Basic Machine Design and the Physics of Motion
Transcript of Basic Machine Design and the Physics of Motion
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Basic Machine Design and the Physics of Motion
Marissa K Tucker Controls Product Marketing Manager
Parker Hannifin
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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
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Are these principles applicable in the βreal worldβ?
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.
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Two types of motion
Linear Rotary
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What is an actuator?
Device or mechanism that converts rotational motion of a motor into linear motion.
Body
Carriage
Motor End
Travel (Stroke)
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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
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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.
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Laws of Motion
How does the 1st law apply?
How does the 2nd?
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Mass vs Weight
Mass [kg, slugs, lbsmass]
a given amount of matter
Weight [N, lbsForce, kgForce]
the force applied to a mass by gravity
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Mass versus Weight
Take a 10 kg objectβ¦
Mass of Object
10 ππ Earth = 10 ππ Moon
Weight of Object
WeightEarth = 10 ππ * 9.81 π
π 2= 98.10 π
WeightMoon = 10 ππ * 1.622 π
π 2= 16.22 π
WeightEarth β WeightMoon
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Weight
60 kg
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Weight
200 kg
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Center of Mass (Center of Gravity)
Masses in the world arenβt evenly distributedβ¦
Center of mass (or gravity) allows us to lump a mass to one location.
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Shape
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Shape
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Center of Gravity example
1. Divide the shape into easier sub-components
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5
10
15
20
10 15
Y
X
30 kg
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Center of Gravity example
1. Divide the shape into easier sub-components
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5
10
15
20
10 15
Y
X
30 kg
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Center of Gravity example
1. Divide the shape into easier sub-components
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5
10
15
20
10 15
Y
X
30 kg
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Center of Gravity example
2. Assume uniform density
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20
10 15
Y
X
30 kg
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Center of Gravity example
2. Assume uniform density
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5
10
15
20
10 15
Y
X
20 kg
10 kg
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Center of Gravity example
2. Find the center point of each object
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20
10 15
Y
X
20 kg
10 kg
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Center of Gravity example
3. Calculate the center of mass for each axis
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5
10
15
20
10 15
Y
X
20 kg
10 kg
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Center of Gravity example
3a. Calculate the center of mass for the x-axis
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CGravity = πβπ
π
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Center of Gravity example
3a. Calculate the center of mass for the x-axis
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CGravity = πβπ
π
20 ππ β 10 ππ + 10 ππ β 15 ππ
20 ππ + 10 ππ
= 11.6 ππ
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Center of Gravity example
3a. Calculate the center of mass for the x-axis
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CGravity = πβπ
π
20 ππ β 10 ππ + 10 ππ β 15 ππ
20 ππ + 10 ππ
= 11.6 ππ
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Center of Gravity example
3a. Calculate the center of mass for the x-axis
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CGravity = πβπ
π
20 ππ β 10 ππ + 10 ππ β 15 ππ
20 ππ + 10 ππ
= 11.6 ππ
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Center of Gravity example
3a. Calculate the center of mass for the x-axis
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CGravity = πβπ
π
20 ππ β 10 ππ + 10 ππ β 15 ππ
20 ππ + 10 ππ
= 11.60 ππ
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Center of Gravity example
3b. Calculate the center of mass for the y-axis
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CGravity = πβπ
π
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Center of Gravity example
3b. Calculate the center of mass for the y-axis
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CGravity = πβπ
π
20 ππ β 5ππ + 10 ππ β 15 ππ
20 πππ + 10 πππ
= 8.33 ππ
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Center of Gravity example
3b. Calculate the center of mass for the y-axis
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CGravity = πβπ
π
20 ππ β 5ππ + 10 ππ β 15 ππ
20 πππ + 10 πππ
= 8.33 ππ
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Center of Gravity example
3b. Calculate the center of mass for the y-axis
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CGravity = πβπ
π
20 ππ β 5ππ + 10 ππ β 15 ππ
20 ππ + 10 ππ
= 8.33 ππ
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Center of Gravity example
3b. Calculate the center of mass for the y-axis
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CGravity = πβπ
π
20 ππ β 5ππ + 10 ππ β 15 ππ
20 ππ + 10 ππ
= 8.33 ππ
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Center of Gravity example
5. Youβve found the center of gravity!
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5
10
15
20
10 15
Y
X
20 kg
10 kg
8.33
11.6
0
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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)
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ForceNormal
Direction of Travel
ForceAxial
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Modeling Forces
Free Body Diagram A means of modeling forces relative to the point of attachment to an actuator.
