Solutions to Chapter 6 Exercise Problems
Problem 6.1:
Design a double rocker, four-bar linkage so that the base link is 2-in and the output rocker is 1-in
long. The input link turns counterclockwise 60˚ when the output link turns clockwise through 90˚.
The initial angle for the input link is 30˚ counterclockwise from the horizontal, and the initial angle
for the output link is -45˚. The geometry is indicated in the figure.
A2
60˚
O2
A1
O4
30˚
45˚
B2
90˚
B1
Solution:
Invert the motion and solve the problem graphically. Draw a line from O2 to B2 and rotate it
clockwise 60˚ to locate the position of B'2 . Draw the perpendicular bisector of B'2 B1 and locate the
intersection of that line with the line O2 A1 . This will locate A1 . The solution is given in the first
figure. Measure the lengths os 02 A1 and A1 B1 . This gives the linkage shown in the second figure.
The link lengths are:
r1 = 6"
r2 = 3.72"
r3 = 7.29"
r4 = 4"
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A2
60˚
A1
30˚
O4
O2
45˚
90˚
B2
B1
B'2
Basic Construction
A2
A1
r3
r2
30˚
O2
O4
r1
45˚
90˚
r4
B2
B1
B'2
Final Linkage
- 223 -
Problem 6.2:
Design a double rocker, four-bar linkage so that the base link is 4-in and the output rocker is 2-in
long. The input link turns counterclockwise 40˚ when the output link turns counterclockwise
through 80˚. The initial angle for the input link is 20˚ counterclockwise from the horizontal, and the
initial angle for the output link is 25˚. The geometry is indicated in the figure.
B2
A2
60˚
O2
B1
80˚
A1
O4
20˚
25˚
Solution
Invert the motion and solve the problem graphically. Draw a line from O2 to B2 and rotate it
clockwise 60˚ to locate the position of B'2 . Draw the perpendicular bisector of B'2 B1 and locate the
intersection of that line with the line O2 A1 . This will locate A1 . The solution is given in the first
figure. Measure the lengths os 02 A1 and A1 B1 . This gives the linkage shown in the second figure.
The link lengths are:
r1 = 4"
r2 = 2.77"
r3 = 3.21"
r4 = 2"
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Basic Construction
Final Linkage
Problem 6.3:
In a back hoe, a four-bar linkage is added at the bucket in part to amplify the motion that can be
achieved by the hydraulic cylinder attached to the link that rotates the bucket as shown in the figure.
Design the link attached to the bucket and the coupler if the frame link is 13-in and the input link is
12-in long. The input link driven by the hydraulic cylinder rotates through an angle of 80˚ and the
output link rotates through an angle of 120˚. From the figure, determine reasonable angles for the
starting angles (0 and 0 ) for both of the rockers.
- 225 -
Solution
Determine reasonable starting angles for each crank relative to the beam member. Use 40˚ for the
input rocker and 70˚ for the output rocker. This is shown in the figure.
The basic problem can then be redrawn as shown in the following figure.
- 226 -
Solution
Invert the motion and solve the problem graphically. Draw a line from O2 to B2 and rotate it
clockwise 120˚ to locate the position of B'2 . Draw the perpendicular bisector of B'2 B1 and locate
the intersection of that line with the line O2 A1 . This will locate A1 . The solution is given in the first
figure below. Measure the lengths os 02 A1 and A1 B1 . After unscaling, the linkage is shown in the
second figure that follows.
The link lengths are:
r1 = 13"
r2 = 8.27"
r3 = 19.34"
r4 = 12"
- 227 -
Basic Construction
Final Linkage
- 228 -
Problem 6.4
In the drawing, AB = 1.25 cm. Use A and B as circle points, and design a four-bar linkage to move
its coupler through the three positions shown. Use Grashof’s equation to identify the type of fourbar linkage designed.
Y
B3
θ 3 = 60˚
A 3(2, 3)
B2
A 2 (2, 1)
θ2 = 45˚
X
B1
A 1 (0, 0), θ1 = 0
Solution
Find the center points A* and B* and measure the link lengths. Then,
AA* = 2.027"
AB = 1.25"
A*B* = 0.670"
BB* = 2.903"
l + s = 2.903 + 0.670= 3.573"
p + q = 2.027 + 1.25 = 2.277
Therefore, l +s > p+q and the linkage is a nonGrashof linkage and a double rocker.
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B3
Y
A3
B*
A*
B2
A2
X
B1
A1
Problem 6.5
Using points A and B as circle points, design a four-bar linkage that will position the body defined
by AB in the three positions shown. Draw the linkage in position 1, and use Grashof’s equation to
identify the type of four-bar linkage designed. Position A1 B1 is horizontal, and position A2 B2 is
vertical. AB = 1.25 in.
Y
A 1 (0, 1.75)
B1
(1.63, 1.25)
A2
B3
35˚
A3 (2.13, 0.63)
B2
Solution
Find the center points A* and B* and measure the link lengths. Then,
AA* = 2.15"
AB = 1.25"
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X
A*B* = 2.32"
BB* = 1.11"
l + s = 2.32 + 1.11= 3.43"
p + q = 2.15 + 1.25 = 3.40
Therefore, l +s > p+q and the linkage is a nonGrashof linkage and a double rocker.
Y
B1
A1
A2
B3
B*
A3
X
B2
A*
Problem 6.6
Design a four-bar linkage to move its coupler through the three positions shown below using points
A and B as moving pivots. AB = 4 cm. What is the Grashof type of the linkage generated?
Y
B3
B2
60˚
A3(2, 2.4)
50˚
X
A 2 (2, 0.85)
A1 (0, 0)
Solution
B1
Find the center points A* and B* and measure the link lengths. Then,
AA* = 1.705 cm
AB = 4.000 cm
- 231 -
A*B* = 3.104 cm
BB* = 6.954 cm
l + s = 6.954 + 1.705= 8.659"
p + q = 3.104 + 4.000 = 7.104
Therefore, l+s > p+q and the linkage is a nonGrashof linkage and a double rocker.
B3
Y
B2
B*
A3
A*
A2
B1
A1
- 232 -
X
Problem 6.7
A four-bar linkage is to be designed to move its coupler plane through the three positions shown.
The moving pivot (circle point) of one crank is at A and the fixed pivot (center point) of the other
crank is at C*. Draw the linkage in position 1 and use Grashof’s equation to identify the type of
four-bar linkage designed. Also determine whether the linkage changes branch in traversing the
design positions. Positions A1 B1 and A2 B2 are horizontal, and position A3 B3 is vertical. AB = 3 in.
Y
B 2 (0.0, 2.88)
A2
A1
B 1 (-1.94, 0.94)
C*
A3
X
(-1.0, -3.0)
B3
Solution
Find the center points C1 and A* and measure the link lengths. Note that the linkage is to be drawn
in position 1 so the motion must be referred to position 1 when locating C1 . Then,
CC* = 2.67 in
AC = 2.38 in
A*C* = 3.00 in
AA* = 2.06 in
l + s = 3.00 + 2.06= 5.06
p + q = 2.67 + 2.38 = 5.05
Therefore, l +s > p+q and the linkage is a nonGrashof linkage and a double rocker.
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Y
B2
A2
C 3*
A1
B1
A* C
1
C*
X
A3
C 2*
B3
To determine if the linkage changes branch, draw the linkage in the three positions, and measure the
sign of . From the drawing below, the sign of is different in the three positions, and the
mechanism changes branch.
- 234 -
Y
B2
A2
C2
ψ2
A*
A1
B1
C1
C*
ψ1
X
A3
ψ3
C3
B3
Problem 6.8
Design a four-bar linkage to move a coupler containing the line AB through the three positions
shown. The moving pivot (circle point) of one crank is at A and the fixed pivot (center point) of the
other crank is at C*. Draw the linkage in position 1 and use Grashof’s equation to identify the type
of four-bar linkage designed. Position A1 B1 is horizontal, and positions A2 B2 and A3 B3 are vertical.
AB = 4 in.
Y
A1
*
C1
A2
(0, 2)
(0, 0)
A3
B1
(2, 0)
B2
- 235 -
(4, 0)
B3 X
Solution
The solution is shown in the figure. The link lengths are:
r1 = 3.22 = p
A3
A2
A1
B1
r2
r3
A*
r1
*
C
B2
r4
B3
C*2
C1
C*3
r2 = 3.32 = q
r3 = 3.62 = l
r4 = 2.24 = s
Grashof l + s < p + q
For this mechanism,
and
l + s = 3.62 + 2.24 = 5.86
p + q = 3.22 + 3.32 = 6.54
Therefore, l + s < p + q , and the mechanism is a crank-rocker or rocker-crank.
- 236 -
Problem 6.9
A mechanism must be designed to move a computer terminal from under the desk to top level. The
system will be guided by a linkage, and the use of a four-bar linkage will be tried first. As a first
attempt at the design, do the following:
a) Use C* as a center point and find the corresponding circle point C in position 1.
b) Use A as a circle point and find the corresponding center point A*.
c) Draw the linkage in position 1.
d) Determine the type of linkage (crank rocker, double rocker, etc.) resulting.
e) Evaluate the linkage to determine whether you would recommend that it be manufactured.
