Review Questions: Computer Numerical Control [PDF]

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Ch07 CNC-4e-A&S 04-12, 04-17-2013



Chapter 7 COMPUTER NUMERICAL CONTROL REVIEW QUESTIONS 7.1



What is numerical control? Answer: As defined in the text, numerical control (NC) is a form of programmable automation in which the mechanical actions of a machine tool or other equipment are controlled by a program containing coded alphanumeric data.



7.2



What are the three basic components of an NC system? Answer: The three components are (1) part program of instructions, (2) machine control unit, and (3) processing equipment (e.g., machine tool) that accomplishes the operation.



7.3



What is the right-hand rule in NC and where is it used? Answer: The right-hand rule is used to distinguish positive and negative directions for the rotational axes in NC. Using the right hand with the thumb pointing in the positive linear axis direction (+x, +y, or +z), the fingers of the hand are curled in the positive rotational direction for the a, b, and c axes.



7.4



What is the difference between point-to-point and continuous path control in a motion control system? Answer: Point-to-point systems move the worktable to a programmed location without regard for the path taken to get to that location. By contrast, continuous path systems are capable of continuous simultaneous control of two or more axes, thus providing control of the tool trajectory relative to the workpart.



7.5



What is linear interpolation, and why is it important in NC? Answer: Linear interpolation is the capability to machine along a straight-line trajectory that may not be parallel to one of the worktable axes. It is important in NC because many workpiece geometries require cuts to be made along straight lines to form straight edges and flat surfaces, and the angles of the lines are not parallel to one of the axes in the coordinate system.



7.6



What is the difference between absolute positioning and incremental positioning? Answer: In absolute positioning, the workhead locations are always defined with respect to the origin of the NC axis system. In incremental positioning, the next workhead position is defined relative to the present location.



7.7



How is computer numerical control (CNC) distinguished from conventional NC? Answer: CNC is an NC system whose machine control unit is a dedicated microcomputer rather than a hard-wired controller, as in conventional NC.



7.8



Name five of the features of a computer numerical control that distinguish it from conventional NC. Answer: The features identified in Table 7.2 are (1) storage of more than one part program, (2) program editing at the machine tool, (3) fixed cycles and programming subroutines, (4) adaptive control, (5) interpolation, (6) positioning features for setup, (7) acceleration and 7-1



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Ch07 CNC-4e-A&S 04-12, 04-17-2013



deceleration calculations when the cutter path changes abruptly, (8) communications interface, and (9) diagnostics to detect malfunctions and diagnose system breakdowns. 7.9



What is distributed numerical control (DNC)? Answer: Distributed numerical control is a distributed computer system in which a central computer communicates with multiple CNC machine control units. It evolved from direct numerical control in which the central computer played the role of the tape reader, downloading part programs one block at a time. In a modern distributed NC system, entire part programs are downloaded to the MCUs. Also shop floor data is collected by the central computer to measure shop performance.



7.10 What are some of the machine tool types to which numerical control has been applied? Answer: NC has been applied to nearly all material-removal machine tools types, including lathes, boring mills, drill presses, milling machines, and cylindrical grinders. Other CNC metalworking machines mentioned in the text are punch presses, metal-bending presses, welding machines, thermal-cutting machines, tube-bending and wire-bending machines, and wire EDM machines. 7.11 What is a machining center? Answer: As defined in the text, a machining center is a machine tool capable of performing multiple machining operations on a single workpiece in one setup. The operations involve rotating cutters, such as milling and drilling, and the feature that enables more than one operation to be performed in one setup is automatic tool-changing. 7.12 Name six part characteristics that are most suited to the application of numerical control. Answer: The six part characteristics identified in the text are the following: (1) batch production, (2) repeat orders, (3) complex part geometry, (4) much metal needs to be removed, (5) many separate machining operations on the part, and (6) the part is expensive. 7.13 Although CNC technology is most closely associated with machine tool applications, it has been applied to other processes also. Name three examples. Answer: The non-metalworking NC applications listed in the text are (1) rapid prototyping and additive manufacturing, (2) water jet cutters and abrasive water jet cutters, (3) component placement machines, (4) coordinate measuring machines, (5) wood routers and granite cutters, (6) wood cutting lathes, (7) tape laying machines for polymer composites, and (8) filament winding machines for polymer composites. 7.14 What are four advantages of numerical control when properly applied in machine tool operations? Answer: The text lists the following advantages: (1) nonproductive time is reduced, (2) greater accuracy and repeatability, (3) lower scrap rates, (4) inspection requirements are reduced, (5) more-complex part geometries are possible, (6) engineering changes can be accommodated more gracefully, (7) simpler fixtures, (8) shorter manufacturing lead times, (9) reduced parts inventory, (10) less floor space, and (11) operator skill requirements are reduced. 7.15 What are three disadvantages of implementing NC technology? Answer: Four disadvantages are identified in the text: (1) higher investment cost because NC machines are more expensive than conventional machine tools, (2) higher maintenance 7-2 © 2015 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458



