solution manual Design of Machinery: An Introduction to the Synthesis and Analysis of Mechanisms and Machines Norton 5th Edition

DESIGN OF MACHINERY - 5th Ed. SOLUTION MANUAL 3-1-1
PROBLEM 3-1
Statement: Define the following examples as path, motion, or function generation cases.
a. A telescope aiming (star tracking) mechanism
b. A backhoe bucket control mechanism
c. A thermostat adjusting mechanism
d. A computer printing head moving mechanism
e. An XY plotter pen control mechanism
Solution: See Mathcad file P0301.
a. Path generation. A star follows a 2D path in the sky.
b. Motion generation. To dig a trench, say, the position and orientation of the bucket must be controlled.
c. Function generation. The output is some desired function of the input over some range of the input.
d. Path generation. The head must be at some point on a path.
e. Path generation. The pen follows a straight line from point to point.
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PROBLEM 3-2
Statement: Design a fourbar Grashof crank-rocker for 90 deg of output rocker motion with no quick
return. (See Example 3-1.) Build a cardboard model and determine the toggle positions
and the minimum transmission angle.
Given: Output angle θ 90 deg
Solution: See Example 3-1 and Mathcad file P0302.
Design choices: Link lengths: Link 3 L3 6.000 Link 4 L4 2.500
1. Draw the output link O4B in both extreme positions, B1 and B2, in any convenient location such that the
desired angle of motion 4 is subtended. In this solution, link 4 is drawn such that the two extreme
positions each make an angle of 45 deg to the vertical.
2. Draw the chord B1B2 and extend it in any convenient direction. In this solution it was extended to the left.
3. Layout the distance A1B1 along extended line B1B2 equal to the length of link 3. Mark the point A1.
4. Bisect the line segment B1B2 and layout the length of that radius from point A1 along extended line B1B2.
Mark the resulting point O2 and draw a circle of radius O2A1 with center at O2.
5. Label the other intersection of the circle and extended line B1B2, A2.
6. Measure the length of the crank (link 2) as O2
A1
or O2
A2
. From the graphical solution, L2 1.76775
7. Measure the length of the ground link (link 1) as O2
O4
1.7677
6.2550
2
O
A2
2
3.5355
6.0000
90.00°
A1
1
3 B2
4
O4
B1
8. Find the Grashof condition.
Condition a b c d( ) S min a b c d( )
L max a b c d( )
SL S L
PQ a b c d SL
"Grashof"return SL PQif
"Special Grashof"return SL PQ=if
"non-Grashof"return otherwise

Condition L1 L2 L3 L4  "Grashof"

PROBLEM 3-3
Statement: Design a fourbar mechanism to give the two positions shown in Figure P3-1 of output rocker
motion with no quick-return. (See Example 3-2.) Build a cardboard model and determine the
toggle positions and the minimum transmission angle.
Given: Coordinates of A1, B1, A2, and B2 (with respect to A1):
xA1 0.00 xB1 1.721 xA2 2.656 xB2 5.065
yA1 0.00 yB1 1.750 yA2 0.751 yB2 0.281
Solution: See Figure P3-1 and Mathcad file P0303.
Design choices: Link length: Link 3 L3 5.000 Link 4 L4 2.000
1. Following the notation used in Example 3-2 and Figure 3-5, change the labels on points A and B in Figure
P3-1 to C and D, respectively. Draw the link CD in its two desired positions, C1D1 and C2D2, using the
given coordinates.
2. Draw construction lines from C1 to C2 and D1 to D2.
3. Bisect line C1C2 and line D1D2 and extend their perpendicular bisectors to intersect at O4.
4. Using the length of link 4 (design choice) as a radius, draw an arc about O4 to intersect both lines O4C1 and
O4C2. Label the intersections B1 and B2.
5. Draw the chord B1B2 and extend it in any convenient direction. In this solution it was extended to the left.
6. Layout the distance A1B1 along extended line B1B2 equal to the length of link 3. Mark the point A1.
7. Bisect the line segment B1B2 and layout the length of that radius from point A1 along extended line B1B2.
Mark the resulting point O2 and draw a circle of radius O2A1 with center at O2.
8. Label the other intersection of the circle and extended line B1B2, A2.
A1
or O2
A2
O4
D1
5.0000
5.3013
C
3
A2
0.9469
O
A1
2
12
R2.0004
1
B1
B2
C2
O4
D2

L max a b c d( )
SL S L


PROBLEM 3-4
Statement: Design a fourbar mechanism to give the two positions shown in Figure P3-1 of coupler
motion. (See Example 3-3.) Build a cardboard model and determine the toggle positions and
the minimum transmission angle. Add a driver dyad. (See Example 3-4.)
Given: Position 1 offsets: xA1B1 1.721 in yA1B1 1.750 in
Solution: See figure below for one possible solution. Input file P0304.mcd from the solutions manual disk
to the Mathcad program for this solution, file P03-04.4br to the program FOURBAR to see the
fourbar solution linkage, and file P03-04.6br into program SIXBAR to see the complete sixbar
with the driver dyad included.
1. Connect the end points of the two given positions of the line AB with construction lines, i.e., lines from A1 to
A2 and B1 to B2.
2. Bisect these lines and extend their perpendicular bisectors in any convenient direction. In the solution below
the bisector of A1A2 was extended downward and the bisector of B1B2 was extended upward.
3. Select one point on each bisector and label them O4 and O6, respectively. In the solution below the distances
O4A and O6B were each selected to be 4.000 in. This resulted in a ground-link-length O4O6 for the fourbar of
6.457 in.
4. The fourbar stage is now defined as O4ABO6 with link lengths
Link 5 (coupler) L5 xA1B1
2
yA1B1
2
 L5 2.454 in
Link 4 (input) L4 4.000 in Link 6 (output) L6 4.000 in
Ground link 1b L1b 6.457 in
5. Select a point on link 4 (O4A) at a suitable distance from O4 as the pivot point to which the driver dyad will be
connected and label it D. (Note that link 4 is now a ternary link with nodes at O4, D, and A.) In the solution
below the distance O4D was selected to be 2.000 in.
6. Draw a construction line through D1D2 and extend it to the left.
7. Select a point on this line and call it O2. In the solution below the distance CD was selected to be 4.000 in.
8. Draw a circle about O2 with a radius of one-half the length D1D2 and label the intersections of the circle with
the extended line as C1 and C2. In the solution below the radius was measured as 0.6895 in.
9. The driver fourbar is now defined as O2CDO4 with link lengths
Link 2 (crank) L2 0.6895 in Link 3 (coupler) L3 4.000 in
Link 4a (rocker) L4a 2.000 in Link 1a (ground) L1a 4.418 in
10. Use the link lengths in step 9 to find the Grashoff condition of the driving fourbar (it must be Grashoff and
the shortest link must be link 2).
Shortest link S L2 S 0.6895 in
Longest link L L1a L 4.4180 in
Other links P L3 P 4.0000 in
Q L4a Q 2.0000 in

L max a b c d( )
SL S L

Condition S L P Q( ) "Grashof"
O2
1C
2C
Ground Link 1a
2
3
A
O4
1D 3
5
4
B 1
2D
4
47.893°
2
Ground Link 1b
1A
O
6
6
50.231°
5
6
2B
11. Using the program FOURBAR and the link lengths given above, it was found that the fourbar O4ABO6 is
non-Grashoff with toggle positions at 2 = -71.9 deg and +71.9 deg. The minimum transmission angle is 35.5
deg. The fourbar operates between 2 = +21.106 deg and -19.297 deg.

PROBLEM 3-5
Statement: Design a fourbar mechanism to give the three positions of coupler motion with no quick return
shown in Figure P3-2. (See also Example 3-5.) Ignore the points O2 and O4 shown. Build a
cardboard model and determine the toggle positions and the minimum transmission angle. Add
a driver dyad.
Design choices:
Length of link 5: L5 4.250 Length of link 4b: L4b 1.375
1. Draw link CD in its three design positions C1D1, C2D2, C3D3 in the plane as shown.
2. Draw construction lines from point C1 to C2 and from point C2 to C3.
3. Bisect line C1C2 and line C2C3 and extend their perpendicular bisectors until they intersect. Label their
intersection O2.
4. Repeat steps 2 and 3 for lines D1D2 and D2D3. Label the intersection O4.
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 (ground link for the fourbar). The fourbar is now defined as O2CDO4
and has link lengths of
Ground link 1a L1a 0.718 Link 2 L2 2.197
Link 3 L3 2.496 Link 4 L4 3.704
1 2
2.496
3.704
4.328
D D
2.197
0.718
O
B
3
4
2
1C
2C
2
1a
O4
5 A
3C
3D
1.230
1b
O6
6
7. Check the Grashof condition. Note that any Grashof condition is potentially acceptable in this case.

L max a b c d( )
SL S L

Condition L1a L2 L3 L4  "Grashof"
8. Select a point on link 4 (O4D) at a suitable distance from O4 as the pivot point to which the driver dyad will be
connected and label it B. (Note that link 4 is now a ternary link with nodes at O4, C, and B.) In the solution
above the distance O4
B was selected to be L4b 1.375 .
9. Draw a construction line through B1B3 and extend it up to the right.
10. Layout the length of link 5 (design choice) along the extended line. Label the other end A.
11. Draw a circle about O6 with a radius of one-half the length B1B3 and label the intersections of the circle
with the extended line as A1
and A3
. In the solution below the radius was measured as L6 1.230 .
12. The driver fourbar is now defined as O4BAO6 with link lengths
Link 6 (crank) L6 1.230
Link 5 (coupler) L5 4.250
Link 1b (ground) L1b 4.328
Link 4b (rocker) L4b 1.375
Condition L6 L1b L4b L5  "Grashof"

PROBLEM 3-6
Statement: Design a fourbar mechanism to give the three positions shown in Figure P3-2 using the fixed
pivots O2 and O4 shown. Build a cardboard model and determine the toggle positions and the
minimum transmission angle. Add a driver dyad.
Design choices:
2. Draw the ground link O2O4 in its desired position in the plane with respect to the first coupler position C1D1.
3. Draw construction arcs from point C2 to O2 and from point D2 to O2 whose radii define the sides of triangle
C2O2D2. This defines the relationship of the fixed pivot O2 to the coupler line CD in the second coupler
position.
position.
5. 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.
6. Repeat the process for the third coupler position and transfer the third relative ground link position to the first,
or reference, position.
7. The three inverted positions of the ground link that correspond to the three desired coupler positions are
labeled O2O4, O2'O4', and O2"O4" in the first layout below and are renamed E1F1, E2F2, and E3F3,
respectively, in the second layout, which is used to find the points G and H.
O ''
2
1
O '4
4
O '2
O2
1
2C
D D
3C
O ''2
C
3D
O4
8. Draw construction lines from point E1 to E2 and from point E2 to E3.

9. Bisect line E1E2 and line E2E3 and extend their perpendicular bisectors until they intersect. Label their
intersection G.
10. Repeat steps 2 and 3 for lines F1F2 and F2F3. Label the intersection H.
11. Connect E1 with G and label it link 2. Connect F1 with H and label it link 4. Reinverting, E1 and F1 are the
original fixed pivots O2 and O4, respectively.
12. Line GH is link 3. Line O2O4 is link 1a (ground link for the fourbar). The fourbar is now defined as O2GHO4
Link 3 L3 1.711 Link 4 L4 7.921
O O
2 4
2F
3F
2E
2
1
E 1a
4
3E
H 3
1F
G
L max a b c d( )
SL S L

The fourbar that will provide the desired motion is now defined as a Grashof double crank in the crossed
configuration. It now remains to add the original points C1 and D1 to the coupler GH and to define the
driving dyad.

14. Select a point on link 2 (O2G) at a suitable distance from O2 as the pivot point to which the driver dyad will be
connected and label it B. (Note that link 2 is now a ternary link with nodes at O2, B, and G.) In the solution
below, the distance O2
and A3
O
3A
D
1
2
56
O
6
1A
1D
3 1
B B
1a
2C
3C
C
H1
H3
H2
2
3
GG3 2
3D
O4
4
2
G1

PROBLEM 3-7
Statement: Repeat Problem 3-2 with a quick-return time ratio of 1:1.4. (See Example 3.9). Design a
fourbar Grashof crank-rocker for 90 degrees of output rocker motion with a quick-return time
ratio of 1:1.4.
Given: Time ratio Tr
1
1.4

Solution: See figure below for one possible solution. Also see Mathcad file P0307.
1. Determine the crank rotation angles  and , and the construction angle  from equations 3.1 and 3.2.
Tr
α
β
= α β 360 deg=
Solving for , and  β
360 deg
1 Tr
 β 210 deg
α 360 deg β α 150 deg
δ β 180 deg δ 30 deg
2. Start the layout by arbitrarily establishing the point O4 and from it layoff two lines of equal length, 90 deg
apart. Label one B1 and the other B2. In the solution below, each line makes an angle of 45 deg with the
horizontal and has a length of 2.000 in.
3. Layoff a line through B1 at an arbitrary angle (but not zero deg). In the solution below, the line is 30 deg to
the horizontal.
4. Layoff a line through B2 that makes an angle  with the line in step 3 (60 deg to the horizontal in this case).
The intersection of these two lines establishes the point O2.
5. From O2 draw an arc that goes through B1. Extend O2B2 to meet this arc. Erect a perpendicular bisector to
the extended portion of the line and transfer one half of the line to O2 as the length of the input crank.

3.8637 = b
1.0353 = a
LAYOUT
B2
O2 3.0119 = d
B1
O4
A2
O2
2
2.0000 = c
LINKAGE DEFINITION
3
A1
O4
4
B1
90.0000°
B2

6. For this solution, the link lengths are:
Ground link (1) d 3.0119 in
Crank (2) a 1.0353 in
Coupler (3) b 3.8637 in
Rocker (4) c 2.000 in

PROBLEM 3-8
Statement: Design a sixbar drag link quick-return linkage for a time ratio of 1:2, and output rocker motion
of 60 degrees. (See Example 3-10.)
1
2

1. Determine the crank rotation angles  and  from equation 3.1.
Tr
α
β
= α β 360 deg=
Solving for and  β
360 deg
1 Tr
 β 240 deg
α 360 deg β α 120 deg
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 through O2.
4. Draw a circle of convenient radius O2A about center O2. In the solution below, the length of O2A is
a 1.000 in .
5. Lay out angle  with vertex at O2, symmetrical about quadrant one.
6. Label points A1 and A2 at the intersections of the lines subtending angle  and the circle of radius O2A.
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 both A1 and A2. Label the intersections C1 and C2. In the solution below, the length of AC is
b 1.800 in .
8. The line O2A is the driver crank, link 2, and the line AC is the coupler, link 3.
9. The distance C1C2 is twice the driven (dragged) crank length. Bisect it to locate the fixed pivot O4.
10. The line O2O4 now defines the ground link. Line O4C is the driven crank, link 4. In the solution below, O4C
measures c 2.262 in and O2O4 measures d 0.484 in .
11. Calculate the Grashoff condition. If non-Grashoff, repeat steps 7 through 11 with a shorter radius in step 7.
L max a b c d( )
SL S L

Condition a b c d( ) "Grashof"
12. Invert the method of Example 3-1 to create the output dyad using XX as the chord and O4C1 as the driving
crank. The points B1 and B2 will lie on line XX and be spaced apart a distance that is twice the length of
O4C (link 4). The pivot point O6 will lie on the perpendicular bisector of B1B2 at a distance from XX which
subtends the specified output rocker angle, which is 60 degrees in this problem. In the solution below, the
length BC was chosen to be e 5.250 in .

LAYOUT OF SIXBAR DRAG LINK QUICK RETURN
WITH TIME RATIO OF 1:2
a = 1.000 b = 1.800 c = 2.262 d = 0.484
e = 5.250 f = 4.524
13. For the design choices made (lengths of links 2, 3 and 5), the length of the output rocker (link 6)
was measured as f 4.524 in .

