basic machine design and the physics of motion

Basic Machine Design and the Physics of Motion

Marissa K Tucker Controls Product Marketing Manager

Parker Hannifin

Agenda

Laws of Motion

Center of Gravity (Center of Mass)

Free Body Diagram

Forces and Reactions Axial V. Normal Moments

Kinematic Equations (Dynamics) Velocity, Acceleration, Jerk

Are these principles applicable in the “real world’?

Short Answer: YES

Long Answer: Max Acceleration/Deceleration Max Velocity Peak Force and Forcerms Moment Loading

Are all critical to specifying the motor, actuator, and how you program your move.

Two types of motion

Linear Rotary

What is an actuator?

Device or mechanism that converts rotational motion of a motor into linear motion.

Body

Carriage

Motor End

Travel (Stroke)

What do we know

Need to know the size and mass of the load acting on the linear actuator

Need to know if the moment applied to the actuator is within the permissible range

Determine Operating Conditions

What speed, acceleration, and positional accuracy do I require

What type of power/force will I need to perform the move

Laws of Motion

1st Law Objects at rest or a constant velocity, want to remain in that state unless acted upon by a force.

2nd Law The sum of forces acting on an object are equal to the mass multiplied by the acceleration.

Laws of Motion

How does the 1st law apply?

How does the 2nd?

Mass vs Weight

Mass [kg, slugs, lbsmass]

a given amount of matter

Weight [N, lbsForce, kgForce]

the force applied to a mass by gravity

Mass versus Weight

Take a 10 kg object…

Mass of Object

10 𝒌𝒈 Earth = 10 𝒌𝒈 Moon

Weight of Object

WeightEarth = 10 𝑘𝑔 * 9.81 𝑚

𝑠2= 98.10 𝑁

WeightMoon = 10 𝑘𝑔 * 1.622 𝑚

𝑠2= 16.22 𝑁

WeightEarth ≠ WeightMoon

Weight

60 kg

Weight

200 kg

Center of Mass (Center of Gravity)

Masses in the world aren’t evenly distributed…

Center of mass (or gravity) allows us to lump a mass to one location.

