Guide to motor selection - suitable power?

[MattyB]

@Peter Beeney, sorry, this PS somehow got cut off in my original post earlier...

 

12 hours ago, Peter Beeney said:

It would seem that no one has come back on my query about the motor's magnetic fields earlier so now I’ll take that as a yes then. Actually I think there is a little clue somewhere right in the middle very much related to the motor’s performance and it all just revolves around this. Sorry about the puny pun but it would just keep on going round and round….

 

Since you issued us a challenge (and I think we all agree that your statement around in interaction of magnetic fields is correct), here is one for you around your thoughts on how wattmeters work. 

 

Scenario:

1. Model (secured safely) on ground, battery plugged in, wattmeter in series, throttle at zero. Prop not turning, wattmeter reading total pack voltage, current 0A, power 0W. 
2. Throttle is then advanced to full, prop turning at high speed. Wattmeter reads total pack voltage, instantaneous current being pulled and total watts at that moment (VxI, voltage multiplied by current). 

 

Questions:

• According to your thinking the wattmeter is only measuring the heat being dissipated by the motor. If that is the case what source of power is turning the prop and accelerating the air?
• Given the wattmeter is sitting between the battery and ESC and the battery is the only obvious power source in the system, why do you believe the energy required to turn the prop and accelerate the air is not being measured by the wattmeter?

 

Edited September 23, 2021 by MattyB

[Peter Beeney]

Hi Matty,

 

Thanks for your reply and I’m quite happy to try and answer your questions. Also thanks for the concurrence re the magnetic fields, this may eventually prove to be quite important.

 

But with the greatest respect there is one tiny detail I’d like to rectify first. I don’t think I’ve ever said that it’s just the motor that’s dissipating heat, my quote was the entire power train so of course I can’t really answer question one properly as it stands but I will try to answer it if I just rewrite it slightly like this:

 

  1  According to your thinking the wattmeter is only measuring the heat being dissipated by the entire power train. If that is the case what source of power is turning the prop and accelerating the air?

 

So just to qualify this a little further I consider the basic circuit might be the battery, wattmeter, ESC and motor all connected in series. the meter’s ammeter, a low resistance, is the connection in series and the voltmeter, a high resistance, is connected across the battery supply. Because this is a closed series loop when the motor is running the same current value is flowing all round the circuit and the ammeter has no option other than to measure the amps flowing, the voltmeter has no option other than to measure the buttery voltage and because the watt is a derived function the wattmeter has no option other than to multiply the volts and the amps together. One way of describing the watt might be to say that it’s a unit of work done in a given time.

 

 All the parts of this loop have resistance on one level or another and when the circuit is closed the potential difference between the battery terminals causes a current to flow. At ambient temperatures the resistance acts as a friction to the current flow and that creates heat in the process; an effect that in this case is undesirable. The meter cannot measure this directly but if the watt can be said to be a measure of work done in a given time then that might well be called heat. A topical example of this might be the ubiquitous 3kw heater soon to be firing up again where the heat (watts) will be fully appreciated.   

 

Also accompanying the current flow wherever it goes is the magnetic field which surrounds the conductor. This is an absolutely crucial element required in creating the power of the motor but like the temperature the wattmeter cannot measure magnetic field strength. Therefore it cannot measure the rotary action of the rotor to which the propeller is attached either.

 

So in my opinion what we are seeing in the wattmeter readout in Scenario 2 are the total watts being expended by just the closed loop circuit within the model. I’m including the motor’s coils wound on the stator as well because they are part of the circuit. When the motor is running normally it’s heat dissipation, or watts, is relatively speaking actually quite a minor part of the whole picture;   ....and is there a cryptic clue in there as well……    

                           

So to try and answer question one as I rewrote it.

 

The source of power driving the prop is the action taking place between the magnetic fields; though perhaps a kind of half transfer of power might just be a more appropriate way of describing it.

