Guide to motor selection - suitable power?

[Dad_flyer]

I do record rpm, partly because I have a range of budget props as well as the good ones. There can be a remarkable difference: 9x4.7 with same current and rpm as the same brand 9x6 for example. Also I notice that without perfect connection the resistance of putting in the wattmeter reduces the rpm and current.

[Mike Blandford]

On 08/10/2021 at 22:51, Peter Beeney said:

. . .

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. . . .

   

 

Ask by all means. I've been flying electric for over 40 years, I still have my Astro Flight, ferrite motor I first use that was powered from 16, sub-C, 1Ah cells.

 

Mike

[Geoff S]

1 hour ago, Dad_flyer said:

I do record rpm, partly because I have a range of budget props as well as the good ones. There can be a remarkable difference: 9x4.7 with same current and rpm as the same brand 9x6 for example. Also I notice that without perfect connection the resistance of putting in the wattmeter reduces the rpm and current.

 

On models with Frsky Neuron ESCs most (all?) of the necessary parameters can be monitored without needing to fit a wattmeter.  When I've had both fitted they seem to cross-correlate quite well but, in reality, I'm not sure of the absolute accuracy but I think probably 'good enough for our purposes' covers it.  

[Geoff S]

1 hour ago, Mike Blandford said:

Ask by all means. I've been flying electric for over 40 years, I still have my Astro Flight, ferrite motor I first use that was powered from 16, sub-C, 1Ah cells.

 

Mike

 

I still have my Astroflight Wattmeter (the first?).  It cost an eye-watering at the time £50.  I now use a much cheaper HK one which also serves as a LiPo checker and seems to give very similar values - it cost less than £10!  A book that helped me a lot in learning about electric motors as 'black boxes' (ie regardless of their construction, brushless or brushed, in or out runner) was the Astroflight Electric Motor Handbook but it's probably long out of print as much of it is dedicated to their high quality brushed motors.

 

Geoff

[Outrunner]

I use a Astroflight Wattmeter too, given me 21 years sterling service so far. I always check a new setup with my Wattmeter. It's amazing how many people have no idea how many Amps their setup is pulling and wonder why the ESC does not last long or motor get over cooked.

 

 

[Trevor]

I’ve still got mine too but it gets very little use nowadays since I invested in a clamp meter. Okay, it doesn’t give a voltage reading but it’s much easier to use, especially at the field on other people’s models, because you don’t have to worry about connector compatibility.

[Geoff S]

2 hours ago, Outrunner said:

I use a Astroflight Wattmeter too, given me 21 years sterling service so far. I always check a new setup with my Wattmeter. It's amazing how many people have no idea how many Amps their setup is pulling and wonder why the ESC does not last long or motor get over cooked.

 

 

 

There is an enthusiastic group of F3A pilots at Ashbourne, many of them using very expensive electric power systems.  I've asked several times over the years how much current their motor draws and none of them had ever checked.  They simply buy an established set up and assume all is well and, to be fair, I haven't heard of any electrical disasters but I'd certainly like to know if it were me spending the money.

 

Geoff

 

[PatMc]

4 hours ago, Trevor said:

I’ve still got mine too but it gets very little use nowadays since I invested in a clamp meter. Okay, it doesn’t give a voltage reading but it’s much easier to use, especially at the field on other people’s models, because you don’t have to worry about connector compatibility.

I bench check with a wattmeter & tacho with every new set up but if I want to try a different prop at the field I just use the tacho to make sure that the revs are near the same or greater than prop I'm changing from. This ensures that the current is lower or hasn't changed significantly.

 

When bench testing I also check the rpm/v (AKA Kv) using the tacho & voltage reading from the wattmeter.

 

BTW I'm surprised no one has picked up that the definition of Kv in @Mike Blandford post on September 9th isn't the same as the definition in the link provided by @MattyB on September 24th.  Matty's definition is the one commonly used for model motors but I believe Mike's definition, which refer's to the motor as a generator, is the correct one.

Edited October 10, 2021 by PatMc

[Peter Beeney]

Hi Mike,

 

Thanks for your answer; much appreciated. I think I’m about the same with e-flying I flew an electric model for a club mate a fair few times circa.1980. Really underpowered at the time so I can remember a lot of concentration; fortunately they were short duration flights! I also have an Astro Flight motor, I also seem to remember they were considered amongst the best back then.