14

Center of Gravity example

1. Divide the shape into easier sub-components

17

5

10

15

20

10 15

Y

X

30 kg



18

5

10

15

20

10 15

Y

X

30 kg



19

5

10

15

20

10 15

Y

X

30 kg


2. Assume uniform density

20

5

10

15

20

10 15

Y

X

30 kg


2. Assume uniform density

21

5

10

15

20

10 15

Y

X

20 kg

10 kg


2. Find the center point of each object

22

5

10

15

20

10 15

Y

X

20 kg

10 kg


3. Calculate the center of mass for each axis

23

5

10

15

20

10 15

Y

X

20 kg

10 kg


3a. Calculate the center of mass for the x-axis

24

CGravity = 𝑚∗𝑑

𝑚



25


𝑚

20 𝑘𝑔 ∗ 10 𝑚𝑚 + 10 𝑘𝑔 ∗ 15 𝑚𝑚

20 𝑘𝑔 + 10 𝑘𝑔

= 11.6 𝑚𝑚



26


𝑚

20 𝑘𝑔 ∗ 10 𝑚𝑚 + 10 𝑘𝑔 ∗ 15 𝑚𝑚

20 𝑘𝑔 + 10 𝑘𝑔

= 11.6 𝑚𝑚



27


𝑚

20 𝑘𝑔 ∗ 10 𝑚𝑚 + 10 𝑘𝑔 ∗ 15 𝑚𝑚

20 𝑘𝑔 + 10 𝑘𝑔

= 11.6 𝑚𝑚



28


𝑚

20 𝑘𝑔 ∗ 10 𝑚𝑚 + 10 𝑘𝑔 ∗ 15 𝑚𝑚

20 𝑘𝑔 + 10 𝑘𝑔

= 11.60 𝑚𝑚


3b. Calculate the center of mass for the y-axis

29


𝑚



30


𝑚

20 𝑘𝑔 ∗ 5𝑚𝑚 + 10 𝑘𝑔 ∗ 15 𝑚𝑚

20 𝑘𝑔𝑠 + 10 𝑘𝑔𝑠

= 8.33 𝑚𝑚



31


𝑚

20 𝑘𝑔 ∗ 5𝑚𝑚 + 10 𝑘𝑔 ∗ 15 𝑚𝑚

20 𝑘𝑔𝑠 + 10 𝑘𝑔𝑠

= 8.33 𝑚𝑚



32


𝑚

20 𝑘𝑔 ∗ 5𝑚𝑚 + 10 𝑘𝑔 ∗ 15 𝑚𝑚

20 𝑘𝑔 + 10 𝑘𝑔

= 8.33 𝑚𝑚



33


𝑚

20 𝑘𝑔 ∗ 5𝑚𝑚 + 10 𝑘𝑔 ∗ 15 𝑚𝑚

20 𝑘𝑔 + 10 𝑘𝑔

= 8.33 𝑚𝑚


5. You’ve found the center of gravity!

34

5

10

15

20

10 15

Y

X

20 kg

10 kg

8.33

11.6

0

Axial Versus Normal Forces

There are 2 applied forces we are always interested in

Normal Maximum compressive or tension load perpendicular to the carriage

Axial Maximum load in the direction of travel (Force Force)

36

ForceNormal

Direction of Travel

ForceAxial

Modeling Forces

Free Body Diagram A means of modeling forces relative to the point of attachment to an actuator.

37

ForceWeight ForceExtern

al

30˚

Free Body Diagram

We want to convert all forces either into nominal or axial forces so that we can Σ all the forces in each direction—a useful form of simplification.

ForceWeight is already a nominal force, so we don’t need to convert it.

38

ForceWeight ForceExtern

al

30˚

Free Body Diagram

ForceExternal can be broken down into it’s nominal and axial components.

39

ForceExtern

al

30˚

Free Body Diagram


40

ForceExtern

al

30˚

ForceExternal(A)

ForceExternal(N)

Free Body Diagram


To calculate these components, simply remember soh, cah, toa!

41

ForceExtern

al

30˚

ForceExternal(A)

ForceExternal(N)

Free Body Diagram


To calculate these components, simply remember soh, cah, toa!

42

ForceExtern

al

30˚

ForceExternal(A)

𝐹𝑜𝑟𝑐𝑒𝐸𝑥𝑡𝑒𝑟𝑛𝑎𝑙 𝑁 = 𝐹𝑜𝑟𝑐𝑒𝐸𝑥𝑡𝑒𝑟𝑛𝑎𝑙 ∗ sin(30)

ForceExternal(N)

𝐹𝑜𝑟𝑐𝑒𝐸𝑥𝑡𝑒𝑟𝑛𝑎𝑙 𝐴 = 𝐹𝑜𝑟𝑐𝑒𝐸𝑥𝑡𝑒𝑟𝑛𝑎𝑙 ∗ cos(30)

Free Body Diagram

Our new body diagram begins to look much more simplified…

43

ForceExternal(A)

ForceExternal(N)

ForceWeight

𝐹𝑜𝑟𝑐𝑒𝑁𝑜𝑟𝑚𝑎𝑙 = 𝐹𝑜𝑟𝑐𝑒𝑊𝑒𝑖𝑔ℎ𝑡 + 𝐹𝑜𝑟𝑐𝑒𝐸𝑥𝑡𝑒𝑟𝑛𝑎𝑙 ∗ sin(30)

𝐹𝑜𝑟𝑐𝑒𝐴𝑥𝑖𝑎𝑙 = 𝐹𝑜𝑟𝑐𝑒𝐸𝑥𝑡𝑒𝑟𝑛𝑎𝑙 ∗ cos(30)

Final Result

44

ForceNormal

ForceAxial

High moment application

45

24”

500 lbsForce 500 lbsForce

36% of maximum rated load 520% of maximum rated load

What is a moment?

