school of engineering (mechanical and automotive)

0 | P a g e

School of Engineering (Mechanical and Automotive)

Major Assessment Course: Mechanics of Machines (MIET1077)

Experiment: Six Bar Linkage Mechanism

Course Coordinator: Prof. Firoz Alam ([email protected])

Student Information Family Name Given Name

Pace Samuel

Student Number: S3659265

Due Date: 25/10/20

mailto:[email protected]

1 | P a g e

Contents Summary ........................................................................................................................................................................................................... 2

Objectives .......................................................................................................................................................................................................... 2

Related Theory .................................................................................................................................................................................................. 2

Calculating Lengths of Links .............................................................................................................................................................................. 3

Graphical Position Analysis ................................................................................................................................................................................ 4

Graphical Velocity Analysis ................................................................................................................................................................................ 5

Complete Vector Polygon ............................................................................................................................................................................ 7

Graphical Velocity Results ............................................................................................................................................................................ 7

Graphical Acceleration Analysis ........................................................................................................................................................................ 8

Complete Vector Polygon .......................................................................................................................................................................... 15

Graphical Acceleration Results .................................................................................................................................................................. 16

Graphical Instantaneous Centre Analysis ........................................................................................................................................................ 16

Comlete Instantaneous Centre Diagram .................................................................................................................................................... 22

Results for Instantaneous Centre of Zero Velocity Method ....................................................................................................................... 23

Analytical Method for Position, Velocity and Acceleration ............................................................................................................................. 23

Position Analysis ........................................................................................................................................................................................ 23

Part A - Crank ........................................................................................................................................................................................ 23

Part B – Coupler & Rocker .................................................................................................................................................................... 24

Part C – Slider ....................................................................................................................................................................................... 29

Velocity Analysis ........................................................................................................................................................................................ 30

Part A - Crank ........................................................................................................................................................................................ 30


Part C - Slider ........................................................................................................................................................................................ 32

Acceleration Analysis ................................................................................................................................................................................. 34

Part A – Crank ....................................................................................................................................................................................... 34


Part C – Slider ....................................................................................................................................................................................... 37

Summary of Analytical Values .................................................................................................................................................................... 38

Dynamic Forces ............................................................................................................................................................................................... 39

Equations of Motion .................................................................................................................................................................................. 39

Equation Matrix ......................................................................................................................................................................................... 45

Working Model Simulation .............................................................................................................................................................................. 45

Balancing Strategy Development ...................................................................................................................................................................... 0

Results ............................................................................................................................................................................................................... 1

Discussion .......................................................................................................................................................................................................... 2

Conclusion ......................................................................................................................................................................................................... 3

References ........................................................................................................................................................................................................ 3

Attachments ...................................................................................................................................................................................................... 3

2 | P a g e

Summary This report details the methods for calculating the position, velocity and acceleration of a 6 bar linkage mechanism. The graphical method of calculating the position, velocity, acceleration and instantaneous centres of zero velocity use vector polygons and position diagrams to calculate and display graphically the magnitude and direction of each component which can be measured to provide an accurate result. The values are then cross checked using the analytical method for calculating each component. A simulation was made using Working Model to provide further analysis and verification of the results. Each of these methods provided results that would be used to calculate the dynamic forces of the system and generate a balancing strategy for reducing the vibration in the system. From this report, the positions, velocity, and accelerations of the linkages have been accurately calculated and verified using multiple methods of calculating the results.

Objectives The objectives of this major assessment are to:

• Apply the understanding of Mechanics of Machines course content • Calculate position, velocity, acceleration, and instantaneous centres of the system using

graphical methods • Calculate position, velocity and acceleration using analytical methods • Develop and analyse a working model simulation to evaluate the outputs • Calculate the dynamic forces on the system and develop a balancing strategy development

Related Theory A linkage mechanism has a grounded link, driver link (crank) and a slave link (rocker). The grounded (link g) link throughout the whole motion of the system will always have no velocity or acceleration. This link is the link which the system revolves around and defines the joints which the links rotate around. The crank (a and α) is what drives the motion of the system and determines the limits of motion for each link and joint. The rocker (b and β) has its motion driven by the crank and that motion is dependent on the size of the crank. The floating link (link f) has complex motion dependant on the crank and rocker and is what connects them together.1

Figure 1 4 Bar Linkage Mechanism Source: http://dynref.engr.illinois.edu/aml.html

Many different linkage mechanisms can be seen in our daily life, some examples can be seen below.

3 | P a g e

Figure 2 Bike Pedalling Source:

http://dynref.engr.illinois.edu/aml.html

Figure 3 Knee Joint Source

:http://dynref.engr.illinois.edu/aml.html

Figure 4 Suspension with Watt's linkage Source: http://dynref.engr.illinois.edu/aml.html

These different linkages seen in our daily lives are made up of different types of 4 bar linkage mechanisms and can have the same principles applied to finding out their function and ranges of movement.

Calculating Lengths of Links To begin the following values were given to begin calculating the links.

OA (m) 0.2 B (deg) 12 θ1 (deg) 6 n2 (rpm) 480

The following lengths are calculated using the provided equations

𝐻𝐻 =𝑂𝑂𝑂𝑂4

=0.24

= 0.05𝑚𝑚

𝑂𝑂𝑂𝑂 = 𝐶𝐶𝐶𝐶 = 4 ∗ 𝑂𝑂𝑂𝑂 = 4 ∗ 0.2 = 0.8𝑚𝑚

The following values are left to be calculated through the simultaneous equations provided.

𝑂𝑂𝐶𝐶 = ?

𝑂𝑂𝐴𝐴 = ?




4 | P a g e

𝐴𝐴𝑂𝑂 = ?

Trial and error was used to find the following values for the unknowns.

𝑂𝑂𝐶𝐶 = 0.3𝑚𝑚

𝑂𝑂𝐴𝐴 = 0.5𝑚𝑚

𝐴𝐴𝑂𝑂 = 0.6𝑚𝑚

These values were checked against the equations to test their validity.

𝑂𝑂𝑂𝑂 + 𝑂𝑂𝐴𝐴 < 𝑂𝑂𝑂𝑂 + 𝐴𝐴𝑂𝑂

0.2 + 0.5 < 0.8 + 0.6

0.7 < 1.4

𝑂𝑂𝑂𝑂 + 𝑂𝑂𝑂𝑂 < 𝑂𝑂𝐴𝐴 + 𝑂𝑂𝐴𝐴

0.2 + 0.8 < 0.5 + 0.6

1.0 < 1.1

Since both these statements are true, the system may continue with the following values used for each of the links.

Graphical Position Analysis

Figure 5 Graphical Position Analysis

5 | P a g e

The positions of 16 different angles of 𝜃𝜃2 and the subsequent positions of each of the other positions. The paths are 𝑆𝑆𝐴𝐴, 𝑆𝑆𝐵𝐵, 𝑆𝑆𝐶𝐶 and 𝑆𝑆𝐷𝐷 with the limits of the motion for B being 𝐴𝐴𝐿𝐿 and 𝐴𝐴𝑅𝑅 and the limits of D being 𝐶𝐶𝐿𝐿 and 𝐶𝐶𝑅𝑅.

Graphical Velocity Analysis To begin the graphical velocity analysis, the velocity of point A is calculated as it has a known rotational speed and direction and so can be found.

𝑛𝑛2 = 480 𝑟𝑟𝑟𝑟𝑚𝑚 =48060

𝑟𝑟𝑟𝑟𝑟𝑟 = 8 𝑟𝑟𝑟𝑟𝑟𝑟 = 8 ∗ 2𝜋𝜋 𝑟𝑟𝑟𝑟𝑟𝑟 𝑟𝑟⁄ = 50.26 𝑟𝑟𝑟𝑟𝑟𝑟 𝑟𝑟⁄

𝑉𝑉𝐴𝐴 = 𝜔𝜔𝐴𝐴𝑟𝑟𝑂𝑂𝐴𝐴 = 50.26 ∗ 200 = 10053.1𝑚𝑚𝑚𝑚/𝑟𝑟

The direction of 𝑉𝑉𝐴𝐴 is tangential to the position circle and the velocity is going anticlockwise.

The next component to be found is 𝑉𝑉𝐵𝐵 which can be found using the following equation

𝑉𝑉𝐵𝐵⊥OE = 𝑉𝑉𝐴𝐴⊥OA + 𝑉𝑉𝐵𝐵/𝐴𝐴⊥BE

This equation produces the following vector polygon.

Figure 6 Finding VB

The same technique is used to find 𝑉𝑉𝐶𝐶 , but there are 2 equations which need to be used. 𝑉𝑉𝐶𝐶 = 𝑉𝑉𝐴𝐴⊥OA + 𝑉𝑉𝐶𝐶/𝐴𝐴⊥AC

𝑉𝑉𝐶𝐶 = 𝑉𝑉𝐵𝐵⊥BE + 𝑉𝑉𝐶𝐶/𝐵𝐵⊥BC

Drawing out the known components and the relative components, the following vector polygon is found.

6 | P a g e

Figure 7 Finding VC

The lines can then be connected to find the relative velocities and the point which they intersect is the arrowhead of 𝑉𝑉𝐶𝐶 .

Figure 8 Graphically Solving for VC

Now that 𝑉𝑉𝐶𝐶 is known, 𝑉𝑉𝐷𝐷 can be found using it in the following equation.

𝑉𝑉𝐷𝐷||𝑋𝑋−𝑋𝑋 = 𝑉𝑉𝐶𝐶 + 𝑉𝑉𝐷𝐷/𝐶𝐶⊥CD

The following vector polygon is produced, and the connection points define the vectors of the velocities.

7 | P a g e

Figure 9 Finding VD

Figure 10 Solving for VD

Complete Vector Polygon Now that all the individual components are found, the entire velocity vector polygon can be seen below.

Figure 11 Velocity Vector Polygon

The program which the vector polygon was drawn in had a scale which was used at 𝑉𝑉𝐴𝐴 of 1 point = 100mm/s. Using this scale, the following magnitudes and angles for the rest of the vectors can be solved.

Graphical Velocity Results Width (pts) Height (pts) Length (pts) Velocity (mm/s) Velocity (m/s)

𝑉𝑉𝐴𝐴/𝑂𝑂 72.649 61.685 95.30434 9530.434 9.530434

8 | P a g e

𝑉𝑉𝐶𝐶/𝑂𝑂 35.039 45.354 57.31245 5731.245 5.731245

𝑉𝑉𝐶𝐶 52.023 4.911 52.25429 5225.429 5.225429

𝑉𝑉𝐴𝐴 14.413 11.42 18.38888 1838.888 1.838888

𝑉𝑉𝐴𝐴/𝐶𝐶 37.674 16.331 41.06132 4106.132 4.106132

𝑉𝑉𝐴𝐴/𝐶𝐶 54.92 0 54.92 5492 5.492

𝑉𝑉𝐶𝐶/𝐶𝐶 2.896 4.911 5.701293 570.1293 0.570129 Table 1 Graphical Velocity Analysis Results

Graphical Acceleration Analysis To begin the graphical acceleration analysis, first the known values are calculated to calibrate the scale and distance. The following equation is used to find the magnitude of 𝑟𝑟𝐴𝐴𝑛𝑛.

𝑟𝑟𝐴𝐴𝑛𝑛 =𝑉𝑉𝐴𝐴2

𝑟𝑟𝐴𝐴=

(𝜔𝜔𝐴𝐴𝑟𝑟𝑂𝑂𝐴𝐴)2

𝑟𝑟𝐴𝐴=

(50.265 ∗ 200)2

200= 505323.6 𝑚𝑚𝑚𝑚/𝑟𝑟2

Since it is known that the direction is perpendicular to link OA, 𝑟𝑟𝐴𝐴𝑛𝑛 can be sketched as follows. Since the velocity is known to be constant, the tangential acceleration of the support is known to be 0. Because there is no tangential component to the acceleration, the normal component of the acceleration is equal to the overall acceleration of the link.

𝑟𝑟𝐴𝐴 = 𝑟𝑟𝐴𝐴𝑛𝑛||𝑂𝑂𝐴𝐴 + 𝑟𝑟𝐴𝐴𝑡𝑡⊥𝑂𝑂𝐴𝐴

𝑟𝑟𝐴𝐴𝑡𝑡⊥𝑂𝑂𝐴𝐴 = 0

Figure 12 Calibrating Scale with aA

Using the following graphic, the directions for each component of the accelerations can be found.

9 | P a g e

Figure 13 Directions of Acceleration Components

The first component to be found is 𝑟𝑟𝐵𝐵/𝐴𝐴𝑛𝑛 using the following calculation.

𝑟𝑟𝐵𝐵/𝐴𝐴𝑛𝑛 =

𝑉𝑉𝐵𝐵/𝐴𝐴2

𝑟𝑟𝐵𝐵/𝐴𝐴=

9530.4342

500= 181658.3 𝑚𝑚𝑚𝑚/𝑟𝑟2

This is graphed as follows with 𝑟𝑟𝐵𝐵/𝐴𝐴𝑛𝑛 off the end of 𝑟𝑟𝐴𝐴 and the line of 𝑟𝑟𝐵𝐵/𝐴𝐴

𝑡𝑡 being tangential.

𝑟𝑟𝐵𝐵/𝐴𝐴 = 𝑟𝑟𝐵𝐵/𝐴𝐴𝑛𝑛

||𝐴𝐴𝐵𝐵+ 𝑟𝑟𝐵𝐵/𝐴𝐴

𝑡𝑡⊥𝐴𝐴𝐵𝐵

Figure 14 Finding aB/A

10 | P a g e

The next part to be calculated is 𝑟𝑟𝐵𝐵𝑛𝑛 using the following equation.

𝑟𝑟𝐵𝐵𝑛𝑛 =𝑉𝑉𝐵𝐵2

𝑟𝑟𝐵𝐵=

1838.882

600= 5635.85 𝑚𝑚𝑚𝑚/𝑟𝑟2

𝑟𝑟𝐵𝐵𝑛𝑛 is graphed from the origin with the 𝑟𝑟𝐵𝐵𝑡𝑡 being tangential.

𝑟𝑟𝐵𝐵 = 𝑟𝑟𝐵𝐵𝑛𝑛||𝐴𝐴𝐵𝐵 + 𝑟𝑟𝐵𝐵𝑡𝑡⊥𝐴𝐴𝐵𝐵

Figure 15 Finding aB

Since 𝑟𝑟𝐵𝐵𝑡𝑡 and 𝑟𝑟𝐵𝐵/𝐴𝐴𝑡𝑡 intersection defines their lengths, the values can be sketched accordingly.