Ch07 CNC-4e-A&S 04-12, 04-17-2013



effort due to greater technological sophistication of NC, (3) part programming is required, and (4) equipment utilization must be high to justify the higher investment, and this might mean that additional work shifts are required in the machine shop. 7.16 Briefly describe the differences between the two basic types of positioning control systems used in NC? Answer: The two types of positioning control systems used in NC systems are open loop and closed loop. An open-loop system operates without verifying that the actual position achieved in the move is the same as the programmed position. A closed-loop system uses feedback measurements to confirm that the final position of the worktable is the location specified in the program. 7.17 What is an optical encoder, and how does it work? Answer: An optical encoder is a device for measuring rotational speed that consists of a light source and a photodetector on either side of a disk. The disk contains slots uniformly spaced around the outside of its face. These slots allow the light source to shine through and energize the photodetector. The disk is connected to a rotating shaft whose angular position and velocity are to be measured. As the shaft rotates, the slots cause the light source to be seen by the photocell as a series of flashes. The flashes are converted into an equal number of electrical pulses. 7.18 With reference to precision in a positioning system, what is control resolution? Answer: Control resolution is defined as the distance separating two adjacent addressable points in the axis movement. Addressable points are locations along the axis to which the worktable can be specifically directed to go. It is desirable for control resolution to be as small as possible. 7.19 What is the difference between manual part programming and computer-assisted part programming? Answer: In manual part programming, the programmer prepares the NC code using a lowlevel machine language. In computer-assisted part programming, the part program is written using English-like statements that are subsequently converted into the low-level machine language. 7.20 What is postprocessing in computer-assisted part programming and CAD/CAM part programming? Answer: Postprocessing converts the cutter location data and machining commands in the CLDATA file into low-level code that can be interpreted by the CNC controller for a specific machine tool. The output of postprocessing is a part program consisting of Gcodes, x-, y-, and z-coordinates, S, F, M, and other functions in word address format. A unique postprocessor must be written for each machine tool system. 7.21 What are some of the advantages of CAD/CAM-based NC part programming compared to computer-assisted part programming? Answer: The text lists the following advantages of CAD/CAM part programming: (1) the part program can be simulated off-line to verify its accuracy; (2) the time and cost of the machining operation can be determined; (3) the most appropriate tooling can be automatically selected for the operation; (4) the CAD/CAM system can automatically insert the optimum values for speeds and feeds; (5) in constructing the geometry or the tool 7-3 © 2015 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458



Ch07 CNC-4e-A&S 04-12, 04-17-2013



path, the programmer receives immediate visual feedback on the CAD/CAM monitor; (6) the CAD database containing the part design can be used to construct the tool path rather than redefining the part geometry; and (7) some of the steps in the tool path construction can be automated. 7.22 What is manual data input of the NC part program? Answer: Manual data input is when the machine operator manually enters the part program data and motion commands directly into the MCU prior to running the job.



PROBLEMS Answers to problems labeled (A) are listed in the Appendix at the back of the book. CNC Machining Applications 7.1



(A) A machinable grade of aluminum is to be milled on a CNC milling machine with a 25mm diameter four-tooth end mill. Cutting speed is 100 m/min and feed is 0.075 mm/tooth. To program the machine tool, convert these values to (a) rev/min and (b) mm/min, respectively.



v 100 = 1273 rev/min = π D 25π (10−3 ) (b) Feed rate in mm/min fr = Nntf = 1273(4)(0.075) = 382 mm/min Solution: (a) N = 7.2



A cast iron workpiece is to be face milled on a CNC machine using cemented carbide inserts. The cutter has 12 teeth and its diameter is 100 mm. Cutting speed is 180 m/min and feed is 0.08 mm/tooth. To program the machine tool, convert these values to (a) rev/min and (b) mm/min, respectively.



v 180 = 573 rev/min = π D 100π (10−3 ) (b) Feed in mm/min fr = Nntf = 573(12)(0.008) = 55 mm/min Solution: (a) N = 7.3



An end milling operation is performed on a CNC machining center. The total length of travel is 800 mm along a straight path. Cutting speed is 1.5 m/s and chip load is 0.09 mm. The end mill has two teeth and its diameter = 12.5 mm. Determine (a) feed rate in rev/min and (b) time to complete the cut.



v 1.5 = 38.2 rev/sec = 2292 rev/min = π D 12.5π (10−3 ) fr = Nntf = 2292(2)(0.09) = 412.6 mm/min L+ A (b) Tm = , where L = length of cut, A = allowance for overtravel (A = cutter diameter) fr 800 + 12.5 = 1.97 min Tm = 412.6 Solution: (a) N =



7.4



A turning operation is to be performed on a CNC lathe. Cutting speed = 2.2 m/s, feed = 0.25 mm/rev, and depth of cut = 3.0 mm. Workpiece diameter = 90 mm and length = 550 mm. Determine (a) rotational speed of the workpiece, (b) feed rate, and (c) time to travel from one end of the part to the other.



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Ch07 CNC-4e-A&S 04-12, 04-17-2013



2.2(60) = 467 rev/min 90π (10−3 ) (b) fr = Nntf = 467(0.25) = 116.7 mm/min Solution: (a) N =



(c) Tm = 550/116.7 = 4.71 min 7.5



A CNC drill press drills four 10.0 mm diameter holes at four locations on a flat aluminum plate in a production work cycle. Although the plate is only 12 mm thick, the drill must travel a full 20 mm vertically at each hole location to allow for clearance above the plate and breakthrough of the drill on the underside of the plate. Time to retract the drill from each hole is one-half the feeding time. Cutting speed = 0.5 m/sec and feed = 0.10 mm/rev. Coordinates of the hole locations are: Hole 1 at x = 25 mm, y = 25 mm; hole 2 at x = 25 mm, y = 150 mm; hole 3 at x = 150 mm, y = 150 mm; and hole 4 at x = 150, y = 25 mm. The drill starts out at point (0,0) and returns to the same position after the work cycle is completed. Travel rate of the table in moving from one coordinate position to another is 600 mm/min. Owing to acceleration and deceleration, and time required for the control system to achieve final positioning, a time loss of 3 sec is experienced at each stop of the table. All moves are made so as to minimize total cycle time. If loading and unloading the plate take 20 sec (total handling time), determine the time required for the work cycle.