Which wrench would you rather use?

46

What is a moment?

Moment = F x d

47

Force

Distance

What is a moment?

Torque (or Moment) to get the nut to turn is the same…

48

8 lbf

10”

?

5"

M

M

M = 80 in-lbf

M = 5” * ? in-lbf

What is a moment?

Torque (or Moment) to get the nut to turn is the same…

49

8 lbf

10”

16 lbf

5"

M

M

Roll, Pitch, and Yaw

moment about the longitudinal axis (axis of travel)

moment about the lateral axis that causes the carriage to rise or fall in the direction of travel

moment about the vertical axis relative to the stage

50

Direction of Travel


51

12 cm

7.5 cm

1.5 cm


52

12 cm

7.5 cm

1.5 cm

Fg

Force of Gravity


53

12 cm

7.5 cm

1.5 cm Fa

Force of Gravity Force of Acceleration

Fg


54

12 cm

7.5 cm

1.5 cm Fa

Fg


55

12 cm

7.5 cm

1.5 cm Fa

Fg


56

12 cm

7.5 cm

1.5 cm Fa

Fg

MR = F * d MR = Fg * 12cm


57

12 cm

7.5 cm

1.5 cm Fa

Fg

MR = F * d MR = Fg * 12cm Fa has no affect on roll


58

12 cm

7.5 cm

1.5 cm Fa

Fg


59

12 cm

7.5 cm

1.5 cm Fa

Fg


60

12 cm

7.5 cm

1.5 cm Fa

Fg

MP = F * d MP = Fa * 7.5 cm


61

12 cm

7.5 cm

1.5 cm Fa

Fg

MP = F * d MP = Fa * 7.5 cm + Fg * 1.5 cm


62

12 cm

7.5 cm

1.5 cm Fa

Fg


63

12 cm

7.5 cm

1.5 cm Fa

Fg


64

12 cm

7.5 cm

1.5 cm Fa

Fg

MY = F * d MY = Fa * 12cm


65

12 cm

7.5 cm

1.5 cm Fa

Fg

MY = F * d MY = Fa * 12cm Fg has no affect on Yaw


66

12 cm

7.5 cm

1.5 cm Fa

MR = Fg * 12cm MP = Fa * 7.5 cm + Fg * 1.5 cm MY = Fa * 12cm

Fg

Static versus Dynamic

67

Static Moment The moment applied to a linear action while at standstill

Dynamic Moment

The moment applied to the linear actuator while transporting a load

What about making things move?

Newtown’s 2nd Law

𝐹𝑜𝑟𝑐𝑒𝑠 = 𝑚𝑎𝑠𝑠 ∗ 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛

𝐹𝑜𝑟𝑐𝑒𝑠 = 𝑚𝑎𝑠𝑠 ∗Δ𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦

Δ𝑡𝑖𝑚𝑒

We need to calculate the maximum acceleration (Factors into the total force or Force)

We also need to calculate maximum speed required based upon drive train technology

68

Application Example:

“I need to move a bowling ball 500 mm in 1 second, V=d/t, so my speed is 500 mm/sec”

Not quite…

This assumes your acceleration is constant throughout the move which is never true.

69

Trapezoidal Profile

Most moves require an acceleration and deceleration period.

70

Kinematic Equations

Assuming acceleration is constant.

71

𝑑 = 𝑣𝑖𝑡 + 1

2𝑎𝑡2

𝑣𝑓 = 𝑣𝑖 + 𝑎𝑡

𝑣𝑓2 = 𝑣𝑖

2 + 2𝑎𝑑

𝑑 = 𝑣𝑖+ 𝑣𝑓

2t

Equation 1

Equation 2

Equation 3

Equation 4

These equations only apply to one segment at a time!