𝑟𝑟𝐵𝐵𝑛𝑛||𝐴𝐴𝐵𝐵 + 𝑟𝑟𝐵𝐵𝑡𝑡⊥𝐴𝐴𝐵𝐵 = 𝑟𝑟𝐵𝐵/𝐴𝐴𝑛𝑛

||𝐴𝐴𝐵𝐵+ 𝑟𝑟𝐵𝐵/𝐴𝐴

𝑡𝑡⊥𝐴𝐴𝐵𝐵

Figure 16 Solving for aB and aB/A

11 | P a g e

From this graph, the connection between the vector between the tangential and normal components of the vector can be resolved to find the full vectors for 𝑟𝑟𝐵𝐵 and 𝑟𝑟𝐵𝐵/𝐴𝐴.

Figure 17 Finding aB and aB/A

Now the acceleration at C is to be calculated starting with 𝑟𝑟𝐶𝐶/𝐴𝐴. First 𝑟𝑟𝐶𝐶/𝐴𝐴𝑛𝑛 is to be calculated using

the following.

𝑟𝑟𝐶𝐶/𝐴𝐴𝑛𝑛 =

𝑉𝑉𝐶𝐶/𝐴𝐴2

𝑟𝑟𝐴𝐴𝐶𝐶=

5731.242

300= 109490.7 𝑚𝑚𝑚𝑚/𝑟𝑟2

𝑟𝑟𝐶𝐶/𝐴𝐴𝑛𝑛 can be sketched from 𝑟𝑟𝐴𝐴 with 𝑟𝑟𝐶𝐶/𝐴𝐴

𝑡𝑡 being perpendicular to the normal component as follows.

𝑟𝑟𝐶𝐶/𝐴𝐴 = 𝑟𝑟𝐶𝐶/𝐴𝐴𝑛𝑛

||𝐴𝐴𝐶𝐶+ 𝑟𝑟𝐶𝐶/𝐴𝐴

𝑡𝑡⊥𝐴𝐴𝐶𝐶

12 | P a g e

Figure 18 Finding aC/A

The same principle can be applied for finding 𝑟𝑟𝐶𝐶/𝐵𝐵𝑛𝑛 and 𝑟𝑟𝐶𝐶/𝐵𝐵

𝑡𝑡 as follows.

𝑟𝑟𝐶𝐶/𝐵𝐵𝑛𝑛 =

𝑉𝑉𝐶𝐶/𝐵𝐵2

𝑟𝑟𝐵𝐵𝐶𝐶=

4106.132

215.77= 78141.1 𝑚𝑚𝑚𝑚/𝑟𝑟2

With 𝑟𝑟𝐶𝐶/𝐵𝐵𝑛𝑛 coming from the end of 𝑟𝑟𝐵𝐵 and 𝑟𝑟𝐶𝐶/𝐵𝐵

𝑡𝑡 being perpendicular to 𝑟𝑟𝐶𝐶/𝐵𝐵𝑛𝑛, the following

vector polygon can be drawn.

𝑟𝑟𝐶𝐶/𝐵𝐵 = 𝑟𝑟𝐶𝐶/𝐵𝐵𝑛𝑛

||𝐵𝐵𝐶𝐶+ 𝑟𝑟𝐶𝐶/𝐵𝐵

𝑡𝑡⊥𝐵𝐵𝐶𝐶

13 | P a g e

Figure 19 Finding aC/B

From here, the 𝑟𝑟𝐶𝐶/𝐴𝐴 and 𝑟𝑟𝐶𝐶/𝐵𝐵 values are combined to find the tangential components using the following equation and vector polygon.

𝑟𝑟𝐶𝐶/𝐵𝐵𝑛𝑛 + 𝑟𝑟𝐶𝐶/𝐵𝐵

𝑡𝑡 = 𝑟𝑟𝐶𝐶/𝐴𝐴𝑛𝑛 + 𝑟𝑟𝐶𝐶/𝐴𝐴

𝑡𝑡

Figure 20 Solving for aC/B and aC/A

14 | P a g e

The lines can be connected accordingly and the vector component of 𝑟𝑟𝐶𝐶/𝐵𝐵, 𝑟𝑟𝐶𝐶/𝐴𝐴 and 𝑟𝑟𝐶𝐶 can be found as follows.

Figure 21 Calculating aC/B and aC/A

The final acceleration to be found is 𝑟𝑟𝐷𝐷 which required 𝑟𝑟𝐶𝐶 and since the directions are known the following vector polygon can be created using the same principles as before.

𝑟𝑟𝐷𝐷/𝐶𝐶𝑛𝑛 =

𝑉𝑉𝐷𝐷/𝐶𝐶2

𝑟𝑟𝐶𝐶𝐷𝐷=

570.132

800= 406.31 𝑚𝑚𝑚𝑚/𝑟𝑟2

The 𝑟𝑟𝐷𝐷/𝐶𝐶𝑛𝑛 is sketched with 𝑟𝑟𝐷𝐷/𝐶𝐶

𝑡𝑡perpendicular and 𝑟𝑟𝐷𝐷 parallel to the x-axis.

𝑟𝑟𝐷𝐷||𝑋𝑋−𝑋𝑋 = 𝑟𝑟𝐷𝐷/𝐶𝐶𝑛𝑛

||𝐷𝐷𝐶𝐶+ 𝑟𝑟𝐷𝐷/𝐶𝐶

𝑡𝑡⊥𝐷𝐷𝐶𝐶

15 | P a g e

Figure 22 Finding aD

This vector polygon can be solved for 𝑟𝑟𝐷𝐷/𝐶𝐶𝑡𝑡 and 𝑟𝑟𝐷𝐷.

Figure 23 Calculating aD

Complete Vector Polygon All the vector polygons are combined to produce the overall vector polygon for the acceleration.

16 | P a g e

Figure 24 Acceleration Vector Polygon

Graphical Acceleration Results Using the initial scale, the vector polygon can be resolved to find the rest of the components in the system. The magnitudes for each of the accelerations calculated using the graphical method can be observed in the table below.

Width (pts)

Height (pts)

Length (pts)

Acceleration (mm/s2)

Angle (deg)

Aa 252.662 437.623 505.323635 505323.635 239.999976

Ab/a 289.362 6.329 289.431206 289431.206 181.252988

Ab 545.44 426.877 692.62455 692624.55 218.047753

Ac/a 168.412 39.139 172.900153 172900.153 193.083313

Ac 421.074 398.485 579.735809 579735.809 223.421191

Ad/c 233.83 398.485 462.024636 462024.636 239.595696

Ad 187.244 0 187.244 187244 180 Table 2 Graphical Acceleration Analysis Results

Graphical Instantaneous Centre Analysis For the graphical instantaneous centre of zero velocities analysis, we will be using Kennedy’s theorem. To begin Kennedy’s theorem, the number of IC’s must be found using the following.

17 | P a g e

𝐶𝐶 =𝑛𝑛(𝑛𝑛 − 1)

2 =

6(6 − 1)2

= 15

From observation of the model, the position of the following instantaneous centres can be found.

𝐼𝐼𝐶𝐶12, 𝐼𝐼𝐶𝐶23, 𝐼𝐼𝐶𝐶34, 𝐼𝐼𝐶𝐶14, 𝐼𝐼𝐶𝐶35, 𝐼𝐼𝐶𝐶56, 𝐼𝐼𝐶𝐶16

The 8 unknown instantaneous centres left to be found can be observed in Kennedy’s circle.

Figure 25 Kennedy's Circle with Unknown and Known Instantaneous Centre

The following instantaneous centres to be found are seen as dotted lines in the Kennedy’s circle.

𝐼𝐼𝐶𝐶13, 𝐼𝐼𝐶𝐶15, 𝐼𝐼𝐶𝐶24, 𝐼𝐼𝐶𝐶45, 𝐼𝐼𝐶𝐶46, 𝐼𝐼𝐶𝐶36, 𝐼𝐼𝐶𝐶25, 𝐼𝐼𝐶𝐶26

The instantaneous centres are found in the following order using the lines between the respective instantaneous centres.

𝐼𝐼𝐶𝐶13 is found using the intersection of lines produced from 𝐼𝐼𝐶𝐶12 to 𝐼𝐼𝐶𝐶23 and 𝐼𝐼𝐶𝐶14 to 𝐼𝐼𝐶𝐶34.

18 | P a g e

Figure 26 Finding IC13

𝐼𝐼𝐶𝐶15 is found using the lines 𝐼𝐼𝐶𝐶13 to 𝐼𝐼𝐶𝐶35 and 𝐼𝐼𝐶𝐶16 to 𝐼𝐼𝐶𝐶56.



19 | P a g e





20 | P a g e





21 | P a g e




All the instantaneous centres can be seen placed onto the 6 bar linkage mechanism as seen below.

22 | P a g e

Comlete Instantaneous Centre Diagram

Figure 34 Graphical Inspection of All Instantaneous Centres

23 | P a g e

𝐼𝐼𝐶𝐶15 is off the top of the graph but can be seen below.

Figure 35 Graphical IC15

Using these graphs, the velocities can be calculated with 𝑉𝑉𝐴𝐴 calibrating 𝛷𝛷1 using 𝐼𝐼𝐶𝐶13. Once 𝛷𝛷1 is calculated 𝑉𝑉𝐵𝐵 and 𝑉𝑉𝐶𝐶 can be calculated. Using 𝑉𝑉𝐶𝐶 and 𝐼𝐼𝐶𝐶15, 𝛷𝛷2 can be calibrated and used to calculated 𝑉𝑉𝐷𝐷. From the graph, the following values were calculated

Results for Instantaneous Centre of Zero Velocity Method Height

(pts) Width (pts)

Length (pts)

Velocity (mm/s)

Graphical Velocity (mm/s)

vA 8.706 5.027 10.05311718 10053.12 10053.09

vB 1.441 1.142 1.83865304 1838.653 1838.89

vC 5.202 0.491 5.225120573 5225.121 5225.43

vD 5.308 0 5.308 5308 5492 Figure 36 Joint Velocities Calculated Using Kennedy's Theorem

Analytical Method for Position, Velocity and Acceleration Position Analysis Part A - Crank To start the analytical method, the calculations are completed for the crank (link OA). The position of the crank is first calculated.

𝒙𝒙𝑨𝑨 = 𝑙𝑙2 cos(𝜃𝜃2) = 200 cos(60) =𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏

𝒚𝒚𝑨𝑨 = 𝑙𝑙2 sin(𝜃𝜃2) = 200 sin(60) =𝟏𝟏𝟏𝟏𝟏𝟏.𝟐𝟐𝟏𝟏𝟏𝟏

24 | P a g e

Part B – Coupler & Rocker To begin calculating the coupler and rocker, the vector diagram for the system is sketched as follows.

Figure 37 Position Vector of Crank, Rocker and Coupler

Starting by using the vector form of the components to create a vector polygon.

𝑅𝑅𝐴𝐴 + 𝑅𝑅3 = 𝑅𝑅2 + 𝑅𝑅4

𝑅𝑅3 = 𝑅𝑅𝐴𝐴𝐴𝐴 + 𝑅𝑅4

𝑅𝑅𝐴𝐴𝐴𝐴 = 𝑅𝑅𝐴𝐴 − 𝑅𝑅𝐴𝐴 = 𝑙𝑙1𝑒𝑒𝑖𝑖𝑖𝑖1 − 𝑙𝑙2𝑒𝑒𝑖𝑖𝑖𝑖2

𝑅𝑅𝐴𝐴𝐴𝐴 + 𝑙𝑙2𝑒𝑒𝑖𝑖𝑖𝑖2 − 𝑙𝑙1𝑒𝑒𝑖𝑖𝑖𝑖1 = 0

Using this equation, the Euler identity can be used to solve the equation.

𝑒𝑒𝑖𝑖𝑖𝑖 = cos (𝜃𝜃) + 𝑖𝑖 ∗ sin (𝜃𝜃)

𝑅𝑅𝐴𝐴𝐴𝐴 = 𝑙𝑙1(cos(𝜃𝜃1) + 𝑖𝑖 ∗ sin(𝜃𝜃1)) − 𝑙𝑙2(cos(𝜃𝜃2) + 𝑖𝑖 ∗ sin(𝜃𝜃2))

Splitting the equation into the real and imaginary component.

𝑅𝑅𝐴𝐴𝐴𝐴 = (𝑙𝑙2 cos(𝜃𝜃2) − 𝑙𝑙1 cos(𝜃𝜃1)) + 𝑖𝑖(𝑙𝑙1 sin(𝜃𝜃) − 𝑙𝑙2 sin(𝜃𝜃2))

Solving for the components.

𝑅𝑅𝑒𝑒 = 𝑅𝑅𝑒𝑒𝑟𝑟𝑙𝑙 𝑐𝑐𝑐𝑐𝑚𝑚𝑟𝑟𝑐𝑐𝑛𝑛𝑒𝑒𝑛𝑛𝑐𝑐

𝐼𝐼𝑚𝑚 = 𝐼𝐼𝑚𝑚𝑟𝑟𝐼𝐼𝑖𝑖𝑛𝑛𝑟𝑟𝑟𝑟𝐼𝐼 𝑐𝑐𝑐𝑐𝑚𝑚𝑟𝑟𝑐𝑐𝑛𝑛𝑒𝑒𝑛𝑛𝑐𝑐

25 | P a g e

𝑹𝑹𝑹𝑹 = 𝑙𝑙1 cos(𝜃𝜃1) − 𝑙𝑙2 cos(𝜃𝜃2) = 800 ∗ cos(6) − 200 ∗ cos(60) = 𝟔𝟔𝟔𝟔𝟔𝟔.𝟔𝟔𝟏𝟏𝟏𝟏𝟏𝟏

𝑰𝑰𝟏𝟏 = 𝑙𝑙1 sin(𝜃𝜃1) − 𝑙𝑙2 sin(𝜃𝜃2) = 800 ∗ sin(6) − 200 ∗ sin(60) = −𝟖𝟖𝟔𝟔.𝟔𝟔𝟖𝟖𝟏𝟏𝟏𝟏

Using the vector graph of the linkages, the following equations can be used to calculate 𝜃𝜃3 and 𝜃𝜃4.