PROBLEM 3-9
Statement: Design a crank-shaper quick-return mechanism for a time ratio of 1:3 (Figure 3-14, p. 112).
Given: Time ratio TR
1
3

Solution: See Figure 3-14 and Mathcad file P0309.
Design choices:
Length of link 2 (crank) L2 1.000 Length of stroke S 4.000
Length of link 5 (coupler) L5 5.000
1. Calculate  from equations 3.1.
TR
α
β
 α β 360 deg α
360 deg
1
1
TR

 α 90.000 deg
2. Draw a vertical line and mark the center of rotation of the crank, O2, on it.
3. Layout two construction lines from O2, each making an angle /2 to the vertical line through O2.
4. Using the chosen crank length (see Design Choices), draw a circle with center at O2 and radius equal to the
crank length. Label the intersections of the circle and the two lines drawn in step 3 as A1 and A2.
5. Draw lines through points A1 and A2 that are also tangent to the crank circle (step 2). These two lines will
simultaneously intersect the vertical line drawn in step 2. Label the point of intersection as the fixed pivot
center O4.
6. Draw a vertical construction line, parallel and to the right of O2O4, a distance S/2 (one-half of the output
stroke length) from the line O2O4.
7. Extend line O4A1 until it intersects the construction line drawn in step 6. Label the intersection B1.
8. Draw a horizontal construction line from point B1, either to the left or right. Using point B1 as center, draw
an arc of radius equal to the length of link 5 (see Design Choices) to intersect the horizontal construction
line. Label the intersection as C1.
9. Draw the slider blocks at points A1 and C1 and finish by drawing the mechanism in its other extreme position.
3
AA
O
1
2
4
2C
4.000
STROKE
C6 5B21
2.000
4
2
O2
B1

PROBLEM 3-10
Statement: Find the two cognates of the linkage in Figure 3-17 (p. 116). Draw the Cayley and Roberts
diagrams. Check your results with program FOURBAR.
Given: Link lengths: Coupler point data:
Ground link L1 2 Crank L2 1 A1P 1.800 δ 34.000 deg
Coupler L3 3 Rocker L4 3.5 B1P 1.813 γ 33.727 deg
1. Draw the original fourbar linkage, which will be cognate #1, and align links 2 and 4 with the coupler.
3 1
1
OB
2
P
OA
4
A1 B 3 12
OB
OA
P
A1 B 4
2. Construct lines parallel to all sides of the aligned fourbar linkage to create the Cayley diagram (see Figure 3-24)
A
A
B
2
2
10
OC
7
8
P
5
A3
B3
6
9
O 2 A1 3 1B 4 BO
4
B3
OC6
5
OB
A
7
A3
2
O
A1
10 BP
8 9
A2
2
3 B1
3. Return links 2 and 4 to their fixed pivots OA and OB and
establish OC as a fixed pivot by making triangle OAOBOC
similar to A1B1P.
4. Separate the three cognates. Point P has the same path
motion in each cognate.
5. Calculate the cognate link lengths based on the geometry
of the Cayley diagram (Figure 3-24c, p. 114).
L5 B1P L5 1.813
L6
L4
L3
B1P L6 2.115

B3
5
6
OB
C
7
A3
O
P
OC
OA
B
A2
210
8
9
P
Cognate #2
Cognate #3
L10 A1P L10 1.800 L9
L2
L3
A1P L9 0.600
L7 L9
B1P
A1P
 L7 0.604
L8 L6
A1P
B1P
 L8 2.100
From the Roberts diagram, calculate the ground link lengths for cognates #2 and #3
L1BC
L1
L3
B1P L1BC 1.209
L1AC
L1
L3
A1P L1AC 1.200
Calculate the coupler point data for cognates #2 and #3
A3P L8 A3P 2.100 A2P L2 A2P 1.000
δ 180 deg δ γ  δ 247.727 deg δ δ δ 34.000 deg
SUMMARY OF COGNATE SPECIFICATIONS:
Cognate #1 Cognate #2 Cognate #3
Ground link length L1 2.000 L1AC 1.200 L1BC 1.209
Crank length L2 1.000 L10 1.800 L7 0.604
Coupler length L3 3.000 L9 0.600 L6 2.115
Rocker length L4 3.500 L8 2.100 L5 1.813
Coupler point A1P 1.800 A2P 1.000 A3P 2.100
Coupler angle δ 34.000 deg δ 34.000 deg δ 247.727 deg
6. Verify that the three cognates yield the same coupler curve by entering the original link lengths in program
FOURBAR and letting it calculate the cognates.

Note that cognate #2 is a Grashof double rocker and, therefore, cannot trace out the entire coupler curve.

PROBLEM 3-11
Statement: Find the three equivalent geared fivebar linkages for the three fourbar cognates in Figure
3-25a (p. 125). Check your results by comparing the coupler curves with programs FOURBAR
and FIVEBAR.
Ground link L1 39.5 Crank L2 15.5 A1P 26.0 δ 63.000 deg
Coupler L3 14.0 Rocker L4 20.0
Solution: See Figure 3-25a and Mathcad file P0311.
1. Calculate the length BP and the angle using the law of cosines on the triangle APB.
B1P L3
2
A1P
2
 2 L3 A1P cos δ 



0.5
 B1P 23.270
γ acos
L3
2
B1P
2
 A1P
2

2 L3 B1P








 γ 84.5843 deg
2. Use the Cayley diagram (see Figure 3-24) to calculate the link lengths of the two cognates. Note that the
diagram is made up of three parallelograms and three similar triangles
L5 B1P L5 23.270 L6
L4
L3
B1P L6 33.243
L10 A1P L10 26.000 L9
L2
L3
A1P L9 28.786
L7 L9
B1P
A1P
 L7 25.763 L8 L6
A1P
B1P
 L8 37.143
A3P L4 A3P 20.000 A2P L2 A2P 15.500
δ γ δ 84.584 deg δ δ δ 63.000 deg
L1BC
L1
L3
B1P L1BC 65.6548 L1AC
L1
L3
A1P L1AC 73.3571
3. Using the calculated link lengths, draw the Roberts diagram (see next page).
Crank length L2 15.500 L10 26.000 L7 25.763

Coupler angle δ 63.000 deg δ 63.000 deg δ 84.584 deg
OC
A
1
3
42
5
10
O
A
B
A
1
1
A
O
3
2
B
8
9
P
B2
6
7
B3
4. The three geared fivebar cognates can be seen in the Roberts diagram. They are: OAA2PA3OB, OAA1PB3OC, and
OBB1PB2OC. They are shown individually below with their associated gears.
O
O
B
C
A
OA
A3
2
10
5
P

OC
2
O
A
A
1
P
OD
7
B3
OC
4
B1
OB
8
P
B2
OE

SUMMARY OF GEARED FIVEBAR COGNATE SPECIFICATIONS:
Crank length L10 26.000 L2 15.500 L4 20.000
Coupler length A2P 15.500 A1P 26.000 L5 23.270
Rocker length A3P 20.000 L8 37.143 L7 25.763
Crank length L5 23.270 L7 25.763 L8 37.143
Coupler point A2P 15.500 A1P 26.000 B1P 23.270
Coupler angle δ 0.00 deg δ 0.00 deg δ 0.00 deg
5. Enter the cognate #1 specifications into program FOURBAR to get a trace of the coupler path.
6. Enter the geared fivebar cognate #1 specifications into program FIVEBAR to get a trace of the coupler path for
the geared fivebar (see next page).

PROBLEM 3-12
Statement: Design a sixbar, single-dwell linkage for a dwell of 90 deg of crank motion, with an output rocker
motion of 45 deg.
Given: Crank dwell period: 90 deg.
Output rocker motion: 45 deg.
Solution: See Figures 3-20, 3-21, and Mathcad file P0312.
Design choices:
Ground link ratio, L1/L2 = 2.0: GLR 2.0
Common link ratio, L3/L2 = L4/L2 = BP/L2 = 2.5: CLR 2.5
Coupler angle, γ 72 deg
Crank length, L2 2.000
1. For the given design choices, determine the remaining link lengths and coupler point specification.
Coupler link (3) length L3 CLR L2 L3 5.000
Rocker link (4) length L4 CLR L2 L4 5.000
Ground link (1) length L1 GLR L2 L1 4.000
Angle PAB δ
180 deg γ
2
 δ 54.000 deg
Length AP on coupler AP 2 L3 cos δ  AP 5.878
2. Enter the above data into program FOURBAR, plot the coupler curve, and determine the coordinates of the
coupler curve in the selected range of crank motion, which in this case will be from 135 to 225 deg..

FOURBAR for Windows File P03-12.DAT
Angle Coupler Pt Coupler Pt Coupler Pt Coupler Pt
Step X Y Mag Ang
Deg
135 -1.961 7.267 7.527 105.099
140 -2.178 7.128 7.453 106.992
145 -2.393 6.977 7.375 108.930
150 -2.603 6.813 7.293 110.911
155 -2.809 6.638 7.208 112.933
160 -3.008 6.453 7.119 114.994
165 -3.201 6.257 7.028 117.093
170 -3.386 6.052 6.935 119.228
175 -3.563 5.839 6.840 121.396
180 -3.731 5.617 6.744 123.595
185 -3.890 5.389 6.646 125.822
190 -4.038 5.155 6.548 128.075
195 -4.176 4.915 6.450 130.351
200 -4.302 4.671 6.351 132.646
205 -4.417 4.424 6.252 134.955
210 -4.520 4.175 6.153 137.274
215 -4.610 3.924 6.054 139.598
220 -4.688 3.673 5.956 141.921
225 -4.753 3.424 5.858 144.235
3. Layout this linkage to scale, including the coupler curve whose coordinates are in the table above. Use the
points at crank angles of 135, 180, and 225 deg to define the pseudo-arc. Find the center of the pseudo-arc
erecting perpendicular bisectors to the chords defined by the selected coupler curve points. The center will
lie at the intersection of the perpendicular bisectors, label this point D. The radius of this circle is the length
of link 5.
225
O2
O4
3
A
2
D 4
B
PSEUDO-ARCP
180
135
y
x

4. The position of the end of link 5 at point D will remain nearly stationary while the crank moves from 135 to 225
deg. As the crank motion causes the coupler point to move around the coupler curve there will be another
extreme position of the end of link 5 that was originally at D. Since a symmetrical linkage was chosen, the
other extreme position will be located along a line through the axis of symmetry (see Figure 3-20) a distance
equal to the length of link 5 measured from the point where the axis of symmetry intersects the coupler curve
near the 0 deg coupler point. Establish this point and label it E.
Step X Y Mag Ang
Deg
300 -4.271 0.869 4.359 168.495
310 -4.054 0.926 4.158 167.133
320 -3.811 1.165 3.985 162.998
330 -3.526 1.628 3.883 155.215
340 -3.159 2.343 3.933 143.437
350 -2.651 3.286 4.222 128.892
0 -1.968 4.336 4.762 114.414
10 -1.181 5.310 5.440 102.534
20 -0.441 6.085 6.101 94.142
30 0.126 6.654 6.656 88.914
40 0.478 7.068 7.085 86.129
50 0.631 7.373 7.400 85.111
60 0.617 7.598 7.623 85.354
225
O2
O4
AXIS OF SYMMETRY
5
3
A
2
D
E
4
B
PSEUDO-ARCP
180
135
y
x
5. The line segment DE represents the maximum displacement that a link of the length equal to link 5, attached
at P, will reach along the axis of symmetry. Construct a perpendicular bisector of the line segment DE and
extend it to the right (or left, which ever is convenient). Locate fixed pivot O6 on the bisector of DE such
that the lines O6D and O6E subtend the desired output angle, in this case 45 deg. Draw link 6 from D
through O6 and extend it to any convenient length. This is the output link that will dwell during the
specified motion of the crank. See next page for the completed layout and further linkage specifications.

225
O2
O4
BISECTOR
5
3
A
2
D
E
4
B
PSEUDO-ARCP
180
135
y
45.000°
x
O6
SUMMARY OF LINKAGE SPECIFICATIONS
Original fourbar:
Ground link L1 4.000
Crank L2 2.000
Coupler L3 5.000
Rocker L4 5.000
Coupler point AP 5.878 δ 54.000 deg
Added dyad:
Coupler L5 6.363
Output L6 2.855
Pivot O6 x 3.833 y 3.375

PROBLEM 3-13
Statement: Design a sixbar double-dwell linkage for a dwell of 90 deg of crank motion, with an output of
rocker motion of 60 deg, followed by a second dwell of about 60 deg of crank motion.
Given: Initial crank dwell period: 90 deg
Final crank dwell period: 60 deg (approx.)
Output rocker motion between dwells: 60 deg
Design choices:
Ground link length L1 5.000 Crank length L2 2.000
Coupler link length L3 5.000 Rocker length L2 5.500
Coupler point data: AP 8.750 δ 50 deg
1. In the absence of a linkage atlas it is difficult to find a coupler curve that meets the specifications. One
approach is to start with a symmetrical linkage, using the data in Figure 3-21. Then, using program
FOURBAR and by trial-and-error, adjust the link lengths and coupler point data until a satisfactory
coupler curve is found. The link lengths and coupler point data given above were found this way. The
resulting coupler curve is shown below and a printout of the coupler curve coordinates taken from
FOURBAR is also printed below.

Angle Cpler Pt Cpler Pt Cpler Pt Cpler Pt
Step X Y Mag Ang
Deg
0.000 9.353 4.742 10.487 26.886
10.000 9.846 4.159 10.688 22.900
20.000 10.167 3.491 10.750 18.951
30.000 10.286 2.840 10.671 15.437
40.000 10.226 2.274 10.476 12.537
50.000 10.031 1.815 10.194 10.257
60.000 9.746 1.457 9.854 8.503
70.000 9.406 1.180 9.480 7.152
80.000 9.039 0.963 9.090 6.081
90.000 8.665 0.787 8.701 5.187
100.000 8.301 0.637 8.325 4.391
110.000 7.958 0.507 7.974 3.644
120.000 7.647 0.391 7.657 2.928
130.000 7.376 0.291 7.382 2.256
140.000 7.151 0.209 7.154 1.671
150.000 6.977 0.151 6.978 1.242
160.000 6.853 0.126 6.854 1.051
170.000 6.778 0.140 6.779 1.182
180.000 6.748 0.201 6.751 1.708
190.000 6.755 0.316 6.763 2.678
200.000 6.792 0.488 6.809 4.110
210.000 6.847 0.719 6.885 5.996
220.000 6.912 1.008 6.985 8.300
230.000 6.976 1.351 7.105 10.963
240.000 7.031 1.741 7.243 13.911
250.000 7.073 2.170 7.398 17.057
260.000 7.099 2.626 7.569 20.302
270.000 7.112 3.098 7.757 23.536
280.000 7.120 3.570 7.965 26.632
290.000 7.137 4.030 8.196 29.448
300.000 7.184 4.458 8.455 31.819
310.000 7.288 4.834 8.746 33.555
320.000 7.481 5.131 9.072 34.446
330.000 7.792 5.312 9.430 34.286
340.000 8.233 5.332 9.809 32.931
350.000 8.779 5.147 10.177 30.384
360.000 9.353 4.742 10.487 26.886
2. Layout this linkage to scale, including the coupler curve whose coordinates are in the table above. Fit tangent
lines to the nearly straight portions of the curve. Label their intersection O6. The coordinates of O6 are (6.729,
0.046).
3. Design link 6 to lie along these straight tangents, pivoted at O6. Provide a slot in link 6 to accommodate slider
block 5, which pivots on the coupler point P. (See next page).
4. The beginning and ending crank angles for the dwell portions of the motion are indicated on the layout and in
the table above by boldface entries.

A
y
3
4
2
O2
B
60.000°
5
170
x
O4
150
O6
P
260
6
90

PROBLEM 3-14
Statement: Figure P3-3 shows a treadle-operated grinding wheel driven by a fourbar linkage. Make a
cardboard model of the linkage to any convenient scale. Determine its minimum transmission
angles. Comment on its operation. Will it work? If so, explain how it does.
Given: Link lengths: Link 2 L2 0.60 m Link 3 L3 0.75 m
Link 4 L4 0.13 m Link 1 L1 0.90 m
Grashof condition function:
L max a b c d( )
SL S L

1. Determine the Grashof condition of the mechanism from inequality 2.8 and its Barker classification from
Table 2-4.
Grashof condition: Condition L1 L2 L3 L4  "Grashof"
Barker classification: Class I-4, Grashof rocker-rocker-crank, GRRC, since the shortest link
is the output link.
2. As a Grashof rocker-crank, the minimum transmission angle will be 0 deg, twice per revolution of the output
(link 4) crank.
3. Despite having transmission angles of 0 deg twice per revolution, the mechanism will work. That is, one
will be able to drive the grinding wheel from the treadle (link 2). The reason is that the grinding wheel will
act as a flywheel and will carry the linkage through the periods when the transmission angle is low.
Typically, the operator will start the motion by rotating the wheel by hand.

