Graphical Linkage

SynthesisChapter 3

Synthesis

• Qualitative Synthesis

– The creation of potential solution in

the absence of a well-defined

algorithm that configures or predicts

the solution

• Type Synthesis

– The definition of the proper

mechanism best suitable to the

problem

Synthesis• Quantitative Synthesis

– Generation of one or more solution of

a particular type that you know to be

suitable to the problem, and

importantly, one for which there is a

synthesis algorithm defined

• Dimensional Synthesis

– Is the determination of the

proportions (length) of the links

necessary to accomplish the desired

motions

Function, Path and Motion Generation• Function

– Defined as the correlation of an input

motion with an output motion in a

mechanism

• Path

– Control of a point in the plane such that is

follows some prescribed path

• Motion Generation

– Control of a line in the plane such that is

follows some prescribed set of sequential

positions

Limiting Conditions• Toggle/ Stationary Position

Limiting Conditions• Toggle/Stationary Position

Limiting Conditions• Transmission angle

– Angle between the output link and the

coupler

Dimensional Synthesis• The determination of the proportions

(lengths) of the links necessary to

accomplish the desired motions

• Two-Position synthesis

– Rocker

– Coupler

Example 1• Rocker Output- Two Position with

Angular Displacement (Function)

– Design a four bar Grashof crank-rocker

speed motor input to give 45° of rocker

motion with equal time forward and back,

from a constant speed motor input.

Example 1• 1. Draw the output

link O4B in both

extreme positions,

B1 and B2 in any

convenient location,

such that the

desired angle of

motion θ4 is

subtended.

• 2. Draw the chord

B1B2 and extended

it in either direction.

• 3. Select a

convenient point O2

on the line B1B2

extended.

• 4. Bisect line

segment B1B2, draw

a circle of that

radius about O2.

• 5. Label the two

intersection of the

circle and B1B2

extended, A1 and

A2.

• 6. Measure the

length of the coupler

as A1 to B1 or A2 to

B2.

• 7. Measure ground

length 1, crank

length 2, and rocker

length 4.

• 8. Find the Grashof

condition. If non-

Grashof, redo steps

3 to 8 with O2 further

from O4.

• 9. Make model of

the linkage and

check its function

and transmission

angles.

Example 1

• 10. You can input

the file F03-04.4br

Example 2• Rocker Output- Two Position with

Complex Displacement (Motion)

– Design a fourbar linkage to move link CD

from C1D1 to C2D2.

Example 2• 1. Draw the link CD

in its two desired

positions, C1D1 and

C2D2 in plane as

shown.

Example • 2. Draw

construction line

from point C1 to C2

and from point D1 to

D2.

Example • 3. Bisect line C1C2

and line D1D2 and

extend their

perpendicular

bisectors to

intersect at O4.

Their intersection is

the rotopole.

Example • 4. Select a

convenient radius

and draw an arc

about the rotopole

to intersect both

lines O4C1 and

O4C2. Label the

intersection B1 and

B2.

Example • 5. Do steps 2 to 8 of

example 1 to

complete the

linkage.

• 6. Make a model of

the linkage and

articulate it to check

its function and its

transmission

angles.

Example 2 • 2. Draw the chord B1B2 and extended it in either

direction.

• 3. Select a convenient point O2 on the line B1B2

extended.

• 4. Bisect line segment B1B2, draw a circle of that

radius about O2.

• 5. Label the two intersection of the circle and B1B2

extended, A1 A2.

• 6. Measure the length of the coupler as A1 to B1 or A2

to B2.

• 7. Measure ground length 1, crank length 2, and

rocker length 4.

• 8. Find the Grashof condition. If non-Grashof, redo

steps 3 to 8 with O2 further from O4.

Example 3• Coupler Output- Two Position with

Complex Displacement (Motion)

– Design a fourbar linkage to move link CD

from C1D1 to C2D2 (with moving pivots at

C and D).

