AME 352 KINEMATIC SYNTHESIS

P.E. Nikravesh 1

KINEMATIC SYNTHESIS

NOT COMPLETE Design 1

Design a crank-rocker four-bar (Grashof) where the input link rotates completely and the output link (the follower) rocks back and forth with a prescribed angle. The design requires equal time forward and back for the rocker assuming a constant speed motor turning the crank. As shown in the diagram, link 2 is the crank and link 4 is

the rocker with a prescribed angle θ4 .

Since there are no other requirements, there are infinite numbers of designs that would fulfill the stated requirements. The design process is summarized in the following steps.

θ4

A

B

O 2 O 4

1. Select a length L4 for the rocker (follower).

Place the pin joint O4 at a convenient

location on the ground. Construct the rocker at its limit positions, based on the

given angle θ4 . This process gives us the

position of the pin joint B at its two limit positions, B1 and B2 .

2. Draw an axis through B1 B2 .

3. Place the pin joint O2 at a convenient

location on this axis (infinite choices). This established the ground link and the length L1 = O2O4 .

4. Measure the distance B1B2 . This is twice the

length of the crank, L2 .

5. Draw a circle with its center at O2 and the

radius L2 .

6. Position the pin joint A at two positions, A1

and A2 , at the intersections of the axis and

the circle. This establishes the length of the coupler L3 = A1B1 = A2B2 .

7. Test for Grashof’s condition. If non-Grashof,

repeat the process from step 3 by locating O2 at a different location on the axis.

θ4

1 B

2 B

O 4

L 4

L 4

2 B

O 4

1 B

2 L 2

O 2

2 L 2

2 B

O 4

1 B

O 2

A 1

A 2

AME 352 KINEMATIC SYNTHESIS

P.E. Nikravesh 2

This is the designed four-bar.

B

O 4

O 2

A

Example 1

Design an equal time forward and back four-bar mechanism to operate the windshield wiper of the rear window of an automobile. The dimensions are given in the figure. The range of motion of the

rocker (blade) is 105o . The designed four-bar must fit within a compartment shown as a shaded area.

38” 36”

23”

6” 17”

5”

16” blade

Time Ratio

In Design-1 it is required that the rocker to have equal time forward and back assuming a constant speed motor as the input. We achieved this requirement by assuring the crank to rotate

180o when the rocker moved forward, and 180o when the rocker moved backward. For such

mechanism, the time ratio is one-to-one; i.e., TR = 1:1 .

In another design, it may be required for the output link to move slow in one direction and fast in the other direction; for example, the time ratio could be TR = 1:1.5 (slow:fast). This is

normally called a quick-return mechanism. To design such a mechanism, we must assure that the

crank circle is split into two portions, α and β , such that α / β = TR . Since α + β = 360o , we

can determine the two angles:

α =TR

1+ TR

360o , β =1

1+ TR

360o (a)

The difference between any of these two angles from 180o is denoted as δ :

δ = 180o −α = β −180o (b)

α

β

A 1

A 2

O 2

Design 2

This is similar to Design 1, but here the time ratio is not one-to-one; e.g., TR = 1:1.2 .

AME 352 KINEMATIC SYNTHESIS

P.E. Nikravesh 3

1. (Same as step 1 in Design 1) Select a length L4 for

the rocker (follower). Place the pin joint O4 at a

convenient location. Construct the rocker at its limit

positions, knowing the angle θ4 . This process gives

us the position of the pin joint B at its two limit positions, B1 and B2 .

2. Compute α , β , and δ .

3. Draw two axes from B1 and B2 such that the angle

between them is δ . For this purpose, you may draw

an axis through B2 first (infinite choices), and draw a

line making an angle δ with the first axis. Then from

B1 draw an axis parallel to the line. The intersection

of the two axes is O2 . This establishes the ground

link and the length L1 = O2O4 .

4. With its center at O2 , draw a circle arc from B2 until

it intersects the O2B1 axis at ′B2 . Measure the

distance between B1 and ′B2 . This is twice the length

of the crank, L2 .

5. Draw a circle with radius L2 , centered at O2 .

6. Position the pin joint A at two positions, A1 and A2 ,

at the intersections of the two axes and the circle. This establishes the length of the coupler L3 = A1B1 = A2B2 .

7. Test for Grashof’s condition. If non-Grashof, repeat the process from step 3 by drawing the axis through B2 in a different orientation.

Note that:

O2B2 = L3 + L2 , O2B1 = L3 − L2

Therefore,

L2 =O2B2 −O2B1

2 , L3 =

O2B2 + O2B1

2

This is the designed four-bar.

B

O 4

O 2

A

2 B

O 4

O 2

δ

δ

1 B

2 B

O 4

1 B

2 L 2

O 2

B’ 2

δ

2 B

O 4

1 B

2 L 2

O 2

B’ 2

δ

A 1

A 2

Note: This method is adequate for

time ratios up to 1:1.5 . For larger time ratios we need to design a six-bar or a more complex mechanism.

AME 352 KINEMATIC SYNTHESIS

P.E. Nikravesh 4

Example 2

Design 3

This problem requires designing a crank-rocker four-bar, where an extension to the rocker link should find two limiting positions, C1D1 and C2D2 , as shown.

This design can easily be turned into Design 1 or Design 2 depending on the required time-ratio.

1. Draw the perpendicular bi-sector to C1C2 .

2. Draw the perpendicular bi-sector to D1D2 .

3. The intersection of the two perpendicular bi-sectors is

the location of the pin joint O4 .

