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.

Carriage Motor End

Travel (Stroke)

Body 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 풌품 Earth = 10 풌품 Moon

Weight of Object 푚 Weight = 10 푘푔 * 9.81 = 98.10 푁 Earth 푠2 푚 Weight = 10 푘푔 * 1.622 = 16.22 푁 Moon 푠2

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 Shape Shape Center of Gravity example

1. Divide the shape into easier sub-components

20 30 kg

X 10 15

17 Center of Gravity example

1. Divide the shape into easier sub-components

20 30 kg

X 10 15

18 Center of Gravity example

1. Divide the shape into easier sub-components

20 30 kg

X 10 15

19 Center of Gravity example

2. Assume uniform density

20 30 kg

X 10 15

20 Center of Gravity example

2. Assume uniform density

15 10 kg

5 20 kg

X 10 15

21 Center of Gravity example

2. Find the center point of each object

20 10 kg

10 20 kg 5

X 10 15

22 Center of Gravity example

3. Calculate the center of mass for each axis

20 10 kg

10 20 kg 5

X 10 15

23 Center of Gravity example

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

푚∗푑 C = Gravity 푚

24 Center of Gravity example

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

푚∗푑 C = Gravity 푚

20 푘푔 ∗ 10 푚푚 + 10 푘푔 ∗ 15 푚푚

20 푘푔 + 10 푘푔

= 11.6 푚푚

25 Center of Gravity example

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

푚∗푑 C = Gravity 푚

20 푘푔 ∗ 10 푚푚 + 10 푘푔 ∗ 15 푚푚

20 푘푔 + 10 푘푔

= 11.6 푚푚

26 Center of Gravity example

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

푚∗푑 C = Gravity 푚

20 푘푔 ∗ 10 푚푚 + 10 푘푔 ∗ 15 푚푚

20 푘푔 + 10 푘푔

= 11.6 푚푚

27 Center of Gravity example

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

푚∗푑 C = Gravity 푚

20 푘푔 ∗ 10 푚푚 + 10 푘푔 ∗ 15 푚푚

20 푘푔 + 10 푘푔

= 11.60 푚푚

28 Center of Gravity example

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

푚∗푑 C = Gravity 푚

29 Center of Gravity example

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

푚∗푑 C = Gravity 푚

20 푘푔 ∗ 5푚푚 + 10 푘푔 ∗ 15 푚푚

20 푘푔푠 + 10 푘푔푠

= 8.33 푚푚

30 Center of Gravity example

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

푚∗푑 C = Gravity 푚

20 푘푔 ∗ 5푚푚 + 10 푘푔 ∗ 15 푚푚

20 푘푔푠 + 10 푘푔푠

= 8.33 푚푚

31 Center of Gravity example

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

푚∗푑 C = Gravity 푚

20 푘푔 ∗ 5푚푚 + 10 푘푔 ∗ 15 푚푚

20 푘푔 + 10 푘푔

= 8.33 푚푚

32 Center of Gravity example

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

푚∗푑 C = Gravity 푚

20 푘푔 ∗ 5푚푚 + 10 푘푔 ∗ 15 푚푚

20 푘푔 + 10 푘푔

= 8.33 푚푚

33 Center of Gravity example

5. You’ve found the center of gravity!

20 10 kg

10 8.3 20 kg 53

10 15

11.6 0

34 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)

ForceNormal

ForceAxial

Direction of Travel

36 Modeling Forces

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

ForceWeight ForceExtern al

37 Free Body Diagram

ForceWeight ForceExtern al

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 Free Body Diagram

ForceExtern al

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

39 Free Body Diagram

ForceExtern al

ForceExternal(N)

30˚

ForceExternal(A)

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

40 Free Body Diagram

ForceExtern al

ForceExternal(N)

30˚

ForceExternal(A)

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

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

41 Free Body Diagram

ForceExtern al

ForceExternal(N)

30˚

ForceExternal(A)

