Guide to motor selection - suitable power?

[Peter Christy]

Tony: If your proposed model was originally designed for an i/c engine, then the calculations become much simpler!

 

Phil Green (of Arduino encoder and retro modelling fame) has pointed out that a wattage of 100x(i/c engine capacity in cc) is a good match. IE, if the model was designed for a 2.5cc engine, you need a 250 watt motor. The nice thing is that this also works for helicopters, where the watts per pound formula doesn't! (Helis tend to be a lot heavier for a given engine capacity than aeroplanes!) My old Schluters were designed for 10cc engines, and I'm using 1000 watt motors. They are very adequately powered!

 

However, that is not the end of the story. If you intend using the original i/c propeller sizes, you also need it to be turning at the same rpm. Most small i/c engines peak at around 12,000 rpm. Confusingly, the kV rating of electric motors has nothing to do with KiloVolts, but refers to the motor's speed - a 1 kV motor would turn 1 rpm for each volt applied. So, to achieve 12,000 rpm on a 3S pack (roughly 12 V), you would need a 1000 kV motor.

 

Now that kV rating is under no load, so its probably best to go a little higher. For example, my helicopters fly on 6S (roughly 24V), so to achieve 12,000 rpm, I needed a 500kV motor. The 500kV motor did indeed fly it, but it was a bit marginal. It really needed a bit more head-speed, so I replaced it with a 580kV motor, which was perfect for the job. (Unlike an aeroplane, you can't fiddle with the prop size on a helicopter!)

 

In other words, work out the kV rating needed from the battery size you intend to use (3S, 4S, whatever) and then get a motor of the next highest kV rating above that for a bit of "elbow room".

 

Most models designed for electric power use bigger props than their i/c equivalent. This can be inconvenient when converting a model designed for i/c, however, you can always increase the pitch, or use a 3-blade prop (if available) to increase the current draw - and hence power - if required. A watt meter is well worth the investment when experimenting!

 

The Veron Robot of mine that you flew at Larcombe flies very nicely on an old OS15 (2.5cc). To convert that, I would be looking for a motor in the 250-300 watt range, and a kV rating of 1100 to 1300 for a 3S battery. However, I think 3S would be pushing it a bit for a model of that size, and 4S would be more appropriate. This would yield a kV requirement of around 800 plus a bit for safety - say 850-900.

 

Under-propping an electric motor won't do much harm. It simply won't draw as much current or deliver as much power. Over-propping one can exceed the current rating of either the motor or ESC, hence the desirability of a watt-meter when experimenting!

 

Hope this helps!

 

--

Pete

 

[Tony Harrison 2]

1 hour ago, David Ovenden said:

Yes these wattmeters work very well. 

Hi David - if you use one of these, it's a good recommendation. Will order one. Thanks for the bits & pieces.

Best, Tony

[Tony Harrison 2]

39 minutes ago, Peter Christy said:

Tony: If your proposed model was originally designed for an i/c engine, then the calculations become much simpler!

 

(snip)

 

Under-propping an electric motor won't do much harm. It simply won't draw as much current or deliver as much power. Over-propping one can exceed the current rating of either the motor or ESC, hence the desirability of a watt-meter when experimenting!

 

Hope this helps!

 

--

Pete

 

Certainly it helps, Peter, and thanks. I'm back in Devon at the weekend, will bring along the plane and the bits for it (supplied by David O of this list, whom you know via the forum) to show you when we meet at the field.

rgds Tony

[Peter Christy]

Hi Tony,

 

I'm hoping to be at the BMFA Scale Helicopter event at Buckminster this week-end - weather permitting - so I won't be around. I'll be traveling back on Monday. How long are you back for?

 

--

Pete

 

[Tony Harrison 2]

1 hour ago, Peter Christy said:

Hi Tony,

 

I'm hoping to be at the BMFA Scale Helicopter event at Buckminster this week-end - weather permitting - so I won't be around. I'll be traveling back on Monday. How long are you back for?