PROBLEM 3-15
Statement: Figure P3-4 shows a non-Grashof fourbar linkage that is driven from link O2A. All dimensions
are in centimeters (cm).
(a) Find the transmission angle at the position shown.
(b) Find the toggle positions in terms of angle AO2O4.
(c) Find the maximum and minimum transmission angles over its range of motion.
(d) Draw the coupler curve of point P over its range of motion.
Given: Link lengths:
Link 1 (ground) L1 95 mm Link 2 (driver) L2 50 mm
Link 3 (coupler) L3 44 mm Link 4 (driven) L4 50 mm
1. To find the transmission angle at the position shown, draw the linkage to scale in the position shown and
measure the transmission angle ABO4.
77.097°
3
50.000°
1
O2
2
A 4
y
B
P
O
x
4
The measured transmission angle at the position shown is 77.097 deg.
2. The toggle positions will be symmetric with respect to the O2O4 axis and will occur when links 3 and 4 are
colinear. Use the law of cosines to calculate the angle of link 2 when links 3 and 4 are in toggle.
L3 L4 2
L1
2
L2
2
 2 L1 L2 cos θ 
where 2 is the angle AO2O4. Solving for 2,
θ acos
L1
2
L2
2
 L3 L4 2

2 L1 L2








 θ 73.558 deg
The other toggle position occurs at θ 73.558 deg

3. Use the program FOURBAR to find the maximum and minimum transmission angles.
FOURBAR for Windows File P03-15 Design # 1
Angle Theta2 Theta3 Theta4 Trans Ang
Step Mag Mag Mag Mag
Deg degrees degrees degrees degrees
-73.557 -73.557 30.861 -149.490 0.352
-58.846 -58.846 64.075 -176.312 60.387
-44.134 -44.134 77.168 170.696 86.472
-29.423 -29.423 83.147 157.514 74.367
-14.711 -14.711 80.604 142.103 61.499
0.000 0.000 68.350 125.123 56.773
14.711 14.711 50.145 111.644 61.499
29.423 29.423 32.106 106.473 74.367
44.134 44.134 16.173 109.701 86.472
58.846 58.846 0.566 120.179 60.387
73.557 73.557 -30.486 149.159 0.355
A partial output from FOURBAR is shown above. From it, we see that the maximum transmission angle is
approximately 86.5 deg and the minimum is zero deg.
4. Use program FOURBAR to draw the coupler curve with respect to a coordinate frame through O2O4.

PROBLEM 3-16
Statement: Draw the Roberts diagram for the linkage in Figure P3-4 and find its two cognates. Are they
Grashof or non-Grashof?
Ground link L1 9.5 Crank L2 5 A1P 8.90 δ 56.000 deg
Coupler L3 4.4 Rocker L4 5
B1P L3
2
A1P
2
 2 L3 A1P cos δ 



0.5
 B1P 7.401
γ acos
L3
2
B1P
2
 A1P
2

2 L3 B1P








 γ 94.4701 deg
L5 B1P L5 7.401 L6
L4
L3
B1P L6 8.410
L10 A1P L10 8.900 L9
L2
L3
A1P L9 10.114
L7 L9
B1P
A1P
 L7 8.410 L8 L6
A1P
B1P
 L8 10.114
A3P L4 A3P 5.000 A2P L2 A2P 5.000
δ γ δ 94.470 deg δ δ δ 56.000 deg
L1BC
L1
L3
B1P L1BC 15.9793 L1AC
L1
L3
A1P L1AC 19.2159
Crank length L2 5.000 L10 8.900 L7 8.410

9
3
2
1
OA
A1
10 4
B1
A2
A3
5
B2
P
8
7
OC
OB
6
B3
6. Determine the Grashof condition of each of the two additional cognates.
L max a b c d( )
SL S L

Cognate #2: Condition L10 L1AC L8 L9  "non-Grashof"
Cognate #3: Condition L5 L1BC L6 L7  "non-Grashof"

PROBLEM 3-17
Statement: Design a Watt-I sixbar to give parallel motion that follows the coupler path of point P of the
linkage in Figure P3-4.
B1P L3
2
A1P
2
 2 L3 A1P cos δ 



0.5
 B1P 7.401
γ acos
L3
2
B1P
2
 A1P
2

2 L3 B1P








 γ 94.4701 deg
L5 B1P L5 7.401 L6
L4
L3
B1P L6 8.410
L10 A1P L10 8.900 L9
L2
L3
A1P L9 10.114
L7 L9
B1P
A1P
 L7 8.410 L8 L6
A1P
B1P
 L8 10.114
A3P L4 A3P 5.000 A2P L2 A2P 5.000
δ γ δ 94.470 deg δ δ δ 56.000 deg
L1BC
L1
L3
B1P L1BC 15.9793 L1AC
L1
L3
A1P L1AC 19.2159
Crank length L2 5.000 L10 8.900 L7 8.410

9
3
2
1
OA
A1
10 4
B1
A2
A3
5
B2
P
8
7
OC
OB
6
B3
3A
5
BO'
q
P'
7
6
B3
2
1
OA
3
A1
OB
4
B1
q
P
4. All three of these cognates are non-Grashof and will,
therefore, have limited motion. However, following
Example 3-11, discard cognate #2 and retain cognates
#1 and #3. Draw line qq parallel to line OAOC and
through point OB. Without allowing links 5, 6, and 7 to
rotate, slide them as an assembly along lines OAOC and
qq until the free end of link 7 is at OA. The free end of
link 5 will then be at point O'B and point P on link 6 will
be at P'. Add a new link of length OAOC between P
and P'. This is the new output link 8 and all points on
it describe the original coupler curve.
5. Join links 2 and 7, making one ternary link. Remove
link 5 and reduce link 6 to a binary link. The result is a
Watt-I sixbar with links numbered 1, 2, 3, 4, 6, and 8
(see next page). Link 8 is in curvilinear translation and
follows the coupler path of the original point P.

A1
OA
P'
8
6
2
B3
OB1
3
P
B1
4

PROBLEM 3-18
Statement: Design a Watt-I sixbar to give parallel motion that follows the coupler path of point P of the
linkage in Figure P3-4 and add a driver dyad to drive it over its possible range of motion
with no quick return. (The result will be an 8-bar linkage).
B1P L3
2
A1P
2
 2 L3 A1P cos δ 



0.5
 B1P 7.401
γ acos
L3
2
B1P
2
 A1P
2

2 L3 B1P








 γ 94.4701 deg
L5 B1P L5 7.401 L6
L4
L3
B1P L6 8.410
L10 A1P L10 8.900 L9
L2
L3
A1P L9 10.114
L7 L9
B1P
A1P
 L7 8.410 L8 L6
A1P
B1P
 L8 10.114
A3P L4 A3P 5.000 A2P L2 A2P 5.000
δ γ δ 94.470 deg δ δ δ 56.000 deg
L1BC
L1
L3
B1P L1BC 15.9793 L1AC
L1
L3
A1P L1AC 19.2159
Crank length L2 5.000 L10 8.900 L7 8.410

9
3
2
1
OA
A1
10 4
B1
A2
A3
5
B2
P
8
7
OC
OB
6
B3
3A
5
BO'
q
P'
7
6
B3
2
1
OA
3
A1
OB
4
B1
q
P
4. All three of these cognates are non-Grashof and will,
therefore, have limited motion. However, following
Example 3-11, discard cognate #2 and retain cognates
#1 and #3. Draw line qq parallel to line OAOC and
through point OB. Without allowing links 5, 6, and 7 to
rotate, slide them as an assembly along lines OAOC and
qq until the free end of link 7 is at OA. The free end of
link 5 will then be at point O'B and point P on link 6 will
be at P'. Add a new link of length OAOC between P
and P'. This is the new output link 8 and all points on
it describe the original coupler curve.
5. Join links 2 and 7, making one ternary link. Remove
link 5 and reduce link 6 to a binary link. The result is a
Watt-I sixbar with links numbered 1, 2, 3, 4, 6, and 8
(see next page). Link 8 is in curvilinear translation and
follows the coupler path of the original point P.
6. Add a driver dyad following Example 3-4.

A1
OA
P'
8
6
2
B3
OB1
3
P
B1
4
A1
OA
P'
8
6
2
B3
OB1
3
P
B1
4
OC

PROBLEM 3-19
Statement: Design a pin-jointed linkage that will guide the forks of the fork lift truck in Figure P3-5 up and
down in an approximate straight line over the range of motion shown. Arrange the fixed pivots
so they are close to some part of the existing frame or body of the truck.
Given: Length of straight line motion of the forks: Δx 1800 mm
Design choices:
Use a Hoeken-type straight line mechanism optimized for straightness.
Maximum allowable error in straightness of line: ΔCy 0.096 %
1. Using Table 3-1 and the required length of straight-line motion, determine the link lengths.
Link ratios from Table 3-1 for
ΔCy 0.096 % :
L1overL2 2.200 L3overL2 2.800 ΔxoverL2 4.181
Link lengths:
Crank L2
Δx
ΔxoverL2
 L2 430.5 mm
Coupler L3 L3overL2 L2 L3 1205.5 mm
Ground link L1 L1overL2 L2 L1 947.1 mm
Rocker L4 L3 L4 1205.5 mm
Coupler point BP L3 BP 1205.5 mm
2. Calculate the distance from point P to pivot O4 (Cy).
Cy 2 L3 2
L1 L2 2
 Cy 1978.5 mm
3. Draw the fork lift truck to scale with the mechanism defined in step 1 superimposed on it..
2
3
4
P
B
A
O4
O2
487.1mm
947.1mm
1978.5mm
900.0mm
620.0mm

PROBLEM 3-20
Statement: Figure P3-6 shows a "V-link" off-loading mechanism for a paper roll conveyor. Design a pin-
jointed linkage to replace the air cylinder driver that will rotate the rocker arm and V-link
through the 90 deg motion shown. Keep the fixed pivots as close to the existing frame as
possible. Your fourbar linkage should be Grashof and be in toggle at each extreme position of
the rocker arm.
Given: Dimensions scaled from Figure P3-6:
Rocker arm (link 4) distance between pin centers: L4 320 mm
Design choices:
1. Use the same rocker arm that was used with the air cylinder driver.
2. Place the pivot O2 80 mm to the right of the right leg and on a horizontal line with the center
of the pin on the rocker arm.
3. Design for two-position, 90 deg of output rocker motion with no quick return, similar to
Example 3-2.
1. Draw the rocker arm (link 4) O4B in both extreme positions, B1 and B2, in any convenient location such that
the desired angle of motion 4 is subtended. In this solution, link 4 is drawn such that the two extreme
positions each make an angle of 45 deg to the vertical.
2. Draw the chord B1B2 and extend it in any convenient direction. In this solution it was extended horizontally
to the left.
3. Mark the center O2 on the extended line such that it is 80 mm to the right of the right leg. This will allow
sufficient space for a supporting pillow block bearing.
4. Bisect the line segment B1B2 and draw a circle of that radius about O2.
5. Label the two intersections of the circle and extended line B1B2, A1 and A2.
6. Measure the length of the coupler (link 3) as A1
B1
or A2
B2
. From the graphical solution, L3 1045 mm
A1
or O2
A2
. From the graphical solution, L2 226.274 mm
O4
. From the graphical solution, L1 1069.217 mm
320.000
1045.000 226.274
80.000
1045.000
1069.217
1
3 2
4

L max a b c d( )
SL S L


PROBLEM 3-21
Statement: Figure P3-7 shows a walking-beam transport mechanism that uses a fourbar coupler curve,
replicated with a parallelogram linkage for parallel motion. Note the duplicate crank and coupler
shown ghosted in the right half of the mechanism - they are redundant and have been removed
from the duplicate fourbar linkage. Using the same fourbar driving stage (links 1, 2, 3, 4 with
coupler point P), design a Watt-I sixbar linkage that will drive link 8 in the same parallel motion
using two fewer links.
B1P L3
2
A1P
2
 2 L3 A1P cos δ 



0.5
 B1P 1.674
γ acos
L3
2
B1P
2
 A1P
2

2 L3 B1P








 γ 109.6560 deg
L5 B1P L5 1.674 L6
L4
L3
B1P L6 1.893
L10 A1P L10 3.060 L9
L2
L3
A1P L9 1.485
L7 L9
B1P
A1P
 L7 0.812 L8 L6
A1P
B1P
 L8 3.461
A3P L8 A3P 3.461 A2P L2 A2P 1.000
δ 180 deg δ γ   δ δ δ 31.000 deg
δ 39.344 deg
L1BC
L1
L3
B1P L1BC 1.8035 L1AC
L1
L3
A1P L1AC 3.2977

Crank length L2 1.000 L10 3.060 L7 0.812
Coupler angle δ 31.000 deg δ 31.000 deg δ 39.344 deg
3
P
B
A1
1
A
B 3
O
2
1
10
OA
3
4 9
A2
8
5
B
7
6
OC
B 2
4. Determine the Grashof condition of each of the two additional cognates.
L max a b c d( )
SL S L

Cognate #2: Condition L8 L9 L10 L1AC  "Grashof"
Cognate #3: Condition L5 L6 L7 L1BC  "Grashof"
5. Both of these cognates are Grashof but cognate #3 is a crank rocker. Following Example 3-11, discard
cognate #2 and retain cognates #1 and #3. Draw line qq parallel to line OAOC and through point OB.
Without allowing links 5, 6, and 7 to rotate, slide them as an assembly along lines OAOC and qq until the
free end of link 7 is at OA. The free end of link 5 will then be at point O'B and point P on link 6 will be at P'.
Add a new link of length OAOC between P and P'. This is the new output link 8 and all points on it
describe the original coupler curve.

P
A1
A3
5
B 1
8
P'
O'B
q
3
O
1
OA
7
B 3
2
6
4
B
q
6. Join links 2 and 7, making one ternary link. Remove link 5 and reduce link 6 to a binary link. The result is a
Watt-I sixbar with links numbered 1, 2, 3, 4, 6, and 8 (see next page). Link 8 is in curvilinear translation and
follows the coupler path of the original point P. The walking-beam (link 8 in Figure P3-7) is rigidly attached to
link 8 below.
P
A1
A3
B 1
8
P'
3
6
O
1
OA
2
4
B

PROBLEM 3-22
Statement: Find the maximum and minimum transmission angles of the fourbar driving stage (links L1, L2,
L3, L4) in Figure P3-7 (to graphical accuracy).
Given: Link lengths: Link 2 L2 1.00 Link 3 L3 2.06
Link 4 L4 2.33 Link 1 L1 2.22
L max a b c d( )
SL S L

Table 2-4.
Barker classification: Class I-2, Grashof crank-rocker-rocker, GCRR, since the shortest link
is the input link.
2. It can be shown (see Section 4.10) that the minimum transmission angle for a fourbar GCRR linkage occurs
when links 2 and 1 (ground link) are colinear. Draw the linkage in these two positions and measure the
transmission angles.
B
A
85.843°
B
31.510°
A
O2
4
O2
O
4O
3. As measured from the layout, the minimum transmission angle is 31.5 deg. The maximum is 90 deg.

PROBLEM 3-23
Statement: Figure P3-8 shows a fourbar linkage used in a power loom to drive a comb-like reed against the
thread, "beating it up" into the cloth. Determine its Grashof condition and its minimum and
maximum transmission angles to graphical accuracy.
Given: Link lengths: Link 2 L2 2.00 in Link 3 L3 8.375 in
Link 4 L4 7.187 in Link 1 L1 9.625 in
L max a b c d( )
SL S L

Table 2-4.
Barker classification: Class I-2, Grashof crank-rocker-rocker, GCRR, since the shortest link
is the input link.
2. It can be shown (see Section 4.10) that the minimum transmission angle for a fourbar GCRR linkage occurs
when links 2 and 1 (ground link) are colinear. Draw the linkage in these two positions and measure the
transmission angles.
58.078°
83.634°
3. As measured from the layout, the minimum transmission angle is 58.1 deg. The maximum is 90.0 deg.