 

So then leading on to question 2, the reason the meter cannot read the the energy required to turn the prop and accelerate the air is because there is no electrical connection between the stator and the rotor, only a magnetic link; and thus no current to read. The meter can only  measure the current flowing in a closed circuit that it’s a part of. The mechanical power output of the motor can measured in watts or any other unit but it has to be an entirely different instrument to do it; and that’s another story for another time perhaps.

 

I think this is about as minimalistic as I can get at the moment, there are many more aspects to cover but it all takes time and space. Also Tony’s thread to think about but if he doesn’t mind too much perhaps…. As always this is only just my ramblings and certainly should not be considered in any way as factual. But it's always worked for me.

 

Just as an afterthought, if we supercooled the circuit to the point where the resistance disappeared then our wattmeter would always read zero. But as the magnetic field is unaffected by the cold our prop would keep turning just the same. I think this might go some way to indicating too that whilst the magnetic field and the current are inexorably linked together it’s also a fact that what I said when I came in, ‘the wattmeter is just telling me how many volts and amps are being converted into heat due to the resistance of the power train’ is correct. But all hypothetical anyway though, we may just get away with a supercooled ESC, I don’t know, but certainly not the battery, it wouldn’t work. So this is always going to have resistance resulting in some watts on the meter. 

 

 Now I’m quite sure we shall have to carry on gently agreeing to disagree for some time to come yet.

 

Best regards.

 

PB

[Martin Harris - Moderator]

In simple terms, you have a finite amount of potential energy stored in the battery.  All of this energy is supplied to the motor via the ESC and associated wiring.  A small amount of this energy is converted to heat by being "pushed" through the resistance of the wiring, ESC circuitry and motor coils - no significant part of this energy is converted to thrust.

 

Interaction between the magnetic fields generated by the current flow and permanent magnets, controlled by the ESC, turns the motor and propeller against air resistance, generating the thrust we require. This is where we want to use the majority of the potential energy - generating heat is simply an unavoidable waste.

 

The Wattmeter - as you say,  a combined current/voltage measuring instrument with a calculator, is displaying the amount of stored electrical energy being converted to heat and thrust.  If it's only measuring energy conversion to heat, where is the energy to turn the propeller and generate thrust coming from?  Short of there being a mystical unknown and as yet undiscovered alternative force, the Wattmeter MUST be measuring that energy!

[Shaun Walsh]

8 hours ago, Martin Harris - Moderator said:

In simple terms, you have a finite amount of potential energy stored in the battery.  All of this energy is supplied to the motor via the ESC and associated wiring.  A small amount of this energy is converted to heat by being "pushed" through the resistance of the wiring, ESC circuitry and motor coils - no significant part of this energy is converted to thrust.

 

Interaction between the magnetic fields generated by the current flow and permanent magnets, controlled by the ESC, turns the motor and propeller against air resistance, generating the thrust we require. This is where we want to use the majority of the potential energy - generating heat is simply an unavoidable waste.

 

The Wattmeter - as you say,  a combined current/voltage measuring instrument with a calculator, is displaying the amount of stored electrical energy being converted to heat and thrust.  If it's only measuring energy conversion to heat, where is the energy to turn the propeller and generate thrust coming from?  Short of there being a mystical unknown and as yet undiscovered alternative force, the Wattmeter MUST be measuring that energy!

To be strictly accurate, the battery has a finite amount of chemical energy which is released in the form of a flow of electrons. The battery's potential energy is determined by how far it would fall if you released it from its resting position and is a function of the battery's mass and the value of G (9.8 metres per second per second)

[Dickw]

Peter

It is quite frustrating because you seem to know the basics of electric systems, and are so close to understanding electric motors, but keep missing some vital points. It is also odd that you keep stating your ideas as if they were facts and then finish up by saying "this is only just my ramblings and certainly should not be considered in any way as factual."?

Anyway, on we go ?

 

12 hours ago, Peter Beeney said:

Because this is a closed series loop when the motor is running the same current value is flowing all round the circuit and the ammeter has no option other than to measure the amps flowing, the voltmeter has no option other than to measure the buttery voltage and because the watt is a derived function the wattmeter has no option other than to multiply the volts and the amps together.