Kinematic Equations

Let’s calculate a problem where:

1.) The total move must take 5 seconds

2.) Acceleration and deceleration limited to taking .5 seconds each

3.) The total distance of the move is 1 meter

Find:

Maximum Velocity

Acceleration and Deceleration

72

Segment 1: Acceleration


73

𝑑 = 𝑣𝑖𝑡 + 1

2𝑎𝑡2

𝑑 = 0(0.5) + 1

2𝑎(0.5)2

𝑑 =1

2𝑎(0.5)2

Two unknowns. Let’s try another equation.

Result 1

Equation 1

𝑣𝑖

𝑣𝑓

𝑡

𝑎



74

𝑣𝑓 = 𝑣𝑖 + 𝑎𝑡

𝑣𝑓 = 𝑎(0.5)

𝑎 = 𝑣𝑓

0.5

Let’s substitute Result 2 into Result 1.

Equation 4

Result 2

𝑣𝑖

𝑣𝑓

𝑡

𝑎



75

𝑑 =1

2𝑎(0.5)2

𝑑 =1

2

𝑣𝑓

0.5(0.5)2

𝑑 =1

2

𝑣𝑓

0.5(0.5)2

𝑑1 = 0.25𝑣𝑓

Result 1

Result 3

𝑣𝑖

𝑣𝑓

𝑡

𝑎

Why is Result 3 Better for Us?

76

𝑑1 = 𝑑3 = 0.25𝑣𝑓

𝑑1 + 𝑑2 + 𝑑3 = 1 meter

We know the total distance of the move.

Thus, distance is a known parameter.

We also know that acceleration and deceleration are the same, thus:

Segment 2: Constant Velocity


77

𝑑 = 𝑣𝑓(4) + 1

2(0)(4)2

𝑑 = 𝑣𝑖𝑡 + 1

2𝑎𝑡2

𝑑2 = 𝑣𝑓(4)

Equation 1 𝑣𝑓 = 𝑣𝑖

𝑡

𝑎 = 0

Finding 𝑣𝑓


78

𝑑1 + 𝑑2 + 𝑑3 = 1 meter

0.25𝑣𝑓 + 4𝑣𝑓 + 0.25𝑣𝑓 = 1

𝑣𝑓 = .22 𝑚

𝑠

Segment 1/3: Finding a and d


79

𝑎 = .44 𝑚

𝑠2

𝑣𝑓 = 𝑣𝑖 + 𝑎𝑡 Equation 4

.22 = (0) + 𝑎(0.5)

