More kV = More power? What are the trade-offs?

[liteCycles]

Right now I am working with Bob from Astro to wind me two 3220s with only 2 turns. This will give me 15,000 RPM (the mechanical safe limit of these motors). That is 50% more RPM than my 3 turn 3220s I have been running. I will gear down to match and should have CRAZY power (torque), while the actual wattage remains roughly the same.

Hey Matt, I know you can easily pick up another 3:1 between your two stage and the rear wheel, so would you only need about 7:1 from your drive unit? And will your 1st stage belt be OK with 15,000 RPM?

And on a related note what is the maximum reduction you can reasonably get out of just your two stage drive? I'm thinking of an inefficent through the crank setup. I just would really, really like to keep the rear disk and have a 11T cog at the back for my cranks.

 

[Miles]

I just would really, really like to keep the rear disk and have a 11T cog at the back for my cranks.

You can do that with the set-up that I used....

 

[mwkeefer]

Hey all,

I've been following (loosely, some of this I still just don't get) along best as I can but I have a question, more the inverse question.

"Less kV = less power" I get, I even understand why (rotating mass, higher RPM, more reduction = higher power output.

The question I have is if going the other way, IE: reducing kv - for instance (the old standin) my tower pro 5330 is rated for a kv of 135 and ships in delta mode... rewiring to wye provides me a rough kV of 70.

Since the rated V is 42 for this motor (10S Lipo) and the currrent is max around 60-80A and continuous rated for I think 2900w... what is the effect (mathmatically) of the reduction of top end.

I have measured the RPM under a very small (just a little 12" prop for some resistance) load at various voltage levels and recorded the following rotor speeds:

HVC (41.5v) = 2800 RPM (I expect 2500 under eBike loads)
Nominal (37v) = 2600 RPM (I expect 2200 under loads)
LVC (33v) = 2300 RPM (expect 2000 RPM under loads)

Obviously I can test (and have) with EagleTree... without much load I don't pull more than 20A steady at 2800 RPM but that's a mere 830w - eBike will cause much higher loads... the question I can't figure or find an answer to (that I can understand) is how to calculate the power loss.

Is it really as simple as dividing by the factor of decreased speed, surely it can't be linear ?

If anyone can shed light on this, I will be forever in your debt (seriously).

Im sure its similar to my grasping how transistors work many years ago... it just didn't make sense until it did.

-Mike

 

[Miles]

Is it really as simple as dividing by the factor of decreased speed, surely it can't be linear

Yes, it is.

Kv is just the velocity constant
More velocity equals more power...

Higher Kv = more rpm per volt
Lower Kv = more torque per amp (which doesn't mean a higher continuous torque rating).

 

[liveforphysics]

Mwkeefer- When a brushless motor lists some number as "max voltage", it means essentially nothing. It has the potential to mean a mechanical limit, but then often you see other winds of the same motor with double or triple the KV listing the same max voltage. If your motor says "42v max" on the spec sheet, don't worry about running 100v (or whatever) into it. I think the side of the 9C motor Methy sent me says "36v, 500w max" printed right on the side lol. My HXT outrunners from HC said "42v max" on them, I hit them with 100v, and they run cooler, have a much better RPM range, and seem to perform worlds better than they did at 48v.


On paper, it really shouldn't make a bit of difference what the wind is on a motor that you can change the gearing to achieve the same wheel speed. It should come down to just motor efficiency% and the power you're dumping into it as to the work you get out of it. In reality, it seems best performance always comes from things that you can feed as high of voltage as possible. I know on a motor calculator it doesn't make a bit of difference if you have 4 turns and run 50v or 8 turns and run 100v as long as the copper fill percentage is the same, the motor calculator predicts identical performance. In personal experience, the higher voltage stuff just always seem to function better and run cooler etc.

It seem the real-world optimization is to figure out what you can do for max controller voltage, then figure out what the mechanical RPM limit is for the motor, and choose a wind that lets you reach as close as possible to the mechanical RPM limit, and then choose your gearing with your intended max speed to occur as the same motor mechanical RPM limit.

Off-Topic, but related to motor mechanical limits. I once took the rotor from a very high KV little helicopter outrunner motor, which only had 6 magnets total in the bell. It was about 1.25" across, and I mounted it onto a 2 turn stator of a motor that was only 1" across, so it ended up with a huge air gap. The motor the stator was from was 5,000KV, and only intended to be fed 11v. I hooked it up at 48v with that giant air gap, and spun it up. It had so little starting torque that I had to help it to get it spinning. lol. It spun up and sounded like a turbine engine, then made a series of strange noises above that, finally mostly silent, then grenaded and stuck little parts of rotor and magnet in the wall. I was thankful for the big aluminum block I set between myself and the motor. It was pretty cool. I've still got the stator, and I'm thinking of buying another of those little $10 HC motors to take the rotor and do it again, but this time on video.

 

[mud2005]

hi guys thanks for this thread I'm really trying to understand how Kv relates to power.

astro 3210 5 turn Kv = 270
astro 3210 10 turn Kv = 135
astro 3220 10 turn Kv = 67

I have a dumb question for y'all, as I'm not sure I'm right with my thinking yet about this.

take an [astro 3210 10 turn geared down 2:1] and an [astro 3220 10 turn not geared] and run them at the same volts and amps.

will these 2 motors produce the same power then?

