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Old Jan 05, 2012, 05:40 PM
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yomgui's Avatar
France, RA, Cormaranche-en-Bugey
Joined Feb 2011
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Hey! sorry I didn't meant to start such a controversy!

I was just surprised by the "Im-Io".

Because I think Io as a way to take "rotor torque" into account... so nothing different than just I on a motor with no load...
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Old Jan 05, 2012, 06:09 PM
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Lnagel's Avatar
Moab, Utah, USA
Joined Apr 2003
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I agree with you, yomgui. Io is current that overcomes the motor's built in losses so it cannot contribute to the torque required to drive a load. Therefore it has to be subtracted from motor current when calculating output power. Pout = (Im - Io) * (V - (Im * Rm)). But Io has the same effect on motor RPM as load current does. Therefore as the motor's load increases the load current is just an extention of Io so Io is not subtracted in the RPM calculation. RPM = kv * (V - ( Im * Rm)).

Larry
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Old Jan 05, 2012, 06:14 PM
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Moab, Utah, USA
Joined Apr 2003
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Quote:
Originally Posted by Martyn McKinney View Post
We seem to be having a communication problem. I have been on these threads many years and recognize both you and yomgui as posters who "get it".
I like to think I get it and from what I've seen of your previous posts you get it too. I'm just having trouble interpreting some of what you are describing. I guess we can chalk it up to semantics. In any case, thanks for taking the time to attempt to unconfuse me.

Larry
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Old Jan 05, 2012, 07:39 PM
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Toronto Canada
Joined Dec 2002
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Quote:
Originally Posted by yomgui View Post
Hey! sorry I didn't meant to start such a controversy!
It's actually my mistake.

When I first read your post about Io, it was on a cellphone and I didn't see the name of the poster. I thought the question was simply about Io itself, not realizing that there was an error in the formula in the previous post.

After posting my first few responses from my cellphone, it was not until I got to my laptop and Larry began asking pertinent questions that I realized my error.

Regardless it was an interesting exercise for me.
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Old Jan 06, 2012, 11:49 AM
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Fourdan's Avatar
Antony (France)
Joined Sep 2003
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Quote:
Originally Posted by Lnagel View Post
I agree with you, yomgui. Io is current that overcomes the motor's built in losses so it cannot contribute to the torque required to drive a load. Therefore it has to be subtracted from motor current when calculating output power. Pout = (Im - Io) * (V - (Im * Rm)). But Io has the same effect on motor RPM as load current does. Therefore as the motor's load increases the load current is just an extention of Io so Io is not subtracted in the RPM calculation. RPM = kv * (V - ( Im * Rm)).

Larry
Hi
I agree with Larry

In fact Io corresponds to several losses factors
a) Hysteresis magnetic losses
b) Eddy currents losses
c) ESC (MOSFETs) losses
d) Bearings friction
e) Aerodynamics drag of rotating rotor

a) and b) are major parts generally
Louis
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Old Jan 07, 2012, 12:33 AM
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Bruce Abbott's Avatar
Hastings, New Zealand
Joined Jan 2001
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Quote:
Originally Posted by Fourdan View Post
In fact Io corresponds to several losses factors
Unless you try to run your brushless motor by connecting two wires directly to the battery, in which case you only have to worry about resistive losses.
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