𝑙𝑙3𝑒𝑒𝑖𝑖𝑖𝑖3 = 𝑅𝑅𝑒𝑒 + 𝑖𝑖𝐼𝐼𝑚𝑚 + 𝑙𝑙4𝑒𝑒𝑖𝑖𝑖𝑖4

𝑙𝑙3𝑒𝑒−𝑖𝑖𝑖𝑖3 = 𝑅𝑅𝑒𝑒 − 𝑖𝑖𝐼𝐼𝑚𝑚 + 𝑙𝑙4𝑒𝑒𝑖𝑖𝑖𝑖4

𝑙𝑙32 = 𝑅𝑅𝑒𝑒2 + 𝐼𝐼𝑚𝑚2 + 𝑅𝑅𝑒𝑒𝑙𝑙4(cos(𝜃𝜃4) + 𝑖𝑖 sin(𝜃𝜃4)) − 𝑖𝑖𝐼𝐼𝑚𝑚𝑙𝑙4(cos(𝜃𝜃4) + 𝑖𝑖 sin(𝜃𝜃4))

+ 𝑅𝑅𝑒𝑒𝑙𝑙4(cos(𝜃𝜃4) + 𝑖𝑖 sin(𝜃𝜃4)) + 𝑖𝑖𝐼𝐼𝑚𝑚𝑙𝑙4(cos(𝜃𝜃4) + 𝑖𝑖 sin(𝜃𝜃4)) + 𝑙𝑙42 = 0

This equation can be simplified and used for finding 𝜃𝜃4.

𝑅𝑅𝑒𝑒2 + 𝐼𝐼𝑚𝑚2 + 2𝑅𝑅𝑒𝑒𝑙𝑙4 cos(𝜃𝜃4) + 2𝐼𝐼𝑚𝑚𝑙𝑙4 sin(𝜃𝜃4) + 𝑙𝑙42 − 𝑙𝑙3

2 = 0

695.622 + (−89.58)2 + 2(695.62)(600) cos(𝜃𝜃4) + 2(−89.58)(600) sin(𝜃𝜃4) + (600)2 − (500)2

= 0

483883.73 + 8024.99 + 834741.02 cos(𝜃𝜃4) − 107498.78 sin(𝜃𝜃4) + 360000 − 250000 = 0

601908.72 + 834741.02 cos(𝜃𝜃4) − 107498.78 sin(𝜃𝜃4) = 0

834741.02 cos(𝜃𝜃4) − 107498.78 sin(𝜃𝜃4) = −601908.72

Solving this equation has 2 answers for different angles of 𝜃𝜃4.

𝜽𝜽𝟒𝟒 = −𝟏𝟏𝟒𝟒𝟐𝟐.𝟔𝟔𝟔𝟔𝟔𝟔 𝒅𝒅𝑹𝑹𝒅𝒅𝒅𝒅𝑹𝑹𝑹𝑹𝒅𝒅 𝑐𝑐𝑟𝑟 𝜽𝜽𝟒𝟒 = 𝟏𝟏𝟐𝟐𝟖𝟖.𝟏𝟏𝟏𝟏 𝒅𝒅𝑹𝑹𝒅𝒅𝒅𝒅𝑹𝑹𝑹𝑹𝒅𝒅

The values of 𝜃𝜃4 are checked using the following equation at 𝜃𝜃4 = −142.995 to make sure that the equations are correct.

sin(𝜃𝜃4) =2 tan �𝜃𝜃42 �

1 + tan2 �𝜃𝜃42 �

sin(−142.995) =2 tan �−142.995

2 �

1 + tan2 �−142.9952 �

−0.6019 = −0.6019

Since these values are true, the equation using cos(𝜃𝜃4) should be true as well, proving that this is the angle of 𝜃𝜃4.

cos(𝜃𝜃4) =1 − tan2 �𝜃𝜃42 �

1 + tan2 �𝜃𝜃42 �

26 | P a g e

cos(−142.995) =1 − tan2 �−142.995

2 �

1 + tan2 �−142.9952 �

−0.7986 = −0.7986

The other calculated value of 𝜃𝜃4 = 128.31 is checked against the same equations to validate the results.

sin(128.31) =2 tan �128.31

2 �

1 + tan2 �128.312 �

0.7847 = 0.7847

cos(128.31) =1 − tan2 �128.31

2 �

1 + tan2 �128.312 �

−0.6199 = −0.6199

The following equation is used to find C which can be also used to find 𝜃𝜃4 and validate the results.

𝑪𝑪 =𝑅𝑅𝑒𝑒2 + 𝐼𝐼𝑚𝑚2 + 𝑙𝑙4

2 − 𝑙𝑙32

2𝑙𝑙4=

695.622 + (−89.58)2 + 6002 − 5002

2 ∗ 600= 𝟔𝟔𝟏𝟏𝟏𝟏.𝟔𝟔𝟔𝟔

𝜽𝜽𝟒𝟒 = 2 arctan �−𝐼𝐼𝑚𝑚 ± √𝑅𝑅𝑒𝑒2 + 𝐼𝐼𝑚𝑚2 − 𝐶𝐶2

𝐶𝐶 − 𝑅𝑅𝑒𝑒�

= 2 arctan�−(−89.58) ± �695.622 + (−89.58)2 − 501.592

501.59 − 695.62�

= −2.496 𝑐𝑐𝑟𝑟 2.239 𝑟𝑟𝑟𝑟𝑟𝑟 = −𝟏𝟏𝟒𝟒𝟐𝟐.𝟔𝟔𝟔𝟔𝟔𝟔 𝒐𝒐𝒅𝒅 𝟏𝟏𝟐𝟐𝟖𝟖.𝟏𝟏𝟏𝟏𝟔𝟔 𝒅𝒅𝑹𝑹𝒅𝒅

Using these equations, the following vector field is used to calculate 𝜃𝜃3

27 | P a g e

Figure 38 Diagram to Find AB'B

Distance between B and B’ real component for 𝜃𝜃4 = −142.995.

𝑙𝑙3 + 𝑙𝑙4 cos(𝜃𝜃4) + 𝑅𝑅𝑒𝑒 = 500 + 600 cos(−142.995) + 695.62 = 716.467𝑚𝑚𝑚𝑚

Distance between B and B’ imaginary component for 𝜃𝜃4 = −142.995.

𝐼𝐼𝑚𝑚 + 𝑙𝑙4 sin(𝜃𝜃4) = −89.58 + 600 sin(−142.995) = −450.713𝑚𝑚𝑚𝑚

Calculating 𝜃𝜃3

𝜃𝜃3 = 2 arctan �𝐼𝐼𝑚𝑚 ± 𝑙𝑙4 sin(𝜃𝜃4)

𝑙𝑙3 + 𝑙𝑙4 cos(𝜃𝜃4) + 𝑅𝑅𝑒𝑒� = 2 arctan�

−89.58 ± 600 sin(−142.995)500 + 600 cos(−142.995) + 695.62

�

𝜽𝜽𝟏𝟏 = −𝟔𝟔𝟒𝟒.𝟏𝟏𝟒𝟒𝟔𝟔 𝒅𝒅𝑹𝑹𝒅𝒅 𝑐𝑐𝑟𝑟 𝜃𝜃3 = 41.514 𝑟𝑟𝑒𝑒𝐼𝐼

Since the link is below the x-axis, the link will not work for 𝜃𝜃3 = 41.514 𝑟𝑟𝑒𝑒𝐼𝐼 so therefore 𝜃𝜃3 =−64.346 𝑟𝑟𝑒𝑒𝐼𝐼.

From this the X and Y position of B can be solved using the following equations.

𝑿𝑿𝑩𝑩 = 𝑋𝑋𝐴𝐴 + 𝑙𝑙4 cos(𝜃𝜃4) = 800 cos(6) + 600 cos(−142.995) = 𝟏𝟏𝟏𝟏𝟔𝟔.𝟒𝟒𝟔𝟔𝟏𝟏𝟏𝟏𝟏𝟏

𝒀𝒀𝑩𝑩 = 𝑌𝑌𝐴𝐴 + 𝑙𝑙4 sin(𝜃𝜃4) = 800 sin(6) + 600 sin(−142.995) = −𝟐𝟐𝟏𝟏𝟏𝟏.𝟔𝟔𝟏𝟏𝟏𝟏𝟏𝟏

28 | P a g e

Figure 39 4 Bar Linkage with the First 𝜃𝜃3

Distance between B and B’ real component for 𝜃𝜃4 = 128.32.

𝑙𝑙3 + 𝑙𝑙2 cos(𝜃𝜃4) + 𝑅𝑅𝑒𝑒 = 500 + 200 cos(128.32) + 695.62 = 823.60𝑚𝑚𝑚𝑚

Distance between B and B’ imaginary component for 𝜃𝜃4 = 128.32.

𝐼𝐼𝑚𝑚 + 𝑙𝑙4 sin(𝜃𝜃4) = −89.58 + 600 sin(128.32) = 381.16𝑚𝑚𝑚𝑚

Calculating 𝜃𝜃3 using the above values.

𝜃𝜃3 = 2 arctan �𝐼𝐼𝑚𝑚 ± 𝑙𝑙4 sin(𝜃𝜃4)

𝑙𝑙3 + 𝑙𝑙4 cos(𝜃𝜃4) + 𝑅𝑅𝑒𝑒� = 2 arctan�

−89.58 ± 600 sin(128.32)500 + 600 cos(128.32) + 695.62

�

𝜽𝜽𝟏𝟏 = 𝟒𝟒𝟔𝟔.𝟔𝟔𝟏𝟏𝟏𝟏 𝒅𝒅𝑹𝑹𝒅𝒅 𝑐𝑐𝑟𝑟 𝜃𝜃3 = −68.458 𝑟𝑟𝑒𝑒𝐼𝐼

Since the link is above the x-axis, the link will not work for 𝜃𝜃3 = −68.458 𝑟𝑟𝑒𝑒𝐼𝐼 so therefore 𝜃𝜃3 =49.670 𝑟𝑟𝑒𝑒𝐼𝐼.

From this the X and Y position of B can be solved using the following equations.

𝑿𝑿𝑩𝑩 = 𝑋𝑋𝐴𝐴 + 𝑙𝑙4 cos(𝜃𝜃4) = 800 cos(6) + 600 cos(49.670) = 𝟒𝟒𝟐𝟐𝟏𝟏.𝟔𝟔𝟏𝟏𝟏𝟏𝟏𝟏

𝒀𝒀𝑩𝑩 = 𝑌𝑌𝐴𝐴 + 𝑙𝑙4 sin(𝜃𝜃4) = 800 sin(6) + 600 sin(49.670) = 𝟔𝟔𝟔𝟔𝟒𝟒.𝟒𝟒𝟐𝟐𝟏𝟏𝟏𝟏

29 | P a g e

Figure 40 4 Bar Linkage with the Second 𝜃𝜃3

After visually producing each of the scenarios, it can be deduced that the angle used for the system will be when𝜃𝜃4 = 128.32 𝑟𝑟𝑒𝑒𝐼𝐼 and 𝜃𝜃3 = 49.670 𝑟𝑟𝑒𝑒𝐼𝐼. For calculating the 𝐶𝐶 position, the β angle will need to be used along with the previous results.

β = −12 𝑟𝑟𝑒𝑒𝐼𝐼

𝑿𝑿𝑪𝑪 = 𝑙𝑙2 cos(𝜃𝜃2) + 𝑙𝑙𝐴𝐴𝐶𝐶 cos(𝜃𝜃3 + β) = 200 cos(60) + 300 cos�49.67 + (−12)� = 𝟏𝟏𝟏𝟏𝟏𝟏.𝟒𝟒𝟔𝟔 𝟏𝟏𝟏𝟏

𝒀𝒀𝑪𝑪 = 𝑙𝑙2 sin(𝜃𝜃2) + 𝑙𝑙𝐴𝐴𝐶𝐶 sin(𝜃𝜃3 + β) = 200 sin(60) + 300 sin�49.67 + (−12)� = 𝟏𝟏𝟔𝟔𝟔𝟔.𝟔𝟔𝟏𝟏 𝟏𝟏𝟏𝟏

Part C – Slider From the data calculated, the following values are known.

𝐻𝐻 = −50 𝑚𝑚𝑚𝑚

𝑋𝑋𝐶𝐶 = 337.46 𝑚𝑚𝑚𝑚

𝑌𝑌𝐶𝐶 = 356.54 𝑚𝑚𝑚𝑚

𝑙𝑙5 = 𝐶𝐶𝐶𝐶 = 800 𝑚𝑚𝑚𝑚

From these given values, 𝜃𝜃5 can be calculated through the following.

𝜽𝜽𝟔𝟔 = arcsin �𝐻𝐻 − 𝑌𝑌𝐶𝐶𝑙𝑙5

� = arcsin �(−50) − 356.54

800� = −0.533 𝑟𝑟𝑟𝑟𝑟𝑟 = −𝟏𝟏𝟏𝟏.𝟔𝟔𝟒𝟒 𝒅𝒅𝑹𝑹𝒅𝒅

From this, the position of 𝑋𝑋𝐷𝐷 and 𝑌𝑌𝐷𝐷 can be found.

𝑿𝑿𝑫𝑫 = 𝑺𝑺 = 𝑋𝑋𝐶𝐶 + 𝑙𝑙5 cos(𝜃𝜃3) = 337.46 + 800 cos(−30.54) = 𝟏𝟏𝟏𝟏𝟐𝟐𝟔𝟔.𝟒𝟒𝟏𝟏𝟏𝟏𝟏𝟏

30 | P a g e

𝒀𝒀𝑫𝑫 = 𝑯𝑯 = −𝟔𝟔𝟏𝟏𝟏𝟏𝟏𝟏

A summary of all the calculated values can be seen in the table below.

Joint X Component (mm)

Y Component (mm)

A 100 173.2051

B 423.6679 554.4237

C 337.4643 356.5373

D 1026.469 -50

E 795.6175 83.62277 Table 3 Analytical Position Results

Angle Value (degrees)

θ1 6 θ2 60 θ3 49.66962 θ4 128.31 θ5 -30.5421

Table 4 Analytical Angles

Velocity Analysis Part A - Crank To begin the velocity analysis, the velocity of the crank is to be calculated.

