# Motion and Machine Automation ( design assignment)

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Ngg505

Engineering

## Description

See the all attachment carefully to understand the assignment. read the instructor for the details information. Demonstrate a proof of concept of working system, Mechanics Selection & Motor Sizing, calculations, drawing, a name of parts. either handwritten or documents is accepted.

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we have a trim board = 1*4 or 1*6 all boards are white. we need to design a machine that can do things to the board. first thing screw some screws. on the door frame, you may have a piece of trim and the door frame about 36" up if you have a door jamb. we could put brackets on the trim. the board we got to stack one against another and they're being shoved into this machine so we need to build a machine over here between the boards to stop and start the board and puts in a bracket. it might have bracket here to screw here screw there. ( same as the figure) between we need a motion control stuff. part of that we going need a coupler nip rollers (coupler, motor,..). will squeezing the board and one of them probably has a motor and one is an ipoare and we've got some friction right here so we can drive the boards. then registration sensor, and we need to sensor that can detect the board. then we will have basically a plate we going to amount have two screwdrivers into this plate. we need motion go up and down. and each one has to exist motor, Dc motor ( to perform better torque control in it). ( 1/2 hp brushed motor) two screws ( 2" and .5") during travel 2.22" and 0.75" -constant force - fix time 1.2 to driver screw - when sensor red should move 36" and stop. - accy +1.5mm -cycle time 5 second. - diameter the roller > 2.5" - contact whole board. - pushed with 3lb force from the beginning. - need to take care of rolling and the tension of rolling. - length of board 72"-90" -force 1lb from the middle to the end. - between the rolling can be bushed, 2 board. - screw collaboration should be 1.2second to drive screw. -power supply 110, 230... no limitation. - the screwdrivers = 20lb. Introduction to Motor Sizing INTRODUCTION TO MOTOR SIZING Nick Repanich Adjunct Research Professor Department of Mechanical and Mechatronic Engineering and Sustainable Manufacturing California State University, Chico Chico, CA 95929-0789 (530) 520-2548 Introduction to Motor Sizing 1 SEVEN STEPS OF SIZING AND SELECTION Step 1: Develop the torque and inertia equations that model the system mechanics. 1a: Draw/diagram the system to establish the relative location of the load mechanics. 1b: Develop the acceleration (Ta=), friction (Tf=), gravity (Tg=), and thrust (Tth=) torque equations. Since Ta= (J)(α), this will also involve developing the inertial model (Jload=). Step 2: • Determine the load motion profile(s) and calculate peak values. 2 X 2 xa = tm ta - a simple triangular profile minimizes torque (lowers cost), but uses higher speeds First find the max velocity reached (V) in a triangular move. If OK, use it. V= • If V is too high for any reason, then find the optimized acceleration time (ta) in a trapezoidal X move using overall move time, distance & the new constrained velocity. ta = tm − V - trapezoidal profiles are useful where rated motor torque drops with speed - for steppers, set maximum speed where the motor changes from the constant torque range into the constant power range, otherwise reduction may be necessary. V • If there are multiple move profiles, find the worst-case acceleration (a), where a= ta Step 3: Calculate the mass moment of inertias of the load mechanics. • Anything that moves with the motor is part of the inertia (J). Step 4: Determine the peak torque (Tp) at maximum speed excluding motor inertia. • determine worst case combination of Ta , Tf , Tg , Tth ( T peak = Ta ± T f ± Tg ± Tth ) • calculate power required to move the load (to find a system in the right power range) Step 5: Choose an approximate motor/drive system. • find a motor/drive with more than the required speed and about double the power • look at speed-torque curves and price list simultaneously - with servos, use most of the speed available or you waste the power available Step 6: Determine the peak torque at speed including motor inertia. • calculate RMS torque to evaluate motor heating issues (necessary for servo systems) • give 50-100% torque margin (only use lower margins if measurable mechanics exist) • check for 10-20% velocity margin Step 7: Optimize the system. • Check the load-to-motor inertia ratio, and compare to the machine’s stiffness to the performance desired. A ratio higher than 10:1 is an indicator of potential instability problems with non-stiff systems. (Note: It is not the cause of the problem.) • Check the need for a power dump circuit (high inertia ratios or vertical load) • It may be necessary to adjust mechanics, add reduction, and start over. Remember, iteration is crucial to successful design! Introduction to Motor Sizing Direct Drive (Rotary) Formulas and Examples MOTION PROFILE FORMULAS for triangular profiles: V= for trapezoidal profiles: 2 X 2 xa = tm ta ta = tm − Where: X V V = maximum velocity X = total move distance tm = total move time xa = acceleration distance ta = acceleration time Note: V is used here rather that ω, as these formulas work with either linear or rotary units. GENERAL TORQUE & POWER FORMULAS T peak = Ta ± T f ± Tg ± Tth Torque Equations: T12 t1 + T22 t 2 + T32 t 3 + T42 t 4 t1 + t 2 + t 3 + t 4 Motor torque required to accelerate the inertial load Motor torque required to overcome the frictional forces Motor torque required to overcome the gravitational forces load Motor torque required to overcome any additional thrust forces Torque to accelerate the load from zero speed to max speed (Tf + Ta) Torque to keep the motor moving once it reaches max speed (Ta = 0) Torque required to decelerate from max speed to a stop (Ta - Tf) Torque required while motor is sitting still at zero speed time spent accelerating the load time spent while motor is turning at constant speed time spent decelerating the load time spent while motor is at rest Trms = Where: Ta Tf Tg Tth T1 T2 T3 T4 t1 t2 t3 t4 = = = = = = = = = = = = Torque Unit Conversions: kg ⋅ m N= 2 s 2 ∴ kg ⋅ cm 2 ⎛ 1 m ⎞ ⎜ ⎟ = N ⋅ m s 2 ⎝ 100 cm ⎠ Power Unit Conversions: 2π (N ⋅ m )(rpm ) (oz ⋅ in)(rps ) (in ⋅ lb)(rpm ) ( ft ⋅ lb)(rpm ) ( ft ⋅ lb)(rps ) Watts = 2π (N ⋅ m )(rps ) = = = = = 60 22.52 84.45 7.038 0.1173 1 hp = 746 W 2 Introduction to Motor Sizing Direct Drive (Rotary) Formulas and Examples DIRECT DRIVE (ROTARY) FORMULAS Inertia (Step 3): Solid Cylinder Hollow Cylinder Where ρ, the density is known, use πLρR 4 J load = 2 Where mass and radius are known use mR 2 J load = 2 m = πLρR 2 Where ρ, the density, is known use πLρ 4 J load = R2 − R14 2 Where m, the mass, is known use ( J load = ) m 2 R1 + R22 2 ( ( ) m = πLρ R22 − R12 ) Torque (Step 2, 4-7) ⎛ ω ⎞ Ta = (J load + J cp + J m )⎜⎜ m ⎟⎟ ⎝ t a ⎠ T f = (F )(R ) Where: Jcp Jm ta ωm F R = = = = = = inertia of the coupler (g·cm2) inertia of the motor's rotor (g·cm2) acceleration time (s) maximum motor velocity reached (rad/s) frictional force (N) radius at which the frictional force acts (cm) Note 1: When using Imperial units, specifically inertia in convenient units of oz·in2, the acceleration torque formula must be modified to correct the units. The units of oz·in2 for inertia are not proper inertial units, yet they are quite convenient to calculate inertia based on real world data. To use the acceleration torque formula with inertia in oz·in2, divide by the gravitational constant g (1/386 in/s2), which converts oz·in2 to proper units of oz·in·s2 as follows: ⎛ ω ⎞ 1 ( Ta = J load + J cp + J m )⎜⎜ m ⎟⎟ in ⎝ t a ⎠ 386 2 s Note 2: Though the correct symbol for mass moment of inertia is “I”, typically in mechatronics we (incorrectly) use the symbol “J” for inertia, as “I” is also used for electrical current. 3 Introduction to Motor Sizing Direct Drive (Rotary) Formulas and Examples DIRECT DRIVE WITH REDUCTION Gear Drive Inertia Formulas: m R2 J load = load 2load 2N or 4 πLload ρ load Rload J load = 2N 2 J gear1 = J gear 2 = Where: Jx m R N L ρ 2 mgear1 Rgear 1 2N 2 2 m gear 2 Rgear 2 2 = = = = = = inertia "as seen by the motor" mass radius reduction ratio (R1/R2) length density Note: Most gearhead manufacturers provide the reflected inertia of the reducer in their user guide. This will help eliminate the need to calculate the individual gear inertias. ⎛ J ⎞⎛ ω N ⎞ Ta = ⎜ load + J r + J m ⎟⎜⎜ m ⎟⎟ ⎝ e ⎠⎝ t a ⎠ Tf = (F )(R ) eN Where: e N Jr Jm ωm ta F R = efficiency of transmission (reduction) = reduction ratio = effective inertia of reducer, gear box, belts and pulleys, or gears = inertia of the motor's rotor = maximum motor velocity reached = acceleration time = frictional force = radius at which the frictional force acts 4 Introduction to Motor Sizing Direct Drive (Rotary) Formulas and Examples EXAMPLE #1 - Fluted-Bit Cutting Machine Parameters for Axis 4: • • • • • • • • Move distance Move time Maximum velocity allowed Length of chuck Diameter of steel chuck Length of bit Diameter of steel bit Torque required during cutting V= Step 1: 0.2 0.15 20 4.0 3.0 14 0.8 0.050 rev s rps cm cm cm cm N·m 2 X (2)(0.2) = = 2.67 rps tm 0.15 (OK, since not above "knee" on most stepper curves) Step 2: Step 3: J bit ⎛ ω ⎞ Ta = (J bit + J chuck + J motor )⎜⎜ m ⎟⎟ ⎝ t a ⎠ 4 π g ⎛ 0.8 cm ⎞ 2 = (14 cm )(7.75 3 )⎜ ⎟ = 4.36 g ⋅ cm 2 cm ⎝ 2 ⎠ J chuck = π 2 4 (4 cm )(7.75 g ⎛ 3.0 cm ⎞ 2 ) ⎟ = 246.5 g ⋅ cm 3 ⎜ cm ⎝ 2 ⎠ 5 Introduction to Motor Sizing Direct Drive (Rotary) Formulas and Examples Step 4: 2 2 cm m ⎛ 2.67 rps ⎞ Ta = 4.36 g ⋅ cm + 246.5 g ⋅ cm (2π )⎜ ⎟ = 56,112 g ⋅ 2 = 0.0056 kg ⋅ 2 = 0.0056 N ⋅ m s s ⎝ 0.15 s / 2 ⎠ ( 2 2 ) Tf = 0.050 N·m Tpeak = Ta + Tf = 0.0056 + 0.050 N·m = 0.0556 N·m @ 2.67 rps P = 2π (0.0556 N ⋅ m)(2.67 rps) = 0.93 W 2 Jload = 4.36 + 246.5 = 250.86 g·cm Step 5: Try Nema 23 half-stack step motor, Step 6: Jm = 0.070 kg·cm2 = 70 g·cm2 cm 2 ⎛ 2.67 ⎞ Ta = (4.36 + 246.5 + 70 )(2π )⎜ ⎟ = 71,770 g ⋅ 2 = 0.0072 N ⋅ m s ⎝ 0.15/ 2 ⎠ Tpeak = Ta + Tf = 0.0072 + 0.050 N·m = 0.0572 N·m @ 2.67 rps Step 7: Inertia Ratio = Torque Margin (@speed) = Jl 250.86 g ⋅ cm 2 = = 3.6 (fine) Jm 70 g ⋅ cm 2 Tavailable − T peak T peak = 0.26 N ⋅ m - 0.0572 N ⋅ m = 355 % (fine) 0.00572 N ⋅ m 6 Introduction to Motor Sizing Direct Drive (Rotary) Formulas and Examples PROBLEM #1 - Grinding Of Bicycle Rim Braking Surface Parameters of Axis 3: • • • • • • • • • • Outside diameter of steel bicycle rim 33.0 cm Move distance 1/36 rev Move time 0.10 s Inertia of rim-holding fixture 1686 kg·cm2 Width of rim 4.5 cm Inside diameter of rim 31.5 cm Friction torque during grinding 1.6 N·m Dwell time between moves 0.05 s Velocity is not limited in any way This is not the final grinding move, but is the worst case move, making repeated passes over the weld. Answer: approximately 20 N·m @ 0.56 rps 7 Introduction to Motor Sizing Leadscrew Formulas and Examples LEADSCREW FORMULAS J load = Inertia (Step 3) Where: m L R ρ p = = = = = m (2πp )2 J screw = πLρR 4 2 mass (kg) length (cm) radius (cm) density (g/cm3) pitch of screw (revs/unit length) (inverse of lead) The formula for Jload converts linear mass into the rotational inertia as reflected to the motor shaft by the lead screw. ⎛ J ⎞⎛ ω ⎞ Ta = ⎜ load + J screw + J cp + J m ⎟⎜⎜ m ⎟⎟ ⎝ e ⎠⎝ t a ⎠ Torque (Step 2, 4-7) T f = Tbr + Where: Jscrew Tbr F p e ωm g Jcp Jm ta Tg µd = = = = = = = = = = = = F 2πpe Tg = mg 2πpe F = µ d mg inertia of lead screw or ball screw (kg·cm2) breakaway torque of nut on screw (N·m) force or thrust required pitch of screw (revs/unit length) efficiency of nut and screw maximum motor speed (rev/s). (watch critical speed of screw) acceleration due to gravity inertia of the coupling (kg·cm2) inertia of the motor's rotor (kg·cm2) acceleration time (s) torque required to overcome gravity (vertical) coefficient of dynamic friction 8 Introduction to Motor Sizing Leadscrew Formulas and Examples EXAMPLE #3 - "Smart" Battery Inspection Parameters for Axis 1: • • • • • • • • • • Length of steel leadscrew Pitch of leadscrew Efficiency of leadscrew Radius of leadscrew Move distance Move time Maximum allowed velocity Weight of load Coefficient of friction Breakaway Torque 36 in 5 0.65 0.5 in 1.5 in 0.5 s 5 in/s 50 lbs. 0.01 25 oz·in 9 Introduction to Motor Sizing Leadscrew Formulas and Examples Step 1: V= (2) (1. 5in) = 6 in / sec, too high 0.5sec ta = tm − V=5 Step 2: Ta = in rev rev ×5 = 25 sec in sec 1 # Jload & 2πV + Jleadscrew + Jmotor + Jcp ( % 386\$ e ' ta Jload = Step 3: X 1.5 = 0. 5 − = 0.2 sec V 5 W ( 2 πp ) 2 = 50(16) ( 2π5) 2 = 0. 81 oz ⋅in2 πLρ R4 π 4 = (36)(4. 48)(. 5) = 15.8 oz ⋅ in2 2 2 Jleadscrew = Jcp = 0 Ta = 1 ! . 81 2 π (25 ) + 15 .8 \$& = 34.69 oz⋅ in @ 25 rps # 386 " . 65 % .2 Step 4: T f = T br + F 0.01(50 )(16 ) = 25 + = 25.39 oz⋅ in 2πp e 2 π 5(.65 ) Tpeak = 34.69 + 25.39 = 60.08 oz·in hp = Step 5: ! (60. 08)25 watts \$ = 0. 0894# 746 & = 66. 7 watts 16, 800 hp % " Choose motor with about 0.2 hp Try S83-93, Jm = 6.70 10 Introduction to Motor Sizing Step 6: Ta = Leadscrew Formulas and Examples 1 ! .81 2π( 25) + 15. 8 + 6.70\$& = 48. 3 oz ⋅ in # 386" .65 % .2 T peak = 48. 3 + 25. 39 = 73. 7 oz ⋅ in @ 25 rps Jl 16.61 = = 2. 48, O.K. Jm 6. 7 Step 7: Pick S83-93 11 Introduction to Motor Sizing Tangential Formulas and Examples TANGENTIAL DRIVES FORMULAS Inertia Formulas: 2 J load = ml Rdp J dp = J fp = m p R p2 2 = πL p ρR p4 2 2 dp J b = mb R Where: Jdp Jfp Jb Rdp ml mp mb = inertia of the driving pulley = inertia of the follow pulley = inertia of the belt = radius of the driving pulley = mass of the load = mass of the pulley = mass of the belt 12 Introduction to Motor Sizing Appendix A Belt Drive Torque Formulas: Direct Drive: With a gearhead: ⎛ J + J belt + J mechanics ⎞⎛ ω ⎞ Ta = ⎜⎜ load + J dp + J m ⎟⎟⎜⎜ m ⎟⎟ eb ⎝ ⎠⎝ t a ⎠ T f = FR dp ⎛ J ⎞⎛ ω m N gh J dp + J belt + J mechanics ⎟⎜ Ta = ⎜ load + + J + J gh m 2 2 ⎜ ⎟⎜ t a e e N e N b gh gh gh gh ⎝ ⎠⎝ FRdp Tf = e gh N gh Where: ta eb egh Ngh Jmechanics Rdp Jgh ωm Jm F = acceleration time = efficiency of belt over pulleys = efficiency of gearhead = reduction ratio of gearhead = inertia of other moving masses = radius of driving pulley = effective inertia of reducer = maximum motor speed = inertia of the motor's rotor = force or thrust or friction forces at the drive pulley ⎞ ⎟⎟ ⎠ 13 Appendix A Introduction to Motor Sizing Rack & Pinion Torque Formulas: Note: In the following formulas, the rack is stationary and the motor moves with the load. If the rack moves and the motor is stationary, then use the belt and pulley formulas, where the rack acts essentially like the belt. ⎛ J load + J m + J p ⎞⎛ ω m ⎞ ⎟⎜ ⎟ Ta = ⎜ Direct Drive: ⎜ ⎟⎜ t ⎟ e g ⎝ ⎠⎝ a ⎠ FR p Tf = eg With gearhead on motor: ⎛ J load + J p J gh + J m Ta = ⎜ + ⎜ e N 2 e eg gh ⎝ g FR p Tf = e g N 2 e gh ⎞⎛ ω m N ⎞ ⎟⎜ ⎟ ⎟⎜ t a ⎟ ⎠ ⎠⎝ Where: Jload eg Jp Rp F Jm ωm ta Jgh egh N = = = = = = = = = = = inertia of moving load efficiency of transmission between rack and pinion inertia of pinion gear and shafts and couplers and outboard bearings (if any) effective radius of pinion gear force, thrust or friction inertia of the motor's rotor maximum motor speed acceleration time effective inertia of reducer efficiency of reducer reduction ratio 14 Appendix A Introduction to Motor Sizing APPENDIX A STANDARD END FIXTURING METHODS The method of end fixturing has a direct effect on critical speed, column load bearing capacity, and system stiffness. Four common methods are shown below. Fixed-Free One end held in a duplex (preloaded) bearing, the other end is free Fixed-Simple One end held in a duplex (preloaded) bearing, one end held in a single bearing. Simple-Simple Both ends held in a single bearing. Fixed-Fixed Both ends held in duplex (preloaded) bearings. DENSITIES OF COMMON MATERIALS Material oz/in3 gm/cm3 Aluminum Alloys 1.54 2.8 Brass, Bronze 4.80 8.6 Copper 5.15 8.9 Plastics 0.64 1.1 Steel (carbon,alloys,stainless) 4.48 7.8 Hard Wood 0.46 0.80 KINEMATIC EQUATIONS (straight-line motion w/ constant acceleration) Equation Contains x vx ax t ü ü ü vx = vx + axt 0 x = x0 + 12 v x0 + v x t ü x = x0 + v x0 t + 12 a x t 2 ü ( 2 2 ) v x = v x0 + 2a x (x − x0 ) ü ü ü ü ü ü ü COEFFICIENTS OF FRICTION Materials (dry contact unless noted) µs µd Steel on Steel 0.74 0.58 Steel on Steel (lubricated) 0.23 0.15 Aluminum on Steel 0.61 0.45 Copper on Steel 0.22 Brass on Steel 0.19 Teflon on Steel 0.04 0.04 Round rails w/ball bearings 0.002 0.002 Linear guides - radius groove 0.003 0.002 Linear guides - gothic