4. Construct the rocker link at its two limits O2C1D1 and

O2C2D2 .

5. Select a proper value for L4 and place the pin joint B

at its two limits B1 and B2 .

6. Continue with either Design-1 or Design-2 procedure.

C 1

D 1

C 2

D 2

C 1

D 1

C 2

D 2

O 4

θ4 1 B

2 B

L 4

L 4

C 1

D 1

C 2

D 2

O 4

Example 3

Design 4

In some four-bar mechanisms, the output link could be the coupler, not the follower. In this design problem two positions of the coupler links are provided. These are not necessarily the limiting positions. We assume that A1 and A2

are two positions of the pin joint A, and B1 and

B2 are two positions of the pin joint B; i.e., L3 is

known.

2 B

1 B

A 1 A 2

AME 352 KINEMATIC SYNTHESIS

P.E. Nikravesh 5

1. Draw the perpendicular bi-sector to A1A2

2. Place the pin joint O2 at a convenient location

on this line (infinite choices). This establishes

the length L2 .

3. Draw the perpendicular bi-sector to B1B2 .

4. Place the pin joint O4 at a convenient location

on this line (infinite choices). This establishes the length L4 and L1 .

If it is required for the four-bar to be

Grashof, we must check for that. If the designed mechanism is not Grashof, we repeat steps 2 and

4 by placing O2 and O4 at different locations on

their corresponding bi-sectors. We repeat this process until we find a Grashof four-bar.

2 B

O 4

1 B

O 2

A 1 A 2

O 4

B

O 2

A

If (a) the problem statement did not require a Grashof four-bar, or (b) we cannot find a

Grashof four-bar, or (c) A1B1 and A2B2 should be the limiting positions of the coupler, we

should do the followings: (1) we must assure that the designed four-bar can move from position 1 to position 2 and back continuously (without a need to disassemble/re-assemble the mechanism); and (2) design a dyad to drive this four-bar (dyads are discussed later in this chapter). Example 4

Design 5

A variation of Design 4 could be that three orientations of the coupler link are provided.

1. Draw the perpendicular bi-sectors to

A1A2 and A2A3 . The intersection of these

two bi-sectors is the pin joint O2 . This

establishes the length L2 .

2. Draw the perpendicular bi-sector to B1B2

and B2B3 . The intersection of these two

bi-sectors is the pin joint O4 . This

establishes the lengths L4 and L1 .

We observe that the design is unique. Most likely, the design is non-Grashof. As long as the designed four-bar can move between the three configurations continuously, we can add a dyad to drive this four-bar.

2 B

1 B

A 1 A 2

A 3

B 3

2 B

O 4

1 B

O 2

A 1 A 2

A 3

B 3

AME 352 KINEMATIC SYNTHESIS

P.E. Nikravesh 6

O 4

B

O 2

A

Example 5

Design 6

Another variation of Design-4 or Design-5 is that two or three positions of an extension of the coupler link are provided. We are given the freedom to shape the coupler link as desired. Assume the extension is given at three EF positions as shown.

1. We can decide on the shape of the coupler

link and construct it at one of the positions. This step establishes the location of the pin joints A and B at that configuration, and the length of the coupler L3 = AB .

2. Construct the same shape for the coupler at the other configurations. This step establishes the location of the joints A and B at the other configurations.

The design can be continued as either

Design-4 or Design-5.

E 1

F 1

E 2 F 2

E 3

F 3

1 B

A 1

E 1

F 1

2 B

1 B

A 1 A 2 A 3

B 3

E 1

F 1

Example 6

AME 352 KINEMATIC SYNTHESIS

P.E. Nikravesh 7

Dyad Most six-bar mechanisms are

combinations of two four-bars or a four-bar and a slider-crank. Most often one of the four-bars is the driving mechanism; i.e., it is a Grashof four-bar with a rotational motor (input). The second mechanism, whether a four-bar or a slider-crank, contains the output link.

A

B

O 2

C

D

O 4 O 6 (2)

(3)

(4)

(5) (6)

Original four-bar Dyad

As an example, consider the six-bar mechanism shown containing two four-bars in series. The original four-bar O4CDO6 was designed first to perform certain task. Link 6 (or link 5) is the

output link. This four-bar may or may not be Grashof. Regardless of that, the four-bar O2ABO4

has been added to drive the original four-bar. This second four-bar is called a dyad. The output link of the dyad becomes the input link of the original mechanism. Design 7

A four-bar has been designed in such a way that its follower link can rock back and forth by an angle of θ6 . The input link of

this four-bar negotiates in a range of θ4 .

Design a dyad to drive this four-bar with a given time ratio (TR could be 1:1 or ≠ 1:1 ).

1. Select an appropriate position for the pin

joint B on the input link of the original four-bar. This point could be along the axis of the link, could coincide with pin joint C, or the link could be made

triangular; i.e., C, B, and O4 form a

triangle. 2. Construct link 4 in its two limiting

positions in order to determine the locations of B1 and B2 .

O 4

O 6

θ4

θ6

(6) (5)

(4)

2 B 1 B

C 1 D 1

C 2 D 2 O 4

O 6

θ4

θ6

2 B 1 B

O 4

θ4

3. Follow the procedure of Design-1 or

Design-2, depending on the time ratio, to complete the dyad. In this example, we have assumed TR = 1:1 .

The resulting six-bar is depicted in

the form of an animation.

2 B 1 B

O 4

θ4

A 1 A 2

O 2

AME 352 KINEMATIC SYNTHESIS

P.E. Nikravesh 8

Example 7