CH-7 Computer Numerical Control – Automation, Production Systems, and Computer-Integrated Manufacturing : Book Guide

This document contains Chapters 7 to 8 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 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 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 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. Answer: (b) Feed rate in mm/min fr = Nntf = 1273(4)(0.075) = 382 mm/min 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. Answer: (b) Feed in mm/min fr = Nntf = 573(12)(0.08) = 550 mm/min 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. Answer: 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. Answer: (a) N = 2.2(60)/90π (10-3) = 467 rev/min fr = Nntf = 467(0.25) = 116.7 mm/min 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. Answer: 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 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? Answer: (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 Motor rotational speed Nm = (1200 mm/min)/(7.5 mm/rev) = 160 rev/min 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. Answer: (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 Pulse frequency fp = 20,000/25 = 800 pulses/sec = 800 Hz 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. Answer: (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), (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? Answer: ns = 360/7.2 = 50 steps/rev Given rg = 1.0, vt = fpp/ns = (400)(5.0)/50 = 40 mm/sec Time to move from (0, 0) to (25, 25) = (25 mm)/(40 mm/sec) = 0.625 sec Time to move from (25, 25) to (50, 150) = (150 – 25)/40 = 3.125 s 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 Cycle rate (assumed equal to production rate) Rc = 60 × .25 = 294 units/hr 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)? Answer: (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)? Answer: (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) ∆x = 300 - 20 = 280 mm, ∆y = 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. Answer: (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. p Answer: (a) CR1 = p/nsrg = 7.5/(120 × 8) = 0.0078 mm (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? Answer: (a) np = xns/p = 350(200)/8 = 8750 pulses fp = frns/60p = 450(200)/60(8) = 187.5 Hz Nm = fr/p = 450/8 = 56.25 rev/min 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? Answer: (a) np = xns/p = 250(120)/6 = 5000 pulses fp = fr ns /60p = 300(120)/60(6) = 100 Hz 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. Answer: (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. Answer: (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 Nm = rgfr /p = 5(238.7)/6 = 198.9 rev/min 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. Answer: (a) CR1x = p/nsrg = 7.5/(75 × 8) = 0.0125 mm Move from (202.5, 35.0) to (25.0, 250.0) at fr = 200 mm/min ∆x = 25.0 - 202.5 = -177.5, ∆y = 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 fp = frxns/60p = 190.99(75)/60(7.5) = 31.83 Hz Angle of rotation As = 360∆x/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. Answer: (a) For each axis, ns = 360/3.6 = 100 step angles CR1 = p/nsrg = 7.5/(100 × 6) = 0.0125 mm for both axes x-axis: CR2 = 600/ 216 – 1 = 600/65,536 – 1 = 0.0091 mm 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 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 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. Answer: (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 (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? Answer: 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. Answer: (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)? Answer: (a) CR1 = p/rgns = 6/(10 × 60) = 0.01 mm CR2 = L/2B - 1 = 300/216 – 1 = 300/65,535 = 0.00458 mm CR = Max{0.01, 0.00458} = 0.01 mm (b) vt = 500 mm/min from (30, 30) to (100, 200) ∆x = 100 - 30 = 70 mm, ∆y = 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 = 70.6 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. Answer: 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
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. 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. Answer: 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 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; Comments 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. Answer: 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 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 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; Comments 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. Answer: As stated, the two holes will be used to clamp the work part 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. Chapter 8 INDUSTRIAL ROBOTICS REVIEW QUESTIONS 8.1 What is an industrial robot? Answer: The definition given in ISO 8373 is the following: An industrial robot is “an automatically controlled, reprogrammable, multipurpose manipulator programmable in three or more axes, which may be either fixed in place or mobile for use in industrial automation applications.” 8.2 What was the first application of an industrial robot? Answer: According to Historical Note 8.1, the first application of an industrial robot was in a die casting operation at a General Motors plant in 1961. 8.3 What are the five joint types used in robotic arms and wrists? Answer: The five joint types are the following: (1) Linear joint (type L joint), in which the relative movement between the input link and the output link is a translational sliding motion, with the axes of the two links being parallel; (2) orthogonal joint (type O joint), which is also a translational sliding motion, but the input and output links are perpendicular to each other during the move; (3) rotational joint (type R joint), which provides rotational relative motion, with the axis of rotation perpendicular to the axes of the input and output links; (4) twisting joint (type T joint), which also involves rotary motion, but the axis of rotation is parallel to the axes of the two links; and (5) revolving joint (type V joint), in which the axis of the input link is parallel to the axis of rotation of the joint, and the axis of the output link is perpendicular to the axis of rotation. 