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<div>Hi All,</div><div>This discussion, reminds me of 25 years ago I had used Walmart 12vdc marine batteries.</div><div>I solved the lift issue by running the lift on 12vdc. Yes the lift was slow, but I can wait.</div><div>The marine batteries at best give 2 years of mowing and plowing.</div><div><br></div><div>For the last 20 years I use only Trojan T-105s that last 10 years. <br></div><div><br></div><div>I ready for another set of Trojans now, can anyone suggest a better battery than T-105s.</div><div><br></div><div>Thanks,</div><div>Paul Holzschuher</div><div>C-185 Cincinnati </div><div><br></div>
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On Wednesday, January 30, 2019, 10:40:24 PM EST, Chris Zach <cz@alembic.crystel.com> wrote:
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<div><div dir="ltr">hPower is what matters. Power is what runs the lift motor and power is <br clear="none">what lifts the snowblower.<br clear="none"><br clear="none">Power=Current*Voltage. Or P=I*E.<br clear="none"><br clear="none">So for a given amount of power you can double the current, half the <br clear="none">voltage, and off you go. Or you can double the voltage and halve the <br clear="none">current. Wires are rated to carry current, not so much voltage so <br clear="none">reducing the current and boosting voltage means less load on the wires <br clear="none">for a given amount of power.<br clear="none"><br clear="none">It also means the motor can develop way more power than it should, and <br clear="none">if loaded will dissipate more power as heat. So you need to limit the <br clear="none">amount of power the motor can produce. One way is with a fuse, if you <br clear="none">half the size of the fuse it will blow when the motor exceeds the same <br clear="none">amount of power as with lower voltage and higher current.<br clear="none"><br clear="none">I think the lift motor fuse was 30a. At 18v that's 30*18v that's 540 <br clear="none">watts of power before it goes. Dropping the fuse to 15 and doubling the <br clear="none">voltage gives you 15*36=540 so the world is good.<br clear="none"><br clear="none">If we didn't derate the fuse the motor would produce over 1kw of power, <br clear="none">which is insane, and will either break things or (if the motor is <br clear="none">stalled and all power is being dissipated in the motor) will put a 1kw <br clear="none">heat load in the armature of the motor. Melted armature and such.<br clear="none"><br clear="none">The one theoretical problem would be in the lift interrupt switch. <br clear="none">That's designed to shut off the motor if the load gets too high. Because <br clear="none">it operates on current it will open a lot later in the power curve, <br clear="none">possibly damaging the motor. And with the increased voltage there is a <br clear="none">chance it could arc over on opening and weld shut. Thus again the <br clear="none">smaller fuse. And a higher voltage can arc over on the commutator, but <br clear="none">probably not 36v.<br clear="none"><br clear="none">Fun stuff. Really the motor was designed for <a shape="rect" href="mailto:12v@30a" rel="nofollow" target="_blank">12v@30a</a> peak or 360 watts <br clear="none">but that's for decades of duty cycles. Bumping it up a bit will not <br clear="none">cause the world to end IMO.<br clear="none"><br clear="none">C<br clear="none"><br clear="none"><br clear="none">On 1/30/2019 9:23 PM, David Roden wrote:<br clear="none">> On 30 Jan 2019 at 18:44, Chris Zach wrote:<br clear="none">> <br clear="none">>> remember, twice the voltage means half the amps<br clear="none">> <br clear="none">> I'm not an EE, just a hobbyist, but that's not what I learned. Ohm's law<br clear="none">> says that I = E/R, so for a given load, current is directly proportional to<br clear="none">> voltage.<br clear="none">> <br clear="none">> The situation is more complicated when the motor is running at normal load<br clear="none">> and speed (back EMF and all that), but my concern would be with what happens<br clear="none">> when you get to the end of travel, or have a very heavy load that keeps RPM<br clear="none">> way down. Under stall or near-stall conditions, double the voltage means<br clear="none">> about double the current.<br clear="none">> <br clear="none">> It's the current that makes the heat that burns out your armature. That's<br clear="none">> part of what the fuse is supposed to protect against. But since it's still<br clear="none">> current that concerns you, not power, I don't see the need to halve the<br clear="none">> fuse's rating. I'd think that would just cause nuisance opens. What am I<br clear="none">> missing here?<br clear="none">> <br clear="none">> <br clear="none">> David Roden - Akron, Ohio, USA<br clear="none">> <br clear="none">> = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =<br clear="none">> Note: mail sent to the "etpost" address will not reach me. 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