SAVAGEJIM

I am more mechanically minded, so all the EM stuff of physics that I learned long ago, I have all but forgotten.

Why is it tat a motor with more turns will produce more force magnitude for a given radius than a motor with less turns? Also, are you assuming the same wire gauge in both motors?

I assume it all has to do, of course, with EM flux, I am guessing that more wire turns presens a higher flux than a less wire turns.
Now, as for which force is the major acting force to provide the magnitude force of your moment, I have no clue if it is electical force or magnetic force. I am guessing magnetic force.

Correct me where I am wrong in this, and please feel free to explain in detail, it will all come back to me if I saw some more info.

Access

It all has to do with the way an electromagnet behaves, again, proportional to current and number of turns. It's magnetic force, not electrical. Wire gauge is going to affect resistance, and that's it. I've shown with the Neu motor I cited above, the resistance of the wire is far from being significant. I prefer to stick to the applied physics, not theoretical.

Notice how voltage is completely absent in this formula. Normally voltage helps you overcome the resistance, and hence determines how many amps flow through the coil. But if this resistance is deemed insignificant, the only difference voltage makes is how quickly I can turn the coil on and off (due to inductance / impedance).

mattnin

Access, if you want to compare torques of different motors, you have to do it such that:

Torque = (magnitude of force) x (lever arm), and since the lever arm will be equal, you have to compare by the force created by the motors. Trying to describe it by different voltages and rpms and not taking into account loading is meaningless and is much too abstract.

Compare torques of different motors like I have, with everything being equal except motors. This means battery is equal, mass of vehicle is equal, voltages are equal, current is equal, etc..... You say that the only thing that matters is number of turns and current, and current is equal in these scenarios as well Access.

I do not feel like disputing all your caveats because your level of abstraction you present is too great, eg. "perfect" electromagnetic field, burning up your controller at 100 % duty cycle at WOT because in reality, these caveats do not matter. At wide open throttle, the ESC is @ 100% duty cycle, and I am at WOT much of the time, and I have yet to burn up a speed controller.

SavageJim, a motor with more turns in its coil will produce more force because the magnetic field created by current passing through the wire add together in a coil to give a net magnetic field N times greater than that of a single loop. N = number of turns. Actually, looking back at just this statement is enough to prove that a motor with higher turns has more torque Access.

Access

I am talking about equal-RPM systems. The RPM stays the same, but the voltages and KVs change. This is since the first post of this series.
When I talk of torque, I always talk of torque '@ (some) RPM'. I've always said something like "a 5000KV motor @10V, and a 10000KV motor @5V", both are 50,000rpm systems. Both have very close to the same amount of torque.

Above you're talking of something totally different. You've got one setup that has a much higher RPM than the other. I touched on this in one of my above posts. It can be looked at as a conservation of energy problem, and yes the faster spinning motor will have less torque, the slower spinning motor more torque, both will ultimately have the same power output if the power input is the same (or close, disregarding efficiency). I don't dispute this. Now if you gear both setups to go the same speed, any advantage in torque is offset by RPM, and vice versa. The torque at the wheels ends up being more or less the same.

To summarize:
KV alone means nothing, other than that it's use to calculate RPM. RPM is what influences torque, not the raw KV value. I can have two setups, running at 50,000rpm, equivelent in every way except for voltage and KV, and the torque curves will match almost exactly.

I posted this thread quite some time ago to respond to the types of posts where people would claim a 5700KV motor running off 3s LiPo, or better yet a 4600KV motor running off a 4s LiPo had significantly better torque than a 7700KV motor running off a 2s LiPo. All these setups had RPM roughly 'in the same ballpark'. There's no reason why one would have significantly more torque than another. Choosing the higher voltage will get you better efficiency (to a point), but it's not going to get you more torque.

