Hypothetical question about kv, gearing, and performance

[simplechamp]

So I'm looking at a certain motor, and trying to decide which wind to get. Gearing will be adjusted to provide approximately same max headspeed with either wind. Voltage and all other aspects of heli will be the same.

470kv at 44.4v, 8.9 gear reduction, approx. 2300 max calculated headspeed
560kv at 44.4v, 10.7 gear reduction, approx. 2300 max calculated headspeed

As I understand it lower kv motor trades some speed for increased torque, having more "electrical gear reduction" so to speak, while the higher kv has more speed and less torque, but will be using more mechanical gear reduction.

Given this basic info, is either setup preferable over the other? Will they perform about the same? Or is there too much more at play to determine based only on kv and gear reduction?

[Andy01]

I think your headspeeds may be a bit high if you are planning to use a governor.

RPM = 470 x 44.4 x 0.9 /8.9 = 2110rpm at 100% towards the end of a flight, so max governed speed would be 95% of this, or 2004rpm.

Likewise for the other option, rpm at 95% governed would be 1987rpm.

You may be assuming 100% efficiency - I use 90% in my calculation.

I am not sure you would really notice much difference in performance between the two, but a larger pinion is probably less likely to strip a main gear because it should have a little more contact area.

Colin

[simplechamp]

Thanks for the reply.

For simplicity the 2300 is assuming 100% efficiency. I know I can't be running at that, wasn't really trying to portray what the setups would really run at, but that the 2 would be equal in terms of headspeed and ESC output. Either setup would be set to govern at 1900 or 2000.

My main question is if there's a preference towards "electrical gearing" i.e. lower kv motor or mechanical gearing. Or if it doesn't make much difference providing both are within reasonable ranges. Maybe there are too many "moving parts" involved to have a simple answer?

I ended up buying the 470kv 3Y. This is the Xnova 4030 series, and the website said about the 560kv 2.5Y "For sport flying or longer flight times, the 470kv 3Y is more suited." Sport flying and mild 3D describe my flight style, so hopefully it will be a good match.

[Andy01]

I have 2 of the original Xera 4030 -470kV motors that the Xnovas are based on. A great motor, smooth, quiet and runs reasonably cool. Far better than the Scorpion 4025-550kV I tried briefly in my Seahawk.

Colin

[redsox]

I know what you're trying to ask and I'd like to know as well. Im not sure many people actually know or have reliable knowledge about it. You will basically get a bunch of responses that do not answer your question unfortunately. Anyone know who would have this info because its a question I see asked many times but the only answer giving is " oh yeah, I had that motor..works great".... Now I'm very curious about this

[extrapilot]

Guys

There are some variables here. It is not that some people don’t know or have reliable info- it is that you don’t define the parameters required to solve the query.

You can get lost in theoretical discussions about torque vs RPM at the gear interface (i.e. higher torque implies more tooth load/deformation, higher RPM implies higher interface speed, which has downsides, etc). There are also concerns with frame loads, where high torque does not just affect teeth- the motor/mount/frame has to resist that force- and it can be significant. Once you get some bend/offset, you now have bad tooth interface geometry, and things get very inefficient/hot/melted, etc. There are other losses- windage (bearing and aero drag from motor rotation), ESC (which tend to be less efficient at higher currents), etc.

Most of this is just banter between armchair engineers who fail to even inquire as to the materials in question, the frame properties, ESC loss, etc etc.

There is a very simple guideline for selection of motor/gearing:
Gear ratio = (Motor kV * nominal battery voltage (3.8v for a Lipo is typical here) * 80%) / Target headspeed

This process works well-
- Select a pack cell count based on whatever criteria you have (cost, charger, legacy ESC you will reuse, etc).
- Write down the min and max mechanical gear ratios available to your machine.
- Select a target headspeed
- Given the equation above, the high and low gear ratios, the headspeed, and the pack voltage, you can get the high and low Kv range for motor selection.
- Once you know the specific Kv (post motor selection), figure out which mechanical gear ratio will get your headspeed closest to your target.


This approach will get you to a place where your motor/ESC operate in a realm which is normally efficient, with some headroom for voltage sag in the pack. It also gives you a bit of flexibility with mechanical gearing and throttle to tweak things a bit as required.

