hPower is what matters. Power is what runs the lift motor and power is
what lifts the snowblower.
Power=Current*Voltage. Or P=I*E.
So for a given amount of power you can double the current, half the
voltage, and off you go. Or you can double the voltage and halve the
current. Wires are rated to carry current, not so much voltage so
reducing the current and boosting voltage means less load on the wires
for a given amount of power.
It also means the motor can develop way more power than it should, and
if loaded will dissipate more power as heat. So you need to limit the
amount of power the motor can produce. One way is with a fuse, if you
half the size of the fuse it will blow when the motor exceeds the same
amount of power as with lower voltage and higher current.
I think the lift motor fuse was 30a. At 18v that's 30*18v that's 540
watts of power before it goes. Dropping the fuse to 15 and doubling the
voltage gives you 15*36=540 so the world is good.
If we didn't derate the fuse the motor would produce over 1kw of power,
which is insane, and will either break things or (if the motor is
stalled and all power is being dissipated in the motor) will put a 1kw
heat load in the armature of the motor. Melted armature and such.
The one theoretical problem would be in the lift interrupt switch.
That's designed to shut off the motor if the load gets too high. Because
it operates on current it will open a lot later in the power curve,
possibly damaging the motor. And with the increased voltage there is a
chance it could arc over on opening and weld shut. Thus again the
smaller fuse. And a higher voltage can arc over on the commutator, but
probably not 36v.
Fun stuff. Really the motor was designed for
12v@30a peak or 360 watts
but that's for decades of duty cycles. Bumping it up a bit will not
cause the world to end IMO.
C
On 1/30/2019 9:23 PM, David Roden wrote:
> On 30 Jan 2019 at 18:44, Chris Zach wrote:
>
>> remember, twice the voltage means half the amps
>
> I'm not an EE, just a hobbyist, but that's not what I learned. Ohm's law
> says that I = E/R, so for a given load, current is directly proportional to
> voltage.
>
> The situation is more complicated when the motor is running at normal load
> and speed (back EMF and all that), but my concern would be with what happens
> when you get to the end of travel, or have a very heavy load that keeps RPM
> way down. Under stall or near-stall conditions, double the voltage means
> about double the current.
>
> It's the current that makes the heat that burns out your armature. That's
> part of what the fuse is supposed to protect against. But since it's still
> current that concerns you, not power, I don't see the need to halve the
> fuse's rating. I'd think that would just cause nuisance opens. What am I
> missing here?
>
>
> David Roden - Akron, Ohio, USA
>
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