BASICS

he following information helps Where: Hollow cylinder rotating about you solve technical problems P ϭ Power, hp its own axis: frequently encountered in de- Q ϭ Flow rate, gpm T WR2+ R2 ϭ ()1 2 signing and selecting motion control H Head, ft WK 2= ()10 components and systems. S ϭ Specific gravity of fluid 2 ␮ ϭ efficiency Where: 2 2 Fans and blowers: WK ϭ Moment of inertia, lb-ft T ϭ FR (1) W ϭ Weight of object, lb Qp ϭ P = ()6 R1 Outside radius, ft ␮ Where: 229 R2 ϭ Inside radius, ft T ϭ Torque, lb-ft F ϭ , lb Where: R ϭ Radius, or distance that the P ϭ Power, hp force is from the pivotal point, ft Q ϭFlow rate, cfm p ϭ Pressure, psi ␮ ϭ Efficiency   2 2 = V Linear to rotary motion WKL W  ()11  2␲N  V N = ()2 Accelerating torque and force 0. 262 D Of rotating objects Material in linear motion with Where: a continuous fixed relation to a 2⌬ N ϭ Speed of shaft rotation, rpm ()WK N rotational speed, such as a con- ϭ T = ()7 V Velocity of material, fpm 308t veyor system: D ϭ Diameter of pulley or , 2  V  in. Where: WK2 = W  ()11 T ϭTorque required, lb-ft L  2␲N  WK2 ϭ Total inertia of load to be ac- Horsepower 2 celerated, lb-ft . (See Formulas 9, 10, Where: 2 2 Rotating objects: 11, and 12.) WKL ϭ Linear inertia, lb-ft ⌬N ϭ Change in speed, rpm W ϭ Weight of material, lb TN P = ()3 t ϭTime to accelerate load, sec V ϭ Linear velocity, fpm 5, 250 N ϭ Rotational speed of shaft, rpm Objects in linear motion: Where: WV⌬ Reflected inertia of a load P ϭPower, hp F = ()8 through a speed reduction means T ϭ Torque, lb-ft 1, 933t — , chain, or system: N ϭ Shaft speed, rpm WK 2 Where: WK 2 = L ()12 R 2 Objects in linear motion: F ϭForce required, lb Rr W ϭ Weight, lb = FV ⌬V ϭChange in velocity, fpm Where: P ()4 2 2 33, 000 t ϭ Time to accelerate load, sec WKR ϭ Reflected inertia, lb-ft 2 2 WKL ϭ Load inertia, lb-ft ϭ Where: Moment of inertia Rr Reduction ratio P ϭ Power, hp ϭ F Force, lb Solid cylinder rotating about Duty cycle calculation V ϭ Velocity, fpm its own axis: The RMS (root mean square) value 2 ϭ 1 2 : WK ( /2)WR (9) of a load is one of the quantities often Where: used to size PT components. WK2 ϭ Moment of inertia, lb-ft2 = QHS Lt2+++ Lt2... Lt2 P ()5 W ϭ Weight of object, lb L = 1122 nn ()13 3, 960␮ ϭ RMS +++ R Radius of cylinder, ft tt12... tn