8.4 Name the common body-and-arm configurations identified in the text. Answer: The common body-and-arm configurations are (1) articulated robot, (2) polar configuration, (3) SCARA, which stands for Selective Compliance Assembly Robot Arm, (4) Cartesian coordinate robot, and (5) delta robot. 8.5 What is the work volume of a robot manipulator? Answer: The work volume of a robot manipulator is the envelope or three-dimensional space within which the robot can manipulate the end of its wrist. 8.6 Robotic sensors are classified as internal and external. What is the distinction? Answer: Internal sensors are components of the robot and are used to control the positions and velocities of the various joints of the robot. These sensors form a feedback control loop with the robot controller. External sensors are additional components in the cell (external to the robot). They are used to coordinate the operation of the robot with the other equipment in the cell. 8.7 What is a playback robot with point-to-point control? Answer: Playback control means that the robot controller has a memory to record the sequence of motions in a given work cycle as well as the locations and other parameters (such as speed) associated with each motion and then to subsequently play back the work cycle during execution of the program. 8.8 What is an end effector? Answer: An end effector is the special tool that is attached to the robot’s wrist that enables the robot to perform a given task. End effectors are either grippers (to grasp parts) or tools (e.g., spot welding gun). 8.9 In a machine loading and unloading application, what is the advantage of a dual gripper over a single gripper? Answer: With a single gripper, the robot must reach into the production machine twice, once to unload the finished part and place it outside the machine, and then to pick up the next part and load it into the machine. With a dual gripper, the robot picks up the next part while the machine is still processing the current part; when the machine cycle is finished, the robot reaches into the machine only once: to remove the finished part and then load the next part. This reduces the cycle time per part. 8.10 What are the general characteristics of industrial work situations that tend to promote the substitution of robots for human workers? Answer: The general characteristics of industrial work situations that tend to promote the substitution of robots for human workers are (1) hazardous work for humans, (2) repetitive work cycle, (3) difficult handling for humans, (4) multi-shift operation, (5) infrequent changeovers, and (6) part position and orientation are established in the work cell. 8.11 What are the three categories of robot industrial applications, as identified in the text? Answer: The three categories are (1) material handling, (2) processing, and (3) assembly and inspection. 8.12 What is a palletizing operation? Answer: Palletizing is a material handling application in which the robot must retrieve parts, cartons, or other objects from one location and deposit them onto a pallet or other container at multiple positions on the pallet. 8.13 What is a robot program? Answer: A robot program can be defined as a path in space to be followed by the manipulator, combined with peripheral actions that support the work cycle. 8.14 What is the difference between powered leadthrough and manual leadthrough in robot programming? Answer: The difference between powered leadthrough and manual leadthrough is the manner in which the manipulator is moved through the motion cycle during programming. Powered leadthrough involves the use of a teach pendant with toggle switches and/or contact buttons for controlling the movement of the manipulator joints. Using the toggle switches or buttons, the programmer power drives the robot arm to the desired positions, in sequence, and records the positions into memory. Manual leadthrough requires the operator to physically grasp the end-of-arm or the tool that is attached to the arm and move it through the motion sequence, recording the path into memory. 8.15 What is control resolution in a robot positioning system? Answer: As defined in the text, control resolution refers to the capability of the robot’s positioning system to divide the range of the joint into closely spaced points to which the joint can be moved by the controller. 8.16 What is the difference between repeatability and accuracy in a robotic manipulator? Answer: Repeatability is the robot’s ability to position its end-of-wrist at a previously taught point in the work volume. It is concerned with the errors associated with the robot’s attempts to return to the same addressable point and is usually defined as ± 3 standard deviations of the mechanical errors of the manipulator. Accuracy is the robot’s ability to position the end of its wrist at a desired location in the work volume. It is concerned with the errors associated with the robot’s attempts to move to a specified location that may not be located at an addressable point. PROBLEMS Answers to problems labeled (A) are listed in the Appendix at the back of the book. Robot Anatomy 8.1 Using the joint notation system for defining manipulator configurations (Section 8.1.2), sketch diagrams (similar to Figure 8.1) of the following robots: (a) TRT, (b) VVR, (c) VROT. Answer: 8.2 Using the joint notation system for defining manipulator configurations (Section 8.1.2), sketch diagrams (similar to Figure 8.1) of the following robots: (a) TRL, (b) OLO, (c) LVL. Answer: 8.3 Using the joint notation system for defining manipulator configurations (Section 8.1.2), sketch diagrams (similar to Figure 8.1) of the following robots: (a) TRT:R, (b) TVR:TR, (c) RR:T. Answer: 8.4 Using the robot configuration notation scheme discussed in Section 8.1, write the configuration notations for some of the robots in your laboratory or shop. Answer: Answer depends on robots in the laboratory or shop of interest. 