𝑎 = 𝑑 = .44 𝑚

𝑠2

Final Result

80

𝑎 = 𝑑 = .44 𝑚

𝑠2

Total distance: 1 m

𝑣𝑖=0 𝑚𝑠

𝑣𝑓= .22

𝑚𝑠

𝑡 = 0.5

𝑎 𝑑

𝑡 = 4 𝑡 = 0.5

1/3 (Cheater) Trapezoidal Move

𝑽𝒆𝒍𝒐𝒄𝒊𝒕𝒚𝑴𝒂𝒙 =1.5𝐷

𝑡

𝑨𝒄𝒄𝒆𝒍𝒆𝒓𝒂𝒕𝒊𝒐𝒏𝑴𝒂𝒙 =4.5𝐷

𝑡2

81

½ (Cheater) Triangular Move

𝑽𝒆𝒍𝒐𝒄𝒊𝒕𝒚𝑴𝒂𝒙 =2𝐷

𝑡

𝑨𝒄𝒄𝒆𝒍𝒆𝒓𝒂𝒕𝒊𝒐𝒏𝑴𝒂𝒙 =4𝐷

𝑡2

82

Application Example

Triangular Move

𝑉𝑀𝑎𝑥 =2 ∗ 500 𝑚𝑚

1 𝑠𝑒𝑐= 1000 mm/sec

𝐴𝑀𝑎𝑥 =4∗500 𝑚𝑚

1 𝑠𝑒𝑐2= 2000 mm/sec2

83

Trapeziodal Move

𝑉𝑀𝑎𝑥 =1.5 ∗ 500 𝑚𝑚

1 𝑠𝑒𝑐= 750 mm/sec

𝐴𝑀𝑎𝑥 =4.5∗500 𝑚𝑚

1 𝑠𝑒𝑐2= 2250 mm/sec2

“I need to move a bowling ball 500 mm in 1 second, V=d/t, so my speed is 500 mm/sec”

Application Check

Q: Which profile is better when I need move something quick, but have limited force capacity in my mechanics

Q: Which profile is better when I’m making long moves, and would like to make them quickly.

84

S-Curve Profile

Sometimes, you don’t want acceleration to be constant.

85

A Quick Note: What is jerk?

The derivative of acceleration

Controlling jerk can create very smooth motion

This is important for move and settle applications where you don’t want to ‘shake’ the payload

Commonly referred to as “S” curve profiles

86

Calculating Total Force Load

𝐹𝑜𝑟𝑐𝑒𝑇𝑜𝑡𝑎𝑙 = 𝐹𝑜𝑟𝑐𝑒𝐴𝑐𝑐𝑒𝑙 + 𝐹𝑜𝑟𝑐𝑒𝐺𝑟𝑎𝑣𝑖𝑡𝑦 + 𝐹𝑜𝑟𝑐𝑒𝐹𝑟𝑖𝑐𝑡𝑖𝑜𝑛…

𝐹𝑜𝑟𝑐𝑒𝐴𝑐𝑐𝑒𝑙. = 𝐴𝑚𝑎𝑥 ∗ 𝑚 𝐹𝑜𝑟𝑐𝑒𝐺𝑟𝑎𝑣. = 𝑚 ∗ 𝐺 ∗ sin 𝜃 𝐹𝑜𝑟𝑐𝑒𝐹𝑟𝑖𝑐𝑡𝑖𝑜𝑛 = 𝜇 ∗ 𝑚 ∗ 𝐺 ∗ cos𝜃

87

𝜃

Brining it all together…

Peak Versus Continuous

Assume a load translating over a horizontal surface using a trapezoidal motion profile.

88


Acceleration (thus force) varies throughout the move.

89


First in the period of acceleration….

90

ForceAccel

+ ForceFriction


Then a time of constant acceleration…

91

ForceAccel

+ ForceFriction

ForceFriction


Lastly, a period of deceleration.

92

ForceAccel

+ ForceFriction

ForceFriction

ForceDecel

- ForceFriction


In a vertical load application, ForceGravity would affect all sections.

93

ForceAccel

+ ForceFriction

+ ForceGravity

ForceFriction

+ ForceGravity

ForceDecel

- ForceFriction

-

ForceFriction

Calculating an RMS Force

But this is “Peak Force”

Often RMS (effective force) is required for determining: Appropriately sizing a system (actuator, motor, and drive)

Accurately modeling the life of a system

94


1. Begin by breaking down your motion profile and examine the forces

95

ForceAccel

+ ForceFriction

ForceFriction

ForceDecel

- ForceFriction

Calculating RMS Force

𝐹𝑜𝑟𝑐𝑒𝑅𝑀𝑆 = 𝐹𝑖

2𝑑𝑖𝑖=𝑛𝑖=1

𝑑𝑖𝑖=𝑛𝑖=1

Where

F = is the force require per stroke segment (di)

di = is the stroke segment a force is seen

n = the total number of segments in a motion profile.

For determining the RMS force for a motor you would instead use time instead of distance. This is important in motor sizing.

96

2. Use the equation below to determine the Force RMS

RMS Force Example

97

The ‘de-nester’ lowers a stack of 4 x 200 kg (~440

lb) pallets 1 m (~40 in) in 3 seconds.

