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FRC ® Pneumatics Workshop It’s not a bunch of hot air!. November 2011 Tim Serge tim.serge@gmail.com. growing stems .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!
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FRC® Pneumatics WorkshopIt’s not a bunch of hot air! November 2011 Tim Serge tim.serge@gmail.com 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
Section 1 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!)
Section 2 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
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
Section 3 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
Section 4 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
Section 5 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