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

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

퐹표푟푐푒퐸푥푡푒푟푛푎푙 푁 = 퐹표푟푐푒퐸푥푡푒푟푛푎푙 ∗ sin(30)

퐹표푟푐푒퐸푥푡푒푟푛푎푙 퐴 = 퐹표푟푐푒퐸푥푡푒푟푛푎푙 ∗ cos(30)

42 Free Body Diagram

ForceWeight ForceExternal(N)

ForceExternal(A)

Our new body diagram begins to look much more simplified…

퐹표푟푐푒푁표푟푚푎푙 = 퐹표푟푐푒푊푒𝑖푔ℎ푡 + 퐹표푟푐푒퐸푥푡푒푟푛푎푙 ∗ sin(30)

퐹표푟푐푒퐴푥𝑖푎푙 = 퐹표푟푐푒퐸푥푡푒푟푛푎푙 ∗ cos(30)

43 Final Result

ForceNormal

ForceAxial

44 High moment application

500 lbs Force 500 lbsForce

24”

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

45 What is a moment?

Which wrench would you rather use?

46 What is a moment?

Moment = F x d

Distance

Force

47 What is a moment?

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

M = 80 in-lbf 10

” 8

lbf M

M = 5” * ? in-lbf 5"

48 What is a moment?

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

” 8

lbf M

lbf

49 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

Direction of Travel

50 Roll, Pitch, and Yaw

1.5 cm

7.5 cm

12 cm

51 Roll, Pitch, and Yaw

1.5 cm

Fg 7.5 cm Force of Gravity 12 cm

52 Roll, Pitch, and Yaw

1.5 cm Fa

Fg 7.5 cm Force of Gravity 12 cm Force of Acceleration

53 Roll, Pitch, and Yaw

1.5 cm Fa

Fg 7.5 cm

12 cm

54 Roll, Pitch, and Yaw

1.5 cm Fa

Fg 7.5 cm

12 cm

55 Roll, Pitch, and Yaw

1.5 cm Fa

Fg 7.5 cm MR = F * d 12 cm MR = Fg * 12cm

56 Roll, Pitch, and Yaw

1.5 cm Fa

Fg 7.5 cm MR = F * d 12 cm MR = Fg * 12cm

Fa has no affect on roll

57 Roll, Pitch, and Yaw

1.5 cm Fa

Fg 7.5 cm

12 cm

58 Roll, Pitch, and Yaw

1.5 cm Fa

Fg 7.5 cm

12 cm

59 Roll, Pitch, and Yaw

1.5 cm Fa

Fg 7.5 cm MP = F * d 12 cm MP = Fa * 7.5 cm

60 Roll, Pitch, and Yaw

1.5 cm Fa

Fg 7.5 cm MP = F * d 12 cm MP = Fa * 7.5 cm + Fg * 1.5 cm

61 Roll, Pitch, and Yaw

1.5 cm Fa

Fg 7.5 cm

12 cm

62 Roll, Pitch, and Yaw

1.5 cm Fa

Fg 7.5 cm

12 cm

63 Roll, Pitch, and Yaw

1.5 cm Fa

Fg 7.5 cm MY = F * d 12 cm MY = Fa * 12cm

64 Roll, Pitch, and Yaw

1.5 cm Fa

Fg 7.5 cm MY = F * d 12 cm MY = Fa * 12cm

Fg has no affect on Yaw

65 Roll, Pitch, and Yaw

1.5 cm Fa

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

66 Static versus Dynamic

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

67 What about making things move?

Newtown’s 2nd Law

퐹표푟푐푒푠 = 푚푎푠푠 ∗ 푎푐푐푒푙푒푟푎푡푖표푛 Δ푣푒푙표푐푖푡푦 퐹표푟푐푒푠 = 푚푎푠푠 ∗ Δ푡푖푚푒

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.

