Lecture 15-18 Ch09 Complex Numbers

Dr. Jagadeep Thota

MAE 364Kinematics and Dynamics of Machines

Chapter 9:Kinematics Using Complex

Numbers

MAE 364: Complex Numbers JT

In statics and dynamics unit vectors in the axes direction are commonly used to define point positions.

Unit Vector

𝑉=𝑉 𝑥 ∙ �⃗�+𝑉 𝑦 ∙ �⃗�

is a unit vector in the x-direction. is a unit vector in the y-direction.

(3 ,2 )𝑉

x

y

𝑉=3∙ �⃗�+2 ∙ �⃗�


If the vector has a magnitude of , and is oriented at an angle with respect to the x-axis, then

The horizontal component of this vector is,

The vertical component is,

Then the vector can be written as,

Vector Components

𝑉 𝑥=𝑀∗ cos𝜃

𝑉 𝑦=𝑀∗𝑠𝑖𝑛𝜃

𝑉=𝑀∗cos𝜃 ∙ �⃗�+𝑀∗ sin 𝜃 ∙ �⃗�

𝑉=𝑀 (cos𝜃 ∙ �⃗�+sin 𝜃 ∙ �⃗� )=𝑀∗ λWhere,

λ=cos𝜃 ∙ �⃗�+sin 𝜃 ∙ �⃗�

𝑉

x

y

𝑀

𝜃


A vector can be plotted in Cartesian coordinates using the complex number approach.

In the complex number approach, the vertical axis represents the imaginary component and the horizontal axis represents the real component of the vector.

Complex Number Approach: Vector

𝑉=𝑉 𝑥+ 𝑖∙𝑉 𝑦

is the unit vector in the x-direction. is a unit vector in the y-direction = .

(3 ,2 )𝑉

Real ()

Imaginary ()

𝑉=3+𝑖 ∙2Lot easier to solve using a computer program (Mathcad) than by hand.


If the vector has a magnitude of , and is oriented at an angle with respect to the real-axis, then




Complex Numbers: Vector Components



𝑉=𝑀∗cos𝜃+𝑖∙𝑀∗ sin 𝜃

𝑉=𝑀∗𝑒𝑖 𝜃

Where,

𝑒𝑖𝜃=cos𝜃+𝑖∙ sin 𝜃

𝑉

𝑀

𝜃

Imaginary ()

Real ()

;

;

Using Complex Numbers Approach

Mathematical Analysis of Mechanisms:


1. Define VectorsMust be a vector polygonVectors must be related to link movementVector directions must be carefully defined

2. Construct vector equationBe careful with signs

3. Write the complex number equationDefine vector angles carefully

Procedure


All the links need to be defined as vectors.

Vector directions must be carefully defined.

The vector polygon (diagram) obtained needs to be closed.

Four Bar Mechanism: Vector Diagram

2

3

42

3

4 B

O4

A

O4 O2

B

A

O2

Changes directionFixed





Procedure


One of the below two methods may be used:Method one:

The vectors forming CW direction is Path 1, and the vectors forming CCW direction is Path 2.

Then, Path 1 = Path 2

Method two:Start at any point and travel around the polygonVectors in direction of travel are +, opposite to direction are –

(the starting point and ending point are the same)

Vector Equation Methods


Four Bar Mechanism: Vector Equations

2

3

4


A

O4 O2

B

Method 1:Start at O4 and end at B

2424 BOOOBAAO

Method 2:Start at O4 and go clockwise

02424 BOOOBAAO

2424 BOOOBAAO or

11





Procedure


If the vector has a magnitude of , and is oriented at an angle with respect to the real-axis, then




Complex Numbers: Vector Components



𝑉=𝑀∗𝑒𝑖 𝜃

Where,𝑒𝑖𝜃=cos𝜃+𝑖∙ sin 𝜃

𝑉

𝑀

𝜃

Imaginary ()

Real ()

Angle is always measured CCW from the +ve real axis


Four Bar Mechanism: Complex Numbers Equation

2

3

4


A

O4O2

B

q2

q3

q4

44

ieAO

2424 BOOOBAAO

3 ieBA 142 OO 22

ieBO

Real ()

Imaginary ()

+


Do a mathematical analysis of the given kinematic linkage by using the complex number approach?