One pallet is then removed, and dwells for 2 seconds.

It then raises .3 m (~12 in) in 1 second to allow another pallet to be removed and again dwells for 2 seconds. This step is repeated until all the pallets are removed.

Assumptions • Neglect friction

• All trapezoidal motion profiles

• Max. acceleration/deceleration 7.85 𝑚

𝑠2

Move 1: Lowering The Stack

Acceleration ForceAcc = 800 kg * 7.58

𝑚

𝑠2 + 800 kg * 9.81

𝑚

𝑠2 = 14,128 N

Total distance: 0.25 m

Constant velocity ForceCVel = 800 kg* 9.81

𝑚

𝑠2 = 7848 N

Total Distance: 0.50 m

Decelerating ForceDec = -(800 kg* 7.58

𝑚

𝑠2) + 800 kg* 9.81

𝑚

𝑠2 = 1568 N


98

Move 2: Indexing to 2nd Pallet


𝑚

𝑠2 + 600kg * 9.81

𝑚

𝑠2 = 10,596 N


Constant velocity ForceCVel = 600 kg * 9.81

𝑚

𝑠2 = 5886 N


Decelerating ForceDec = -(600 kg *7.58

𝑚

𝑠2) + 600 kg* 9.81

𝑚

𝑠2 = 1176 N


99

Move 3: Indexing to 3rd Pallet


𝑚

𝑠2 + 400 kg * 9.81

𝑚

𝑠2 = 7,064 N



𝑚

𝑠2 = 3,924 N


Decelerating ForceDec= -(400 kg * 7.58

𝑚

𝑠2) + 400 kg * 9.81

𝑚

𝑠2 = 784 N


100

Move 4: Indexing to the Final Pallet


𝑚

𝑠2 + 200 kg * 9.81

𝑚

𝑠2 = 3,532 N



𝑚

𝑠2 = 1,962 N


Decelerating ForceDec= -(200 kg * 7.58

𝑚

𝑠2) + 200 kg * 9.81

𝑚

𝑠2 = 392 N


101


102

Adding it all up…

𝐹𝑜𝑟𝑐𝑒𝑅𝑀𝑆 = 𝐹𝑖

2𝑑𝑖𝑖=𝑛𝑖=1

𝑑𝑖𝑖=𝑛𝑖=1


𝐹𝑜𝑟𝑐𝑒𝑅𝑀𝑆 =(14128 𝑁)2∗ 0.25 𝑚 + 7848 𝑁 2 ∗ 0.5 𝑚 + 1568 𝑁 2 ∗ 0.25 𝑚 + …

1.9 𝑚

103

Adding it all up…


104

Σ𝐹𝑀𝑜𝑣𝑒1,𝐴𝑐𝑐

Adding it all up…

𝐹𝑜𝑟𝑐𝑒𝑅𝑀𝑆 =(14128 𝑁)2∗ 0.25 𝑚 + 7848 𝑁 2 ∗ 0.5 𝑚 + 1568 𝑁 2 ∗ 0.25 𝑚 + …

1.9 𝑚


𝐹𝑜𝑟𝑐𝑒𝑅𝑀𝑆 =(14128 𝑁)2∗ 0.25 𝑚 + 7848 𝑁 2 ∗ 0.5 𝑚 + 1568 𝑁 2 ∗ 0.25 𝑚 + …

1.9 𝑚

105

∆𝐷𝑀𝑜𝑣𝑒1,𝐴𝑐𝑐

Adding it all up…

Σ𝐹𝑀𝑜𝑣𝑒1,𝐴𝑐𝑐


