Re: mill servo conversion ?'s

some last questions and summations to make sure I've got this all
together:

Using Camtronics motors as an example, they state their motors are 38V
at 900RPM.

That means Kv = 38v / 0.9 KRPM = 42.2V/Krpm
Kt = Kv * 1.35 = 57in oz/amp
You state the terminal resistance (Rt) is high at 4ohm, and on the 36V
power supply inclued in the kit, maximum torque occurs at stall with 9
amps supplied. 9 A * 57 inoz/A = 513 in oz total.
Because of the back emf, torque will decrease at a rate of
Kv / Rt * Kt for every 1000 RPMs or 42.2V/Krpm / 4ohm * 57inoz = 601
in oz for every 1000RPMs (coincidence?)

Maximum torque can be achieved over a wider range of speeds by simply
using a higher voltage power supply, and limiting the amount of amps
supplied to the motor through the gecko drive through adjustment of
the pot in order to avoid damaging the motors.

Finally, you stated that the motors in this auction:

http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&category=26261&item=3831771088&rd=1&ssPageName=WDVW

are the same ones as camtronics sells. However, it lists the terminal
resistance as 2.6ohm which I assume is what you're referring to when
you say camtronics' motors are 4-5 ohm. Does that mean these motors
will run closer to peak torque on a lower voltage power supply? Follow
my math here if you will. 2.6 ohm motors with a kt of 44ozin/amp and a
peak torque of 600inoz means max stall amps are ~13.5amps, which means
in order to see that torque, one needs a 13.5*2.6=35V power supply.

I guess, restated another way: it seems these motors are better suited
for the power supply camtronics supplies.

How does one determine the capabilities of a power supply? They have
continuous current supplied and peak current. How long can peak
current be supplied? Is there any set relationship, or is it just a
function of the type of components used? What value is normally given
with a power supply; peak current or continuous?