The reason you don't burn up your controller is b'cos it only applies 100% duty cycle when the motor is spinning near full throttle. If it actually did 100% duty cycle when the motor was at rest, it would burn up. That's like the example I gave of the brushed motor, connected directly to the battery, and with you holding the rotor in place (or only allowing the rotor to turn very slowly). Same thing.

SAVAGEJIM

Okay, alot of this is coming back to me, mostly from poking around on the net, Wiki and all, and also memory of magnetic force, since magnetic force is the primary force to provide the force vector of the moment equation.

I remember that the magnetic force, I will sa Fmag for short is as follows: Fmag = charge(velocity of charge) X magnetic field. "X" being a cross product, just like in torque (moment). For the sake of simplicity, I am assuming that the velocity vector is always perpendicular to the mag field vector, so the cross product becomes a simple multiplication. So for the moment calculation in this case, we do not need to worry about cross products by considering the vector components. Enough of cross products, I am getting to deep for purposes here.

Now, I do not know what the strengths of the mag fields are in our RC motors (in terms of Tesla), but I did run into something called Amphere-turn concerning electrical motors. The higher the Amp-Turns (lets use "N" in this case), the more charged particles moving through a given space within a given mag field. The more charged particles moving from the Fmag equations creates a higher Fmag, thise a higher force to apply to the same moment arm. End result is more mechanical torque.

Now, I have no idea just how many electrons are moving for a given Amp, but if one turn = one Amp-Turn in our RC motors, that means a higher Fmag for a higher mechanical torque.

Again, correct me where I im incorrect.

Now I am curious why the torque curve would drop considerably at RPM. I.e. why the Fmag is lost at higher RPMs.

mattnin

Access, you are saying a whole lot of nothing. Equal RPM systems? You mean something like a 3S Lipo @ 11.1V using a 5405kV motor to get 60,000 RPM and a 2S Lipo @ 7.4V using a 8108kV motor to get 60,000 RPM? If you are saying this, you are certainly not comparing like to like because the battery voltages are different, and you haven't had a single explanation thus far explaining 'torque' in terms of actual physics.

mattnin

I think I am done in this thread, I can't keep explaining actual physics of until I am blue in the face. It is up to you and everyone else to decide who is right.

Access, if you want to dispute Faraday, be my guest.

dacaur

Exactly.... Access, all your points are valid in theory and "in the lab", but in the real world, they are are either completely irrelevant, or im some situations, wrong. Because you have to consider the battery into the equation, and that changes everything... You are looking at the motor as a standalone system, when its really just one component in a system, and to get the full picture, you have to look at the whole system, not just one component... Consider the following...

Fact. A lower KV motor produces more torque with less power than a higher kv motor. (power=battery power)

Put two motors in identical cars with identical gearing.

One is 5000kv
one is 10000kv (just examples)
Same gearing, same esc, same battery.

You say they will both produce the same torque, so should accelerate at the same speed. But a huge variable that you cannot discount in the real world is the battery, and how many amps it can supply.

The reason that matters is voltage sag under load.

Say the battery can supply 50 amps without sagging.... Lets say the 5000kv motor with the gearing we have chosen draws 50 amps when you peg the throttle. Its a given that the 10000kv motor is going to draw lots more current in the same situation with the same gearing, but since the battery cant easily supply the amps its asking for but still tries, the voltage sags, and less voltage = less power. The lower kv motor produces more tourqe. with the available power.

Its easy to say "well that's not a valid test because you just need a better battery" But that's the whole point. It IS a valid test, because in the real world, most people cant/don't just go out and buy the best $500 battery they can get ahold of.... in fact most people buy just enough battery, or, in many cases, even not quite enough battery.....

