The Basics of Motion Control—Part 1

ⅥPRECISION MOTION CONTROL The basics of motion control—Part 1

JOHN MAZURKIEWICZ, Baldor Electric Co. Nomenclature: Here are easy-to-follow equations for determining drive mechanics — friction and inertia — for any electromechanical positioning e = Efficiency Fl = Load force, lb application. Then, using this information, a following article will Ff = Friction force, lb show you how to determine motor and control requirements. Fpf = Preload force, lb g = Gravitational constant, 386 in./sec2 he first step in determining the re- Based on density, radius, and length: J = Inertia, lb-in.-sec2 quirements of a motion-control 2 πρLR4 Jls = Leadscrew inertia, lb-in.-sec system is to analyze the mechanics = 2 T J ()2 Jl = Load inertia, lb-in.-sec 2g 2 — including friction and inertia — of the Jm = Motor inertia, lb-in.-sec 2 load to be positioned. Load friction can Hollow cylinder, Figure 2. Jt = Total inertia, lb-in.-sec 2 easily be determined either by estimating Based on weight and radius: Jp = Pulley inertia, lb-in.-sec or by simply measuring with a torque L = Length, in. =+W 22 wrench. J ()RRoi ()3 ␮ = Coefficient of friction 2g Inertia — the resistance of an object N = Gear ratio to accelerate or decelerate — defines Based on density, radius, and length: Nl = Number of load gear teeth the torque required to accelerate a load πρL Nm = Number of motor gear teeth =−44 3 from one speed to another, but it ex- J ()RRoi ()4 p = Density, lb/in. cludes frictional forces. Inertia is calcu- 2g P= Pitch, rev/in. lated by analyzing the mechanical link- With these equations, the inertia of R = Radius, in. age system that is to be moved. Such mechanical components (such as shafts, Ri= Inner radius, in. systems are categorized as one of four ba- gears, drive rollers) can be calculated. Ro= Outer radius, in. sic drive designs: direct, gear, tangential, Then, the load inertia and friction are re- Sl = Load speed, rpm or leadscrew. flected through the mechanical linkage Sm = Motor speed, rpm In the following analyses of mechani- system to determine motor require- Tf = Friction torque, lb-in. cal linkage systems, the equations reflect ments. Tl = Load torque, lb-in. the load parameters back to the motor Example: If a cylinder is a leadscrew Tm = Motor torque, lb-in. shaft. A determination of what the motor with a radius of 0.312 in. and a length of Tr = Torque reflected to motor, “sees” is necessary for selecting both mo- 22 in., the inertia can be calculated by us- lb-in. tor and control. ing Table 1 and substituting in equation 2: Vl = Load velocity, ipm 4 W= Weight, lb πρLR4 π()()(22 0.. 28 0 312 ) Cylinder inertia J == Wlb= Weight of load plus belt, lb 2g 2() 386 The inertia of a cylinder can be calculated based on its weight and radius, or =0.000237 lb- in.- sec2 its density, radius, and length. Solid cylinder, Figure 1. Based on weight and radius: WR2 J = ()1 2g

John Mazurkiewicz is servo product man- Figure 1 — Solid cylinder. Figure 2 — Hollow cylinder. ager at Baldor Electric Co., Fort Smith, Ark.