𝜃𝜃2̇ = 𝜔𝜔2 = 50.26 𝑟𝑟𝑟𝑟𝑟𝑟/𝑟𝑟

�̇�𝑥𝐴𝐴 = −𝑙𝑙2𝜃𝜃2̇ cos(𝜃𝜃2) = −200 ∗ 50.26 cos(60) = −8706.24

�̇�𝐼𝐴𝐴 = −𝑙𝑙2 𝜃𝜃2̇sin(𝜃𝜃2) = −200 ∗ 50.26sin(60) = −5026.55

𝑽𝑽𝑨𝑨 = ��̇�𝑥𝐴𝐴2 + �̇�𝐼𝐴𝐴2 = �(−8706.24)2 + (−5026.55)2 = 𝟏𝟏𝟏𝟏𝟏𝟏𝟔𝟔𝟏𝟏.𝟏𝟏𝟏𝟏𝟏𝟏/𝒅𝒅

𝜹𝜹𝑽𝑽𝑨𝑨 = 𝜃𝜃2 +𝜋𝜋2

=𝜋𝜋3

+𝜋𝜋2

=5𝜋𝜋6𝑟𝑟𝑟𝑟𝑟𝑟 = 𝟏𝟏𝟔𝟔𝟏𝟏 𝒅𝒅𝑹𝑹𝒅𝒅𝒅𝒅𝑹𝑹𝑹𝑹𝒅𝒅

Part B – Coupler & Rocker To begin finding the velocities at each point, first the rotational velocity for link 3 and 4 must be found (𝜔𝜔3 and 𝜔𝜔4). Using the closed vector loop method, these values can be found.

𝑅𝑅2 + 𝑅𝑅3 − 𝑅𝑅1 − 𝑅𝑅4 = 0

Using the Euler identity:

𝑙𝑙2𝑒𝑒𝑖𝑖𝑖𝑖2 + 𝑙𝑙3𝑒𝑒𝑖𝑖𝑖𝑖3 − 𝑙𝑙1𝑒𝑒𝑖𝑖𝑖𝑖1 − 𝑙𝑙4𝑒𝑒𝑖𝑖𝑖𝑖4 = 0

200 ∗ 𝑒𝑒𝑖𝑖𝑖𝑖2 + 500 ∗ 𝑒𝑒𝑖𝑖𝑖𝑖3 − 800 ∗ 𝑒𝑒𝑖𝑖𝑖𝑖1 − 600 ∗ 𝑒𝑒𝑖𝑖𝑖𝑖4 = 0

Differentiate with respect to time to find the velocity.

31 | P a g e

𝑟𝑟𝑟𝑟𝑐𝑐

(200𝑒𝑒𝑖𝑖𝑖𝑖2 + 500𝑒𝑒𝑖𝑖𝑖𝑖3 − 800𝑒𝑒𝑖𝑖𝑖𝑖1 − 600𝑒𝑒𝑖𝑖𝑖𝑖4) =𝑟𝑟𝑟𝑟𝑐𝑐

(0)

200𝑖𝑖𝑒𝑒𝑖𝑖𝑖𝑖2𝑟𝑟𝜃𝜃2𝑟𝑟𝑐𝑐

+ 500𝑖𝑖𝑒𝑒𝑖𝑖𝑖𝑖3𝑟𝑟𝜃𝜃3𝑟𝑟𝑐𝑐

− 800𝑖𝑖𝑒𝑒𝑖𝑖𝑖𝑖1𝑟𝑟𝜃𝜃1𝑟𝑟𝑐𝑐

− 600𝑖𝑖𝑒𝑒𝑖𝑖𝑖𝑖4𝑟𝑟𝜃𝜃4𝑟𝑟𝑐𝑐

= 0

200𝑖𝑖𝑒𝑒𝑖𝑖𝑖𝑖2𝜔𝜔2 + 500𝑖𝑖𝑒𝑒𝑖𝑖𝑖𝑖3𝜔𝜔3 − 800𝑖𝑖𝑒𝑒𝑖𝑖𝑖𝑖1𝜔𝜔1 − 600𝑖𝑖𝑒𝑒𝑖𝑖𝑖𝑖4𝜔𝜔4 = 0

Using Euler’s formula, the equation can be solved.

𝑒𝑒𝑖𝑖𝑖𝑖 = cos (𝜃𝜃) + 𝑖𝑖sin (𝜃𝜃)

200𝑖𝑖𝜔𝜔2(cos (𝜃𝜃2) + 𝑖𝑖sin (𝜃𝜃2)) + 500𝑖𝑖𝜔𝜔3(cos (𝜃𝜃3) + 𝑖𝑖sin (𝜃𝜃3)) − 800𝑖𝑖𝜔𝜔1(cos(𝜃𝜃1) + 𝑖𝑖 sin(𝜃𝜃1))− 600𝑖𝑖𝜔𝜔4(cos (𝜃𝜃4) + 𝑖𝑖sin (𝜃𝜃4)) = 0

This equation can be divided into the real and imaginary components to solve for 𝜔𝜔4.

200𝜔𝜔2 sin(𝜃𝜃2) + 500𝜔𝜔3 sin(𝜃𝜃3) − 800𝜔𝜔1 sin(𝜃𝜃1) − 600𝜔𝜔4sin (𝜃𝜃4) = 0

500𝜔𝜔3 =800𝜔𝜔1 sin(𝜃𝜃1) + 600𝜔𝜔4 sin(𝜃𝜃4) − 200𝜔𝜔2 sin(𝜃𝜃2)

sin(𝜃𝜃3)

200𝜔𝜔2 cos(𝜃𝜃2) + 500𝜔𝜔3 cos(𝜃𝜃3) − 800𝜔𝜔1 cos(𝜃𝜃1) − 600𝜔𝜔4𝑐𝑐𝑐𝑐𝑟𝑟(𝜃𝜃4) = 0

500𝜔𝜔3 =800𝜔𝜔1 cos(𝜃𝜃1) + 600𝜔𝜔4 cos(𝜃𝜃4) − 200𝜔𝜔2 cos(𝜃𝜃2)

cos(𝜃𝜃3)

These 2 equations are equal, so simultaneous equations can be used to find 𝜔𝜔4.

800𝜔𝜔1 sin(𝜃𝜃1) + 600𝜔𝜔4 sin(𝜃𝜃4) − 200𝜔𝜔2 sin(𝜃𝜃2)sin(𝜃𝜃3)

=800𝜔𝜔1 cos(𝜃𝜃1) + 600𝜔𝜔4 cos(𝜃𝜃4) − 200𝜔𝜔2 cos(𝜃𝜃2)

cos(𝜃𝜃3)

Since 𝜔𝜔1 = 0, the equation can be further simplified.

600𝜔𝜔4 sin(𝜃𝜃4) − 200𝜔𝜔2 sin(𝜃𝜃2)sin(𝜃𝜃3) =

600𝜔𝜔4 cos(𝜃𝜃4) − 200𝜔𝜔2 cos(𝜃𝜃2)cos(𝜃𝜃3)

600𝜔𝜔4(cos(𝜃𝜃4) sin(𝜃𝜃3) − sin(𝜃𝜃4) cos(𝜃𝜃3)) = 200𝜔𝜔2(cos(𝜃𝜃2) sin(𝜃𝜃3) − sin(𝜃𝜃2) cos(𝜃𝜃3))

𝜔𝜔4 =200𝜔𝜔2

600∗

cos(𝜃𝜃2) sin(𝜃𝜃3) − sin(𝜃𝜃2) cos(𝜃𝜃3)cos(𝜃𝜃4) sin(𝜃𝜃3) − sin(𝜃𝜃4) cos(𝜃𝜃3)

𝜔𝜔4 =200(50.26)

600∗

cos(60) sin(49.67) − sin(60) cos(49.67)cos(128.31) sin(49.67) − sin(128.31) cos(49.67)

𝝎𝝎𝟒𝟒 = 𝟏𝟏.𝟏𝟏𝟔𝟔𝟒𝟒 𝒅𝒅𝒓𝒓𝒅𝒅/𝒅𝒅

Using this value of 𝜔𝜔4, it can be substituted back into one of the original equations to find 𝜔𝜔3.

500𝜔𝜔3 =600 ∗ 3.064 sin(128.31) − 200 ∗ 50.26 sin(60)

sin(49.67)

32 | P a g e

𝝎𝝎𝟏𝟏 = −𝟏𝟏𝟔𝟔.𝟏𝟏𝟔𝟔 𝒅𝒅𝒓𝒓𝒅𝒅/𝒅𝒅

Now that the angular velocities have been calculated, the velocities of the links can be calculated.

𝑉𝑉𝐵𝐵/𝐴𝐴 = 𝑖𝑖𝑙𝑙3𝜔𝜔3(cos(𝜃𝜃3) + 𝑖𝑖 sin(𝜃𝜃3)) = 𝑙𝑙3𝜔𝜔3(𝑖𝑖 cos(𝜃𝜃3) − sin(𝜃𝜃3))

𝑉𝑉𝐵𝐵/𝐴𝐴 = 500 ∗ (−19.05)(𝑖𝑖 cos(49.67) − sin(49.67)) = (−6166.44) + 𝑖𝑖(−7263.40)

𝑽𝑽𝑩𝑩/𝑨𝑨 = �(−6166.44)2 + (−7263.40)2 = 𝟔𝟔𝟔𝟔𝟐𝟐𝟏𝟏.𝟔𝟔𝟔𝟔𝟏𝟏𝟏𝟏/𝒅𝒅

Using the same principles, 𝑉𝑉𝐵𝐵 can be found.

𝑉𝑉𝐵𝐵 = 𝑖𝑖𝑙𝑙4𝜔𝜔4(cos(𝜃𝜃4) + 𝑖𝑖 sin(𝜃𝜃4)) = 𝑙𝑙4𝜔𝜔4(𝑖𝑖 cos(𝜃𝜃4) − sin(𝜃𝜃4))

𝑉𝑉𝐵𝐵 = 600 ∗ (3.064)(𝑖𝑖 cos(128.31) − sin(128.31)) = (−1442.83) + 𝑖𝑖(−1139.89)

𝑽𝑽𝑩𝑩 = �(−1442.83)2 + (−1139.89)2 = 𝟏𝟏𝟖𝟖𝟏𝟏𝟖𝟖.𝟏𝟏𝟖𝟖𝟏𝟏𝟏𝟏/𝒅𝒅

The direction of 𝑉𝑉𝐵𝐵 can be found using the following equation.

𝛿𝛿𝑉𝑉𝐵𝐵 = arctan �𝐼𝐼𝑚𝑚𝑅𝑅𝑒𝑒� + 𝜋𝜋 = arctan �

−1139.89−1442.83

� + 𝜋𝜋 = 3.810 𝑟𝑟𝑟𝑟𝑟𝑟

𝜹𝜹𝑽𝑽𝑩𝑩 = 𝟐𝟐𝟏𝟏𝟖𝟖.𝟏𝟏𝟏𝟏 𝒅𝒅𝑹𝑹𝒅𝒅

Part C - Slider Using the currently calculated values, 𝜔𝜔5 can be found.

𝑆𝑆 = 𝑙𝑙2 cos(𝜃𝜃2) + 𝑙𝑙𝐴𝐴𝐶𝐶 cos(𝜃𝜃3 − 𝛽𝛽) + 𝑙𝑙5 cos(𝜃𝜃5)

𝑙𝑙2𝑒𝑒𝑖𝑖𝑖𝑖2 + 𝑙𝑙𝐴𝐴𝐶𝐶𝑒𝑒𝑖𝑖(𝑖𝑖3−𝛽𝛽) − 𝑙𝑙5𝑒𝑒𝑖𝑖𝑖𝑖5 − 𝑆𝑆𝑒𝑒𝑖𝑖𝑖𝑖3 − 𝐻𝐻𝑒𝑒𝑖𝑖𝑖𝑖4 = 0


(𝑙𝑙2𝑒𝑒𝑖𝑖𝑖𝑖2 + 𝑙𝑙𝐴𝐴𝐶𝐶𝑒𝑒𝑖𝑖(𝑖𝑖3−𝛽𝛽) − 𝑙𝑙5𝑒𝑒𝑖𝑖𝑖𝑖5 − 𝑆𝑆𝑒𝑒𝑖𝑖𝑖𝑖3 − 𝐻𝐻𝑒𝑒𝑖𝑖𝑖𝑖4) =𝑟𝑟𝑟𝑟𝑐𝑐

(0)


(200𝑒𝑒𝑖𝑖𝑖𝑖2 + 300𝑒𝑒𝑖𝑖(𝑖𝑖3−𝛽𝛽) − 800𝑒𝑒𝑖𝑖𝑖𝑖5 − 1026.47𝑒𝑒𝑖𝑖𝑖𝑖3 − 50𝑒𝑒𝑖𝑖𝑖𝑖4) = 0

200𝑖𝑖𝜔𝜔2(cos(𝜃𝜃2) + 𝑖𝑖 sin(𝜃𝜃2)) + 300𝑖𝑖𝜔𝜔3(cos(𝜃𝜃3 − 𝛽𝛽) + 𝑖𝑖 sin(𝜃𝜃3 − 𝛽𝛽)) − 800𝑖𝑖𝜔𝜔5(cos(𝜃𝜃5)+ 𝑖𝑖 sin(𝜃𝜃5)) = 0

This equation can be split into its real and imaginary components to solve for 𝜔𝜔5.

200𝜔𝜔2 sin(𝜃𝜃2) + 300𝜔𝜔3 sin(𝜃𝜃3 − 𝛽𝛽) − 800𝜔𝜔5 sin(𝜃𝜃5) = 0

200𝑖𝑖𝜔𝜔2 cos(𝜃𝜃2) + 300𝑖𝑖𝜔𝜔3 cos(𝜃𝜃3 − 𝛽𝛽) − 800𝑖𝑖𝜔𝜔5 cos(𝜃𝜃5) = 0

One of these equations can be solved to find 𝜔𝜔5 since the rest of the values are known.

200 ∗ 50.26 cos(60) + 300 ∗ −19.05 cos(49.67 − 12) − 800𝜔𝜔5 cos(−30.54) = 0

𝝎𝝎𝟔𝟔 = −𝟏𝟏.𝟏𝟏𝟐𝟐𝟖𝟖 𝒅𝒅𝒓𝒓𝒅𝒅/𝒅𝒅

𝑉𝑉𝐶𝐶/𝐴𝐴 is now calculated using the same principles as previous velocity calculations.

33 | P a g e

𝑉𝑉𝐶𝐶/𝐴𝐴 = 𝑙𝑙𝐴𝐴𝐶𝐶𝜔𝜔3(𝑖𝑖 cos(𝜃𝜃3 − 𝛽𝛽) − sin(𝜃𝜃3 − 𝛽𝛽))

𝑉𝑉𝐶𝐶/𝐴𝐴 = 300 ∗ (−19.06)(𝑖𝑖 cos(49.67 − 12) − sin(49.67 − 12))

𝑉𝑉𝐶𝐶/𝐴𝐴 = (3493.6) + 𝑖𝑖(−4525.1)

𝑽𝑽𝑪𝑪/𝑨𝑨 = �(3493.6)2 + (−4525.1)2 = 𝟔𝟔𝟏𝟏𝟏𝟏𝟔𝟔.𝟏𝟏𝟔𝟔𝟏𝟏𝟏𝟏/𝒅𝒅

𝑉𝑉𝐶𝐶 is calculated using velocity addition.