 

The first thought I had when I read ‘On testing you find the motor is spinning at 10000 RPM……was now I have to get the rev counter out first, which takes me straight back to my first post here…whoever would’ve thunk it……

 

But much more to the point perhaps, I’m ok down to the ‘The power used to drive the propellor is the back emf voltage times the current, so 10V times 20A = 200W.’ sentence, I’m afraid I cannot get this one to work at all. In my view the back emf is simply just a standing voltage which can never equal or exceed the applied battery voltage and is also only ever opposing and in effect cancelling the battery voltage by it’s exact value, which in your case is 10 volts leaving just 1 volt applied to the motor’s windings. If I connect two power sources with equal output voltages together in a mirror image, neg to neg and a voltmeter between the positives there will be a zero reading. If one output is higher the only current that can flow will be from higher to lower, the extent of which will be decided by the voltage difference and the resistance of the complete circuit, any diode blocking aside. Here we have 1 volt difference resulting in a current of 20A.

 

I’m convinced it’s just the 1 volt driving the 20 amps of current together with it’s ever present magnetic field though the windings that is in fact sufficient to be the driving force of the motor. The magnetic field is perhaps the ‘fuel’ and the rotary reaction could be a sort of ‘power stroke’. There is no radiator to disperse the unwanted heat from the power train but we can arrange good air cooling; better nowadays than it used to be because I believe it may have caused more than one or three unexplained problems back in the day. An engine’s radiator is designed to channel fast moving air through small spaces for a max cooling rate, maybe we could exploit that a bit more. The windings concentrate the current’s field strength and the ‘displacement’, ‘deflection’ or ‘twisting’, mechanical rotary motion, let’s just call it the torque, of the rotor created by the interaction between this and the permanent magnets is the power conversion tool and no-contact link between battery and propeller. But it’s not possible to be able to measure this moving rotary power by gazing at the electrical power input from any angle so this requires a dynamometer connected to the motor’s drive shaft, an instrument that can measure the physical mechanical force in units of lbs - foot, Newton - metre, kg - metre or whatever. But even this measurement is still not the power in terms of watts so it’s multiplied by the rpm to establish the time element. Rightly or wrongly, I’ve long held the view that the very close and often very (relatively) powerful relationship between the magnetic fields truly is the beating heart of the motor. And it’s fraternal twin the alternator too, as it so happens….        

Now leading on from this to another tiny sticking point, your meter will be in the power circuit, the 20A will be indicated on the ammeter, the 11V from the battery on the voltmeter and the current will be flowing round the complete circuit. So if you going to attribute all of the 220W to the motor and propeller, (and bear in mind this now appears to be saying it’s a measure of the output power and not the input power) may I please ask where are the watts being displayed that are emanating  from the battery and ESC? Dare I suggest that in fact they might actually be the 220 watts showing on the meter, with as you say 20W credited to the motor and the remaining 200 given to the battery and ESC. Also your statement: The power lost as heat is (applied voltage - back emf) times current, so (11-10) times 20 = 20W I do entirely agree with so therefore for exactly the same reasons can we now say that the remaining 200 watts are power lost as heat from the battery and ESC etc.       I believe we can…..

 

A quick thought, do you think your motor has something like an 8 x 5 prop on the front? Sounds about right for an around 40 mph little airborne jaunt for a smallish model. Lovely jubbly. Just a guess.

 

Also standing a bit closer to: ‘To find a good approximation for the mechanical losses, run the motor with no propellor. Measure the voltage and current and the resulting power used approximates the mechanical loss.’ may be of some interest as well. The start up sequence operated in slow motion and ‘Power out / Power in’ may be a clue as to where I’m going with this. But we’ll leave one that for another day. 

 

Sincerely hope all this makes some sort of sense and is also of interest; and maybe I’ll post my daydreams about the magnets later.

 

Best regards.

 

PB

[Denis Watkins]

Just to help your magnetic day dream PB 

Each magnet identifies the position, and next position, of the armature and so on.