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ForceWeight ForceExtern
al
30Λ
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Free Body Diagram
We want to convert all forces either into nominal or axial forces so that we can Ξ£ all the forces in each directionβa useful form of simplification.
ForceWeight is already a nominal force, so we donβt need to convert it.
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ForceWeight ForceExtern
al
30Λ
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Free Body Diagram
ForceExternal can be broken down into itβs nominal and axial components.
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ForceExtern
al
30Λ
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Free Body Diagram
ForceExternal can be broken down into itβs nominal and axial components.
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ForceExtern
al
30Λ
ForceExternal(A)
ForceExternal(N)
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Free Body Diagram
ForceExternal can be broken down into itβs nominal and axial components.
To calculate these components, simply remember soh, cah, toa!
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ForceExtern
al
30Λ
ForceExternal(A)
ForceExternal(N)
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Free Body Diagram
ForceExternal can be broken down into itβs nominal and axial components.
To calculate these components, simply remember soh, cah, toa!
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ForceExtern
al
30Λ
ForceExternal(A)
πΉπππππΈπ₯π‘πππππ π = πΉπππππΈπ₯π‘πππππ β sin(30)
ForceExternal(N)
πΉπππππΈπ₯π‘πππππ π΄ = πΉπππππΈπ₯π‘πππππ β cos(30)
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Free Body Diagram
Our new body diagram begins to look much more simplifiedβ¦
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ForceExternal(A)
ForceExternal(N)
ForceWeight
πΉππππππππππ = πΉππππππππβπ‘ + πΉπππππΈπ₯π‘πππππ β sin(30)
πΉπππππ΄π₯πππ = πΉπππππΈπ₯π‘πππππ β cos(30)
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Final Result
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ForceNormal
ForceAxial
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High moment application
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24β
500 lbsForce 500 lbsForce
36% of maximum rated load 520% of maximum rated load
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What is a moment?
Which wrench would you rather use?
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What is a moment?
Moment = F x d
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Force
Distance
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What is a moment?
Torque (or Moment) to get the nut to turn is the sameβ¦
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8 lbf
10β
?
5"
M
M
M = 80 in-lbf
M = 5β * ? in-lbf
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What is a moment?
Torque (or Moment) to get the nut to turn is the sameβ¦
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8 lbf
10β
16 lbf
5"
M
M
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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
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Direction of Travel
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Roll, Pitch, and Yaw
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12 cm
7.5 cm
1.5 cm
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Roll, Pitch, and Yaw
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12 cm
7.5 cm
1.5 cm
Fg
Force of Gravity
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Roll, Pitch, and Yaw
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12 cm
7.5 cm
1.5 cm Fa
Force of Gravity Force of Acceleration
Fg
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Roll, Pitch, and Yaw
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12 cm
7.5 cm
1.5 cm Fa
Fg
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Roll, Pitch, and Yaw
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12 cm
7.5 cm
1.5 cm Fa
Fg
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Roll, Pitch, and Yaw
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12 cm
7.5 cm
1.5 cm Fa
Fg
MR = F * d MR = Fg * 12cm
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Roll, Pitch, and Yaw
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12 cm
7.5 cm
1.5 cm Fa
Fg
MR = F * d MR = Fg * 12cm Fa has no affect on roll
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Roll, Pitch, and Yaw
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12 cm
7.5 cm
1.5 cm Fa
Fg
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Roll, Pitch, and Yaw
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12 cm
7.5 cm
1.5 cm Fa
Fg
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Roll, Pitch, and Yaw
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12 cm
7.5 cm
1.5 cm Fa
Fg
MP = F * d MP = Fa * 7.5 cm
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Roll, Pitch, and Yaw
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12 cm
7.5 cm
1.5 cm Fa
Fg
MP = F * d MP = Fa * 7.5 cm + Fg * 1.5 cm
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Roll, Pitch, and Yaw
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12 cm
7.5 cm
1.5 cm Fa
Fg
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Roll, Pitch, and Yaw
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12 cm
7.5 cm
1.5 cm Fa
Fg
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Roll, Pitch, and Yaw
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12 cm
7.5 cm
1.5 cm Fa
Fg
MY = F * d MY = Fa * 12cm
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Roll, Pitch, and Yaw
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12 cm
7.5 cm
1.5 cm Fa
Fg
MY = F * d MY = Fa * 12cm Fg has no affect on Yaw
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Roll, Pitch, and Yaw
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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
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Static versus Dynamic
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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
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What about making things move?