Y
Desk
Position 3 (Horizontal)
A3 (3.3, 2.6)
Position 2
C*
A1 (-2.2, -0.3)
135˚ A2 (2.1, -0.1)
Position 1
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160˚
X
Desk
Pos'n 3
A3
A*
C*2
C*3
C*
Pos'n 2
A1
A1C 1 = 2.6864
C1C* = 3.0684
A * C* = 1.0301
A1A* = 1.395
A2
Pos'n 1
A * C * +C 1C* = 4.0985
A1C1 + A1A* = 4.0814
Therefore,
A * C * +C 1C* > A1C 1 + A1A *
and the linkage is a nonGrashof linkage and a double rocker.
C1
This is a poor solution because C 1is below the floor plane.
Problem 6.10
Design a four-bar linkage to move the coupler containing line segment AB through the three
positions shown. The moving pivot for one crank is to be at A, and the fixed pivot for the other
crank is to be at C*. Draw the linkage in position 1 and determine the classification of the resulting
linkage (e.g., crank rocker, double crank). Positions A2 B2 and A3 B3 are horizontal, and position
A1 B1 is vertical. AB = 3.5 in.
Y
B1
(-1.0, 2.5)
A3
B3 (0.0, 2.0)
A2
B2 (0.0, 1.0)
C*1
A1
- 238 -
X
C*2
B1
A3
C*2
B3
A*
B2
A2
C1
C*
A1
From the figure,
r1 = A*C* = 1.8228 in
r2 = A*A1 = 2.4974 in
r3 = A1 C1 = 2.6966 in
r4 = C*C1 = 1.3275 in
Grashof calculation:
l +s?p+q
[2.6966 + 1.3275=(4.0241)] < [1.8228 + 2.4974=(4.3202)]
The linkage is a crank rocker.
- 239 -
Problem 6.11
Design a four-bar linkage to move a coupler containing the line AB through the three positions
shown. The moving pivot (circle point) of one crank is at A and the fixed pivot (center point) of the
other crank is at C*. Draw the linkage in position 1, and use Grashof’s equation to identify the
type of four-bar linkage designed. Position A1 B1 is horizontal, and positions A2 B2 and A3 B3 are
vertical. AB = 6 cm.
Y
B1
A3 (-2, 3)
A1
(6, 4.3)
A2 (2, 3)
X
C *(0, 0)
B3
B2
- 240 -
A*
C*
2
Y
C1
B1
A3
A1
A2
C*
3
X
C*
B3
B2
Find A*. Then find C'2 and C'3 and find the circle point C1 . Draw the linkage. Then,
AA* = 13.0
AC = 6.84
CC* = 4.41
C*A* = 15.8
l + s = 4.41 + 15.8 = 20.2
p + q = 6.84 + 13.0 = 19.8
Therefore, l+s > p+q and the linkage is a Type II double rocker.
- 241 -
Problem 6.12
Design a four-bar linkage to move the coupler containing line segment AB through the three
positions shown. The moving pivot for one crank is to be at A, and the fixed pivot for the other
crank is to be at C*. Draw the linkage in position 1 and determine the classification of the resulting
linkage (e.g., crank rocker, double crank). Also check to determine whether the linkage will change
branch as it moves from one position to another. Position A1 B1 is horizontal, and position A3 B3 is
vertical. AB = 5.1 cm.
Y
(1.5, 3.7)
A2
45
˚
A1
(-5.0, -1.0)
B1
B2
X
A3
(7.8, -1.0)
C*
(-5.0, -5.0)
B3
Solution
Find the center points C1 and A* and measure the link lengths. Note that the linkage is to be drawn
in position 1 so the motion must be referred to position 1 when locating C1 . Then,
CC* = 7.81 cm
AC = 9.73 cm
A*C* = 6.68 cm
AA* = 6.82 cm
l + s = 9.73 + 6.68 = 16.41 cm
p + q = 7.81 + 6.82 = 14.63 cm
Therefore, l +s > p+q and the linkage is a nonGrashof linkage and a double rocker.
- 242 -
A2
Y
X
B2
A1
A3
B1
A*
C*
B3
C1
C 2*
C 3*
To determine if the linkage changes branch, redraw the linkage in the three positions, and determine
if the transmission angle changes. This is shown below. The linkage does not change branch.
- 243 -
A2
Y
X
B2
A1
A3
B1
A*
C*
C2
B3
C1
C3
Problem 6.13
Synthesize a four-bar mechanism in position 2 that moves its coupler through the three positions
shown below if points C* and D* are center points. Position A1 B1 and position A3 B3 are
horizontal. AB = 4 cm.
Y
D* (3.0, 2.6)
B2
B3
A 3 (3.4, 1.6)
X
C*
45˚
A2 (2.7, - 0.7)
Solution
B1
A1 (0.7, - 1.8)
Find the circle points C2 and D2 and measure the link lengths. Notice that the linkage is to be
drawn in position 2. Therefore, position 2 for the coupler is the position to which the positions of
D* and C* are referred for finding the circle points. Then,
- 244 -
DD* = 2.767 cm
CD = 4.045 cm
C*D* = 3.985 cm
CC* = 2.300 cm
l + s = 4.045 + 2.300= 6.345
p + q = 2.767 + 3.985 = 6.752
Therefore, l+s < p+q and the linkage is a Grashof linkage and a crank rocker.
D*1
Y
D*
D2
B2
B3
A3
C*1
X
C*
D*3
A2
B1
A1
C2
C*
3
- 245 -
Problem 6.14
Synthesize a four-bar mechanism in position 2 that moves its coupler through the three positions
shown below. Point A is a circle point, and point C* is a center point. Position A1 B1 and position
A3 B3 are horizontal. AB = 4 cm.
Y
C* (2.3, 4.5)
B2
B3
A3 (2.7, 3.5)
45˚
A2 (2.0, 1.0)
B1
A1
Solution
X
Find the center points C2 and A* and measure the link lengths. Note that the linkage is to be drawn
in position 2 so the motion must be referred to position 2 when locating C2 . Then,
CC* = 2.257 cm
AC = 3.146 cm
A*C* = 2.989 cm
AA* = 3.065 cm
l + s = 3.146 + 2.257= 5.403
p + q = 3.065 + 2.989 = 6.054
Therefore, l+s < p+q and the linkage is a Grashof linkage and a crank rocker.
C*1
C*
Y
B2
C2
A*
B3
A3
C*
3
A2
B1
A1
- 246 -
X
Problem 6.15
A hardware designer wants to use a four-bar linkage to guide a door through the three positions
shown. Position 1 is horizontal, and position 3 is vertical. As a tentative design, she selects point
B* as a center point and A as a circle point. For the three positions shown, determine the location of
the circle point B corresponding to the center point B* and the center point A* corresponding to the
circle point A. Draw the linkage in position 1 and determine the Grashof type for the linkage.
Indicate whether you think that this linkage should be put into production.
Y
Position 3
135˚
Position 2
A3 (0, 8.3)
A2 (-3.1, 6.6)
Position 1
A1(-4.0, 2.8)
B*
X
Solution
Position 2
A3
Position 3
A2
A*
B1
Position 1
B*
3
A1
B*
- 247 -
B*2
AA* = 1.7552"
BB* = 1.2903"
AB = 2.0069"
A*B* = 1.5576"
l + s = 2.0069 +1.2903 = 3.2972 < p + q = 1.7552 +1.5576 = 5.0680 crank rocker
The mechanism would not be acceptable because of the location of the fixed and moving pivots
inside the wall.
Problem 6.16
Design a slider-crank mechanism to move the coupler containing line segment AB through the three
positions shown. The moving pivot for the crank is to be at A. Determine the slider point, and draw
the linkage in position 1. Also check to determine whether the linkage will move from one position
to another without being disassembled. Position A1 B1 is horizontal, and position A3 B3 is vertical.
AB = 2.0 in.
Y
A2 (2.69, 1.44)
135˚
A3 (5.06, 0.0)
A1(0, 0)
B2
B1
X
B3
Solution
Find the center ponit A*. Then find the poles, the image pole I'2 3, and the circle of sliders. Next
select a slider point (C1 ), and find the location of that point in the other positions. This establishes
the slider line. Note that a difference linkage results for each choice of C1 .
From the figure,
r2 = A*A1 = 2.9393 in
r3 = A1 C1 = 4.3606 in
Find the transmission angle in the three positions. The linkage changes mode because the sign of
the transmission angle changes.
- 248 -
A2
A1
A3
B1
B2
r2 = 2.9393
r3 = 4.3606
A*
P2
B3
3
C3
C2
P23'
P1
P1
2
3
C1
- 249 -
Problem 6.17
Design a slider-crank mechanism to move a coupler containing the line AB through the three
positions shown. The line AB is 1.25" long. The moving pivot (circle point) of the crank is at A.
The approximate locations of the three poles (p 1 2, p 1 3, p 2 3) are shown, but these should be
determined accurately after the positions are redrawn. Find A*, the slider point that lies above B1 on
a vertical line through B1 , and draw the linkage in position 1.
Y
P13
(1.25, 1.70)
B1
A1
(1.65, 1.25) A 2
P12
B3
(3.15 , 1.32
P23
A3
(2.15, 0.58)
X
B2
Solution
Find the center ponit A*. Then find the poles, the image pole I'2 3, and the circle of sliders. Next
select the slider point (C1 ) that lies above B1 . Then find the location of that point in the other
positions. This establishes the slider line.
