The Picture is the example the problem is talking about

2. Suppose the mechanical inaccuracies in the open-loop positioning system of Example 7.1 are described by a normal distribution with standard deviation = 0.006 mm. The range of the worktable axis is 1600 mm, and there are 16 bits in the binary register used by the digital controller to store the programmed position. Other relevant parameters from Example 7.1 are the following: pitch = 6.0 mm gear ratio between motor shaft and screw = 8.0, and number of step angles in the stepper motor = 28. Determine the (a) control resolution, (b) accuracy, and (c) repeatability of the positioning system.

AMPLE 7.1 NC Open-Loop Positioning The worktable of a positioning system is driven by a ball screw whose pitch =6.0mm. The screw is connected to the output shaft of a stepper motor through a gearbox whose ratio is 5:1 (five turns of the motor to one turn of the screw). The stepper motor has 48 step angles. The table must move a distance of 250mm from its present position at a linear velocity =500mm/min. Determine (a) how many pulses are required to move the table the specified distance and (b) the required motor speed and pulse rate to achieve the desired table velocity. (a) Rearranging Equation (7.7) to find the screw rotation angle As corresponding to a distance x=250mm, As=p360x=6.0360(250)=15,000 With 48 step angles, each step angle is =48360=7.5 Thus, the number of pulses to move the table 250mm is np=p360xrg=Asrg=7.515,000(5)=10,000pulses (b) The rotational speed of the screw corresponding to a table speed of 500mm/min is determined from Equation (7.10): Ns=pvt=6500=83.333rev/min Equation (7.6) is used to find the motor speed: Nm=rgNs=5(83.333)=416.667rev/min The applied pulse rate to drive the table is given by Equation (7.11): fp=60pvtnsrg=60(6)500(48)(5)=333.333zz