PROBLEM 3-24
Statement: Draw the Roberts diagram and find the cognates for the linkage in Figure P3-9.
Ground link L1 2.22 Crank L2 1.0 A1P 3.06 δ 31.00 deg
B1P L3
2
A1P
2
 2 L3 A1P cos δ 



0.5
 B1P 1.674
γ acos
L3
2
B1P
2
 A1P
2

2 L3 B1P








 γ 109.6560 deg
L5 B1P L5 1.674 L6
L4
L3
B1P L6 1.893
L10 A1P L10 3.060 L9
L2
L3
A1P L9 1.485
L7 L9
B1P
A1P
 L7 0.812 L8 L6
A1P
B1P
 L8 3.461
A3P L8 A3P 3.461 A2P L2 A2P 1.000
δ 180 deg δ γ δ 39.344 deg δ δ δ 31.000 deg
L1BC
L1
L3
B1P L1BC 1.8035 L1AC
L1
L3
A1P L1AC 3.2977
Crank length L2 1.000 L10 3.060 L7 0.812

1AC
7
6
A
B3
3 OC
B
3
A
2
1
10
O
1
A
4
A
8
5
2
OB
1BC
9
P
1
B2

PROBLEM 3-25
Statement: Find the equivalent geared fivebar mechanism cognate of the linkage in Figure P3-9.
B1P L3
2
A1P
2
 2 L3 A1P cos δ 



0.5
 B1P 1.674
γ acos
L3
2
B1P
2
 A1P
2

2 L3 B1P








 γ 109.6560 deg
L5 B1P L5 1.674 L6
L4
L3
B1P L6 1.893
L10 A1P L10 3.060 L9
L2
L3
A1P L9 1.485
L7 L9
B1P
A1P
 L7 0.812 L8 L6
A1P
B1P
 L8 3.461
A3P L8 A3P 3.461 A2P L2 A2P 1.000
δ 180 deg δ γ δ 39.344 deg δ δ δ 31.000 deg
L1BC
L1
L3
B1P L1BC 1.8035 L1AC
L1
L3
A1P L1AC 3.2977
Crank length L2 1.000 L10 3.060 L7 0.812

1AC
7
6
A
B3
3 OC
B
3
A
2
1
10
O
1
A
4
A
8
5
2
OB
1BC
9
P
1
B2
4. The three geared fivebar cognates can be seen in the Roberts diagram. They are: OAA2PB3OB, OAA1PA3OC,
and OBB1PB2OC. The three geared fivebar cognates are summarized in the table below.
Crank length L10 3.060 L2 1.000 L4 2.330
Crank length L5 1.674 L7 0.812 L8 3.461
5. Enter the cognate #1 specifications into program FOURBAR to get a trace of the coupler path (see next page)

PROBLEM 3-26
Statement: Use the linkage in Figure P3-9 to design an eightbar double-dwell mechanism that has a rocker
output through 45 deg.
1. Enter the given data into program FOURBAR and print out the resulting coupler point coordinates (see table
below).
Step X Y Mag Ang
Deg
0.000 2.731 2.523 3.718 42.731
10.000 3.077 2.407 3.906 38.029
20.000 3.350 2.228 4.023 33.626
30.000 3.515 2.032 4.060 30.035
40.000 3.576 1.855 4.028 27.412
50.000 3.554 1.708 3.943 25.672
60.000 3.473 1.592 3.820 24.635
70.000 3.350 1.499 3.671 24.107
80.000 3.203 1.420 3.503 23.915
90.000 3.040 1.348 3.326 23.915
100.000 2.872 1.278 3.144 23.988
110.000 2.706 1.207 2.963 24.039
120.000 2.548 1.135 2.789 24.001
130.000 2.403 1.062 2.627 23.834
140.000 2.274 0.990 2.480 23.533
150.000 2.164 0.925 2.354 23.134
160.000 2.075 0.869 2.249 22.719
170.000 2.005 0.826 2.168 22.404
180.000 1.953 0.802 2.111 22.326
190.000 1.917 0.798 2.076 22.614
200.000 1.892 0.817 2.061 23.365
210.000 1.875 0.860 2.063 24.632
220.000 1.862 0.925 2.079 26.417
230.000 1.848 1.011 2.107 28.678
240.000 1.832 1.115 2.145 31.340
250.000 1.810 1.235 2.192 34.306
260.000 1.784 1.367 2.248 37.463
270.000 1.754 1.508 2.313 40.683
280.000 1.723 1.654 2.388 43.826
290.000 1.698 1.804 2.477 46.730
300.000 1.687 1.955 2.582 49.207
310.000 1.702 2.105 2.707 51.038
320.000 1.761 2.251 2.858 51.965
330.000 1.883 2.386 3.040 51.715
340.000 2.088 2.494 3.253 50.064
350.000 2.380 2.550 3.488 46.967
360.000 2.731 2.523 3.718 42.731

2. Layout this linkage to scale, including the coupler curve whose coordinates are in the table above. Fit
tangent lines to the nearly straight portions of the curve. Label their intersection O6.
3. Design link 6 to lie along these straight tangents, pivoted at O6. Provide a guide on link 6 to accommodate
slider block 5, which pivots on the coupler point P.
8
45.000°
A
F
B
70.140°
O
2
O2
3
4
O4
6
6
5
P
D
E
7
8O
C
4. Extend link 6 a convenient distance to point C. Draw an arc through point C with center at O6. Label the
intersection of the arc with the other tangent line as point D. Attach link 7 to the pivot at C. The length of
link 7 is CE, a design choice. Extend line CDE from point E a distance equal to CD. Label the end point F.
Layout two intersecting lines through E and F such that they subtend an angle of 45 deg. Label their
intersection O8. The link joining O8 and point E is link 8. The link lengths and locations of O6 and O8 are:
Link 6 L6 2.330 Link 7 L7 3.000 Link 8 L8 3.498
Fixed pivot O6: x 1.892 Fixed pivot O8: x 1.379
y 0.762 y 6.690

PROBLEM 3-27
Statement: Use the linkage in Figure P3-9 to design an eightbar double-dwell mechanism that has a slider
output stroke of 5 crank units.
1. Enter the given data into program FOURBAR and print out the resulting coupler point coordinates (see table
below).
Step X Y Mag Ang
Deg
0.000 2.731 2.523 3.718 42.731
10.000 3.077 2.407 3.906 38.029
20.000 3.350 2.228 4.023 33.626
30.000 3.515 2.032 4.060 30.035
40.000 3.576 1.855 4.028 27.412
50.000 3.554 1.708 3.943 25.672
60.000 3.473 1.592 3.820 24.635
70.000 3.350 1.499 3.671 24.107
80.000 3.203 1.420 3.503 23.915
90.000 3.040 1.348 3.326 23.915
100.000 2.872 1.278 3.144 23.988
110.000 2.706 1.207 2.963 24.039
120.000 2.548 1.135 2.789 24.001
130.000 2.403 1.062 2.627 23.834
140.000 2.274 0.990 2.480 23.533
150.000 2.164 0.925 2.354 23.134
160.000 2.075 0.869 2.249 22.719
170.000 2.005 0.826 2.168 22.404
180.000 1.953 0.802 2.111 22.326
190.000 1.917 0.798 2.076 22.614
200.000 1.892 0.817 2.061 23.365
210.000 1.875 0.860 2.063 24.632
220.000 1.862 0.925 2.079 26.417
230.000 1.848 1.011 2.107 28.678
240.000 1.832 1.115 2.145 31.340
250.000 1.810 1.235 2.192 34.306
260.000 1.784 1.367 2.248 37.463
270.000 1.754 1.508 2.313 40.683
280.000 1.723 1.654 2.388 43.826
290.000 1.698 1.804 2.477 46.730
300.000 1.687 1.955 2.582 49.207
310.000 1.702 2.105 2.707 51.038
320.000 1.761 2.251 2.858 51.965
330.000 1.883 2.386 3.040 51.715
340.000 2.088 2.494 3.253 50.064
350.000 2.380 2.550 3.488 46.967
360.000 2.731 2.523 3.718 42.731

2. Layout this linkage to scale, including the coupler curve whose coordinates are in the table above. Fit
tangent lines to the nearly straight portions of the curve. Label their intersection O6.
3. Design link 6 to lie along these straight tangents, pivoted at O6. Provide a guide on link 6 to accommodate
slider block 5, which pivots on the coupler point P.
D
E
A
B
70.140°
O
2
O2
3
4
O4
6
5
P
8
F
6
C
7
4. Extend link 6 and the other tangent line until points C and E are 5 units apart. Attach link 7 to the pivot at C.
The length of link 7 is CD, a design choice. Extend line CDE from point D a distance equal to CE. Label the
end point F. As link 6 travels from C to E, slider block 8 will travel from D to F, a distance of 5 units. The
link lengths and location of O6:
Link 6 L6 4.351 Link 7 L7 2.000
Fixed pivot O6: x 1.892
y 0.762

PROBLEM 3-28
Statement: Use two of the cognates in Figure 3-26 (p. 126) to design a Watt-I sixbar parallel motion
mechanism that carries a link through the same coupler curve at all points. Comment on
its similarities to the original Roberts diagram.
Ground link L1 45 Crank L2 56 A1P 11.25 δ 0.000 deg
B1P L3
2
A1P
2
 2 L3 A1P cos δ 



0.5
 B1P 11.250
γ acos
L3
2
B1P
2
 A1P
2

2 L3 B1P








 γ 0.0000 deg
L5 B1P L5 11.250 L6
L4
L3
B1P L6 28.000
L10 A1P L10 11.250 L9
L2
L3
A1P L9 28.000
L7 L9
B1P
A1P
 L7 28.000 L8 L6
A1P
B1P
 L8 28.000
A3P L4 A3P 56.000 A2P L2 A2P 56.000
δ δ δ 0.000 deg δ δ δ 0.000 deg
L1BC
L1
L3
B1P L1BC 22.5000 L1AC
L1
L3
A1P L1AC 22.5000
Crank length L2 56.000 L10 11.250 L7 28.000

Coupler angle δ 0.000 deg δ 0.000 deg δ 0.000 deg
A
2 3
2 4
78
9
10
OA
O
2
C
3
B A
P
1 1
B B
5
6
A
O
3
B
4. Both of these cognates are identical. Following Example 3-11, discard cognate #2 and retain cognates #1
and #3. Without allowing links 5, 6, and 7 to rotate, slide them as an assembly along line OAOC until the free
end of link 7 is at OA. The free end of link 5 will then be at point O'B and point P on link 6 will be at P'. Add a
new link of length OAOC between P and P'. This is the new output link 8 and all points on it describe the
original coupler curve.
B
2 4
OA
5 A3
6
O'B
7
3
3
A
P
1
B1
8
P'
OB

5. Join links 2 and 7, making one ternary link. Remove link 5 and reduce link 6 to a binary link. The result is a
Watt-I sixbar with links numbered 1, 2, 3, 4, 6, and 8. Link 8 is in curvilinear translation and follows the
coupler path of the original point P. Link 8 is a binary link with nodes at P and P'. It does not attach to
link 4 at B1.
B
2 4
OA
3
3
A
P
1
B1
8
P'
6
OB

PROBLEM 3-29
Statement: Find the cognates of the Watt straight-line mechanism in Figure 3-29a (p. 131).
Ground link L1 4 Crank L2 2 A1P 0.500 δ 0.00 deg
Coupler L3 1 Rocker L4 2 B1P 0.500 γ 0.00 deg
Solution: See Figure 3-29a and Mathcad file P0329.
1. Input the link dimensions and coupler point data into program FOURBAR.
2. Use the Cognate pull-down menu to get the link lengths for cognates #2 and #3 (see next page). Note that,
for this mechanism, cognates #2 and #3 are identical. All three mechanisms are non-Grashof with limited
crank angles.

PROBLEM 3-30
Statement: Find the cognates of the Roberts straight-line mechanism in Figure 3-29b.
Ground link L1 2 Crank L2 1 A1P 1.000 δ 60.0 deg
Coupler L3 1 Rocker L4 1 B1P 1.000 γ 60.0 deg
Solution: See Figure 3-29b and Mathcad file P0330.
1. Input the link dimensions and coupler point data into program FOURBAR.
2. Note that, for this mechanism, cognates #2 and #3 are identical with cognate #1 because of the symmetry of
the linkage (draw the Cayley diagram to see this). All three mechanisms are non-Grashof with limited crank
angles.

PROBLEM 3-31
Statement: Design a Hoeken straight-line linkage to give minimum error in velocity over 22% of the cycle
for a 15-cm-long straight line motion. Specify all linkage parameters.
Given: Length of straight line motion: Δx 150 mm
Percentage of cycle over which straight line motion takes place: 22%
Link ratios from Table 3-1 for 22% cycle:
Link lengths:
Crank L2
Δx
ΔxoverL2
 L2 81.30 mm
Rocker L4 L3 L4 200.24 mm
Coupler point AP 2 L3 AP 400.49 mm
2. Calculate the distance from point P to pivot O4 (Cy) when crank angle is 180 deg.
Cy 2 L3 2
L1 L2 2
 Cy 319.20 mm
3. Enter the link lengths into program FOURBAR to verify the design (see next page for coupler point curve).
Using the PRINT facility, determine the x,y coordinates of the coupler curve and the x,y components of the
coupler point velocity in the straight line region. A table of these values is printed below. Notice the small
deviations over the range of crank angles from the y-coordinate and the x-velocity at a crank angle of 180
deg.
FOURBAR for Windows File P03-31.DOC
Angle Cpler Pt Cpler Pt Veloc CP Veloc CP
Step X Y X Y
Deg mm mm mm/sec mm/sec
140 235.60 319.95 -1,072.61 -10.73
150 216.84 319.72 -1,076.20 -14.74
160 198.06 319.46 -1,075.51 -13.54
170 179.31 319.27 -1,073.75 -7.99
180 160.58 319.20 -1,072.93 0.02
190 141.85 319.27 -1,073.75 8.03
200 123.09 319.47 -1,075.52 13.58
210 104.31 319.72 -1,076.22 14.78
220 85.55 319.95 -1,072.63 10.76

PROBLEM 3-32
Statement: Design a Hoeken straight-line linkage to give minimum error in straightness over 39% of the
cycle for a 20-cm-long straight line motion. Specify all linkage parameters.
Given: Length of straight line motion: Δx 200 mm
Percentage of cycle over which straight line motion takes place: 39%
Link ratios from Table 3-1 for 39% cycle:
Link lengths:
Crank L2
Δx
ΔxoverL2
 L2 55.20 mm
Rocker L4 L3 L4 179.41 mm
Coupler point AP 2 L3 AP 358.82 mm
2. Calculate the distance from point P to pivot O4 (Cy) when crank angle is 180 deg.
Cy 2 L3 2
L1 L2 2
 Cy 302.36 mm
3. Enter the link lengths into program FOURBAR to verify the design (see next page for coupler point curve).
Using the PRINT facility, determine the x,y coordinates of the coupler curve and the x,y components of the
coupler point velocity in the straight line region. A table of these values is printed below. Notice the small
deviations over the range of crank angles from the y-coordinate and the x-velocity from a crank angle of 180
deg.
Angle Coupler Pt Coupler Pt Veloc CP Veloc CP
Step X Y X Y
Deg mm mm mm/sec mm/sec
110 237.992 302.408 -696.591 -6.416
120 225.289 302.361 -755.847 -0.019
130 211.710 302.378 -797.695 1.426
140 197.521 302.398 -826.217 0.664
150 182.927 302.399 -844.774 -0.483
160 168.076 302.385 -856.043 -1.052
170 153.076 302.368 -861.994 -0.800
180 138.010 302.360 -863.841 0.000
190 122.944 302.368 -861.994 0.800
200 107.944 302.385 -856.043 1.052
210 93.093 302.399 -844.774 0.483
220 78.499 302.398 -826.217 -0.664
230 64.311 302.378 -797.695 -1.426
240 50.731 302.361 -755.847 0.019
250 38.028 302.408 -696.591 6.416

PROBLEM 3-33
Statement: Design a linkage that will give a symmetrical "kidney bean" shaped coupler curve as shown in
Figure 3-16 (p. 114 and 115). Use the data in Figure 3-21 (p. 120) to determine the required link
ratios and generate the coupler curve with program FOURBAR.
Design choices:
Angle PAB δ
180 deg γ
2
 δ 54.000 deg
2. Enter the above data into program FOURBAR and plot the coupler curve.