1. Robot Arm: •Base Joint: R1 •Shoulder Joint: R2 •Elbow Joint: R3 •Wrist Pitch Joint: R4 •Wrist Roll Joint: R5 Configuration Notation: R1 R2 R3 R4 R5 2. Cartesian Robot: •X-axis Position: X1 •Y-axis Position: X2 •Z-axis Position: X3 •Roll: X4 •Pitch: X5 •Yaw: X6 Configuration Notation: X1 X2 X3 X4 X5 X6 3. SCARA Robot: •Base Rotation: S1 •Arm Extension: S2 •Elbow Rotation: S3 •Wrist Rotation: S4 Configuration Notation: S1 S2 S3 S4 4. Delta Robot: •Lower Arm Angle: D1 •Upper Arm Angle: D2 •Vertical Arm Angle: D3 Configuration Notation: D1 D2 D3 These notations provide a concise representation of the joint configurations or positions of different types of robots in the laboratory or shop. 8.5 Describe the differences in orientation capabilities and work volumes for a TR and a RT wrist assembly. Use sketches as needed. Answer: The work volume of the TR wrist is the intersection of a sphere and a cone. The work volume of the RT wrist is a circular arc, provided by the R joint, with the capability to rotate around the line established by the angle of the R joint. Cycle Time and Cost Analysis 8.6 (A) An articulated robot loads and unloads a CNC machine tool. The cell is scheduled to produce a batch of 300 parts. Setting up the cell for this part style takes 30 min, programming the robot takes 75 min, and programming the CNC machine tool takes 55 min. For safety reasons, these three setup activities must be done sequentially. The programmed machining cycle takes 3.75 min. Cutting tools wear out and must be periodically changed, which takes 6.0 min every 20 cycles and is performed by a human worker. At the end of each machining cycle, the robot reaches into the machine and removes the just-completed part, places it in a parts storage carousel, then reaches for a starting work part from the same carousel and places it in the machine tool fixture. This sequence of handling activities takes 45 sec. The storage carousel holds 25 parts. At the beginning of the production run, it is full of raw work parts. As each part is retrieved by the robot, the carousel indexes one position to present a new raw part. The robot is programmed to place the completed part in the empty position in the carousel and take the next raw part. Periodically, workers visit the carousel to collect finished parts and replace them with starting work parts. This is accomplished without loss of production time. Determine (a) the average production time, (b) production rate of the cell, and (c) how many hours are required to complete the production run. (d) What is the proportion of the total time that the robot is working? Answer: (a) Changing cutting tools involves lost production time. Per cycle, Tt = 6.0/20 = 0.30 min Operation cycle time Tc = To + Th + Tt = 3.75 + 45/60 + 0.30 = 4.80 min/pc Setup time Tsu = 30 + 75 + 55 = 160 min Average production time Tp = (160 + 300 × 4.80)/300 = 5.333 min/pc Production rate Rp = 60/5.333 = 11.25 pc/hr Hours required to produce 300 parts TT = 300(5.333) = 1600 min = 26.667 hr The robot works 45 sec = 0.75 min during the production of each part. For 300 parts, that is 300(0.75) = 225 min = 3.75 hr. The proportion of total time including setup = 3.75/26.667 = 0.1406 = 14.06% The proportion of time not including setup = 0.75/4.80 = 0.1563 = 15.63% 8.7 Solve the previous problem, except that the following changes apply: (1) Setting up the cell, programming the robot, and programming the CNC machine can be performed at the same time rather than sequentially; (2) an automatic tool changer and tool storage unit are used so worn tools can be exchanged from the tool storage unit with no lost production time; and (3) a dual gripper is used rather than a single gripper, which reduces the part handling time from 45 sec to 15 sec. Answer: (a) Changing cutting tools involves no lost production time, so Tt = 0 Operation cycle time Tc = To + Th + Tt = 3.75 + 15/60 + 0 = 4.00 min/pc Setup time Tsu = Max{30, 75, 55} = 75 min Average production time Tp = (75 + 300 × 4.00)/300 = 4.25 min/pc Production rate Rp = 60/4.25 = 14.12 pc/hr Hours required to produce 300 parts TT = 300(4.25) = 1275 min = 21.25 hr Comment: Note that the gain in productivity compared to the previous problem has been achieved by reducing the non-productive elements of the job. The machining cycle remains the same 3.75 min The robot works 15 sec = 0.25 min during the production of each part. For 300 parts, that is 300(0.25) = 75 min = 1.25 hr. The proportion of total time including setup = 1.25/21.25 = 0.0588 = 5.88% The proportion of time not including setup = 0.25/4.00 = 0.0625 = 6.25% Comment: The use of the dual gripper has reduced proportion of time that the robot is working. This might suggest that the robot is underutilized, but the bottom line is that productivity has been increased by using the dual gripper. 8.8 A large overhead gantry robot loads and unloads three CNC lathes in an automated work cell. The three lathes are dedicated to the mass production of the same part style, so their semi-automatic turning cycles are the same: 3.30 min. Assume setup time can be ignored. The cell includes an automated parts storage unit from which the robot retrieves starting work units and deposits finished parts using a dual gripper. The storage unit has an indexing system that presents starting work units at one position and accepts finished parts at another position. To perform the loading and unloading sequence for each lathe requires 48 sec of the robot’s time; however, only 18 sec of lost production time are experienced by each lathe. The loading and unloading sequence for the three lathes is synchronized so there is no machine interference. The tooling of each lathe must be changed once each hour, and this takes 6.0 min, which is lost production time. Determine (a) the average production time for each lathe, (b) production rate of the cell, and (c) the proportion of time that the robot is working? Answer: (a) Changing cutting tools takes 6 min each hour. That leaves 54 min of production time. Without tool change time, Tc = To + Th = 3.30 + 18/60 = 3.60 min/pc Number of parts produced each hour = 54/3.6 = 15 pc/hr Tool change time Tt = 6.0/15 = 0.4 min Including tool change time, Tp = Tc = To + Th + Tt = 3.30 + 18/60 + 0.4 = 4.00 min/pc Production rate of the cell = 3(15) = 45 pc/hr The robot works 48 sec = 0.8 min during the production of each part. With a cell production rate of 45 pc/hr, total time = 45(0.8) = 36 min each hour. The proportion of time robot is working = 36/60 = 0.60 = 60% 8.9 A robot performs a loading and unloading operation for a machine tool. The work cycle consists of the following sequence of activities:
Seq. Activity Time
1 Robot reaches and picks part from incoming conveyor and loads into fixture on machine tool. 5.5 sec
2 Machining cycle (automatic) 33.0 sec
3 Robot reaches in, retrieves part from machine tool, and deposits it onto outgoing conveyor. 4.8 sec
4 Move back to pickup position 1.7 sec
The activities are performed sequentially as listed. Every 30 work parts, the cutting tools in the machine must be changed. This irregular cycle takes 3.0 min to accomplish. The uptime efficiency of the robot is 97%; and the uptime efficiency of the machine tool is 98%, not including interruptions for tool changes. These two efficiencies are assumed not to overlap (i.e., if the robot breaks down, the cell will cease to operate, so the machine tool will not have the opportunity to break down; and vice versa). Downtime results from electrical and mechanical malfanctions of the robot, machine tool, and fixture. Determine the hourly production rate, taking into account the lost time due to tool changes and the uptime efficiency. Answer: Tc = 5.5 + 33.0 + 4.8 + 1.7 = 45 sec/cycle Tool change time Tt = 180 sec/30 pc = 6 sec/pc Tc + Tt = 45 + 6 = 51 sec = 0.85 min/pc Rc = 60/0.85 = 70.59 pc/hr Robot uptime E = 0.97, lost time = 0.03. Machine tool uptime E = 0.98, lost time = 0.02. These two inefficiencies are assumed not to overlap in the following solution: Accounting for uptime efficiencies, Rp = 70.59(1.0 − 0.03 − 0.02) = 67.06 pc/hr 8.10 In the previous problem, suppose that a double gripper is used instead of a single gripper as indicated in that problem. The activities in the cycle would be changed as follows: Steps 1, 4, and 5 are performed simultaneously with the automatic machining cycle. Steps 2 and 3 must be performed sequentially. The same tool change statistics and uptime efficiencies are applicable. Determine the hourly production rate when the double gripper is used, taking into account the lost time due to tool changes and the uptime efficiency. Answer: Tc = 5.0 + 33.0 = 38 sec/cycle Tt = 180/30 = 6 sec/pc as in previous problem Tc + Tt = 38 + 6 = 44 sec = 0.733 min/pc Rp = (60/0.733) = 81.86 pc/hr Accounting for uptime efficiencies, Rp = 81.86(1.0 − 0.03 − 0.02) = 77.77 pc/hr 8.11 Because the robot's portion of the work cycle requires much less time than the machine tool in Problem 8.9, the possibility of installing a cell with two machines is being considered. The robot would load and unload both machines from the same incoming and outgoing conveyors. The machines would be arranged so that distances between the fixture and the conveyors are the same for both machines. Thus, the activity times given in Problem 8.9 are valid for the two machine cell. The machining cycles would be coordinated so that the robot would be servicing only one machine at a time. The tool change statistics and uptime efficiencies in Problem 8.9 are applicable. Determine the hourly production rate for the two-machine cell. Assume that if one of the two machine tools is down, the other machine can continue to operate, but if the robot is down, cell operation is stopped. Answer: Robot cycle time Tc = 5.5 + 4.8 + 1.7 = 12 sec Tool change time Tt = 180 sec/30 pc = 6 sec/pc Machine Tc = 33 (machining) + 6 (tool change) + 12 (unload/load) = 51 sec = 0.85 min Because robot cycle is significantly smaller than machine tool cycle, assume there is little or no risk of machine interference when robot services two machines. (The machine interference issue is considered in Chapter 14, Section 14.4.2). The production rate for each machine tool, disregarding for the moment consideration of the uptime efficiency of the robot: Rp = (60/0.85)(1.0 − 0.02) = 69.176 pc/hr Because there are two machines, production rate will be double this value. However, the uptime efficiency of the robot will reduce production rate: Rp = 2(69.176)(1 − .03) = 134.2 pc/hr 8.12 Determine the hourly production rate for a two-machine cell as in Problem 8.11, only the robot is equipped with a double gripper as in Problem 8.10. Assume the activity times from Problem 8.10 apply here. Answer: Robot cycle time Tc = 3.3 + 5.0 + 3.0 + 1.7 = 13 sec Machine cycle = 33 (machining) + 6 (tool change) + 5 (unload/load) = 44 sec = 0.733 min Rp = (60/0.7333)(1.0 − 0.02) = 80.182 pc/hr With two machines, production rate is doubled, adjusted for robot efficiency. Rp = 2(80.182)(1 − .03) = 155.55 pc/hr 8.13 Arc-on time is a measure of efficiency in an arc welding operation. Typical arc-on times in manual welding range between 20% and 30%. Suppose that a certain welding operation is currently performed using a welder and a fitter. Production requirements are steady at 500 units per week. The fitter's job is to load the component parts into the fixture and clamp them in position for the welder. The welder then welds the components in two passes, stopping to reload the welding rod between the two passes. Some time is also lost each cycle for repositioning the welding rod on the work. The fitter's and welder's activities are done sequentially, with times for the various elements as follows:
Seq. Worker and activity Time
1 Fitter: load and clamp parts 4.2 min
2 Welder: weld first pass 2.5 min
3 Welder: reload weld rod 1.8 min
4 Welder: weld second pass 2.4 min
5 Welder: repositioning time 2.0 min
6 Delay time between work cycles 1.1 min
