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FRC ® Pneumatics Workshop It’s not a bunch of hot air!PowerPoint Presentation

FRC ® Pneumatics Workshop It’s not a bunch of hot air!

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### FRC® Pneumatics WorkshopIt’s not a bunch of hot air!

November 2011

Tim Serge

growingstems.org robobees.org

Agenda

Poll Audience: “What do you want to know?”

Introduction to Pneumatics

Pneumatics in FRC®

Actuators and How They Work

Back to Basics: Science!

Examples

Introduction to Pneumatics

Concept: what are pneumatics?

- Pneumatics vs. Hydraulics
- Both systems are similar in concept
- Pneumatics use a gas as the working fluid
- Hydraulics use a liquid as the working fluid

- Pneumatics use compressed gas to perform mechanical work

Examples of Pneumatics

- Air brakes on heavy trucks and busses
- Air tools
- Air hammer
- Air rachet
- Impact driver

- FIRST® Robotics Competition
- Robot claw
- Shifting transmissions
- Deployment mechanisms

Trade Study: Advantages

Can be used in compact environments

High reliability

Actuation can be very fast

Performance is consistent

Easy to set up (several FRC® resources)

Simple command and control support

One-way distribution (exhaust air)

Trade Study: Disadvantages

Excessive air requirement may lead to a higher compressor drain on battery

Weight and space of supporting hardware

Vibration in system

Possible compressor overheating

Possibility for sudden movement (air is compressible!)

Pneumatics in FRC®

FRC® Components

- Air Compressor
- Pressure Release Valve
- Air Receiver
- Pressure Gauge
- Pressure Switch
- Pressure Regulator

- Actuator (Air Cylinder, Rotary, Linear)
- Brass Fittings
- Solenoid Valves
- Plug Valve
- Tubing

Shameless Copy + Paste from FIRST® Manual

General Pneumatics Hints

Use Teflon® tape on all fittings, leaving first two threads bare. Wrap tape in the direction of thread so tape stays on when tightening

Use short runs of hose

Use large diameter hose

Pressure test system to ensure no leaks (pressurize and let sit – listen and look for pressure drops using gauges)

Control actuation speed using needle valves on the inlet and outlet of actuators

FRC® NI cRIO Connectivity

- NI cRIO contains solenoid control module
- Spike Relay must be connected to pressure switch
- Identify actuator position using magnetic reed switch(es) in conjunction with a magnet on the actuator shaft (special order actuator)
- Account for time to actuate (delay) with the control system code

Actuators and How They Work

Basic Actuator Types

Single Acting Spring Return

Double Acting

Air pressure compression

Air pressure extension

Use a Two Port Solenoid

- Spring pressure compression
- Air pressure extension
- Use a Single Port Solenoid

Telescoping

Double Ended

- Air pressure compression & extension
- Three Ports

- Air pressure compression & extension
- Moves single shaft in two directions

Basic Actuator Types

Rotary

- Air spins a turbine
- Output shaft spins
- Often referred to as an “air motor”

How does a double acting actuator work?

Actuator in Extension

The double solenoid valve acts as a “switch”, directing the pressurized system air to one actuator port

As air enters the system, pressure is exerted on a sealed piston

In turn, air exits the system

Solenoid Valve

Regulated pressure

How does a double acting actuator work?

Actuator in Compression

As air enters the system, pressure is exerted on a sealed piston

The double solenoid valve acts as a “switch”, directing the pressurized system air to one actuator port

In turn, air exits the system

Solenoid Valve

Regulated pressure

Actuator Notes

There are minimum pressure requirements for solenoids. Check the specs!

Find the stroke and bore required to size the actuator

Remember to leave “wiggle room” to overcome friction and system dynamics

Size to operate between the minimum and maximum allowable pressures

Use needle valves on actuators to control minimum and maximum actuation speed

Back to Basics: Science!

Back to Basics: Science!

Combined Gas Law

Newton’s Laws of Motion

Force and Pressure

Moments and Torque

Back to Basics: Combined Gas Law

Mass of gas is conserved in a closed system

- Boyle’s Law
- Constant temperature
- ↓V proportional with ↑p

- Charles’s Law
- Constant pressure
- ↑T proportional with ↑V

- Gay-Lussac’s Law
- Constant volume
- ↑T proportional to ↑p

Back to Basics: Newton’s 1st Law of Motion

ΣF = 0

More info: http://en.wikipedia.org/wiki/Newton%27s_laws

Newton’s 1st Law

Inertia: Sum of Forces equals zero

Unless acted upon by an unbalanced force, an object’s velocity will not change

Back to Basics: Newton’s 2nd Law of Motion

Δ p = mΔv

Newton’s 2nd Law

Conservation of Momentum

“The change of momentum of a body is proportional to the impulse impressed on the body, and happens along the straight line on which that impulse is impressed.”