𝑉𝑉𝐶𝐶 = 𝑉𝑉𝐶𝐶/𝐴𝐴 + 𝑉𝑉𝐴𝐴

𝑉𝑉𝐶𝐶 = �3493.6 + (−8706.24)� + 𝑖𝑖(−4525.1 + (−5026.55))

𝑉𝑉𝐶𝐶 = (−5212.67) + 𝑖𝑖(501.45)

𝑽𝑽𝑪𝑪 = �(−5212.67)2 + (501.45)2 = 𝟔𝟔𝟐𝟐𝟏𝟏𝟔𝟔.𝟏𝟏𝟒𝟒𝟏𝟏𝟏𝟏/𝒅𝒅

𝛿𝛿𝑉𝑉𝐶𝐶 = arctan �𝐼𝐼𝑚𝑚𝑅𝑅𝑒𝑒� + 𝜋𝜋 = arctan �

501.45−5212.67

� + 𝜋𝜋 = 3.046 𝑟𝑟𝑟𝑟𝑟𝑟

𝜹𝜹𝑽𝑽𝑪𝑪 = 𝟏𝟏𝟏𝟏𝟒𝟒.𝟔𝟔𝟏𝟏 𝒅𝒅𝑹𝑹𝒅𝒅

𝑉𝑉𝐷𝐷/𝐶𝐶 is calculated using the same technique as shown in 𝑉𝑉𝐶𝐶/𝐴𝐴.

𝑉𝑉𝐷𝐷/𝐶𝐶 = 𝑙𝑙5𝜔𝜔5(𝑖𝑖 cos(𝜃𝜃5) − sin(𝜃𝜃5))

𝑉𝑉𝐷𝐷/𝐶𝐶 = 800 ∗ (−0.73)(𝑖𝑖 cos(−30.54) − sin(−30.54))

𝑉𝑉𝐷𝐷/𝐶𝐶 = (−295.87) + 𝑖𝑖(−501.45)

𝑽𝑽𝑫𝑫/𝑪𝑪 = �(−295.87)2 + (−501.45)2 = 𝟔𝟔𝟖𝟖𝟐𝟐.𝟐𝟐𝟏𝟏𝟏𝟏𝟏𝟏/𝒅𝒅

𝑉𝑉𝐷𝐷 is calculated using velocity addition, the same way as 𝑉𝑉𝐶𝐶 .

𝑉𝑉𝐷𝐷 = 𝑉𝑉𝐷𝐷/𝐶𝐶 + 𝑉𝑉𝐶𝐶

𝑉𝑉𝐷𝐷 = �−295.87 + (−5212.67)� + 𝑖𝑖(−501.45 + 501.45) = (−5508.55) + 𝑖𝑖(0)

𝑽𝑽𝑫𝑫/𝑪𝑪 = �(−5508.55)2 + (0)2 = 𝟔𝟔𝟔𝟔𝟏𝟏𝟖𝟖.𝟔𝟔𝟔𝟔𝟏𝟏𝟏𝟏/𝒅𝒅

𝛿𝛿𝑉𝑉𝐷𝐷 = arctan �𝐼𝐼𝑚𝑚𝑅𝑅𝑒𝑒� + 𝜋𝜋 = arctan �

0−5508.55

� + 𝜋𝜋 = 3.142 𝑟𝑟𝑟𝑟𝑟𝑟

𝜹𝜹𝑽𝑽𝑫𝑫 = 𝟏𝟏𝟖𝟖𝟏𝟏 𝒅𝒅𝑹𝑹𝒅𝒅

Now that all the values are calculated, their values are compared to the velocities calculated in the graphical method to evaluate their differences.

Graphical

(mm/s) Analytical

(mm/s) Difference

(%) vA 10053.09 10053.10 -3.497E-05

34 | P a g e

vB/A 9530.43 9527.96 0.02597

vC/A 5731.25 5716.78 0.2525

vC 5225.43 5236.74 -0.2164

vB 1838.89 1838.78 0.005762

vD 5492 5508.55 -0.3013

vD/C 570.13 582.23 -2.122 Table 5 Analytical Velocity Results

Observing these values obtained from the graphical and analytical methods shows that the numbers obtained from the calculations are within 3% of each other at every calculation step. This agreement between both methods proves that both the methods are effective in evaluating the numbers.

Acceleration Analysis Part A – Crank To begin the acceleration analysis, the crank acceleration is found using the following equations.

𝜃𝜃2̈ = 𝛼𝛼2 = 0 𝑟𝑟𝑟𝑟𝑟𝑟/𝑟𝑟2

𝑥𝑥�̈�𝐴 = −𝑙𝑙2𝜃𝜃2̈ sin(𝜃𝜃2)−𝑙𝑙2𝜃𝜃2̇ cos(𝜃𝜃2) = −200 ∗ 0 sin(60) − 200 ∗ 50.26 cos(60)= −252661.87𝑚𝑚𝑚𝑚 𝑟𝑟2⁄

𝐼𝐼�̈�𝐴 = 𝑙𝑙2𝜃𝜃2̈ cos(𝜃𝜃2)−𝑙𝑙2𝜃𝜃2̇ sin(𝜃𝜃2) = 200 ∗ 0 cos(60) − 200 ∗ 50.26 sin(60) = −437623.20𝑚𝑚𝑚𝑚 𝑟𝑟2⁄

𝒓𝒓𝑨𝑨 = �𝑥𝑥�̈�𝐴2 + 𝐼𝐼�̈�𝐴2 = �(−252661.87)2 + (−437623.20)2 = 𝟔𝟔𝟏𝟏𝟔𝟔𝟏𝟏𝟐𝟐𝟏𝟏.𝟏𝟏𝟔𝟔𝟏𝟏𝟏𝟏 𝒅𝒅𝟐𝟐⁄

Part B – Coupler & Rocker Differentiating the velocity expression is used to obtain the acceleration.

𝑙𝑙2𝑖𝑖𝜔𝜔2𝑒𝑒𝑖𝑖𝑖𝑖2 + 𝑙𝑙3𝑖𝑖𝜔𝜔3𝑒𝑒𝑖𝑖𝑖𝑖3 − 𝑙𝑙4𝑖𝑖𝜔𝜔4𝑒𝑒𝑖𝑖𝑖𝑖4 − 𝑙𝑙1𝑖𝑖𝜔𝜔1𝑒𝑒𝑖𝑖𝑖𝑖1 = 0

𝑟𝑟𝑟𝑟𝑐𝑐�𝑙𝑙2𝑖𝑖𝜔𝜔2𝑒𝑒𝑖𝑖𝑖𝑖2 + 𝑙𝑙3𝑖𝑖𝜔𝜔3𝑒𝑒𝑖𝑖𝑖𝑖3 − 𝑙𝑙4𝑖𝑖𝜔𝜔4𝑒𝑒𝑖𝑖𝑖𝑖4 − 0� =


(0)

200𝑖𝑖𝜔𝜔2𝑒𝑒𝑖𝑖𝑖𝑖2𝑟𝑟𝜃𝜃22

𝑟𝑟𝑐𝑐+ 500𝑖𝑖𝜔𝜔3𝑒𝑒𝑖𝑖𝑖𝑖3

𝑟𝑟𝜃𝜃32

𝑟𝑟𝑐𝑐− 600𝑖𝑖𝜔𝜔4𝑒𝑒𝑖𝑖𝑖𝑖4

𝑟𝑟𝜃𝜃42

𝑟𝑟𝑐𝑐= 0

Therefore, the acceleration is equal to the following equation.

�200𝑖𝑖𝛼𝛼2𝑒𝑒𝑖𝑖𝑖𝑖2 − 200𝜔𝜔22𝑒𝑒𝑖𝑖𝑖𝑖2� + �500𝑖𝑖𝛼𝛼3𝑒𝑒𝑖𝑖𝑖𝑖3 − 500𝜔𝜔3

2𝑒𝑒𝑖𝑖𝑖𝑖3� + (−600𝑖𝑖𝛼𝛼4𝑒𝑒𝑖𝑖𝑖𝑖4 + 600𝜔𝜔42𝑒𝑒𝑖𝑖𝑖𝑖4) = 0

𝑂𝑂𝐴𝐴 = 200𝑖𝑖𝛼𝛼2𝑒𝑒𝑖𝑖𝑖𝑖2 − 200𝑖𝑖𝜔𝜔22𝑒𝑒𝑖𝑖𝑖𝑖2

𝑂𝑂𝐵𝐵/𝐴𝐴 = 500𝑖𝑖𝛼𝛼3𝑒𝑒𝑖𝑖𝑖𝑖3 − 500𝑖𝑖𝜔𝜔32𝑒𝑒𝑖𝑖𝑖𝑖3

𝑂𝑂𝐵𝐵/𝐴𝐴 = −600𝑖𝑖𝛼𝛼4𝑒𝑒𝑖𝑖𝑖𝑖4 + 600𝑖𝑖𝜔𝜔42𝑒𝑒𝑖𝑖𝑖𝑖4

35 | P a g e

Using Euler’s formula, the equation can split into the real and imaginary component and solved for 𝛼𝛼3 and 𝛼𝛼4.


𝛼𝛼2 = 0

�−200𝜔𝜔22(cos(𝜃𝜃2) + 𝑖𝑖 sin(𝜃𝜃2))� + �500𝑖𝑖𝛼𝛼3(cos(𝜃𝜃3) + 𝑖𝑖 sin(𝜃𝜃3))�

− �500𝜔𝜔32(cos(𝜃𝜃3) + 𝑖𝑖 sin(𝜃𝜃3))�

+ �−600𝑖𝑖𝛼𝛼4(cos(𝜃𝜃4) + 𝑖𝑖 sin(𝜃𝜃4)) + 600𝜔𝜔42(cos(𝜃𝜃4) + 𝑖𝑖 sin(𝜃𝜃4))� = 0

The real component is extracted from the equation.

(−200𝜔𝜔22 cos(𝜃𝜃2)) − (500𝛼𝛼3 sin(𝜃𝜃3)) − (500𝜔𝜔3

2 cos(𝜃𝜃3)) + (600𝛼𝛼4 sin(𝜃𝜃4))+ (600𝜔𝜔42 cos(𝜃𝜃4)) = 0

(−200(50.26)2 cos(60)) − (500𝛼𝛼3 sin(49.67)) − (500(−19.06)2 cos(49.67))+ (600𝛼𝛼4 sin(128.31)) + (600(3.06)2 cos(128.31)) = 0

The imaginary component is extracted from the equation.

(−200𝜔𝜔22𝑖𝑖 sin(𝜃𝜃2)) + (500𝑖𝑖𝛼𝛼3 cos(𝜃𝜃3)) − (500𝜔𝜔3

2𝑖𝑖 sin(𝜃𝜃3)) + (−600𝑖𝑖𝛼𝛼4 cos(𝜃𝜃4))+ (600𝜔𝜔42𝑖𝑖 sin(𝜃𝜃4)) = 0

(−200(50.26)2𝑖𝑖 sin(60)) + (500𝑖𝑖𝛼𝛼3 cos(49.67)) − (500(−19.06)2𝑖𝑖 sin(49.67))+ (−600𝑖𝑖𝛼𝛼4 cos(128.31)) + (600(3.06)2𝑖𝑖 sin(128.31)) = 0

Solving both the real and imaginary component simultaneously will solve for 𝛼𝛼3 and 𝛼𝛼4.

𝜶𝜶𝟏𝟏 = 𝟒𝟒𝟒𝟒𝟐𝟐.𝟒𝟒𝟒𝟒 𝒅𝒅𝒓𝒓𝒅𝒅/𝒅𝒅𝟐𝟐

𝜶𝜶𝟒𝟒 = 𝟏𝟏𝟏𝟏𝟔𝟔𝟏𝟏.𝟖𝟖𝟖𝟖 𝒅𝒅𝒓𝒓𝒅𝒅/𝒅𝒅𝟐𝟐

Now that the alpha values are obtained, 𝑂𝑂𝐵𝐵/𝐴𝐴, 𝑂𝑂𝐵𝐵 and 𝑂𝑂𝐴𝐴 can be calculated.