Example 1: Four Bar Mechanism Analysis

2

3

4

A

O4O2

B

Parameter Value

Fixed link (O2O4) 3.50 in

Link 2 (BO2) 1.25 in

Link 3 (BA) 5.25 in

Link 4 (AO4) 3.50 in

Link 2 orientation ()

75⁰


Solution:

Example 1: Four Bar Mechanism Analysis

2

3

4

A

O4O2

B


Vector Diagram

�⃗�𝑂4+�⃗�𝐴=�⃗�2𝑂4+�⃗�𝑂2

𝐴𝑂4 ∙𝑒𝑖 ∙𝜃4+𝐵𝐴 ∙𝑒𝑖 ∙𝜃3=𝑂2𝑂4+𝐵𝑂2 ∙𝑒

𝑖 ∙𝜃 2

Vector Equation

Complex Number Equation

2

3

4

A

O4O2

B

q2

q3

q4


Solution: The complex number equation obtained describes the Position (or

displacement) of the links. Solve this Position Equation, using Mathcad software, to determine the

unknown parameters ( and ).In-class demonstration shown by the Instructor.

In order to get the velocity parameters, the Position Equation is to be differentiated with respect to time once.Since velocity is defined as the change in displacement per unit time.

Example 1: Position Equation

𝐴𝑂4 ∙𝑒𝑖 ∙𝜃4+𝐵𝐴 ∙𝑒𝑖 ∙𝜃3=𝑂2𝑂4+𝐵𝑂2 ∙𝑒

𝑖 ∙𝜃 2Position Equation

Velocity Equation

𝑑𝑑𝑡

[𝑃𝑜𝑠𝑖𝑡𝑖𝑜𝑛𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 ]


Solution: Differentiate each vector individually,

Example 1: Velocity Equation

𝐼=𝑑𝑑𝑡 ( 𝐴𝑂4 ∙𝑒

𝑖 ∙𝜃4 )=𝐴𝑂4 ∙𝑑𝑑𝑡

(𝑒𝑖 ∙𝜃4 )=𝐴𝑂4 ∙𝑒𝑖 ∙𝜃 4∙

𝑑𝑑𝑡 (𝑖 ∙𝜃4 )=𝐴𝑂4 ∙𝑒

𝑖 ∙𝜃4 ∙ 𝑖 ∙𝑑𝑑𝑡 (𝜃4 )

𝐼=𝑖 ∙ 𝐴𝑂4 ∙ �̇�4 ∙𝑒𝑖 ∙𝜃 4

𝐼=𝑑𝑑𝑡

(𝐵𝐴 ∙𝑒𝑖 ∙𝜃3 )=𝐵𝐴 ∙𝑑𝑑𝑡

(𝑒𝑖 ∙𝜃 3 )=𝐵𝐴 ∙𝑒𝑖 ∙𝜃3 ∙𝑑𝑑𝑡 (𝑖 ∙𝜃3 )=𝐵𝐴 ∙𝑒𝑖 ∙𝜃3 ∙𝑖 ∙

𝑑𝑑𝑡 (𝜃3 )

𝐼𝐼=𝑖∙𝐵𝐴 ∙ �̇�3 ∙𝑒𝑖 ∙𝜃3

𝐼𝐼𝐼=𝑑𝑑𝑡 (𝐵𝑂2 ∙𝑒

𝑖 ∙𝜃2 )=𝐵𝑂2 ∙𝑑𝑑𝑡

(𝑒𝑖 ∙𝜃 2 )=𝐵𝑂2 ∙𝑒𝑖 ∙𝜃2 ∙

𝑑𝑑𝑡 (𝑖 ∙𝜃2)=𝐵𝑂2 ∙𝑒

𝑖 ∙𝜃2 ∙𝑖 ∙𝑑𝑑𝑡 (𝜃2 )

𝐼𝐼𝐼=𝑖 ∙𝐵𝑂2 ∙ �̇�2 ∙𝑒𝑖 ∙𝜃2

𝐼𝑉=𝑑𝑑𝑡 (𝑂2𝑂4 )=0


Solution:

Solve the above Velocity Equation using Mathcad. Since link 2 is the input crank; assume link 2 is rotating at some constant

speed angular velocity of link 2 (BO2) = 120 rpm

Find out the unknown velocity parameters: angular velocity of link 3 (BA)

angular velocity of link 4 (AO4)

Example 1: Velocity Equation

Velocity Equation

𝐼+𝐼𝐼=𝐼𝐼𝐼 +𝐼𝑉

Velocity Equation:

𝑖 ∙𝐴𝑂4 ∙ �̇�4 ∙𝑒𝑖 ∙𝜃4+ 𝑖∙𝐵𝐴 ∙ �̇�3 ∙𝑒

𝑖 ∙𝜃3=𝑖∙𝐵𝑂2 ∙ �̇�2 ∙𝑒𝑖 ∙𝜃2+0

𝑖 ∙𝐴𝑂4 ∙ �̇�4 ∙𝑒𝑖 ∙𝜃4+ 𝑖∙𝐵𝐴 ∙ �̇�3 ∙𝑒

𝑖 ∙𝜃3=𝑖∙𝐵𝑂2 ∙ �̇�2∙𝑒𝑖 ∙𝜃2


Solution: Differentiate the Velocity Equation with respect to time to get the

Acceleration Equation,Since, acceleration is defined as the change in velocity per unit time.

