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Peter Beeney

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Everything posted by Peter Beeney

  1. Yes indeed GG, my thoughts exactly; and particularly in view of the fact I can't think of a way of permanently preventing the voltage regulator working even with the battery connected. With great respect all round of course but I think I would be having a little chat with Overlander pronto to see what their thoughts were and then I think I would be very surprised if in turn they didn't have to consult the manufacturer. I suspect you meant to say ‘on plugging in a fresh battery.........nothing’. Very odd might just prove to be something of an understatement eventually……. PB
  2. Just to fill in a little more time here with a bit of the idle rambling stuff and getting back to the BEC I'd say it’s actually quite difficult to permanently zap a voltage regulator, at least in the normal run of events anyway. They normally have a bit of built-in protection circuitry, such as current limiting and a thermal cut-off point specifically to prevent self destruction. I did a spot of tinkering with a couple of 6V 1A regulators a very long time ago now, I hooked up a 12V SLA to the input, attached a suitable resistor on the output to well overload it and couple of meters in circuit to check out what was going on. It fired up okay, got finger blistering hot… although I didn't try it… the cut-off temperature is 150°C…, it then shutdown with very little output, a few erratic milliamps that just jiggled about, then cooled down in a fairly short time, it was out in the open air after all, reset itself and started to heat up again. It carried on like this for a while so I just left it talking to itself for at least 24 hours or more. When I disconnected the resistor there was the normal steady 6V as usual so I just left it like that for a couple of days and then did it all over again using the other one with exactly the same result. It certainly did not appear to do any lasting damage to them whatsoever. Of course if you were to shove reversed polarity at it or seriously apply over volts, (the ‘never exceed limit’ is 35, don’t even go anywhere near there) or hit it very hard with a small club hammer you probably might render it rather mortuus elektus but that's only pure vandalism. That doesn't really count. This seems to explain a couple of things that I’ve pondered on in the distant past. Firstly the sheer and utter tenacity of some of the solid-state electronics to insist on keeping going in sometimes the most hazardous and hostile of circumstances, but it’s also a very plausible explanation too perhaps of another rather mysterious calamity that used to happen on the very odd occasion. When electric power first started to become popular someone would be zipping about quite happily and then suddenly have an unexplained crash due to apparent loss of control but then when the pilot and others reached the splinters they usually found the radio to be working okay; resulting in some scratching of heads....and even more splinters perhaps… However, when I explained what I thought might be happening that would be accepted plus some ideas about adding a bit of ventilation to help keep things cool also went down as a good idea. Fortunately a good cooling through draught is now the norm… Although 150° does seem to be quite hot it occurred to me that when you place a piece of kit, namely the battery and the ESC, in a polystyrene foam box with little or no ventilation then they may well get hotter faster than the heat can escape resulting in the ESC particularly getting very hot indeed; and with the BEC right in the middle of it, (or probably towards the front end I guess) it's not long before the temperature gets that high. Cooling down would be slower too, perhaps I should have noted the time and temperature fall before it restarted… Hope this helps. I have to admit I wouldn't know where to start on an ESC to prevent the BEC working even with the flight pack connected, let alone when it's not… One for the manufacturer I reckon. Just an opinion, kick it to death as much as you like. Good luck. PB
  3. Hi Toto, Just so that I can reiterate a little here as well, you now have four ESCs which you cannot use because there is no 5 volt supply on the throttle lead. If this statement is correct then I think in the first instance I would have to at least consider these items to be well in front to be the potential problem; and more specifically the BEC because this is what supplies the 5 volts. This is what I might do to establish the exact conditions in these circumstances: Firstly I’d find a flight pack with some charge in it together with an ESC that I know to be good to go. Then any receiver that's just handy. I’d plug the throttle lead into the receiver, any channel, and a red flashing LED will appear. That's my benchmark indicating that I've got power okay. I’d then disconnect that lot and just swap it for my faulty ESC. Now I will either have a flashing LED or not. In the case of it flashing then I’d contemplate that I've got a problem that's currently undiagnosed; and unexpected to boot.. In the case of it not flashing then I would