1997 Power Design A19 Mechanical properties of common materials Ultimate strength, psi Modulus of elasti- Modulus Yield city, of point, tension elasti- Weight ten- or com- city, (lb Equiva- Com- sion pression shear per Material lent Tension pression* Shear (psi) (psi) (psi) in.3) Steel, forged-rolled C, 0.10-0.20 ...... SAE 1015 60,000 39,000 48,000 39,000 30,000,000 12,000,000 0.28 C, 0.20-0.30 ...... SAE 1025 67,000 43,000 53,000 43,000 30,000,000 12,000,000 0.28 C, 0.30-0.40 ...... SAE 1035 70,000 46,000 56,000 46,000 30,000,000 12,000,000 0.28 C, 0.60-0.80 ...... 125,000 65,000 75,000 65,000 30,000,000 12,000,000 0.28 Nickel ...... SAE 2330 115,000 ...... 92,000 ...... 30,000,000 12,000,000 0.28 Cast iron: Gray ...... ASTM 20 20,000 80,000 27,000 ...... 15,000,000 6,000,000 0.26 Gray ...... ASTM 35 35,000 125,000 44,000 ...... 0.26 Gray ...... ASTM 60 60,000 145,000 70,000 ...... 20,000,000 8,000,000 0.26 Malleable ...... SAE 32510 50,000 120,000 48,000 ...... 23,000,000 9,200,000 0.26 Wrought iron ...... 48,000 25,000 38,000 25,000 27,000,000 ...... 0.28 Steel cast: Low C ...... 60,000 ...... 0.28 Medium C ...... 70,000 ...... 0.28 High C ...... 80,000 45,000 ...... 45,000 ...... 0.28 Aluminum alloy: Structural, No. 350 ...... 16,000 5,000 11,000 5,000 10,000,000 3,750,000 0.10 Structural, No. 17ST ...... 58,000 35,000 35,000 35,000 10,000,000 3,750,000 0.10 Brass: Cast ...... 40,000 ...... 0.30 Annealed ...... 54,000 18,000 ...... 18,000 ...... 0.30 Cold-drawn ...... 96,700 49,000 ...... 49,000 15,500,000 6,200,000 0.30 Bronze: Cast ...... 22,000 ...... 0.31 Cold-drawn ...... 85,000 ...... 15,000,000 6,000,000 0.31 Brick, clay...... ASTM ...... 1,500 3,000 ...... 0.72 Concrete 1:2:4 (28 days) ...... 2,000 ...... 3,000,000 ...... 0.087 Stone ...... 8,000 ...... 0.092 Timber ...... 300 4,840 860 550 ...... 1,280,000 0.015

*The ultimate strength in compression for ductile materials is usually taken as the yield point. The bearing value for pins and rivets may be much higher, and for structural steel is taken as 90,000 psi.Source: S.I. Heisler, The Wiley Engineer’s Desk Reference, 1984. Used with permission of John Wiley & Sons, New York.

Where: Where: ϭ LRMS ϭ RMS value of the load E Modulus of elasticity, 2 which can be in any unit, hp, amp, etc. lb/in. P ϭ Axial load, lb L1 ϭ Load during time of period 1

L2 ϭ Load during time of period 2, L ϭ Length of object, in. etc. A ϭ Area of object, in.2

t1 ϭ Duration of time for period 1 ⌬d ϭ Increase in length re-

t2 ϭ Duration of time for period 2, sulting from axial load, in. etc. General technical references 1. S.I. Heisler, The Wi- ley Engineer’s Desk Ref- erence, John S. Wiley & Sons, New York, 1984. 2. Hindehide, Zimmer- man, Design Funda- Reston, Va., 1984. mentals, John S. Wiley & Sons, 4. ASM Handbook of Engineering New York, 1983. Mathematics, American Society of Modulus of elasticity 3. K.M. Walker, Applied Mechanics Metals, Metals Park, Ohio, 1983. PL for Engineering , Third 5. The Smart Motion Cheat Sheet, E = (14) Edition, Reston Publishing Co. Inc., Amechtron Inc., Denton, Texas 1995. A⌬d

1997 Power Transmission Design A21 CONVERSION FACTORS Unless otherwise stated, pounds are U.S. avoirdupois, feet are U.S. standard, and seconds are mean solar. Multiply By To obtain

Length Angstrom units 3.937 ϫ 10-9 in. cm 0.3937 in. ft 0.30480 m in. (U.S.) 2.5400058 cm in. (British) 0.9999972 in. (U.S.) m 1010 Angstrom units m 3.280833 ft m 39.37 in. m 1.09361 yd m 6.2137 ϫ 10Ϫ4 miles (U.S. statute) yd 0.91440 m miles (U.S. statute) 5,280 ft