Example 3• 1. Draw the link CD

in its two desired

positions, C1D1 and

C2D2 in plane as

shown.

Example • 2. Draw

construction line

from point C1 to C2

and from point D1 to

D2.

Example • 3. Bisect line C1C2

and line D1D2 and

extend their

perpendicular

bisectors in

convenient

directions.The

rotopole will not be

used in this solution.

Example • 4. Select any

convenient point on

each bisector as the

fixed pivots O2 and

O4 , respectively.

Example • 5. Connect O2 with C1

and call it link 2.

Connect O4 with D1 and

call it link 4.

• 6. Line C1D1 is link 3.

Line O2O4 is link 1.

• 7. Check the Grashof

condition, and repeat

steps 4 to 7 if

unsatisfied. Note that

any Grashof condition

is potentially acceptable

in this case.

Example • 8. Construct a model

and check its function

to be sure it can get

from the initial to final

position without

encountering ant limit

(toggle) positions.

• Check the transmission

angles.

Example • Review Example 3-4

– Design a dyad to control

and limits the extremes

of motion of the linkages

in the previous example

to its two design

positions

Three-Position Synthesis• Example 1 – Coupler

Output – 3 position with

Complex Displacement

– Design a fourbar linkage to

move the link CD shown

from position C1D1 to C2D2

and then to position C3D3.

Moving pivots are C and D.

Find the fixed pivot locations.

Three-Position Synthesis• 1. Draw link CD in its three position

C1D1, C2D2 , C3D3 in the plane as shown.

Three-Position Synthesis• 2. Draw construction lines from point C1 to C2

and from C2 to C3.

Three-Position Synthesis• 3. Bisect line C1C2 and line C2C3 and extend

their perpendicular bisector until they

intersect. Label their intersection O2.

Three-Position Synthesis• 4. Repeat steps 2 and 3 for lines D1D2 and

D2D3. Label the intersection O4.

Three-Position Synthesis• 5. Connect O2 with C1 and call link 2. Connect

O4 with D1 and call link 4.

Three-Position Synthesis• 5. Line C1D1 is link 3. Line O2O4 is link 1.

Three-Position Synthesis• 7. Check the Grashof

condition. Note that any

Grashof condition is

potentially acceptable in this

case.

• 8.Construct a model and

check its function to be sure it

can get from initial to final

position without encountering

any limits positions.

• 9.Construct a driver dyad

using an extension of link 3

attach the dyad.

Three-Position Synthesis – Example 2• Example 2 – Coupler

Output – 3 position with

Complex Displacement –

Alternate Attachment Points

for Moving Pivots

– Design a fourbar linkage to

move the link CD shown

from position C1D1 to C2D2

and then to position C3D3.

Use different moving pivot

than CD. Find the fixed pivot

locations.

Three-Position Synthesis – Example 2• 1. Draw link CD in its three

position C1D1, C2D2 , C3D3 in the

plane as shown.

• 2. Define new attachment points

E1 and F1 that have a fixed

relationship between C1D1 and

E1F1 within the link. Now use

E1F1 to define the three position

of the link.

• 3. Draw construction lines from

point E1 to E2 and from E2 to E3.

Example 2• 4. Bisect line E1E2 and line E2E3

and extend their perpendicular

bisector until they intersect.

Label their intersection O2.

• 5. Repeat steps 2 and 3 for lines

F1F2 and F2F3. Label the

intersection O4.

• 6. Connect O2 with E1 and call

link 2. Connect O4 with F1 and

call link 4.

• 7. Line E1F1 is link 3. Line O2O4 is

link 1

Example 2• 8. Check the Grashof

condition. Note that any

Grashof condition is

potentially acceptable in this

case.

• 9.Construct a model and

check its function to be sure it

can get from initial to final

position without encountering

any limits positions. If not,

change locations of point E

and F and repeat steps 3 to 9.