There is a huge problem with all your "examples" you are assuming only 10-20% throttle, but forgetting the other 80-90% also exists.. From your first post...
This is plain wrong. You seem to be unclear on how an ESC works, despite your claimed knowledge. If the ESC is delivering 100amps to the motor, you seem to be trying to say its drawing 10amps from the battery 100% of the time. But that's incorrect, you have to say its drawing 100amps from the battery 10% of the time. The ESC cannot and does not provide 100amps@ 1v. It provides 100Amps at 10v 10% of the time. In your example, if the esc were 100% efficient, then at 100% throttle the ESC is providing 1000amps to the motor and drawing 1000amps from the battery 100% of the time.... unrealistic and just plain silly.

misinformation like this is a big part of the reason people destroy a lot of lipo batteries do to trying to draw too much current. They think "OK, my battery can provide 20 amps without danger to the battery. My motor draws 40 amps at full throttle, but only 20 amps at 50% throttle, so if I just keep the throttle below 50%, I should be ok".... But in reality, if you are prividing "20 amps" to the motor at 50% throttle, you are drawing 40 amps out of the battery 50% of the time, damaging the battery....


You are spouting numbers, but only giving half of the info required to prove what you are stating. What is the amp draw in the above statement? The big problem here is again, you are looking at it from a "laboratory standpoint" and not a "real world standpoint". You are giving out theoretical numbers, taking into consideration ONLY the motor, without considering a battery. IF you take a 5000kv motor and a 10,000kv motor, and run them both in a vehicle with the same "10 volt" battery and the same gearing, you will not get the above results (10000RPM@20% throttle on the 5000KV motor, and 10% throttle on the 10000kv motor) Because of things like voltage sag, which will be higher on the 1000KV motor, because while both motors (in a perfect world) may be drawing the same average amps, the 10000kv motor is actually drawing double the amps as the 5000kv motor, and that means the voltage will sag more, and less voltage = less power....

So to put your above example in a real world context..

Lets plug in 50 amps as the "average" amp draw required to get to 10,000rpm on either motor, which is the amps the motor "sees", since you say the required power will be the same on either motor...

So you are saying to achieve 10000rpm, the 5000KV motor is at 20% throttle, so the motor "sees" 50 amps, but its actually drawing from the battery 250 amps 20% of the time.

The 10,000kv motor only needs 10% throttle to get to 10,000rpm, so to get the motor to "see" 50 amps, it needs to draw 500amps 10% of the time.

Now, you also have to take into consideration voltage sag under load The higher the load, the more the voltage will sag. the 10000kv motor is putting a higher load on the battery, so the voltage sags more, and yep, you know it, less voltage = less power (torque) so the 5000kv motor is putting out more torque in the real world, despite having no advantage "in the lab"

In all your "examples" You are assuming that either 1) there is no load on either motor, or 2) you have a battery that can deliver unlimited power, neither of which will be the case in the real world.

Even if you plug in a realistic PWM, the results are still the same. lets aim for 40000Rpm...

The 5000kv motor needs 80% throttle to get to 40000RPM. Lets use the 100 amps as the "average" amp draw required to get to 40000rpm. SO that means the 5000kv motor is drawing 80 amps 80% of the time.

The 10,000kv motor is drawing 140 amps 40% of the time. The double amp draw of the 10,000kv motor, even though it is being drawn for less total time, still sags the voltage of the battery more than the amp draw of the 5000kv motor... And that's not a "lab theory" that's something I know from actual experience.....

Also, one last point...
I don't know where you got that, but its 100% wrong. The definition of KV is the unloaded RPM per volt.

Access

Okay, this started as a reactionary post to things I saw repeated commonly a year ago, you can't take it out of context for something it wasn't. 'Equal RPM systems' (or approximations thereof) have been a part of the hobby since Castle introduced the 3s capable Mamba Max. People had to choose between the 7700KV motor / 2s LiPo and the 5700KV motor / 3s LiPo. Arguments like 'what is the benefit of one over the other' and the common response 'oh the 5700KV / 3s setup definately has more torque' is what led me to post this.