Because of fatigue, the welder must take a 20 min rest at mid-morning and mid-afternoon, and a 40 min lunch break around noon. The fitter joins the welder in these rest breaks. The nominal time of the work shift is eight hours, but the last 20 min of the shift is non-productive time for clean-up at each workstation. A proposal has been made to install a robot welding cell to perform the operation. The cell would be set up with two fixtures, so that the robot could be welding one job while the fitter is unloading the previous job and loading the next job. In this way, the welding robot and the human fitter could be working simultaneously rather than sequentially. Also, a continuous wire feed would be used rather than individual welding rods. It has been estimated that the continuous wire feed must be changed only once every 40 weldments and the lost time will be 20 min to make the wire change. The times for the various activities in the regular work cycle are as follows:
Seq. Fitter and robot activities Times
1 Fitter: Load and clamp parts 4.2 min
2 Robot: Weld complete 4.0 min
3 Repositioning time 1.0 min
4 Delay time between work cycles 0.3 min
A 10 min break would be taken by the fitter in the morning and another in the afternoon, and 40 min would be taken for lunch. Clean-up time at the end of the shift is 20 min. In your calculations, assume that the proportion uptime of the robot will be 98%. Determine the following: (a) arc-on times (expressed as a percent, using the eight-hour shift as the base) for the manual welding operation and the robot welding station, and (b) hourly production rate on average throughout the eight-hour shift for the manual welding operation and the robot welding station. Answer: (a) Manual cycle arc-on time: Cycle time Tc = 4.2 + 2.5 + 1.8 + 2.4 + 2.0 + 1.1 = 14.0 min Time arc is on = 2.5 + 2.4 = 4.9 min Out of 8 hr, time available = 480 - 20 - 20 - 40 - 20 = 380 min/shift Number of cycles in 380 min = 380/14 = 27.143 → 27 cycles/shift Time arc is on during shift = 27(4.9) = 132.3 min/shift Arc-on time = 132.3/480 = 0.276 = 27.6% Robot cycle arc-on time: Cycle time Tc = Max{(4.2 + .3), (4.0 + 1.0 + 0.3)} = 5.3 min/cycle Time arc is on each cycle = 4.0 min Out of 8 hr, time available = 480 - 10 - 10 - 40 - 20 = 400 min/shift Number of cycles in 400 min = 400/5.3 = 75.472 → 75 cycles/shift Time arc is on during shift = 75(4.0) = 300 min/shift Arc-on time = 300/480 = 0.625 = 62.5% (b) Manual cycle production rate Rp = (27 units/shift)/(8 hr) = 3.375 units/hr Robot cycle production rate Rp = (75 units/shift)/(8 hr) = 9.375 units/hr 8.14 (A) A work cell is currently operated 2000 hr/yr by a human worker who is paid an hourly rate of $23.00, which includes applicable overhead costs. One work unit is produced in a cycle time of 4.8 min. Management would like to increase output to meet increasing demand and a robot cell is being considered as a replacement for the present manual cell. The cycle time of the proposed cell would be reduced to 4.0 min. The installed cost of the robot plus supporting equipment is $120,000. Power and other utilities to operate the robot will be $0.30/hr, and annual maintenance costs are $2500. Determine (a) the number of parts produced annually by the manually-operated cell and (b) cost per part produced. (c) How does the cost per part of the robot cell compare with your answer in part (b), given a 4-year service life, 10% rate of return, and no salvage value. Answer: (a) Given Tc = 4.8 min, annual Q = 2000(60)/4.8 = 25,000 pc/yr Annual cost of the manually-operated cell = $23.00(2000) = $46,000/yr Cpc = $46,000/25,000 = $1.84/pc Robot cell: From Equation (3.29), (A/P,10%,4) = 0.10(1 + 0.10)4/(1 + 0.10)4 – 1 = 0.3155 TC = 120,000(0.3155) + 2000(0.30) + 2500 = $40,956.50/yr Given Tc = 4.0 min, annual Q = 2000(60)/4.0 = 30,000 pc Cpc = $40,956.50/30,000 = $1.365/pc 8.15 A manual arc-welding cell uses a welder and a fitter. The cell operates 2000 hr/yr. The welder is paid $30/hr and the fitter is paid $25.00/hr. Both rates include applicable overheads. The cycle time to complete one welded assembly is 15.4 min. Of this time, the arc-on time is 25%, and the fitter’s participation in the cycle is 30% of the cycle time. A robotic arc-welding cell is being considered to replace this manual cell. The new cell would have one robot, one fitter, and two workstations, so that while the robot is working at the first station, the fitter is unloading the other station and loading it with new components. The fitter’s rate would remain at $25.00/hr. For the new cell, the production rate would be eight welded assemblies per hour. The arc-on time would increase to almost 52%, and the fitter’s participation in the cycle would be about 62%. The installed cost of the robot and workstations is $158,000. Power and other utilities to operate the robot and arc welding equipment will be $3.80/hr, and annual maintenance costs are $3500. Given a 3-year service life, 15% rate of return, and no salvage value, (a) determine the annual quantity of welded assemblies that would have to be produced to reach the breakeven point for the two methods. (b) What is the annual quantity of welded assemblies produced by the two methods working 2000 hr/yr? Answer: For the manual arc-welding cell, given Tc = 15.4 min, Variable cost Cv = ($30 + $25)(15.4/60) = $14.117/pc Total cost as a function of Q is TC = 14.117Q, assuming no fixed costs. Robot cell: From Equation (3.29), (A/P,15%,3) = 0.15(1 + 0.15)3/(1 + 0.15)3 – 1 = 0.2919 3 Annual fixed costs Cf = 158,000(0.2919) + 3500 = $49,626.30/yr Variable cost Cv = $3.80/8 = $0.475/pc Total cost as a function of Q is TC = 49,626.30 + 0.475Q Setting TC for the manual cell = TC for the robot cell, 14.117Q = 49,626.30 + 0.475Q 14.117Q – 0.475Q = 13.642Q = 49,626.30 Q = 3638 pc (b) For the manual cell, annual Q = 2000(60)/15.4 = 7792 pc For the robotic cell, annual Q = 2000(8) = 16,000 pc Robot Programming Exercises Note: The problems in the following group are all programming exercises to be performed on robots available to students. The solutions depend on the particular programming methods or languages used. They represent suggestions for laboratory exercises to instructors using the book. 8.16 The setup for this exercise requires a felt-tipped pen mounted to the robot's end-of-arm (or held securely in the robot's gripper). Also required is a thick cardboard, mounted on the surface of the worktable. Pieces of plain white paper will be pinned or taped to the cardboard surface. The exercise is the following: Program the robot to write your initials on the paper with the felt-tipped pen. Answer: To tackle this exercise, the following steps can be followed: 1. Initial Setup: Ensure that the robot is equipped with a felt-tipped pen attached securely to its end effector or gripper. Additionally, affix a thick cardboard to the worktable's surface where the robot will operate. Pin or tape plain white paper onto the cardboard. 