Correct.

 

12 hours ago, Peter Beeney said:

One way of describing the watt might be to say that it’s a unit of work done in a given time.

Work done in a given time would be watt-hours (Wh). Watts (W) is the power doing that work at any instant.

 

12 hours ago, Peter Beeney said:

Just as an afterthought, if we supercooled the circuit to the point where the resistance disappeared then our wattmeter would always read zero.

Definitely not true for an electric motor system, and that is the major point you are missing. The watt-meter would still be measuring the power required to generate the constantly changing magnetic fields - i.e. the power required to run the motor and make it do work.

 

Watts input into the system, as you correctly point out, is battery volts x current (VI).

The heating effect in the system due to resistance in the circuit is current squared x resistance (I^2R).

Please note that for a running electric motor system VI does NOT equal I^2R.

 

So for a supercooled system the I^2R component (the losses) would indeed be zero, but the VI would not.

 

Dick

[Nigel R]

11 minutes ago, Dickw said:

for a supercooled system the I^2R component (the losses) would indeed be zero

 

We should note that the copper in a motor would never reach zero resistance, even at absolute zero, thus some heating losses would still occur.

 

 

[Dickw]

6 minutes ago, Nigel R said:

 

We should note that the copper in a motor would never reach zero resistance, even at absolute zero, thus some heating losses would still occur.

I was just trying to keep it simple (or as simple as possible ?)

 

Dick

[Martin Harris - Moderator]

3 hours ago, Shaun Walsh said:

To be strictly accurate, the battery has a finite amount of chemical energy which is released in the form of a flow of electrons. The battery's potential energy is determined by how far it would fall if you released it from its resting position and is a function of the battery's mass and the value of G (9.8 metres per second per second)

To be even more strictly accurate, you’re referring to the battery’s gravitational potential energy whereas I was referring to its electro-chemical potential energy. Like Dick, I was trying to express the concept in simple terms in order to help understanding of the point we’re trying to make. 

[Nigel R]

16 minutes ago, Dickw said:

(or as simple as possible ?)

 

that boat sailed a while back ?

[MattyB]

14 hours ago, Peter Beeney said:

...The source of power driving the prop is the action taking place between the magnetic fields; though perhaps a kind of half transfer of power might just be a more appropriate way of describing it.

 

So then leading on to question 2, the reason the meter cannot read the the energy required to turn the prop and accelerate the air is because there is no electrical connection between the stator and the rotor, only a magnetic link; and thus no current to read. The meter can only measure the current flowing in a closed circuit that it’s a part of. The mechanical power output of the motor can measured in watts or any other unit but it has to be an entirely different instrument to do it; and that’s another story for another time perhaps.

 

The piece you are forgetting is that there is a prop on the end of the motor shaft, and that prop is doing a huge amount of work accelerating air. This is by far the most dominant load in the system, not the I^2R powertrain losses. It also has a huge bearing on the current flowing (and therefore power measured) in the closed circuit.

 

Imagine this.... I have motor of 1000Kv (Kv definition) with no prop attached and a unwaivering 10V bench power supply that will never move irrelevent of the curent pulled. If I go to full throttle the motor will try and turn at 10 x 1000 = 10,000RPM. Since the motor has no load on it the wattmeter will only measure the I^2R losses in the powertrain (essentially the energy needed to accelerate the rotor and overcome any internal friction in the bearings). Current and power readings will therefore be very low. 

 

Now I bolt a 12x6 prop onto the front and do the same test. The ESC knows nothing about the load, so at full throttle it will still try to run the motor at 10,000 RPM (as the Kv and input voltage remain unchanged). The difference now is that takes far, far more power because in addition to the I^2R losses we are also accelerating a massive volume of air via the prop.

