EV Warrior motor: stall torque of 691 oz-in = X operating torque ?

You can't.  The torque you can safely use in continuous operation depends
primarily on how much heat is generated inside the motor and how quickly it
dissipates.  You can calculate operating torque from the motor's torque
constant (torque/current) and either the continuous horsepower or continuous
current rating.

--
Matt

One web site gives the following specs for the EV Warrior motor. But the
term "ozf.in/A" is confusing to me.
It's the "f" in there that is confusing ... what's it mean ? ( I understand
oz-in. )

Thanks once again !

      Specifications for clockwise rotation: Motor only performance.
     Typical system performance.

      Voltage: 12 V

      Angular-velocity constant: 229 rpm/V

      Torque constant: 5.91 ozf·in/A

      Termaina resistance: 0.121 ohm

      No-load current: 3.9 A

      Peak efficiency: 64.3 %

      Power source resistance: 0 ohm
     0.121 ohm

      Peak power: 0.368 hp
     0.168 hp

      No-load angular velocity: 2640 rpm
     2530 rpm

      Stall current: 99.2 A
     49.6 A

      Stall torque: 563 ozf·in
     270 ozf·in

One presumes ozf is analogous to lbf, pound force, as distinguished from
pound mass. 'ozf' would then be ounce force, which is the same as ounce.
ozf.in/A is then oz-in/A.

Note that they specify for "clockwise rotation". Brushed DC motor
performance is strongly dependent on commuatation timing, advance or retard.
These values likely don't apply if your application is bi-directional.


=======

Matt, given this motor's Kt, Kv, and no-load specs, what do you think about
gearing for this motor for:

]]> - For example, to get 1Kg up a 1m, 45 degree hill in 1s takes 7N of
force at
]]> a speed of 1.44 m/s.
]]>
]]> - Applying a 10 Kg push through a distance of 1 m in 10 seconds is 100N
at
]]> 0.1m/s

Aha! thanks - that makes much more sense. I was thinking it meant lb-feet
somehow, which didn't make any sense to me.

Now on to the "/A" part --- does that mean per Amp ? so it would mean x lbs
force - in per one Amp ? I find several quotes on the net using this
nomenclature but no explicit definition yet.

The short answer is "Yes." The longer answer is "Sort of."

Kt, or torque constant, is one of the primary defining characteristics of
the motor. In an ideal motor -- ie, a theoretical electric motor -- torque
is directly proportional to current. Real motors have non-linear losses that
make this not precisely true, but the value is still useful in defining and
estimating performance.

Similarly, voltage is directly proportional to speed in an ideal motor. The
corresponding parameter is often denoted Kv, often expressed as RPM or
radians per volt. There is a fixed relationship between Kt and Kv,
accounting only for measurement units.

In practice, you might measure RPM, current, and voltage at two load
conditions near your operating points. Combined with the no-load parameters,
you can estimate fairly accurately motor performance near that range. The
values will likely differ from the specs, sometimes only trivially, other
times quite dramatically, depending on how and what was measured. Lacking
other information, published data is a reasonably good starting point for
selection.

Last, efficiency is sometimes denoted Eta (Greek letter) in specs. It is
actually a curve, not the single value specified. Not knowing how it was
measured, and what it represents (probably peak, at some unspecified RPM,
voltage, and loading condition), its value is not immediately useful.
However, knowing the likely range of efficiency of about 40% to 60% in this
case is perhaps meaningful in itself.

Extrapolating the rest of the performance graphs is not particularly
complex, but certainly more than will fit comfortably in a newgroup message
(or even a series of messages). A web search will certainly reap useful
information. In addition to extracts from college texts, don't forget the
R/C hobby sites. They also have strong vested interests in extracting high
levels of performance from their motors, and often presented without the
(IMO, unnecessary) higher math.

Mike, keeper stuff here! You should put some of this on a Web page!

-- Gordon

Well, thanks, Gordon. But others beat me to it :), and go much much deeper.
By now, I'm certain I've forgotten more than I recall.

I am certainly copying it for *my* notes! Thanks again guys.

Well, I really can’t say without the motor’s current rating.  Guessing 20A
(see reply to pogo), which translates into 95oz-in of torque, I would say:



95oz-in is 4.215 J/rev, and that 100N push in the second scenario is 100J/m,
so your gearing has to give you <= 4.215/100 = 0.04215 m/rev.