The communication in logic terms at each Mark is

Move, 

Have arrived at that position

Move to next magnet

Have arrived at that position

 

12 or 14 magnets as an example, would indicate 12 or 14 positions during one rotation, 

Communication at Very High Speed of data exchange.

That's it, all there is to it.

 

All assisted and sorted by a microprocessor of course.

 

Edited October 12, 2021 by Denis Watkins

[EarlyBird]

Having resisted buying a Wattmeter until reading the advice given here I am thinking about this one.

Anyone bought one and any feedback?

[Andy Gates]

Hi EB,

 

Yeah, I have one of those and use it regularly.

I still have my Irvine meter but the tool kit one does hold the data for a while so it will catch current peak which the Irvine can't.

Both meters agree with each other and with my telemetry units which is great.

It also does not have to be connected to a receiver so it can be used to experiment with motors / battery pack / prop combinations on a test bench with no connection to the model or radio gear.

At the moment I can't fault it.

 

 

[Mike Blandford]

Peter: Perhaps think of it this way. The back emf is providing a voltage (a bit like a battery) but we are driving a current Into that voltage (like charging a battery). So, in my example, we have a point in the circuit where we see 10V and 20A, which is 200W. If that was a battery providing the 10V, we would be charging the battery with this 200W, instead the 200W is going "into" the back emf. But where is the back emf coming from? It is coming from the motor rotating, so the 200W is going into the rotation of the motor. If the motor is not rotating, then no back emf, so no output power.

 

Mike

[Peter Beeney]

Hi Mike,

Thanks for the reply. I’m afraid I still have just as any problems with this. In your first motor example I’m assuming your wattmeter is connected in the circuit and you are quoting your values from this. So it would be reading 220W. If not I’m not at all sure where to go with this anyway. I consider the 10V of back emf as being effectively cancelled out by 10V of the battery supply which in turn is also being cancelled out, leaving the 1 volt to provide the 20 amps total circuit current which results in the 20 watts.  I believe that the back emf cannot create it’s own current flow, it’s p.d. can never equal the incoming battery voltage, also the existing 20A current is flowing the ‘wrong way’ anyway so just how we can say this is is producing any watts I cannot grasp at all; at the moment of starting there is no back emf so how did how the motor get started without it and how does 200 watts go into powering the motor, where and how is the connection made and what is the exact technique used. Finally if the motor’s own back emf is actually supplying watts back to the motor then it would appear to be powering itself…. I think it’s gonna take me a whole month of Sundays and then a whole year more as well to get even just half a handle on this lot…

And it still begs the question, where did the the watts from the rest of the circuit disappear to…..

Regarding the battery charger, if we saw 10 volts and 200 watts on a meter between the charger and battery we would be charging the pack at 20 amps. That’s ok if it’s a 200Ah lump, it’s a rate of 1C, but if it’s a model pack I think it’s probably a mite ambitious. To use this battery charging analogy I think it really makes my point - just like the motor powering itself, it would appear to be charging itself…

                      

To enable a current to flow into the battery the charger voltage has to be raised above the battery voltage and at least up to the fully charged voltage of the battery. The charging voltage is simply the difference between that of the discharged pack, say 9V, and that of the max output voltage of the charger, say 12.6V. The charger output will be constant voltage with current limiting until the charge is nearly completed at 12.6 volts because charging needs to be a relatively slow process. The charging current flowing backwards though the battery and it’s voltage is just reversing the chemical action that took place to provide the electricity when the pack was being discharged. Because of the current limiting a wattmeter in the charging circuit will always read the rising battery voltage, let’s say it’s at 11.1V at some point, the current is flowing at a constant 2A and so then the meter would read 22.2 watts.

   

I certainly agree that when the motor isn’t turning there is no back emf but a number of possible reasons. Maybe the battery is just disconnected. Or still connected with the throttle fully closed. But if the throttle is fully open, the prop shaft prevented from turning and the full battery voltage is reaching the motor then this may soon become apparent by the sound of the blues and twos coming down the road on the hurry up to put the resultant fire out…. Definitely no output power but certainly maximum input power now though.

Now I’m off to make a toasted cheese and onion sandwich. There’s nothing quite like toast burnt over some hot embers……

Sorry about the delay, for some reason my password seems to have been changed.