It seems they would which means I was thinking all wrong. I thought bigger motor = more power at same volts and amps.

So it looks like the advantage of the 3220 over the 3210 is just less reduction needed for the same amount of power?

so if this is true, an [astro 3210 5 turn geared 4:1] would give me more acceleration on a bike than a [3220 10 turn not geared]?
even though they have the same no load rpm?

damn I've been lazy in learning how this all works

 

[Miles]

I thought bigger motor = more power at same volts and amps.

That would be a miracle

Bigger motor equals greater torque levels.

 

[mud2005]

Bigger motor gives more torque.

take an [astro 3210 10 turn geared down 2:1] and an [astro 3220 10 turn not geared] and run them at the same volts and amps.

will these 2 motors produce the same power then?

I guess I should have said torque here. since the 3210 has to be geared 2:1 to match the 3220 no load rpm, the torque is the same right?

 

[Miles]

I guess I should have said torque here. since the 3210 has to be geared 2:1 to match the 3220 no load rpm, the torque is the same right?

Pretty much. There'll obviously be losses in the reduction.

 

[mud2005]

Thanks Miles, it's FINALLY "sinking in" how these motors work

It seems to me the best benefit for getting a low Kv motor would be that it would be quieter and easier to gear down.

and the benefit of higer Kv = more torque at the wheel

I'm already rethinking my next build

 

[Miles]

It seems to me the best benefit for getting a low Kv motor would be that it would be quieter and easier to gear down.

Yes.

You save money and weight by going for a smaller and faster motor. Even with the additional gearing, the total weight and cost is usually less but the downside is the extra losses in the reduction and the complexity..... The important thing is to minimise the number of reduction stages...

 

[mwkeefer]

Miles and Luke,

Thanks for clearing this up for me... it makes much better sense finally (and clarifies about a dozen other questions I've had which I can skip now!)

-Mike

 

[recumpence]

What took me years to figure out, because no-one ever explained it to me, is the difference between torque and actual output power (wattage). Torque is a measure of static pressure, only, while power or wattage is a measure of work that can be accomplished by a given device (or how much electrical energy pulled from a power source, but I will not cover that right now).

You can have a hugely efficient motor that has very little torque and requires tons of reduction and not really have a decent system because you do need some static torque capability in the system. However, you can have a hugely torquey motor with terrible efficiency as well. You really need both good efficiency AND torque. Even though torque is not a measure of how much work a system can produce, it is still felt by the rider as output power. This is why the big mod for cars in the 60s was to gear the rear end lower (ususally 4:11). You would recieve more engine revolutions per wheel revolution, thereby increasing torque. However, that increase of torque was a reduction in total wheel revolutions per minute. So, the wattage did not go up (total power). However, you "Felt" more kick in the pants and you can get to 60 much faster. Yet, your top speed is hampered.

What I do is figure out how much speed I need, then I gear only that tall. Any taller and I am wasting power to heat.

Then you have to throw in motor efficiency too. Every motor has its sweet-spot. When I built the first twin 3220 Typhoon, it ran pretty strong geared for 40mph. When we geared it for 50, the accelleration increased. Hmm. When wer geared it for 60 the accelleration went up even higher! That is because the motors were finally being run at the RPM and wattage where they are most efficient.

There are many factors at play in setting up a system like this.

Matt

 

[Miles]

In SI units:

Ke = Kt = 1/Kv

Ke is the back EMF constant

Kt is the torque constant

Kv is the velocity constant

So, rpm per Volt and torque per Amp are inversely proportional.

As Kv reduces Kt increases

A low Kv motor will draw less current to produce the same amount of torque, which is great for the battery and the controller but, as the motor resistance increases with Kt, pretty much the same amount of heat is created in the motor, for a given amount of torque.

 

[liveforphysics]

Force x distance = work.
A billion pounds of force pushing against an object that doesn't move is zero work done.



Power is the rate at which work is delivered.

You can think of it as torque * rpm = power, or the rate at which work (getting your bike from one point to another) can occur.

1hp = 1 ft-lbs of torque at 5252rpm
1hp = 550lbs lifted up at 1ft per second
1hp = 746watts.
1hp = a zillion other things which all express an amount of power.

Nothing on earth can create power. We only convert existing energy, and we always lose some in each step.

Torque on the other hand can be created. Any amount of power, even 0.0000001w, can create any quanity of torque. You could take the second hand on your wrist watch and lift a trillion pound weight with it. Its going to require something like a (10^20):1 reduction, which means its going to lift the weight at 1/(10^20) the speed that it moves the second hand, but it can do it, as unlimited torque can always be created by any power source. Absolutely zero power can be created, only lost, and the more you mess with it, the more of it you lose. Power is the only factor which can determine the rate at which work is done, as power = the rate at which work is done.