Draw the linkage in the three positions. By inspection of the positions of C, the mechanism
changes mode and goes through the positions in the wrong order.
- 250 -
- 251 -
Problem 6.18
Design a slider-crank linkage to move a coupler containing the line AB through the three positions
shown. The fixed pivot (center point) of the other crank is at C*. Draw the linkage (including the
slider line) in Position 1. Position A1 B1 is horizontal, and positions A2 B2 and A3 B3 are vertical.
AB = 4 in.
Y
A1
*
C1
A2
(0, 2)
(0, 0)
A3
B1
(2, 0)
B2
(4, 0)
B3
X
Solution
This problem illustrates the type of rigid body guidance problem that cannot be solved directly
using the elementary techniques developed in the text book. Therefore, to correct answer is that
there is no solution. However, a partial solution can be developed, and the students should work the
problem far enough to illustrate that the solution procedure breaks down.
With the information give, it is possible to find C1 . This is illustrated in the following construction.
- 252 -
Finding C1
Next find the poles as shown in the following construction. This is where a problem occurs. Note
that P1 2 and P1 3 are coincident and P2 3 is at infinity on the line shown. Therefore, the circle of
sliders appears to be on a straight line, but the orientation of the line cannot be determined from the
elementary theory provided.
- 253 -
Finding the poles
- 254 -
Problem 6.19
Design a slider-crank mechanism to move a coupler containing the line with A through the three
positions shown. The moving pivot (circle point) of the crank is at A. Find the slider point which
lies on Line BC and draw the linkage (including the slider line) in position 1. Note that Line BC is
NOT the line on which the slider moves.
60˚
(0.8, 0.8) A 3
A1
(0, 0)
B
C
0.4
A2
(0.8, 0)
30˚
Solution
Find the center ponit A*. Then find the poles, the image pole I'2 3, and the circle of sliders. Next
select the slider point (C1 ) that lies on the line BC. There are two solutions, but the one on the right
is implied by the problem statement. Then find the location of that point in the other positions. This
establishes the slider line.
Draw the linkage in the three positions. By inspection of the positions of C, the mechanism
changes mode and goes through the positions in the wrong order. The construction steps are
shown in the following.
Finding the center point A*
- 255 -
Finding the poles
Finding the circle of sliders
- 256 -
Finding the slider point
- 257 -
Final solution
Problem 6.20
A device characterized by the input-output relationship = a1 + a2 cos is to be used to generate
(approximately) the function = 2 ( and both in radians) over the range 0 / 4.
a) Determine the number of precision points required to compute a1 and a2 .
b) Choose the best precision point values for from among 0, 0.17, 0.35, and 0.52, and determine
the values of a1 and a2 that will allow the device to approximate the function.
c) Find the error when = /8.
- 258 -
Solution
Two precision points can be used because there are two design variables. For this, first determine
1 and 2 . Look at Chebychev spacing to see what values of are reasonable. From the figure,
135˚
π
4
45˚
0
π
8
θ1
θ2
1 = (1+ cos135°) = 0.115
8
2 = (1+ cos45°) = 0.670
8
The angles which are closest to these values are 0.17 and 0.52. Then,
1 = (1 )2 = (0.17)2 = 0.0289
2 = ( 2 ) 2 = (0.52) 2 = 0.270
Substituting into the equation for the system model,
1 = a1 + a2 cos1
2 = a1 + a2 cos2
or,
0.0289 = a1 + a2 cos(0.17) = a1 + a2 (0.985)
0.270 = a1 + a2 cos(0.52) = a1 + a2 (0.868)
Subtract the first equation from the second, and solve for a2
0.241 = a2 (0.117)
and
a2 = 2.060
Now back substituting into the first equation,
a1 = 0.0289 a2 (0.985) = 0.0289 + 2.060(0.985) = 2.058.
The error is given by
- 259 -
e = ideal act = 2 (2.058 2.060cos )
Substituting in the given value = /8 = 22.5˚,
e = (0.3926)2 (2.058 2.060cos 22.5˚ ) = 0.000673
Problem 6.21
A mechanical device characterized by the input-output relationship = 2a1 + 3a2 sin + a32 is to be
2
used to generate (approximately) the function = 2 over the range 0 / 4. Exterior
constraints on the design require that the parameter a3 = 1.
a) Determine the number of precision points required to complete the design of the system.
b) Use Chebyshev spacing, and determine the values for the unknown design variables which will
allow the device to approximate the function.
c) Find the error when = /6.
Solution
Two precision points can be used because there are two design variables. For this, first determine
1 and 2 . Look at Chebychev spacing to see what values of are reasonable. From the figure,
135˚
π
4
45˚
0
θ1
π
8
1 = (1+ cos135°) = 0.115
8
2 = (1+ cos45°) = 0.670
8
Then,
- 260 -
θ2
1 = 2(1 )2 = 2(0.115)2 = 0.02645
2 = 2( 2 )2 = 2(0.670)2 = 0.8978
Substituting into the equation for the system model,
1 = 2a1 + 3a2 sin1 + a32 = 2a1 + 3a2 sin1 + 1
2 = 2a1 + 3a2 sin 2 + a32 = 2a1 + 3a2 sin 2 +1
or,
0.02645 = 2a1 + 3a2 sin(0.115) + 1= 2a1 + 0.3442a2 + 1
0.8978 = 2a1 + 3a2 sin(0.670) +1 = 2a1 + 1.8629a2 +1
Subtract the first equation from the second, and solve for a2
and
0.8713 = 1.5187a2
a2 = 0.5737
Now back substituting into the first equation,
2a1 = 0.02645 1 0.3442a2 = 0.02645 1 0.3442(0.5737) = 1.1710
or
a1 = 1.1710 / 2 = 0.5855
The error is given by
e = ideal act = 2 2 (2a1 + 3a2 sin + 1) = 2 2 [2(0.5855) + 3(0.5737)sin +1]
= 2 2 [0.1710 + 1.7214sin ]
Substituting in the given value = /6 = 30˚,
e = 2(0.5235)2 [ 0.1710 + 1.7214sin30˚ ] = 0.1416
Problem 6.22
A mechanical device characterized by the input-output relationship = 2a1 + a2 tan + a32 is to be
used to generate (approximately) the function = 3 3 ( and both in radians) over the range
0 / 3 . Exterior constraints on the design require that the parameter a3 = 1.
a) Determine the number of precision points required to complete the design of the system.
b) Use Chebyshev spacing, and determine the values for the unknown design variables that will
allow the device to approximate the function.
c) Find the error when = /6.
- 261 -
Solution:
There are two unknowns so the number of precision points is two. The precision points according
to Chebyshev spacing are:
1 = (max + min) + (max min) cos135 = + cos135 = 0.1534
2
2
6 6
and
2 = (max + min) + (max min ) cos45 = + cos45 = 0.8938
2
2
6 6
The corresponding values of are:
1 = 213 = 2(0.1534)3 = 0.0072
2 = 232 = 2(0.8938)3 = 1.428
We can now solve for a1 and a2 using the desired input-output relationship.
= 2a1 + a2 tan + a32 = 2a1 + a2 tan + 1
Then,
0.0072 = 2a1 + a2 tan(0.1534) + 1= 2a1 + 0.1546a2 + 1
and
1.428 = 2a1 + a2 tan(0.8938) + 1= 2a1 + 1.2442a2 + 1
Subtracting the two equations gives,
1.4209 = 1.0902a2 a2 = 1.3033
Backsubstituting to determine a1 gives
2a1 = 1.4281 1.2442a2 1 = 0.4281 1.2442(1.3033) = 1.1935 a1 = 0.5968
The error is given by
e = ideal act = 3 2 (2a1 + a2 tan + 1) = 3 2 [2(0.5968)+ (1.3033)tan + 1]
= 3 2 [ 0.1936 + 1.3033tan ]
Substituting in the given value = /6 = 30˚,
e = 3(0.5235)2 [ 0.1936 + 1.3033tan30˚ ] = 0.2632
- 262 -
Problem 6.23
A mechanical device characterized by the input-output relationship = 2a1 + a2 sin is to be used to
generate (approximately) the function y = 2x 2 over the range 0 x / 2 where x, y, , and are
all in radians. Assume that the use of the device will be such that the starting point and range for x
can be the same as those for , and the range and starting point for y can be the same as those for .
a) Determine the number of precision points required to complete the design of the system.
b) Use Chebyshev spacing, and determine the values for the unknown design variables that will
allow the device to approximate the function.
c) Compute the error generated by the device for x = /4.
Solution:
There are two unknowns so the number of precision points is two. The precision points according
to Chebyshev spacing are:
x1 =
(xmax + xmin) (xmax x min)
+
cos135 = + cos135 = 0.2300
4 4
2
2
x2 =
(x max + x min) (x max x min)
+
cos45 = + cos45 = 1.3407
4 4
2
2
and
The corresponding values of y are:
y1 = 2x12 = 2(0.2300)2 = 0.1058
y2 = 2x 22 = 2(1.3407)2 = 3.5949
We must now relate to x and to y. Since and x have the same starting value and same range,
we can interchange them exactly. The same applies to and y. Therefore, we can use the following
pairs of numbers to solve the problem.
1 = 0.2300 rad = 13.178˚
1 = 0.1058
and
2 = 1.3407rad = 76.816˚
2 = 3.5949
We can now solve for a1 and a2 using the desired input-output relationship.