𝑂𝑂𝐵𝐵/𝐴𝐴 = 𝑖𝑖𝑙𝑙3𝛼𝛼3𝑒𝑒𝑖𝑖𝑖𝑖3 − 𝑙𝑙3𝜔𝜔32𝑒𝑒𝑖𝑖𝑖𝑖3

𝑂𝑂𝐵𝐵/𝐴𝐴 = 𝑖𝑖𝑙𝑙3𝛼𝛼3�cos(𝜃𝜃3) + 𝑖𝑖𝑟𝑟𝑖𝑖𝑛𝑛(𝜃𝜃3)� − (𝑙𝑙3𝜔𝜔32(cos(𝜃𝜃3) + 𝑖𝑖𝑟𝑟𝑖𝑖𝑛𝑛(𝜃𝜃3))

𝑂𝑂𝐵𝐵/𝐴𝐴 = 𝑖𝑖(𝑙𝑙3𝛼𝛼3cos(𝜃𝜃3)−𝑙𝑙3𝜔𝜔32𝑟𝑟𝑖𝑖𝑛𝑛(𝜃𝜃3)) − 𝑙𝑙3𝜔𝜔3

2 cos(𝜃𝜃3) − 𝑙𝑙3𝛼𝛼3𝑟𝑟𝑖𝑖𝑛𝑛(𝜃𝜃3)

𝑂𝑂𝐵𝐵/𝐴𝐴 = 𝑖𝑖(500(442.44) cos(49.67) − 500(−19.06)2𝑟𝑟𝑖𝑖𝑛𝑛(49.67)) − 500(−19.06)2 cos(49.67)− 500(442.44)𝑟𝑟𝑖𝑖𝑛𝑛(49.67)

𝑂𝑂𝐵𝐵/𝐴𝐴 = (−286148.97) + 𝑖𝑖(4761.69)

𝑨𝑨𝑩𝑩/𝑨𝑨 = �(−286148.97)2 + (4761.69)2 = 𝟐𝟐𝟖𝟖𝟔𝟔𝟏𝟏𝟖𝟖𝟖𝟖.𝟔𝟔𝟔𝟔𝟏𝟏𝟏𝟏 𝒅𝒅𝟐𝟐⁄

Calculating 𝑂𝑂𝐴𝐴

𝑂𝑂𝐴𝐴 = 𝑖𝑖𝑙𝑙2𝛼𝛼2𝑒𝑒𝑖𝑖𝑖𝑖2 − 𝑙𝑙2𝜔𝜔22𝑒𝑒𝑖𝑖𝑖𝑖2

36 | P a g e

𝑂𝑂𝐴𝐴 = 𝑖𝑖𝑙𝑙2𝛼𝛼2�cos(𝜃𝜃2) + 𝑖𝑖𝑟𝑟𝑖𝑖𝑛𝑛(𝜃𝜃2)� − (𝑙𝑙2𝜔𝜔22(cos(𝜃𝜃2) + 𝑖𝑖𝑟𝑟𝑖𝑖𝑛𝑛(𝜃𝜃2))

𝑂𝑂𝐴𝐴 = 𝑖𝑖(𝑙𝑙2𝛼𝛼2cos(𝜃𝜃2)−𝑙𝑙2𝜔𝜔22𝑟𝑟𝑖𝑖𝑛𝑛(𝜃𝜃2)) − 𝑙𝑙2𝜔𝜔2

2 cos(𝜃𝜃2) − 𝑙𝑙2𝛼𝛼2𝑟𝑟𝑖𝑖𝑛𝑛(𝜃𝜃2)

𝜶𝜶𝟐𝟐 = 𝟏𝟏

𝑂𝑂𝐴𝐴 = 𝑖𝑖�−𝑙𝑙2𝜔𝜔22𝑟𝑟𝑖𝑖𝑛𝑛(𝜃𝜃2)� − 𝑙𝑙2𝜔𝜔2

2 cos(𝜃𝜃2)

𝑂𝑂𝐴𝐴 = 𝑖𝑖�−200(50.26)2𝑟𝑟𝑖𝑖𝑛𝑛(60)� − 200(50.26)2 cos(60)

𝑂𝑂𝐴𝐴 = (−252661.87) + 𝑖𝑖(−437623.20)

𝑨𝑨𝑨𝑨 = �(−252661.87)2 + (−437623.20)2 = 𝟔𝟔𝟏𝟏𝟔𝟔𝟏𝟏𝟐𝟐𝟏𝟏.𝟏𝟏𝟔𝟔𝟏𝟏𝟏𝟏 𝒅𝒅𝟐𝟐⁄

𝛿𝛿𝐴𝐴𝐴𝐴 = arctan �𝐼𝐼𝑚𝑚𝑅𝑅𝑒𝑒� + 𝜋𝜋 = arctan �

−437623.20−252661.87

� + 𝜋𝜋 = 4.189 𝑟𝑟𝑟𝑟𝑟𝑟

𝜹𝜹𝑨𝑨𝑨𝑨 = 𝟐𝟐𝟒𝟒𝟏𝟏 𝒅𝒅𝑹𝑹𝒅𝒅

Calculating 𝑂𝑂𝐵𝐵

𝑂𝑂𝐵𝐵 = 𝑖𝑖𝑙𝑙4𝛼𝛼4𝑒𝑒𝑖𝑖𝑖𝑖4 − 𝑙𝑙4𝜔𝜔42𝑒𝑒𝑖𝑖𝑖𝑖4

𝑂𝑂𝐵𝐵 = 𝑖𝑖𝑙𝑙4𝛼𝛼4�cos(𝜃𝜃4) + 𝑖𝑖𝑟𝑟𝑖𝑖𝑛𝑛(𝜃𝜃4)� − (𝑙𝑙4𝜔𝜔42(cos(𝜃𝜃4) + 𝑖𝑖𝑟𝑟𝑖𝑖𝑛𝑛(𝜃𝜃4))

𝑂𝑂𝐵𝐵 = 𝑖𝑖(𝑙𝑙4𝛼𝛼4cos(𝜃𝜃4)−𝑙𝑙4𝜔𝜔42𝑟𝑟𝑖𝑖𝑛𝑛(𝜃𝜃4)) − 𝑙𝑙4𝜔𝜔42 cos(𝜃𝜃4) − 𝑙𝑙4𝛼𝛼4𝑟𝑟𝑖𝑖𝑛𝑛(𝜃𝜃4)

𝑂𝑂𝐵𝐵 = 𝑖𝑖(600 (1151.88)cos(128.31)−600(3.06)2𝑟𝑟𝑖𝑖𝑛𝑛(128.31)) − 600(3.06)2 cos(128.31)− 600(1151.88)𝑟𝑟𝑖𝑖𝑛𝑛(128.31)

𝑂𝑂𝐵𝐵 = (−545797.55) + 𝑖𝑖(−432861.51)

𝑨𝑨𝑩𝑩 = �(−545797.55)2 + (−432861.51)2 = 𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟏𝟏𝟖𝟖.𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟏𝟏𝟏𝟏 𝒅𝒅𝟐𝟐⁄

𝛿𝛿𝐴𝐴𝐵𝐵 = arctan �𝐼𝐼𝑚𝑚𝑅𝑅𝑒𝑒� + 𝜋𝜋 = arctan �

−432861.51−545797.55

� + 𝜋𝜋 = 3.812 𝑟𝑟𝑟𝑟𝑟𝑟

𝜹𝜹𝑨𝑨𝑩𝑩 = 𝟐𝟐𝟏𝟏𝟖𝟖.𝟒𝟒𝟐𝟐 𝒅𝒅𝑹𝑹𝒅𝒅

Calculating 𝑂𝑂𝐶𝐶/𝐴𝐴

𝑂𝑂𝐶𝐶/𝐴𝐴 = 𝑖𝑖𝑙𝑙3𝛼𝛼3𝑒𝑒𝑖𝑖𝑖𝑖3 − 𝑙𝑙3𝜔𝜔32𝑒𝑒𝑖𝑖𝑖𝑖3

𝑂𝑂𝐶𝐶/𝐴𝐴 = 𝑖𝑖𝑙𝑙𝐴𝐴𝐶𝐶𝛼𝛼3�cos(𝜃𝜃𝐴𝐴𝐶𝐶) + 𝑖𝑖𝑟𝑟𝑖𝑖𝑛𝑛(𝜃𝜃𝐴𝐴𝐶𝐶)� − (𝑙𝑙𝐴𝐴𝐶𝐶𝜔𝜔32(cos(𝜃𝜃𝐴𝐴𝐶𝐶) + 𝑖𝑖𝑟𝑟𝑖𝑖𝑛𝑛(𝜃𝜃𝐴𝐴𝐶𝐶))

𝑂𝑂𝐶𝐶/𝐴𝐴 = 𝑖𝑖(𝑙𝑙𝐴𝐴𝐶𝐶𝛼𝛼3cos(𝜃𝜃𝐴𝐴𝐶𝐶)−𝑙𝑙𝐴𝐴𝐶𝐶𝜔𝜔32𝑟𝑟𝑖𝑖𝑛𝑛(𝜃𝜃𝐴𝐴𝐶𝐶)) − 𝑙𝑙𝐴𝐴𝐶𝐶𝜔𝜔3

2 cos(𝜃𝜃𝐴𝐴𝐶𝐶) − 𝑙𝑙𝐴𝐴𝐶𝐶𝛼𝛼3𝑟𝑟𝑖𝑖𝑛𝑛(𝜃𝜃𝐴𝐴𝐶𝐶)

𝑂𝑂𝐶𝐶/𝐴𝐴 = 𝑖𝑖(300(442.44)cos(43.67)−300(−19.06)2𝑟𝑟𝑖𝑖𝑛𝑛(43.67)) − 300(−19.06)2 cos(43.67)− 300(442.44)𝑟𝑟𝑖𝑖𝑛𝑛(43.67)

𝑂𝑂𝐶𝐶/𝐴𝐴 = (−170450.21) + 𝑖𝑖(20787.79)

𝑨𝑨𝑪𝑪/𝑨𝑨 = �(−170450.21)2 + (20787.79)2 = 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏.𝟏𝟏𝟔𝟔𝟏𝟏𝟏𝟏 𝒅𝒅𝟐𝟐⁄

37 | P a g e

Calculating 𝑂𝑂𝐶𝐶

𝑂𝑂𝐶𝐶 = 𝑂𝑂𝐶𝐶/𝐴𝐴 + 𝑉𝑉𝐴𝐴

𝑂𝑂𝐶𝐶 = (−170450.21 + (−252661.87)) + 𝑖𝑖(20787.79 + (−437623.20))

𝑂𝑂𝐶𝐶 = (−423112.08) + 𝑖𝑖(−416835.41)

𝑨𝑨𝑪𝑪 = �(−423112.08)2 + (−423112.08)2 = 𝟔𝟔𝟔𝟔𝟏𝟏𝟔𝟔𝟒𝟒𝟔𝟔.𝟏𝟏𝟔𝟔𝟏𝟏𝟏𝟏 𝒅𝒅𝟐𝟐⁄

𝛿𝛿𝐴𝐴𝐶𝐶 = arctan �𝐼𝐼𝑚𝑚𝑅𝑅𝑒𝑒� + 𝜋𝜋 = arctan �

−416835.41−423112.08

� + 𝜋𝜋 = 3.920 𝑟𝑟𝑟𝑟𝑟𝑟

𝜹𝜹𝑨𝑨𝑪𝑪 = 𝟐𝟐𝟐𝟐𝟒𝟒.𝟔𝟔𝟏𝟏𝒅𝒅𝑹𝑹𝒅𝒅

Part C – Slider In order to calculate 𝑂𝑂𝐷𝐷, firstly 𝛼𝛼5 must be calculated as follows. Firstly, the velocity equation is differentiated.

𝑖𝑖𝑙𝑙2𝜔𝜔2𝑒𝑒𝑖𝑖𝑖𝑖2𝑟𝑟𝜃𝜃2𝑟𝑟𝑐𝑐

+ 𝑖𝑖𝑙𝑙𝐴𝐴𝐶𝐶𝜔𝜔3𝑒𝑒𝑖𝑖𝑖𝑖𝐴𝐴𝐴𝐴𝑟𝑟𝜃𝜃3𝑟𝑟𝑐𝑐

− 𝑖𝑖𝑙𝑙5𝜔𝜔5𝑒𝑒𝑖𝑖𝑖𝑖5𝑟𝑟𝜃𝜃5𝑟𝑟𝑐𝑐

= 0

This produces the following acceleration equation.

�𝑖𝑖𝑙𝑙2𝛼𝛼2𝑒𝑒𝑖𝑖𝑖𝑖2−𝑙𝑙2𝜔𝜔22𝑒𝑒𝑖𝑖𝑖𝑖2� + �𝑖𝑖𝑙𝑙𝐴𝐴𝐶𝐶𝛼𝛼3𝑒𝑒𝑖𝑖𝑖𝑖𝐴𝐴𝐴𝐴−𝑙𝑙3𝜔𝜔3

2𝑒𝑒𝑖𝑖𝑖𝑖𝐴𝐴𝐴𝐴� − (𝑖𝑖𝑙𝑙5𝛼𝛼5𝑒𝑒𝑖𝑖𝑖𝑖5−𝑙𝑙5𝜔𝜔52𝑒𝑒𝑖𝑖𝑖𝑖5) = 0

Euler’s formula can be applied to this equation and divided into the imaginary and real components.


𝛼𝛼2 = 0

−(𝑙𝑙2𝜔𝜔22(cos(𝜃𝜃2) + 𝑖𝑖 sin(𝜃𝜃2))) + (𝑖𝑖𝑙𝑙𝐴𝐴𝐶𝐶𝛼𝛼3(cos(𝜃𝜃𝐴𝐴𝐶𝐶) + 𝑖𝑖 sin(𝜃𝜃𝐴𝐴𝐶𝐶)))−(𝑙𝑙𝐴𝐴𝐶𝐶𝜔𝜔3

2(cos(𝜃𝜃𝐴𝐴𝐶𝐶)+ 𝑖𝑖 sin(𝜃𝜃𝐴𝐴𝐶𝐶))) − (𝑖𝑖𝑙𝑙5𝛼𝛼5(cos(𝜃𝜃5) + 𝑖𝑖 sin(𝜃𝜃5)))+(𝑙𝑙5𝜔𝜔5

2(cos(𝜃𝜃5) + 𝑖𝑖 sin(𝜃𝜃5))) = 0

Using the imaginary component, 𝛼𝛼5 can be found.

−(𝑙𝑙2𝜔𝜔22𝑖𝑖 sin(𝜃𝜃2)) + (𝑖𝑖𝑙𝑙𝐴𝐴𝐶𝐶𝛼𝛼3 cos(𝜃𝜃𝐴𝐴𝐶𝐶))−(𝑙𝑙𝐴𝐴𝐶𝐶𝜔𝜔3

2𝑖𝑖 sin(𝜃𝜃𝐴𝐴𝐶𝐶)) − (𝑖𝑖𝑙𝑙5𝛼𝛼5 cos(𝜃𝜃5))+(𝑙𝑙5𝜔𝜔52𝑖𝑖 sin(𝜃𝜃5))

= 0

−(200(50.26)2𝑖𝑖 sin(60)) + (300𝑖𝑖(442.44) cos(43.67)) − (300(−19.06)2𝑖𝑖 sin(43.67))− (800𝑖𝑖𝛼𝛼5 cos(−30.54)) + (800(−0.73)2𝑖𝑖 sin(−30.54)) = 0

𝜶𝜶𝟔𝟔 = 𝟔𝟔𝟏𝟏𝟔𝟔.𝟐𝟐𝟔𝟔𝒅𝒅𝒓𝒓𝒅𝒅 𝒅𝒅𝟐𝟐⁄

Using 𝛼𝛼5, 𝑂𝑂𝐷𝐷/𝐶𝐶 and 𝑂𝑂𝐷𝐷 can now be calculated.

𝑂𝑂𝐷𝐷/𝐶𝐶 = 𝑙𝑙5𝛼𝛼5(cos(𝜃𝜃5) + 𝑖𝑖 sin(𝜃𝜃5)) + 𝑙𝑙5𝜔𝜔52(cos(𝜃𝜃5) + 𝑖𝑖 sin(𝜃𝜃5))

𝑂𝑂𝐷𝐷/𝐶𝐶 = 800(605.29)(cos(−30.54) + 𝑖𝑖 sin(−30.54)) + 800(−0.73)2(cos(−30.54)+ 𝑖𝑖 sin(−30.54))

𝑂𝑂𝐷𝐷/𝐶𝐶 = (245709.84) + 𝑖𝑖(416835.41)

38 | P a g e

𝑨𝑨𝑪𝑪 = �(245709.84)2 + (416835.41)2 = 𝟒𝟒𝟖𝟖𝟏𝟏𝟖𝟖𝟔𝟔𝟒𝟒.𝟏𝟏𝟒𝟒𝟏𝟏𝟏𝟏 𝒅𝒅𝟐𝟐⁄

Using vector addition, 𝑂𝑂𝐷𝐷 can be found.

