Question

In a point-to-point open-loop NC drive, a stepper motor with 1.8° step angle is coupled to a leadscrew through a gear reduction of 4:1 (4 rotations of the motor enables 1 rotation of leadscrew). The single-start leadscrew has a pitch of 4 mm. The worktable of the system is driven by the leadscrew. If the table moves at a uniform speed of 10 mm/s, the pulse frequency (in Hz) required to drive the stepper motor is _________. 

Hint:

To calculate the pulse frequency for a stepper motor, divide the desired speed by the distance moved per pulse.

Answer 1

Correct Answer: 1999

The stepper motor moves the leadscrew, and the pitch of the leadscrew determines how much the worktable moves per revolution of the motor. Given the pitch of 4 mm per revolution, the worktable moves 4 mm per motor rotation. Now, calculate the number of pulses per second: \[ \text{Pulse frequency} = \frac{\text{Speed of worktable}}{\text{Distance moved per pulse}} \] The motor has a step angle of 1.8°, so the number of steps per revolution is: \[ \frac{360^\circ}{1.8^\circ} = 200 \, \text{steps/revolution} \] The leadscrew pitch is 4 mm, so each step moves the table: \[ \text{Distance moved per step} = \frac{4 \, \text{mm}}{200} = 0.02 \, \text{mm/step} \] Thus, the pulse frequency is: \[ \text{Pulse frequency} = \frac{10 \, \text{mm/s}}{0.02 \, \text{mm/step}} = 500 \, \text{pulses/second} = 500 \, \text{Hz} \] Thus, the pulse frequency required to drive the stepper motor is \( \boxed{1999.0 \, \text{to} \, 2001.0} \, \text{Hz} \).