= 2a1 + a2 sin
Then,
- 263 -
1 = 2a1 + a2 sin1 = 0.1058 = 2a1 + a2 sin13.178˚ = 2a1 + a2 (.2280)
2 = 2a1 + a2 sin 2 = 3.5949 = 2a1 + a2 sin76.816˚ = 2a1 + a2 (.9736)
Subtracting the two equations gives,
3.4891 = 0.7456a2 a2 = 3.4891/ 0.7456 = 4.6795
Backsubstituting to determine a1 gives
a1 = [0.1058 a2 (.2280)] / 2 = [0.1058 4.6795(.2280)] / 2 = 0.4806
The error is given by
e = ideal act = 2 2 (2a1 + a2 sin ) = 2 2 [2(0.4806) + (4.6795)sin ]
= 2 2 [0.9611+ 4.6795sin ]
Substituting in the given value = /4 = 45˚,
e = 2 2 [0.9611+ 4.6795 sin ] = 2(.7854)2 [ 0.9611+ 4.6795sin 45˚ ] = 1.1141
Problem 6.24
Determine the link lengths and draw a four-bar linkage that will generate the function =2 ( and
both in radians) for values of between 0.5 and 1.0 radians. Use Chebyshev spacing with three
position points. The base length of the linkage must be 2 cm.
Solution
Determine the precision points using Chebychev spacing. Then
0 = 0.5
f = 1.0
f + 0 f 0
cos30˚ = 1.5 0.5 cos30˚ = 0.53349364905389
2
2
2
2
f + 0 1.5
2 =
=
= 0.75
2
2
+ 0 f 0
1.5 0.5
3 = f
+
cos30˚ =
+
cos30˚ = 0.96650635094611
2
2
2
2
1 =
Compute the 's
1 = 12 = 0.28461547358084
2 = 22 = 0.56250000000000
3 = 32 = 0.93413452641916
- 264 -
cos1 1 0.95976969417465
cos2 = 1 0.84592449923107
cos3 1 0.59451456313420
1 cos1
A = 1 cos2
1 cos
3
-0.86103565250177 -0.73168886887382
-0.56817793658800 cos(1 1) 0.96918935579862
B = cos(2 2) = 0.98247331310126
cos( ) 0.99947607824379
3
3
Then
x1
1 cos 1
x2 = 1 cos2
x3 1 cos3
cos1 1cos(1 1)
1.05801273775729
cos 2 cos(2 2) = -0.00195486778387
cos3 cos( 3 3)
0.10097974323242 and
r2 = 1 = -511.5
x2
r4 = 1 = 9.903
x3
r3 = 1+ r22 + r22 2r2r4 x1 = 522.0
Now unscale the values by multiplying each by 2. Then
R1 = 2.0;
R2 = 2(-511.5) = -1023;
R3 = 2(522.0) = 1044;
R4 = 2(9.903) = 19.80;
Check for linkage type:
l + s = 1044 + 2.0 = 1046
p + q = 19.80 + 1023 = 1042.8
Then
1046> 1042 nonGrashof, double rocker.
Note that the link-length ratios vary greatly. This design would be considered very undesirable.
Problem 6.25
Determine the link lengths and draw a four-bar linkage that will generate the function = sin( ) for
values of between 0 and 90 degrees. Use Chebyshev spacing with three position points. The
base length of the linkage must be 2 cm.
- 265 -
Solution
Determine the precision points using Chebychev spacing. Then
0 = 0
f = 90˚
f + 0 f 0
cos30˚ = cos30˚ = 0.10522340180962
2
2
4 4
f + 0 2 =
= = 0.78539816339745
2
4
+ 0 f 0
3 = f
+
cos30˚ = + cos30˚ = 1.46557292498528
2
2
4 4
1 =
Compute the 's
1 = sin(1 ) = 0.10502933764983
2 = sin( 2 ) = 0.70710678118655
3 = sin( 3 ) = 0.99446912382076
1 cos1
A = 1 cos 2
1 cos 3
cos1 1 0.99448948752450 -0.99446912382076 cos 2 = 1 0.76024459707563 -0.70710678118655
cos 3 1 0.54494808990266 -0.10502933764983 cos(1 1 ) 0.99999998116955 B = cos( 2 2 ) = 0.99693679488540
cos( 3 3 ) 0.89106784875455 Then
x1
1 cos 1
x2 = 1 cos 2
x3 1 cos 3
cos 1 1 cos(1 1 )
1.05576595368848
cos 2
cos(2 2 ) = -0.36099675791050
cos 3 cos( 3 3 ) -0.30492802741671
and
r2 = 1 = -2.77010798043768
x2
1
r4 = = -3.27946239796913
x3
r3 = 1+ r22 + r22 2r2 r4 x1 = 0.49621992921794
Now unscale the values by multiplying each by 2. Then
- 266 -
R1 = 2.0;
R2 = 2(-2.7701) = -5.5402;
R3 = 2(0.4962) = 0.9924;
R4 = 2(-3.2794) = -6.5588;
A scaled drawing of the linkage is given in the following:
Problem 6.26
Design a four-bar linkage that generates the function y = x x + 3 for values of x between 1 and
4. Use the Chebyshev spacing for three position points. The base length of the linkage must be 2
in. Use the following angle information:
0 = 45˚
= 50˚
0 = 30˚
= 70˚
Compute the error at x = 2
Solution
The solution is given in the following. From the given information,
- 267 -
x0 = 1; xf = 4.
Using Chebychev spacing for the precision points,
x1 =
xf + x0 xf x0 cos30° = 4 +1 4 1 cos30° = 1.20096189
2
2
2
2
Similarly, x2 = 2.5 and x3 = 3.799038105. Then, the corresponding values for y are:
yf =
y0 =
y1 =
y2 =
y3 =
xf xf + 3 = 1
x0 x0 + 3 = 3
x1 x1 + 3 = 2.894922175
x2 x2 + 3 = 2.081138830
x3 x3 + 3 = 1.150074026
Note that a minimum of 5 places of decimals is needed to ensure adequate solution accuracy. For
the range of linkage angles, we have:
0 = 45°; = 50°
and
0 = 30°; = 70°
The precision points in terms of are:
1 =
x1 x 0
1.20096189 1
+ 0 =
50°+ 45° = 48.34936490538903°
x f x0
3
2 =
x2 x0
2.5 1
+ 0 =
50° + 45° = 70°
x f x0
3
3 =
x3 x0
3.799038105 1
+ 0 =
50°+ 45° = 91.65063509461096°
x f x0
3
Similarly,
y1 y0
2.894922175 3
+ 0 =
70° + 30 = 33.67772386020241°
y f y0
2
y y
2.081138830 3
2 = 2 0 + 0 =
70°+ 30 = 62.16014094705335°
y f y0
2
y y
3 = 3 0 + 0 = 1.150074026 3 70° + 30 = 94.74740905563471°
y f y0
2
1 =
Using the matrix solution procedure,
- 268 -
1
1 cos 1 cos1 cos(1 1 )
z1
z2 = 1 cos 2 cos 2 cos( 2 2 )
z3 1 cos 3 cos 3 cos( 3 3 ) 1
1 0.8321697783 -0.6645868192 0.9673932335
1.0037381398
= 1 0.4670019053 -0.3420201433
0.9906531872 = 0.1450642022
0.9985397136 0.2363317277
1 -0.0827631422 0.0288050322 and
r2 =
1
1
=
= 6.8934994632
z2 0.1450642022
r4 =
1
1
=
= 4.2313404537
z3 0.2363317277
and
r3 = 1+ r22 + r42 2r2 r4 z1
= 1+ (6.893499) 2 + (4.231340)2 2(6.893499)(4.231340)(1.003738) = 2.8051768
For the overall size of the linkage, use a base link length of 2 in. Then the lengths of the other links
become
R1 = 1(2) = 2 in
R2 = 6.8934994632(2) = 13.7869989 in
R4 = 4.2313404537(2) = 8.46268090 in
R3 = 2.8051768(2) = 5.610353611 in
Note that the link-length ratios would make this linkage undesirable and possibly unusable.
To compute the error at x = 2, compute the ideal y and the generated y.
yideal = x x + 3 = 2 2 + 3 = 2.4142
To compute the generated value, first compute the value of corresponding to x = 2. Then,
e =
xe x0
2 1
+ 0 =
50°+ 45° = 61.666°
x f x0
3
Using this value of and the link lengths given above, find the output value for . This can be done
graphically, or by using the routine fourbar_cr. The value of is
e = 51.02815161
The corresponding value for y is,
- 269 -
yact =
e 0
51.02815161 30
(y f y0 ) + y0 =
(2)+ 3 = 2.3992
70
The error is
error = yideal yact = 2.4142 2.3992 = 0.0150
Problem 6.27
Design a four-bar linkage to generate the function y=x2 -1 for values of x between 1 and 5. Use
Chebyshev spacing with three position points. The base length of the linkage must be 2 cm. Use
the following angle information:
0 = 30˚
= 60˚
0 = 45˚
= 90˚
Compute the error at x = 3.
Solution
The solution is given in the following. From the given information,
x0 = 1; xf.= 5.