0.5(60) = 955 rev/min 10π (10−3 ) Feed rate fr = Nf = 955(0.10) = 95.5 mm/min For each hole, feeding time = 20/95.5 = 0.21 min Retraction time of drill at each hole 0.21/2 = 0.105 min Total time/hole = 0.315 min For four holes, Tm = 4(0.315) = 1.26 min Solution: Drilling operations: N =



Workpart and axis system with tool path is shown below: y



(25,150) (150,150)



(25,25)



(150,25) x



(0,0) Total distance traveled =



(150,0)



252 + 252 + 125 + 125 + 125 +



Time to move between positions =



252 + 1502 = 562.43 mm



562.43 5(3) = 1.19 min + 600 60



Cycle time = Th + Tm + move time = 20/60 + 1.26 + 1.19 = 2.78 min Analysis of Open Loop Positioning Systems 7.6



(A) One axis of the worktable in a CNC positioning system is driven by a ball screw with a 7.5-mm pitch. The screw is powered by a stepper motor which has 200 step angles using a 3:1 gear reduction (three turns of the motor for each turn of the ball screw). The worktable is 7-5



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Ch07 CNC-4e-A&S 04-12, 04-17-2013



programmed to move a distance of 400 mm from its present position at a travel speed of 1200 mm/min. (a) How many pulses are required to move the table the specified distance? (b) What is the required motor rotational speed and (c) pulse rate to achieve the desired table speed? Solution: (a) Angular rotation of screw to move 300 mm As = 360(300)/6 = 18,000° Total angular rotation of motor to move 300 mm Am = 3(18,000) = 54,000° Number of motor revolutions = 54,000/360 = 150 rev Number of pulses to motor np = 150(200) = 30,000 pulses (b) Motor rotational speed Nm = (1200 mm/min)/(7.5 mm/rev) = 160 rev/min (c) Pulse frequency fp = (160 rev/min)(200 pulses/rev)/60 = 533.33 Hz 7.7



One axis of an open-loop positioning system is driven by a stepper motor, which is connected to a ball screw with a gear reduction of 2:1 (two turns of the motor for each turn of the screw). The ball screw drives the positioning table. Step angle of the motor is 3.6°, and pitch of the ball screw is 6.0 mm. The table is required to move along this axis a distance of 600 mm from its current position in exactly 25 sec. Determine (a) the number of pulses required to move the specified distance, (b) pulse frequency, and (c) rotational speed of the motor to make the move. Solution: (a) Number of step angles ns = 360/3.6 = 100 steps/rev = 100 pulses/rev Total angular rotation of screw to move 600 mm As = 360(600)/6 = 36,000° Total angular rotation of motor to move 600 mm Am = 2(36,000) = 72,000° Number of motor revolutions = 72,000/360 = 200 rev Number of pulses to motor np = 200(100) = 20,000 pulses (b) Pulse frequency fp = 20,000/25 = 800 pulses/sec = 800 Hz (c) Motor rotational speed Nm = 200 rev/25 sec = 8 rev/sec = 480 rev/min



7.8



A stepper motor with 50 step angles is coupled to a leadscrew through a gear reduction of 5:1 (5 rotations of the motor for each rotation of the leadscrew). The leadscrew has 1.25 threads/cm. The worktable driven by the leadscrew must move a distance = 40.0 cm at a feed rate = 90 cm/min. Determine (a) the number of pulses required to move the table, (b) required motor speed, and (c) pulse rate to achieve the desired table speed. Solution: (a) With 1.25 threads/cm, pitch p = 1/1.25 = 0.8 cm = 8.0 mm Angular rotation of screw to move 40 cm As = 360(400)/8 = 18,000° Total angular rotation of motor to move 40 cm Am = 5(18,000) = 90,000° Number of motor revolutions = 90,000/360 = 250 rev Number of pulses to motor np = 250(50) = 12,500 pulses (b) Motor rotational speed Nm = (900 mm/min)/(8 mm/rev) = 112.5 rev/min (c) Pulse frequency fp = (112.5 rev/min)(50 pulses/rev)/60 = 93.75 Hz



7.9



A component placement machine takes 0.5 sec to position a component onto a printed circuit (PC) board, once the board has been positioned under the placement head. The x-y table that positions the PC board uses a stepper motor directly linked to a ball screw for each axis (no gear reduction). Screw pitch = 5.0 mm. The motor step angle = 7.2°, and the pulse frequency = 400 Hz. Two components are placed on the PC board, one each at positions (25, 25) and (50, 150), where coordinates are mm. The sequence of positions is (0,0), (25, 25), (50, 150), 7-6



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Ch07 CNC-4e-A&S 04-12, 04-17-2013



(0,0). Time required to unload the completed board and load the next blank onto the machine table = 3.0 sec. Assume that 0.25 sec is lost due to acceleration and deceleration on each move. What is the hourly production rate for this PC board?