 

--

Pete

 

Back for the winter, Peter, so I'll be up for any Larcombe meetings when weather permits.

rgds Tony

[PatMc]

5 hours ago, Peter Beeney said:

I’m reluctant about the wattmeter, in my view this is just telling me how many volts and amps are being converted into heat due to the resistance of the power train

 

A wattmeter doesn't tell you that.

[Peter Beeney]

That’s interesting, Pat, maybe I should do some revision perhaps. My only excuse is that I haven’t got a wattmeter so can I just plead  extenuating circumstances……  

 

PB

[MattyB]

6 hours ago, Peter Beeney said:

I’m reluctant about the wattmeter, in my view this is just telling me how many volts and amps are being converted into heat due to the resistance of the power train

 

Long answer: A wattmeter is telling you the instantaneous current draw in amps and total pack voltage at that point in time under load. Amps and Volts are multiplied together to give the total power drawn. The magnitude of the load is determined (in simple terms) by a combination of the motor Kv, prop and input voltage, but the wattmeter doesn’t know anything about that or indeed what is generating the load. What is measured by the wattmeter includes energy lost within the system from the ESC forward that doesn’t go to generating thrust (i.e. heat, noise etc), but it’s not limited to just the losses. I seem to remember reading most brushless power trains are in the 80-85% efficiency range end to end, so if you are measuring (say) 300W at the wattmeter probably ~250W is going into the prop.
 

Short answer: Every electric flyer I know has a wattmeter, and they all use them to optimise power train performance and avoid a overloading their components. If you are right, they are all wrong… ?

Edited September 8, 2021 by MattyB

[PatMc]

1 hour ago, Peter Beeney said:

That’s interesting, Pat, maybe I should do some revision perhaps. My only excuse is that I haven’t got a wattmeter so can I just plead  extenuating circumstances……  

 

PB

Why the post if you don't know what a wattmeter actually measures ?

[PatMc]

37 minutes ago, MattyB said:

 

Long answer: A wattmeter is telling you the instantaneous current draw in amps and total pack voltage at that point in time under load. Amps and Volts are multiplied together to give the total power drawn. The magnitude of the load is determined (in simple terms) by a combination of the motor Kv, prop and input voltage, but the wattmeter doesn’t know anything about that or indeed what is generating the load. What is measured by the wattmeter includes energy lost within the system from the ESC forward that doesn’t go to generating thrust (i.e. heat, noise etc), but it’s not limited to just the losses. I seem to remember reading most brushless power trains are in the 80-85% efficiency range end to end, so if you are measuring (say) 300W at the wattmeter probably ~250W is going into the prop.
 

Short answer: Every electric flyer I know has a wattmeter, and they all use them to optimise power train performance and avoid a overloading their components. If you are right, they are all wrong… ?

Remember that a wattmeter measure the voltage applied to the ESC not the voltage at the motor therefore it only gives a true reading of watts at WOT.

If you want to measure the heat lost in the motor you need to know the resistance of the coils circuit that's presented to the ESC output, then calculate using current sq * resistance.

[Peter Beeney]

Pat,

 

My first post was simply to try and explain how I might use a tacho to look at how an electric motor was performing and I was just simply stating what I think the wattmeter is measuring. Also I’m not really sure that I said or implied anywhere in the second post that I didn’t know what the meter was actually measuring, sorry if I gave that impression; just an attempt at a friendly answer to your statement. 

     

I don’t think I’ll be swapping the tacho anytime soon…

 

Stay cool…

 

PB

 

[MattyB]

3 hours ago, PatMc said:

Remember that a wattmeter measure the voltage applied to the ESC not the voltage at the motor therefore it only gives a true reading of watts at WOT.

Agreed, I simplified the example a tad. I was just trying to get the point across that the wattmeter is not just somehow magically measuring heat being dissipated by the motor, but all the power being consumed by the system as a whole.

 

3 hours ago, PatMc said:

If you want to measure the heat lost in the motor you need to know the resistance of the coils circuit that's presented to the ESC output, then calculate using current sq * resistance.

Agreed again. ?

 

2 hours ago, Peter Beeney said:

…I was just simply stating what I think the wattmeter is measuring. Also I’m not really sure that I said or implied anywhere in the second post that I didn’t know what the meter was actually measuring,

Sorry, I am totally confused by that sentence tbh! 
 