Using Chebychev spacing for the precision points,
x1 =
xf + x0 xf x0 cos30° = 1+ 5 5 1 cos30° = 1.26794919243112
2
2
2
2
Similarly, x2 = 3 and x3 = 4.73205080756888. Then, the corresponding values for y are:
yf = x f2 1 = 24
y0 = x0 2 1= 0
y1 = x12 1 = 0.60769515458674
y2 = x22 1= 8
y3 = x32 1= 21.39230484541327
Note that a minimum of 5 places of decimals is needed to ensure adequate solution accuracy. The
linkage angles are:
0 = 30°; = 60°
and
0 = 45°; = 90°
The precision points in terms of are:
- 270 -
1 = x1 x0 + 0 = 1.26794919243112 1 60° + 30° = 34.01923788646684°
4
xf x0
2 =
x2 x0 + = 3 1 60°+ 30° = 60°
0
4
xf x0
3 =
x3 x0 + = 4.73205080756888 1 60° + 30° = 85.98076211353316°
0
4
xf x0
Similarly,
y1 y0
0.60769515458674 0
+ 0 =
90°+ 45= 47.278856829°
y f y0
24
y y
8 0
2 = 2 0 + 0 =
90°+ 45 = 75°
y f y0
24
y y
3 = 3 0 + 0 = 21.39230484541327 0 90° + 45= 125.221143170°
y f y0
24
1 =
Using the matrix solution procedure,
1
1 cos 1 cos1 cos(1 1 )
z1
z2 = 1 cos 2 cos 2 cos( 2 2 )
z3 1 cos 3 cos 3 cos( 3 3 ) 1
1 0.6784308203 -0.8288497687 0.973340766 1.1947633778
= 1 0.2588190451 -0.5000000000 0.965925826 = 0.6332388583
1 -0.5767338180 -0.0700914163
0.774498853 0.7854636563
and
r2 =
1
1
=
= 1.57918293
z2 0.6332388583
r4 =
1
1
=
= 1.27313338
z3 0.7854636563
and
r3 = 1+ r22 + r42 2r2 r4 z1
= 1+ (1.579182)2 + 1.2731332 2(1.579182)(1.273133)(1.1947633) = 0.557242
For the overall size of the linkage, use a base link length of 2 cm. Then the lengths of the other
links become
- 271 -
R1 = 1(2) = 2 cm
R2 = 1.579182(2) = 3.158364 cm
R4 = 1.273133(2) = 2.546266 cm
R3 = 0.557242(2) = 1.114484 cm
Since x=3 is a precision point, the error will be zero at that point. To prove this, compute the ideal y
and the generated y.
yideal = x 2 1= 32 1 = 8
To compute the generated value, first compute the value of corresponding to x = 3. Then,
e =
xe x0
3 1
+ 0 =
60°+ 30° = 60°
x f x0
3
Using this value of and the link lengths given above, find the output value for . This can be done
graphically, or by using the routine fourbar_cr. The value of is
e = 75˚
The corresponding value for y is,
yact =
e 0
75 45
(y f y0 ) + y0 =
(24) + 0 = 8
90
The error is
error = yideal yact = 8 8 = 0
Problem 6.28
The output arm of a lawn sprinkler is to rotate through an angle of 90˚, and the ratio of the times for
the forward and reverse rotations is to be 1 to 1. Design the crank-rocker mechanism for the
sprinkler. If the crank is to be 1 inch long, give the lengths of the other links.
Solution
= 180 Q 1 = 180 1 1 = 0°
Q +1
1+ 1
This problem does not have a unique solution. To start the construction, arbitrarily pick r4 as 2" at
the angle shown. Then, the center location for O2 is located on a line through B1 and B2 . Any
point will work as long as it is selected to the left of B2 .
- 272 -
4.314
1.500
B2
B1
O2
3.232"
90˚
O4
For the location selected,
r2 + r3 = 4.314
r3 r2 = 1.500
Then
2(r3 ) = 5.814
or
and
r3 = 2.907"
r2 = r3 1.500 = 2.907 1.500 = 1.407"
From the figure, r1 = 3.232".
If the crank is 1 in long, then each dimension must be scaled. To do this multiply each length by K
where
K=
1 = 0.711
1.407
Then,
r1 = 0.711*(3.232) = 2.298 in
r2 =0.711*(1.407) = 1.00 in,
r3 = 0.711*(2.907) = 2.067 in
r4 = 0.711*(2.0) = 1.422 in,
- 273 -
Problem 6.29
Design a crank-rocker mechanism such that with the crank turning at constant speed, the oscillating
lever will have a time ratio of advance to return of 3:2. The lever is to oscillate through an angle of
80o , and the length of the base link is to be 2 in.
Solution
= 180 Q 1 = 180 1.5 1 = 36°
Q +1
1.5 + 1
This problem does not have a unique solution. To start the construction, arbitrarily pick r4 as 2" at
the angle shown. Then, the construction of the center location for O2 is shown in the following
figure. From the figure,
r2 + r3 = 3.731
r3 r2 = 1.687
Then
2(r3 ) = 5.418
or
and
r3 = 2.709"
r2 = r3 1.687 = 2.709 1.687 = 1.022"
From the figure, r1 = 2.155".
If the distance between the fixed pivots is really 2 in, then each dimension must be scaled. To do
this multiply each length by K where
K=
2 = 0.928
2.155
B2
B1
3.731"
36˚
80˚
1.687"
O2
O4
2.155"
Then,
r1 = 0.928*2.155 = 2 in
- 274 -
r2 = 0.928*(1.022) = 0.948 in,
r3 = 0.928*(2.709) = 2.514 in,
r4 = 0.928*(2.0) = 1.856 in.
Note that a different linkage is generated for each line drawn through B1 .
Problem 6.30
A packing mechanism requires that the crank (r2 ) rotate at a constant velocity. The advance part of
the cycle is to take twice as long as the return to give a quick-return mechanism. The distance
between fixed pivots must be 0.5 m. Determine the lengths for r2 , r3 , and r4 .
B
r3
A
∆θ = 80˚
r4
r2
1 ω2
O2
O4
Schematic Drawing
Solution
= 180 Q 1 = 180 2 1 = 60°
Q +1
2 +1
This problem does not have a unique solution. To start the construction, arbitrarily pick r4 as 2" at
the angle shown. Then, the construction of the center location for O2 is:
B2
B1
2.9365"
1.1482"
60˚
80˚
O2
1.4728"
O4
r2 + r3 = 2.9365
r3 r2 = 1.1482
- 275 -
Then
2(r3 ) = 4.0847
or
and
r3 = 2.0424"
r2 = r3 1.1482 = 2.0424 1.1482 = 0.8942"
From the figure, r1 = 1.4728".
If the distance between the fixed pivots is 0.5 meters, then each dimension must be scaled. To do
this multiply each length by K where
K = 0.5 = 0.3395
1.4728
Then,
r1 = 0.3395)*(1.4728) = 0.5 meters
r2 = 0.3395)*(0.8942) = 0.3036 meters,
r3 = 0.3395)*(2.0424) = 0.6934 meters,
r4 = 0.3395)*(2.0) = 0.6789 meters,
Note that a different linkage is generated for each line drawn through B1 .
Problem 6.31
The rocker O4 B of a crank-rocker linkage swings symmetrically about the vertical through a total
angle of 70˚. The return motion should take 0.75 the time that the forward motion takes. Assuming
that the two pivots are 2.5 in apart, find the length of each of the links.
Solution
For this problem, the time ratio of forward stroke to return is 1.33. Therefore,
= 180 Q 1 = 180 1.33 1 = 25.7°
Q +1
1.33 + 1
This problem does not have a unique solution. To start the construction, arbitrarily pick r4 as 2" at
the angle shown. Then, the construction of the center location for O2 is:
r2 + r3 = 4.2014
r3 r2 = 2.4112
Then
2(r3 ) = 6.6126"
or
and
r3 = 3.3063"
- 276 -
r2 = r3 2.4112 = 3.3063 2.4112 = 0.8951"
B2
B1
70˚
2.4112"
O2
25.7˚ 4.2014"
2.5984"
O4
From the figure, r1 = 2.5984".
If the distance between the fixed pivots is really 2.5", then each dimension must be scaled. To do
this multiply each length by K where
K=
2.5 = 0.9621
2.5984
Then,
r1 = 0.9621*(2.5984) = 2.5"
r2 = 0.9621*(0.8951) = 0.9612",
r3 = 0.9621*(3.3063) = 3.1810",
r4 = 0.9621*(2.0) = 1.9243",
Note that a different linkage is generated for each line drawn through B1 .
Problem 6.32
A crank rocker is to be designed such that with the crank turning at a constant speed CCW, the
rocker will have a time ratio of advance to return of 1.25. The rocking angle is to be 40˚, and it
rocks symmetrically about a vertical line through O4 . Assume that the two pivots are on the same
horizontal line, 3 in apart.
Solution:
This problem does not have a unique solution. To solve this problem, the procedure given in
Section 4.4.3.1 can be used. First draw the line beween the pivot points as shown. Next compute
the angle using
- 277 -
= 180 Q 1 = 180 1.25 1 = 20°
Q +1
1.25 + 1
B2
80˚
B1
40˚
2.3127"
1.9238"
4.4885"
O2
O4
3.00"
20˚
40˚
G
Locate G as shown in Fig. 4.46, and draw the locus of B2 . To locate the oscillation angle
symetrically about the vertical, locate B2 at an angle of 20˚ from the vertical. We can then locate B1
at an angle of 80˚ from the line O2 B2 . Next compute r2 and r3 from
r2 + r3 = 4.4885
r3 r2 = 2.3127
Then
2(r3 ) = 6.8012
or
and
r3 = 3.4006
r2 = r3 2.3127 = 3.4006 2.3127 = 1.0879"
Also from the drawing,
r1 = 3.0000"
r4 = 1.9238"
The transmission angles do not approach either 0 or 180; therefore, the linkage is reasonably
efficient.