definitely know that there is every chance that the regulator/BEC was certainly blown. Now I’d disconnect that lot and reconnect the original back again to prove everything is still good. Lastly I’d find the other three miscreants and try those in exactly the same way. I certainly don’t think this is in any way “manufacturer bashing” or indeed anything that you are doing wrong; there’s not really much in the way of a lot of other kit that you can point the finger at either. But I think you would have to agree that from the general tone of some of the recent posts this is certainly a very very unusual set of circumstances……. You may well be surprised how little time this actually takes to do if you decide to try it and I’d like to wish you all the very best of good luck with whatever you choose to do, I'm sure you will soon be able to tell us exactly what is going on. Just my take on this problem. PB
  4. Yes indeed Dale, may I say that's a perfect answer to what on the face of it is apparently an impossible question. I have to admit I was also beginning to think along the same lines but I couldn't think of any situation where I would want to pull the throttle lead out of the rx. Also could I have done this at least four times without somebody else noticing either?? It seems to me that just this simple point is seriously questionable too. I think if Toto does prove beyond any shadow of doubt that there is no 5 volt supply from the ESC and the others are exactly the same then Overlander should at least be advised of this problem. Or, of course, that being the case, they may already have an inkling anyway. Really like the ‘then you are the only person in the history of RC flying that this has happened to.’ bit; Toto’s claim to fame perhaps; however, should it transpire that if this particular ESC version is prone to failure for whatever reason then maybe someone in a dark corner of a model flying field faraway is doing their level best to play catch up... I feel we are close to getting an answer….. or more likely some further interesting questions….. hopefully….. PB
  5. Philip, My apologies, I guess I was a bit too vague there. This is only my opinion of course but I would consider that if the regulator (BEC) is blown then there is no power to some of the components and therefore the ESC is unable to function at all. Neither can the radio because there is no power supply there as well. I very much suspect it would've failed open circuit, if it failed short-circuit it would then most likely destroy itself fairly quickly and then become open circuit anyway, leaving the pos and neg wires in place between the rx and ESC provides the complete circuit for the battery connected to the rx to then provide the 5 volts for the necessary components on the ESA. Admittedly the supply is now a bit back to front but that doesn't make any difference at all. I think I can say with some certainty though that disconnecting the red wire will indeed definitely prevent this cobbled up arrangement from working at all. Sincerely hope this makes better sense…. PB
  6. Hi Philip, With total respect but it may be worth sounding a tiny note of caution here. The BEC is a regulator and if that has rather popped its clogs then I’d consider that the whole ESC is in something of a major malfunction too, it’s going to be a bit inert to say the very least. However, if the red wire is left in place and still connected then plugging a battery in will provide the necessary 5 volts to power the radio and ESC for more sweet and harmonious loops and rolls…. Having said that I think my strong preference though would certainly be to return it to the supplier and get his angle on it…and hopefully some sort of recompense as well… Although this arrangement might work for a long time and indeed never give any problems I feel that because this is not how it’s designed to work I would not have sufficient faith in this. Not something I’ll be doing anytime soon, at least in the normal run of events anyway. Cheers PB
  7. Toto, If you would rather not check this with the ESC plugged into the receiver have you got some other means of checking the voltage output on the throttle lead from the ESC? Voltmeter perhaps if you have one. From your description of the fault it does seem difficult to consider it as anything else other than that the regulator has gone open circuit for whatever reason. I'd of thought that for this to happen just once is bordering on virtually unprecedented, the frequency at which it appears to be happening to you is pretty well right off the clock… If it does so happen that the regulator has gone open circuit then it may be well worth considering that the others have also done the same….. which is quite remarkable but I have to say very difficult to see that it's maybe anything that you are doing incorrectly either. I think you are getting close to cracking this! PB
  8. Toto, Thanks for your prompt answer and I have to say straight away that I can't give you an absolutely definitive answer without having a quick look at the bits and pieces but if in fact your Domino motor does run when you connect the receiver battery I would have to say this points inexorably to the voltage regulator a.k.a. the BEC. This regulator supplies a steady five volts to some components on the ESC which would seriously object to being connected to the flight pack voltage. It also provides the supply to the receiver so if this suddenly ceases to function for some reason when you connect the 5V battery that simply replaces the regulator supply and back feeds the said components. For the moment establish beyond doubt the motor run, or not run, condition and we'll take it on from there….. I'm sure you can't do any harm here but as usual please consider the prop turning without warning and also I'd have to say I'd find it very difficult to find any other sort of explanation external to the ESC that might cause this situation. Take it easy. PB
  9. Hi Toto, Very briefly, I've just been reading some of the later posts of this thread, I've by no means read it all I'm afraid, but with regard to the failing Overlander ESCs can I respectfully ask if they are all the same version, i.e. same current supply and price or have you used a variety of types and sizes? Also in your post at 16:45 on Saturday where you mention the ESC failing in the Domino and then when you plugged in a NICD???? battery everything worked, did this also include the throttle/motor as well do you remember? An equally brief answer will be fine. Many thanks. PB
  10. Indeed, I couldn't agree more Don. Disconnect the push rod from the server arm and just anchor with a 5A electrical ‘chocbloc’ . If I were a learner at this stage I’m sure would be asking the instructor for some takeoff practice as well, flying basic square circuit, left or right as the site orientation dictates, overflying the strip and the next stage is a landing approach. I think that getting used to flying near the ground would really inspire my confidence, and indeed, I too have to abide by the KISS rule, Keep It Strictly Simple, anything else would be a bit too clever by half…. A very long term, from single channel days, flying colleague had a couple of maxims he always quoted to beginners, ‘If you think you are going to get upset crashing a model aeroplane, don't even start,’ and secondly ‘If you don't fly in a wind you will never do any flying’ and I must admit I don't think I've ever had any occasion to argue with that. Over the years I don't think there is anything else that quite comes close to sometimes turning normal rational human beings, usually the more senior type, into more or less quivering imbeciles than learning to fly a model aeroplane. One little bit of advice I tend to throw out right from the start is ‘Whatever happens. try to appreciate and enjoy all this scary stuff because you won't be able to experience this twice’ and of course the the same applies to the first solo flight, you only ever do it once. Blue skies PB
  11. Another attempt at an answer to this conundrum and to be fair all round I’ve considered the original question. Reproduced below. But to be honest I think Paul's O/P is near enough to make no difference anyway so this is an answer to that as well. If indeed this is an answer, though, it may well be just rubbish.. Imagine a 747 is sitting on a conveyor belt, as wide and long as a runway. The conveyor belt is designed to exactly match the speed of the wheels, moving in the opposite direction. Can the plane take off? And now imagine this sentence is a free hand sketch of a 747 standing on the conveyor belt!! I would agree with Martin, providing we keep absolutely to the spec. in the statement above then the plane cannot move forward. With regard to the timing would say a picosecond lead/lag time be sufficient and would this actually be obtainable in practice? In today's technical world I think it may well be yes to both… The plane is standing on the conveyor belt. We make a chalk mark on the side of the tyre on the bottom of the wheel and another one to match this on the the conveyor belt. Then a 3rd at the side of the conveyor belt on the ground to mark the position of the axle in the centre of the wheel in relation to the ground. At this stage the wheel is held firmly in contact with the conveyor belt by the weight of the plane bearing down vertically through the axle. The pilot opens the throttle, the thrust increases and the plane starts to move slowly forward due to the pressure of the axle moving at right angles to the vertical forcing the wheel to turn forwards against the friction of the belt. One picosecond later said belt starts to move backwards and from then on in keeping strictly with the spec. it's speed is inexorably linked exactly with the speed of the accelerating wheel. Normally the wheel would of course move forward on a stationary runway but now the wheel’s point of contact moving forward is meeting the belt’s point of contact moving backwards at exactly the same speed, the nett result causing the wheel to revolve around the axle but keeping it in a stationary position relative to the ground beneath the belt. If the circumference of the wheel is 15 feet after the first revolution we note the chalk mark on the belt is 15 feet behind the wheel but the chalk mark on the ground is still exactly in line with the axle…. and probably much more obvious, the plane itself has not gone anywhere either! The second revolution is the same as the first except that the speed of the whole process is accelerating due to the forward pressure on the wheel’s axle and will continue to do so. Therefore, again like Martin, I think this is going to eventually end in one almighty tangle. Also I think crucially not only does the speed of both items have to match but also the power output as well, it seems to me that the two forces have to equalise. Maybe the power supplied to the belt would need to be in excess to that of the jet! So now is it possible that before the wheels start to come off the wagon natural inertia becomes great enough to keep the process at a steady speed at some point? I've no idea, maybe it’s the next silly question perhaps… I would consider that as long as the belt speed stays in exact sync. with the speed of the wheel then the axle and thus the plane is never ever going to move. If the pilot shuts the throttle, say to around half, maybe that would still be enough force to keep the wheel turning and then if the belt stayed in sync going backwards too you could sit and watch the wheels going round with the plane going nowhere. But it sure would cost a bob or three I reckon. PB
  12. 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
  13. 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
  14. 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
  15. 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
  16. 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
  17. 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
  18. 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
  19. 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
  20. Hi John, Many thanks for your reply but I do think I would very much prefer to remain strictly in a neutral corner on this one. Earlier in the thread I was handed a few rather salty crisps simply just for posting what I thought was a simple might-be-of-some-interest idea. In the event I couldn’t have been more mistaken; and as I’ve not added any salt to my lunch as a seasoning flavour for many years now I couldn’t but help noticing. I don’t have any particular problems with this, rightly or wrongly I’m more than prepared to defend what I said, but I do feel that some sort of explanation might not come amiss as well. Although had I thought for one moment it would have resulted in that I definitely would not have bothered………… 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…. PB
  21. Before this thread finally disappears into a misty oblivion may I just take this opportunity to bump it up and say that there are still one or two questions I’d like to ask regarding the wattmeter and the useful information that can be given by it’s readouts. I’m not very familiar with it as I’ve never seen one, at least not the kind that modellers use anyway, so I’m thinking it may it may help to avoid making mistakes in any future posts. To show the greatest respect to Tony’s Original Post and taking Bob C’s Good Idea firmly on board as well, would it then also be an alternative plan if maybe one of the moderators could be persuaded to fire up a new thread, say something along the lines of 'Watt difference does the odd volt or amp here and there make to anything’ perhaps, so that all this stuff can be discussed at leisure in an amicable fashion and any folks not interested needn’t even bother to open it. I”d be more than happy to include a disclaimer with any of my posts saying that these are just my ideas and may not necessarily be the true facts etc. PB
  22. Very sorry for any distress caused by my posts folks, that’s definitely not what I intended. Indeed far from it, I’ve often said in the past that my ramblings should not be considered as any form of advice or instruction. I just thought it might be interesting to suggest another idea for checking prop speed etc. Now it seems that tachometers have gone out of fashion big time. Also I have no issues whatsoever with whatever anyone chooses to do their checking and testing with, it’s simply a personal choice. I’m just an old dinosaur I guess, way back in the day when i/c was king in the modelling world everyone had a tacho; a quick rev check before flying could often be seen. But i/c also had it’s own set of problems as well… However, I firmly reject the suggestion that the information I’ve posted is incorrect. I realise that won’t count for anything at all and I know I only tinker around with this stuff but as a for instance if anyone can show me with proof that the rotary action of a motor is not the result of mechanical interaction between two magnetic fields then I’d be pleased to read it and admit that I’ve got it all a bit wrong… Until then we shall have to gently agree to disagree on the whole subject… and I don’t think I can get any closer to Keeping It Strictly Simple than that… Again humble apologies, I might well be looking at my toy tacho in a whole new light now; I’ll probably give it a good kicking for starters.. PB
  23. 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
  24. 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
  25. 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
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