Three-Position Synthesis – Example 3• Example 3 – Three –Position Synthesis

with Specified Fixed Pivots - Inverting

the 3-position problem

– Invert a fourbar linkage which move the

link CD shown from position C1D1 to C2D2

and then to position C3D3. Use specified

fixed pivots O2 and O4.

Example 3• 1. Draw link CD in its three position C1D1,

C2D2 , C3D3 in the plane as shown.

• 2. Draw the ground link O2O4 in its desired

position in the plane with respect to the first

coupler position C1D1

Example 3• 3. Draw construction arc from point C2 to O2

and from D2 to O2 whose radii define the side

of triangle C2O2D2 This define the

relationship of the fixed pivot O2 to the

coupler line CD in the second coupler

position.

Example 3• 4. Draw construction arc from point C2 to O4

and from D2 to O4 whose radii define the side

of triangle C2O4D2 This define the

relationship of the fixed pivot O4 to the

coupler line CD in the second coupler

position.

Example 3• 5. Now transfer this relationship back

to the first coupler position C1D1 so

that the ground plane position O2’O4’

bears the same relationship to C1D1

as O2O4 bore to the second coupler

position C2D2. In effect, you are

sliding C2 along the dotted line C2C1

and D2 along the dotted D2D1. By

doing this, we have pretended that

ground plane moved from O2O4 to

O2’O4’ instead of the coupler moving

from C1D1 to C2D2. We have inverted

the problem.

Example 3• 6. Repeat the process for the third coupler

position as shown in the figure and transfer

the third relative ground link position to the

first, or reference, position.

Example 3• 7. The three inverted position of the ground

plane that correspond to the three desired

coupler positions are labeled O2O4,O2’O4’ ,

and O2’’O4’’ and have also been renamed

E1F1, E2F2 and E3F3 as shown in the figure

Three-Position Synthesis – Example 4• Example 4 – Finding the Moving Pivots

for Three Positions and Specified Fixed

Pivots

– Design a fourbar linkage to move the link

CD shown from position C1D1 to C2D2 and

then to position C3D3. Use specified fixed

pivots O2 and O4. Find the required moving

pivot location on the coupler by inversion.

Example 4• 1. Start with inverted three positions plane as

shown in the figures. Lines E1F1, E2F2 and

E3F3 define the three positions of the inverted

link to be moved.

Example 4• 2. Draw construction lines from point E1 to E2

and from point E2 to E3.

Example 4• 3. Bisect line E1E2 and line E2E3 extend the

perpendicular bisector until they intersect.

Label the intersection G.

Example 4• 4. Repeat steps 2 and 3 for lines F1F2 and

line F2F3. Label the intersection H.

Example 4• 5. Connect G with E1 and call it link 2.

Connect H with F1 and call it link 4.

Example 4• 6. In this inverted linkage, line E1F1 is the

coupler, link 3. Line GH is the “ground” link1.

Example 4• 7. We must now reinvert the linkage to return to

the original arrangement. Line E1F1 is really the

ground O2O4 and GH is really the coupler. The

figure shows the reinversion of the linkage in

which points G and H are now the moving pivots

on the coupler and E1F1 has resumed its real

identity as ground link O2O4.

Example 4• 8. The figure reintroduces the original line C1D1

in its correct relationship to line O2O4 at the initial

position as shown in the original example 3. This

form the required coupler plane and defines a

minimal shape of link 3.

Example 4• 9. The angular motions required to

reach the second and third position

of line CD shown in the figure are

the same as those defined in figure

b for the linkage inversion. The

angle F1HF2 in the figure b is the

same as the angle H1O4H2 in the

figure and F2HF3 is the same as

angle H2O4H3. The angular

excursions of link 2 retain the same

between figure b and e as well. The

angular motions of links 2 and 4 are

the same for both inversion as the

link excursions are relative to one

another.

