RE: [CAD_CAM_EDM_DRO] reality check...

les,
running a 2.38(motor) to 1 (gear box ratio), hence the division....


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

<mailto:signworks@cond>

Visit our website at: http://www.shelbyvilledesign.com



-----Original Message-----
From: Leslie M. Watts [mailto:leswatts@...]
Sent: Wednesday, August 20, 2003 4:00 PM
To: CAD_CAM_EDM_DRO@yahoogroups.com
Subject: RE: [CAD_CAM_EDM_DRO] reality check...


Hi Erie

Math looks ok up to the point of pinion torque and force.

At peak motor torque of 37.51x20=750.2 oz in you would get
750.2 x 2.38 x 40 = 71420 oz in at the pinion. I think you
divided rather than multiplied but gearing down of course
increases torque.

With your 3.926 effective diameter you get an effective radius
of 3.926/(2 x Pi)= .625 in

Which would give you a linear force of
71420/.625= 114272 oz or 114272/16= 7142 lbs

Seems a bit much... looks like you could easily run much
higher rapids. I run with a linear force of about 2300 lb
to accelerate a gantry effective mass of 2000 lbm at about
1 g.

That is if I didn't make a mistake....


Les

Leslie M.Watts
L M Watts Furniture
Tiger Georgia USA
http://home.alltel.net/leswatts/wf.html
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-----Original Message-----
From: Erie Patsellis [mailto:signworks@...]
Sent: Wednesday, August 20, 2003 1:17 PM
To: CAD_CAM_EDM_DRO@yahoogroups.com
Subject: [CAD_CAM_EDM_DRO] 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@...>

<mailto:signworks@cond>

Visit our website at: http://www.shelbyvilledesign.com






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