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or Load × = Sl Nl Sm (7) Nm Motor torque: Motor Load T T = l (8) m Ne R Figure 3 — Direct drive. Load is Reflected load inertia: Pulley coupled directly to motor without any speed changing device. = Jl Motor Jr (9) N 2 Direct drive Total inertia at motor: Figure 5 — Tangential drive. The total The simplest drive system is a direct = Jl + load (belt plus load) is moved with a Jt Jm (10) drive, Figure 3. Because there are no me- N 2 lever arm with a radius, R. chanical linkages involved the load pa- Example: To calculate the reflected rameters are directly transmitted to the inertia for a 6-lb, solid cylinder with a 4- Friction torque: motor. The speed of the motor is the in. diameter, connected through a 3:1 T = F R (13) same as that of the load, so the load fric- gear set, first use equation 1 to determine f f tion is the friction the motor must over- the load inertia. Load inertia: 2 come, and load inertia is what the motor 2 W R “sees.” Therefore, the total inertia is the 2 6(2) J = lb (14) = WR = l g load inertia plus the motor inertia. Jl 2g 2(386) J = J + J (5) Total inertia: t l m = 2 0.031 lb- in.- sec W R2 Gear drive J = lb + J + J + J (15) To reflect this inertia through the gear t g p1 p2 m The mechanical linkage between the set to the motor, substitute in equation 9. Example: A belt and pulley arrange- load and motor in a gear drive, Figure 4, 0.031 ment will be moving a weight of 10 lb. The requires reflecting the load parameters J = = 0.0034 lb- in.- sec2 pulleys are hollow cylinders, 5-lb each, r 2 back to the motor shaft. As with any 3 with an outer radius of 2.5 in. and an in- speed changing system, the load inertia For high accuracy, the inertia of the ner radius of 2.3 in. reflected back to the motor is a squared gears should be included when determin- To calculate the inertial for a function of the speed ratio. ing total inertia. This value can be ob- hollow, cylindrical pulley, substitute in Motor speed: tained from literature or calculated using equation 3: S = S × N (6) the equations for the inertia of a cylinder. m l = W 2 + 2 = 5 2 + 2 Gearing efficiencies should also be con- J p (Ro Ri ) (2.5 2.3 ) sidered when calculating required torque 2g 2(386) values. = 2 N 0.0747 lb- in.- sec m Tangential drive Substitute in equation 14 to determine load inertia: Motor Consisting of a timing belt and pulley, 2 WR2 10(2.5) chain and sprocket, or rack and pinion, a J = = tangential drive, Figure 5, also requires l g 386 2 Load reflecting load parameters back to the = 0.1619 lb- in.- sec N l motor shaft. Total inertia reflected to the motor Figure 4 — Speed changer between Motor speed: shaft is the sum of the two pulley inertias load and motor. Any speed changing V plus the load inertia: S = l (11) device — gearing, belt, or chain — m 2πR J = J + J + J alters the reflected inertia to the l p1 p2 motor by the square of the speed ratio. Load torque: = 0.1619 + 0.0747 + 0.0747 = 2 Tl Fl R (12) = 0.3113 lb- in.- sec.

44 POWER TRANSMISSION DESIGN SEPTEMBER 1995 Table 1—Material densities

Material Density, lb per cu in. Aluminum 0.096 Copper 0.322 Plastic 0.040 Steel 0.280 Wood 0.029 Also, the inertia of pulleys, sprockets or pinion gears must be included to determine the total inertia. Table 2—Typical leadscrew efficiencies Leadscrew drive Type Efficiency Illustrated in Figure 6, a leadscrew drive also requires reflecting the load pa- Ball-nut 0.90 rameters back to the motor. Both the Acme (plastic nut) 0.65 leadscrew and the load inertia have to be Acme (metal nut) 0.40 considered. If a leadscrew inertia is not readily available, the equation for a cylinder may be used. Table 3—Leadscrew For precision positioning, the lead- coefficients of friction screw may be preloaded to eliminate or reduce backlash. Such preload torque Steel on steel (dry) 0.58 can be significant and must be included, Steel on steel as must leadscrew efficiency. (lubricated) 0.15 Motor speed: Teflon on steel 0.04 = × Sm Vl P (16) Ball bushing 0.003 Load torque reflected to motor: F   2   2 = 1 Fl + 1 pf × µ W 1 200 1 Tr (17) J =   =   2π Pe 2π P l g  2πP  386  2π5 For typical values of leadscrew effi- = 0.00052 lb- in.- sec2 ciency (e) and coefficient of friction (␮), see Tables 2 and 3. Leadscrew inertia is based on the Friction force: equation for inertia of a cylinder: F = µ × W (18) 4 f π ρ 4 π(44)0.28(0.5) = L R = Jls Friction torque: 2g 2(386) 1 F = f 2 Tf (19) = 0.00313 lb- in.- sec 2π Pe Total inertia: Total inertia to be connected to the   2 motor shaft is: = W 1 + + Jt   Jls Jm (20) = + = + g  2πP  J Jl Jls 0.00052 0.00313 = 0.00365 lb- in.- sec2 Example: A 200-lb load is positioned by a 44-in. long leadscrew with a 0.5-in. radius and a 5-rev/in. pitch. The reflected load inertia is: The next article will show you how to how to use this information to determine torque and power requirements for selecting motor and control. ■

Motor Load

Figure 6 — Leadscrew drive.

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