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Brushless motor sizing

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Sulaiman

Wed Oct 07 2009, 05:17PM

Registered Member #162 Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3140

Just to add to the confusion;
1) Power is roughly proportional to speed cubed. i.e. 2x faster = 8x more power.
2) Current is proportional to torque
3) Torque is proportional to speed squared
4) Voltage is proportional to rpm
5) rpm is proportional to speed

Dr. Slack

Wed Oct 07 2009, 06:40PM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

Here I lost you. Why is it appropriate to use the "rated current" to estimate output power at any RPM? I thought rated current was just a thermal dissipation concern?

It's appropriate to estimate the maximum power, because the rated current gives you the max continuous current that won't cause the motor to overheat (just an important thermal disspiation concern, the short term output can be much higher). So when you multiply that max current by the input voltage (which is implied by the rpm through kv) you get the max continuous input power (60*12 = 720 watts). When you knock off a realistic amount for losses, you end up with a plausible 500w outupt shaft power.

I think Saluiman's confusion generator could do with clarifying a little

1) For a propellar or fan yes, not for a motor
2) motor current and torque are proportional with a ratio ki
3) This is for the propellar or fan load again, and is implied by 1)
4) <-> 5) these are equivalent, motor voltage and rpm are proportional with a ratio kv

The important thing is that volts * current is input power (watts), torque * radians/second is output power (watts). If a motor has stronger magnets, it will generate more torque for any given input current, but will turn more slowly as it generates a higher back emf. Weakening the field makes a motor speed up, up to a point, until the torque losses becomes significant.

The kv*ki product is always 1 (for a lossless motor, implied by conservation of energy) when both are expressed in consistent units, so rad/s/V and Nm/A torque, or rpm/V and some bizarre imperial torque per current unit that I can't be bothered to dimension).

If you are happy that you understand transformers well, then consider a motor as a type of transformer, their correspondance is complete (apart from the obvious moving or not moving thang)

a) power out ~< power in (there's a very good reason for this restriction **)
b) lossless model is easier to start with
c) apply input voltage, input current is zero with no output current (torque) assuming lossless, relatively small assuming losses
d) output voltage (speed) is governed by the input voltage
e) as you consume output power, by drawing current (or putting a torque load on the output), the input current rises to draw input power, with input current governed by the output current (torque)
f) current in the windings causes it to get hot, you can overcurrent for a short while

(** it's the site rules, section II, rule I, no pseudoscience, if output power > input power, then a mod would lock the thread for discussion of over-unity

)

AndrewM

Thu Oct 08 2009, 02:54AM

Registered Member #49 Joined: Thu Feb 09 2006, 04:05AM
Location: Bigass Pile of Penguins
Posts: 362

Funny, once upon a time I considered myself smart, this thread is seriously impacting that belief. What you say make sense, but I can't figure out how to use it - I'm not using a current-controlled power supply, so I have to size my motor so as not to exceed its rated current at my design point.

Is the following not true, ignoring losses?:

I=(V-RPM/kv)/R

Dr. Slack

Thu Oct 08 2009, 07:41AM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

I=(V-RPM/kv)/R

RPM/kv gives the back emf, so if V is the supply voltage, then the difference is what drives I through the internal resistance of the motor R. So that's true. However, it's more useful for system analysis than for synthesis. The reason being that design doesn't change a whole lot if we assume a lossless motor, which would have R=zero in the above equation, so all it would say is that for a lossless motor, RPM/kv = V.

It's possibly better to synthesise from the point of view of power.

Let's say you want to deliver 500 (shaft) watts to the prop. If you have a few points on your prop curve, then read off the RPM that will do this.

You have a 12v power supply, and a motor which can tolerate 60A continuous. 12v * 60A is 720 electrical watts, which for a reasonably efficient motor of this size is spot-on for 500w mechanical output. So if your supply can handle this (*), there is a design which will work.

Now all you have to do is gear the motor to the prop, so that the 12v (loaded) speed of the motor is geared down to the 500watt speed of the prop.

At that design speed, the prop will demand 500 watts of power from the gearbox, which will throw (500W+gearbox losses)/speed of torque onto the motor, which will demand (500W+gearbox losses + motor losses)/volts of input current from the supply.

Note that the 12v loaded speed of the motor will be somewhat lower than the 12v off-load speed. If your motor curves do not give a speed/load curve for the right voltage, then you can estimate it from your quoted equation above, or assume it's in the 10-20% reduction region. As you see, the difference between a lossy motor design and a lossless one is whether we account for the relatively small difference in loaded and unloaded speed.

* If your supply can't handle 60A, then what can it do? Say it's 40A. The max electrical power from the supply is 40*12 = 480 watts, which you could expect to turn into maybe 300 shaft watts. Find the prop speed that will absorb 300 watts, and do a new gearbox design so that the 12v motor speed will be geared to the 300 watt prop speed. Same motor, different gearbox. The slower prop will demand less torque, which will require less current.

Steve Conner

Thu Oct 08 2009, 10:21AM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

AM wrote ...

I'm not using a current-controlled power supply, so I have to size my motor so as not to exceed its rated current at my design point.

Is the following not true, ignoring losses?:

I=(V-RPM/kv)/R

It's true, but useless, because R is a loss term. If you allowed it to control anything, the system would be inefficient. (It's resistive ballast.) In a good motor, R is negligibly small, so I tends to infinity.

In practice, your ESC should have an electronic current limit that you can set to whatever you want. This keeps the current under control, even under the assumption that R=0. Current limiting is vital with "good" motors, it stops batteries being destroyed and shafts snapped, and going into current limit is a normal part of operation.

Dennis Rogers

Fri Oct 09 2009, 04:38AM

Registered Member #1837 Joined: Tue Dec 02 2008, 02:20PM
Location: NYC
Posts: 65

Another point to consider is putting a slow-blow / time delay fuse or overload ahead of your motor. This way you can conduct carefree experimentation without risk of wrecking your hard earned motor. This should be included in the final design as well. (I'm guessing you knew this already, but just in case.)

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