𝑂𝑂𝐷𝐷 = 𝑂𝑂𝐷𝐷/𝐶𝐶 + 𝑂𝑂𝐶𝐶

𝑂𝑂𝐷𝐷 = (245709.84 − 423112.08) + 𝑖𝑖(416835.41 + (−416835.41))

𝑂𝑂𝐷𝐷 = (−177402.24) + 𝑖𝑖(0)

𝑨𝑨𝑫𝑫 = 𝟏𝟏𝟏𝟏𝟏𝟏𝟒𝟒𝟏𝟏𝟐𝟐.𝟐𝟐𝟒𝟒𝟏𝟏𝟏𝟏 𝒅𝒅𝟐𝟐⁄

𝛿𝛿𝐴𝐴𝐷𝐷 = arctan �𝐼𝐼𝑚𝑚𝑅𝑅𝑒𝑒� + 𝜋𝜋 = arctan �

0−177402.24

� + 𝜋𝜋 = 3.142 𝑟𝑟𝑟𝑟𝑟𝑟

𝜹𝜹𝑨𝑨𝑫𝑫 = 𝟏𝟏𝟖𝟖𝟏𝟏𝒅𝒅𝑹𝑹𝒅𝒅

Comparing the graphical method for calculating acceleration to the analytical method, similar to the velocities, both the methods have very similar values.

Graphical (mm/s2)

Analytical (mm/s2)

Difference (%)

Aa 505323.6 505323.7 2.178E-05

Ab/a 289431.2 286188.6 -1.133

Ab 692624.5 696609 0.5720

Ac/a 172900.2 171713.2 -0.6913

Ac 579735.8 593949.2 2.393

Ad/c 462024.6 483864.7 4.514

Ad 187244 177402.2 -5.548 Table 6 Analytical Acceleration Results Compared to Graphical

The variation in the methods can be attributed to the compounding small errors between calculations and difference in velocity values to calculate acceleration.

Summary of Analytical Values

Joint X Component (mm)

Y Component (mm)

A 100 173.21

B 423.67 554.42

C 337.46 356.54

D 1026.47 -50

E 795.62 83.62 Table 7 Analytical Position Results

39 | P a g e

Angle Value (degrees)

θ1 6 θ2 60 θ3 49.66962 θ4 128.31 θ5 -30.5421

Table 8 Analytical Angles

Joint Linear Velocity (mm/s)

Linear Acceleration (mm/s2)

A 10053.1 505323.7

B 1838.78 696609

C 5236.74 593949.2

D 5508.55 177402.2 Table 9 Analytical Linear Velocity and Acceleration

Link Angular Velocity ω (rad/s)

Angular Acceleration α (rad/s2)

2 50.26548 0

3 -19.0559 442.4406

4 3.064637 1151.876

5 -0.72779 605.2945 Table 10 Analytical Angular Velocity and Acceleration

Dynamic Forces Equations of Motion For link 2:

Figure 41 Link 2 Free Body Diagram

40 | P a g e

First the variables used are found.

𝑟𝑟𝐺𝐺2 = 252.66𝑚𝑚/𝑟𝑟2

𝛼𝛼2 = 0

𝑅𝑅12 = 𝑅𝑅32 = 0.1𝑚𝑚

𝜃𝜃𝐺𝐺2 = 240 𝑟𝑟𝑒𝑒𝐼𝐼

𝑟𝑟𝐺𝐺2𝑋𝑋 = 𝑟𝑟𝐺𝐺2cos(𝜃𝜃) = 252.66cos(240) = −126.22 𝑚𝑚/𝑟𝑟2

𝑟𝑟𝐺𝐺2𝑌𝑌 = 𝑟𝑟𝐺𝐺2sin(𝜃𝜃) = 252.66sin(240) = −218.88 𝑚𝑚/𝑟𝑟2

The x and y components of each of the radii are found.

𝑅𝑅12𝑋𝑋 = 𝑅𝑅12 cos(𝜃𝜃12) = 0.1 cos(240) = −0.05𝑚𝑚

𝑅𝑅12𝑌𝑌 = 𝑅𝑅12 sin(𝜃𝜃12) = 0.1 sin(240) = −0.086𝑚𝑚

𝑅𝑅32𝑋𝑋 = 𝑅𝑅32 cos(𝜃𝜃32) = 0.1 cos(60) = 0.05𝑚𝑚

𝑅𝑅32𝑌𝑌 = 𝑅𝑅32 sin(𝜃𝜃32) = 0.1 sin(60) = 0.086𝑚𝑚

The forces and moment equation are found for link 2.

𝛴𝛴𝛴𝛴𝑋𝑋 = 𝑚𝑚2𝑟𝑟𝐺𝐺2𝑥𝑥 = 1 ∗ (−126.22) = −126.22𝑁𝑁

𝛴𝛴𝛴𝛴𝑋𝑋 = 𝛴𝛴12𝑋𝑋 + 𝛴𝛴32𝑋𝑋 … 1

𝛴𝛴𝛴𝛴𝑌𝑌 = 𝑚𝑚2𝑟𝑟𝐺𝐺2𝑌𝑌 = 1 ∗ (−218.88) = −218.88𝑁𝑁

𝛴𝛴𝛴𝛴𝑌𝑌 = 𝛴𝛴12𝑌𝑌 + 𝛴𝛴32𝑌𝑌 … 2

𝛴𝛴𝛴𝛴 = 𝑇𝑇12 + (𝑅𝑅12𝑋𝑋𝛴𝛴12𝑌𝑌 − 𝑅𝑅12𝑌𝑌𝛴𝛴12𝑋𝑋) + (𝑅𝑅32𝑋𝑋𝛴𝛴32𝑌𝑌 − 𝑅𝑅32𝑌𝑌𝛴𝛴32𝑋𝑋) = 𝐼𝐼2𝛼𝛼2 … 3

𝐼𝐼2𝛼𝛼2 = 0.002 ∗ 0 = 0

For link 3:

41 | P a g e




𝛼𝛼3 = 442.44 𝑟𝑟𝑟𝑟𝑟𝑟/𝑟𝑟2

𝑅𝑅23 = 0.2731𝑚𝑚

𝑅𝑅43 = 0.2299𝑚𝑚

𝑅𝑅43 = 0.03811𝑚𝑚

𝜃𝜃𝐺𝐺3 = 226.36 𝑟𝑟𝑒𝑒𝐼𝐼

𝑟𝑟𝐺𝐺3𝑋𝑋 = 𝑟𝑟𝐺𝐺3cos(𝜃𝜃) = 584.89cos(226.36) = −403.62 𝑚𝑚/𝑟𝑟2

𝑟𝑟𝐺𝐺3𝑌𝑌 = 𝑟𝑟𝐺𝐺3sin(𝜃𝜃) = 252.66sin(226.36) = −423.30 𝑚𝑚/𝑟𝑟2


𝑅𝑅23𝑋𝑋 = 𝑅𝑅23 cos(𝜃𝜃23) = 0.2731 cos(226.90) = −0.1994𝑚𝑚

𝑅𝑅23𝑌𝑌 = 𝑅𝑅23 sin(𝜃𝜃23) = 0.2731 sin(226.90) = −0.1866𝑚𝑚

𝑅𝑅43𝑋𝑋 = 𝑅𝑅43 cos(𝜃𝜃43) = 0.229 cos(57.48) = 0.1236𝑚𝑚

𝑅𝑅43𝑌𝑌 = 𝑅𝑅43 sin(𝜃𝜃43) = 0.229 sin(57.48) = 0.1938𝑚𝑚

𝑅𝑅53𝑋𝑋 = 𝑅𝑅53 cos(𝜃𝜃53) = 0.03811 cos(−5.026) = 0.03796𝑚𝑚

𝑅𝑅53𝑌𝑌 = 𝑅𝑅53 sin(𝜃𝜃53) = 0.03811 sin(−5.026) = −0.003339𝑚𝑚

42 | P a g e


𝛴𝛴𝛴𝛴𝑋𝑋 = 𝑚𝑚3𝑟𝑟𝐺𝐺3𝑋𝑋 = 2.5 ∗ (−403.62) = −1009.05𝑁𝑁

𝛴𝛴𝛴𝛴𝑋𝑋 = 𝛴𝛴23𝑋𝑋 + 𝛴𝛴43𝑋𝑋 + 𝛴𝛴53𝑋𝑋 … 4

𝛴𝛴𝛴𝛴𝑌𝑌 = 𝑚𝑚3𝑟𝑟𝐺𝐺3𝑌𝑌 = 2.5 ∗ (−423.30) = −1058.25𝑁𝑁

𝛴𝛴𝛴𝛴𝑌𝑌 = 𝛴𝛴23𝑌𝑌 + 𝛴𝛴43𝑌𝑌 + 𝛴𝛴53𝑌𝑌 … 5

𝛴𝛴𝛴𝛴 = −(𝑅𝑅32𝑋𝑋𝛴𝛴32𝑌𝑌 − 𝑅𝑅32𝑌𝑌𝛴𝛴32𝑋𝑋) + (𝑅𝑅53𝑋𝑋𝛴𝛴53𝑌𝑌 − 𝑅𝑅53𝑌𝑌𝛴𝛴53𝑋𝑋) + (𝑅𝑅43𝑋𝑋𝛴𝛴43𝑌𝑌 − 𝑅𝑅43𝑌𝑌𝛴𝛴43𝑋𝑋) = 𝐼𝐼3𝛼𝛼3 … 6

𝐼𝐼3𝛼𝛼3 = 0.008 ∗ 442.44 = 3.5395

For link 4:




𝛼𝛼4 = 1151.88𝑟𝑟𝑟𝑟𝑟𝑟/𝑟𝑟2

𝑅𝑅14 = 𝑅𝑅34 = 0.3𝑚𝑚

𝜃𝜃𝐺𝐺4 = 218.80𝑟𝑟𝑒𝑒𝐼𝐼

𝑟𝑟𝐺𝐺4𝑋𝑋 = 𝑟𝑟𝐺𝐺4cos(𝜃𝜃) = 345.48cos(218.80) = −269.27𝑚𝑚/𝑟𝑟2

𝑟𝑟𝐺𝐺4𝑌𝑌 = 𝑟𝑟𝐺𝐺3sin(𝜃𝜃) = 345.48sin(218.80) = −216.44𝑚𝑚/𝑟𝑟2



43 | P a g e









𝛴𝛴𝛴𝛴 = (𝑅𝑅34𝑋𝑋𝛴𝛴34𝑌𝑌 − 𝑅𝑅34𝑌𝑌𝛴𝛴34𝑋𝑋) + (𝑅𝑅14𝑋𝑋𝛴𝛴14𝑌𝑌 − 𝑅𝑅14𝑌𝑌𝛴𝛴14𝑋𝑋) = 𝐼𝐼4𝛼𝛼4 … 9

𝐼𝐼4𝛼𝛼4 = 0.005 ∗ 1151.88 = 5.7594

For link 5:



𝑟𝑟𝐺𝐺5 = 362.21 𝑚𝑚/𝑟𝑟2

𝛼𝛼5 = 605.29 𝑟𝑟𝑟𝑟𝑟𝑟/𝑟𝑟2

𝑅𝑅35 = 𝑅𝑅65 = 0.4𝑚𝑚

𝜃𝜃𝐺𝐺5 = 213.44𝑟𝑟𝑒𝑒𝐼𝐼

𝑟𝑟𝐺𝐺5𝑋𝑋 = 𝑟𝑟𝐺𝐺5cos(𝜃𝜃) = 362.21cos(213.44) = −302.24𝑚𝑚/𝑟𝑟2

𝑟𝑟𝐺𝐺5𝑌𝑌 = 𝑟𝑟𝐺𝐺5sin(𝜃𝜃) = 362.21sin(213.44) = −199.63𝑚𝑚/𝑟𝑟2

44 | P a g e











𝛴𝛴𝛴𝛴 = (𝑅𝑅35𝑋𝑋𝛴𝛴35𝑌𝑌 − 𝑅𝑅35𝑌𝑌𝛴𝛴35𝑋𝑋) + (𝑅𝑅65𝑋𝑋𝛴𝛴65𝑌𝑌 − 𝑅𝑅65𝑌𝑌𝛴𝛴65𝑋𝑋) = 𝐼𝐼5𝛼𝛼5 … 12

𝐼𝐼5𝛼𝛼5 = 0.006 ∗ 605.29 = 3.6318

For link 6:




𝛼𝛼6 = 0 𝑟𝑟𝑟𝑟𝑟𝑟/𝑟𝑟2

𝜃𝜃𝐺𝐺6 = 180 𝑟𝑟𝑒𝑒𝐼𝐼

𝑟𝑟𝐺𝐺6𝑋𝑋 = 𝑟𝑟𝐺𝐺6cos(𝜃𝜃) = 177.40cos(180) = −177.40𝑚𝑚/𝑟𝑟2

𝑟𝑟𝐺𝐺6𝑌𝑌 = 𝑟𝑟𝐺𝐺6sin(𝜃𝜃) = 177.40sin(180) = 0𝑚𝑚/𝑟𝑟2

45 | P a g e


𝛴𝛴𝛴𝛴𝑋𝑋 = 𝑚𝑚6𝑟𝑟𝐺𝐺6𝑋𝑋 = 0.9 ∗ (177.40) = −159.66𝑁𝑁

𝛴𝛴𝛴𝛴𝑋𝑋 = −𝛴𝛴65𝑋𝑋 + 𝛴𝛴16𝑋𝑋 … 13

𝛴𝛴𝛴𝛴𝑋𝑋 = −𝛴𝛴65𝑋𝑋 + 𝜇𝜇𝛴𝛴16𝑌𝑌 … 14

𝜇𝜇 = 0.3


𝛴𝛴𝛴𝛴𝑌𝑌 = 𝑚𝑚6𝑟𝑟𝐺𝐺6𝑌𝑌 = 0.9 ∗ (0) = 0𝑁𝑁

𝛴𝛴𝛴𝛴𝑌𝑌 = −𝛴𝛴65𝑌𝑌 + 𝛴𝛴16𝑌𝑌

𝐼𝐼6𝛼𝛼6 = 𝐼𝐼6 ∗ 0 = 0

Equation Matrix Using the 14 equations found from the dynamic forces equations, the following matrix can be used to represent each of the equations.

⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎡

1 0 1 0 0 0 0 0 0 0 0 0 0 00 1 0 1 0 0 0 0 0 0 0 0 0 0

0.086628 −0.04996 −0.08663 0.04996 0 0 0 0 0 0 0 0 0 10 0 −1 0 1 0 1 0 0 0 0 0 0 00 0 0 −1 0 1 0 1 0 0 0 0 0 00 0 −0.18666 0.199477 −0.19389 0.123602 0.003339 0.037966 0 0 0 0 0 00 0 0 0 −1 0 0 0 1 0 0 0 0 00 0 0 0 0 −1 0 0 0 1 0 0 0 00 0 0 0 0.235141 0.186303 0 0 0.235141 0.186303 0 0 0 00 0 0 0 0 0 −1 0 0 0 1 0 0 00 0 0 0 0 0 0 −1 0 0 0 1 0 00 0 0 0 0 0 0.203346 0.344456 0 0 0.20335 0.344456 0 00 0 0 0 0 0 0 0 0 0 −1 0 0.3 00 0 0 0 0 0 0 0 0 0 0 −1 1 0⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎤

∗

⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎡𝛴𝛴12𝑋𝑋𝛴𝛴12𝑌𝑌𝛴𝛴32𝑋𝑋𝛴𝛴32𝑌𝑌𝛴𝛴43𝑋𝑋𝛴𝛴43𝑌𝑌𝛴𝛴53𝑋𝑋𝛴𝛴53𝑌𝑌𝛴𝛴14𝑋𝑋𝛴𝛴14𝑌𝑌𝛴𝛴65𝑋𝑋𝛴𝛴65𝑌𝑌𝛴𝛴16𝑌𝑌𝑇𝑇12 ⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎤

=

⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎡−126.221−218.875

0−1009.05−1058.253.539525−403.912−324.6655.759379−544.025−359.3313.631767−159.662

0 ⎦⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎤

This matrix is solved to produce the following results for each of the unknown variables.

⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎡𝛴𝛴12𝑋𝑋𝛴𝛴12𝑌𝑌𝛴𝛴32𝑋𝑋𝛴𝛴32𝑌𝑌𝛴𝛴43𝑋𝑋𝛴𝛴43𝑌𝑌𝛴𝛴53𝑋𝑋𝛴𝛴53𝑌𝑌𝛴𝛴14𝑋𝑋𝛴𝛴14𝑌𝑌𝛴𝛴65𝑋𝑋𝛴𝛴65𝑌𝑌𝛴𝛴16𝑌𝑌𝑇𝑇12 ⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎤

=

⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎡−1791.45−1626.381665.231407.50

61.89354.57594.29−5.32−342.02

29.9150.27

−364.65−364.65147.89 ⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎤

Working Model Simulation

0 | P a g e

Analysis of the positions of each of the joints.

Figure 46 Working Model Position Analysis

1 | P a g e

Analysis of all the velocities at each joint.

Figure 47 Working Model Velocity Analysis

2 | P a g e

Analysis of all the accelerations at all of the joints.

Figure 48 Working Model Acceleration of Joints Analysis

3 | P a g e

Analysis of the acceleration at the mass centre of each of the links.

Figure 49 Working Model Acceleration at Centre of Mass Analysis

4 | P a g e

Analysis of each of the forces at each of the joints and torque moment at the motor.

Figure 50 Working Model Force Analysis

0 | P a g e

Balancing Strategy Development To find the shaking forces, the following equation is used.

𝛴𝛴𝑆𝑆 = 𝛴𝛴21 + 𝛴𝛴41 + 𝛴𝛴6

𝛴𝛴𝑆𝑆 = 𝛴𝛴21𝑋𝑋 + 𝛴𝛴21𝑌𝑌 + 𝛴𝛴41𝑋𝑋 + 𝛴𝛴41𝑌𝑌 + 𝛴𝛴61𝑋𝑋 + 𝛴𝛴61𝑌𝑌

Where the values are found from the dynamic forces’ matrix.

⎣⎢⎢⎢⎢⎡𝛴𝛴21𝑋𝑋𝛴𝛴21𝑌𝑌𝛴𝛴41𝑋𝑋𝛴𝛴41𝑌𝑌𝛴𝛴61𝑋𝑋𝛴𝛴61𝑌𝑌⎦

⎥⎥⎥⎥⎤

=

⎣⎢⎢⎢⎢⎡1791.451626.38342.02−29.91109.395364.65 ⎦

⎥⎥⎥⎥⎤

These forces can be split into their x and y components of the resultant force respectively.

Firstly, the x axis resultant force is found.

𝛴𝛴𝛴𝛴𝑅𝑅𝑋𝑋 = 𝛴𝛴21𝑋𝑋 + 𝛴𝛴41𝑋𝑋 + 𝛴𝛴61𝑋𝑋

𝛴𝛴𝛴𝛴𝑅𝑅𝑋𝑋 = 1791.45 + 342.02 + 109.40 = 2242.873𝑁𝑁

Following with the y axis resultant force.

𝛴𝛴𝛴𝛴𝑅𝑅𝑌𝑌 = 𝛴𝛴21𝑌𝑌 + 𝛴𝛴41𝑌𝑌 + 𝛴𝛴61𝑌𝑌

𝛴𝛴𝛴𝛴𝑅𝑅𝑌𝑌 = 1626.38 − 29.91 + 364.65 = 1961.118𝑁𝑁

𝛴𝛴𝑆𝑆 = �(𝛴𝛴𝛴𝛴𝑅𝑅𝑌𝑌)2 + (𝛴𝛴𝛴𝛴𝑅𝑅𝑋𝑋)2 = 2979.34𝑁𝑁

The shaking moment is found for the system.

𝑇𝑇𝑆𝑆 = 𝑇𝑇21 = −𝑇𝑇12 = −147.89 𝑁𝑁𝑚𝑚

Figure 51 Shaking Moment Diagram

Finally, the shaking moment is developed from the diagram above.

𝛴𝛴𝑆𝑆 = 𝑇𝑇21 + (𝑅𝑅1𝛴𝛴41) + (𝑅𝑅2𝛴𝛴61)

1 | P a g e

𝛴𝛴𝑆𝑆 = 𝑇𝑇21 + (𝑅𝑅1𝑋𝑋𝛴𝛴41𝑌𝑌) − (𝑅𝑅1𝑌𝑌𝛴𝛴41𝑋𝑋) − (𝑅𝑅2𝑋𝑋𝛴𝛴61𝑌𝑌) − (𝑅𝑅2𝑌𝑌𝛴𝛴61𝑋𝑋)

𝛴𝛴𝑆𝑆 = −147.89 + �0.7956 ∗ (−29.91)� − (0.0836 ∗ 342.02) − (1.0264 ∗ 364.65)− (0.05 ∗ 109.40) = −580.02 𝑁𝑁𝑚𝑚

In order to reduce the shaking moment transferred to the ground, some actions could be made.

Firstly, if the shaking moment isn’t too great, a damper can be used to absorb the energy being transferred to the ground to help reduce the overall effect on the ground.

If the shaking moment is too great or a damper is unable to be installed, another solution is to remove or add mass to the system at points which can reduces the overall moments at each of the ground points.

Results Graphical Method Analytical Method Computational

Method Linear Velocity Vector

(𝟏𝟏𝟏𝟏/𝒅𝒅) IC

(𝟏𝟏𝟏𝟏/𝒅𝒅) θ

(𝒅𝒅𝑹𝑹𝒅𝒅) Vector

(𝟏𝟏𝟏𝟏/𝒅𝒅) θ

(𝒅𝒅𝑹𝑹𝒅𝒅) Vector

(𝟏𝟏𝟏𝟏/𝒅𝒅) 𝑽𝑽𝑨𝑨 10053.09298 10053.1171

8 150.00 10053.09649 150 10053.1

𝑽𝑽𝑩𝑩 1838.888167 1838.65304 218.39 1838.782206 218.31 2520.1

𝑽𝑽𝑪𝑪 5225.428643 5225.120573

174.61 5236.737183 174.5051768

5617.9

𝑽𝑽𝑫𝑫 5492 5308 180 5508.545817 180 5676.2

Angular Velocity Vector (𝒅𝒅𝒓𝒓𝒅𝒅/𝒅𝒅) Vector (𝒅𝒅𝒓𝒓𝒅𝒅/𝒅𝒅) Vector (𝒅𝒅𝒓𝒓𝒅𝒅/𝒅𝒅)

𝝎𝝎𝟐𝟐 50.26548246 50.26548246

𝝎𝝎𝟏𝟏 -19.0608557 -19.05591673

𝝎𝝎𝟒𝟒 3.064816259 3.064637009

𝝎𝝎𝟔𝟔 -0.712661395 -0.727786137

Linear Acceleration

Vector (𝟏𝟏𝟏𝟏/𝒅𝒅𝟐𝟐)

θ (𝒅𝒅𝑹𝑹𝒅𝒅) Vector (𝟏𝟏𝟏𝟏/𝒅𝒅𝟐𝟐)

θ (𝒅𝒅𝑹𝑹𝒅𝒅) Vector (𝟏𝟏𝟏𝟏/𝒅𝒅𝟐𝟐)

𝒓𝒓𝑨𝑨 505323.6353 239.9999761 505323.7453 240 505323.7

𝒓𝒓𝑩𝑩 692624.5496 218.0477535 696608.9655 218.4172561

671303.7

𝒓𝒓𝑪𝑪 579735.8094 223.4211914 593949.1532 224.571854 563438.9

𝒓𝒓𝑫𝑫 187244 180 177402.2398 180 160192.2

Angular Acceleration

Vector (𝟏𝟏𝟏𝟏/𝒅𝒅𝟐𝟐) Vector (𝟏𝟏𝟏𝟏/𝒅𝒅𝟐𝟐) Vector (𝟏𝟏𝟏𝟏/𝒅𝒅𝟐𝟐)

𝜶𝜶𝟐𝟐 0 0

𝜶𝜶𝟏𝟏 445.6566542 442.4406331

𝜶𝜶𝟒𝟒 1154.375247 1151.875838

𝜶𝜶𝟔𝟔 577.5306514 605.2945123

Table 11 Velocity and Acceleration Results for 𝜃𝜃2 = 60 degrees

2 | P a g e

Analytical Method Computational Method

Raw Data (N) Joint Forces (N) Software Generated (N)

𝑭𝑭𝟏𝟏𝟐𝟐𝑿𝑿 -1791.454367 𝛴𝛴12 2419.5905

𝑭𝑭𝟏𝟏𝟐𝟐𝒀𝒀 -1626.379177

𝑭𝑭𝟏𝟏𝟐𝟐𝑿𝑿 1665.233433 𝛴𝛴32 2180.383 𝛴𝛴𝑨𝑨 2195.65

𝑭𝑭𝟏𝟏𝟐𝟐𝒀𝒀 1407.504103

𝑭𝑭𝟒𝟒𝟏𝟏𝑿𝑿 61.88841423 𝛴𝛴𝟒𝟒𝟏𝟏 359.93553 𝛴𝛴𝑩𝑩 438.99

𝑭𝑭𝟒𝟒𝟏𝟏𝒀𝒀 354.5749709

𝑭𝑭𝟔𝟔𝟏𝟏𝑿𝑿 594.2925285 𝛴𝛴𝟔𝟔𝟏𝟏 594.31632 𝛴𝛴𝑪𝑪 709.01

𝑭𝑭𝟔𝟔𝟏𝟏𝒀𝒀 -5.317468002

𝑭𝑭𝟏𝟏𝟒𝟒𝑿𝑿 -342.0237588 𝛴𝛴14 343.32907 𝛴𝛴𝑬𝑬 433.53

𝑭𝑭𝟏𝟏𝟒𝟒𝒀𝒀 29.90989642

𝑭𝑭𝟔𝟔𝟔𝟔𝑿𝑿 50.26751607 𝛴𝛴𝟔𝟔𝟔𝟔 368.09677

𝑭𝑭𝟔𝟔𝟔𝟔𝒀𝒀 -364.6483324

𝑭𝑭𝟏𝟏𝟔𝟔𝒀𝒀 -364.6483324 𝛴𝛴16 380.70404 𝛴𝛴𝑺𝑺 2642.40

𝑭𝑭𝟏𝟏𝟔𝟔𝑿𝑿 -109.3944997

𝑻𝑻𝟏𝟏𝟐𝟐 (𝑵𝑵𝟏𝟏) 147.8827066

𝑇𝑇12 (𝑁𝑁𝑚𝑚) 157.32

Table 12 Force Results for 𝜃𝜃2 = 60 degrees

The stroke distance was determined as seen below:

Stroke Distance = 1026.4mm

The shaking force, shaking torque and shaking moment at 𝜃𝜃2 = 60 degrees were found as seen below:

𝛴𝛴𝑠𝑠 = 2979.34𝑁𝑁

𝑇𝑇𝑠𝑠 = −147.883𝑁𝑁𝑚𝑚

𝛴𝛴𝑠𝑠 = −580.02𝑁𝑁𝑚𝑚

Discussion The results at every stage of the process produced values which were within an acceptable tolerance of other values found using different methods. Most values were within 5% of other calculated values for other methods, however, the values which were outside this calculation were still relatively close to the other methods calculations.

This discrepancy between some of the values can be attributed to measurement inaccuracy for the different methods. Some of the errors in the graphical method can be attributed to rounding of numbers and measurement tools resolution being unable to 100% accurately determine values. For the Working Model results, the snapshot which the results were taken may not have been at exactly when 𝜃𝜃2 = 60 degrees causing a slightly different situation to be analysed. Across all methods, compounding differences and rounding errors in the methods will attribute to some of the larger differences in results obtained towards the end of each section. Since all the results have relatively

3 | P a g e

similar magnitude and results, it can be assumed that these values are correct for if this mechanism was to be created in the real world.

Conclusion Overall the values that were produced from all the methods may not be completely accurate but most of the values are close enough that any of them can be used as an approximate used for a system and thus confirms that the methods are relatively accurate in determining the values for the position, velocity, acceleration and forces on the system.

References 1. Matthew West 2015, Four-Bar Linkages, Dynamics, Viewed September 12 2020,

<http://dynref.engr.illinois.edu/aml.html> 2. Course material of MIET1077: Mechanics of Machines provided by Prof. Firoz Alam

Attachments Attached to this document is a zip folder containing all the Working Model files used.


school of engineering (mechanical and automotive)

Documents