Back to Basics: Newton’s 3rd Law of Motion

Must exert 5lbs of force to hold the object stationary

Object weighs 5lbs

- Newton’s 3rd Law
- Forces of b acting on a= Forces of a acting on b
- “To every action there is always an equal and opposite reaction: or the forces of two bodies on each other are always equal and are directed in opposite directions.”

Back to Basics: Force and Pressure

- F = Force, the force normal to the surface is of interest
- A = Area, references the piston bore 2-dimensional area (d = diameter).
- p= pressure of air entering actuator, controlled by pressure regulator

Double Acting Actuator Pressure

D

d

Small bore:

Abore = π * ((D-d)/2)2

Large bore:

Abore = π * (D/2)2

Given the same pressure, the actuator requires more force to close than open

Gauge Pressure vs. Absolute Pressure

pabsolute = pgauge + patmosphere

In this instruction, gauge pressure will be used

Gauge pressure is pressure inside a system and does not include atmospheric pressure

Absolute pressure includes atmospheric pressure

Moments and Torque

M

r

Denotes positive moment direction about an axis

F

- Moments are a cross product of distance (r) and Force (F)
- Moments occur around a defined point
- Moments have a positive (clockwise) and negative (counterclockwise) value

Examples

Mounting Actuators

final position

r

α

initial position

r

linitial

lfinal

y

gravity

x

Know the arc!

Find final length of initial and final position of base

Determine arc of length

Place base of actuator pivot (clevis) at intersect of arcs

Example: Force and Pressure

Draw a free-body diagram!

5lb

Apply Newton’s 3rd Law of Motion!

5lb

1

Factuator≥ 5 lb

2

gravity

pregulated= ?

Fmaxis limited by pmax!

Factuator

Factuator= ?

3

y

x

pregulated = 5lb / 0.1 in2

Abore = 0.1 in2

pregulated≥ 50 psi

Assumptions

No friction loss

Known actuator stroke

Known actuator size

Example: Moments

Draw free-body diagrams!

2”

6”

5lb

2”

Actuator Extended

α

6”

Fy

αmax = 30°

5lb

Fx

Factuator

Actuator Compressed

αmax = 30°

αmin = 0°

y

gravity

Fy

6”

2”

Factuator, min = ?

x

Fx

Factuator

5lb

Assumptions

No friction loss

Known actuator stroke

Known actuator size

Ignore weight of bar

Example: Moments

Apply Newton’s Laws of Motion!

ΣFx = 0

ΣFy = 0

ΣM = 0

Factuator=?

2”

6”

Fy

αmax = 30°

5lb

Fx

Factuator

ΣM = 0

Mcouterclockwise = Mclockwise

Factuator* 6 in = 5lb * 8 in

Factuator= (5lb*8in) / 6 in

ΣM = 0

Mcouterclockwise= Mclockwise

Factuator* 6 in*sin(30°) = 5lb * 8 in*sin(30°)

Factuator= (5lb*8in) / 6 in

6”

2”

Factuator

5lb

Factuator= 6.7 lb

Example: Air Receivers

- Determine or estimate the number of actuations per each actuator required in a match
- Use Boyle’s Law to calculate the volume of air required pre-regulator.
- Adiabatic system because isentropic (no additional heat added to system) and throttling (significant pressure change – regulator)

- Sum all individual requirements to get the total

Example: Sizing Air Receivers

to compressor

Abore,rear = 1 in2

Abore,front = 0.8 in2

stroke = 4 in

p1= 115 psi

V1=?

regulator

solenoid

actuator

p2= 30 psi

V2 = V2,open + V2,close = (Abore * stroke)rear,front

V2= 1 in2 * 4 in * 2 act. + 0.8 in2 * 4 in * 2 act.

V2= 14.4 in3 /match

Nactuations, open = 2 actuations/match

Nactuations, close = 2 actuations/match

p1V1 = p2V2

V1 = 30 psi * 14.4 in3 / 115 psi

V1 = 3.8 in3

Exercise: Build a system

- Given variables:
- Calculate force requirement of piston
- Calculate bore diameter
- Determine sizing of piston

- Lay out system
- Match component to slot in diagram
- Identify pressure regulation as required

Exercise: Build a System

p = 115 psi

p = 60 psi

p = ?

p = 50 psi

A = 10 in2

F=?

F = 400 lb

A = 5 in2

F = 250 lb

A = ?

References

http://www.pneumaticsfirst.org/Kickoff/PneuAndFIRSTFinal.ppt

http://en.wikipedia.org/wiki/Pneumatic_actuator

http://www.usfirst.org/uploadedFiles/Robotics_Programs/FRC/Game_and_Season__Info/2011_Assets/Kit_of_Parts/2011_FIRST_Robotics_Competition_Pneumatics_Manual_Rev_B.pdf

http://en.wikipedia.org/wiki/Newton's_laws_of_motion

http://www.mdfirst.org/images/stories/documents2010/Presentations/Pneumatics-basics.pdf

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