Example 4

Example 4• 10. Check the Grashof condition. Note that any

Grashof condition is potentially acceptable in this

case provided that the linkage has mobility

among all three position. This solution is a non-

Grashof linkage.

• 11. Construct a model and check its function to

be sure it can get from initial to final position

without encountering any limit (toggle) positions.

In this case link 3 and 4 reach a toggle position

between points H1 and H2. This means that this

linkage cannot be driven from link 2 as it will

hang up at that toggle position. It must driven

from link 4.

Instructional Videos• Professor Robert Norton’s Instructional

Videos are included on the book’s

DVD. Please watch the following

videos;

– Quick-Return Mechanisms

– Coupler Curves

– Cognates

– Parallel Motion

– Dwell Mechanisms

Quick – Return Mechanism• Fourbar Quick-Return

– Time ratio (TR) defines the degree of quick-return

of the linkage.

– Works well for time ratios down to about 1.5

RT 360

180180

Quick-Return– Example 1• Example 1 – Fourbar

Crank-Rocker Quick-

Return Linkage for

Specified Time Ratio

– Redesign Example 1 (two

position) to provide a time

1:125 with 45° output

rocker motion.

Example 1• 1. Draw the output link O4B in both extreme

position, in any convenient location, such that

the desired angle of motion, θ4 , is

subtended.

Example 1• 2. Calculate α, β, δ using the

equations. In this example

α=160°, β=200°, δ=20°.

• 3. Draw a construction line

through point B1 at any

convenient angle.

• 4. Draw a construction line

through point B2 at an angle δ

from the first line.

• 5. Label the intersection of the

two construction line O2.

• 6. The line O2O4 define ground.

Example 1• 7. Calculate the length of crank

and coupler by measuring O2B1

and O2B2 and solve

simultaneously;

coupler + crank=O2B1

coupler - crank=O2B2

or you can construct the crank

length by swinging an arc centered

at O2 from B1 to cut line O2B2

extended. Label that intersection

B1’. The line B2B1’ is twice the

crank length. Bisect this line

segment to measure crank length

O2A1.

Example 1• 8. Calculate the Grashof condition.

If non-Grashof, repeat steps 3 to 8

with O2 further O4.

• 9. Make a model of the linkage

and articulate it to check its

function.

• 10. Check the transmission angle.

Quick-Return– Example 2• Example 2 – Sixbar

Drag Link Quick-

Return Linkage for

Specified Time Ratio

– Provide a time ratio of

1:1.4 with 90 degree

rocker motion

Example 2• 1. Calculate α, β, δ using the

equations. In this example

α=150°, β=210°.

• 2. Draw a line of centers XX at

any convenient location.

• 3. Choose a crank pivot location

O2 on line XX and draw an axis

YY perpendicular to XX.

Example 2• 4. Draw a line of convenient

radius O2A about center O2

• 5. Lay out angle α with vertex at

O2, symmetrical about quadrant

one

• 6. Label points A1 and A2 at the

intersection of the lines

subtending angle α and the

circle of radius O2A.

Example 2• 7. Set the compass to a

convenient radius AC long

enough to cut XX in two places

on either side of O2 when

swung from A1 and A2. Label

the intersection C1 and C2.

• 8. The line O2A1 id the driver

crank, link 2, and line A1C1 is

the coupler, link 3.

• 9. The distance C1C2 is twice

the driven (dragged) crank

length. Bisect it to locate the

fixed pivot O4.

Example 2• 10. The line O2O4 now defines

the ground link. Link O4C1 is

the driven crank, link 4.

• 11. Calculate the Grashof

condition. If Non-Grashof,

repeat steps 7-11 with a shorter

radius in step 7.

Example 2• 12. invert the method of example 1 (two

positions) to create the output dyad using XX

as the chord and O4C1 as the driving crank.

The point B1 and B2 will lie on line XX and be

spaced apart a distance 2O4C1. The pivot O6

will lie on the perpendicular bisector of B1B2,

at a distance from line XX which subtends the

specified output rocker angle.