360 = 50 steps/rev 7.2 f p (400)(5.0) Given rg = 1.0, vt = p = = 40 mm/sec 50 ns Time to move from (0, 0) to (25, 25) = (25 mm)/(40 mm/sec) = 0.625 sec 150 − 25 Time to move from (25, 25) to (50, 150) = = 3.125 s 40 Time to move from (50, 150) to (0, 0) = (150 mm)/(40 mm/sec)= 3.75 s Cycle time Tc = 3.0 + (0.625+0.25+0.5) + (3.125+0.25+0.5) + (3.75+0.25) = 12.25 sec 60 × 60 Cycle rate (assumed equal to production rate) Rc = = 294 units/hr 12.25 Solution: ns =



7.10 Two stepper motors are used in an open loop system to drive the leadscrews for x-y positioning. The range of each axis is 550 mm. The shafts of the motors are connected directly to the leadscrews (no gear reduction). The leadscrew pitch is 5.0 mm, and the number of step angles on each motor is 120. (a) How closely can the position of the table be controlled, assuming there are no mechanical errors in the positioning system? (b) What are the required rotational speeds of each stepper motor and corresponding pulse train frequencies to drive the table at 300 mm/min in a straight line from point (x = 0, y = 0) to point (x = 330 mm, y = 220 mm)? Solution: (a) Table position can be controlled to (5 mm)/120 = 0.04167 mm (b) Travel speed vt = 300 mm/min from (x = 0, y = 0) to (x = 330 mm, y = 220 mm) Angle = tan-1(220/330) = 33.69° vtx = 300 cos 33.69 = 249.6 mm/min Nmx = rgvtx/p = 1(249.6)/5 = 49.92 rev/min fpx = Nmxns/60 = 49.92(120)/60 = 99.8 Hz vty = 300 sin 33.69 = 166.4 mm/min Nmy = rgvty/p = 1(166.4)/5 = 33.3 rev/min fpy = Nmyns/60 = 33.3(120)/60 = 66.6 Hz 7.11 (A) The two axes of an x-y positioning table are each driven by stepper motors connected to ball screws with a 4:1 gear reduction (four turns of the motor for each turn of the ball screw). The number of step angles on each stepper motor is 100. Each screw has a pitch = 7.5 mm and provides an axis range = 600.0 mm. There are 16 bits in each binary register used by the controller to store position data for the two axes. (a) What is the control resolution of each axis? (b) What are the required rotational speeds of each stepper motor and corresponding pulse frequencies to drive the table at 800 mm/min in a straight line from point (20, 20) to point (350, 450)? Solution: (a) CR1 = p/rgns = 7.5/(4 × 100) = 0.01875 mm CR2 = L/(2B – 1)= 400/(216 – 1) = 400/65,535 = 0.0061 mm CR = Max{0.01875, 0.00610} = 0.01875 mm (b) vt = 800 mm/min from (20, 20) to (300, 450) 7-7 © 2015 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458



Ch07 CNC-4e-A&S 04-12, 04-17-2013



Δx = 300 − 20 = 280 mm, Δy = 450 − 20 = 430 mm Angle A = tan-1(430/280) = 56.93° vtx = 800 cos 56.93 = 436.54 mm/min Nmx = rgvtx/p = 4(436.54)/7.5 = 232.8 rev/min fpx = Nmxns/60 = 232.8(100)/60 = 388 Hz vty = 800 sin 56.93 = 670.40 mm/min Nmy = rgvty/p = 4(670.4)/7.5 = 357.55 rev/min fpy = Nmyns/60 = 357.55(100)/60 = 595.9 Hz Analysis of Closed Loop Positioning Systems 7.12 In a CNC milling machine, the axis corresponding to the feed rate uses a dc servomotor as the drive unit and a rotary encoder as the feedback sensing device. The motor is geared to a leadscrew with a 10:1 reduction (10 turns of the motor for each turn of the leadscrew). If the leadscrew pitch is 6 mm, and the encoder emits 60 pulses per revolution, determine (a) the rotational speed of the motor and (b) pulse rate of the encoder to achieve a feed rate of 300 mm/min. Solution: (a) Nm = rgfr/p = 10(300)/6 = 500 rev/min (b) fp = frns/60p = 300(60)/60(6) = 50 Hz 7.13 A dc servomotor is used to drive one of the table axes of a CNC milling machine. The motor is coupled to a ball screw for the axis using a gear reduction of 8:1 (eight turns of the motor for each turn of the screw). The ball screw pitch is 7.5 mm. An optical encoder attached to the screw emits 120 pulses per revolution of the screw. The motor rotates at a top speed of 1000 rev/min. Determine (a) control resolution of the system, based on mechanical limits of each axis, (b) frequency of the pulse train emitted by the optical encoder when the servomotor operates at full speed, and (c) travel rate of the table at the top speed of the motor. Solution: (a) CR1 =



p = 7.5/(120×8) = 0.0078 mm ns rg



(b) Ns = Nm/rg = 1000/8 = 125 rev/min fp = nsNs = 120(125/60) = 250 Hz (c) vt = 60pfp/ns = 60(7.5)(250)/120 = 937.5 mm/min Check: vt = Nsp = 125(7.5) = 937.5 mm/min 7.14 (A) The worktable of a CNC machine tool is driven by a closed-loop positioning system which consists of a servomotor, leadscrew, and rotary encoder. The leadscrew pitch = 8 mm and is coupled directly to the motor shaft (gear ratio = 1:1). The encoder generates 200 pulses per leadscrew revolution. The table has been programmed to move a distance of 350 mm at a feed rate = 450 mm/min. (a) How many pulses are received by the control system to verify that the table has moved the programmed distance? What are (b) the pulse rate and (c) motor speed that correspond to the specified feed rate? Solution: (a) np = xns/p = 350(200)/8= 8750 pulses (b) fp = frns/60p = 450(200)/60(8) = 187.5 Hz (c) Nm = fr/p = 450/8 = 56.25 rev/min 7-8 © 2015 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458