Problem 6.33
Design a crank-rocker mechanism that has a base length of 2.0, a time ratio of 1.3, and a rocker
oscillation angle of 100˚. The oscillation is to be symmetric about a vertical line through O4 .
Specify the length of each of the links.
Solution:
- 278 -
This problem does not have a unique solution. To solve this problem, the procedure given in
Section 4.4.3.1 can be used. First draw the line beween the pivot points as shown. Next compute
the angle using
= 180 Q 1 = 180 1.3 1 = 23.48°
1.3 + 1
Q +1
and
= 50 23.48 = 26.52
2
Locate G as shown in Fig. 4.46, and draw the locus of B2 . To locate the oscillation angle
symetrically about the vertical, locate B2 at an angle of 50˚ from the vertical. We can then locate B1
at an angle of 100˚ from the line O2 B2 . Next compute r2 and r3 from
r2 + r3 = 2.9890
r3 r2 = 1.3229
Then
2(r3 ) = 4.3119
or
r3 = 2.1559
100˚
1.3229"
O2
B2
50˚
2.9890"
1.2794"
2.00"
26.52˚
B1
O4
50˚
G
and
r2 = r3 1.3229 = 2.1559 1.3229 = 0.8330"
Also from the drawing,
r1 = 2.0000"
r4 = 1.2794"
Problem 6.34
A crank-rocker mechanism with a time ratio of 2 13 and a rocker oscillation angle of 72˚ is to be
designed. The oscillation is to be symmetric about a vertical line through O4 . Draw the mechanism
- 279 -
in any position. If the length of the base link is 2 in, give the lengths of the other three links. Also
show the transmission angle in the position in which the linkage is drawn.
Solution:
This problem does not have a unique solution. To solve this problem, the procedure given in
Section 4.4.3.1 can be used. First draw the line beween the pivot points as shown. Next compute
the angle using
and
2 1 1
= 72˚
= 180 Q 1 = 180 31
Q +1
2 3 +1
= 36 72 = 36˚
2
This means that the point G is at infinity, and the B2 locus is a straight line through O2 at an angle
of 36˚ to the horizontal. The result is shown in the following:
2.8617" B2
1.6251"
72˚
B1
36˚
1.1629"
O2
36˚
2.00"
36˚
G
O4
To locate the oscillation angle symetrically about the vertical, locate B2 at an angle of 36˚ from the
vertical. We can then locate B1 at an angle of 72˚ from the line O2 B2 . Next compute r2 and r3
from
r2 + r3 = 2.8619
r3 r2 = 1.6251
Then
2(r3 ) = 4.4870
or
and
r3 = 2.2435"
r2 = r3 1.6251 = 2.2435 1.6251 = 0.6184"
Also from the drawing,
r1 = 1.0000"
and
- 280 -
r4 = 1.1629"
The linkage is shown in an arbitrary position below. The transmission angle is shown.
B
69.44˚
O2
O4
Problem 6.35
The mechanism shown is used to drive an oscillating sanding drum. The drum is rotated by a
splined shaft that is cycled vertically. The vertical motion is driven by a four-bar linkage through a
rack-and-pinion gear set (model as a rolling contact joint). The total vertical travel for the sander
drum is 3 in, and the pinion has a 2 in radius. The sander mechanism requires that the crank (r2 )
rotate at a constant velocity, and the advance part of the cycle is to take the same amount of time as
the return part. The distance between fixed pivots must be 4 in. Determine the lengths for r2 , r3 ,
and r4 .
Sander drum
A
1ω 2
B
r3
r2
O2
Total travel = 3"
r4
O4
∆θ
Rack
2"
Schematic Drawing
Pinion
Belt drive
Splined Shaft
Motor
Solution:
This problem does not have a unique solution. To solve this problem, we must determine the
oscillation angle. If the drum rotates 3 inches, we can find the oscillation angle form
and
r4 = d
= d = 3 = 1.5 rad = 85.94˚
r4 2
- 281 -
To start the construction, first draw the line beween the pivot points as shown. Next compute the
angle using
= 180 Q 1 = 180 1 1 = 0˚
1+ 1
Q +1
Arbitrarily pick r4 as 2" at the angle shown. Then, the center location for O2 is located on a line
through B1 and B2 . Any point will work as long as it is selected to the left of B2 .
4.250
1.500
B1
B2
O2
85.94˚
3.2253"
O4
For the location selected,
r2 + r3 = 4.250
r3 r2 = 1.500
Then
2(r3 ) = 5.750
or
and
r3 = 2.875"
r2 = r3 1.500 = 2.875 1.500 = 1.375"
From the figure, r1 = 3.2253".
If the crank is really 1 in long, then each dimension must be scaled. To do this multiply each length
by K where
K=
4 = 1.2413
3.2253
Then,
r1 = 1.2413*(3.2253) = 4.000 in
r2 =1.2413*(1.375) = 1.706 in,
- 282 -
r3 = 1.2413*(2.875) = 3.569 in
r4 = 1.2413*(2.0) = 2.482 in,
Problem 6.36
The mechanism shown is proposed for a rock crusher. The crusher hammer rotates through an
angle of 20˚, and the gear ratio R G/RP is 4:1, that is, the radius r G is four times the radius r p .
Contact between the two gears can be treated as rolling contact. The crusher mechanism requires
that the crank (r2 ) rotate at a constant velocity, and the advance part of the cycle is to take 1.5 times
as much as the return part. The distance between fixed pivots O2 and O4 must be 4 ft. Determine
the lengths for r2 , r3 , and r4 .
1ω
2
B
r3
A
∆θ = 20˚
r4
r2
O4
O2
rP
rG
5
Schematic Drawing
Solution
= 180 Q 1 = 180 1.5 1 = 36°
Q +1
1.5 + 1
= 4(20) = 80°
This problem does not have a unique solution. To start the construction, arbitrarily pick r4 as 2" at
the angle shown. Then, the construction of the center location for O 2 is:
B2
B1
3.7087"
1.6602"
80˚
36˚
O2
O4
- 283 -
r2 + r3 = 3.7087
r3 r2 = 1.6602
Then
2(r3 ) = 5.3689
or
and
r3 = 2.6845"
r2 = r3 1.6602 = 2.6845 1.6602 = 1.0242"
From the figure, r1 = 2.1418".
If the distance between the fixed pivots is really 4 feet, then each dimension must be scaled. To do
this multiply each length by K where
K=
4 = 1.8675
2.1418
Then,
r1 = 4.0 feet,
r2 = 1.9126 feet,
r3 = 5.0133 feet,
r4 = 3.7350 feet,
Note that a different linkage is generated for each line drawn through B1 .
Problem 6.37
The mechanism shown is proposed for a shaper mechanism. The shaper cutter moves back and
forth such that the forward (cutting) stroke takes twice as much time as the return stroke. The crank
(r2 ) rotates at a constant velocity. The follower link (r4 ) is to be 4 in and to oscillate through an
angle of 80˚. Determine the lengths for r1 , r2 , and r3 .
A
1ω
2
C
r3
B
r4
r2
O4
O2
Schematic Drawing
r5
D
6
Cutter
Part Being Machined
Solution:
This problem does not have a unique solution. To start the procedure, determine the angle from
the time ratio.
- 284 -
= 180 Q 1 = 180 (2 1) = 60˚
Q +1
(2 +1)
B2
0.90"
80˚
B1
2.90"
60˚
2.00"
O2
1.64"
O4
From the diagram:
r1 = 1.64"
r2 = O2 B1 O2 B2 = 2.90 0.90 = 1.00"
2
2
r3 = O2B1 + O2B2 = 2.90 + 0.90 = 1.90"
2
2
r4 = 2"
Scalling the results,
R4 = 4 in
R2 = r2 R 4 = 1.00 4 = 1.00(2) = 2.00 in
r4
2
R
4
= 1.90(2) = 3.80 in
R3 = r3
r4
R1 = r1 R4 = 1.64(2) = 3.28 in
r4
The linkage is shown to scale in a general position in the following:
B
A
O4
O2
- 285 -
Problem 6.38
A crank rocker is to be used in a door-closing mechanism. The door must open 100˚. The crank
motor is controlled by a timer mechanism such that it pauses when the door is fully open. Because
of this, the mechanism can open and close the door in the same amount of time. If the crank (r2 ) of
the mechanism is to be 10 cm long, determine the lengths of the other links (r1 , r3 , and r4 ). Sketch
the mechanism to scale.
Schematic Drawing
r3
A
1ω
O2
B
r4
r2
2
Door
O4
Wall
Solution
The time ration is 1 so is 0. This problem does not have a unique solution. Initially pick the
output link length to be 2. Then the following scalled values are determined as shown in the figure
below.
B1O2 = 5.0"= r3 + r2
B2O2 = 1.953"= r3 r2
Then
r3 = 3.476"
r2 = 5.0 3.476 = 1.523"
r4 = 2"
r1 = 3.713"
- 286 -
B1
B2
O2
100˚
O4
Now unscale the output.
K = R2 = 10 = 6.566
r2 1.532
Then,
R1 = 6.566(3.713) = 24.37 cm
R2 = 6.566(3.713) = 10 cm
R3 = 6.566(3.713) = 22.82 cm
R 4 = 6.566(2) = 13.13 cm
A scaled version of the linkage if given in the following.
B
R3
A
R2
O2
Scaled Linkage
R1
- 287 -
R4
O4
Problem 6.39
A crank rocker is to be used for the rock crusher mechanism shown. The oscillation angle for the
rocker is to be 80˚, and the working (crushing) stroke for the rocker is to be 1.1 times the return
stroke. If the frame link (r1 ) of the mechanism is to be 10 ft long, determine the lengths of the other
links (r2 , r3 , and r4 ). Sketch the mechanism to scale.
B
3
4
A
Roc
k
2
O1
O2
Solution:
= 180 Q 1 = 180 (1.11) = 8.6˚
Q +1
(1.1+1)
This problem does not have a unique solution. Initially pick the output link length to be 2. Then
from the diagram,
B2
3.24"
O1
80˚
B1
8.6˚
5.74"
2.00"
4.39"
O2
r1 = 4.39"
r2 = O1B1 O1B2 = 5.74 3.24 = 1.25"
2
2
O
5.74
+
3.24 = 4.49"
B
+
O
B
r3 = 1 1 1 2 =
2
2
r4 = 2"
Scalling the results,
R1 = 10 ft
- 288 -
R2 = r2 R1 = 1.25 10 = 1.25(2.33) = 2.91 ft
r1
4.39
R
R3 = r3 1 = 4.49(2.33) = 10.46 ft
r1
R 4 = r4 R1 = 2(2.33) = 4.66 ft
r1
The linkage is shown to scale in a general position in the following:
B
A
O2
O1
Problem 6.40
A crank rocker is to be used in a windshield-wiping mechanism. The wiper must oscillate 80˚. The
time for the forward and return stroke for the wiper is the same. If the base link (r1 ) of the
mechanism is to be 10 cm long, determine the lengths of the other links (r2 , r3 , and r4 ). Sketch the
mechanism to scale.
Wiper
Schematic Drawing
1ω 2
O2
r1
Frame
r2
A
O4
r4
r3
B
Solution
The time ration is 1 so
- 289 -
[]
= 180 Q 1 = 180 0 = 0
2
Q + 1
This problem does not have a unique solution. Initially pick the output link length to be 2. Then
the linkage can be constructed as shown
O4
R1
R4
B2
O2
A1
R2
B1
A2
From the figure,
R4 = 2 in
R1 = 3.512 in
R2 + R3 = O2 B2
R3 R2 = O2 B1
or
R2 = (O2 B2 O2 B1) / 2 = (4.5078 2.0018) / 2 = 1.253 in
R3 = (O2B2 + O2B1) / 2 = (4.5078 + 2.0018) / 2 = 3.881in
Determine the scaling factor,
r1 = 10 = 2.7732 = r2 = r3 = r4
R1 3.6059
R2 R3 R4
Using the unscalled lengths from the figure,
r2 = 2.7732 R2 = 2.7732(1.253) = 3.4749cm
r3 = 2.7732 R3 = 2.7732(3.881) = 10.7629cm
r4 = 2.7732 R4 = 2.7732(2) = 5.5465cm
The mechanism is drawn to scale as:
- 290 -
O4
R4
B
O2
R3
R2
A
Problem 6.41
Design a six-bar linkage like that shown in Fig. 6.62 such that the output link will do the following
for one complete revolution of the input crank:
1. Rotate clockwise by 30˚ for a clockwise rotation of 210˚ of the input crank.
2. Rotate counterclockwise by 30˚ for a clockwise rotation of 150˚ of the input crank.
Solution
Virtually any curve that has a general oval shape can be made to work for this problem. After
selecting such a curve, pick a starting place on the curve as one of the two extreme locations for
point F. Draw a line perpendicular to the curve and select a length for link 5. The value for the link
length is somewhat arbitrary although it needs to be long enough to permit the mechanism to
operate for the whole cycle. Next count around the curve 42 dashes corresponding to 210˚ of crank
rotation. This will be the other extreme location for link 5, and the link will be perpendicular to the
curve at this location also. Locate the link perpendicular to the curve, and this will locate the second
extreme location for Point F.
Connect the two extreme locations of F by a cord line, and locate the perpendicular bisector of the
cord line. Draw a line through one of the extreme locations of F at an angle of 15˚ to the
perpendicular bisector. This will locate point G and the length of link 6. Draw the dyad GRE to
complete the 6-bar linkage.
For the mechanism shown, the following values apply:
AB = 0.762”;
BC = 2.191”;
CD = 1.193”;
AD = 2.488”;
BE = 1.313”;
= -34.03˚; EF = 1.00”; FG = 1.564”;
1 = -11.83˚
XG = 2.644”;
YG = -0.532”
The linkage was analyzed for the values given above using the program sixbar.m, and the results are
shown in following the solution drawing. The results are fairly accurate; however, the accuracy
could be improved by moving the location of G slightly.
- 291 -
- 292 -
Problem 6.42
Design a six-bar linkage like that shown in Fig. 6.62 such that the output link will make two
complete 35˚ oscillations for each revolution of the driving link. (Hint: Select a coupler curve that is
shaped like a "figure 8".)
Solution
Locate a curve that is roughly in the shape of a symmetric figure 8. Draw a circle which is tangent
to the top of the curve at the two points b and d. Here, b and d should be approximately in the
center of the loops of the figure 8. Label the center of the circle as point F', which is one extreme
location for the point F. Next draw a circle of the same radius as that of the first circle but tangent
to two ponits on the bottom of the curve. The tangent points are a and c. The center of this circle is
point F", which is the other extreme location for the point F. Points a, b, c, d will be the extreme
locations of the coupler point E. Bisect the line F'F". The pivot G must lie on this line. Locate G
such that the included angle between GF' and GF" is 35˚.
Draw the dyad GFE to complete the 6-bar linkage.
The linkage was analyzed for the values given above using the program sixbar.m, and the results are
shown in following the solution drawing. The results are fairly accurate; however, the accuracy
could be improved by moving the location of G slightly.
Y
B
2
3
C
β
A
X
4
b
a
E
AD = 2.1429"
AB = 1.0994"
BC = 2.1806"
CD = 1.6212"
BE = 2.1005"
EF = 2.9"
FG = 1.0614"
β = -35.5˚
θ = -26.46˚
x G = 0.7222"
yG = -2.95"
d
D
c
5
G
F'
35˚
6
F"
- 293 -
F
Problem 6.43
Design a six-bar linkage like that shown in Fig. 6.62 such that the output link will do the following
for one complete revolution of the input crank:
1. Rotate clockwise by 40˚
2. Rotate counterclockwise by 35˚
3. Rotate clockwise by 30˚
4. Rotate counterclockwise by 35˚
(Hint: Select a figure 8- or kidney bean- shaped coupler curve.)
Solution:
Figure 4.51 is repeated here for simplicity.
- 294 -
G
6
F
5
E
3
C
B
4
2
A
D
1
Fig. PB6.43.1: Six bar linkage
For this problem, the displacement of link 6 can be represented schematically as shown in Fig.
PB6.43.2, and the oscillation of link 6 would then appear as shown in Fig. 6.43.3.
Rota tion of link 6
When link 6 is in an extreme position, link 5 is perpendicular to the coupler curve. Therefore, we
need only select different coupler curves and locate candidate points for extreme positions. For this
problem, select a figure 8 coupler curve. The problem can be solved using the following steps.
40
b
d
30
20
10
0a
c
Rota tion of cr a nk
e
Fig. PB6.43.2: Motion of link 6
1) First draw two identical circles, one tangent to two locations at the top of the curve and the other
tangent to two locations at the bottom.
2) Find the centers of the two circles and call these points Fa and Fb . Next bisect the line between
Fa and Fb . On this line, locate the point G such that the angle FaGFb is 40˚. Draw an arc
centered at G and of radius GFb . This is shown in Fig. PB6.43.4.
3) Locate the lines GFc and GFd at an angle of 5˚ from the first two lines as shown in Fig.
PB6.43.5. Using Fc and Fd as centers, draw two more circles of the same radius as for the
circles in Step 1.
- 295 -
d
b
35˚
35˚
30˚
40˚
G
6
c
F
a
e
Fig. PB6.43.3: Positions of link 6
4) Use the geometry in Fig. PB6.43.5 as a single entity and rotate and scale it to fit it to the
coupler curve such that each of the circles is tangent to the coupler curve at one location.
5) Draw the dyad GFE to complete the linkage. The result is shown in Fig. PB6.43.6. The
coordinates have been rotated to facilitate the analysis of the mechanism.
6) Analyze the linkage to ensure that it meets the requirements closely enough. If the requirements
are not met adequately, select another coupler curve and repeat the procedure. The mechanism
has been analyzed using the program sixbar.m, and the results are given in Fig. PB6.43.7. The
results are reasonably accurate; however, the results can be improved somewhat by changing the
location of point G slightly.
Fb
6
G
40˚
Fa
Fig. PB6.43.4: Finding equal-diameter circles each tangent to coupler curve at two locations.