1 푑 = 푣 푡 + 푎푡2 Equation 1 𝑖 2

2 2 푣푓 = 푣𝑖 + 2푎푑 Equation 2

푣 + 푣 푑 = 푖 푓t 2 Equation 3

푣푓 = 푣𝑖 + 푎푡 Equation 4 These equations only apply to one segment at a time!

71 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

Assuming acceleration is constant.

1 푣푓 푑 = 푣 푡 + 푎푡2 Equation 1 𝑖 2 1 푑 = 0(0.5) + 푎(0.5)2 2 푎 1 푑 = 푎(0.5)2 Result 1 2 푣𝑖

Two unknowns. Let’s try another equation. 푡

73 Segment 1: Acceleration

Assuming acceleration is constant.

푣푓 푣푓 = 푣𝑖 + 푎푡 Equation 4

푣푓 = 푎(0.5)

푣푓 푎 푎 = Result 2 0.5 푣𝑖

Let’s substitute Result 2 into Result 1. 푡

74 Segment 1: Acceleration

Assuming acceleration is constant.

1 푣 푑 = 푎(0.5)2 Result 1 푓 2

1 푣푓 푑 = (0.5)2 2 0.5 푎

1 푣푓 푣 푑 = (0.5)2 𝑖 2 0.5 푡

푑1 = 0.25푣푓 Result 3

75 Why is Result 3 Better for Us?

We know the total distance of the move. Thus, distance is a known parameter.

푑1 + 푑2 + 푑3 = 1 meter

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

푑1 = 푑3 = 0.25푣푓

76 Segment 2: Constant Velocity

Assuming acceleration is constant.

1 2 푑 = 푣𝑖푡 + 푎푡 Equation 1 푣푓 = 푣𝑖 2 푎 = 0 1 푑 = 푣 (4) + (0)(4)2 푓 2

푑2 = 푣푓(4)

푡

77 Finding 푣푓

Assuming acceleration is constant.

푑1 + 푑2 + 푑3 = 1 meter

0.25푣푓 + 4푣푓 + 0.25푣푓 = 1

푚 푣 = .22 푓 푠

78 Segment 1/3: Finding a and d

Assuming acceleration is constant.

푣푓 = 푣𝑖 + 푎푡 Equation 4

.22 = (0) + 푎(0.5)

푚 푎 = .44 푠2 푚 푎 = 푑 = .44 푠2

79 Final Result

푣 푚 푓= .22 푠 푚 푎 = 푑 = .44 푠2 푎 푑 Total distance: 1 m

푣 푚 𝑖=0 푠

푡 = 0.5 푡 = 4 푡 = 0.5

80 1/3 (Cheater) Trapezoidal Move

1.5퐷 푽풆풍풐풄풊풕풚 = 푴풂풙 푡

4.5퐷 푨풄풄풆풍풆풓풂풕풊풐풏 = 푴풂풙 푡2

81 ½ (Cheater) Triangular Move

2퐷 푽풆풍풐풄풊풕풚 = 푴풂풙 푡

4퐷 푨풄풄풆풍풆풓풂풕풊풐풏 = 푴풂풙 푡2

82 Application Example

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

Triangular Move 2 ∗ 500 푚푚 푉 = = 1000 mm/sec 푀푎푥 1 푠푒푐 4∗500 푚푚 퐴 = = 2000 mm/sec2 푀푎푥 1 푠푒푐2

Trapeziodal Move 1.5 ∗ 500 푚푚 푉 = = 750 mm/sec 푀푎푥 1 푠푒푐 4.5∗500 푚푚 퐴 = = 2250 mm/sec2 푀푎푥 1 푠푒푐2

83 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

Brining it all together…

퐹표푟푐푒푇표푡푎푙 = 퐹표푟푐푒퐴푐푐푒푙 + 퐹표푟푐푒퐺푟푎푣𝑖푡푦 + 퐹표푟푐푒퐹푟𝑖푐푡𝑖표푛 …

퐹표푟푐푒퐴푐푐푒푙. = 퐴푚푎푥 ∗ 푚 퐹표푟푐푒퐺푟푎푣. = 푚 ∗ 퐺 ∗ sin 휃 퐹표푟푐푒퐹푟𝑖푐푡𝑖표푛 = 휇 ∗ 푚 ∗ 퐺 ∗ cos 휃

휃

87 Peak Versus Continuous

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

88 Peak Versus Continuous

Acceleration (thus force) varies throughout the move.