Ch07 CNC-4e-A&S 04-12, 04-17-2013



7.15 A CNC machine tool table is powered by a servomotor, ball screw, and optical encoder. The ball screw has a pitch = 6.0 mm and is connected to the motor shaft with a gear ratio of 16:1 (16 turns of the motor for each turn of the screw). The optical encoder is connected to the ball screw and generates 120 pulses/rev of the screw. The table must move a distance of 250 mm at a feed rate = 300 mm/min. (a) Determine the pulse count received by the control system to verify that the table has moved exactly 250 mm. Also, what are (b) the pulse rate and (c) motor speed that correspond to the specified feed rate? Solution: (a) np = xns/p = 250(120)/6 = 5000 pulses (b) fp = fr ns /60p = 300(120)/60(6) = 100 Hz (c) Nm = rg fr /p = 16(300)/6 = 800 rev/min 7.16 A dc servomotor coupled to a leadscrew with a 4:1 gear reduction is used to drive one of the table axes of a CNC milling machine. The leadscrew has 1.5 threads/cm. An optical encoder attached to the leadscrew emits 100 pulses/rev. The motor rotates at a maximum speed of 800 rev/min. Determine (a) the control resolution of the system, expressed in linear travel distance of the table axis and (b) the frequency of the pulse train emitted by the optical encoder when the servomotor operates at maximum speed. Solution: (a) p = 1/1.5 = 0.6667 cm = 6.667 mm CR1 = p/nsrg = 6.667/(100×4) = 0.0167 mm (b) fr = Nmp/rg = 800(6.667)/4 = 1333.3 mm/min fp = frns/60p = 1333.3(100)/60(6.667) = 333.3 Hz 7.17 A milling operation is performed on a CNC machining center. Total travel distance = 430 mm in a direction parallel to one of the axes of the worktable. Cutting speed = 1.25 m/s and chip load = 0.05 mm. The end milling cutter has four teeth and its diameter = 20.0 mm. The axis uses a dc servomotor whose output shaft is coupled to a leadscrew with a 5:1 gear reduction (five turns of the motor for each turn of the leadscrew). The leadscrew pitch is 6.0 mm. An optical encoder which emits 80 pulses per revolution is attached to the leadscrew. Determine (a) feed rate and time to complete the cut, (b) rotational speed of the motor and (c) pulse rate of the encoder at the feed rate indicated. Solution: (a) Spindle N = v/πD = 1.25/π20(10-3) = 19.89 rev/s = 1193.4 rev/min fr = Nfnt = 1193.4(0.05)(4) = 238.7 mm/min Tm = (L + A)/fr, where A = allowance for tool overtravel (let A = tool diameter) Tm = (430 + 20)/238.7 = 1.89 min (b) Nm = rgfr /p = 5(238.7)/6 = 198.9 rev/min (c) fp = frns/60p = 238.7(80)/60(6) = 53.04 Hz 7.18 A dc servomotor drives the x-axis of a CNC milling machine table. The motor is coupled to a ball screw, whose pitch = 7.5 mm, using a gear reduction of 8:1 (eight turns of the motor to one turn of the ball screw). An optical encoder is connected to the ball screw. The optical encoder emits 75 pulses per revolution. To execute a certain programmed instruction, the table must move from point (x = 202.5 mm, y = 35.0 mm) to point (x = 25.0 mm, y = 250.0 mm) in a straight-line path at a feed rate = 300 mm/min. Determine (a) the control resolution of the system for the x-axis, (b) rotational speed of the motor, and (c) frequency of the pulse train emitted by the optical encoder at the desired feed rate. (d) How many pulses are emitted by the x-axis encoder during the move. 7-9 © 2015 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458