Differentiate the velocity terms of each link (or vector) separately.

Example 1: Acceleration Equation

Velocity Equation

𝑖 ∙𝐴𝑂4 ∙ �̇�4 ∙𝑒𝑖 ∙𝜃4+ 𝑖∙𝐵𝐴 ∙ �̇�3 ∙𝑒

𝑖 ∙𝜃3=𝑖∙𝐵𝑂2 ∙ �̇�2 ∙𝑒𝑖 ∙𝜃2

Acceleration Equation

𝑑𝑑𝑡

[𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 ]

𝐼= 𝑑𝑑𝑡

( 𝑖 ∙𝐴𝑂4 ∙ �̇�4 ∙𝑒𝑖 ∙𝜃4 )=𝑖∙ 𝐴𝑂4 ∙

𝑑𝑑𝑡

( �̇�4 ∙𝑒𝑖 ∙𝜃4 )=𝑖 ∙𝐴𝑂4 ∙[�̇�4 ∙ 𝑑𝑑𝑡 (𝑒𝑖 ∙𝜃4 )+𝑒𝑖 ∙𝜃4 ∙𝑑𝑑𝑡 ( �̇�4 )]

𝐼=𝑖 ∙ 𝐴𝑂4 ∙[ �̇�4 ∙𝑒𝑖 ∙𝜃4 ∙𝑑𝑑𝑡 (𝜃4 )+𝑒𝑖 ∙𝜃4 ∙

𝑑𝑑𝑡 ( �̇�4 )]=𝑖 ∙𝐴𝑂4 ∙𝑒

𝑖 ∙𝜃 4∙ [( �̇�4 ∙ �̇�4 )+�̈�4 ]


Solution:

Since all the moving links comprise of the same type of vector (i.e., only the vector direction is a function of time),


𝐼=𝑖 ∙ 𝐴𝑂4 ∙𝑒𝑖 ∙𝜃 4 ∙ [ �̇�42+ �̈�4 ]

𝐼𝐼=𝑑𝑑𝑡 (𝑖 ∙𝐵𝐴 ∙ �̇�3 ∙𝑒𝑖 ∙𝜃 3 )=𝑖 ∙𝐵𝐴 ∙𝑒𝑖 ∙𝜃3 ∙ [ �̇�32+�̈�3 ]

𝐼𝐼𝐼=𝑑𝑑𝑡 ( 𝑖∙𝐵𝑂2 ∙ �̇�2 ∙𝑒

𝑖 ∙𝜃 2 )=𝑖 ∙𝐵𝑂2 ∙𝑒𝑖 ∙𝜃2 ∙ [ �̇�22+ �̈�2 ]

Acceleration Equation

𝐼+𝐼𝐼=𝐼𝐼𝐼

Acceleration Equation:𝑖 ∙𝐴𝑂4 ∙𝑒

𝑖 ∙𝜃 4∙ [ �̇�42+ �̈�4 ]+𝑖 ∙𝐵𝐴 ∙𝑒𝑖 ∙𝜃3 ∙ [�̇�32+ �̈�3 ]=𝑖 ∙𝐵𝑂2 ∙𝑒𝑖 ∙𝜃2 ∙ [ �̇�22+ �̈�2 ]


Solution: Solve the below shown Acceleration Equation using Mathcad.

Earlier in the problem link 2 (input crank) was assumed to be rotating at a constant speed (). Hence, angular acceleration of link 2 (BO2) = 0 rad/s2 (since a link rotating with

constant velocity will have no acceleration).

Find out the unknown acceleration parameters: angular acceleration of link 3 (BA)

angular acceleration of link 4 (AO4)


𝑖 ∙𝐴𝑂4 ∙𝑒𝑖 ∙𝜃 4∙ [ �̇�42+ �̈�4 ]+𝑖 ∙𝐵𝐴 ∙𝑒𝑖 ∙𝜃3 ∙ [�̇�32+ �̈�3 ]=𝑖 ∙𝐵𝑂2 ∙𝑒

𝑖 ∙𝜃2 ∙ [ �̇�22+ �̈�2 ]

With lower pair joints (revolute & prismatic)

Mechanisms


Drawing Vectors

1. Links with two revolute joints Use a vector pointing from one joint to the other

2. Prismatic joints on ground Use a vector pointing from a fixed point towards

the sliding link (along the path)

3. Prismatic joints on a moving link (Straight path)

Use a vector pointing from a fixed point on the link with the path to the prismatic joint (along the path)