89 Peak Versus Continuous

First in the period of acceleration….

ForceAccel +

ForceFriction

90 Peak Versus Continuous

Then a time of constant acceleration…

ForceAccel + ForceFriction ForceFriction

91 Peak Versus Continuous

Lastly, a period of deceleration.

ForceAccel ForceDecel + ForceFriction - ForceFriction ForceFriction

92 Peak Versus Continuous

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

ForceAccel ForceDecel + ForceFriction - ForceFriction + ForceFriction + ForceGravity - ForceGravity ForceFriction

93 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 Peak Versus Continuous

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

ForceAccel ForceDecel + ForceFriction - ForceFriction ForceFriction

95 Calculating RMS Force

2. Use the equation below to determine the Force RMS

𝑖=푛 2 𝑖=1 퐹𝑖 푑𝑖 퐹표푟푐푒푅푀푆 = 𝑖=푛 𝑖=1 푑𝑖

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 RMS Force Example

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 푚 • Max. acceleration/deceleration 7.85 푠2

97 Move 1: Lowering The Stack

Acceleration 푚 푚 Force = 800 kg * 7.58 + 800 kg * 9.81 = 14,128 N Acc 푠2 푠2 Total distance: 0.25 m

Constant velocity 푚 Force = 800 kg* 9.81 = 7848 N CVel 푠2 Total Distance: 0.50 m

Decelerating 푚 푚 Force = -(800 kg* 7.58 ) + 800 kg* 9.81 = 1568 N Dec 푠2 푠2 Total distance: 0.25 m

98 Move 2: Indexing to 2nd Pallet

Acceleration 푚 푚 Force = 600 kg * 7.58 + 600kg * 9.81 = 10,596 N Acc 푠2 푠2 Total distance: 0.075 m

Constant velocity 푚 Force = 600 kg * 9.81 = 5886 N CVel 푠2 Total Distance: 0.15 m

Decelerating 푚 푚 Force = -(600 kg *7.58 ) + 600 kg* 9.81 = 1176 N Dec 푠2 푠2 Total distance: 0.075 m

99 Move 3: Indexing to 3rd Pallet

Acceleration 푚 푚 Force = 400 kg * 7.58 + 400 kg * 9.81 = 7,064 N Acc 푠2 푠2 Total distance: 0.075 m

Constant velocity 푚 Force = 400 kg * 9.81 = 3,924 N CVel 푠2 Total Distance: 0.15 m

Decelerating 푚 푚 Force = -(400 kg * 7.58 ) + 400 kg * 9.81 = 784 N Dec 푠2 푠2 Total distance: 0.075 m

100 Move 4: Indexing to the Final Pallet

Acceleration 푚 푚 Force = 200 kg * 7.58 + 200 kg * 9.81 = 3,532 N Acc 푠2 푠2 Total distance: 0.075 m

Constant velocity 푚 Force = 200 kg * 9.81 = 1,962 N CVel 푠2 Total Distance: 0.15 m

Decelerating 푚 푚 Force = -(200 kg * 7.58 ) + 200 kg * 9.81 = 392 N Dec 푠2 푠2 Total distance: 0.075 m

101 Calculating RMS Force

Adding it all up…

𝑖=푛 2 𝑖=1 퐹𝑖 푑𝑖 퐹표푟푐푒푅푀푆 = 𝑖=푛 𝑖=1 푑𝑖

102 Calculating RMS Force

Adding it all up…