 

Remember the first law of thermodynamics - the total energy of an isolated system is constant; energy can be transformed from one form to another, but can be neither created nor destroyed. Since air is being accelerated we must be converting energy from somewhere into that kinetic energy; it can't come for free. So what happens? The power and current consumption measured by the wattmeter in the closed circuit both go up dramatically (in this example probably to ~400-450W), because turning the motor at 10k is now far harder work. To be clear, the I^2R losses are still measured as part of the wattmeter reading, but they will are now a very small percentage, maybe 5-10% of the total reading.

 

One final point - The prop itself is not 100% efficient, so the mechanical power driving the model forward is undoubtedly less than the total wattage measured by the wattmeter minus the I^2R powertrain losses. However this is not normally something we need to to worry about as all the props that you'd run on a given model and powertrain will tend to have roughly the same efficiency*.

 

(* - Outside of specialist use cases such as very high pitch props on high RPM models that might be stalled or on the borderline during a static test)

Edited September 24, 2021 by MattyB

[Peter Beeney]

Hi Matty,

 

Thanks for your answer again and I do apologise for not being able to get back any sooner.

 

But with the greatest respect I’m afraid I still have a few problems with it so perhaps I can explain how I think an electric motor operates.

 

When a conductor is moved through a magnetic field a voltage appears on the conductor. One situation that affects this voltage is the speed at which either the field or the conductor is moving, in general the faster the speed, the higher the voltage. So this voltage appears on the motor’s winding all the time it’s turning, it’s called the back emf - electromotive force or sometimes motional force, I guess because it’s facing back toward the incoming supply voltage in a mirror image. The back emf directly opposes or cancels out the supply voltage all the time to a greater or lesser degree but can never equal or exceed it.

 

If I might borrow your example for a moment (with no prop on) then at the moment we apply the full supply voltage the motor is stationary and this voltage will appear on the motor’s coils, the current will be at a maximum and so will it’s accompanying magnetic field. Therefore the deflecting action between the magnetic fields is also at a max., the motor starts to turn and as the speed increases the back emf also increases proportionally on the coils and opposes and reduces the supply voltage that is applied to them and in turn also reduces the current flow and it’s ever present magnetic field; eventually to the point where the magnetic field interaction  can just hold the unloaded motor’s speed steady. If it were able go any faster the back emf would equal the supply voltage, there wouldn’t be any current and the motor couldn’t run, an impossible situation to arrive at. Checking the wattmeter, we find the ammeter is reading this small amount of current, the least it will ever be. But to check the speed of the motor at this point I have to use a rev counter, my problem throughout this saga has been that I’ve never been able to understand how I can use a wattmeter to do this. When I do check it I find it is invariably reading the advised unloaded revs. But I have found more than one that is completely different.

 

If I now fit a prop on this will load the motor by some definite amount and therefore it will just simply reduce the amount of rpm the motor can now achieve. These will be definitely less than the unloaded rpm so this in turn reduces the back emf and thus increases the voltage applied to the coils and therefore the current and magnetic field strength. That increases the deflecting force of the magnetic fields by the exact amount to maintain this new speed. This action is completely self regulating for any increase or decrease in load however small. We now have three separate voltages, the bench supply, the back emf and the bench supply minus the back emf, let’s call this one the power voltage. It’s this power voltage that’s always applied to the motor’s windings; if the motor is running within it’s normal limits the power voltage will generally be quite low because the back emf is fairly close to the supply voltage and therefore the current flow will be within spec. But if we were to fit larger and larger props, steadily increasing the load, the motor's rpm would become less and less, the back emf would also become less and less allowing more and more power voltage to be applied to the motor’s windings until ultimately if the motor couldn’t turn at all it gets the full supply voltage with maybe smoke signals to indicate that this has happened. I’m sure very few few folk will ever get into this situation though, but a crash with the prop stuck in the mud and the throttle still full up can, and does, occasionally happen……

 

I find it very difficult indeed to see how the back emf can be related to the rotary output other than how I’ve described. It has no circuit so no current or associated magnetic field. It simply reduces or increases the supply voltage which reduces or increases the current flow to the motor. One point to note perhaps, every time the increased load reduces the back emf by a factor of two this doubles the power voltage and thus the amps flowing so the power dissipated goes up by a factor of 4. But for the rest of the circuit the supply voltage doesn’t change so these watts only double. This can only go so far though, when everything comes to a standstill but it’s still powered up it’s the total circuit resistance that’s the final limiting factor.