PROBLEM 3-34
Statement: Design a linkage that will give a symmetrical "double straight" shaped coupler curve as shown
in Figure 3-16. Use the data in Figure 3-21 to determine the required link ratios and generate
the coupler curve with program FOURBAR.
Design choices:
Angle PAB δ
180 deg γ
2
 δ 36.000 deg

PROBLEM 3-35
Statement: Design a linkage that will give a symmetrical "scimitar" shaped coupler curve as shown in
Figure 3-16. Use the data in Figure 3-21 to determine the required link ratios and generate
the coupler curve with program FOURBAR. Show that there are (or are not) true cusps on
the curve.
Design choices:
Angle PAB δ
180 deg γ
2
 δ 18.000 deg

3. The points at the ends of the "scimitar" will be true cusps if the velocity of the coupler point is zero at these
points. Using FOURBAR's plotting utility, plot the magnitude and angle of the coupler point velocity vector.
As seen below for the range of crank angle from 50 to 70 degrees, the magnitude of the velocity does not
quite reach zero. Therefore, these are not true cusps.

PROBLEM 3-36
Statement: Find the Grashof condition, inversion, any limit positions, and the extreme values of the
transmission angle (to graphical accuracy) of the linkage in Figure P3-10.
Link 4 L4 0.950 Link 1 L1 0.544
L max a b c d( )
SL S L

Table 2-4.
Barker classification: Class I-3, Grashof rocker-crank-rocker, GRCR, since the shortest link
is the coupler link.
2. A GRCR linkage will have two toggle positions. Draw the linkage in these two positions and measure the
input link angles.
O
158.286°
158.286°A
B
A
B
O
O2
2
4
O4
3. As measured from the layout, the input link angles at the toggle positions are: +158.3 and -158.3 deg.
4. Since the coupler link in a GRCR linkage can make a full rotation with respect to the input and output
rockers, the minimum transmission angle is 0 deg and the maximum is 90 deg.

PROBLEM 3-37
B1P L3
2
A1P
2
 2 L3 A1P cos δ 



0.5
 B1P 0.734
γ acos
L3
2
B1P
2
 A1P
2

2 L3 B1P








 γ 180.0000 deg
L5 B1P L5 0.734 L6
L4
L3
B1P L6 1.959
L10 A1P L10 1.090 L9
L2
L3
A1P L9 2.404
L7 L9
B1P
A1P
 L7 1.619 L8 L6
A1P
B1P
 L8 2.909
A3P L4 A3P 0.950 A2P L2 A2P 0.785
δ 180 deg δ δ 180.000 deg δ δ δ 0.000 deg
L1BC
L1
L3
B1P L1BC 1.1216 L1AC
L1
L3
A1P L1AC 1.6656
Crank length L2 0.785 L10 1.090 L7 1.619

6
7
B3
2
3
4
1
510
A
OA
1
A
OB
2
3A
8
9
B
P
1
OC
B2

PROBLEM 3-38
Statement: Find the three geared fivebar cognates of the linkage in Figure P3-10.
B1P L3
2
A1P
2
 2 L3 A1P cos δ 



0.5
 B1P 0.734
γ acos
L3
2
B1P
2
 A1P
2

2 L3 B1P








 γ 180.0000 deg
L5 B1P L5 0.734 L6
L4
L3
B1P L6 1.959
L10 A1P L10 1.090 L9
L2
L3
A1P L9 2.404
L7 L9
B1P
A1P
 L7 1.619 L8 L6
A1P
B1P
 L8 2.909
A3P L4 A3P 0.950 A2P L2 A2P 0.785
δ 180 deg δ δ 180.000 deg δ δ δ 0.000 deg
L1BC
L1
L3
B1P L1BC 1.1216 L1AC
L1
L3
A1P L1AC 1.6656
Crank length L2 0.785 L10 1.090 L7 1.619

6
7
B3
2
3
4
1
510
A
OA
1
A
OB
2
3A
8
9
B
P
1
OC
B2
4. The three geared fivebar cognates can be seen in the Roberts diagram. They are: OAA2PA3OB, OAA1PB3OC,
and OBB1PB2OC. They are specified in the summary table below.
Crank length L10 1.090 L2 0.785 L4 0.950
Crank length L5 0.734 L7 1.619 L8 2.909
5. Enter the cognate #1 specifications into program FOURBAR to get a trace of the coupler path (see next page).

PROBLEM 3-39
Statement: Find the Grashof condition, any limit positions, and the extreme values of the transmission
angle (to graphical accuracy) of the linkage in Figure P3-11.
Link 4 L4 0.86 Link 1 L1 2.22
L max a b c d( )
SL S L

Table 2-4.
Grashof condition: Condition L1 L2 L3 L4  "non-Grashof"
Barker classification: Class II-1, non-Grashof triple rocker, RRR1, since the longest link is
the ground link.
2. An RRR1 linkage will have two toggle positions. Draw the linkage in these two positions and measure the
input link angles.
116.037°
116.037°
A
88.2°
67.3°
O2
A
O4
B
A
O2
B
O4 O2
B
O4
3. As measured from the layout, the input link angles at the toggle positions are: +116 and -116 deg.
4. Since the coupler link in an RRR1 linkage cannot make a full rotation with respect to the input and output
rockers, the minimum transmission angle is 0 deg and the maximum is 88 deg.

PROBLEM 3-40
B1P L3
2
A1P
2
 2 L3 A1P cos δ 



0.5
 B1P 0.520
γ acos
L3
2
B1P
2
 A1P
2

2 L3 B1P








 γ 0.0000 deg
L5 B1P L5 0.520 L6
L4
L3
B1P L6 0.242
L10 A1P L10 1.330 L9
L2
L3
A1P L9 0.618
L7 L9
B1P
A1P
 L7 0.242 L8 L6
A1P
B1P
 L8 0.618
A3P L8 A3P 0.618 A2P L7 A2P 0.242
δ 180 deg δ 180.000 deg δ 180 deg δ 180.000 deg
L1BC
L1
L3
B1P L1BC 0.6240 L1AC
L1
L3
A1P L1AC 1.5960
Crank length L2 0.860 L10 1.330 L7 0.242

A B1 3
9
10
7
2
8
O O
A2
A C
B2
B
3
4
56
O
P
A
1
B3

PROBLEM 3-41
B1P L3
2
A1P
2
 2 L3 A1P cos δ 



0.5
 B1P 0.520
γ acos
L3
2
B1P
2
 A1P
2

2 L3 B1P








 γ 0.0000 deg
L5 B1P L5 0.520 L6
L4
L3
B1P L6 0.242
L10 A1P L10 1.330 L9
L2
L3
A1P L9 0.618
L7 L9
B1P
A1P
 L7 0.242 L8 L6
A1P
B1P
 L8 0.618
A3P L8 A3P 0.618 A2P L7 A2P 0.242
δ 180 deg δ 180.000 deg δ 180 deg δ 180.000 deg
L1BC
L1
L3
B1P L1BC 0.6240 L1AC
L1
L3
A1P L1AC 1.5960
Crank length L2 0.860 L10 1.330 L7 0.242

A B1 3
9
10
7
2
8
O O
A2
A C
B2
B
3
4
56
O
P
A
1
B3
4. The three geared fivebar cognates can be seen in the Roberts diagram. They are: OAB2PB3OB, OAA1PA3OC,
and OBB1PA2OC. The three geared fivebar cognates are summarized in the table below.
Crank length L10 1.330 L2 0.860 L4 0.860
Coupler length L2 0.860 A1P 1.330 L5 0.520
Crank length L5 0.520 L7 0.242 L8 0.618
Coupler point L2 0.860 A1P 1.330 B1P 0.520
5. Enter the cognate #1 specifications into program FOURBAR to get a trace of the coupler path (see next page)
6. Enter the geared fivebar cognate #1 specifications into program FIVEBAR to get a trace of the coupler path
for the geared fivebar (see next page).

PROBLEM 3-42
Statement: Find the Grashof condition, any limit positions, and the extreme values of the transmission
angle (to graphical accuracy) of the linkage in Figure P3-12.
Link 4 L4 0.85 Link 1 L1 1.82
L max a b c d( )
SL S L

Table 2-4.
Grashof condition: Condition L1 L2 L3 L4  "non-Grashof"
Barker classification: Class II-1, non-Grashof triple rocker, RRR1, since the longest link
is the ground link.
2. An RRR1 linkage will have two toggle positions. Draw the linkage in these two positions and measure the
input link angles.
O2
B
O4
55.4°
88.8°
A
A
55.4°
O2
A
B O4
O2
B
O4
3. As measured from the layout, the input link angles at the toggle positions are: +55.4 and -55.4 deg.
4. Since the coupler link in an RRR1 linkage it cannot make a full rotation with respect to the input and output
rockers, the minimum transmission angle is 0 deg and the maximum is 88.8 deg.

PROBLEM 3-43
B1P L3
2
A1P
2
 2 L3 A1P cos δ 



0.5
 B1P 0.792
γ acos
L3
2
B1P
2
 A1P
2

2 L3 B1P








 γ 82.0315 deg
L5 B1P L5 0.792 L6
L4
L3
B1P L6 0.990
L10 A1P L10 0.970 L9
L2
L3
A1P L9 1.027
L7 L9
B1P
A1P
 L7 0.839 L8 L6
A1P
B1P
 L8 1.212
A3P L4 A3P 0.850 A2P L2 A2P 0.720
δ γ δ 82.032 deg δ δ δ 54.000 deg
L1BC
L1
L3
B1P L1BC 2.1208 L1AC
L1
L3
A1P L1AC 2.5962
Crank length L2 0.720 L10 0.970 L7 0.839

A2
2
3
4
1
10
5
1AC
A
B
OA
1
1
A3
9
8
P
B
OC
2
1BC
OB
7
6
B3

PROBLEM 3-44
B1P L3
2
A1P
2
 2 L3 A1P cos δ 



0.5
 B1P 0.792
γ acos
L3
2
B1P
2
 A1P
2

2 L3 B1P








 γ 82.0315 deg
L5 B1P L5 0.792 L6
L4
L3
B1P L6 0.990
L10 A1P L10 0.970 L9
L2
L3
A1P L9 1.027
L7 L9
B1P
A1P
 L7 0.839 L8 L6
A1P
B1P
 L8 1.212
A3P L4 A3P 0.850 A2P L2 A2P 0.720
δ γ δ 82.032 deg δ δ δ 54.000 deg
L1BC
L1
L3
B1P L1BC 2.1208 L1AC
L1
L3
A1P L1AC 2.5962
Crank length L2 0.720 L10 0.970 L7 0.839

A2
2
3
4
1
10
5
1AC
A
B
OA
1
1
A3
9
8
P
B
OC
2
1BC
OB
7
6
B3
4. The three geared fivebar cognates can be seen in the Roberts diagram. They are: OAA2PA3OB, OAA1PB3OC, and
OBB1PB2OC.
Crank length L10 0.970 L2 0.720 L4 0.850
Crank length L5 0.792 L7 0.839 L8 1.212
5. Enter the cognate #1 specifications into program FOURBAR to get a trace of the coupler path (see next page).

PROBLEM 3-45
Statement: Prove that the relationships between the angular velocities of various links in the Roberts
diagram as shown in Figure 3-25 (p. 125) are true.
Given: OAA1PA2, OCB2PB3, and OBB1PA3 are parallelograms for any position of link 2..
Proof:
1. OAA1 and A2P are opposite sides of a parallelogram and are, therefore, always parallel.
2. Any change in the angle of OAA1 (link 2) will result in an identical change in the angle of A2P.
3. Angular velocity is the change in angle per unit time.
4. Since OAA1 and A2P have identical changes in angle, their angular velocities are identical.
5. A2P is a line on link 9 and all lines on a rigid body have the same angular velocity. Therefore, link 9 has the
same angular velocity as link 2.
6. OCB3 (link 7) and B2P are opposite sides of a parallelogram and are, therefore, always parallel.
7. B2P is a line on link 9 and all lines on a rigid body have the same angular velocity. Therefore, link 7 has the
same angular velocity as links 9 and 2.
8. The same argument holds for links 3, 5, and 10; and links 4, 6, and 8.

PROBLEM 3-46
Statement: Design a fourbar linkage to move the object in Figure P3-13 from position 1 to 2 using points A
and B for attachment. Add a driver dyad to limit its motion to the range of positions shown,
making it a sixbar. All fixed pivots should be on the base.
Given: Length of coupler link: L3 52.000
Design choices:
Length of link 2 L2 130 Length of link 4 L4 110
Length of link 2b L2b 40
1. Connect the end points of the two given positions of the line AB with construction limes, i.e., lines from A1
to A2 and B1 to B2.
2. Bisect these lines and extend their perpendicular bisectors into the base.
3. Select one point on each bisector and label them O2 and O4, respectively. In the solution below the
distances O2
A was selected to be L2 130.000 and O4
B to be L4 110.000 . This resulted in a
ground-link-length O2O4 for the fourbar of 27.080.
Ground link 1a L1a 27.080 Link 2 (input) L2 130.000
Link 3 (coupler) L3 52.000 Link 4 (output) L4 110.000
5. Select a point on link 2 (O2A) at a suitable distance from O2 as the pivot point to which the driver dyad will be
connected and label it C. (Note that link 2 is now a ternary link with nodes at O2, C, and A.) In the solution
below the distance O2
C was selected to be L2b 40.000 .
6. Draw a construction line through C1C2 and extend it to the left.
7. Select a point on this line and call it O6. In the solution below O6 was placed 20 units from the left edge of the
base.
40.000
111.764
106.866
23.003
6
D O61 D
5
2
2
A
3
1
27.080
O
2O
4
4
C1
C2
B1
A
2B
2
8. Draw a circle about O6 with a radius of
one-half the length C1C2 and label the
intersections of the circle with the
extended line as D1 and D2. In the
solution below the radius was measured as
23.003 units.
9. The driver fourbar is now defined as
O2CDO6 with link lengths
Link 6 (crank) L6 23.003
Link 5 (coupler) L5 106.866

L max a b c d( )
SL S L

Condition L1b L2b L5 L6  "Grashof"
min L1b L2b L5 L6  23.003

PROBLEM 3-47
Design choices:
to A3 and B2 to B3.
distances O2
5. Select a point on link 4 (O4B) at a suitable distance from O4 as the pivot point to which the driver dyad
will be connected and label it C. (Note that link 4 is now a ternary link with nodes at O4, C, and B.) In
the solution below the distance O4
6. Draw a construction line through C2C3 and extend it downward.
7. Select a point on this line and call it O6. In the solution below O6 was placed 20 units from the bottom of the
base.
6
1a
111.758
92.425 O
C
83.977
1b
10.480
D3
D2
O4
C3
5
2 4
6
O2
B
A
2
A
B
3
10.480 units.
Link 6 (crank) L6 10.480

L max a b c d( )
SL S L

min L1b L4b L5 L6  10.480

PROBLEM 3-48
Statement: Design a fourbar linkage to move the object in Figure P3-13 through the three positions shown
using points A and B for attachment. Add a driver dyad to limit its motion to the range of
positions shown, making it a sixbar. All fixed pivots should be on the base.
Design choices:
1. Draw link AB in its three design positions A1B1, A2B2, A3B3 in the plane as shown.
2. Draw construction lines from point A1 to A2 and from point A2 to A3.
3. Bisect line A1A2 and line A2A3 and extend their perpendicular bisectors until they intersect. Label their
intersection O2.
4. Repeat steps 2 and 3 for lines B1B2 and B2B3. Label the intersection O4.
5. Connect O2 with A1 and call it link 2. Connect O4 with B1 and call it link 4.
6. Line A1B1 is link 3. Line O2O4 is link 1 (ground link for the fourbar). The fourbar is now defined as O2ABO4
Link 3 L3 52.000 Link 4 L4 120.254
O
O 1
6
C3
2
5
D3
6
2
D
A
3
1
C
O
C2
4
1
B3
3A
2
4
B1
A
2B