Regards.

PB

[Martin Harris - Moderator]

Trying to follow this discussion.

 

Are you assuming that a motor acting as a generator at 10 000 rpm is producing the same current as that same motor driving a prop at 10 000 rpm consumes, Peter?  The voltage produced may be similar to the power source but its energy isn't providing the thrust that the motor is using most of the supplied energy for.

[Outrunner]

About time this thread was locked. Such stupid off topic ramblings.

[EarlyBird]

3 minutes ago, Outrunner said:

About time this thread was locked. Such stupid off topic ramblings.

Agree. That's what the report option is for.

[MattyB]

2 hours ago, Outrunner said:

About time this thread was locked. Such stupid off topic ramblings.

Yep, agreed. I started to respond again to a challenge issued a week or so ago, but reflected and decided it was pointless. It is clear PB is very happy being in a minority of one and is not open to doing any further research or learning on this topic, despite having zero citeable sources that support his views. Personally if it were me I’d have climbed down at the point when more knowledgeable posters pointed out that those views contravene all accepted electrical theory and the first law of thermodynamics, but hey ho, we’re all different. I’m out!

[Dickw]

I agree this thread has become slightly silly in an attempt to correct the unfounded comment about what a watt-meter measures :-

On 08/09/2021 at 09:20, 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

 

Continuing with the example now being discussed i.e. 11v supply to a 1000kv motor running at 10,000rpm and drawing 20 amps.

 

Back emf is indeed 10v leaving 1v to push the 20amps through the motor.

V/I=R therefore the winding resistance R cannot be greater than 1/20= 0,05 ohms.

So, the heating effect of the current (I^2*R) cannot be greater than 20x20x.05 = 20 watts

Yet the watt-meter is showing 11x20 = 220 watts going into the motor.

What is the remaining 220-20  = 200 watts doing if it is not powering the motor?

 

You can do this sort of calculation with any motor under any conditions, and we all know that the bulk of what a watt-meter is showing is the power turning the motor and prop.

 

Perhaps it is time the thread was closed.

 

Dick

 

 

Edited October 17, 2021 by Dickw

[Gary Manuel]

I don't agree with locking a thread just because someone is struggling to understand the concepts and is frankly making a bit of a fool of himself in the process. PB has been given plenty of pointers on where he's going wrong and where to look if he wants to improve his knowledge. If he choses to ignore the advice, then that's his loss. Others can similarly choose to ignore parts of this or any other thread that they don't like. That's what I'll be doing.

 

P.S. Dickw's previous post uses simple maths, which pretty much explain what PB has been trying to work out. Hopefully he'll read it and the penny will drop.

 

 

[Peter Beeney]

Sorry if I was giving that impression Martin, the problem I have is this;

 

Mike is multiplying the 10 volts of back emf by the 20 amps of current flowing  round the complete circuit to give the motor 200 watts of power and I seriously cannot understand this. For a start the back emf can never equal or exceed the battery voltage so the back emf can never complete it’s own circuit anyway; so therefore no current will ever flow. I believe that the 20 amps is established by the 11 - 10 = 1 volt applied to the motor’s resistance. The motor’s rotation is creating the back emf, not the other way round, so how can any power developed from the back emf be used to power the motor?

 

May I ask if you agree with this so far; and if not exactly why it’s incorrect and therefore how we actually can develop watts from the back emf.

 

Also I made a boo boo, or perhaps yesterday's deliberate mistake… I think the battery charging rate should be: 20 amps to 200Ah is 0.1C, to a 20Ah battery it’s 1C and to a model’s 2Ah pack it’s 10C.

 

Many thanks.

 

PB       

[Gary Manuel]

Peter, I am answering this purely because I am logged in and your message popped up.

 

It is NOT the 10v Back EMF that generates the current. The Back EMF actually opposes the current. It is the Power Supply minus the back EMF i.e. 11 - 10 =1volt that generates the current. Have another look at Dickw's maths above.

[Martin Harris - Moderator]

I’m having a little trouble following your argument Peter but I think DickW’s explanation is very clear and logical. 
 