Then there’s the question of how fast we can run the motor in the other
scenario.  We need 10W of power, which we can deliver at 2615 rpm with 12V,
which is probably safe.



The required 1.44 m/s is 86.4 m/min, so you need gearing at >= 86.4/2615 =
0.033 m/rev.



So the motor seems just big enough to do the job.  Splitting the difference
with the geometric mean, I’d try setting my gearing at 0.37 m/rev.



The 25% difference between the bounds is a little too close for my comfort,
though -- I would have to build and test this before being sure that this
1/5hp motor could meet the spec.

--

Matt

Good analysis in your other message, BTW.

In regards your comments here, the motor will be much happier on 24V, even
though it isn't unhappy on 12v. At which point it becomes a 1/2 hp motor,
assuming you can take the heat away fast enough. (IOW: It's the wrong motor
for the job; too big. Kv can be higher, and it can do with much less copper
in the windings.)

Only if it's expendable.  Otherwise, you still have to keep under the
current (torque) rating, and now you have to do something special to keep it
down to a safe RPM.   The maximum safe RPM is usually a bit higher than the
no-load speed at the rated voltage, but not that much higher.  Minimal load
at twice the rated voltage will certainly break something after a while.


You can't.  The cooling rate for a brushed DC motor is mostly determined by
how fast heat can get from the rotor to the housing, which you have no
control over.  That heat has only bearings, brushes, and air to travel
through, so it goes pretty slowly, and anything you do outside the motor
will help less than you think.  To make matters worse, though, when they
test the current rating, they usually bolt the motor to a big aluminum plate
that can quickly dissipate any heat that makes it to the housing, so you
really can't expect to do any better at all in practice.

--
Matt

To add to Mike's excellent summary, the "f" stands for force, and it's
actually the more accurate description of the measurement. This style of
reference has picked up momementum during the last few years. Except for
engineering texts, most older books and articles won't use this
nomenclature, but a lot of the newer stuff does. The nomenclature is
designed to reduce ambiguity.

-- Gordon

Thanks Gordon. What about the "/A" part ? Does that mean "per amp" ?
Thanks again !

Yes. Many motor manufacturers use a traditional value of
torque-per-ampere for a torque constant, though the way they represent
it is not always the same. The factor they provide in the specs you
printed corrolates (more or less) with the other figures.

Recall that a DC motor produces torque linearly with applied current.
You can fairly easily determine the torque for a given current with
simple math -- Kt (the symbol they use for torque constant) x I (er,
amps).

The other way is to power the motor and put the shaft between your
teeth. I've become quite good at guestimating torque using this simple,
if not dentist-disapproved, method!

-- Gordon

Yet even more informative stuff --- thanks again!

Perhaps an "Enamel Abrasion Factor" may be adopted as the universal
measurement of motor torque in the future?  :-)

Well, you don't really have enough here to calculate the motor's continuous
torque rating.  A quick check shows that the "peak horsepower" they give was
just calculated as 0.25(stall_torque*no_load_rpm), which is a pretty good
estimate of the maximum power you'll be able to get with a 12V supply, but
is probably a lot more than you should run it at continuously.



One thing we can use to take a guess is peak efficiency.  Since these motors
come from electric bikes with 12V batteries, it is reasonable to guess that
they are designed to be most efficient when delivering their rated
horsepower at 12V.



From the stats given, we know that at 12V, with t=torque:



RPM = 2640-t*4.67



Current = 3.9A +t*0.17



Input power is current*12 watts:



Pin=(3.9+t*0.17)*12



Output power is torque * RPM oz-in-rpm, or torque * RPM/1350 watts:



Pout=(2640t-t*t*4.67)/1350



Efficiency = Pout/Pin, and it is maximized when t“.2 oz-in.



Speed at that torque is 2205 RPM, and output power is 152W, or 1/5 hp.

Current draw at that torque is 19.7A and input power is 237W.  That makes
efficiency 64%, which agrees with the number quoted in your specs (they will
have calculated it the same way).



Since 19.7 is close to 20, I'd guess that the motor's current rating is 20A,
which translates to 95oz-in continuous torque.



All of those guesses seem pretty reasonable to me, given the size of the
motor.



--

Matt

Wow --- thanks for the detailed explanation. I am going to save this thread
in my notes for future reference. I really do appreciate your taking the
extra time needed to explain all of this.

Many thanks again.