Is it a 'like versus like' system? The answer to that question is ultimately a matter of opinion, not something that can be argued scientifically. I say it is, b'cos it's an available and real choice we have to make in this hobby, and they both have near the same RPM, same motor family, etc. All else I can really say to you is take it for what it is, I've always specified up front in each post what the parameters are. If it confused you, sorry, I don't take it personally and neither should anyone else IMO.

dacaur

in theory sure, but In the real world there are more factors than rpm and kv. I would bet my paycheck they wouldn't match. And here is the reason: Voltage sag under load... the setup that's drawing more amps, which will be the higher KV setup, will also be producing less tourqe, becuase the motor will be reciving less voltage, because the more amps you pull from any battery, the lower its voltage will sag..... Fact.

Are you trying to say there is no advantage to a higher voltage setup? Do you even own a brushless motor?

Take a bunch of identical lipo cells, and assemble them into 2, 3 and 4 cell packs. then set up 3 of your "matched RPM systems" so that each pack produces the same theoretical RPM according to KV and battery voltage.

Now start increasing the gear ratio (numerically lower) on each setup (keeping the same ratio across all setups so the theoretical rpm says the same) and before you long you will see that the high voltage/lower KV systems are producing much more torque than the lower voltage/higher KV systems due to the voltage sag of the lower volt/higher amp setups... using identical lipo cells, you can get SIGIFICANTLY more torque out of a higher voltage/lower KV setup than you can out of a lower voltage/higher KV setup. And that's a fact, not some "theory"

Access

This isn't the system I'm talking about. One motor is going to go twice the RPM of the other. You might as well have two 5000KV motors and run one at twice the voltage of the other. Either way one is probably going to either suck more power than the other, or if they both are limited to the same amount of power, one is going to have less torque.

Conceptually I don't think we're disagreeing here, we're just talking about completely different things.

One note is that any battery will 'sag' at least a little with any amp draw, if it sags a measurable amount at 50A, it will sag some portion of that measurable amount at 10A. You generally model batteries with some fixed, finite internal resistance so anytime you have a battery and a load, you always have a voltage divider. But that's just a detail.

The ESC has capacitors on the input end, capacitors on the output end, and the current is being fed into an inductor. I said it above I don't know how else to say it. All these things act to either level the voltage, or flatten out delta-current over time. If the ESC had none of these things, you'd be right. But it does. Any good PWM system needs capacitors on the input and output to work well and efficiently.

I'm aware of this and it's only true if the ESC is poorly designed. An ESC has to work at a sufficiently high PWM frequency, and have enough capacitance to not do these things under normal operation.

I know the technical definition is unloaded. But that is not what all the manufactuers are using. Some use unloaded, and some don't.

"Are you trying to say there is no advantage to a higher voltage setup? Do you even own a brushless motor?"
No. The difference is a few percentage points of efficiency. Which in itself is very valuable. But it is not torque. The torque difference is minimal.

I've done the gear ratio thing for my speed setups and the first thing that always hit me or stopped me from going higher was overheating. The torque or acceleration was always fine.

mattnin

It is funny to watch Access try to explain his way out of this one.... You are wrong and too proud to admit it.

Torque cannot be described in terms of RPMs, duty cycle, or whatever else you are gonna make up, and you surely aren't going to fool myself and other educated folk with your long, drawn out and confusing arguments, sorry!

Torque can only be described in a few ways..

Torque = (magnitude of force) x (lever arm)

or in case of an electromagnetic coil,

Torque = N (number of turns) x I (current) x A (area of loop) x B (magnitude of the magnetic field) sin O (orientation of the normal to the coil with respect to the direction of the field)

I took this all out of my physics text, and no where do I see RPMs here, duty cycle, and other things like you have described.

I am sure you are gonna make some post where I am wrong, or I didn't fully understand your explanations, but I think you are trying to grasp at anything to give you even a little bit of credibility here.

Access

No, I'm just going to say take it easy and don't get so personal, it's becoming an argument of ad-hominem attacks and I'm not going to take it that far myself.

redfisher1974

Very good read.........I bet Access could talk his way out of speeding ticket...