Ch07 CNC-4e-A&S 04-12, 04-17-2013



Solution: (a) CR1x = p/nsrg = 7.5/(75×8)= 0.0125 mm (b) Move from (202.5, 35.0) to (25.0, 250.0) at fr = 200 mm/min Δx = 25.0 − 202.5 = -177.5, Δy = 250.0 − 35.0 = 215.0 Angle A = tan-1(215/-177.5) = 129.54° frx = 300 cos 129.54 = 300(-0.6366) = -190.99 mm/min Nmx = rgfrx/p = 8(-190.99)/7.5 = -203.7 rev/min (c) fp = frxns/60p = 190.99(75)/60(7.5) = 31.83 Hz (d) Angle of rotation As = 360Δx/p = 360(177.5)/7.5 = 8520° np = Asns/360 = 8520(75)/360 = 1775 pulses Precision of Positioning Systems 7.19 (A) A two-axis positioning system uses a bit storage capacity of 16 bits in its control memory for each axis. To position the worktable, a stepper motor with step angle = 3.6° is connected to a leadscrew with a 6:1 gear reduction (six turns of the motor for each turn of the leadscrew). The leadscrew pitch is 7.5 mm. The range of the x-axis is 600 mm and the range of the y-axis is 500 mm. Mechanical accuracy of the worktable can be represented by a Normal distribution with standard deviation = 0.002 mm for both axes. For each axis of the positioning system, determine (a) the control resolution, (b) accuracy, and (c) repeatability. Solution: (a) For each axis, ns = 360/3.6 = 100 step angles CR1 = p/nsrg = 7.5/(100×6) = 0.0125 mm for both axes 600 600 x-axis: CR2 = 16 = = 0.0091 mm 2 − 1 65,536 − 1 CRx = Max {0.0125, 00091} = 0.0125 mm y-axis: CR2 = 500/65,535 = 0.0076 mm CRy = Max {0.0125, 00076} = 0.0125 mm (b) x-axis: Accuracy = CRx + 3σ = 0.0125/2 + 3(0.002) = 0.01225 mm y-axis: Accuracy = CRy + 3σ = 0.0125/2 + 3(0.002) = 0.01225 mm (c) Repeatability = ± 3σ = ± 3(0.002) = ± 0.006 mm 7.20 Stepper motors are used to drive the two axes of a component placement machine used for electronic assembly. A printed circuit board is mounted on the table and must be positioned accurately for reliable insertion of components into the board. Range of each axis = 700 mm. The lead screw used to drive each of the two axes has a pitch of 3.0 mm. The inherent mechanical errors in table positioning can be characterized by a Normal distribution with standard deviation = 0.005 mm. If the required accuracy for the table is 0.04 mm, determine (a) the number of step angles that the stepper motor must have, and (b) how many bits are required in the control memory for each axis to uniquely identify each control position. Solution: (a) Accuracy = CR/2 + 3σ = CR/2 + 3(0.005) = 0.04 (as specified) Assume CR = CR1 CR1/2 = 0.04 - 0.015 = 0.025 mm CR1 = 0.05 mm CR1 = p/nsrg Rearranging, ns = p/CR1, since rg = 1 ns = 3/0.05 = 60 steps/rev 7-10 © 2015 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458



Ch07 CNC-4e-A&S 04-12, 04-17-2013



(b) For the mechanical errors to be the limiting factor in control resolution, CR2 ≤ CR1. CR2 = L/(2B − 1) Rearranging, 2B − 1 = L/CR2 = 700/0.05 = 14,000 positions 2B = 14,001 B ln 2 = ln 14001 0.6931 B = 9.5469, B = 13.77 → 14 bits 7.21 An open loop positioning system uses a stepper motor connected to a ball screw with a 4:1 gear reduction (four turns of the motor for each turn of the ball screw). The stepper motor has a step angle of 7.2°. The ball screw pitch is 5 mm. Mechanical inaccuracies can be described by a normal distribution whose standard deviation = 0.005 mm. The range of the worktable axis is 500 mm. What is the minimum number of bits that the binary register must have so that the mechanical drive system becomes the limiting component on control resolution? Solution: Number of step angles ns = 360/7.2 = 50 CR1 = p/nsrg = 5/(50×4) = 0.025 mm For the mechanical errors to be the limiting factor in control resolution, CR2 ≤ CR1. CR2 = L/(2B − 1) Rearranging, 2B − 1 = L/CR2 = 500/0.025 = 20,000 positions 2B = 20,001 B ln 2 = ln 20001 0.6931 B = 9.9035 B = 14.29 → 15 bits 7.22 The positioning table for a component placement machine uses a stepper motor and leadscrew mechanism. The design specifications call for a table speed of 0.3 m/s and an accuracy of 0.05 mm. The pitch of the leadscrew is 8.0 mm, and there is no gear reduction. Mechanical errors in the motor, gear box, leadscrew, and table connection are characterized by a Normal distribution with standard deviation = 0.00333 mm. Determine (a) the minimum number of step angles in the stepper motor and (b) the frequency of the pulse train required to drive the table at the desired maximum speed. Solution: (a) Accuracy = CR/2 + 3σ 0.05 = CR/2 + 3(0.00333) = CR/2 + 0.01 0.05 − 0.01 = 0.040 = CR/2 CR = 0.080 mm Assume CR = CR1 CR1 = 0.080 = p/ns = 8/ns ns = 8/(0.08) = 100 step angles (b) fp = vtns/p = (0.3)(100)/(8.0×10-3) = 3,750 Hz 7.23 The two axes of an x-y positioning table are each driven by a stepper motor connected to a leadscrew with a 10:1 gear reduction. The number of step angles on each stepper motor is 60. Each leadscrew has a pitch = 6 mm and provides an axis range = 300 mm. There are 16 bits in each binary register used by the controller to store position data for the two axes. (a) What is the control resolution of each axis? (b) What are the required rotational speeds and corresponding pulse train frequencies of each stepper motor in order to drive the table at 500 mm/min in a straight line from point (30, 30) to point (100, 200)? Solution: (a) CR1 = p/rgns = 6/(10×60) = 0.01 mm 7-11 © 2015 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458



Ch07 CNC-4e-A&S 04-12, 04-17-2013



300 300 L = 16 = = 0.00458 mm 2 − 1 2 − 1 65,535 CR = Max{0.01, 0.00458} = 0.01 mm CR2 =



