reality check...

OK, given the following:
Application: Rack & Pinion Gantry router, work area is 8' x 13'

Motor Specs:

37.51 oz-in/A (Kt)
25.0 - 30.5 V/kRPM (Kv)
Terminal resistance 0.75 - 1.02 Ohms @ 4A
Max terminal voltage 60V
Max continuous speed 2100RPM
Max peak torque 1500 oz-in
Max peak current 44.5A (unsure)
Max continuous stall torque at maximum motor temp 300oz-in
Max continuous current 8.9A

Reduction Calulations:

a 40:1 drive reduction with 24 d.p. rack, 30t 24 d.p. pinion (effective
circumfrence of 3.926 inches)

desired rapid of 400-500 ipm (approx 100-125 rpm pinion speed)

I calculate I would need a 2.38:1 ratio from motor to reduction unit, does
this ((125*40)/2100=2.380)
will give me a torque at 20 amps(gecko 340 max current) of
(37.51*20)/2.38*40=12,608 in oz)or 788 in lbs of force.

>anybody remember how to calculate force at the tool with a rack & pinion

drive?

1024 count quadrature encoders * 4= 4096 counts per rev, 2100 rpm=35 rps,
35*4096=143,360 pulse frequency, in excess of
mach 2's limit of 45 Khz, I'll have to use a pulse multiplier of either
x5(28672 khz) or x10(14336 khz).

If I go with x5, I end up with about 3500 steps per inch, x 10 is around
1750 steps per inch.

any problems with my math???? suggestions as to which way to go (x5, x10)



Erie Patsellis
Shelbyville Design & signworks
217-774-4444
signworks@... <mailto:signworks@...>

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