Drive Sizing Considerations

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INTRODUCTION In the application of industrial servo drives to machines it is important that the drive be large enough to meet all the load torque requirements and be stable. It is necessary to make sure that the servo drive torque rating is large enough to meet the load thrust and acceleration requirements plus the machine friction losses; an organized method to accomplish this is referred to as ‘‘sizing the drive.’’ There are many drive-sizing software programs available from commercial servo drive suppliers. It is of critical importance that machine designers size the servo drive in an organized engineering manner. Either manual...

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13 Drive Sizing Considerations 13.1 INTRODUCTION In the application of industrial servo drives to machines it is important that the drive be large enough to meet all the load torque requirements and be stable. It is necessary to make sure that the servo drive torque rating is large enough to meet the load thrust and acceleration requirements plus the machine friction losses; an organized method to accomplish this is referred to as ‘‘sizing the drive.’’ There are many drive-sizing software programs available from commercial servo drive suppliers. It is of critical importance that machine designers size the servo drive in an organized engineering manner. Either manual or computer software sizing must be used to avoid an unacceptable servo-drive performance. It is also a requirement that the software drive- sizing programs be documented and interactive with the user. Machine design engineers may not have a feedback control background; thus it is important that the programs have complete documentation (sometimes referred to as remark statements). To be of value to the user, the software programs should be interactive with enough description of what is happening during the drive-sizing process. The criteria of the drive sizing are different for hydraulic and electric drive sizing. Hydraulic servo drives usually have more than ample torque to Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved
meet the load requirements. However, the hydraulic ﬂuid has compressi- bility, which is like a spring inside the servo loop and can cause a minor loop instability. This hydraulic spring is represented as a hydraulic resonance. Any hydraulic servo drive with a large volume of hydraulic oil compressed between the actuator and the servo valve has a potential for an unstable minor loop servo drive. Hydraulic servo valve drives that use long-travel piston actuators have a great potential for servo instability. Likewise, hydraulic servo pump drives with large volumes of hydraulic ﬂuid between the pump and actuator are prone to servo instability. The hydraulic resonance is an indicator of what to expect for stability. An industrial index of performance (I.P.) for the hydraulic resonance is that the resonance should be 200 rad/sec or greater. The hydraulic resonance can be calculated from sﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 26b6D2 m oh ¼ ðrad=secÞ (13.1-1) Vc 6JT where: Dm ¼ motor displacement (in:3 =rad) Vc ¼ oil under compression (in:3 ) JT ¼ total inertia at the motor (lb-in.