Area cir mils 7.854 ϫ 10Ϫ7 in.2 cm2 1.07639 ϫ 10Ϫ3 ft2 cm2 0.15499969 in.2 ft2 0.092903 m2 ft2 929.0341 cm2 in.2 6.4516258 cm2

Volume cm3 3.531445 ϫ 10Ϫ5 ft3 cm3 2.6417 ϫ 10Ϫ4 gal (U.S.) cm3 0.033814 oz (U.S. fluid) ft3 (British) 0.9999916 ft3 ft3 (U.S.) 28.31625 L (liter) m3 264.17 gal (U.S.) gal (British) 4,516.086 cm3 gal (British) 1.20094 gal (U.S.) gal (U.S.) 0.13368 ft3 (U.S.) gal (U.S.) 231 in.3 gal (U.S.) 3.78533 L (liter) gal (U.S.) 128 oz (U.S. fluid) oz (U.S. fluid) 29.5737 cm3 oz (U.S. fluid) 1.80469 in.3 yd3 0.76456 m3 yd3 (British) 0.76455 m3

Plane angle radian 57.29578 deg

Weight Dynes 2.24809 ϫ 10Ϫ6 lb kg 35.2740 oz (avoirdupois) kg 2.20462 lb kg 0.001 tons (metric) kg 0.0011023 tons (short) oz (avoirdupois) 28.349527 grams tons (long) 1,106 kg tons (long) 2,240 lb tons (metric) 1,000 kg tons (metric) 2,204.6 lb tons (short) 2,000 lb

A22 1997 Power Transmission Design Multiply By To obtain

Velocities feet/sec (fps) 0.68182 mph meters per sec 2.23693 mph rpm 0.10472 radians/sec mph 44.7041 cm/sec mph 1.4667 fps Temperature deg C ϭ 0.555 (deg F Ϫ 32) deg F = 1.8 (deg F) + 32

Pressure atmosphere 14.696 psi atmosphere 10,333 kg/m2 lb/ft2 4.88241 kg/m2 psi 70.307 grams/cm2 psi 703.07 kg/m2

Force Newton 0.22481 lb Newton 9.80 kg

Torque lb-in. 0.113 (Newton-meters) N-m lb-ft 1.356 N-m lb-ft 1.3558 ϫ 107 dyne-cm oz-in. 0.00706 N-m

Energy lb-in. 0.113 W-sec lb-in. 0.113 j (joule) Btu 251.98 calories Btu 1,055.06 j

Power gram-cm/sec 9.80665 ϫ 10Ϫ5 W hp 2,545.08 Btu (mean)/hr hp 550 lb-ft/sec hp 0.74570 kW hp 5,250 lb-ft/rpm

Inertia Mass inertia: lb-in.2 2.93 ϫ 10Ϫ4 kg-m2 oz-in.2 1.83 ϫ 10Ϫ5 kg-m2 kg-cm2 10Ϫ4 kg-m2 Weight inertia: lb-in.-sec2 1.13 ϫ 10Ϫ4 kg-m2 in-oz-sec2 7.06 ϫ 10Ϫ3 kg-m2 lb-ft-sec2 1.355 kg-m2

Source: S.I. Heisler, The Wiley Engineer’s Desk Reference, 1984. Used with permission of John Wiley & Sons, New York.

The Smart Motion Cheat Sheet, Brad Grant, Amechtron Inc., Denton, Texas.

1997 Power Transmission Design A23 MOTION CONTROL BASICS Nomenclature:

he first step in determining the ␣acc = Rotary acceleration, requirements of a motion-con- rad/sec2 T trol system is to analyze the e = Efficiency mechanics — including friction and Fl = Load force, lb

inertia — of the load to be positioned. Ff = Friction force, lb

Load friction can easily be determined Fpf = Preload force, lb either by estimating or by simply mea- g = Gravitational constant, suring with a torque wrench. 386 in./sec2