L max a b c d( )
SL S L

the solution above the distance O4
10. Select a point on this line and call it O6. In the solution above O6 was placed 20 units from the left edge of
the base.
11. Draw a circle about O6 with a radius of one-half the length C1C3 and label the intersections of the circle
with the extended line as D1
and D3
min L6 L1b L4b L5  45.719

PROBLEM 3-49
Design choices:
to A2 and B1 to B2.
distances O2
5. Select a point on link 4 (O4B) at a suitable distance from O4 as the pivot point to which the driver dyad will be
connected and label it C. (Note that link 4 is now a ternary link with nodes at O4, C, and B.) In the solution
below the distance O4
130.479
137.327
A
6
D
25.808
O61
1b
5
D2
1B
3
1
4
97.195
O4
C1
1a
C2
A
2
O2
B2
2
7. Select a point on this line and call it O6. In
the solution below O6 was placed 20 units
from the left edge of the base.
25.808 units.
Link 6 (crank) L6 25.808

10. Use the link lengths in step 9 to find the Grashof condition of the driving fourbar (it must be Grashof and
L max a b c d( )
SL S L

min L1b L4b L5 L6  25.808

PROBLEM 3-50
Design choices:
to A3 and B2 to B3.
distances O2
5. Select a point on link 2 (O2A) at a suitable distance from O2 as the pivot point to which the driver dyad will
be connected and label it C. (Note that link 4 is now a ternary link with nodes at O2, C, and A.) In the
solution below the distance O2
C was selected to be L2b 50.000 and the link was extended away from A
to give a better position for the driving dyad.
7. Select a point on this line and call it O6. In the solution below O6 was placed 35 units from the bottom of the
base.
107.974
98.822
5
O6
D3
C2
C3
67.395
155°
24.647
6
D2
O4
O
1a
1b
2
4
3B
2 B2
3
A2
A3
24.647 units.
Link 6 (crank) L6 24.647

L max a b c d( )
SL S L

min L1b L2b L5 L6  24.647

PROBLEM 3-51
Statement: Design a fourbar linkage to move the object in Figure P3-14 through the three positions shown
using points A and B for attachment. Add a driver dyad to limit its motion to the range of
positions shown, making it a sixbar. All fixed pivots should be on the base.
Design choices:
intersection O2.
Link 3 L3 86.000 Link 4 L4 124.668
C
O
O
4
C3
1b
D1
5
6
D3
B
O6
1
2
3
A1
2 C1
4 1a
B3
2
B2
A 2
A3

L max a b c d( )
SL S L

will be connected and label it C. (Note that link 4 is now a ternary link with nodes at O4, C, and B.) In the
solution above the distance O4
the base.
11. Draw a circle about O6 with a radius of one-half the length C1C3 and label the intersections of the
circle with the extended line as D1 and D3. In the solution below the radius was measured as
L6 45.178 .
14. Unfortunately, although the solution presented appears to meet the design specification, a simple cardboard
model will quickly demonstrate that it has a branch defect. That is, in the first position shown, the linkage is
in the "open" configuration, but in the 2nd and 3rd positions it is in the "crossed" configuration. The
linkage cannot get from one circuit to the other without removing a pin and reassembling after moving the
linkage. The remedy is to attach the points A and B to the coupler, but not at the joints between links 2 and
3 and links 3 and 4.

PROBLEM 3-52
Design choices:
to A2 and B1 to B2.
3. Select one point on each bisector and label them O2 and O4, respectively. In the solution below
the distances O2
B to be L4 160.000 . This resulted in
a ground-link-length O2O4 for the fourbar of 81.463.
7. Select a point on this line and call it O6. In the solution below O6 was placed 20 units from the left edge of
the base.
2
132.962
138.105
14.351
DD 21
6
O6
1b
5
A
3
1
81.463
O2
4O
C1
1a C2
4
B1
A 2
B2
14.351 units.
Link 6 (crank) L6 14.351

L max a b c d( )
SL S L

min L1b L4b L5 L6  14.351

PROBLEM 3-53
Design choices:
to A3 and B2 to B3.
3. Select one point on each bisector and label them O2 and O4, respectively. In the solution below
the distances O2
B to be L4 200.000 . This resulted in
a ground-link-length O2
O4
for the fourbar of L1a 80.864 .
Ground link 1a L1a 80.864 Link 2 (input) L2 150.000
7. Select a point on this line and call it O6. In the solution below O6 was placed 25 units from the bottom of
the base.
12.763
80.864
122.445
112.498
O4
O6
51b
1a
C2
B
D3
O
D2
2
C3
3
2
4
3 B2
A2
A3
L6 12.763 .

L max a b c d( )
SL S L

min L1b L4b L5 L6  12.763

PROBLEM 3-54
Statement: Design a fourbar linkage to move the object in Figure P3-15 through the three positions
shown using points A and B for attachment. Add a driver dyad to limit its motion to the range
of positions shown, making it a sixbar. All fixed pivots should be on the base.
Design choices:
intersection O2.
Link 3 L3 52.000 Link 4 L4 90.203
1b
O
6
D1
2
5
D3
3
6
A1
1aC
O2
3
B3
C C1 2
4
O4
B2
B1
A2
A3

L max a b c d( )
SL S L

Condition L1a L2 L3 L4  "non-Grashof"
Although this fourbar is non-Grashof, there are no toggle points within the required range of motion.
8. Select a point on link 2 (O2A) at a suitable distance from O2 as the pivot point to which the driver dyad
will be connected and label it C. (Note that link 2 is now a ternary link with nodes at O2, C, and A.) In
the solution above the distance O2
the base.
11. Draw a circle about O6 with a radius of one-half the length C1C3 and label the intersections of the
circle with the extended line as D1 and D3. In the solution below the radius was measured as
L6 29.760 .

PROBLEM 3-55
Statement: Design a fourbar mechanism to move the link shown in Figure P3-16 from position 1 to position
2. Ignore the third position and the fixed pivots O2 and O4 shown. Build a cardboard model
and add a driver dyad to limit its motion to the range of positions designed, making it a sixbar.
Given: Position 1 offsets: xC1D1 3.744 in yC1D1 2.497 in
1. Connect the end points of the two given positions of the line CD with construction lines, i.e., lines from C1
to C2 and D1 to D2.
2. Bisect these lines and extend their perpendicular bisectors in any convenient direction. In the solution
below the bisector of C1C2 was extended downward and the bisector of D1D2 was extended upward.
distances O4D and O6C were each selected to be 7.500 in. This resulted in a ground-link-length O4O6
for the fourbar of 15.366 in.
4. The fourbar stage is now defined as O4CDO6 with link lengths
Link 5 (coupler) L5 xC1D1
2
yC1D1
2
 L5 4.500 in
5. Select a point on link 4 (O4D) at a suitable distance from O4 as the pivot point to which the driver dyad will
be connected and label it B. (Note that link 4 is now a ternary link with nodes at O4, B, and D.) In the
solution below the distance O4B was selected to be 4.000 in.
6. Draw a construction line through B1B2 and extend it to the right.
7. Select a point on this line and call it O2. In the solution below the distance AB was selected to be 6.000 in.
with the extended line as A1 and A2. In the solution below the radius was measured as 1.370 in.
9. The driver fourbar is now defined as O2ABO4 with link lengths
L max a b c d( )
SL S L


Condition L1a L2 L3 L4a  "Grashof"
min L1a L2 L3 L4a  1.370 in
15.366
7.500
2
6
A7.500
4.000
3
1
2
1A O2
D 2
C 1
6
O6
4
5
D 1
B
4
2C
5
4O
B2
11. Using the program FOURBAR and the link lengths given above, it was found that the fourbar O4DCO6
is non-Grashoff with toggle positions at 4 = -49.9 deg and +49.9 deg. The fourbar operates between
4 = +28.104 deg and -11.968 deg.

PROBLEM 3-56
Solution: See figure below for one possible solution. Input file P0356.mcd from the solutions manual
disk to the Mathcad program for this solution, file P03-56.4br to the program FOURBAR to
see the fourbar solution linkage, and file P03-56.6br into program SIXBAR to see the
complete sixbar with the driver dyad included.
to C3 and D2 to D3.
distances O4D and O6C were each selected to be 6.000 in. This resulted in a ground-link-length
O4O6 for the fourbar of 14.200 in.
4. The fourbar stage is now defined as O4DCO6 with link lengths
2
yC2D2
2
 L5 4.500 in
5. Select a point on link 4 (O4D) at a suitable distance from O4 as the pivot point to which the driver dyad
will be connected and label it B. (Note that link 4 is now a ternary link with nodes at O4, B, and D.) In
the solution below the distance O4B was selected to be 4.000 in.
L max a b c d( )
SL S L


min L1a L2 L3 L4a  1.271 in
O
14.200
2C
6.000
6
3
7.099
6
6
C 3
5
D 3
5
O2
A1
2
A2
2
6.000
4.000 O
D 2
4
4
B
1B
4
4 = +26.171 deg and -11.052 deg.

PROBLEM 3-57
Statement: Design a fourbar mechanism to give the three positions shown in Figure P3-16. Ignore the
points O2 and O4 shown. Build a cardboard model and add a driver dyad to limit its motion to
the range of positions designed, making it a sixbar.
Design choices:
intersection O6.
6. Line C1D1 is link 5. Line O6O4 is link 1a (ground link for the fourbar). The fourbar is now defined as
O6CDO4 and has link lengths of
Link 5 L5 4.500 Link 4 L4 6.901
C
D 3
O
4
4
O46.080
6.901
2.616
4
B3
A3
A1
O
10.611
6
6
6
6
3
2
2.765
2
1C
5
B1
D 1
C2
B2
3
5
5
D2
L max a b c d( )
SL S L


will be connected and label it B. (Note that link 4 is now a ternary link with nodes at O4, D, and B.) In the
and A3
Condition L2 L3 L1b L4b  "Grashof"
min L2 L3 L1b L4b  2.765

PROBLEM 3-58
pivots O2 and O4 shown. (See Example 3-7.) Build a cardboard model and add a driver dyad
to limit its motion to the range of positions designed, making it a sixbar.
Design choices: Length of link 5: L5 5.000 Length of link 2b: L2b 2.500
position.
position.
6. Repeat the process for the third coupler position and transfer the third relative ground link position to the
first, or reference, position.
O '2O ''2
O ''4 O '4
O2
1
C
1D
O4
2
2C
3C
D
3D

intersection G.
Link 3 L3 1.711 Link 4 L4 7.921
2EE3
G
FF3 2
O2E1
1a
2
3
H
O4
F1
4
L max a b c d( )
SL S L

The fourbar that will provide the desired motion is now defined as a Grashof double crank in the crossed
configuration. It now remains to add the original points C1 and D1 to the coupler GH and to define the
driving dyad.

15. Draw a construction line through B1B3 and extend it up to the left.
and A3
1a
A
6
1A
O6
1G
1B5
3
O2
2
3
B2 3
1H 2
G2
1C
1
D
3
O4
H
4
4
H
4
2
D
G3
2C 3
3
3C
D3
2

PROBLEM 3-59
to C2 and D1 to D2.
distances O6C and O4D were each selected to be 6.500 in. This resulted in a ground-link-length O4O6
2
yC1D1
2
 L5 2.250 in
L max a b c d( )
SL S L


min L1a L2 L3 L4a  0.645 in
O
4.500
6.500
14.722
6.500
5
6O
1
6
C
6
C 2
A2
D 1
3
5
D 2
4
B2
B1
4
7.472
0.645
A1
2
2
4O
11. Using the program FOURBAR and the link lengths given above, it was found that the fourbar O4CDO6
is non-Grashoff with toggle positions at 4 = -17.1 deg and +17.1 deg. The fourbar operates between 4
= +5.216 deg and -11.273 deg.

PROBLEM 3-60
to C3 and D2 to D3.
distances O4D and O6C were each selected to be 6.000 in. This resulted in a ground-link-length O4O6
2
yC2D2
2
 L5 2.250 in
L max a b c d( )
SL S L


5
C 3
O6
6
6
4.000
5.500
12.933
7.173
D
O4
A
3
C 2
5
D 2
3 3
4
B3
B2
4
A
2
2
2
O
4 = +12.403 deg and -8.950 deg.

PROBLEM 3-61
points O2 and O4 shown. Build a cardboard model and add a driver dyad to limit its motion
to the range of positions designed, making it a sixbar.
Design choices:
intersection O6.
6. Line C1D1 is link 5. Line O2O4 is link 1a (ground link for the fourbar). The fourbar is now defined as O6CDO4
Link 5 L5 2.250 Link 4 L4 3.323
6.347
1
3
3.323
1.835 B1
2.967
O
O6
6
1C
56
6
C24
D
D
B
2
2
B3
5
4 5
C3
4
4
D3
1.403
O2
2 A1
A3

L max a b c d( )
SL S L

will be connected and label it B. (Note that link 4 is now a ternary link with nodes at O4, D, and B.) In the
and A3
Condition L1b L2 L3 L4b  "Grashof"
min L1b L2 L3 L4b  1.403

PROBLEM 3-62
pivots O2 and O4 shown. (See Example 3-7.) Build a cardboard model and add a driver dyad to
limit its motion to the range of positions designed, making it a sixbar.
position.
position.
O ''
O '2
2
O2
3D
C2
O ''4
O '4
O4
1
C
3C
2D
D1

intersection G.
Link 3 L3 1.222 Link 4 L4 1.950
1a
3
2E
O
1
E
2
G
2
3
E
F3
F2
H
O4
4
F1
L max a b c d( )
SL S L

The fourbar that will provide the desired motion is now defined as a non-Grashof double rocker in the
crossed configuration. It now remains to add the original points C1 and D1 to the coupler GH and to
define the driving dyad, which in this case will drive link 4 rather than link 2.
14. Select a point on link 2 (O2G) at a suitable distance from O2 as the pivot point to which the driver dyad
will be connected and label it B. (Note that link 2 is now a ternary link with nodes at O2, B, and G.) In the
solution below, the distance O2
B was selected to be L2b 0.791 . Thus, in this case B and G coincide.

and A3
O2
2
G
35
1b
D3
6
O6
A3
3C
A1
C2
H
3
O4
4
4
4
1
C
3
2D
D1

PROBLEM 3-63
1. Connect the end points of the two given positions of the line CD with construction lines, i.e., lines from C1 to
C2 and D1 to D2.
the bisector of C1C2 was extended downward and the bisector of D1D2 was extended upward.
O4C and O6D were each selected to be 5.000 in. This resulted in a ground-link-length O4O6 for the fourbar of
10.457 in.
2
yC1D1
2
 L5 2.250 in
5. Select a point on link 4 (O4C) at a suitable distance from O4 as the pivot point to which the driver dyad will be
connected and label it B. (Note that link 4 is now a ternary link with nodes at O4, B, and C.) In the solution
below the distance O4B was selected to be 3.750 in.
8. Draw a circle about O2 with a radius of one-half the length B1B2 and label the intersections of the circle with
the extended line as A1 and A2. In the solution below the radius was measured as 0.882 in.
L max a b c d( )
SL S L


4
5
4
3.7505.000 7.020
10.457
C 2
B2
O4
6
D
5
B
C
3
1
1
D 2
OA2
2 2
1
6
O6
A1
11. Using the program FOURBAR and the link lengths given above, it was found that the fourbar O4CDO6 is
non-Grashoff with toggle positions at 4 = -38.5 deg and +38.5 deg. The fourbar operates between 4 =
+15.206 deg and -12.009 deg.

PROBLEM 3-64
1. Connect the end points of the two given positions of the line CD with construction lines, i.e., lines from C2 to
C3 and D2 to D3.
the bisector of C2C3 was extended downward and the bisector of D2D3 was extended upward.
O4D and O6C were each selected to be 5.000 in. This resulted in a ground-link-length O4O6 for the fourbar of
8.773 in.
2
yC2D2
2
 L5 2.250 in
connected and label it B. (Note that link 4 is now a ternary link with nodes at O4, B, and D.) In the solution
below the distance O4B was selected to be 3.750 in.
the extended line as A1 and A2. In the solution below the radius was measured as 0.892 in.
L max a b c d( )
SL S L


D
C
C 2
3
3
O4
4
O66
5
B3
A
7.019
8.773
3.750
5.000 3
3
5
6
D
B2
2
4
2A O 2 2
11. Using the program FOURBAR and the link lengths given above, it was found that the fourbar O4DCO6 is
non-Grashoff with toggle positions at 4 = -55.7 deg and +55.7 deg. The fourbar operates between 4
= -7.688 deg and -35.202 deg.