2. Coordinate System Calibration: Calibrate the robot's coordinate system to correspond with the dimensions of the cardboard surface. This step ensures accurate positioning of the pen tip over the paper. 3. Programming: Develop a program that instructs the robot to move in a sequence that forms your initials on the paper. This involves breaking down the letters of your initials into a series of precise movements and pen strokes. Consider factors such as pen pressure, stroke speed, and directionality for legible writing. 4. Testing and Refinement: Execute the program on the robot and observe the writing process. Make adjustments as needed to improve the clarity and accuracy of the initials. This may involve tweaking movement trajectories, pen angles, or other parameters. 5. Verification: Once satisfied with the programming, verify that the robot consistently writes your initials correctly on the paper. Repeat the process multiple times to ensure repeatability and reliability. 6. Documentation: Document the program and any relevant settings or configurations for future reference. This documentation should include detailed instructions on how to set up the robot for the writing task and execute the program successfully. By following these steps, the robot can be programmed to autonomously write your initials on paper, demonstrating proficiency in robot programming and coordination for practical applications. 8.17 As an enhancement of the previous programming exercise, consider the problem of programming the robot to write any letter that is entered at the alphanumeric keyboard. A textual programming language is required for this exercise. Answer: Expanding on the previous exercise, where the robot was programmed to write specific initials, this enhanced exercise introduces the challenge of dynamically writing any letter entered via an alphanumeric keyboard. Here's a structured approach to tackle this task: 1. System Integration: Ensure seamless integration between the robot's control system and an alphanumeric keyboard. This may involve connecting the keyboard to the robot controller and establishing communication protocols. 2. User Input Handling: Develop a program module to receive user input from the alphanumeric keyboard. This module should capture the entered letter and pass it to the main control algorithm. 3. Alphabet Recognition: Implement a logic module capable of recognizing the entered letter and converting it into a format understandable by the robot's programming language. This module should account for variations in letter size, font, and style. 4. Letter Generation Algorithm: Design an algorithm that translates the recognized letter into a series of robot movements required to write it on paper. This involves breaking down each letter into individual strokes and defining corresponding robot trajectories and pen movements. 5. Path Planning and Optimization: Optimize the robot's path planning to minimize writing time and maximize accuracy. This may involve analyzing the letter's geometry to determine the most efficient stroke sequence and trajectory. 6. Error Handling: Implement error detection and recovery mechanisms to handle potential issues during the writing process, such as pen malfunction or unexpected obstacles on the paper. 7. User Feedback: Incorporate feedback mechanisms to inform the user of the writing progress and any encountered errors. This ensures transparency and enables prompt intervention if necessary. 8. Testing and Validation: Thoroughly test the programmed system using a variety of input letters to ensure robustness and accuracy. Validate the system's performance against predefined criteria, such as writing speed, legibility, and reliability. 9. Documentation and Training: Document the programming logic, algorithms, and system configurations for future reference. Provide comprehensive training materials for users to understand and operate the system effectively. By following this structured approach, the robot can be programmed to dynamically write any letter entered at the alphanumeric keyboard, showcasing advanced capabilities in robot programming and automation. 8.18 Apparatus for this exercise consists of two wood or plastic blocks of two different colors that can be grasped by the robot gripper. The blocks should be placed in specific positions (call the positions A and B on either side of a center location (call it position C). The robot should be programmed to do the following: (1) pick up the block at position A and place it at position C; (2) pick up the block at position B and place it at position A; (3) pick up the block at position C and place it at position B. (4) Repeat steps (1), (2), and (3) continually. Answer: To accomplish the tasks outlined in this exercise, the following steps can be followed: 1. Setup and Calibration: Ensure that the robot's gripper is capable of grasping the wood or plastic blocks securely. Calibrate the robot's coordinate system to accurately locate positions A, B, and C on the work surface. 2. Programming Logic: •Step 1: Program the robot to pick up the block at position A and place it at position C. This involves specifying the precise gripping and releasing motions needed to manipulate the block. •Step 2: Develop instructions for the robot to pick up the block at position B and move it to position A. This requires careful planning to avoid collisions or interference with the other block. •Step 3: Define the actions for the robot to pick up the block at position C and transfer it to position B. Again, ensure smooth execution to prevent disruptions in the sequence. •Step 4: Create a loop structure in the program to continuously repeat steps 1, 2, and 3 in sequence. 