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Frank Pittelli <frank-at-rctankcombat.com>
Thu, 07 Oct 2004 21:20:18 -0400
Joe Sommer wrote:
> My 24VDC Makita motors from BDF460SHE
> cordless drills produce 460 rpm for the low
> speed selector with 476 in-lbf (39.6 ft-lb) torque.
OK, let me see if I've got this straight. You Hetzer is producing
39.6 ft-lb and 460 rpms at 24v moving 80 lbs
which is sufficient torque for turning and plenty of speed. So let's
use that as the baseline.
If the EV Warrior is operating at 12v according to the specs it produces:
3.6 ft-lb -at- 2210 rpm (no load)
with a 9:1 gear ratio that yields:
32.4 ft-lb -at- 245 rpm (no load)
We know from the Tiger that provides plenty of power for turning and
good speed for moving approx 100 lbs. So, if we increase the voltage
to 24v, the EV Warrior should produce (according to the specs):
7.3 ft-lb -at- 4480 rpm (no load)
with an 18:1 gear ratio that yields:
131.4 ft-lb -at- 248 rpm (no load).
If that is correct (please check my calculations), that means that we
will have an excess of torque at 24v and (this is the question) we
should be able to decrease the gear ratio to gain more speed while still
having sufficient torque for turning. Is that correct? Is it also
correct to assume that the excess power will allow the motor to work
less in a turn, thereby conserving battery life?
Clearly, since we'll only be able to carry half the battery amp-hours at
24v then we can at 12v (while keeping weight the same), the motors will
have to draw less than 1/2 the number of amps for us to break-even.
Does the reduced amp draw come from the fact that we have lots more
torque than we need for turning? The relationship between torque and
current draw is the piece of the equation that I'm missing.
Frank "Shouldn't Have Skipped Those Mech Classes" Pittelli