PROBLEM 3-65
points O2 and O4 shown. Build a cardboard model and add a driver dyad to limit its motion to
the range of positions designed, making it a sixbar.
Design choices:
intersection O6.
6. Line C1D1 is link 5. Line O6O4 is link 1a (ground link for the fourbar). The fourbar is now defined as
O6CDO4 and has link lengths of
Link 5 L5 2.250 Link 4 L4 6.953
7.646
1.593
D
6.953
8.869
5
1C
D
5
1.831
O6
6
6
1
2
C 3
C2
O
B
4
4 B
1
3
4
2
D 3
3
O2
A1
A3

L max a b c d( )
SL S L

Condition L6 L1a L4 L5  "non-Grashof"
connected and label it B. (Note that link 4 is now a ternary link with nodes at O4, D, and B.) In the solution
above the distance O4
and A3
Condition L1b L2 L3 L4b  "Grashof"
min L1b L2 L3 L4b  1.593

PROBLEM 3-66
pivots O2 and O4 shown. (See Example 3-7.) Build a cardboard model and add a driver dyad to
limit its motion to the range of positions designed, making it a sixbar.
position.
position.
6. Repeat the process for the third coupler position and transfer the third relative ground link position to the first,
or reference, position.
3
C
O2
3C
D
2
O '2
4O O ''4
O '4
1
D2
C
O ''2
D1
intersection G.

Link 3 L3 6.002 Link 4 L4 7.002
G
2
O2
1E 12E F
4O F3
F2
3
E3
4
H
L max a b c d( )
SL S L

The fourbar that will provide the desired motion is now defined as a non-Grashof crank rocker in the open
configuration. It now remains to add the original points C1 and D1 to the coupler GH and to define the driving
dyad, which in this case will drive link 4 rather than link 2.

B was selected to be L2b 2.000 . Thus, in this case B and G coincide.
the extended line as A1
and A3
G
A3
C
2
6
O6
A1
5
1b
G2
2
O2
C3
3
2
H3
3D
2
3
1a
41
G
4
4
O4
D
C1
2
3
3 D1
1
HH2

PROBLEM 3-67
Statement: Design a fourbar Grashof crank-rocker for 120 degrees of output rocker motion with a
quick-return time ratio of 1:1.2. (See Example 3-9.)
1
1.2

Tr
α
β
= α β 360 deg=
360 deg
1 Tr
 β 196 deg
α 360 deg β α 164 deg
δ β 180 deg δ 16 deg
3. Layoff a line through B1 at an arbitrary angle (but not zero deg). In the solution below the line is 60 deg to
the horizontal.
LAYOUT
B2
O2
O4
B1
LINKAGE DEFINITION
A2
2
A1
3
3.833=d
0.953
=
c
4.491=b
0.255=a
4
B1
B2
O4
O2
16.00°
90.00°

Ground link (1) d 3.833 in Coupler (3) b 4.491 in
Crank (2) a 0.255 in Rocker (4) c 0.953 in

PROBLEM 3-68
1
1.5

Tr
α
β
= α β 360 deg=
360 deg
1 Tr
 β 216 deg
α 360 deg β α 144 deg
δ β 180 deg δ 36 deg
the horizontal.
2.5364 = d
2
3.0524 = b
O2
B2
1.2694 = a
LAYOUT
A2
O2
O4
B1
2.0000 = c
LINKAGE DEFINITION
A1
O4
3
B2
4
B1

PROBLEM 3-69
1
1.33

Tr
α
β
= α β 360 deg=
360 deg
1 Tr
 β 205 deg
α 360 deg β α 155 deg
δ β 180 deg δ 25 deg
the horizontal.
LAYOUT
B2
O2
O4
B1
LINKAGE DEFINITION
A2
2
A1
3
4
B1
B2
O4
O2
25.00°
90.00°
4.763=d
2.000=c
6.232=b
0.435=a

PROBLEM 3-70
Statement: Design a sixbar drag link quick-return linkage for a time ratio of 1:4 and output rocker motion
of 50 degrees. (See Example 3-10.)
1
4

1. Determine the crank rotation angles  and  from equation 3.1.
Tr
α
β
= α β 360 deg=
Solving for and  β
360 deg
1 Tr
 β 288 deg
α 360 deg β α 72 deg
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 through O2.
4. Draw a circle of convenient radius O2A about center O2. In the solution below, the length of O2A is
a 1.000 in .
5. Lay out angle  with vertex at O2, symmetrical about quadrant one.
6. Label points A1 and A2 at the intersections of the lines subtending angle  and the circle of radius O2A.
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 both A1 and A2. Label the intersections C1 and C2. In the solution below, the length of AC is
b 2.000 in .
8. The line O2A is the driver crank, link 2, and the line AC is the coupler, link 3.
9. The distance C1C2 is twice the driven (dragged) crank length. Bisect it to locate the fixed pivot O4.
10. The line O2O4 now defines the ground link. Line O4C is the driven crank, link 4. In the solution below, O4C
measures c 2.282 in and O2O4 measures d 0.699 in .
11. Calculate the Grashoff condition. If non-Grashoff, repeat steps 7 through 11 with a shorter radius in step 7.
L max a b c d( )
SL S L

Condition a b c d( ) "Grashof"
12. Invert the method of Example 3-1 to create the output dyad using XX as the chord and O4C1 as the
driving crank. The points B1 and B2 will lie on line XX and be spaced apart a distance that is twice the
length of O4C (link 4). The pivot point O6 will lie on the perpendicular bisector of B1B2 at a distance
from XX which subtends the specified output rocker angle, which is 50 degrees in this problem. In the
solution below, the length BC was chosen to be e 5.250 in .

LAYOUT OF SIXBAR DRAG LINK QUICK RETURN
WITH TIME RATIO OF 1:4
a = 1.000 b = 2.000 c = 2.282 d = 0.699
e = 5.250 f = 5.400
72.000°
9.000°
13. For the design choices made (lengths of links 2, 3 and 5), the length of the output rocker (link 6)
was measured as f 5.400 in .

PROBLEM 3-71
Statement: Design a crank-shaper quick-return mechanism for a time ratio of 1:2.5 (Figure 3-14, p. 112).
Given: Time ratio TR
1
2.5

Design choices:
Length of link 2 (crank) L2 1.000 Length of stroke S 4.000
Length of link 5 (coupler) L5 5.000
1. Calculate  from equations 3.1.
TR
α
β
 α β 360 deg α
360 deg
1
1
TR

 α 102.86 deg
2. Draw a vertical line and mark the center of rotation of the crank, O2, on it.
3. Layout two construction lines from O2, each making an angle /2 to the vertical line through O2.
4. Using the chosen crank length (see Design Choices), draw a circle with center at O2 and radius equal to the
crank length. Label the intersections of the circle and the two lines drawn in step 3 as A1 and A2.
5. Draw lines through points A1 and A2 that are also tangent to the crank circle (step 2). These two lines will
simultaneously intersect the vertical line drawn in step 2. Label the point of intersection as the fixed pivot
center O4.
6. Draw a vertical construction line, parallel and to the right of O2O4, a distance S/2 (one-half of the output
stroke length) from the line O2O4.
7. Extend line O4A1 until it intersects the construction line drawn in step 6. Label the intersection B1.
8. Draw a horizontal construction line from point B1, either to the left or right. Using point B1 as center, draw
an arc of radius equal to the length of link 5 (see Design Choices) to intersect the horizontal construction
line. Label the intersection as C1.
9. Draw the slider blocks at points A1 and C1 and finish by drawing the mechanism in its other extreme position.
O4
2C
4.000
STROKE
C6
2A
5B21
2.000
4
2
O2
3
A1
B1

PROBLEM 3-72
Statement: Design a sixbar, single-dwell linkage for a dwell of 70 deg of crank motion, with an output rocker
motion of 30 deg using a symmetrical fourbar linkage with the following parameter values:
ground link ratio = 2.0, common link ratio = 2.0, and coupler angle  = 40 deg. (See Example
3-13.)
Design choice: Crank length, L2 2.000
Solution: See Figures 3-20 and 3-21 and Mathcad file P0372.
1. For the given design choice, determine the remaining link lengths and coupler point specification.
Angle PAB δ
180 deg γ
2
 δ 70.000 deg
coupler curve in the selected range of crank motion, which in this case will be from 145 to 215 deg.

FOURBAR for Windows File P03-72
Step X Y Mag Ang
Deg in in in in
145 -2.231 3.818 4.422 120.297
150 -2.368 3.661 4.360 122.895
155 -2.497 3.494 4.295 125.549
160 -2.617 3.319 4.226 128.259
165 -2.728 3.135 4.156 131.025
170 -2.829 2.945 4.083 133.846
175 -2.919 2.749 4.009 136.723
180 -2.999 2.547 3.935 139.655
185 -3.067 2.342 3.859 142.639
190 -3.124 2.133 3.783 145.674
195 -3.169 1.923 3.707 148.757
200 -3.202 1.711 3.631 151.886
205 -3.223 1.499 3.555 155.055
210 -3.232 1.289 3.479 158.261
215 -3.227 1.080 3.403 161.498
lie at the intersection of the perpendicular bisectors, label this point D. The radius of this circle is the length
of link 5.
B
145
2
PSEUDO-ARC
O 2
215
A
180
P
D
3
y
x
O 4
4
4. The position of the end of link 5 at point D will remain nearly stationary while the crank moves from 145 to 215
deg. As the crank motion causes the coupler point to move around the coupler curve there will be another
extreme position of the end of link 5 that was originally at D. Since a symmetrical linkage was chosen, the
other extreme position will be located along a line through the axis of symmetry (see Figure 3-20) a distance
equal to the length of link 5 measured from the point where the axis of symmetry intersects the coupler curve
near the 0 deg coupler point. Establish this point and label it E.

Step X Y Mag Ang
Deg in in n in
340.000 -0.718 0.175 0.739 166.325
345.000 -0.615 0.481 0.781 142.001
350.000 -0.506 0.818 0.962 121.717
355.000 -0.386 1.178 1.240 108.135
0.000 -0.255 1.549 1.570 99.365
5.000 -0.117 1.917 1.920 93.499
10.000 0.022 2.269 2.269 89.434
15.000 0.155 2.598 2.603 86.581
B
145
2
PSEUDO-ARC
O 2
5
215
A
180
P
D
355
5
3
y
x
O 4
AXIS OF
SYMMETRY4
E
5. The line segment DE represents the maximum displacement that a link of the length equal to link 5, attached at
P, will reach along the axis of symmetry. Construct a perpendicular bisector of the line segment DE and
extend it to the right (or left, which ever is convenient). Locate fixed pivot O6 on the bisector of DE such that
the lines O6D and O6E subtend the desired output angle, in this case 30 deg. Draw link 6 from D through O6
and extend it to any convenient length. This is the output link that will dwell during the specified motion of
the crank.
SUMMARY OF LINKAGE
SPECIFICATIONS
B
145
2
O 2
5
215
A
180
P
D
355
5
3
y
x
O 4
4
E
30.000°
O 6
6
Original fourbar:
Crank L2 2.000
Coupler L3 4.000
Rocker L4 4.000
Coupler point AP 2.736
δ 70.000 deg
Added dyad:
Pivot O6 x 3.841
y 5.809

PROBLEM 3-73
Statement: Design a sixbar, single-dwell linkage for a dwell of 100 deg of crank motion, with an output
rocker motion of 50 deg using a symmetrical fourbar linkage with the following parameter
values: ground link ratio = 2.0, common link ratio = 2.5, and coupler angle  = 60 deg. (See
Example 3-13.)
Angle PAB δ
180 deg γ
2
 δ 60.000 deg

Step X Y Mag Ang
Deg in in in in
130 -2.192 6.449 6.812 108.774
140 -2.598 6.171 6.695 112.833
150 -2.986 5.840 6.559 117.078
160 -3.347 5.464 6.408 121.493
170 -3.675 5.047 6.244 126.060
180 -3.964 4.598 6.071 130.765
190 -4.209 4.123 5.892 135.588
200 -4.405 3.631 5.709 140.504
210 -4.551 3.130 5.523 145.482
220 -4.643 2.629 5.336 150.482
230 -4.681 2.138 5.146 155.454
erecting perpendicular bisectors to the chords defined by the selected coupler curve points. The center
will lie at the intersection of the perpendicular bisectors, label this point D. The radius of this circle is the
length of link 5.
PSEUDO-ARC
230
2
O 2
3
A
D
180
P
130
y
x
O 4
4
B
4. The position of the end of link 5 at point D will remain nearly stationary while the crank moves from 130 to
230 deg. As the crank motion causes the coupler point to move around the coupler curve there will be
another extreme position of the end of link 5 that was originally at D. Since a symmetrical linkage was
chosen, the other extreme position will be located along a line through the axis of symmetry (see Figure
3-20) a distance equal to the length of link 5 measured from the point where the axis of symmetry
intersects the coupler curve near the 0 deg coupler point. Establish this point and label it E.

Step X Y Mag Ang
Deg in in in in
340 -2.652 1.429 3.013 151.688
350 -2.262 2.316 3.237 134.332
0 -1.743 3.316 3.746 117.727
10 -1.137 4.265 4.414 104.920
20 -0.564 5.047 5.078 96.371
AXIS OF
SYMMETRY
PSEUDO-ARC
230
2
O 2
3
350
A
340
0
D
5
180 10
P
130
20
y
x
O 4
4
E
B
5. The line segment DE represents the maximum displacement that a link of the length equal to link 5, attached
at P, will reach along the axis of symmetry. Construct a perpendicular bisector of the line segment DE and
extend it to the right (or left, which ever is convenient). Locate fixed pivot O6 on the bisector of DE such
that the lines O6D and O6E subtend the desired output angle, in this case 30 deg. Draw link 6 from D
through O6 and extend it to any convenient length. This is the output link that will dwell during the
specified motion of the crank.
6
PSEUDO-ARC
230
2
O2
3
350
A
340
0
D
5
180 10
P
130
20
y
x
O4
4
E
O6
B
50.000°
SUMMARY OF LINKAGE
SPECIFICATIONS
Original fourbar:
Crank L2 2.000
Coupler L3 5.000
Rocker L4 5.000
δ 60.000 deg
Added dyad:
Pivot O6 x 3.166
y 3.656

PROBLEM 3-74
Statement: Design a sixbar, single-dwell linkage for a dwell of 80 deg of crank motion, with an output
rocker motion of 45 deg using a symmetrical fourbar linkage with the following parameter
values: ground link ratio = 2.0, common link ratio = 1.75, and coupler angle  = 70 deg.
(See Example 3-13.)
Angle PAB δ
180 deg γ
2
 δ 55.000 deg

Step X Y Mag Ang
Deg in in in in
140 -0.676 5.208 5.252 97.395
150 -0.958 4.940 5.032 100.971
160 -1.226 4.645 4.804 104.781
170 -1.480 4.332 4.578 108.860
180 -1.720 4.005 4.359 113.242
190 -1.945 3.668 4.152 117.942
200 -2.153 3.322 3.958 122.946
210 -2.337 2.969 3.779 128.210
220 -2.493 2.613 3.612 133.663
lie at the intersection of the perpendicular bisectors, label this point D. The radius of this circle is the
length of link 5.
4
180
2
O 2
A
PSEUDO-ARC
220
3
B
P
140
y
x
4
D
O
4. The position of the end of link 5 at point D will remain nearly stationary while the crank moves from 140 to
220 deg. As the crank motion causes the coupler point to move around the coupler curve there will be
another extreme position of the end of link 5 that was originally at D. Since a symmetrical linkage was
chosen, the other extreme position will be located along a line through the axis of symmetry (see Figure
3-20) a distance equal to the length of link 5 measured from the point where the axis of symmetry intersects
the coupler curve near the 0 deg coupler point. Establish this point and label it E.