3. Error Handling: Implement error detection mechanisms to handle potential issues such as dropped blocks, misalignment during gripping, or obstacles in the workspace. Incorporate corrective actions to recover from errors and resume the task. 4. Optimization: Fine-tune the program to optimize the robot's movements for efficiency and speed. This may involve minimizing unnecessary motions, optimizing gripper trajectories, and reducing idle time between actions. 5. Testing and Validation: Execute the programmed sequence on the robot and observe its performance. Verify that the robot accurately completes each step of the task without errors or deviations. Iterate on the program as needed based on testing results. 6. Documentation: Document the programming logic, including detailed instructions for each step of the task. Provide information on the setup requirements and any specific considerations for executing the program successfully. By following these steps, the robot can be programmed to autonomously manipulate the blocks between positions A, B, and C in a continuous loop, demonstrating proficiency in robotic manipulation and automation. 8.19 Apparatus for this exercise consists of a cardboard box and a dowel about 4 inches long (any straight thin cylinder will suffice, e.g., pen, pencil, etc.). The dowel is attached to the robot’s end-of-arm or held in its gripper. The dowel is intended to simulate an arc welding torch, and the edges of the cardboard box are intended to represent the seams that are to be welded. The programming exercise is the following: With the box oriented with one of its corners pointing towards the robot, program the robot to weld the three edges that lead into the corner. The dowel (welding torch) must be continuously oriented at a 45° angle with respect to the edge being welded. See Figure P8.19. Answer: To tackle this programming exercise effectively, the following steps can be followed: 1. Initial Setup: Ensure that the cardboard box is positioned with one of its corners pointing towards the robot. Attach the dowel to the robot's end-of-arm or gripper, simulating an arc welding torch. Calibrate the robot's coordinate system to accurately locate the edges of the cardboard box. 2. Orientation Control: Implement a control algorithm to continuously orient the dowel at a 45° angle with respect to the edge being welded. This requires real-time feedback from the robot's sensors to adjust the orientation as the robot moves along the edge. 3. Path Planning: Develop a path planning algorithm to guide the robot along the three edges leading into the corner of the cardboard box. Ensure that the dowel maintains the specified angle while traversing each edge to achieve uniform welding. 4. Welding Motion: Define the welding motion profile, including parameters such as welding speed, arc length, and welding current. Coordinate the movement of the robot with the activation of the simulated welding torch to create consistent and reliable welds along the seams. 5. Collision Avoidance: Incorporate collision detection and avoidance mechanisms to prevent the robot or dowel from colliding with the cardboard box or other obstacles in the workspace. This may involve adjusting the robot's trajectory or pausing the welding process if an obstacle is detected. 6. Quality Assurance: Implement quality control measures to ensure the integrity of the welds. This may include monitoring parameters such as weld bead width, penetration depth, and surface smoothness to verify compliance with welding standards. 7. Error Handling: Develop error detection and recovery routines to address potential issues such as incomplete welds, deviation from the specified angle, or equipment malfunctions. Implement strategies to resume the welding process or alert operators to intervene if necessary. 8. Testing and Validation: Execute the programmed sequence on the robot and observe the welding process. Verify that the dowel maintains the correct orientation and that the welds are applied accurately along the designated edges. Make adjustments as needed based on testing results. 9. Documentation: Document the programming logic, including algorithms, parameters, and setup instructions. Provide detailed guidance for setting up the exercise apparatus and executing the welding task on the robot. By following these steps, the robot can be programmed to simulate arc welding along the edges of a cardboard box, demonstrating proficiency in robotic welding techniques and automation. 8.20 This exercise is intended to simulate a palletizing operation. The apparatus includes: six wooden (or plastic or metal) cylinders approximately 20 mm in diameter and 75 mm in length, and a 20 mm thick wooden block approximately 100 mm by 133 mm. The block is to have six holes of diameter 25 mm drilled in it as illustrated in Figure P8.20. The wooden cylinders represent workparts and the wooden block represents a pallet. (As an alternative to the wooden block, the layout of the pallet can be sketched on a plain piece of paper attached to the work table.) The programming exercise is the following: Using the powered leadthrough programming method, program the robot to pick up the parts from a fixed position on the work table and place them into the six positions in the pallet. The fixed position on the table might be a stop point on a conveyor. (The student may have to manually place the parts at the position if a real conveyor is not available.) Answer: To tackle this palletizing operation simulation exercise, follow these steps: 1. Apparatus Setup: •Arrange six wooden cylinders (or alternative workparts) and a wooden block (or sketched pallet layout) on the worktable as illustrated in Figure P8.20. •Ensure the wooden block has six holes drilled in it, each with a diameter of 25 mm. 2. Position Calibration: •Calibrate the robot's coordinate system to accurately locate the fixed position on the worktable where the workparts are placed. 