4. Prismatic joints on a moving link (Curved path) Use a vector pointing from the center of

curvature (on the link with the path) to the prismatic joint


Slider Crank: Vector Diagram

23

4

1. Links with two revolute joints Use a vector pointing from one joint to the

other

2. Prismatic joints on ground Use a vector pointing from a fixed point

towards the sliding link (along the path)Changes directionChanges magnitude

O2 B

A


Slider Crank: Vector Equation

23

4

Changes directionChanges magnitude

O2 B

A


22 BOBAAO


or022 BOBAAO

22 BOBAAO


Slider Crank: Complex Number Equation

Real (1)

Imaginary (i)

+

23

4

Changes directionChanges magnitude

O2 B

A

22 BOBAAO

q2

q3

22

ieAO 2BO3 ieBA


Offset Slider Crank: Vector Diagram

2

3

4


other

2. Prismatic joints on ground Use a vector pointing from a fixed point

towards the sliding link (along the path)Changes directionChanges magnitudeFixed

Want the vectors on link 1 to be perpendicular

O2

B

A


Offset Slider Crank: Vector Equation

2

3

4

Changes directionChanges magnitudeFixed


O2

B

A


BCCOBAAO 22


or

C

BCCOBAAO 22

022 BCCOBAAO


Offset Slider Crank: Complex # Equation

2

3

4



O2

B

A

BCCOBAAO 22

C

22

ieAOq3

q2

2COi BC3 ieBA

Real (1)

Imaginary (i)

+


Slider Crank Inversion: Vector Diagram1. Links with two revolute joints

Use a vector pointing from one joint to the other


3

3. Prismatic joints on a moving link (Straight path) Use a vector pointing from a fixed point on the link

with the path to the prismatic joint (along the path)

Changes direction & magnitude

O2

O4

A

2

3

4


Slider Crank Inversion: Vector Equation


3Changes direction & magnitude

O2

O4

A

2

3

4

Method 1:Start at O2 and end at A

4242 AOOOAO


or

4242 AOOOAO

04242 AOOOAO


Slider Crank Inversion: Complex # Equ

Real (1)

Imaginary (i)

+


3


O2

O4

A

2

3

4

4242 AOOOAO

22

ieAO

q4q

2

124 OO 44

ieAO





3. Prismatic joints on a moving link (Straight path) Use a vector pointing from a fixed point on the link

with the path to the prismatic joint (along the path)



O2

O4

AB 4 3

2



(3 Revolute, 1 Prismatic)

Changes directionChanges magnitudeFixedChanges direction & magnitude


O2

O4

AB

4

32


4242 AOOOBABO


or

4242 AOOOBABO

04242 AOOOBABO




O2

O4

AB

4

32

124 OO

q4

q2

4242 AOOOBABO 2

2 ieBO 902ieBA 4

4 ieAO

2222 )90sin()90cos(9090 iiiii eiieeee

124 OO22

ieBO 2 ieBAi 44

ieAO

90°


Real (1)

Imaginary (i)

+





4. Prismatic joints on a moving link (Curved path) Use a vector pointing from the center of curvature

(on the link with the path) to the prismatic joint

This vector diagram is very similar to the four bar linkage

O2

O4

A

C

3

2

4




O2

O4

A

C

3

2

4


ACCOOOAO 4242


or

ACCOOOAO 4242

04242 ACCOOOAO




O2

O4

A

C

3

24

ACCOOOAO 4242

22

ieAO

q4

q2

q3

124 OO 3 ieAC44

ieCO

Real (1)

Imaginary (i)

+


Scotch Yoke: Vector Diagram



other2. Prismatic joints on ground

Use a vector pointing from a fixed point towards the sliding link (along the path)

A

B

32

4

O2


Scotch Yoke: Vector Equation


A

B

32

4

O2


ABBOAO 22


or

ABBOAO 22

022 ABBOAO


Scotch Yoke: Complex Number Equation


A

B

32

4

O2

ABBOAO 22

22

ieAO

q2

12 BO ABiReal (1)

Imaginary (i)

+


Scotch Yoke Variation: Vector Diagram



other2. Prismatic joints on ground

Use a vector pointing from a fixed point towards the sliding link (along the path)

A

B

32

4

O2


Scotch Yoke Variation: Vector Equation


A

B

32

4

O2


ABBOAO 22


or

ABBOAO 22

022 ABBOAO


Scotch Yoke Variation: Complex # Equ

12 BO


A

B

32

4

O2

ABBOAO 22

22

ieAO

a

q2

ieAB

Real (1)

Imaginary (i)

+

With higher pair joints (direct contact)

Mechanisms

Lecture 15-18 Ch09 Complex Numbers

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Transcript of Lecture 15-18 Ch09 Complex Numbers