Step X Y Mag Ang
Deg in in in in
340 -1.382 1.658 2.158 129.810
350 -0.995 2.360 2.562 112.856
0 -0.494 3.147 3.185 98.919
10 0.074 3.886 3.887 88.916
20 0.601 4.490 4.530 82.372
180
2
O 2
A
PSEUDO-ARC
220
3
340
B
350
0
10
P
140
20
y
x
4
D
E
AXIS OF
SYMMETRY
O
4
5. The line segment DE represents the maximum displacement that a link of the length equal to link 5, attached at
P, will reach along the axis of symmetry. Construct a perpendicular bisector of the line segment DE and
extend it to the right (or left, which ever is convenient). Locate fixed pivot O6 on the bisector of DE such that
the lines O6D and O6E subtend the desired output angle, in this case 30 deg. Draw link 6 from D through O6
and extend it to any convenient length. This is the output link that will dwell during the specified motion of
the crank.

180
2
O 2
A
PSEUDO-ARC
220
3
340
B
350
0
10
P
140
20
y
x
4 6
D
E
O
4
45.000°
O
SUMMARY OF LINKAGE
SPECIFICATIONS
Original fourbar:
Crank L2 2.000
Coupler L3 3.500
Rocker L4 3.500
δ 55.000 deg
Added dyad:
Pivot O6 x 6.217
y 0.653

PROBLEM 3-75
Statement: Using the method of Example 3-11, show that the sixbar Chebychev straight-line linkage of
Figure P2-5 is a combination of the fourbar Chebychev straight-line linkage of Figure 3-29d
and its Hoeken's cognate of Figure 3-29e. See also Figure 3-26 for additional information
useful to this solution. Graphically construct the Chebychev sixbar parallel motion linkage
of Figure P2-5a from its two fourbar linkage constituents and build a physical or computer
model of the result.
Solution: See Figures P2-5, 3-29d, 3-29e, and 3-26 and Mathcad file P0375.
1. Following Example 3-11and Figure 3-26 for the Chebyschev linkage of Figure 3-29d, the fixed pivot OC is
found by laying out the triangle OAOBOC, which is similar to A1B1P. In this case, A1B1P is a striaght line
with P halfway between A1 and B1 and therefore OAOBOC is also a straightline with OC halfway between
OA and OB. As shown below and in Figure 3-26, cognate #1 is made up of links numbered 1, 2, 3, and 4.
Cognate #2 is links numbered 1, 5, 6, and 7. Cognate #3 is links numbered 1, 8, 9, and 10.
O
Links
Removed
OA
1
5
2A
6
2
B1 P
4
1
OC B
4
P
6
B2
A1
2
3
3A
B A1 1
OA
OC
9
10
1 1
8 7
B 2 4
9
6
3
A
OB
6
5
2
B2
3 3
P
2. Discard cognate #3 and shift link 5 from the fixed pivot OB to OC and shift link 7 from OC to OB. Note that due
to the symmetry of the figure above, L5 = 0.5 L3, L6 = L2, L7 = 0.5 L2 and OCOB = 0.5 OAOB. Thus, cognate #2
is, in fact, the Hoeken straight-line linkage. The original Chebyschev linkage with the Hoeken linkage
superimposed is shown above right with the link 5 rotated to 180 deg. Links 2 and 6 will now have the same
velocity as will 7 and 4. Thus, link 5 can be removed and link 6 can be reduced to a binary link supported and
constrained by link 4. The resulting sixbar is the linkage shown in Figure P2-5.

PROBLEM 3-76
Statement: Design a driver dyad to drive link 2 of the Evans straigh-line linkage in Figure 3-29f from 150
deg to 210 deg. Make a model of the resulting sixbar linkage and trace the couple curve.
Given: Output angle θ 60 deg
Solution: See Figjre 3-29f, Example 3-1, and Mathcad file P0376.
1. Draw the input link O2A in both extreme positions, A1 and A2, at the specified angles such that the desired
angle of motion 2 is subtended.
2. Draw the chord A1A2 and extend it in any convenient direction. In this solution it was extended downward.
3. Layout the distance A1C1 along extended line A1A2 equal to the length of link 5. Mark the point C1.
4. Bisect the line segment A1A2 and layout the length of that radius from point C1 along extended line A1A2.
Mark the resulting point O6 and draw a circle of radius O6C1 with center at O6.
5. Label the other intersection of the
circle and extended line A1A2,
C2.
2
O
C1
2
A1
1
3
B2
4
O4
B ,1
C2
O6
A2
5
6
2.932"
0.922"
L1 = 2.4
L2 = 2
L3 = 3.2
L4 = 2.078
L5 = 3.00
L6 = 1.00
AP = 5.38
P2
P1
3.073"
6. Measure the length of the crank
(link 6) as O6C1 or O6C2. From
the graphical solution,
L6 1.000
7. Measure the length of the ground
link (link 1) as O2O6. From the
graphical solution, L1 3.073
L max a b c d( )
SL S L


PROBLEM 3-77
Statement: Design a driver dyad to drive link 2 of the Evans straigh-line linkage in Figure 3-29g from -40
deg to 40 deg. Make a model of the resulting sixbar linkage and trace the couple curve.
Given: Output angle θ 80 deg
Solution: See Figjre 3-29G, Example 3-1, and Mathcad file P0377.
1. Draw the input link O2A in both extreme positions, A1 and A2, at the specified angles such that the desired
angle of motion 2 is subtended.
2. Draw the line A1C1 and extend it in any convenient direction. In this solution it was extended at a 30-deg
angle from A1O2 (see note below).
3. Layout the distance A1C1 along extended line A1C1 equal to the length of link 5. Mark the point C1.
2
O
C1
2
A1 1
3
B2
4
O4
B1
C2
O6
A2
5
6
L1 = 4.61
L2 = 2
L3 = 2.4
L4 = 2.334
L5 = 3.00
L6 = 1.735
AP = 3.00
P2
P1
3.165"
4. Bisect the line segment A1A2 and
layout the length of that radius from
point C1 along extended line A1C1.
Mark the resulting point O6 and
draw a circle of radius O6C1 with
center at O6.
5. Extend a line from A2 through O6.
Label the other intersection of the
circle and extended line A2O6, C2.
6. Measure the length of the crank
(link 6) as O6C1 or O6C2. From
the graphical solution,
L6 1.735
7. Measure the length of the ground
link (link 1) as O2O6. From the
graphical solution, L1 3.165
Note: If the angle between link 2 and
link 5 is zero the resulting driving
fourbar will be a special Grashof.
For angles greater than zero but
less than 33.68 degrees it is a
Grashof crank-rocker. For angles
greater than 33.68 it is a
non-Grashof double rocker.
L max a b c d( )
SL S L


PROBLEM 3-78
Statement: Figure 6 on page ix of the Hrones and Nelson atlas of fourbar coupler curves (on the book
DVD) shows a 50-point coupler that was used to generate the curves in the atlas. Using
the definition of the vector R given in Figure 3-17b of the text, determine the 10 possible
pairs of values of  and R for the first row of points above the horizontal axis if the
gridpoint spacing is one half the length of the unit crank.
Given: Grid module g 0.5
Solution: See Figure 6 H&N Atlas, Figure 3-17b, and Mathcad file P0378.
1. The moving pivot point is located on the 3rd grid line from the bottom and the third grid line from the
left when the crank angle is  radians. Let the number of horizontal grid spaces from the left end of the
coupler to the coupler point be n 2 1 7 and the number of vertical grid spaces from the coupler
to the coupler point be m 2 1 2
2. For the first row of points above the horizontal axis shown in Figure 6, n 2 1 7 and m 1.
3. The angle, , between the coupler and the line from the coupler/crank pivot to the coupler point is
ϕ m n( ) if n 0 atan2 n m( ) if m 0= 0 if m 0
π
2

π
2



















4. The distance, R, from the pivot to the coupler point along the same line is
R m n( ) g m
2
n
2

ϕ m n( )
deg
153.435
135.000
90.000
45.000
26.565
18.435
14.036
11.310
9.462
8.130

n
-2.000
-1.000
0.000
1.000
2.000
3.000
4.000
5.000
6.000
7.000
 R m n( )
1.118
0.707
0.500
0.707
1.118
1.581
2.062
2.550
3.041
3.536

5. The coupler point distance, R, like the link lengths A, B, and C is a ratio of the given length to the the
length of the driving crank.

PROBLEM 3-79
Statement: The set of coupler curves in the Hrones and Nelson atlas of fourbar coupler curves (on the
book DVD, page 16 of the PDF file) has A = B = C = 1.5. Model this linkage with program
FOURBAR using the coupler point fartherest to the left in the row shown on page 1 and
plot the resulting coupler curve.
Given: A 1.5 B 1.5 C 1.5
Solution: See Figure on page 1 H&N Atlas, Figure 3-17b, and Mathcad file P0379.
2. For the second column of points to the left of the coupler pivot and the second row of points above the
horizontal axis n 2 and m 2. The grid spacing is g 0.5
π
2

π
2


















 ϕ m n( ) 135.000 deg
4. The distance from the pivot to the coupler point, R, along the same line is
R m n( ) g m
2
n
2
 R m n( ) 1.414
5. Determine the values needed for input to FOURBAR.
Link 2 (Crank) a 1
Link 3 (Coupler) b A a b 1.500
Link 4 (Rocker) c B a c 1.500
Link 1 (Ground) d C a d 1.500
Distance to coupler point R m n( ) 1.414
Angle from link 3 to coupler point ϕ m n( ) 135.000 deg
6. Calculate the coordinates of O4. Let the angle between links 2 and 3 be  , then
α acos
A
2
1 C( )
2
 B
2

2 A 1 C( )






 α 33.557 deg
xO4 C cos α  xO4 1.250
yO4 C sin α  yO4 0.829
7. Enter this data into FOURBAR and then plot the coupler curve. (See next page)

PROBLEM 3-80
Statement: The set of coupler curves on page 17 in the Hrones and Nelson atlas of fourbar coupler
curves (on the book DVD, page 32 of the PDF file) has A = 1.5, B = C = 3.0. Model this linkage
with program FOURBAR using the coupler point fartherest to the right in the row shown and
plot the resulting coupler curve.
Given: A 1.5 B 3.0 C 3.0
coupler to the coupler point be n 2 1 7 and the number of vertical grid spaces from the coupler to
the coupler point be m 2 1 2
2. For the fifth column of points to the right of the coupler pivot and the first row of points above
the horizontal axis n 5 and m 1. The grid spacing isg 0.5
π
2

π
2


















 ϕ m n( ) 11.310 deg
R m n( ) g m
2
n
2
 R m n( ) 2.550
α acos
A
2
1 C( )
2
 B
2

2 A 1 C( )






 α 39.571 deg
xO4 C cos α  xO4 2.313
yO4 C sin α  yO4 1.911

PROBLEM 3-81
curves (on the book DVD, page 36 of the PDF file) has A = 1.5, B = C = 3.5. Model this
linkage with program FOURBAR using the coupler point fartherest to the right in the
row shown and plot the resulting coupler curve.
Given: A 1.5 B 3.5 C 3.5
2. For the fourth column of points to the right of the coupler pivot and the second row of points above the
horizontal axis n 4 and m 2. The grid spacing isg 0.5
π
2

π
2


















 ϕ m n( ) 26.565 deg
R m n( ) g m
2
n
2
 R m n( ) 2.236
α acos
A
2
1 C( )
2
 B
2

2 A 1 C( )






 α 40.601 deg
xO4 C cos α  xO4 2.657
yO4 C sin α  yO4 2.278

PROBLEM 3-82
curves (on the book DVD, page 49 of the PDF file) has A = 2.0, B = 1.5, C = 2.0. Model
this linkage with program FOURBAR using the coupler point fartherest to the right in
the row shown and plot the resulting coupler curve.
Given: A 2.0 B 1.5 C 2.0
2. For the sixth column of points to the right of the coupler pivot and the first row of points below
the horizontal axis n 6 and m 1. The grid spacing is g 0.5
π
2

π
2


















 ϕ m n( ) 9.462 deg
R m n( ) g m
2
n
2
 R m n( ) 3.041
Angle from link 3 to coupler point ϕ m n( ) 9.462 deg
α acos
A
2
1 C( )
2
 B
2

2 A 1 C( )






 α 26.384 deg
xO4 C cos α  xO4 1.792
yO4 C sin α  yO4 0.889

PROBLEM 3-83
curves (on the book DVD, page 130 of the PDF file) has A = 2.5, B = 1.5, C = 2.5. Model
this linkage with program FOURBAR using the coupler point fartherest to the right in the
row shown and plot the resulting coupler curve.
Given: A 2.5 B 1.5 C 2.5
2. For the second column of points to the right of the coupler pivot and the second row of points below the
horizontal axis n 2 and m 2. The grid spacing is g 0.5
π
2

π
2


















 ϕ m n( ) 45.000 deg
R m n( ) g m
2
n
2
 R m n( ) 1.414
Angle from link 3 to coupler point ϕ m n( ) 45.000 deg
α acos
A
2
1 C( )
2
 B
2

2 A 1 C( )






 α 21.787 deg
xO4 C cos α  xO4 2.321
yO4 C sin α  yO4 0.928

PROBLEM 3-84
that demonstrates the required movement.
Solution: See figure below and Mathcad file P0384 for one possible solution.
to C2 and D1 to D2.
below the bisector of C1C2 was extended upward and the bisector of D1D2 was also extended upward.
distances O2C and O4D were selected to be 15.000 in. and 8.625 in, respectively. This resulted in a
ground-link-length O2O4 for the fourbar of 9.351 in.
4. The fourbar is now defined as O2CDO4 with link lengths
2
yC1D1
2
 L3 17.197 in
Ground link 1 L1 9.351 in
O2
O4
C1
D1
D2
C2
17.197
9.351
15.000
8.625

PROBLEM 3-85
Statement: Design a fourbar mechanism to move the link shown in Figure P3-19 from position 2 to
position 3. Ignore the first position and the fixed pivots O2 and O4 shown. Build a
cardboard model that demonstrates the required movement.
Solution: See figure below and Mathcad file P0385 for one possible solution.
to C3 and D2 to D3.
below the bisector of C2C3 was extended upward and the bisector of D2D3 was also extended upward.
distances O2C and O4D were selected to be 15.000 in and 8.625 in, respectively. This resulted in a
ground-link-length O2O4 for the fourbar of 9.470 in.
2
yC2D2
2
 L3 17.196 in
O2
O4
C3
D2
C2
D3
17.196
15.000
8.625
9.470

4 = +12.403 deg and -8.950 deg.

PROBLEM 3-86
points O2 and O4 shown. Build a cardboard model that has stops to limit its motion to the
range of positions designed.
intersection O2.
6. Line C1D1 is link 3. Line O2O4 is link 1 (ground link for the fourbar). The fourbar is now defined as O2CDO4 an
has link lengths of
Ground link 1 L1 9.187 Link 2 L2 14.973
Link 3 L3 17.197 Link 4 L4 8.815
O2
O4
C1 D1
C3
D2
C2
D3
2
3
4
14.973
17.197
9.187
8.815

PROBLEM 3-87
pivots O2 and O4 shown. (See Example 3-7.) Build a cardboard model that has stops to limit
its motion to the range of positions designed.
position.
position.
D1C1
O2
O4
O'2
O'4
O"2
O"4
C2
C3
D2
D3
First layout for steps 1 through 7

G H
O2
O4
O'2
O'4
O"2
O"4
E
E
E
F
F
F
1
1
2
2
3
3
2
3
4
Second layout for steps 8 through 12
intersection G.
Link 3 L3 18.017 Link 4 L4 8.786
L max a b c d( )
SL S L


The fourbar that will provide the desired motion is now defined as a non-Grashof double rocker in the
open configuration. It now remains to add the original points C1 and D1 to the coupler GH.
D1C1
O2
O4
G
H
2
3
4
9.216
16.385
8.786
18.017

solution manual Design of Machinery: An Introduction to the Synthesis and Analysis of Mechanisms and Machines Norton 5th Edition

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solution manual Design of Machinery: An Introduction to the Synthesis and Analysis of Mechanisms and Machines Norton 5th Edition