3. Powered Leadthrough Programming: •Utilize the powered leadthrough programming method to guide the robot through the task. •Direct the robot to pick up the workparts from the fixed position on the worktable one by one. 4. Placement Sequence: •Define a sequence for placing the workparts into the six positions in the pallet. •Ensure that the robot places each workpart accurately into its corresponding hole in the pallet, maintaining proper alignment. 5. End Effector Control: •Program the robot's end effector (gripper) to grasp and release the workparts securely. •Coordinate the gripper's actions with the robot's movements to ensure smooth and precise handling of the workparts. 6. Error Handling: •Implement error detection mechanisms to identify and handle potential issues such as dropped parts or misalignment during placement. •Include error recovery procedures to address any deviations from the programmed sequence and resume the palletizing operation. 7. Testing and Optimization: •Test the programmed sequence on the robot to ensure it correctly picks up and places the workparts into the pallet positions. •Optimize the program as needed to improve efficiency and accuracy, adjusting parameters such as approach speed and gripper force. 8. Documentation: •Document the programmed sequence, including step-by-step instructions and any necessary setup details. •Provide guidance on how to execute the palletizing operation using the powered leadthrough programming method. By following these steps, the robot can simulate a palletizing operation by picking up workparts from a fixed position on the worktable and placing them into predefined positions in the pallet, demonstrating proficiency in palletizing automation techniques. Accuracy and Repeatability 8.21 (A) The linear joint (type L) of an industrial robot is actuated by a piston. The length of the joint when fully retracted is 800 mm and when fully extended is 1400 mm. If the robot's controller has a 10-bit storage capacity, determine the control resolution for this joint. Answer: Range L = 1400 mm - 800 mm = 600 mm. Given: B = 10 bit storage capacity Control resolution CR2 = R/210 – 10= 600/1023 = 0.59 mm 8.22 (A) In the previous problem, the mechanical errors associated with the linear joint form a normal distribution in the direction of the joint actuation with standard deviation = 0.11 mm. Determine the accuracy and repeatability for the robot. Answer: Ac = CR/2 + 3σ =0.59/2 + 3(0.11) = 0.625 mm Re = ±3σ = ±3(0.11) = ± 0.33 mm 8.23 The rotational joint (type R) of an industrial robot has a range of 170° rotation. The mechanical errors in the joint and the input/output links can be described by a normal distribution with its mean at any given addressable point and a standard deviation of 0.15°. Determine the number of storage bits required in the controller memory so that the accuracy of the joint is as close as possible to, but less than, its repeatability. Use six standard deviations as the measure of repeatability. Answer: Repeatability = 6σ = 6(0.15) = 0.9° and Accuracy = 0.5CR2 + 3σ Find bit storage capacity B so that 0.5CR2 + 3σ = approx 6σ Set (0.5CR2 + 3σ) - 6σ = 0 0.5CR2 = 3σ CR2 = 6σ = 0.9° 2B = (170°)/(0.9°) = 188.9 B ln(2) = ln(188.9), 0.69315 B = 5.241, B = 7.56 → B = 8 bits CR2 = (170°)/(28 - 1) = 170/255 = 0.667° Accuracy = 0.5(.667) + 3(0.15) = 0.783° which is ≤ repeatability = 0.9° 8.24 An articulated robot has a T-type wrist axis that can be rotated a total of 2 rotations (each rotation is a full 360°). It is desired to be able to position the wrist with a control resolution of 0.25° between adjacent addressable points. (a) Determine the number of bits required in the binary register for that axis in the robot's control memory. (b) Using this number of bits, what is the actual control resolution of the joint? Answer: (a) Range R = 2(360°) = 720° Specification: CR = 0.25° or better, Find B CR = 720/2B - 1 = 0.25°. Rearranging, 2B − 1 = 720/0.25 = 2880, 2B = 2881 B ln(2) = ln(2881), 0.69315 B = 7.966, B = 11.4 → B = 12 bits (b) CR = 720/212 - 1 = 720/4095 = 0.176° 8.25 One axis of an RRL robot is a linear slide with a total range of 750 mm. The robot's control memory has a 10-bit capacity. It is assumed that the mechanical errors associated with the arm are normally distributed with a mean at the given taught point and a standard deviation of 0.10 mm. Determine (a) the control resolution for the axis under consideration, (b) accuracy, and (c) repeatability. Answer: Joint range R = 750 mm CR = (750 mm)/(210-1) = 750/1023 = 0.733 mm Accuracy = 0.5CR + 3σ = 0.666 mm Repeatability = ± 3σ = 6σ = 0.60 mm 8.26 (A) Link 3 of a TRR robot has a rotational joint whose output link is connected to the wrist assembly. Considering the design of this joint only, the output link is 600 mm long, and the total range of rotation of the joint is 60°. The control resolution of this joint is expressed as a linear arc measure at the wrist, and is specified to be 1.0 mm or less. It is known that the mechanical inaccuracies in the joint result in an error of ±0.018° rotation, and it is assumed that the output link is perfectly rigid so as to cause no additional errors due to deflection. (a) Determine the minimum number of bits required in the robot's control memory to achieve the control resolution specified. (b) Using this number of bits, what is the actual control resolution of the joint? Answer: (a) CR = 1.0 mm at end of output link. Converting control resolution to an angular value, CR = 1.0(360)/(2π × 600) = 0.0955° Given joint range R = 60°, CR = (60°)/(2B−1) = 0.0955° 2B − 1 = 60/0.0955 = 628.3 2B = 628.3 + 1 B ln(2) = ln(629.3) 0.6931B = 6.445 B = 9.3 → B = 10 bits (b) CR = (60°)/(210 − 1) = 60/1023 = 0.0587° Comment: The ±0.018° error due to mechanical inaccuracies is a “red herring.” That data is not needed to solve the problem. Solution Manual for Automation, Production Systems, and Computer-Integrated Manufacturing Mikell P. Groover 9780133499612, 9780134605463