         

Not sure about the ESC driving the motor either, in my view rather more like the other way round. If we momentarily call it the Electronic Switching Commutator it may be a bit clearer. The brushed motor’s mechanical commutator is synced to the motor’s speed by simply being part of the armature and turning at the same speed, so it has to do the switching and 180 degree current reversal at the correct time whatever the load and speed. I don’t know very much about ESCs so I’m pretty much guessing here that the motor and ESC are synced and switched by the back emf that’s present on the windings that are not switched on at any one time in the cycle; So by the same token this also has to do the switching and current reversal at the correct time.  
I hope it’s possible to see from this tome why a low resistance all round is so important, the Holy Grail and all that stuff…. 

 

Re the previous ramble about supercooling, there maybe yet some truth in the old proverb - There’s many a true word spoken in jest…

 

First this:

An article I noticed recently said that someone, I think it’s the Danes, have erected a small wind farm using supercooled generators. They are claiming a one third increase in output together with a one third reduction in size as well. Very little other specific detail though; I looked for any indication as to how much of the extra power delivered needed to be fed back to maintain the low temperature, particularly in times such as recently with extended periods of no wind and warm sunshine but they’re not saying. Do they switch the fridge off or just leave it running I wonder… Also it’s one thing to build a magnet with no moving parts, how much thought must have gone into the materials used in making bearings etc. I guess if they get a result we shall see more about it, if not it will probably just quietly fade away.

 

So then a googled copy and paste from Wiki:

A superconducting magnet is an electromagnet made from coils of superconducting wire. They must be cooled to cryogenic temperatures during operation. In its superconducting state the wire has no electrical resistance and therefore can conduct much larger electric currents than ordinary wire, creating intense magnetic fields. Superconducting magnets can produce greater magnetic fields than all but the strongest non-superconducting electromagnets and can be cheaper to operate because no energy is dissipated as heat in the windings. They are used in MRI machines in hospitals, and in scientific equipment such as NMR spectrometers, mass spectrometers, fusion reactors and particle accelerators. They are also used for levitation, guidance and propulsion in a magnetic levitation (maglev) railway system being constructed in Japan.

 

Then I read that a Swedish gentleman had achieved superconductivity at a temperature of about minus 21 degrees C; little more than a freezer. No details on conductor material (or anything else) but if it could ever be forged into an electromagnet think of the possibilities. After all a motor might be called the Siamese’d twin of an alternator.

 

And now Ford have announced they are investing $11.4billion into new electric car and battery making plants, I’d be very much surprised if supercooled motors to power them aren’t at least in the line of consideration at some point. 

      

Finally the last chapter here is left to come, measuring the mechanical rotary action of the motor. I’ll kick it off by just saying that I think this is done by a dynamometer or similar, this measures the force of the deflecting action of the magnetic fields at the same time monitoring the rpm with a rev counter. Like the electrical watt the mechanical watt is also a derived function so it multiplies the two results together at various rpm to plot an output power curve.

 

NB   I think ‘deflecting action’ can probably be translated as an ‘attraction, repulsion or twisting’ force according to the orientation of the magnets at any one time. Or perhaps just simply as torque. 

    

I may have to go AWOL again for a while soon, so reply times could still be erratic, sorry.

 

PB

[Martin Harris - Moderator]

I'm a little lost Peter.  While a skim reading of your treatise gives the impression that your explanation of how an electric motor works and how the back EMF is used to communicate the motor state to the ESC is generally accurate, the fact remains that you appear to be claiming that watts in don't equate to thrust out minus any heating effect in the motor, ESC and wiring - i.e. your previous claim that all of the electrical power used is turned to heat.  This clearly can't be the case as the energy converted to thrust has to come from somewhere!