Putting that aside, we know that current is measured in digital wattmeters and telemetry sensors use Hall effect sensors. They are simple and cheap, measuring the induced voltage in a transverse conductor that is proportional to the current flowing through the main battery wires. They have nothing in them capable of measuring the resistance of a motor or any other parameter of the load. More information here.

 

Example - Frsky 150A current sensor with the loop clearly visible where the ESC wire goes:

 

Edited September 8, 2021 by MattyB

[Dickw]

37 minutes ago, MattyB said:

Putting that aside, we know that current is measured in digital wattmeters and telemetry sensors use Hall effect sensors.

 

Just a minor irrelevant point as I agree that most current sensors probably use Hall effect, but the SM UniLog current sensors use resistance (I can confirm this having taken apart one of my 400A sensors) and the UniSens-E may also (but I have not taken one of those apart yet).

 

Dick

[Peter Beeney]

Matty, I think I was totally confused by Pat’s answer as well tbh; it seemed to me that he thought for some reason I didn’t know what I was talking about. Or perhaps he was just saying I didn’t know what I was talking about… and maybe I don’t… Apologies for inadvertently also confusing you too.

 

Maybe my very simple way of thinking might help…or maybe not….  rightly or wrongly I’ll just carry on carrying on as I always do.       
 
If a voltage is applied to a 4 ohm resistor such that a current of 1 amp flows then 4 watts will be given off as heat. I very much suspect that a wattmeter also in the circuit will display the volts, amps and watts as three separate readings so I can only think that the watts reading is telling me how much heat is being created. If a voltage is applied to a 2 ohm resistor such that the same 1 amp flows there will be 2 watts of heat to lose. If I’m now a bit pedantic here I might also think that the whole circuit has resistance at one level or another and as the 1 amp current is flowing around all of this loop the wattmeter is actually measuring all the heat dissipated, even from the meter itself.

 

So now we have two different wattmeter readings for the same 1 amp current flow.

 

A motor operates by the interaction of two magnetic fields, one created by the flow of current through the windings. This magnetic field doesn’t require or use any energy to exist, it’s just always there as a sort of freebie, lucky for us in this case. It’s strength is also proportional to the amount of current flow; the energy is expended when the current flows through the resistance of the windings and is again all dissipated as heat. From the examples above it seems to me that if these were two different motor windings then for exactly the same current flow and therefore magnetic field strength they would have exactly the same performance.

 

If I were relying on just the meter to tell me this I’m sure I’d very quickly become even more confused than I am now, but my trusty tacho reading the revs would at least help me out a bit. If one motor were supercooled to the extent there was no resistance then there would be no watts reading at all but the tacho will still give me a true reading and I’m sure it would tell me that the prop was still turning normally. 

                 

This is also why I’ve always thought that a low resistance in the motor and the associated other parts of the power supply is the Holy Grail of motor performance.

 

Sorry if this might be going a bit a bit off topic now.

 

PB  

 

 

 

[PatMc]

18 minutes ago, Peter Beeney said:

 

 

 the energy is expended when the current flows through the resistance of the windings and is again all dissipated as heat.

 

No it isn't all expended as heat. Most of it is used to drive the load.

[Dickw]

33 minutes ago, Peter Beeney said:

......................... the energy is expended when the current flows through the resistance of the windings and is again all dissipated as heat. ............................................................

Peter

The only time that statement would be true would be at wide open throttle with the motor stalled (i.e. the shaft clamped stationary).

Once the motor begins to spin ohms law and winding resistance become much less relevant to the relationship between volts and current in the motor.

 

Dick

[MattyB]

36 minutes ago, PatMc said:

No it isn't all expended as heat. Most of it is used to drive the load.

Agreed, this is getting silly. Peter seems to have forgotten that a very high percentage of the electrical energy is converted to kinetic energy by the motor and prop as they accelerate the air, generating thrust.  If it were really true that “…the energy is expended when the current flows through the resistance of the windings and is again all dissipated as heat“ then the prop/EDF/blades would never turn, and the plane would sit stationary at the end of the runway gently warming the air around it. That wouldn’t make for a terribly exciting hobby…? 