B



(b) vt = 500 mm/min from (30, 30) to (100, 200) Δx = 100 − 30 = 70 mm, Δy = 200 − 30 = 170 mm Angle A = tan-1(170/70) = 67.62° vtx = 500 cos 67.62 = 190.38 mm/min Nmx = rgvtx/p = 10(190.38)/6 = 317.3 rev/min fpx = Nmx ns/60 = 317.3(60)/60 = 317.3 Hz vty = 500 sin 67.62 = 462.34 mm/min Nmy = rgvty/p = 10(462.34)/6 = 770.6 rev/min fpx = Nmy ns /60 = 770.6(60)/60 = 770.6 Hz NC Manual Part Programming (Appendix A7) 7.24 Write the part program to drill the holes in the part in Figure P7.24. The part is 12.0 mm thick. Cutting speed = 100 m/min and feed = 0.06 mm/rev. Use the lower left corner of the part as the origin in the x-y axis system. Write the part program in the word address format using absolute positioning. The program style should be similar to Example A7.1. Solution: At the beginning of the job, the drill point will be positioned at a target point located at x = 0, y = 0, and z = + 10. The program begins with the tool positioned at this target point. Feed is given as 0.06 mm/rev. Rotational speed of drill is calculated as follows: N = v/πD = 100/(π×10×10-3) = 3183 rev/min NC part program code Comments N001 G21 G90 G92 X0 Y0 Z010.0; N002 G00 X040.0 Y025.0; N003 G01 G95 Z-20.0 F0.06 S3183 M03; N004 G01 Z010.0; N005 G00 Y100.0; N006 G01 G95 Z-20.0 F0.06; N007 G01 Z010.0; N008 G00 X100.0; N009 G01 G95 Z-20.0 F0.06; N010 G01 Z010.0; N011 G00 X160.0; N012 G01 G95 Z-20.0 F0.06; N013 G01 Z010.0; N014 G00 X125.0 Y060.0; N015 G01 G95 Z-20.0 F0.06; N016 G01 Z010.0; N017 G00 X200.0 Y040.0; N018 G01 G95 Z-20.0 F0.06; N019 G01 Z010.0; N020 G00 X0 Y0 M05; N021 M30;



Define origin of axes. Rapid move to first hole location. Drill first hole. Retract drill from hole. Rapid move to second hole location. Drill second hole. Retract drill from hole. Rapid move to third hole location. Drill third hole. Retract drill from hole. Rapid move to fourth hole location. Drill fourth hole. Retract drill from hole. Rapid move to fifth hole location. Drill fifth hole. Retract drill from hole. Rapid move to sixth hole location. Drill sixth hole. Retract drill from hole. Rapid move to target point, stop spindle. End of program, stop machine.



7.25 The part in Figure P7.25 is to be drilled on a turret-type drill press. The part is 15.0 mm thick. Three drill sizes will be used: 8 mm, 10 mm, and 12 mm, which are to be specified in the part program by tool turret positions T01, T02, and T03. All tooling is high speed steel. 7-12 © 2015 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458



Ch07 CNC-4e-A&S 04-12, 04-17-2013



Cutting speed = 75 mm/min and feed = 0.08 mm/rev. Use the lower left corner of the part as the origin in the x-y axis system. Write the part program in the word address format using absolute positioning. The program style should be similar to Example A7.1. Solution: At the beginning of the job, the drill point will be positioned at a target point located at x = 0, y = 0, and z = + 10. The program begins with the tool positioned at this target point. Feed is given as 0.08 mm/rev. Rotational speeds for the three drill diameters are calculated as follows: For the 8 mm drill, N = 75/(8π x 10-3) = 2984 rev/min For the 10 mm drill, N = 75/(10π x 10-3) = 2387 rev/min For the12 mm drill, N = 75/(12π x 10-3) = 1989 rev/min NC part program code



Comments



N001 G21 G90 G92 X0 Y0 Z010.0; N002 G00 X025.0 Y025.0 T01; N003 G01 G95 Z-20.0 F0.08 S2984 M03; N004 G01 Z010.0; N005 G00 X150.0; N006 G01 G95 Z-20.0 F0.08; N007 G01 Z010.0; N008 G00 X175.0; N009 G01 G95 Z-20.0 F0.08; N010 G01 Z010.0; N011 G00 X100.0 Y075.0 T02; N012 G01 G95 Z-20.0 F0.08; N013 G01 Z010.0; N014 G00 X050.0; N015 G01 G95 Z-20.0 F0.08; N016 G01 Z010.0; N017 G00 X050.0 Y075.0 T03; N018 G01 G95 Z-22.0 F0.08; N019 G01 Z010.0; N020 G00 X0 Y0 M05; N021 M30;



Define origin of axes. Rapid move to first hole location, 8 mm drill. Drill first hole. Retract drill from hole. Rapid move to second hole location. Drill second hole. Retract drill from hole. Rapid move to third hole location. Drill third hole. Retract drill from hole. Rapid move to fourth hole location, 10 mm drill. Drill fourth hole. Retract drill from hole. Rapid move to fifth hole location. Drill fifth hole. Retract drill from hole. Rapid move to sixth hole location, 12 mm drill. Drill sixth hole. Retract drill from hole. Rapid to target point, stop spindle rotation. End of program, stop machine.