[Peter Beeney]

Thanks for that Martin and please hang on for a while when hopefully everything may become clearer. What I think I’m trying to say is that the magnetic field strength might have a significant part to play in all of this. It’s always seemed to me that if everything is tickety-boo in the model’s power department right from the start then a reasonable current flow will provide a powerful enough magnetic field to fly the model in the way that I intend it to fly; I don’t really ever need to change it very much. However, should I ever decide that I did want more power, as in chucking a larger prop on the front say, although most unlikely because as far as I’m concerned this is not always synonymous with a better all round model performance, then I think what might eventually happen is this.

The current flow would increase because the rpm has reduced and the magnetic field strength has increased. This increased current flow has to exert even more energy to overcome the the resistance of the conductor (the complete power train) and in so doing may create a degree of heat rising to a level which we definitely do not want, particularly when it occurs within the ESC and battery etc. but we can at least see this by monitoring it on the wattmeter. But the increased magnetic field strength does not require or take any energy at all from the current. So the wattmeter will only telling us that the smell of burnt toast is even stronger…….

 

It’s just my opinion that the magnetic field does not use energy from the current, I don’t know to be true. But if we get to a point down to where there is no resistance (supercooled) so the meter reading is zero and most importantly there is no heat produced but still current aplenty and as a result a strong magnetic field I can only take this to indicate that the wattmeter was previously telling me how much heat was being produced in normal circumstances. Also another small point, when the wattmeter is indicating the two highest points on the power train’s power output curve, that's max output, the first at the front end of the curve at the very instant of start up and the second at the back end when the motor’s shaft is prevented from moving, the same two points on the motor’s output curve that’s established by the dynamometer are showing the exact opposite, that there is no power (and no thrust) being developed by the motor at all. So I’d think any effort taken in between these points to be able to get a sensible indication of a level of one by simply reading the other is to say the very least titanically tenuous indeed. Having said all that I’m sure there is a point on the curves where they do converge; maybe quite closely. I don’t know this for sure, though, and nowadays I’m certainly not going to try and find out but I think the one thing I can say with considerable certainty is that the power of the interaction between the magnetic fields can never really be overestimated…..

 

Stay tuned…

 

PB

[Denis Watkins]

For lay people, and flyers

I have found that 9 times out of 10

Performance is improved by a smaller prop selection

Heat is reduced, at the ESC, motor, and power pack

And the model stays in the air longer.

[EarlyBird]

On 03/10/2021 at 00:15, Martin Harris - Moderator said:

I'm a little lost Peter

I'm totally lost and the more I read the more lost I become. 

8 hours ago, Peter Beeney said:

The current flow would increase because the rpm has reduced and the magnetic field strength has increased.

The motor uses permanent magnets which have a fixed strength. A larger prop requires a higher current for a given speed which reduces the voltage that the battery supplies resulting in a lower rpm. The ESC will drive the prop at an rpm, dependent on throttle position and battery voltage, the resultant current and watts is all that is of concern to me, which is shown on a watt meter.

 

[Dickw]

10 hours ago, Peter Beeney said:

t’s just my opinion that the magnetic field does not use energy from the current, I don’t know to be true.

Put simply and bluntly, your opinion is unfounded, and it is not a matter of opinion but a matter of fact established during the last century of electrical engineering.

The magnetic field requires energy to create and/or change it and in our motor it is constantly changing so requires a continuous energy input to maintain it. That is the energy being used to power our motors.

 

10 hours ago, Peter Beeney said:

But if we get to a point down to where there is no resistance (supercooled) so the meter reading is zero and most importantly there is no heat produced but still current aplenty and as a result a strong magnetic field I can only take this to indicate that the wattmeter was previously telling me how much heat was being produced in normal circumstances.