The example with shortening up the gears in a car is a perfect example. The car accelerates faster because the shorter gears are able to increase the time the engine is in a more powerful part of the powerband which makes for higher average power applied, and this is why it accelerates faster. Its really neat to dyno a car, study the curve of the powerband, correct gearing to keep the engine in the fat of the powerband, calculate the new area under the curve as it runs through the gears, and go run a new time that exactly reflects the calculated increase in average power. Its one of those times you get to blow peoples minds by guessing there new times with the new gears to 0.1second, and then go see your guess come true at the track.

Examples get crazy with low displacement engines with giant turbos. We've had some examples where a car that could dyno 700whp in a tiny peak and fitted with long gears actually ran slower than cars with 400whp, but had a broad powerband and gearing to match. Once you worked out the shift points overlayed on the dyno curve and calculated area under the curve for each car, it became clear the 400whp car had a higher average power delivered through the quarter mile than the 700whp car.

 

[Miles]

Any more questions while we're on a roll?

 

[Miles]

I'll chuck this in here, as well:

In SI units the formula for rotary power is

Power (Watts) = Torque (Nm) x Angular Velocity (radians per sec)

Angular Velocity = 2pi * 60 * rpm

So:
Watts = Nm * rpm * 2pi / 60
or
Watts = Nm * rpm * 0.105

Solved for torque:
Nm = Watts * 60 / rpm / 2pi
or
Nm = 9.549 * Watts / rpm.

 

[liveforphysics]

There are at least a thousand more combos to express the same power, rate, torque relationships. That's not a joke either.

They all work out the same, some just have more familiar and comfortable units than others, but they all express the same concept. Force * speed = power. Or in identical terms, but more familiar for dealing with shaft power, torque * rpm = power.

You can trade speed for force, you can trade force for speed, but the power you get out can never be more than the power you put in.

 

[Malcolm]

Any more questions while we're on a roll?

Absolutely zero power can be created, only lost, and the more you mess with it, the more of it you lose.

Does this mean that we're accelerating the decay of the universe by messing about with batteries and motors?

 

[Toorbough ULL-Zeveigh]

why is a circle divided into 360 equal parts?
not a j/k, have been searching a long time where this is derived from.

 

[liveforphysics]

Any more questions while we're on a roll?

Absolutely zero power can be created, only lost, and the more you mess with it, the more of it you lose.

Does this mean that we're accelerating the decay of the universe by messing about with batteries and motors?

Yes.

Enthalpy will approach zero on its own just fine if we help it or not though

 

[Miles]

why is a circle divided into 360 equal parts?
not a j/k, have been searching a long time where this is derived from.

I like this explanation:
http://www.newton.dep.anl.gov/askasci/ast99/ast99606.htm

There seem to be plenty of others....

 

[recumpence]

why is a circle divided into 360 equal parts?
not a j/k, have been searching a long time where this is derived from.

I seem to remember running across the answer to this, but it escapes my memory ATM.

I think it had nothing to do with science, but had some spiritual significance. The reason I think that is because I ran across it while researching pagan roots to Christian holidays. I believe it goes back to ancient Babylon.

If I find it, I will get back to you.

A simple Google search on this brought up a quick Wiki definition;

http://en.wikipedia.org/wiki/Babylonian_mathematics

Matt

Doh!

Miles, how do you always beat me to it?

 

[mud2005]

information overload!

ok more dumb questions... I'm looking at the 3210 chart http://www.astroflight.com/pdfs/3210WEB.pdf and each motor has a voltage listed next to it. I'm assuming thats the "best" voltage for a given winding to keep the motor in the most efficient rpm range which seems to be about 7500rpm according to the chart.
also listed is the "best amps" which is for 93% efficiency at 170 in oz of torque and stays above 90% up to 375 in oz.

So then whats the best way to find the proper motor for a build based on this?

Say I have a 48V battery. Seems like the 8 turn would be the best choice and would run above 90% efficiency between 21 and 47 amps.

Say I have a 24V battery. Seems like the 4 turn would be the best choice but the best amps are 42-94 which seems too high.

Say I have a 10 turn motor which lists 60V as the voltage, but I run it at 48V do I lose alot of efficiency at a lower voltage? you could use a 24 turn at 24 volts and it would run at 1600rpm, but I'm assuming that would be inefficient, but I'm not sure why.

On the other end would running a 5 turn at higher voltage lose efficiency? a 5 turn at 48V would be 13,000rpm which seems a bit too high.

I guess I asked alot of questions there, I don't expect answers to all of them I'm mostly just wondering what's the best way to decide which motor to choose.
It seems to me that the 8 turn at 48V would be the best for most builds considering the voltage limit on the controllers being about 50V.
opinions, comments??

 

[Miles]

7,500 rpm is just the nominal speed chosen for the spec., I think. You could probably run it quite a bit faster (higher voltages for a given winding) before the eddy current losses dragged the peak efficiency down significantly.

Best amps is the current at peak efficiency, for a given winding, run at a given voltage. Peak efficiency, for a motor with an iron core, occurs when copper losses and iron losses are equal.

First, you decide on the size of motor and the speed you want to run it at. The operating voltage is limited by the controller choice. Lastly, you choose the winding to get the speed you want at the operating voltage.

More speed equals more power.