Edited September 8, 2021 by MattyB

[SIMON CRAGG]

No wonder Tony was / is getting a bit confused.....reading some of this he will be put off for life. Whatever happened to the "KISS" principle?.

[Peter Beeney]

With regard to the output power of the motor as I said this is the interaction between two magnetic fields. It’s a deflection motion between the poles. It doesn’t require or use any energy to do this. The output power of the motor is calculated by multiplying the torque by the rpm and the torque is measured on a dynamometer although I guess these days there are other devices that do this just as well. The energy requirement comes from the current flowing in the windings, the more current flow the stronger the magnetic field surrounding the windings. The current flowing in the windings creates heat due to the resistance of the conductor and so likewise the more current flow the more the more heat generated. The wattmeter is measuring the amps and volts, multiplying them together and displaying the result as watts. This is the heat generated, not the magnetic field strength.

 

Just borrowing Dick’s example for a mo with the shaft clamped then my take on this would be that be that because that the full battery voltage would be looking at a low resistance a maximum current flow would soon exist and the meter would be demanding more noughts to give a true reading. But at the motor the shaft is stationary, the rpm is zero so the torque x rpm equation also has to be zero. Max electrical watts for zero mechanical watts. Not a good situation to be in I agree but just making the point that the meter is measuring the electrical power which is dissipated as heat.

                          

Them’s some of my apples and I’ll be sticking with ‘em until I see a better tree…

 

PB

[Martin Harris - Moderator]

Surely according to accepted principles, energy can neither be created nor destroyed.  With a conversion of all the available energy of the battery to heat, how does the model get propelled forwards?  Isn't the heat generated simply energy that isn't being used for propulsion and, for our purposes, wasted?  I really can't accept that your magnetic deflection requires no energy to accelerate an object...the potential energy between the magnetic poles has been provided by energy stored in the battery and as it's converted to kinetic energy in rotating the motor against the drag of the propeller, it is replaced continuously from the battery.

[Dickw]

9 hours ago, Peter Beeney said:

..................................

Just borrowing Dick’s example for a mo with the shaft clamped then my take on this would be that be that because that the full battery voltage would be looking at a low resistance a maximum current flow would soon exist and the meter would be demanding more noughts to give a true reading. But at the motor the shaft is stationary, the rpm is zero so the torque x rpm equation also has to be zero. Max electrical watts for zero mechanical watts. Not a good situation to be in I agree but just making the point that the meter is measuring the electrical power which is dissipated as heat.

................................................................

Peter

You seem to be missing my point which was that this is the ONLY circumstance that the meter is measuring just the power that is dissipated in heat.

Once the motor is spinning the watt-meter would also be measuring the power used to spin the motor which would be by far the larger component of the overall power input.

 

The magnetic fields from the coils don't come as free energy. You seem to be forgetting basics such as the back emf being generated and the inductance of the coils. The pure resistance of the windings and their I^2R losses become much less important once the motor is spinning.

 

If your watt-meter shows 12V, 20A, 240W, that does not mean the resistance of the circuit is 0.6 ohms (R=V/I) as your simple explanation would suggest.

 

You seem to have a system that works for you, and that is fine, but publishing partial/incorrect theories on this forum does not help people and could cause them problems.

That is why a number of people are trying to correct some of your statements.

 

Dick

[Graham Davies 3]

Gents, All this is very interesting but is doing nothing to help the OP...

[Dickw]

10 minutes ago, Graham Davies 3 said:

Gents, All this is very interesting but is doing nothing to help the OP...

Agreed, but once incorrect information is posted it has to be countered for the sake of anyone reading this later.

 

Dick

[Graham Davies 3]

Does it? The poor OP was confused, and 2 pages of Mansplaining has just made it all worse...

 

[Shaun Walsh]

I just plug the wattmeter in between the battery and the ESC. If the number of amps drawn at WOT is less than the rating of the ESC and the motor with a margin for safety then I'm good to go.