Motor Acceleration

Estimate from Applied Maths but there are many models available from transmission suppliers to refer to:

Total distance = 20m, so travel time approx 60s at 20m/min.

You will need enough torque to:

1. Accelerate in desired time

2. Overcome resistance when travelling.

1. Assuming total resistance is largely friction and transmission losses and zero gradient, F = m*A. If resistance is estimated at 20% (is it on a track with nylon rollers?), then F*0.8 = m*A. e.g. If 2s accel, then A = 0.166. m/s2, so F= (600*0.166)/0.8 = 125N. At full speed, this equates to 125N * 0.333 = 42W power. (Power = force x speed.

2. Your losses at full speed are 600 * g * 20% = 1200N so this equates to 1200 * 0.333. = 400W.

If a normal motor/gearbox, this would mean a 0.55kW or probably a 0.75kW selection. Losses would mean stopping is easier than starting. You need to ensure gearing is selected according to your max speed required and a standard inverter would also make a good selection if positioning is done by your control system or by hand. Its good to have torque in hand for starting where there may be any irregularities in the track / wheel assembly for trouble free operation, the losses are probably overestimated since these are not given. You can investagate data for better info, if its a one off, better to be more conservative in selection / estimation.

Does this help with your ideas?