7.26 The outline of the part in the previous problem is to be profile milled using a 30 mm diameter end mill with four teeth. The part is 15 mm thick. Cutting speed = 150 mm/min and feed = 0.085 mm/tooth. Use the lower left corner of the part as the origin in the x-y axis system. Two of the holes in the part have already been drilled and will be used for clamping the part during profile milling. Write the part program in the word address format with TAB separation and variable word order. Use absolute positioning. The program style should be similar to Example A7.2. Solution: As stated, two of the holes drilled in Problem 7.25 will be used to clamp the workpart for milling the outside edges. The part will be fixtured so that its top surface is 40 mm above the surface of the machine tool table, and the x-y plane of the axis system will be defined 40 mm above the table surface. As given, a 30 mm diameter end mill with four teeth will be used. The cutter is assumed to have a side tooth engagement length of 40 mm. Throughout the machining sequence the bottom tip of the cutter will be positioned 25 mm below the part top surface, which corresponds to z = -25 mm. Since the part is 15 mm thick, this z position will allow the side cutting edges of the milling cutter to cut the full thickness 7-13 © 2015 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458



Ch07 CNC-4e-A&S 04-12, 04-17-2013



of the part during profile milling. Cutting speed is specified as 150 m/min. Rotational speed of the cutter is N = 150/(30π x 10-3) = 1592 rev/min Given a feed f = 0.085 mm/tooth, feed rate is fr = 1592(4)(0.085) = 541 mm/min Cutter diameter data has been manually entered into offset register 05. At the beginning of the job, the cutter will be positioned so that its center tip is at a target point located at x = -50, y = -50, and z = + 10. The program begins with the tool positioned at this location. NC part program code



Comments



N001 G21 G90 G92 X-050.0 Y-050.0 Z010.0; N002 G00 Z-025.0 S1592 M03; N003 G01 G94 G42 Y0 D05 F541; N004 G01 X200.0; N005 G01 Y050.0; N006 G01 X150.0; N007 G17 G02 X125.0 Y075.0 R025.0; N008 G01 X125.0 Y100.0; N009 G01 Y025.0; N010 G01 X0 Y050.0; N011 G01 Y0; N012 G40 G00 X-050.0 Y-050.0 Z010.0 M05; N013 M30;



Define origin of axes. Rapid to cutter depth, turn spindle on. Bring tool to starting y-value, start cutter offset. Mill lower part edge. Mill right straight edge. Mill horizontal step above two 8 mm holes Circular interpolation around arc. Mill vertical step above arc. Mill top part edge. Mill angled edge at left of part. Mill vertical edge at left of part. Rapid to target point, cancel offset, spindle stop. End of program, stop machine.



7.27 The outline of the part in Figure P7.27 is to be profile milled, using a 20 mm diameter end mill with two teeth. The part is 10 mm thick. Cutting speed = 125 mm/min and feed = 0.10 mm/tooth. Use the lower left corner of the part as the origin in the x-y axis system. The two holes in the part have already been drilled and will be used for clamping the part during milling. Write the part program in the word address format with TAB separation and variable word order. Use absolute positioning. The program style should be similar to Example A7.2. Solution: As stated, the two holes will be used to clamp the workpart during milling. The part will be fixtured so that its top surface is 40 mm above the surface of the machine tool table, and the x-y plane of the axis system will be defined 40 mm above the table surface. As given, a 20 mm diameter end mill with two teeth will be used. The cutter is assumed to have a side tooth engagement length of 30 mm. Throughout the machining sequence the bottom tip of the cutter will be positioned 20 mm below the part top surface, which corresponds to z = -20 mm. Since the part is 10 mm thick, this z position will allow the side cutting edges of the milling cutter to cut the full thickness of the part during profile milling. Cutting speed is specified as 125 m/min. Rotational speed of the cutter is N = 125/(20π x 10-3) = 1989 rev/min Given a feed f = 0.10 mm/tooth, feed rate is fr = 1989(2)(0.10) = 398 mm/min Cutter diameter data has been manually entered into offset register 05. At the beginning of the job, the cutter is positioned so that its center tip is at a target point located at x = -50, y = 50, and z = + 10. The program begins with the tool positioned at this location. NC part program code



Comments



N001 G21 G90 G92 X-050.0 Y-050.0 Z010.0; N002 G00 Z-020.0 S1989 M03;



Define origin of axes. Rapid to cutter depth, turn spindle on.



7-14 © 2015 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458



Ch07 CNC-4e-A&S 04-12, 04-17-2013



N003 G01 G94 G42 Y0 D05 F398; N004 G01 X075.0; N005 G01 X150.0 Y043.02; N006 G01 Y070.0; N007 G01 X080.0; N008 G17 G02 X050.0 Y100.0 R030.0; N009 G01 Y125.0; N010 G01 X0; N011 G01 Y0 N012 G40 G00 X-050.0 Y-050.0 Z010.0 M05; N013 M30;



Bring tool to starting y-value, start cutter offset. Mill lower horizontal edge of part. Mill angled edge at 35 degrees. Mill vertical edge at right of part. Mill horizontal edge leading to arc. Circular interpolation around arc. Mill vertical step above arc. Mill top part edge. Mill vertical edge at left of part. Rapid to target point, cancel offset, spindle stop. End of program, stop machine.



7-15 © 2015 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458