arch (depends 0.008 0.004 on load, pre-load & type) to .05 to .02 15 Appendix A Introduction to Motor Sizing BALL SCREW FORMULAS Critical Speed ⎛ S ⋅ d × 10 6 ⎞ ⎟⎟ N c = ⎜⎜ 2 f ⋅ L ⎝ s ⎠ Nc d L fs S = = = = = = = = critical speed (rpm) root diameter of screw (cm) distance between bearings (cm) Safety factor (recommend 1.25) end support conditions factor 4.22 fixed - free 18.85 fixed - simple 27.31 fixed – fixed Note: The ‘S’ term carries units rev ⋅ cm /min. This equation is based on the natural frequency of the rod, and acquires its time component from that derivation. = 10 6 inches (Fm / dynamic load )3 For Equivalent Load ( ⎛ 14.03 × 10 6 ⋅ S ⋅ d 4 ⎞ ⎟⎟ Fcl = ⎜⎜ L2 ⎝ ⎠ Fcl = maximum load (lb) S = end fixity factor = .250 one end fixed one free = 1.0 both ends supported (simple) = 2.0 one fixed one simple = 4.0 both fixed d = root diameter of screw L = distance between ball nut and load-carrying bearing Note: The ‘S’ term carries hidden units that make the formula simpler and easier to use. Backdriving Torque (mainly used to determine holding brake torque) Tb = = = = = For Other then Rated (Dynamic) Load 10 6 inches Life = (operating load/dynamic load )3 3 3 Fm = 3 Y1 (F1 ) + …Yn (Fn ) Column Load Tb F l e Life Expectancy F ×l ×e 2π torque required to backdrive axial load lead of screw efficiency of screw ) Fm = equivalent load Fn = a particular increment of load Yn = the portion of a cycle (sub cycles) of a particular increment of load expressed as a decimal, i.e. the sum of the sub cycles must equal one. Example if L1 is applied for 20% of the cycle, L2 is applied for 30% of the cycle and L3 is applied for 50% of the cycle, then the associated Y values are Y1 =.2, Y2 = .3, Y3 = .5. 16 X-Axis Z-Axis Completo Complelo Attempted (F3 of PIN) Altompled (S of P/N) Name: Grade: Sketch of Layout Linear Bearings Selected Drive Mechanism Selected Motor Sizing incl. amplifier Motor Sizing RMS Calcs Electrical Load Calcs Coupler / Shaft Load Calcs Sensor Selected Accuracy Predicted Name: X-Axis Z-Axis Complete completo Altempted (F8 or PIN) Altempted (FS of P/N) Grade: work on sensor New Aplication. company name fance Regestration two surowes driver moulon contray 1st stuff i put 3 \$736" 1 lxu vies 1x 6 319 meto (nip rollers) mechaum can do things for the board 36" up The door board The Stor & stop puł Pracket 2 exis Motor A {beteer corge control I bp brushed 2 3 dc Motor hed} motor Z محقر القاسمي المساهم F Two scrows 2.22 Travel -0.75 travel move it up & down Et constant force during travel fix Time o when sensor red should move słop. 1.2 to drive scrow // 36 and 2 sohnall Goles en e pois Two scrows 2.22 Travel -0.75 Travel move ił up & down -constant force during cravel TE fix Time w 1.2 o drive scrow a O when sensor red should move 36 and 1207 stop. a accy + 1,5 mm cycle time 5 second every ss diameter the roller > 2.5 in Contact whole board pushed withe 316 Need Take care of rooling and the tenilen of rolling lingon of the board 72" - 40" force 116 - bushed force jill knell is bełt ween rolling can bushed 2 board من المرايا To ( the rolling
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The required motor sizing from the motor calculations should be a servo motor with torque
control in it to allow for communication with the sen...

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