Inertia — the resistance of an ob- Figure 1 — Solid cylinder. Iacc = Current during ject to accelerate or decelerate — de- acceleration, A fines the torque required to accelerate Irms = Root-mean-squared a load from one speed to another, but current, A it excludes frictional . Inertia is J = Inertia, lb-in.-sec2

calculated by analyzing the mechani- Jls = Leadscrew inertia, cal linkage system that is to be lb-in.-sec2 2 moved. Such systems are categorized Jl = Load inertia, lb-in.-sec 2 as one of four basic drive designs: di- Jm = Motor inertia, lb-in.-sec 2 rect, gear, tangential, or leadscrew. Jt = Total inertia, lb-in.-sec In the following analyses of me- J = Pulley inertia, lb-in.-sec2 Figure 2 — Hollow cylinder. p chanical linkage systems, the equa- Kt = Torque constant, lb-in./A tions reflect the load parameters back L = Length, in. to the motor shaft. A determination of ␮ = Coefficient of friction what the motor “sees” is necessary for N = Gear ratio

selecting both motor and its control. Nl = Number of load gear teeth

Nm= Number of motor gear teeth Cylinder inertia 3 Motor Load p = Density, lb/in. The inertia of a cylinder can be cal- P= Pitch, rev/in. Figure 3 — Direct drive. Load is coupled culated based on its weight and ra- Pdel = Power delivered to the dius, or its density, radius, and directly to motor without any speed load, W changing device. length. Pdiss = Power (heat) dissipated Solid cylinder, Figure 1. by the motor, W Based on weight and radius: motor requirements. Pp = Total power, W − Example: If a cylinder is a lead- R = Radius, in. NNL NFL ×100 with a radius of 0.312 in. and a Ri= Inner radius, in.

NFL length of 22 in., the inertia can be cal- Rm = Motor resistance, ⍀

culated by using Table 1 and substi- Ro= Outer radius, in.

Based on density, radius, and tuting in equation 2: Sl = Load speed, rpm length: 4 Sm = Motor speed, rpm π ρ 4 π(22)(0.28)(0.312) πLρR4 = L R = tacc = Acceleration time, sec = J t = Deceleration time, sec J (2) 2g 2(386) dec 2g tidle = Idle time, sec Hollow cylinder, Figure 2. = 0.000237 lb- in.- sec2 trun = Run time, sec Based on weight and radius: T = Torque, lb-in. Tacc = Acceleration torque, lb-in. W Direct drive = 2 + 2 Tdec = Deceleration torque, lb-in. J (Ro Ri ) (3) Tf = Friction torque, lb-in. 2g The simplest drive system is a di- Tl = Load torque, lb-in.

rect drive, Figure 3. Because there are Tm = Motor torque, lb-in.

Based on density, radius, and no mechanical linkages involved. The Tr = Torque reflected to motor, length: load parameters are directly trans- lb-in.

mitted to the motor. The speed of the Trms = Root-mean-squared πLρ 4 4 J = (R − R ) (4) motor is the same as that of the load, torque, lb-in. 2g o i so the load friction is the friction the Trun = Running Torque, lb-in. With these equations, the inertia of motor must overcome, and load iner- Ts = Stall torque, lb-in.

mechanical components (such as tia is what the motor “sees.” There- Vl = Load speed, ipm shafts, , drive rollers) can be cal- fore, the total inertia is the load iner- W= Weight, lb

culated. Then, the load inertia and tia plus the motor inertia. Wlb= Weight of load plus belt, lb friction are reflected through the me- J = J + J (5) chanical linkage system to determine t l m

A24 1997 Power Transmission Design Gear drive equation 9. ders, 5-lb each, with an outer radius of 2.5 in. and an inner radius of 2.3 in. The mechanical linkages between = 0.031 = 2 To calculate the inertial for a hol- Jr 0.0034 lb- in.- sec the load and motor in a gear drive, 32 low, cylindrical pulley, substitute in Figure 4, requires reflecting the load equation 3: parameters back to the motor shaft. For high accuracy, the inertia of the As with any speed changing system, gears should be included when deter- = W 2 + 2 = 5 2 + 2 J p (Ro Ri ) (2.5 2.3 ) the load inertia reflected back to the mining total inertia. This value can be 2g 2(386) motor is a squared function of the obtained from literature or calculated speed ratio. using the equations for the inertia of a = 0.0747 lb- in.- sec2 cylinder. Gearing efficiencies should also be considered when calculating Substitute in equation 14 to deter- N m required torque values. mine load inertia:

2 Motor 2 10(2.5) Tangential drive = WR = Jl Consisting of a timing belt and pul- g 386 ley, chain and sprocket, or rack and Load pinion, a tangential drive, Figure 5, Total inertia reflected to the motor N l also requires reflecting load parame- shaft is the sum of the two pulley iner- ters back to the motor shaft. tias plus the load inertia: Figure 4 — Speed changer between load Motor speed: = + + and motor. Any speed changing device — J Jl J p1 J p2 gearing, belt, or chain — alters the = Vl Sm (11) = 0.1619 + 0.0747 + 0.0747 reflected inertia to the motor by the 2πR 2 square of the speed ratio. Load torque: = 0.3113 lb- in.- sec. = Motor speed: Tl Fl R (12) Also, the inertia of pulleys, sprock- S = S × N (6) ets or pinion gears must be included m l Friction torque: to determine the total inertia. or = Tf Ff R (13) × = Sl Nl Table 1—Material densities Sm (7) Load inertia: Nm W R2 Material Density, J = lb (14) Motor torque: l g lb per cu in. T Aluminum 0.096 T = l (8) Total inertia: m Ne Copper 0.322 W R2 0.040 J = lb + J + J + J (15) Steel 0.280 Reflected load inertia: t g p1 p2 m Wood 0.029 = Jl Example: A belt and pulley ar- Jr (9) N 2 rangement will be moving a weight of Table 2—Typical leadscrew 10 lb. The pulleys are hollow cylin- efficiencies Total inertia at motor: Type Efficiency J J = l + J (10) Load t 2 m N Ball-nut 0.90 Acme (plastic nut) 0.65 Example: To calculate the re- Acme (metal nut) 0.40 flected inertia for a 6-lb, solid cylinder with a 4-in. diameter, connected through a 3:1 gear set, first use equa- R Table 3—Leadscrew tion 1 to determine the load inertia. Pulley coefficients of friction 2 2 6(2) = WR = Motor Jl Steel on steel (dry) 0.58 2g 2(386) Steel on steel (lubricated) 0.15 = 0.031 lb- in.- sec2 Figure 5 — Tangential drive. The total Teflon on steel 0.04 To reflect this inertia through the load (belt plus load) is moved with a Ball bushing 0.003 gear set to the motor, substitute in arm with a radius, R.

1997 Power Transmission Design A25 Leadscrew drive equation for inertia of a cylinder: control system will either take too long to position the load, or it will be 4 Illustrated in Figure 6, a leadscrew π ρ 4 π(44)0.28(0.5) unnecessarily costly. = L R = drive also requires reflecting the load Jls In a basic motion-control system, parameters back to the motor. Both 2g 2(386) Figure 7, the load represents the me- the leadscrew and the load inertia = 0.00313 lb- in.- sec2 chanics being positioned. The load is have to be considered. If a leadscrew coupled or connected through one of inertia is not readily available, the Total inertia to be connected to the the mechanical linkages previously equation for a cylinder may be used. motor shaft is: described. = + = + The motor may be a traditional J Jl Jls 0.00052 0.00313 PMDC servo motor, a vector motor, or = 0.00365 lb- in.- sec2 a brushless servo motor. Motor start- ing, stopping and speed are dictated by the control unit which takes a low- level incoming command signal and Motor Load Motion control system amplifies it to a higher-power level for controlling the motor. Figure 6 — Leadscrew drive. Once the mechanics of the applica- The programmable motion con- tion have been analyzed, and the fric- troller is the brain of the motion con- tion and inertia of the load are known, trol system and controls the motor For precision positioning, the lead- the next step is to determine the control (amplifier). The motion con- screw may be preloaded to eliminate or torque levels required. Then, a motor troller is programmed to accomplish a reduce backlash. Such preload torque can be sized to deliver the required specific task for a given application. can be significant and must be in- torque and the control sized to power This controller reads a feedback sig- cluded, as must leadscrew efficiency. the motor. If friction and inertia are nal to monitor the position of the load. Motor speed: not properly determined, the motion By comparing a pre-programmed, = × Sm Vl P (16) Load torque reflected to motor: User’s AC power 1 F 1 F T = l + pf × µ (17) interface r 2π Pe 2π P Motion control For typical values of leadscrew effi- Power supply system ciency (e) and coefficient of friction (␮), see Tables 2 and 3. L o Friction force: Programmable Control a motion