Newtownβs 2nd Law
πΉπππππ = πππ π β πππππππππ‘πππ
πΉπππππ = πππ π βΞπ£ππππππ‘π¦
Ξπ‘πππ
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
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Application Example:
βI need to move a bowling ball 500 mm in 1 second, V=d/t, so my speed is 500 mm/secβ
Not quiteβ¦
This assumes your acceleration is constant throughout the move which is never true.
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Trapezoidal Profile
Most moves require an acceleration and deceleration period.
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Kinematic Equations
Assuming acceleration is constant.
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π = π£ππ‘ + 1
2ππ‘2
π£π = π£π + ππ‘
π£π2 = π£π
2 + 2ππ
π = π£π+ π£π
2t
Equation 1
Equation 2
Equation 3
Equation 4
These equations only apply to one segment at a time!
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Kinematic Equations
Letβs 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
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Segment 1: Acceleration
Assuming acceleration is constant.
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π = π£ππ‘ + 1
2ππ‘2
π = 0(0.5) + 1
2π(0.5)2
π =1
2π(0.5)2
Two unknowns. Letβs try another equation.
Result 1
Equation 1
π£π
π£π
π‘
π
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Segment 1: Acceleration
Assuming acceleration is constant.
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π£π = π£π + ππ‘
π£π = π(0.5)
π = π£π
0.5
Letβs substitute Result 2 into Result 1.
Equation 4
Result 2
π£π
π£π
π‘
π
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Segment 1: Acceleration
Assuming acceleration is constant.
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π =1
2π(0.5)2
π =1
2
π£π
0.5(0.5)2
π =1
2
π£π
0.5(0.5)2
π1 = 0.25π£π
Result 1
Result 3
π£π
π£π
π‘
π
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Why is Result 3 Better for Us?
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π1 = π3 = 0.25π£π
π1 + π2 + π3 = 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:
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Segment 2: Constant Velocity
Assuming acceleration is constant.
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π = π£π(4) + 1
2(0)(4)2
π = π£ππ‘ + 1
2ππ‘2
π2 = π£π(4)
Equation 1 π£π = π£π
π‘
π = 0
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Finding π£π
Assuming acceleration is constant.
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π1 + π2 + π3 = 1 meter
0.25π£π + 4π£π + 0.25π£π = 1
π£π = .22 π
π
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Segment 1/3: Finding a and d
Assuming acceleration is constant.
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π = .44 π
π 2
π£π = π£π + ππ‘ Equation 4
.22 = (0) + π(0.5)
π = π = .44 π
π 2
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Final Result
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π = π = .44 π
π 2
Total distance: 1 m
π£π=0 ππ
π£π= .22
ππ
π‘ = 0.5
π π
π‘ = 4 π‘ = 0.5
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1/3 (Cheater) Trapezoidal Move
π½ππππππππ΄ππ =1.5π·
π‘
π¨ππππππππππππ΄ππ =4.5π·
π‘2
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Β½ (Cheater) Triangular Move
π½ππππππππ΄ππ =2π·
π‘
π¨ππππππππππππ΄ππ =4π·
π‘2
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Application Example
Triangular Move
ππππ₯ =2 β 500 ππ
1 π ππ= 1000 mm/sec
π΄πππ₯ =4β500 ππ
1 π ππ2= 2000 mm/sec2
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Trapeziodal Move
ππππ₯ =1.5 β 500 ππ
1 π ππ= 750 mm/sec
π΄πππ₯ =4.5β500 ππ
1 π ππ2= 2250 mm/sec2
βI need to move a bowling ball 500 mm in 1 second, V=d/t, so my speed is 500 mm/secβ
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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 Iβm making long moves, and would like to make them quickly.