The watt-meter reading would not be zero if the motor was running. The resistance might be zero but the impedance is not. You still need volts to drive the current through the motor against the winding impedance and the back emf, so with volts and amps you still have watts input.

 

10 hours ago, Peter Beeney said:

Also another small point, when the wattmeter is indicating the two highest points on the power train’s power output curve, that's max output, the first at the front end of the curve at the very instant of start up and the second at the back end when the motor’s shaft is prevented from moving,

Those are not two different points on the motor power curve they are the same point, i.e. a stationary shaft while under power. Watts in but no output equals zero efficiency. No one is disputing this so I am not sure what point you are making.

 

The relationship between power in and power out is a statement of efficiency and is (relatively!) easy to calculate for an electric motor. I try to run my motors at about 80% efficiency meaning the losses (including heating loss) are about 20% of the power input.

An example of typical power curves for our type of electric motor attached.

 

Dick

 

Edited October 6, 2021 by Dickw
edit - sorry forgot the legend - Amps and Watts LH scale, efficiency RH scale, all against rpm at bottom

[Andy48]

Just to muddy the water, not all the losses are due to heat. The noise the motor/propeller system generates represents a loss of efficiency too.

 

Dick's graph is very appropriate too, though in my experience peak efficiency of a motor/propeller system seems to be generated at about 3/2 max power a bit earlier than the above graph, and drops off significantly above that power. The difference to Dick's graph may be accounted for by the fact that I am looking at the efficiency of the whole system rather than just the motor.

 

Thus when specifying a motor, it is always better to go for something more powerful than needed bearing in mind that running it above that 2/3 peak reduces efficiency, and also considering there is a very small weight penalty fitting a larger motor.

[Geoff S]

I tend to try to reduce things to be as simple as possible, especially when they involve esoteric issues like the magnetism sections of my college education, which occurred nearly 60 years ago and I tried to avoid in exams! ?  I was more interested in electronics or logic design.

 

Peter is struggling with 1^2 R losses in motors and confusing them with the useful power actually used to turn the prop and thus propel the model.  It might be easier to regard the motor as a perfect one.  A perfect motor has no friction, windage or copper losses yet will still provide power.  In fact the motors we use are approaching perfection compared to the run of the mill brushed ones that were the norm 10 years ago. 

 

Obviously copper losses, seen as heat caused by current flow through resistive can't be totally ignored but can be mitigated by regarding maximum current draw specified but, apart from that,  simple calculations will get you in the right ball park to select a suitable power train.  Friction and windage losses are absorbed in the copper losses as they will be overcome by drawing extra current and, I suspect, aren't all that significant.

[MattyB]

12 hours ago, Peter Beeney said:

What I think I’m trying to say…

 

It’s always seemed to me…

 

However, should I ever decide that I did want more power, then I think what might eventually happen is….

 

It’s just my opinion that the magnetic field does not use energy from the current, I don’t know it to be true…
 

…So I’d think any effort taken in between these points to be able to get a sensible indication of a level of one by simply reading the other is to say the very least titanically tenuous indeed.

 

…I don’t know this for sure, though, and nowadays I’m certainly not going to try and find out…

If you don’t know whether what you are posting has any basis in truth and aren’t prepared to do even 5 mins of research to find out whether established science supports your views, why post at all? The content posted by myself, @Dickw, @Mike Blandford, @Martin Harris - Moderator, @EarlyBirdand others in this thread is all well established electrical theory that has stood the test of time and can be found in any decent textbook on the subject or within 5 mins on Google.
 

I humbly suggest that before you post any more theories that could confuse newcomers (and established users of electric power!) you do that bit of research, particularly on brushless motors and ESCs which it’s clear you don’t fully nderstand. Sorry to be blunt, but I’m not sure debating another treatise of unproven guesswork is a good use of anyone’s time.

 

 

Edited October 6, 2021 by MattyB

[Peter Beeney]

Hi Matty,

 

Thanks for your reply and apologies for the intermittent reply time. I’m sorry that you think my theories and ideas don’t hold any water and have no basis in truth etc. but until I’ve thought up some better ones I’ll just stick with what I’ve got.

 

I have not the slightest desire at all to get involved in this sort of debate?, that was completely unintentional, but it’s definite and beyond any question at all that I’m not going to accept any of the comments that seem to continually come my way. So if you would please care to highlight all the areas of unproven guesswork that I’m alleged to have posted then I’ll do my very best to explain them further. The reason I write as as I do is simply because I’m not a suitably qualified person to give instructions or directions on the forum so I’m trying to make it crystal clear that no one has to take any notice at all of anything I write. In my first post I did say ‘in my view’ and I meant exactly that, it’s only my view and I’m not suggesting anyone else should necessarily pay any attention to it at all.  

 

With the greatest respect to you if we are going to speak bluntly then may I take a passing glance at a couple of your quoted forumites? As you say, this is all well established electrical theory so this may well be a case of fools rushing in where angels fear to tread….

 

So also with the utmost respect to EarlyBird and Mike Blandford for starters can I say that I have a couple of queries about their posts on 6th Oct at 08.22 and 9th Sept. I’d most certainly like to ask them first if it’s ok to discuss this on the open forum though. I have no problems with this but if they have any reservations at all then I’ll completely respect and understand that. 

   

In my opinion I think Denis W has got it all just about spot on; in my book 100% right. Main Man with a Top Post indeed. I really wouldn’t want to put it any differently; a flyer who undoubtably much prefers to fly his models as opposed to winding up his wattmeter. He may well own one even, I’ve no idea; but as he takes the direction he takes I’m sure it’s only gathering dust……  If I might just add one small point though, I first noticed the ancient adage - For Success, Add Lightness a very long time ago and that has remained a bright inspiration for me ever since; but nowadays for the modern electric flyer it might well be seen as - For Success, Add Lightness…..And Less Propeller To Boot!

 

As I said before there's an ongoing chance I may not be around for an unknown time so a possible further delay in replying but I will post something about the motor’s rotary action soon which might help. By all means feel free to tear it to shreds if you so wish, that’s fine by me, but please back that up at the same time with some hard facts and figures and explanations as to why it is all unproven guesswork etc.

Cheers for now.

PB

[PatMc]

Peter, perhaps you could back some of your theories up with hard facts and figures but do try to be succinct or your posts won't be more than "speed read". 

[EarlyBird]

As I have said a Watt meter gives me the information I require and a Tacho does not.

For me there is no debate to be had.

Peter you win.

I am out of here as this has become tediously annoying for me.

[leccyflyer]

If you have and are using a wattmeter you are definitely doing the right thing. Adding a tachometer adds useful information to testing for the optimal power train, but, as you say, the wattmeter gives you the most important information that you need. 

 

 

[Outrunner]

For many years now I log data for every motor setup I have onto an excel spreadsheet. I record battery used, ESC, motor, prop, Watts, Amps and sometimes rpm. I refer back to this data when setting up a new model to get my setup into the right ballpark. To measure the data I use a Wattmeter and a tachometer. According to PB this all a complete waste of time as I've used a Wattmeter.

 

Why didn't anybody warn me that my Wattmeter is recording useless information and that I should chuck it in the bin.

 

My models fly nicely though.

[leccyflyer]

My e-flight mentor the hugely missed Peter Wilson used a similar system, recording all of that information on the cut out backs of Corn Flakes packets and I've carried that on, with the Excel spreadsheet same as you.  I don't record prop rpm as religiously as Peter did and haven't done so recently, since my workshop is lit exclusively by LED lighting and the optical tacho I use won't work under those circumstances. The Wattmeter is a tremendous tool and realistically, if you are flying anything but RTF using an electric power system you really should use one. For ease of use at the field especially I prefer a clamp meter for testing whether I risk overworking the power train- I make a basic assumption on the pack voltage of a fully charged pack and it's the amp draw that I'm interested in.