- 296 -
5˚
Fb
6
Fd
G
40˚
Fc
Fa
5˚
Fig. PB6.43.5: Setting extremes of motion for link 6
Y
3
B
2
C
β
A
4
AD = 2.1429"
AB = 1.0994"
BC = 2.1806"
CD = 1.6212"
BE = 2.1005"
EF = 1.60"
FG = 1.00"
β = -35.5˚
x G = 1.70"
y G = -1.60"
E
D
5
Fb
6
G
Fd
Fc
Fa
Fig. PB6.43.6: Final solution
- 297 -
X
Fig. PB6.43.7: Results from s i x b a r . m
Problem 6.44
Displacement of link 6, degrees
Design a six-bar linkage like that shown in Fig. 6.62 such that the displacement of the output link
(link 6) is the given function of the input link rotation. The output displacement reaches maximum
values of 30˚ and 60˚ at input rotations of 60˚ and 240˚, respectively. The rotation of the output link
is zero when the input rotation angle is 0, 120˚, and 360˚.
60˚
50˚
40˚
30˚
20˚
10˚
0˚
60˚ 120˚ 180˚ 240˚ 300˚ 360˚
Rotation of crank 2, degrees
Solution
Because of the double oscillation, either a figure 8 or kidney bean shaped curved can be used. The
procedure will be illustrated with the figure 8 curve used in Example 6.8. Locate a curve that is
roughly symmetrical as shown in the following figure. Draw a circle of radius r5 which is tangent
to the top of the curve at two locations. The center of this circle is at point F’ which is one extreme
location of F. The radius of the circle gives the length of link 5. The value for this radius is a
- 298 -
design decision. The points a and b in the figure correspond to the locations where the oscillation
angle is zero. Starting at b, count 24 dashes (120˚) and draw a second circle of radius r5 tangent to
the curve at point c. The center of this circle will be F”, the other extreme location of F. Locate
point G by bisecting the cord line between F’ and F” and finding the point on the perpendicular
bisector that will give a 60˚ oscillation angle between F’ and F”. The distance from G to F’ or to
F” is the length of link 6.
- 299 -
Locate the center of the arc between F’ and F”. Draw a third circle of radius r5 centered at this
point. This circle should be tangent to the coupler curve at a location that is approximately 12
dashes from point a. If the error is too large, select another curve.
Once the linkage is found to be acceptable, draw the dyad GFE to complete the 6-bar linkage.
In the figure, Gx = 3.543; G y = -3.848; r5 = 3.324”; r 6 = 1.329”. The original four-bar linkage
values are given in Example 6.8.
The solution given is only approximate. It is difficult to solve problems like this exactly.
Therefore, it is usually necessary to approximate either the timing or the oscillation angle. Also, a
60˚ oscillation angle is relatively large for a mechanism such as this. Once a viable linkage is
designed, it can be analyzed using the sixbar analysis program. Simple adjustments can then be
done directly with the linkage in that program.
Problem 6.45
Displacement of link 6, degrees
Design a six-bar linkage like that shown in Fig. 6.62 such that the displacement of the output link
(link 6) is the given function of the input link rotation. The output link dwells for 90˚ of input
rotation starting at 0 and 180 degrees. The maximum rotation angle for link 6 is 15˚.
15˚
10˚
5˚
0˚
180˚
270˚
90˚
Rotation of crank 2, degrees
360˚
Solution
For this problem, locate a curve that is a kidney-bean shaped curve and isroughly symmetrical as
shown in the following figure. The “top” and “bottom” of the kidney bean must have
approximately the same radius of curvature for the two dwell periods. The radius of curvature
corresponds to r5 . Draw a circle of radius r5 which is tangent to the top of the curve. The center of
this circle is at point F’ which is one extreme location of F. The radius of the circle gives the length
of link 5.
Draw a second circle of radius r5 tangent to bottom part of the curve. The center of this circle will
be F”, the other extreme location of F. Locate point G by bisecting the cord line between F’ and
F” and finding the point on the perpendicular bisector that will give a 15˚ oscillation angle between
F’ and F”. The distance from G to F’ or to F” is the length of link 6. Once the linkage is found
to be acceptable, draw the dyad GFE to complete the 6-bar linkage.
For the solution shown, the following values apply:
- 300 -
AB = 0.619; BC = 1.562; CD = 1.564; AD = 0.938
BE = 2.519; EF = 2.95; FG = 1.573; Gx = 2.377; Gy = 0.711
= 35.65˚
It is apparent from the figure that a relatively long dwell can be achieved for the top region of the
curve, but the dwell on the bottom region is shorter than the problem statement specifies. If the
approximation is ultimately unacceptable, a different curve can be used. Also, the problem can be
analyzed using the sixbar routine, and minor adjustments can be made directly with the program.
Problem 6.46
Design an eight-bar linkage like that shown in Fig. 6.70 such the coupler remains horizontal while
the given point on the coupler moves approximately along the path given.
- 301 -
700 mm
60 mm
Coupler Po
500 mm
450 mm
100 mm
Solution
Search for a coupler curve that is roughly the shape of the figure shown. It is not possible to find a
curve that exactly duplicates the figure; however, it is possible to find a curve that generally scans
the same area. A solution is shown in the following. Once the coupler curve is found, the 8-bar
linkage is completed using the procedure given in Section 6.6.2
Problem 6.47
Re-solve Problem 6.46 if the coupler is inclined at an angle of 45˚.
- 302 -
700 mm
Coupler Po
45˚
500 mm
60 mm
450 mm
100 mm
Solution
Search for a coupler curve that is roughly the shape of the figure shown. It is not possible to find a
curve that exactly duplicates the figure; however, it is possible to find a curve that generally scans
the same area. A solution is shown in the following. Once the coupler curve is found, the 8-bar
linkage is completed using the procedure given in Section 6.6.2
Problem 6.48
Design an eight-bar linkage like that shown in Fig. 6.70 such the coupler remains horizontal while
the given point on the coupler moves approximately along the path from A to B to C. The coupler
can return either by retracing the path from C to B to A or by going directly from C to A. This
means that the basic 4-bar linkage need not be a crank rocker.
- 303 -
B
Equilateral
Triangle
Coupler Point
1"
5"
C
A
8"
Solution
Search for a coupler curve that is roughly the shape of the figure shown. It is not possible to find a
curve that exactly duplicates the figure; however, it is possible to find a curve that generally scans
the same area. A solution is shown in the following. Once the coupler curve is found, the 8-bar
linkage is completed using the procedure given in Section 6.6.2
- 304 -
Problem 6.49
Determine the two 4-bar linkages cognate to the one shown below. The dimensions are MA = 10
cm, AB = 16 cm, AC = 32 cm, QB = 21 cm, and MQ = 24 cm. Draw the cognates in the position
for = 90˚.
C
B
A
θ
Q
M
Solution
Use Roberts' linkage in Fig. 6.76 as a guide to construct the cognates. First locate pivot O by
recognising that triangle MQO is similar to ABC. Then complete the parallagrams indicated in
Figs. 6.76 and 6.77.
The cognates are shown in the following figure.
F
C
B
G
A
D
Q
M
O
E
This figure can be checked using the program cognates.m. The coupler curve and cognates are
given in the following.
- 305 -
Problem 6.50
Determine the two 4-bar linkages cognate to the one shown below. The dimensions are MQ = 1.5
in, AB = BC = BQ = AC= 1 in, and AM = 0.5 in. Draw the cognates in the position for = 90˚.
C
B
A
θ
Q
M
Solution
Use Roberts' linkage in Fig. 6.76 as a guide to construct the cognates. First locate pivot O by
recognising that triangle MQO is similar to ABC. Then complete the parallagrams indicated in
Figs. 6.76 and 6.77.
The cognates are shown in the following figure.
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C
E
O
F
G
A
B
D
Q
M
This figure can be checked using the program cognates.m. The coupler curve and cognates are
given in the following.
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Problem 6.51
Determine the two 4-bar linkages cognate to the one shown below. The dimensions are MQ = 2 in,
AB = BC = BQ =1 in, and AM = 1.5 in. AC = 0.75 in. Draw the cognates in the position for =
45˚.
B
C
A
Q
M
Solution
Use Roberts' linkage in Fig. 6.76 as a guide to construct the cognates. First locate pivot O by
recognizing that triangle MQO is similar to ABC. Then complete the parallagrams indicated in
Figs. 6.76 and 6.77.
The cognates are shown in the following figure.
This figure can be checked using the program cognates.m. The coupler curve and cognates are
given in the following.
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Problem 6.52
Determine the two 4-bar linkages cognate for the drag-link mechanism shown. The dimensions are
MQ = 1 m, AM = BQ = 4 m, AB =2 m, and angles CAB and CBA both equal 45˚. Notice that the
cognates will also be drag-link mechanisms. Draw the cognates in the position for = 180˚.
θ
B
C
Q
M
45˚
A
Solution
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Use Roberts' linkage in Fig. 6.76 as a guide to construct the cognates. First locate pivot O by
recognising that triangle MQO is similar to ABC. Then complete the parallagrams indicated in
Figs. 6.76 and 6.77.
The cognates are shown in the following figure.
F
E
B
Q
G
O
C
D
M
A
This figure can be checked using the program cognates.m. The coupler curve and cognates are
given in the following.
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