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S-Curve Profile
Sometimes, you donβt want acceleration to be constant.
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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 donβt want to βshakeβ the payload
Commonly referred to as βSβ curve profiles
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Calculating Total Force Load
πΉπππππππ‘ππ = πΉπππππ΄ππππ + πΉπππππΊπππ£ππ‘π¦ + πΉπππππΉππππ‘πππβ¦
πΉπππππ΄ππππ. = π΄πππ₯ β π πΉπππππΊπππ£. = π β πΊ β sin π πΉπππππΉππππ‘πππ = π β π β πΊ β cosπ
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π
Brining it all togetherβ¦
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Peak Versus Continuous
Assume a load translating over a horizontal surface using a trapezoidal motion profile.
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Peak Versus Continuous
Acceleration (thus force) varies throughout the move.
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Peak Versus Continuous
First in the period of accelerationβ¦.
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ForceAccel
+ ForceFriction
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Peak Versus Continuous
Then a time of constant accelerationβ¦
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ForceAccel
+ ForceFriction
ForceFriction
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Peak Versus Continuous
Lastly, a period of deceleration.
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ForceAccel
+ ForceFriction
ForceFriction
ForceDecel
- ForceFriction
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Peak Versus Continuous
In a vertical load application, ForceGravity would affect all sections.
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ForceAccel
+ ForceFriction
+ ForceGravity
ForceFriction
+ ForceGravity
ForceDecel
- ForceFriction
-
ForceFriction
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Calculating an RMS Force
But this is βPeak Forceβ
Often RMS (effective force) is required for determining: Appropriately sizing a system (actuator, motor, and drive)
Accurately modeling the life of a system
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Peak Versus Continuous
1. Begin by breaking down your motion profile and examine the forces
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ForceAccel
+ ForceFriction
ForceFriction
ForceDecel
- ForceFriction
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Calculating RMS Force
πΉπππππ ππ = πΉπ
2πππ=ππ=1
πππ=ππ=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.
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2. Use the equation below to determine the Force RMS
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RMS Force Example
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The βde-nesterβ 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 β’ Neglect friction
β’ All trapezoidal motion profiles
β’ Max. acceleration/deceleration 7.85 π
π 2
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Move 1: Lowering The Stack
Acceleration ForceAcc = 800 kg * 7.58
π
π 2 + 800 kg * 9.81
π
π 2 = 14,128 N
Total distance: 0.25 m
Constant velocity ForceCVel = 800 kg* 9.81
π
π 2 = 7848 N
Total Distance: 0.50 m
Decelerating ForceDec = -(800 kg* 7.58
π
π 2) + 800 kg* 9.81
π
π 2 = 1568 N
Total distance: 0.25 m
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Move 2: Indexing to 2nd Pallet
Acceleration ForceAcc = 600 kg * 7.58
π
π 2 + 600kg * 9.81
π
π 2 = 10,596 N
Total distance: 0.075 m
Constant velocity ForceCVel = 600 kg * 9.81
π
π 2 = 5886 N
Total Distance: 0.15 m
Decelerating ForceDec = -(600 kg *7.58
π
π 2) + 600 kg* 9.81
π
π 2 = 1176 N
Total distance: 0.075 m
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Move 3: Indexing to 3rd Pallet