Access

1) Electrical Power (Volts * Amps) in Watts
2) Mechanical Power (Torque * RPM) / 5252 in HP
The drive system (battery, ESC, motor) ultimately just consumes (1) from the battery, and creates (2) at the motor shaft, minus efficiency losses (heat, vibration, etc.)

http://en.wikipedia.org/wiki/Energy_transformation
http://en.wikipedia.org/wiki/Conservation_of_energy
There isn't really any energy being 'stored' here, so we're safe in that respect.

There are places where this doesn't hold true, or where it can't be applied, ie. 0 RPM, when the motor is saturated or 'maxed out', etc. But otherwise we can make use of it.

If we are comparing a 20,000RPM system and a 40,000RPM system, and (1) is the same for both, and the efficiencies are reasonably close, of course there will be a big difference in the Torque. Or, if the torque is the same, of course the 40,000RPM system is going to be consuming more electrical power.

If we are comparing two 40,000RPM systems, with motors from the same family, the only thing that differs from one motor to the other is the number of turns, then go figure. It really doesn't matter what voltage or what KV motors we are using, except as it relates to efficiency. If (1) is the same for each, and the RPM is the same, and the efficiency is the same, then torque will also be the same. If the efficiency changes slightly, torque will also change slightly. And so on.

I hope that satisfies the physics buffs among us.
And about the speeding ticket, I've never got one, so I've never had to try.

dacaur

What exactly are you saying in this post? you typed a whole lot of words but really didn't say anything....

If the voltage input is the same for each, and RPM is the same, then they are the same KV, so of course the torque will be the same.........

Ttam Says Blarg

Holy cow I think my head is going to explode

Argess

Hey, hey, hey.....this is a discussion forum. Access is discussing like a gentleman. So should we all. I really appreciate the post Access has made. It's making me think, and I've been bored lately.

Anyway, here's what I get out of it. With EVERYTHING else being equal, and having a 10000kv motor with 1/2 the turns of a 5000kv motor, the torque will be the same. Why?

The current in the 10000 kv motor is double that of the 5000kv motor due to the # of turns (reactance) being 1/2, so it sounds like twice the torque.

But the number of turns in the 5000KV motor is twice that of the 10000kv motor, so that multiplies the torque by 2.

So, in the end, the torque is the same in both motors, howver the 10000kv motor is drawing twice the current.

Guess our low-turn motor isn't too effecient, as it's drawing twice the power from the battery to produce the same torque as the higher turn motor.

Now, if our battery can't deliver twice the current, it appears our 10000kv motor doesn't have the power of our 5000KV motor.

************************************************** ************

Now for the top end....keep in mind all the discussion so far has been either locked rotor, or very low rpms. At higher RPMs, we need more power transferred to the rotor. But the rotor impedance is going up because the frequency of its rotation is going up. In order to get more power to a high impedance, we need a higher air-gap to rotor emf, and to get that, we need the higher current from a low-turn motor.

************************************************** ************

So....can we then conclude that two motors, identical except for # of turns, and hence different KV rating, will behave as follows:

Lower KV = more torque at low rpms dependant solely on the battery and ESC design. Proper ESC and battery choices will show no difference in low end torque

Higher KV =higher rpm than the lower KV motor, and draws the battery power to prove it.

So it's a trade-off to some extent....like an internal combustion engine.....top end hp, or low end torque....dependant on battery and ESC design and applicability

************************************************** *************

By the way...ever see a motor chart.......nice flat horz torque line and a straight (and rising) Hp line. Sure would be nice to get a nitro engine to perform like that.

Willystylz

Since we're talking torque, kv, rpms etc.......i have a somewhat related question to ask.

What influence does power have on a brushless motor. Lets say for example we have a 30,000rpm motor rated at 200W and another 30,000rpm motor rated at 300W. The reason i ask this is because LRP's original Vector brushless motors were rated at very high rpms and lower power. The newer Vector X11 motors seem to have a lower rpm rating but higher power. I'm somewhat confused by this...