= µ × controller (amplifier) Motor d Ff W (18) Encoder Friction torque: Speed and position feedback or resolver

1 F T = f (19) Figure 7 — Basic motion system. f 2π Pe

Total inertia:

2 2,000 W  1  J =   + J + J (20) t g  2πP  ls m

Example: A 200-lb load is posi- tioned by a 44-in. long leadscrew with a 0.5-in. radius and a 5-rev/in. pitch.

The reflected load inertia is: Speed, rpm

2 2 0.12 0.12 0.12 0.3 W  1  200  1  J =   =   l g  2πP  386  2π5 Acceleration Run Deceleration Idle t=0 Time, sec t=1 = 0.00052 lb- in.- sec2

Leadscrew inertia is based on the Figure 8 — Move profile.

A26 1997 Power Transmission Design “desired” position with the feedback = α + Control requirements Tacc Jt ( acc ) Tf (22) position, the controller can take ac- = + + α + tion to minimize an error between the (Jt Jls Jm ) acc Tf (23) Determining a suitable control actual and desired load positions. (amplifier) is the next step. The con- trol must be able to supply sufficient Movement profile Example: Our application, Figure acceleration current (Iacc), as well as 9, requires moving a load through a continuous current (Irms) for the appli- A movement profile defines the de- leadscrew. The load parameters are: cation’s duty-cycle requirements. sired acceleration rate, run time, Weight of load (Wlb) = 200 lb Required acceleration current that speed, and deceleration rate of the Leadscrew inertia (Jls) = 0.00313 must be supplied to the motor is: load. For example, suppose with a sys- lb-in-sec2 Tacc tem at rest (time =0, Figure 8), the mo- Friction torque (Tf) = 0.95 lb-in. I = (25) acc K tion controller issues a command to the Acceleration rate (␣acc) =1745.2 rad t motor (through the servo control) to per sec2. 13.75 = = 2.86 A start motion. At t=0, with full power- Typical motor parameters are: 4.8 supply voltage and current applied, Motor rotor inertia (Jm) = 0.0037 lb- the motor has not yet started to move. in.2

At this instant, there is no feedback Continuous stall torque (Ts) = 14.4 Current over the duty cycle, which signal, but the error signal is large. lb-in. the control must be able to supply to

