Question 37.2: NC Closed-Loop Positioning An NC worktable is driven by a cl...
NC Closed-Loop Positioning
An NC worktable is driven by a closed-loop positioning system consisting of a servomotor, ball screw, and optical encoder. The screw has a pitch = 5.0 mm and is coupled to the motor shaft with a gear ratio of 4:1 (four turns of the motor for each turn of the screw). The optical encoder generates 100 pulses/rev of the screw. The table has been programmed to move a distance of 75.0 mm at a feed rate = 400 mm/min. Determine (a) how many pulses are received by the control system to verify that the table has moved exactly 75.0 mm; and (b) the pulse rate and (c) motor speed that correspond to the specified feed rate.
(a) Rearranging Equation (37.12) to find n_p,
x=\frac{p n_p}{n_s}=\frac{pA_s}{360} (37.12)
n_p=\frac{xn_s}{p}=\frac{75(100)}{5} = 1500 pulses
(b) The pulse rate corresponding to 400 mm/min can be obtained by rearranging Equation (37.13) :
v_t=f_r=\frac{60 p f_p}{n_s} (37.13)
f_p =\frac{f_r n_s}{60 p}=\frac{400(100)}{60(5)} = 133.33 Hz
(c) Screw rotational speed is the table velocity divided by the pitch:
N_s=\frac{f_r}{p}= 80 rev/min
With a gear ratio r_g= 4:1 , the motor speed N_m = 4(80) = 320 rev / min