Acceleration ForceAcc = 400 kg * 7.58
π
π 2 + 400 kg * 9.81
π
π 2 = 7,064 N
Total distance: 0.075 m
Constant velocity ForceCVel = 400 kg * 9.81
π
π 2 = 3,924 N
Total Distance: 0.15 m
Decelerating ForceDec= -(400 kg * 7.58
π
π 2) + 400 kg * 9.81
π
π 2 = 784 N
Total distance: 0.075 m
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Move 4: Indexing to the Final Pallet
Acceleration ForceAcc = 200 kg * 7.58
π
π 2 + 200 kg * 9.81
π
π 2 = 3,532 N
Total distance: 0.075 m
Constant velocity ForceCVel = 200 kg * 9.81
π
π 2 = 1,962 N
Total Distance: 0.15 m
Decelerating ForceDec= -(200 kg * 7.58
π
π 2) + 200 kg * 9.81
π
π 2 = 392 N
Total distance: 0.075 m
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Calculating RMS Force
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Adding it all upβ¦
πΉπππππ ππ = πΉπ
2πππ=ππ=1
πππ=ππ=1
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Calculating RMS Force
πΉπππππ ππ =(14128 π)2β 0.25 π + 7848 π 2 β 0.5 π + 1568 π 2 β 0.25 π + β¦
1.9 π
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Adding it all upβ¦
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Calculating RMS Force
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Ξ£πΉπππ£π1,π΄ππ
Adding it all upβ¦
πΉπππππ ππ =(14128 π)2β 0.25 π + 7848 π 2 β 0.5 π + 1568 π 2 β 0.25 π + β¦
1.9 π
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Calculating RMS Force
πΉπππππ ππ =(14128 π)2β 0.25 π + 7848 π 2 β 0.5 π + 1568 π 2 β 0.25 π + β¦
1.9 π
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βπ·πππ£π1,π΄ππ
Adding it all upβ¦
Ξ£πΉπππ£π1,π΄ππ
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Calculating RMS Force
πΉπππππ ππ =(14128 π)2β 0.25 π + 7848 π 2 β 0.5 π + 1568 π 2 β 0.25 π + β¦
1.9 π
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Adding it all upβ¦
Ξ£πΉπππ£π1,πΆπππ
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Calculating RMS Force
πΉπππππ ππ =(14128 π)2β 0.25 π + 7848 π 2 β 0.5 π + 1568 π 2 β 0.25 π + β¦
1.9 π
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Adding it all upβ¦
Ξ£πΉπππ£π1,πΆπππ
βπ·πππ£π1,πΆπππ
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Calculating RMS Force
πΉπππππ ππ =(14128 π)2β 0.25 π + 7848 π 2 β 0.5 π + 1568 π 2 β 0.25 π + β¦
1.9 π
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Adding it all upβ¦
Ξ£πΉπππ£π1,π·ππ
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Calculating RMS Force
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Adding it all upβ¦
Ξ£πΉπππ£π1,π·ππ
βπ·πππ£π1,π·ππ
πΉπππππ ππ =(14128 π)2β 0.25 π + 7848 π 2 β 0.5 π + 1568 π 2 β 0.25 π + β¦
1.9 π
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Calculating RMS Force
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Adding it all upβ¦
Continue for Moves 2-4
πΉπππππ ππ =(14128 π)2β 0.25 π + 7848 π 2 β 0.5 π + 1568 π 2 β 0.25 π + β¦
1.9 π
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Calculating RMS Force
πΉπππππ ππ =(14128 π)2β 0.25 π + 7848 π 2 β 0.5 π + 1568 π 2 β 0.25 π + β¦
1.9 π
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Adding it all upβ¦
Divide by total distance traveled
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Calculating RMS Force
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Adding it all upβ¦
= 7350.41 ππππ (equivalent load) or 1652.35 lbf
πΉπππππ ππ =(14128 π)2β 0.25 π + 7848 π 2 β 0.5 π + 1568 π 2 β 0.25 π + β¦
1.9 π
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Duty Cycle
Duty cycle is an important thing to note, as it pertains to selecting your actuator as well as your motor
ππ’π‘π¦ ππ¦πππ = ππππ ππ π’π π
πππ‘ππ π‘πππ π₯ 100 [%]
Using our de-nester example:
3 + 1 + 1 + 1
3 + 2 + 1 + 2 + 1 + 2 + 1 + 2π₯ 100 = 42.8%
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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