
c
a
Θ1
d
Θ2
ex
a 2  b 2  c 2  d 2  2ad cos(1 )  2cd cos( 2 )  2ac cos(1   2 )  0
take time derivative of (), assume 1  1  const ,  2  2  const
2ad 1 sin(1 )  2cd2 sin( 2 )  2ac(1  2 ) sin(1   2 )  0
transmission ratio :
 T ad sin(1 )  ac sin(1   2 )
r 2  1 
1 T2 cd sin( 2 )  ac sin(1   2 )
• Θ2 is a complicated trigonometric function of Θ1, Θ2=f(Θ1)
Four Bar Transmission Ratio
a 2  b 2  c 2  d 2  2ad cos(1 )  2cd cos( 2 )  2ac cos(1   2 )  0
substitute :
K1  a 2  b 2  c 2  2ad cos( 1)
K 2  2cd  2ac cos( 1)
K 3  2ac sin( 1)
 K1  K 2 cos( 2)  K 3sin( 2)  0
two solutions :
K1 K 3 2 +K 2 (K 3 2 (K 2 2 -K 12 +K 3 2 ))0.5 -K 2 K1+(K 3 2 (K 2 2 -K 12 +K 3 2 ))0.5
 21  atan(,
)
K 3 (K 2 2 +K 3 2 )
(K 2 2 +K 3 2 )
-K1 K 3 2 +K 2 (K 3 2 (K 2 2 -K 12 +K 3 2 ))0.5 K 2 K 1+(K 3 2 (K 2 2 -K 12 +K 3 2 ))0.5
 22 =atan(
,)
K 3 (K 2 2 +K 3 2 )
(K 2 2 +K 3 2 )
transmission ratio :
 T ad sin(1 )  ac sin(1   2 )
r 2  1 
1 T2 cd sin( 2 )  ac sin(1   2 )
Slider Crank Analysis
x  a cos(1 )  b cos( 2 )
y  a sin(1 )  b sin( 2 )
( x  x0 ) 2  ( y  y0 ) 2  R 2
a 2  b 2  x0 2  y0 2  R 2  2ab cos(1   2 )  2[ x0 (a cos(1 )  b cos( 2 ))  y0 (a sin(1 )  b sin( 2 ))]  0
substitute :
Q0  a  b  x0  y0  R
2
2
2
2
b
2
a
Q1  Q0  2 y0 a sin(1 )  2 x0 a cos(1 )
F, V
Q2  2ab cos(1 )  2 x0b
R
Q3  2ab sin(1 )  2 y0b
T1, ω1
 Q1  Q2 cos( 2 )  Q3 sin( 2 )  0
two solutions :
Q1Q3 2 +Q2 (Q3 2 (Q2 2 -Q12 +Q3 2 ))0.5 -Q2Q1+(Q3 2 (Q2 2 -Q12 +Q3 2 ))0.5
 21  atan(,
)
Q3 (Q2 2 +Q3 2 )
(Q2 2 +Q3 2 )
 22 =atan(
2
θ2
2
2
2
2
0.5
-Q1Q3 +Q2 (Q3 (Q2 -Q1 +Q3 ))
Q3 (Q2 2 +Q3 2 )
2
,-
2
2
2
0.5
θ1
φ
y0
x0
Q2Q1+(Q3 (Q2 -Q1 +Q3 ))
)
(Q2 2 +Q3 2 )
transmission ratio :
a sin(1   2 )
V T1
r
 R 2
2 0.5
1 F
(x0 +y0 ) sin(  2   )  a sin(1   2 )
• F is maximum available force (no friction or other loads
taken into account)
Slider Crank Transmission Ratio
transmission ratio :
b
a
T
a sin(1   2 )
r
 1  R 2 2 0.5
1 F
(x0 +y0 ) sin(  2   )  a sin(1   2 )
θ2
F, V
V
R
T1, ω1
θ1
φ
y0
x0
• The transmission ratio determines the relation of slider
(flapper) position and motor angle.
Power Budget
Pout  Pin
Pin  Tinin  Tin
Pout  Fload
l
t
 in
t
and therefore
Fload  l  Tin in
y
b
a
δl
δΘin
Tin
x
Fload  F friction  Fpressure  Fothers
( F friction  Fpressure  Fothers ) l  Tin in
• Assumes that load forces are constant
• Average and max load forces should be used to check
for safety factors
• δl  arc length between starting point and end point of
slider movement
Slider Friction I
y
at the slider,
forces can only be applied
in the direction of the link:
R  F,
could also be obtained by
Θ2
b
a force balance in x- and y- directions
moment balance around crank pivot:
a
α
T
Θ1
cos( 2 )   a cos(1 ) 
Tez  F  sin( 2 )    a sin(1 ) 
 0  

0
F  F cos( )
x
F
y
F  F sin( )
Θ2
F
after some trigonometry :
T cos( )
F 
a sin(1   2 )
b
a
α
T
Θ1
x
R
F
F 
T sin( )
a sin(1   2 )
Ffriction   F  
T sin( )
a sin(1   2 )
note :  2 is a function of  1,  is a function of 1
 2 =f( 1 ),  =g( 1 )
Slider Friction II
y
Θ2
a
F
α
T
Θ1
x
F┴
F
1

F tan( )
Ffriction   F
sticking if : F   F
and therefore at any point along the slider track it must be :
1

tan( )
F║
Questions