|Sep 05, 2012, 09:53 PM|
Motor Question - Probably a dumb one!
If I have a motor that I only know the Kv and voltage rating. Is it safe to compare another motor which you have all the data and it's stator is within 2 mm in diameter but maybe 30 to 40% longer as having more power capabilities?
My logic says yes. But since I have only been fooling around with this stuff for a couple years I wanted to get reassurance from the more expert electric flyers.
|Sep 05, 2012, 09:59 PM|
Generally, bigger motors will be able to handle more current (and thus power at a given voltage) assuming the Kv values are the same and efficiencies are in the same ballpark (efficiency is determined by PRICE, not what the manufacturer's data tells you ).
Note that the Kv value itself doesn't tell you a whole lot, and most "voltage ratings" are useless numbers.
There are a whole lot of variables involved here.
|Sep 06, 2012, 03:12 AM|
Joined Nov 2003
If it's longer and it weighs more it probably has a correspondingly greater power capacity. The other parameters don't really matter very much as far as power rating goes.
200W is still 200W whether it's (approximately) 3S/18A @ 3000Kv or 2S/27A @ 1000Kv. Just different size props that's all .
|Sep 06, 2012, 11:36 AM|
As far as Kv, if you have a drill (prferably a drill press) a tach and an AC volt meter, you can figure out the Kv. Hook the motor to the drill and run it at a constant speed. Use the tach to get the speed and the volt meter to read the voltages from the three phases. There's a formula to calculating the Kv from the RPM and voltage generated, but Ken Meyers has developed a spread sheet to do the math. Here's a link to Ken's instructions with the spread sheet <http://www.theampeer.org/Kv/kv.html>.
Ken also has a rule of thumb for input power limits for outrunner motors based on weight. A good estimate is between 1.5 watts/gram and 3 watts/gram.
Those two sets of numbers, the Kv and the input power range, should get you n the ball park with any outrunner motor.
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