As friction and torque are over- Torque constant (Kt) = 4.8 ;b-in./A the motor, is: come, the motor and load begin to ac- Motor resistance (Rm) = 4.5 ⍀ = Trms celerate. As the motor approaches the Acceleration torque can be deter- Irms (26) commanded speed, the error signal is mined by substituting in equation 23. Kt reduced and, in turn, voltage applied 7.73 = + + + = = 1.61 A to the motor is reduced. As the system Tacc (.00052 .00313 .0037)1745.2 .95 4.8 stabilizes at running speed, only nom- = 13.75 lb- in. inal power (voltage and current) are Power requirements required to overcome friction and windage. At t=1, the load position ap- Duty cycle torque The control must supply sufficient proaches the desired position and be- power for both the acceleration por- gins to decelerate. In addition to acceleration torque, tion of the movement profile, as well In applications with similar move the motor must be able to provide suf- as for the duty-cycle requirements. profiles, most of the input energy is ficient torque over the entire duty cy- The two aspects of power require- dissipated as heat. Therefore, in such cle or movement profile. This includes ments include (1) power to move the systems, the motor’s power dissipa- a certain amount of constant torque load, (Pdel) and (2) power losses dissi- tion capacity is the limiting factor. during the run phase, and a decelera- pated in the motor, (Pdiss). Thus, basic motor dynamics and tion torque during the stopping phase. Power delivered to move the load is: power requirements must be deter- Running torque is equal to friction T(S )(746) mined to ensure adequate power ca- torque (T ), in this case, 0.95 lb-in. = m f Pdel (27) pability for each motor. During the stopping phase, deceler- 63,025 Determining acceleration rate is ation torque is: the first step. For example, with a Power dissipated in the motor is a = − α + movement profile as shown in Figure Tdec Jt ( acc ) Tf (24) function of the motor current. Thus, 6, the acceleration rate can be deter- during acceleration, the value de- = −(.00052 + .00313 + .0037)1,745.2 + .95 mined from the speed and accelera- pends on the acceleration current tion time. (Dividing the motor speed = −11.85 lb- in. (Iacc); and while running, it is a func- expressed in rpm by 9.55 converts the tion on th rms current (Irms). There- speed to radians per second). fore, the appropriate value is used in Now, the root-mean-squared (rms) place of “I” in the following equation. S α = m (21) value of torque required over the = 2 acc Pdiss I (Rm ) (28) 9.55tacc movement profile can be calcuated:

2,000 2 2 2 2 α = = 1,745.2 rad/sec T (t )+ T (t )+ T (t ) The sum of (Pdel) and (Pdiss) deter- acc T = acc acc run run dec dec 9.55(0.12) rms + + + tacc trun tdec tidle mine total power requirements. Example: Power required during 2 + 2 + 2 = (13.75) (.12) (.95) (.12) (11.85) (.12) the acceleration portion of the move- + + + .12 .12 .12 .3 ment profile can be obtained by sub- = 7.73 lb- in. stituting in equations 27 and 28: Acceleration torque = 13.75(2,000) = The motor tentatively selected for Pdel (746) 325.5 W The torque required to accelerate this application can supply a continu- 63,025 the load and overcome mechanical ous stall torque of 14.4 lb-in., which is P = (2.86)2(4.5)(1.5) = 55.2 W friction is: adequate for the application. diss

1997 Power Transmission Design A27 = + 7.73(2,000) calls for peak power of 380.7 W and Pp Pdel Pdiss P = (746) = 182.9 W del 63.025 continuous power of 200.4 W. = 325.5 + 55.2 = 380.7 W To aid in selecting both motors and = 2 = Pdiss (1.61) (4.5)(1.5) 17.5 W controls (amplifiers), many suppliers = + = Pp 182.9 17.5 200.4 W offer computer software programs to Note: The factor of 1.5 in the Pdiss perform the iterative calculations calculation is a factor used to make In summary necessary to obtain the optimum mo- the motor’s winding resistance “hot.” The control selected must be capa- tor and control. This is a worst-case analysis, assum- ble of delivering (as a minimum) an ing the resistance is a 155 C. acceleration (or peak) current of 2.86 From articles by Baldor Electric, Continous power required for the A, and a continuous (or rms) current published in the September 1995 and duty cycle is: of 1.61 A. The power requirement March 1996 issues of PTD.