Access

I said (1), or the power input is the same. I never said voltage. If the voltage input were the same, you're right that it would be kinda useless to say that. But you can have the same power without having the same voltage.

This relates to "what we do" b'cos, for instance:
one powersource could have a 3s, 4000mAH battery (12,000 Watt*Hours) and the other could have a 2s, 6000mAH battery (12,000 Watt*Hours).
Both these batteries are almost the same size... the 3s will probably be a tad bigger and heavier b'cos of the seperators, wiring, but hopefully not significantly bigger.

The use here is when comparing, say:
The (A) CM36-5700 running on 3s and the (B) CM36-7700 on 2s.

"If we are comparing two XX,XXXRPM systems, " (both systems are about the same RPM, roughly 60,000)
"with motors from the same family" (they are both CM36 motors, the only difference between one motor and the other is the turns of wire)
"voltage and KV don't really matter as it relates to torque... except efficiency" Both systems are _close_ in terms if efficiency.
Therefore torque at a given RPM, should also be _almost_ equal.

Again my original post was a response, almost a year ago, people would often tell the boards that (A) should be chosen b'cos it had significantly more torque than (B). We know (A) is the right choice, but is it b'cos it has significantly more torque than (B)? "(A) is the right choice, b'cos it's a little more efficient. Both will have the same torque."

That's all it is. On the other hand, if voltage is the same, KV is different, you have two different RPM systems. You might run a single KV motor different voltages and compare that. With brushless, you can gear each to go the same top speed, to a point. If you're gearing to go the same speed, the torque advantage you gain through gearing, you lose b'cos of the lack of RPM.

"No physics for physics sake, always try to relate it back to what you do (in RC)"

Access

I'd think that's the manufacturer's guess as to say, "when this motor is used as we recommend it (ie. a certain voltage, a certain ESC, a certain general weight of vehicle, a certain gearing), this is how many watts it's going to draw. If you know the voltage they spec it at, you can then find the Amps in current draw.

Ie. if they say 300W @ 7V, that's a little over 40A of current draw, though probably average current draw. Don't go out and buy 3000mAH, 15C LiPo and run it unless they specifically say 300W is the 'peak' wattage and not the average.

"ever see a motor chart"
http://www.teslamotors.com/performan...and_torque.php
The torque is not just a flat line. It starts out flat, and then at some point it drops. The flatness is b'cos the motor has a physical limit to the torque it can actually make, If the motor was 'perfect', the torque would keep rising as RPM dropped.

Argess

Willy...just think of your Dremel tool.....high speed and no "oompfh". All in the design of the motor.

Generally speaking, if your motor draws 300W, it can generate more power than a 200W motor because we can realistically assume decent efficiency. The speed at which these HPs are generated are again, inherent in the motor design

Willystylz

So from my understanding a motor rated with more power will more easily achieve it's rated rpm? For example when trying to push a heavier monster truck, a higher power motor would (depending on gearing obviously) more easily reach a higher rpm than say a lesser powered motor? I apologize for my basic understanding in this lol!

Argess

Yes Willy...you are correct. More power = faster acceleration

mattnin

Once again for the sake of clarity, Access is comparing a lower voltage higher KV system to a higher voltage lower KV system and claiming they put out equal power. He is not talking about the torque at all and only describing power output, so the title of this thread is misleading. Also, he has a whole new set of problems to explain with his claim because he is wrong about that as well. A heavier vehicle like a monster truck or 1/8 truggy is powered by 4S or 5S lipos and a lower KV motor, not a 2S lipo and higher KV motor. As matter of fact, there is a grass-roots movement underway which is trying to get a higher voltage/lower KV systems used in racing due to the increase in efficiencies.

A heavier vehicle can run a higher KV motor and less voltage, but as we all know, it will run hotter and require a much higher current draw and therefore be much less efficient.