• 13. Check the transmission angle.

Example 2

Coupler Curves

Coupler Curves

Cognates• Roberts-Chebyschev

– Three different planar fourbar linkages will

trace identical coupler curves.

• Cayley Diagram: Construction lines parallel to

all sides of the links

• Roberts Diagram: Three fourbars linkage

cognates which shares the same coupler curve

Cognates- Cayley

Cognates- Roberts

Cognates- Parallel Motion

Cognates- Parallel Motion

Cognates- Geared Fivebar Cognates• Chebyschev discovered that any

fourbar coupler curve can be duplicated

with a geared fivebar mechanism

whose gear ratio is plus one, meaning

that the gears turn with the same speed

and direction.

GFBM

Straight-Line Mechanisms• Watt’s

Straight-Line Mechanisms• Roberts

Straight-Line Mechanisms

• Chebyshev

Straight-Line Mechanisms

• Hoeken

Straight-Line Mechanisms

• Evans

Straight-Line Mechanisms

• Peaucellier

Dwell Mechanisms• Is defined as zero output for some

nonzero input motion

– Single-Dwell: Design a sixbar linkage for

90° rocker motion over 300 crank degree

with dwell for the remaining 60°

Dwell Mechanisms-Example 1• 1. Search the H&N atlas for a

fourbar linkage with a coupler

curve having an approximate

(pseudo) circle arc portion

which occupies 60° of crank

motion (12 dashes). The

chosen fourlink is shown in the

figure.

• 2. Lay out this linkage to scale

including the coupler curve and

find the approximate center of

the chosen coupler curve using

graphical techniques.

Example 1• 2. To do so, draw the chord of the arc and

construct its perpendicular bisector. The center

lie on this bisector. Find it by striking arcs with

your compass point on the bisector, while

adjusting the radius to get best fit to the coupler.

Label the arc center D.

Example 1• 3. Your compass should now be

set to the approximate radius of

the coupler arc. This will be the

length of link 5 which is to be

attached at the coupler point P.

• 4. Trace the coupler curve with

the compass point, while keeping

the compass pencil lead on the

perpendicular bisector, and find

the extreme location along the

bisector that the compass lead

will reach. Label this point E.

Example 1• 5. The line segment DE

represents the maximum

displacement that a link of length

PD, attached at P, will reach

along the bisector.

• 6. Construct a perpendicular

bisector of the line segment DE,

and extend it in a convenient

direction.

Example 1• 7. Locate fixed pivot O6 on the

bisector of DE such line O6D and

O6E subtend the desired output

angle, 90°.

• 8. Draw link 6 from D (or E)

through O6 and extend to any

convenient length. This is the

output link which will dwell for

specified portion of the crank

cycle.

• Check the transmission angles.

Example 1– Single-Dwell: Design a sixbar linkage for

90° rocker motion over 300 crank degree

with dwell for the remaining 60°

Dwell Mechanisms-Example 2– Double-Dwell: Design a sixbar linkage for

80° rocker motion over 20 crank degree

with dwell for the remaining 160°, return

motion over 140° and second dwell for 40°.

Example 2• 1. Search the H&N atlas for a

linkage with a coupler curve

having two approximate straight-

line portions. One should occupy

160° of crank motion (32 dashes),

and the second 40° of crank

motion (8 dashes). This is a

wedge-shaped curve.

• 2. Lay out this linkage to scale

including the coupler curve and

find the intersection of two tangent

lines colinear with the straight

segment. Label this point O6.

Example 2• 3. Design link 6 to lie along

these straight tangent, pivoted

at O6. Provide a slot in link 6 to

accommodate slider block 5.

• 4. Connect slider block 5 to the

coupler point P on link 3 with a

pin joint. The finish sixbar is

shown in the next figure.

• Check transmission angle.

Example 2– Double-Dwell