-sec2 ) b ¼ bulk modulus of oil ¼ 16105 lb=in:2 In general, hydraulic servo drives also have the added complexity of oil contamination, leakage, and changes in viscosity with temperature. Electric drives do not have the hydraulic medium problems of leakage, compressibility, etc. However, electric drives do have some problems obtaining sufﬁcient torque for load requirements. These torque limitations stem mostly from the ampliﬁer. DC silicon controlled rectiﬁer (SCR) ampliﬁers have the problem of excessive phase lag due to the circuit transport lag. Transistor design ampliﬁers have limits on available current. In general, electric servo drives have more than adequate capability for torque requirements and performance for industrial feed drive applications. Section 13.2 consists of a manual hydraulic servo drive-sizing form. This drive sizing will give assurance that the correct size hydraulic servo motor and hydraulic pressure will be used to meet the torque or thrust requirements for a machine feed drive. The hydraulic resonance is of particular importance. To have a stable drive the hydraulic resonance should be at least 200 rad/sec or higher. If this requirement is not met, the drive should be resized with a different ratio, lead, motor, etc. Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved
Section 13.3 consists of a manual electric drive-sizing form. This form can be used for DC or brushless DC servo drives. In addition, most commercial drive suppliers will provide CAD (computer-assisted design) sizing programs on computer disk. As with the hydraulic drive-sizing form, the goal is to provide a servo drive capable of providing enough torque and thrust to meet the load requirements of the machine feed drive. Not only must the drive be large enough to provide the required torque, but it must be stable. Other servo analysis techniques must be used to ascertain if the servo drive is stable. One such technique is the use of simulation software programs. The simulation programs must include the control, drive, and servo plant (machine). Such a machine simulation technique is discussed in Chapter 15. 13.2 HYDRAULIC DRIVE SIZING To illustrate the use of the hydraulic sizing form, it will be presented as applied to a typical machine axis. Blank hydraulic sizing forms are located in the Appendix. The illustrated hydraulic servo drive is a horizontal axis on a machine tool and has a positioning controller. The hydraulic motor is coupled to a ball bearing lead screw with a Thomas coupling. No belt ratio or gear ratio is used. This is called a direct drive. The axis slide uses Turcite way liners having a coefﬁcient of friction of 0.12 lbf/lb. As with all hydraulic servo drives, it is necessary to minimize the effect of the hydraulic spring to have a stable servo drive. Therefore it is a requirement to keep the hydraulic resonance above an I.P. of 200 rad/sec. The machine and servo speciﬁcations are: Weight of the load and machine slide 8000 lb Slide friction coefﬁcient 0.12 lbf/lb Maximum thrust required 10,000 lbf Length of the drive ball screw 90 in. Diameter of the ball screw 2 in. Lead of the ball screw 0.375 in./rev Traverse rate 180 ipm Direct drive (no ratio) Desired servo motor Hartman HT-50 Coupling Thomas 201DBZ Position-loop gain 2 ipm/mil or 34/sec Machine slide will be accelerated exponentially Servo valve Pegasus 160 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved
HYDRAULIC DRIVE SIZING Machine HBM, Axis Horiz., Traverse rate 180 (ipm) Ball screw: Length 90 (in.), Diameter 2.0 (in.), Lead 0.375 (in./rev) Gearbox # —, Ratio 1 Slide weight 8000 (lbs) Inertia calculations: Slide inertia at motor: Jslide ¼ WT ðlead=ratioÞ2 66:5610À5 (lb-in.-sec2 ) Jslide ¼ 8000 ð0 :375 =1Þ2 66 :5 610 À5 ¼ 0 :07312 lb-in :-sec 2 Ball screw inertia at motor: Jscr ¼ DIA4 6LGTH=ðratio2 Þ67:2610À5 (lb-in.-sec2 ) Jscr ¼ 2 4 690 =167 :2 610 À5 ¼ 0 :10368 lb-in :-sec 2 Coupling inertia ¼ 0.144 (lb-in.-sec2) Motor pulley inertia: Pulley number ¼ none Pulley inertia ¼ Screw pulley inertia: Pulley number ¼ none Pulley inertia ¼ Pulley inertias reﬂected to motor: Jp ¼ Jmot:pul: þ Jscrew pul: =Ratio2 (lb-in.-sec2 ) 2 Jp ¼ þ ¼ — = Total reﬂected inertia: Jref ¼ Jslide þ Jscr þ Jcoup þ Jp Jref ¼ 0 :321 (lb-in.-sec2 ) ¼ 0 :0731 þ 0 :103 þ 0 :144 þ — Total inertia: Jtot ¼ Jmot þ Jref ¼ 0 :0152 þ 0 :321 ¼ 0 :3362 (lb-in.-sec2 ) Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved
Coefﬁcient of friction (0.12 for Rulon) ¼ 0.12 (lbf/lb) (0.01 for rollers or hydrostatics) 1 Maximum motor speed ¼ Traverse (ipm)6 Lead 6Gear ratio ¼ VT ðrpmÞ 1 VT ¼ 180 6 61 ¼ 480 ðrpmÞ 0 :375 Drive motor chosen Hartman HT-50 Displacement DM ¼ 5 ðin:3 =revÞ61=6:28 ¼ 0 :796 (in:3 =rad) Torque constant KT ¼ 80 ðin:-lbÞ at 100 psi Torque available 1600 (in.-lbs) at 2000 psi Torque available 2400 (in.-lbs) at 3000 psi 1. Establish the maximum thrust required ¼ FM ¼ 10,000 (lbf). 2. Establish the no-load friction force ¼ FF . FF ¼ Coefficient of friction6slide weight FF ¼ 0 :12 68000 ¼ 960 ðlbf Þ No-load friction torque ¼ TF lead 1 TF ¼ no-load friction force ðFF Þ6 (lb-in.) 6 6:28 ratio 0 :375 1 TF ¼ 960 6 6 ¼ 57 :32 (lb-in.) 1 6:28 3. Stall thrust ¼ FS ¼ 1:56maximum thrust ðFM Þ (lbf). FS ¼ 1:5610,000 ¼ 15,000 ðlbf Þ 4. Stall thrust for efﬁciency ¼ FST ¼ stall90 eff ðFS Þ (lbf). thrust 0: 15,000 FST ¼ ¼ 16,666 (lbf) 0:9 5. Maximum motor torque ¼ TM ¼ max thrust ðFM Þ 6 6:28 6 ratio (lb-in.) lead 1 0:9 eff 10,000 0 :375 1 TM ¼ 6 ¼ 663 (lb-in.) (torque needed) 6 0 :9 1 6:28 Lead ¼ 0 :375 ðin=revÞ Ratio ¼ 1 Torque available ðat 2000 psiÞ ¼ 1600 (lb-in.) Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved
6. Drive acceleration. To avoid excessive acceleration forces, the servo drive position loop must control the acceleration. There are three methods that have been used to control acceleration: velocity ramp, velocity exponential, velocity ‘‘S’’ curve. The velocity exponential acceleration has a maximum acceleration (sometimes referred to as ‘‘jerk’’) at time zero from: Acceleration ¼ Kv 6 max velocity6eÀKv t Max velocity ¼ VT ðrpmÞ Kv ¼ position-loop gain ¼ ðipm=milÞ616:8 ¼ ð1=secÞ at t ¼ 0 6:28 6Kv ð1=secÞ6VT ðrpmÞ ¼ ðrad=sec2 Þ Accelerationmax ¼ 60 6:28 634 6480 ¼ 1708 ðrad=sec2 Þ ¼ Accelerationmax 60 This is a ‘‘worst-case’’ condition for acceleration. To avoid an initial maximum acceleration, the ‘‘S’’ curve velocity control can be used. The initial acceleration is zero. The maximum acceleration occurs when the rate of change in velocity is a maximum. The acceleration for the ‘‘S’’ curve is Traverse feed Acceleration ¼ 60 6 ðipmÞ6C 2 6t6eÀCt ðin:=sec2 Þ where C is a factor determining acceleration rate. Values of C for desired acceleration rates (in./sec2) are: Acceleration (in./sec2) C 9 5 18 10 27 15 36 20 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved
The maximum acceleration occurs at t ¼ 1=C . Final velocity ðVT Þ Ratio 6C 2 6t6eÀCt 66:286 (rad=sec2 ) Accelmax ¼ 60 Lead 6e ð Þ 6 66:286 ¼6 Accelmax 6 60 (rad=sec2 ) ¼ Ratio ¼ Lead ¼ (in:=rev) 7. Torque at maximum acceleration. TA ¼ JT 6accelerationmax Motor inertia ¼ 0 :0152 (lb-in.-sec2 ) Reflected load inertia ¼ 0 :32 (lb-in.-sec2 ) (lb-in.-sec2 ) JT ¼ total inertia at motor ¼ Jmotor þ Jload JT ¼ 0 :0152 þ 0 :32 ¼ 0 :3362 (lb-in.-sec2 ) TA ¼ 0 :3362 (lb-in.-sec2 )61708 (rad=sec2 ) ¼ 574 (lb-in.) 8. Torque requirement summary. Friction torque ðTF Þ ¼ 57 :32 (lb-in:) Thrust load ðTM Þ ¼ 663 (lb-in:) Acceleration torque ðTA Þ ¼ 574 (lb-in:) 9. Hydraulic pressure at the motor. torque6100 Pressure at motor ¼ ðpsiÞ motor torque constant ðKT Þ 100 PFR ¼ Pressure for friction load ðTF Þ ¼ 57 :32 6 KT ¼ 71:65 ðpsiÞ 100 PTH ¼ Pressure for thrust ðTM Þ ¼ 663 6 ¼ 828 ðpsiÞ KT 100 PA ¼ Pressure for max acceleration ðTA Þ ¼ 5746 KT ¼ 717 ðpsiÞ Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved
10. Hydraulic losses. Line loss ¼ 12% of system pressure; approx: 300 psi 300 ðpsiÞ Valve drop ðapprox: 1000 psi at rated flowÞ 1000 ðpsiÞ Pressure for friction load 72 ðpsiÞ No load motor loss at traverse ¼ KPM VT =1000 KPM ¼ 150 psi=1000 rpm ðmotor pressure drop no-loadÞ VT ¼ 480 ðrpmÞ 1506480 No load motor loss ¼ ¼ 72 ðpsiÞ 1000 Total losses 1443 ðpsiÞ 11. Supply pressure needed. (Note: Load pressure ¼ 2 supply pressure, or 1.5 6 load 3 pressure ¼ supply pressure) Pressure for friction ¼ friction psi ðPFR Þ þ losses ¼ 72 þ 1443 ¼ 1514 psi Pressure for max thrust ¼ 1:5 thrust psi ðPTH Þ þ 300 ¼ 828 þ 300 ¼ 1128 psi Pressure for max accel ¼ accel psi ðPA Þ þ losses ¼ 717 þ 1443 ¼ 2160 psi (Select a pressure greater than the highest of these pressures.) Recommended supply pressure ¼ 2500 ðpsiÞ 12. Motor size. Effective drive displacement (from thrust requirement) ¼ EDD Total stall thrust ðFST Þ 16,660 lbf ¼ 2500 psi Supply pressure ¼ 6 :6 in:3 =in: of drive movement Motor displacement ðfrom torque requirementÞ: Eff drive disp ðEDD Þ 6Lead Ratio ¼ motor displacement required ðin:3 =revÞ 6 :6 in:2 60 :375 in:=rev ¼ 2 :47 ðin:3 =revÞ Displacement of motor chosen DM ¼ 5 ðin:3 =revÞ Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved
If displacement of motor chosen is less than required, use the next larger motor. 13. Flow rating. Theoretical ﬂow at traverse ¼ QT (Select a servo valve larger than required ﬂow at traverse.) 1:3VT DM 1:36480 65 QT ¼ ¼ ¼ 13 :5 ðgpmÞ 231 231 VT ¼ max motor speed at traverse ¼ 480 ðrpmÞ DM ¼ motor displacement ¼ 5 ðin:3 =revÞ 14. Maximum accelerating torque TA (lb-in.). PS 6KT TA ¼ À TF 100 PS ¼ hydraulic power supply pressure ¼ 2500 ðpsiÞ KT ¼ motor torque constant ¼ 80 ðlb-in:=100 psiÞ tF ¼ friction torque ¼ 57 :32 ðlb-in:Þ 2500 680 TA ¼ À 57 :32 ¼ 1942 ðlb-in:Þ 100 15. Maximum decelerating torque TD . ðPs þ PR ÞKT TD ¼ þ TF 100 PS ¼ supply pressure ¼ 2500 ðpsiÞ PR ¼ relief pressure setting in manifold ¼ 0 ðpsiÞ ð2500 þ 0 Þ680 TD ¼ þ 57 :32 ¼ 2057 ðlb-in:Þ 100 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved
16. Hydraulic resonance. sﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2 BD2 M wH ¼ ðrad=secÞ Vc JT DM ¼ motor displacement ¼ 0 :796 ðin:3 =radÞ Vc ¼ oil under compression ¼ Vmotor þ Vvalve þ Vmanifold ¼ ðin:3 Þ 1 ð2 total volumeÞ Vvalve ¼ 0:17 in:3 ðapprox:Þ Vmanifold ¼ 0:3 in:3 ðapprox:Þ 2 :91 þ 0 :17 þ 0 :3 ¼ 3 :38 ðin:3 Þ JT ¼ total inertia at the motor ¼ 0 :3362 (lb-in.-sec2 ) B ¼ bulk modulus of oil ¼ 16105 ðlb=in:2 Þ rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 26105 60 :796 oh ¼ ¼ 374 ðrad=secÞ 3 :38 60 :3362 17. Hydraulic damping factor. rﬃﬃﬃﬃﬃﬃﬃﬃﬃ Lt BJT dH ¼ DM 2Vc Lt ¼ motor leakage þ valve leakage (in:3 =sec=psi) Valve leakage ¼ 0:0019 in:3 =sec=psi (approx:) Motor leakage ¼ 0 :00134 (in:3 =sec=psi) Lt ¼ 0 :00134 þ 0 :0019 ¼ 0 :0032 (in:3 =sec=psi) rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ﬃ 0 :0032 16105 60 :3362 dH ¼ ¼ 0 :28 0 :796 263 :38 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved
Typical motor speciﬁcations Compressed Torque Motor Max. Inertia oil constant Leakage (in.3/ Hartman displacement speed (lb-in. volume (in.-lb/ (in.3/rev) -sec2) (in.3) motor (rpm) 100 psi) sec-psi) HT-5 0.5 2000 0.00174 8 HT-10 1.0 2000 0.00198 0.77 16 0.00058 HT-22 2.2 1500 0.006 1.31 35 0.00065 HT-50 5.0 1500 0.0152 2.91 80 0.00134 HT-76 7.6 1200 0.0263 4.81 121 0.00167 Thomas coupling inertias J (lb-in.2) J (lb-in.-sec2) Coupling Used on motor 101DBZ 4.7 0.01216 126DBZ 10 0.02587 163DBZ 22 0.0569 HT22 201DBZ 56 0.1449 HT50 226DBZ 100 0.2587 263DBZ 210 0.5434 301DBZ 420 1.0869 351DBZ 950 2.4585 401DBZ 1090 2.8209 451DBZ 3500 9.0579 101DBZA 4.7 0.01216 126DBZA 10 0.02587 163DBZA 22 0.0569 201DBZA 56 0.1449 226DBZA 100 0.2587 263DBZA 220 0.5434 301DBZA 430 1.0869 351DBZA 970 2.4585 401DBZA 1900 2.8209 451DBZA 3500 9.0579 101DBZB 4.7 0.01216 HT10 126DBZB 11 0.02846 163DBZB 23 0.0592 201DBZB 57 0.1475 HT76 226DBZB 110 0.2846 263DBZB 220 0.5693 301DBZB 440 1.13871 351DBZB 1000 2.5879 401DBZB 1900 4.9171 451DBZB 3600 9.3167 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved
13.3 ELECTRIC DRIVE SIZING To illustrate the use of the drive-sizing form it will be presented as applied to a typical machine axis. Blank electric drive-sizing forms are located in the Appendix. The electric servo-drive design problem is for a horizontal axis on a large machine. The machine slide has Rulon way liners with a coefﬁcient of friction of 0.12 lbf/lb. The servo drive will be a brushless DC electric drive. The machine slide is large (50,000 lbs). This weight was selected to illustrate that the mechanical isolation of the ball screw lead and ratio will minimize the reﬂected inertia of this weight to the servo-motor drive shaft. The servo drive axis will be accelerated exponentially for this drive-sizing problem to illustrate an undersized servo drive. The drive-sizing problem will be followed by two computer drive-sizing program printouts. The ﬁrst drive- sizing printout (see Figure 1) will be for this problem using exponential acceleration. The second drive-sizing printout (see Figure 2) will be for the same problem, but using an ‘‘S’’ curve acceleration to illustrate the advantage of using this type of acceleration. The machine and servo have the following speciﬁcations: Machine Horizontal axis Axis Table Traverse rate 337 ipm Total axis weight 50,000 lbs Slide friction coefﬁcient 0.12 lbf/lb Required cutting thrust 8000 lbf Ball screw diameter 3 in. Ball screw length 70 in. Ball screw lead 0.375 in./rev Motor pulley (Woods pulley) 18H300 Screw pulley (Woods pulley) 60H300 Motor Kollmorgan M607-B Position loop gain 1 ipm/mil Find reﬂected inertias to the drive motor for the following: Machine slide Ball screw Pulley inertias Motor inertia Total reﬂected inertia at the drive motor Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved
Fig. 1 Printout for drive-sizing problem with exponential acceleration. Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved
Fig. 2 Printout for drive-sizing problem with ‘‘S’’ curve acceleration. Size the drive using the manual drive-sizing forms. Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved
INERTIA CALCULATIONS Slide inertia at the motor: lead 2 0:375 2 J slide ¼ weight 60:0000656 ¼ 50,000 60:0000656 3:333 N ¼ 0:0415 lb-in.-sec2 Ball screw inertia at the motor: diameter4 6lead60:000072 34 67060:000072 J screw ¼ ¼ N2 3:33332 ¼ 0:0367 lb-in.-sec2 Motor pulley inertia Pulley number ¼ 18H300 Pulley inertia ¼ 0.0136 lb-in.-sec2 For Woods pulleys 3- and 2-inch inertias, see Section 12.2 Screw pulley inertia Pulley number ¼ 60H300 Pulley inertia ¼ 0.8005 lb-in.-sec2 Pulley inertias reﬂected to the motor Jscrew pulley J pulley ¼ J motor pulley þ ratio2 0:8005 ¼ 0:0857 lb-in.-sec2 J pulley ¼ 0:0136 þ 3:3332 Total reflected inertia ¼ J slide þ J screw þ J pulley ¼ 0:0415 þ 0:0367 þ 0:0857 ¼ 0:1639 lb-in.-sec2 0:0196 Total inertia ¼ J total reflected þ J motor ¼ 0:1639 þ 12 2 ¼ 0:3511 lb-in.-sec ELECTRIC DRIVE SIZING Machine: Horizontal, Axis: Table, Traverse rate: 337 ipm. Ball screw lead: 0.375 in. rev., Length ¼ 70 in., Diameter ¼ 3 in., Gear box: — , Gear ratio ¼ 3.333 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved
Reﬂected load inertia ¼ 0.1639 lb-in.-sec2, Slide weight ¼ 50,000 lbs. Slide coefﬁcient of friction: 0.12 lbf/lb. 1. Establish the maximum load or working thrust. Load thrust ¼ 8000 lbf : 2. Establish no-load thrust. Thrustnl ¼ Coefficient of friction6slide weight Thrustnl ¼ 0:12650,000 ¼ 6000 lbf : 3. Torque requirements at the drive motor. (a) Geared Drives. NOTE: Use speed matching to select the ratio. Motor rated speed ½rpm6lead ½in. / rev Ideal ratio ¼ Traverse rate½ipm 0:375 Ideal ratio ¼ 30506 ¼ 3:394 337 Load thrust torque: lead 1 Motor load torque ¼ load thrust6 in.-lbs 6 2p gear ratio 0:375 1 Motor load torque ¼ 80006 ¼ 143 in.-lbs 6 2p 3:333 No-load torque: lead 1 Motor no-load torque ¼ no-load thrust6 6 2p gear ratio 0:375 1 Motor no-load torque ¼ 60006 ¼ 107 in.-lbs 6 2p 3:333 Total thrust torque: Total thrust torque ¼ load torque þ no-load torque ¼ 143 þ 107 ¼ 250 in.-lbs: Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved
(b) Direct Drives: Load thrust torque: lead Motor load torque ¼ load thrust6 2p Motor load torque ¼ 6¼ in.-lbs: 2p No-load torque: lead Motor no-load torque ¼ no-load thrust6 2p Motor no-load torque ¼ ¼ in.-lbs: 6 2p Total thrust torque: Total thrust torque ¼ load torque þ no-load torque ¼ þ ¼ in.-lbs: 4. Total drive motor torque required must be derated according to the ampliﬁer used. Ampliﬁer Derating Form Factor for Motor a. SCR ampliﬁers Single-phase full wave 1.66 Single-phase full wave/inductor 1.2 Three-phase half wave 1.25 Three-phase half wave/inductor 1.05 b. DC generator 1.0 c. PWM or brushless DC 1.0 Required torque ¼ total torque6form factor ¼ in.-lbs: ¼ 25061 ¼ 250 in.-lbs: 5. Motor traverse speed. Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved
(a) Geared Drives: 1 Maximum motor speed ¼ traverse ipm6 6gear ratio lead 1 Maximum motor speed ¼ 3376 63:333 ¼ 2995 rpm 0:375 (b) Direct Drives: 1 Maximum motor speed ¼ traverse ipm6 lead Maximum motor speed ¼ 61 ¼ rpm 6. Drive motor manufacturer selected: Kollmorgan Motor selected: M607B Rated speed of motor selected 3050 rpm Maximum speed of motor selected 3050 rpm Motor rated speed should be equal or larger than the traverse speed. If this requirement is not satisﬁed, a different gear ratio and/or drive screw lead should be selected to resize the motor for the required torque. 7. Torque for maximum acceleration (based on exponential response to a traverse rate step input of velocity). ipm 1000 mil min 1 Gain ¼ 1 ¼ 16:66 6 6 mil in: ! 60 sec sec ! 2p 1 rev a max: ¼ 6gain 6 max: motor speed 60 sec min ¼ rad/sec2 ¼ VF 6Kv [ ÀKv T 16:66 rad a max: ¼ 0:10456 62995 rpm ¼ 5214 sec2 sec Torquea ¼ J T 6a max : J T ¼ motor inertia + reflected load inertia J T 0:0156612 þ 0:1639 ¼ 0:3511 lb-in.-sec2 T a ¼ 0:3511 lb-in.-sec2 65214 rad/sec2 ¼ 1830 in.-lbs 8. Motor torque requirements summary. Torque for no-load 107 in.-lbs Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved
Torque for load thrust 143 in.-lbs Torque for maximum acceleration 1830 in.-lbs 9. Motor selected M607-B . Rated torque 33 6 12 ¼ 396 in.-lbs Maximum torque 93 6 12 ¼ 1116 in.-lbs Motor rated torque should be greater than required torque for total thrust. NOTE: Refer to motor curves. Motor maximum torque should be approximately the same or greater than the required torque for maximum acceleration ðTaÞ plus no-load thrust. Maximum torque ¼ 1116 in.-lb from motor curves at traverse 10. rpm. Imax xKT Ta ¼ À Tnoload ¼ torque available for acceleration form factor I max ¼ Motor armature maximum current ¼ 118 amps K t ¼ Motor torque constant ¼ 0:826612 ¼ 9:9 in.-lbs/amp T nl ¼ No-load torque ¼ 107 in.-lbs ! 118 Ta ¼ 69:9 À 107 ¼ 1061 in.-lb torque available 1 for acceleration torque available 1061 Allowable acceleration ¼ ¼a¼ total inertia 0:3511 rad rad ¼ 3022 À sec2 sec2 Maximum decelerating torque ¼ Td. 11. I ma 6K T Td ¼ þ T no load form factor I max ¼ Motor armature maximum current ¼ 118 amps ! 118 Td ¼ 69:9 þ 107 ¼ 1275 in.-lb torque available 1 for deceleration Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved
12. Motor time constants. Brushless DC Motors 67:7 v 1 volts Ke ¼ ¼ 0:646 6 1000 rev=min 2prad/rev6 min =60 sec rad/sec KeðlÀLÞ 0:646 K eðPHASEÞ ¼ Motor voltage constant ¼ ¼ ¼ 0:3734 1:73 1:73 volts-sec/rad (see motor data) K t ¼ Motor torque constant ¼ 9:9 in.-lb./amp (see motor data) RM ðLÀLÞ ¼ Motor resistance ¼ 0:14 ohms SRM ðLÀLÞ ¼ Total motor circuit resistance ¼ 1:35RM ðLÀLÞ ¼ 1:3560:14 ¼ 0:189 ohm SRM ðPHASEÞ ¼ SRM ðLÀLÞ 60:5 ¼ 0:18960:5 ¼ 0:0945 ohm LLÀL ¼ Motor inductance ¼ 0:0038 henries J m ¼ Motor armature inertia ¼ 0:0156612 ¼ 0:1872 lb-in.-sec2 J LOAD ¼ Reflected inertia to motor ¼ 0:1639 lb-in.-sec2 J Total ¼ J m þ J LOAD ¼ 0:1872 þ 0:1639 ¼ 0:3511 lb-in.-sec2 0:0038 LLÀL te ¼ Motor electrical time constant ¼ ¼ 0:189 SRM ðLÀLÞ ¼ 0:0201 sec 1 1 oe ¼ ¼ ¼ 49:7 rad/sec te 0:0201 SRM ðPHASEÞ J Total tm ¼ Motor mechanical time constant ¼ K eðPHASEÞ K t 0:094560:3511 tm ¼ ¼ 0:008975 sec 0:373469:9 1 om ¼ ¼ 111 rad/sec tm tm 0:008975 ¼ ¼ 0:4465 0:0201 te DC Motors K e ¼ Motor voltage constant ¼ volts/rad/sec (see motor data) K t ¼ Motor torque constant ¼ in.-lb/amp (see motor data) Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved