Yorkman Posted May 7, 2014 Author Share Posted May 7, 2014 ok, I've just connected the first 50amp esc up, and I've fitted the props that came with the model, which have much less area than the master ones. On 3s a quick 'burst' to full throttle showed 43amps and 308 watts. On 2s not fully charged, 33amps and 200+ watts. Quote Link to comment Share on other sites More sharing options...
Yorkman Posted May 8, 2014 Author Share Posted May 8, 2014 Can we have a go at 'reverse engineering' this? On the original Chinese motors supplied with the kit, turning 3 blade master 10x7s, with a 3s I got a steady- 250 watts 24 amps 10.3 volts .57aH 899 uS (?) what ever that is... all I want is a little more top speed....while keeping the same props, if possible, but certainly 3 bladers, maybe 9" dia and less pitch if necessary. so, what 'spec' motors do I need to look for? Edited By Yorkman on 08/05/2014 11:38:51 Quote Link to comment Share on other sites More sharing options...
Peter Beeney Posted May 8, 2014 Share Posted May 8, 2014 Yorkman, I think I’d go along with your line of thinking and go for a bit of hurry up. I like to see warbirds haring around flat out anyway, and the Mosquito would be no exception; it was primarily designed to go fast and in the beginning it could outrun anything else. I’d be very tempted to try a 9 x 6 three blader on the motors you’ve bought before you do anything else. 10 x 7 x 3 does seem a mite hefty to me. If you can get 12,000 rpm with a 6 inch pitch, that implies a speed of 68 mph; it might not actually go this fast, but it’s fun trying… but, as always, you need to keep an eye on the current. I use a tacho, an ammeter and a contact thermometer for static tests; I’m always keen to get the motor turning as fast as possible, that way it uses less current. Different brand props do seem to make quite a difference here, a selection is always useful; and a flying test always proves the point. A larger size ESC, in current carrying capacity, that is, not necessarily physically bigger or heavier, might be an advantage in terms of a lower internal resistance, perhaps. This is what we need to aim for, a low(er) overall internal resistance is the crux of the whole system. The lower it is, the less power, in heat, is dissipated from the component parts; and maybe, of course, to some extent at least, this is dictated by how much money we throw at it. Your model only weighs 4 lbs so I’d have thought these props are big enough. If it goes too fast, unlikely for you, by the sounds of it, you can always throttle back a bit. Reducing the pitch to say 5 might slow it a bit, too, at 14,000 rpm that’s 66 mph. In fact, at 1500 rpm 5 inch equates to 71 mph and I’d have though also a nice reduced current flow into the bargain; if that’s the sort of figure you get, that is, I’m only guessing. Maybe the current flow will still be far too high anyway; so you’d might have to consider an 8 by xxx. Like you, I’ve found these little scale type models can sometimes be a bit unpredictable, even the lightweight ones, suddenly dropping a wing on landing, say, so I try and keep the speed up by flying them in… and, indeed, plenty on the clock at all times when close to the ground. Cooling can also make a difference. It often seems pretty abysmal, so this is why I’m always checking the temperature, but if you can get the air flowing nicely through a motor I’m sure you can load it to the point to where it starts to get warm. But the ESC and battery would also require equally good cooling, too. I’m not entirely convinced by some of your figures @ 16:12:20 either, if 43 amps are correct on 3S that would suggest 470 watts, if the 308 watts figure is kosher then that hints at 28 amps, and if both are on the button the voltage reading would have be down in the dumps at around the 7 level. Good Luck! PB Quote Link to comment Share on other sites More sharing options...
Yorkman Posted May 9, 2014 Author Share Posted May 9, 2014 thanks for the advice Peter, that's the sort of info I can understand! I think I might try and get some 9x5s and see what sort of current/power they produce. I think those figures you're quoting are probably incorrect (by me) I saw 43 amps briefly, the watts at that time may have been higher but I was using an old 5c 2200 3s so the volts may well have been on the low side. Just as an aside, I was reminded yesterday that if you fit a 'reverse rotation' prop, it has a tendency to undo the prop nut.... Quote Link to comment Share on other sites More sharing options...
Peter Beeney Posted May 10, 2014 Share Posted May 10, 2014 My pleasure, Yorkman, and nice of you to say so, too. Although looking at it again I think I might have been have been a trifle optimistic, I now notice that the resistance is given as 28 milliohms and from that it may be possible to make some sort of random guesstimates. I think you might need to keep the rpm’s up on that motor; this does tend to mean smaller props; however, at least even if they are relatively fine pitch as well if they are really spinning freely you will still have the flying speed you’re after. It sounds as though your your original motor’s rotational speed, using a 7 inch pitch, may have been fairly pedestrian. I know that using a wattmeter is a pretty standard way of getting a handle on how a motor is going to perform, but as it happens I’m sometimes a bit cagey about all this. In my view it doesn’t always translate precisely as to what’s going on at the propeller, or in other words exactly how the model is going to fly. Also, indeed, the propeller itself can also make a measurable difference. I’ve never been entirely convinced that striving to obtain the maximum amount of current flowing is always the right way to go. But, like just about everything else, in the end it has to be a compromise, nothing’s perfect, and you never get something for nothing. Hopefully you will be really impressed when you manage to get the new motor/prop combination sorted. For interest, please post your findings when you get a result. Happy testing……. PB PS I guess you managed to unwind a prop off a bit sharpish, did you? From the ‘reverse rotation remark’…. Edited By Peter Beeney on 10/05/2014 11:56:00 Quote Link to comment Share on other sites More sharing options...
Yorkman Posted May 10, 2014 Author Share Posted May 10, 2014 Posted by Peter Beeney on 10/05/2014 11:52:24: PS I guess you managed to unwind a prop off a bit sharpish, did you? From the ‘reverse rotation remark’…. Edited By Peter Beeney on 10/05/2014 11:56:00 ALWAYS position oneself behind the motor when running it....or you may end up with a very sore finger... Quote Link to comment Share on other sites More sharing options...
Yorkman Posted May 20, 2014 Author Share Posted May 20, 2014 Ok, I'm getting nowhere fast here. Got a pair of 9x5x3 bladers hoping to reduce current draw, it actually went up (saw about 50amps before shutting down) It's obvious I need lower kV motors, I've gone way too high. So-I'm trawling through GiantShark's 70-99g brushless list (original motors weighed 90g, my starting point)-looking for kV around 1000 and watts around 300. Anyone got any thoughts? Don't want to just keep buying motors on the off-chance they'll be ok.... Quote Link to comment Share on other sites More sharing options...
Yorkman Posted May 21, 2014 Author Share Posted May 21, 2014 Have been pondering....if I change the 1700kV motors which are pulling 43amps with my chosen propellors, to, for example, identical spec motors apart from being 1100kV- which is 65% of the original, will the amps reduce to 65% proportionately? Which would give me 28amps? Is that a valid calculation? Quote Link to comment Share on other sites More sharing options...
Mr.B. Posted May 21, 2014 Share Posted May 21, 2014 Yorkman, From the 'fan laws' fan power varies as a cube of fan speed. In other words; Q2/Q1 = (N2/N1)^3 Where Q1 is power 1, Q2 is power 2 N1 is fan speed 1, N2 fan speed 2 Power is volts x amps So Q1 = 10.3 x 43 = 443W Q2/443 = (1100/1700)^3 This assumes input volts is the same for both motors and therefore cancels out Q2 = 120W Again assume volts remains the same 120/10.3 = 11.6 Amps If you look at the graph above it shows the relationship between power and RPM rising as an exponential (to the power 3) curve, not linear. Quote Link to comment Share on other sites More sharing options...
Yorkman Posted May 21, 2014 Author Share Posted May 21, 2014 Mr B. Thanks. I think. I'm afraid I'm not able to follow the theorising above-I was always rubbish at algebra at school! I THINK I deduce that 1100kV would be too little....but unsure* how to do the calculation to insert different values at N2... *absolutely no clue... Quote Link to comment Share on other sites More sharing options...
Yorkman Posted May 21, 2014 Author Share Posted May 21, 2014 *think I might have got it-by my calculations 1500kV gives 300w...?? Quote Link to comment Share on other sites More sharing options...
Mr.B. Posted May 22, 2014 Share Posted May 22, 2014 Yes, 304.3W. An A for your home work. As you've sused N2 is the speed of the new motor. As motors are specified in KV (RPM per volt) and volts will be the same in both cases you can just use the KV of the motor. Quote Link to comment Share on other sites More sharing options...
Yorkman Posted May 23, 2014 Author Share Posted May 23, 2014 2837/10 1480 480 43A 80A 2-3 N/A 79 3.17 28.1 40.8 Yes from Giantshark-Keda Thumrun, has loads of power and can take 43 amps while weighing very little....suitable? Quote Link to comment Share on other sites More sharing options...
Simon Chaddock Posted May 23, 2014 Share Posted May 23, 2014 Yorkman I am sure the 2837/10 will do fine but you are rather doing this the hard way by stipulating a prop and then trying to match motor and current draw to it. The easier way is to decide what power you need in the plane, then select a motor with specification that is likely to give the required power at a suitable revs with the intended battery and then select a prop. Props tend to be cheaper than motors. Quote Link to comment Share on other sites More sharing options...
Yorkman Posted May 23, 2014 Author Share Posted May 23, 2014 Posted by Simon Chaddock on 23/05/2014 14:36:29: . Props tend to be cheaper than motors. Good point, well made Unfortunately, saddo that I am, I hate to see scale warbirds with ridiculously small two blade propellers.... Hence, I'd like to keep the 10" 3 blade ones-incidental to the fact is the spinners wouldn't fit two blade props either. As i said right at the beginning, I was only after a bit more top speed, if possible, as the motors needed replacing anyway. Should have come here first, before buying anything! Edited By Pete B - Moderator on 23/05/2014 15:29:03 Quote Link to comment Share on other sites More sharing options...
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