The basic problem we’re having is in reconciling your original assertion that ALL of the power from the battery is turned into heat making a Wattmeter superfluous. Perhaps we could address this point and come to agreement that this was either a misinterpretation of what you were trying to put across or a mistake?

 

Otherwise we seem to have reached an impasse and we’ll need to consider the request to put the thread out of its misery. 

[Mike Blandford]

10 hours ago, Peter Beeney said:

. . .Mike is multiplying the 10 volts of back emf by the 20 amps of current flowing  round the complete circuit to give the motor 200 watts of power and I seriously cannot understand this. For a start the back emf can never equal or exceed the battery voltage so the back emf can never complete it’s own circuit anyway; so therefore no current will ever flow. I believe that the 20 amps is established by the 11 - 10 = 1 volt applied to the motor’s resistance. The motor’s rotation is creating the back emf, not the other way round, so how can any power developed from the back emf be used to power the motor?

 

May I ask if you agree with this so far; and if not exactly why it’s incorrect and therefore how we actually can develop watts from the back emf.

. . .

The battery is forcing 20 Amps against the back emf.

Hopefully you can agree the motor is providing some mechanical power to turn the propeller. So where is this power coming from? The answer has to be from the battery (otherwise we wouldn't need a battery!), so at least some of the electrical power from the battery is being converted into mechanical power, not all the electrical power is being converted to heat.

All that remains Is to decide how much of the electrical power is converted to mechanical power and how much to heat. This is where the back emf comes in, if you see a "black box" with 20A flowing into it, and measure 10V across it, then you should conclude 200W is flowing into the "black box". In our case, the "black box" is the back emf, so 200W is flowing into the back emf, and this is the electrical power that is being converted to mechanical power.

 

Mike

 

 

 

[Peter Beeney]

Hi Martin,

Thanks for your post but I’m not quite sure exactly what I need to say here. I’d certainly be more than happy to address this specific point about the wattmeter and resistance etc. for as long as you wish but I wouldn’t be making any changes in my narrative. Also I don’t remember that I ever said that the wattmeter was superfluous, as I remember I 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….. I have no problems whatsoever with anyone using a wattmeter.

If you wish I’m more than happy to describe exactly how I view the circuit and it’s resistance just from the most basic point of view. Then you can take it completely  apart right down to the last nut, bolt and spring washer and explain precisely what I’ve got wrong. If that is not a satisfactory solution can you suggest an alternative?

Just for interest in an idle moment I’ve just had a quick Google this morning and in about two seconds flat this was the very first hit. So I didn’t really bother looking for any more….. it’ll do just nicely for starters at least….  It’s a copy and paste as written, the upper case and bold text are the author’s own, not mine.

 

When a current flows through a resistor, electrical energy is converted into HEAT energy. ... The rate at which the heat is dissipated is called POWER, given the letter P and measured in units of Watts (W). The amount of power dissipated can be worked out using any two of the quantities used in Ohms law calculations.

And also for what it’s worth, here’s my take on someone’s ‘I squared R is not always V times A’ thingy from earlier in the thread somewhere.

                       When W = watts,  V = volts,   I = amps,  R = resistance.
        

                         Let W = V x I = V x I
                       From Ohm’s Law V = I x R
        Substituting, W = V x I = I x R x I
                      Or,    W = V x I = I squared R

                                           And

                          Let W = V x I = V x I
                       From Ohm’s Law I = V/ R
          Substituting, W = V x I = V x V/R
                        Or,  W =  V x I = V squared over R

Ironically I can still remember my very first post on the forum, as can many other folk I guess. It was indeed a long time ago, I didn’t even have a computer back then, I often visited the local library and communicated from there. Cutting edge CRT technology at the time, (for the library!), with a Windows OS of course… . which perhaps explains why I’ve used a Mac ever since!  

There was a debate going on even back then about electric motors together with AC and DC current etc and for whatever reason at the time I said that that the true expression for watts was volts x amps x power factor. There was a general consensus of agreement with this but as it was also thought that it didn't very much matter with our motors and as nobody wanted to go anywhere near this anyway we’d best forget all about it. Something with which I’ve totally agreed with ever since…

So now I’m thinking it’s quite possible that my last post on the forum will also be related to watts as well. They say that what goes round comes round….

 

Regards

 

PB