𝐹𝑜𝑟𝑐𝑒𝑅𝑀𝑆 =(14128 𝑁)2∗ 0.25 𝑚 + 7848 𝑁 2 ∗ 0.5 𝑚 + 1568 𝑁 2 ∗ 0.25 𝑚 + …

1.9 𝑚

106

Adding it all up…

Σ𝐹𝑀𝑜𝑣𝑒1,𝐶𝑉𝑒𝑙


𝐹𝑜𝑟𝑐𝑒𝑅𝑀𝑆 =(14128 𝑁)2∗ 0.25 𝑚 + 7848 𝑁 2 ∗ 0.5 𝑚 + 1568 𝑁 2 ∗ 0.25 𝑚 + …

1.9 𝑚

107

Adding it all up…

Σ𝐹𝑀𝑜𝑣𝑒1,𝐶𝑉𝑒𝑙

∆𝐷𝑀𝑜𝑣𝑒1,𝐶𝑉𝑒𝑙


𝐹𝑜𝑟𝑐𝑒𝑅𝑀𝑆 =(14128 𝑁)2∗ 0.25 𝑚 + 7848 𝑁 2 ∗ 0.5 𝑚 + 1568 𝑁 2 ∗ 0.25 𝑚 + …

1.9 𝑚

108

Adding it all up…

Σ𝐹𝑀𝑜𝑣𝑒1,𝐷𝑒𝑐


109

Adding it all up…

Σ𝐹𝑀𝑜𝑣𝑒1,𝐷𝑒𝑐

∆𝐷𝑀𝑜𝑣𝑒1,𝐷𝑒𝑐

𝐹𝑜𝑟𝑐𝑒𝑅𝑀𝑆 =(14128 𝑁)2∗ 0.25 𝑚 + 7848 𝑁 2 ∗ 0.5 𝑚 + 1568 𝑁 2 ∗ 0.25 𝑚 + …

1.9 𝑚


110

Adding it all up…

Continue for Moves 2-4

𝐹𝑜𝑟𝑐𝑒𝑅𝑀𝑆 =(14128 𝑁)2∗ 0.25 𝑚 + 7848 𝑁 2 ∗ 0.5 𝑚 + 1568 𝑁 2 ∗ 0.25 𝑚 + …

1.9 𝑚


𝐹𝑜𝑟𝑐𝑒𝑅𝑀𝑆 =(14128 𝑁)2∗ 0.25 𝑚 + 7848 𝑁 2 ∗ 0.5 𝑚 + 1568 𝑁 2 ∗ 0.25 𝑚 + …

1.9 𝑚

111

Adding it all up…

Divide by total distance traveled


112

Adding it all up…

= 7350.41 𝑁𝑟𝑚𝑠 (equivalent load) or 1652.35 lbf

𝐹𝑜𝑟𝑐𝑒𝑅𝑀𝑆 =(14128 𝑁)2∗ 0.25 𝑚 + 7848 𝑁 2 ∗ 0.5 𝑚 + 1568 𝑁 2 ∗ 0.25 𝑚 + …

1.9 𝑚

Duty Cycle

Duty cycle is an important thing to note, as it pertains to selecting your actuator as well as your motor

𝑑𝑢𝑡𝑦 𝑐𝑦𝑐𝑙𝑒 = 𝑇𝑖𝑚𝑒 𝑖𝑛 𝑢𝑠𝑒

𝑇𝑜𝑡𝑎𝑙 𝑡𝑖𝑚𝑒 𝑥 100 [%]

Using our de-nester example:

3 + 1 + 1 + 1

3 + 2 + 1 + 2 + 1 + 2 + 1 + 2𝑥 100 = 42.8%

113

Speaker Contact Details

Marissa K Tucker Controls and HMI Product Marketing Manager

Parker Hannifin Electromechanical and Drives Division

9225 Forsyth Park Dr

Charlotte, NC 28273

USA

Telephone 707 477 7431

Email [email protected]

LinkedIn https://www.linkedin.com/in/marissatucker/

www.parker.com/emn

mailto:[email protected]

https://www.linkedin.com/in/marissatucker/

http://www.parker.com/emn

basic machine design and the physics of motion

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