COMPUTER TERMINOLOGY

ere are some of the vocabu- of individuals who are signed into a E-mail — Electronic mail sup- lary used in digital commu- particular Chat Room. Usually they ported across the Internet. Requires H nication, whether it is be- have a posted subject for discussion. an address consisting of the Internet tween user and machine, machine Many times the same individuals re- name or number of the recipient at and machine, or user and user. They turn to the same room at the same the specific service provider. A Com- are words used by those involved with time each day. puServe address has a series of num- PCs, PLCs, and control devices. Connectivity — The ability to bers: [email protected]. No have one device connect, attach, or spaces are allowed between letters or BBS — Bulletin board systems of- communicate with another. numbers. fer forums, mostly local or regional. They support all types of communica- Data highway — Another term Fieldbus — A general term used tion, conversation, and postings. for bus or network. Also, a network to describe any bus that connects system created by Allen-Bradley. devices to microprocessor-based Bit — Symbolic representation of an controls. Synonymous with device- “On” or “Off” state of a device. In compu- Distributed — In communica- level bus, sensor- bus, mid- ter code, it is indicated as a “1” or a “0.” tions, a configuration where control, level bus. command execution ability, or intelli- Browsers — Graphics-based soft- gence (such as microprocessor intelli- File Transfer Protocol (FTP) — ware programs that let you reach a gence) is spread among two or more The means by which computer files variety of locations on the Internet devices. and software programs are trans- and move from one to the other (surf). ferred from a host computer to recipi- They also retrieve and display infor- Domain or zone — Part of an In- ent’s computers. mation, both text and graphics, from ternet address. It consists of a two or these locations. Examples include three-letter designation for the type Finger — Search capability for e- Netscape and Mosaic. of organization or geographical loca- mail addresses where location is tion, such as: known but e-mail address is not. Bus — Years ago, bus referred to the path or paths data traveled on the .com...Commercial Organizations Foundation fieldbus — A specifi- backplane of a computer board. The cation for process applications. definition is broadening to include .gov.....Government Departments data traveling within the physical .edu ...... Educational Institutions Gateway — Software on a board or medium of a few wires or cables. .mil...... Military chip that converts one protocol to another. Like converting a .net ...... Networking Units Byte — Eight 1s or 0s grouped to- DOS program to an Apple-based pro- gether, in any combination. Each .org...... Professional Societies gram. Sometimes gateways also con- group of eight bits represents an in- .US...... United States vert cable types. struction, a command, or datum. .JP ...... Japan Gopher — Menu-based system .UK or .GB...... United Kingdom Chat channels — Addresses on that helps find data residing on com- the Internet where real time “conver- .DE...... Germany puters at various locations. Some- sations” take place between groups times a long process down a series of

A28 1997 Power Transmission Design decision-tree paths. Network — All the cabling, World Wide Web — A rapidly wiring, and software parameters and expanding group of home pages that Hub — Hardware interface device control used to connect microproces- provide information on many sub- between different cable types. Con- sor-based devices over long distances. jects; individuals, companies, organi- zations, and governments. nects these cables together into a Distance is less of a factor now. Usually the address looks like: network. http://www.foggy.com/dognews Packet — Several bytes of data /~csmith/collars.html. There are no Integration — Sufficient commu- grouped together in a network spaces and every letter, number, and nication among devices, such that message. punctuation must be exact to reach performance is enhanced to a level the right address. Once there, the not possible as independent devices. Protocol — A specification that user can usually look at sub-pages or Devices become part of a whole as op- defines input signal levels, polari- be linked with other locations of posed to separate pieces of a system. ties, and speeds, and a device’s similar interest. output signals. Internet — Worldwide network of These definitions were taken from computers and computer networks. Search — Software pro- “PLCs bus into the future” (PTD July Begun by the Defense Department in grams that let the user find informa- 1995 p. 19) and “Internet for Engi- 1969 to ensure be- tion sites by definition, subject, or neers” (PTD Aug. 1996, p. 59) tween colleges and universities, con- even key words, then retrieve infor- , and government. Colleges mation from these sites, including and students have used it for a long text, graphics, and sound. Examples time, but companies and individuals include Yahoo, Lycos, and Alta Vista. jumped onboard only recently. Usenet — A worldwide bulletin Kbaud — A transmission rate of board divided into categories on one byte per second. which you can post news, make in- quiries, or add comments.

1997 Power Transmission Design A29