Unless you define a whole bunch of criteria regarding frame, gears, motor, RPM/output power, etc, no one can tell you if you are better off with either of the two setups defined by the OP.

[ahahn]

Originally Posted by simplechamp

So I'm looking at a certain motor, and trying to decide which wind to get. Gearing will be adjusted to provide approximately same max headspeed with either wind. Voltage and all other aspects of heli will be the same.

470kv at 44.4v, 8.9 gear reduction, approx. 2300 max calculated headspeed
560kv at 44.4v, 10.7 gear reduction, approx. 2300 max calculated headspeed

As I understand it lower kv motor trades some speed for increased torque, having more "electrical gear reduction" so to speak, while the higher kv has more speed and less torque, but will be using more mechanical gear reduction.

Given this basic info, is either setup preferable over the other? Will they perform about the same? Or is there too much more at play to determine based only on kv and gear reduction?

The problem is that "it depends".

If all you had were resistive losses in the motor, then the higher Kv motor and gearing down would be more efficient.

However, motors also have magnetic losses whic depend on rpm, and of course mechanical friction that would depend on gearing-- gearing down would be less efficient since for a head revolution, more teeth would interact with the main gear.

That makes it hard to figure what the highest efficiency would be. You would need to know the efficiency at the motor rpm you would be running at. That information isn't usually available.

In your case you have two motors differing by ~20%. I'd say there isn't a big deal of difference between the two cases, at least not enough to be really worried about. My suggestion is to go with the commonly used case. At least it would be known to work satisfactorily.

[Jmann841]

IIRC if two motors are built exactlh the same except KV winding, the higher KV motor is usually more powerful.

So if both had the same copper fill and everything except KV winding the higher KV would be more powerful and running a higher mechanical gearing.

This is a question I wonder as well but it seems power is favored to higher KV. A good example of this is the scorpion 2520 motor. The 1360kv varient is considerably less powerful than the 1880kv (Peak and sustained watts)

[ahahn]

Originally Posted by Jmann841

IIRC if two motors are built exactlh the same except KV winding, the higher KV motor is usually more powerful.

So if both had the same copper fill and everything except KV winding the higher KV would be more powerful and running a higher mechanical gearing.

This is a question I wonder as well but it seems power is favored to higher KV. A good example of this is the scorpion 2520 motor. The 1360kv varient is considerably less powerful than the 1880kv (Peak and sustained watts)

Motors are not "more powerful". Put one down on a table. It just sits there. Put the three wires together--nothing happens. For comparison, try that with a 6s battery pack!

Motors are transformers, they transform input electrical power into mechanical output power. They also waste some of that input power as heat.

Remember, if you want the same head speed, then both motors (with different Kvs) must be making the same output power, which is related to the torque*rpm. Higher torque needs less rpm to make the same output power as a lower torque motor (with higher Kv) but higher rpm. Gearing just matches the torque to what the rotor demands at the desired head speed.

The OPs question is how much of that input power is going into heat in each case. As I said, it isn't completely obvious.

Motors and input voltage are fungible. A low Kv with a high voltage is equivalent to a high Kv motor with a lower voltage (Volts * Kv set to be a content). I am using the same criterion as you, that the motors are basically identical with same copper fill, just a different number of turns.

[Jmann841]

I think you're missing what I am saying.

The motor is not a power source, but the factors of the motor play into its peak power output (in this case we use watts, and peak power is typically at around 50% of max RPM although this varies of corse). So yes a motor can be more or less powerful in the same sense we say an engine is rated in power. An engine does not have power intrinsically, it needs fuel for power just as our motors do. Let's not be silly.

When I say a higher KV motor is more powerful assuming all things equal this is assuming ALL things equal, including voltage. The example of the scorpion 2520 highlights this as the 1360kv motor and the 1880kv motor both run on 6s. If you change voltage than all things are not equal. The OPs original question was assuming both motors are running on 12s

[ahahn]

Originally Posted by Jmann841

I think you're missing what I am saying.

The motor is not a power source, but the factors of the motor play into its peak power output (in this case we use watts, and peak power is typically at around 50% of max RPM although this varies of corse). So yes a motor can be more or less powerful in the same sense we say an engine is rated in power. An engine does not have power intrinsically, it needs fuel for power just as our motors do. Let's not be silly.

When I say a higher KV motor is more powerful assuming all things equal this is assuming ALL things equal, including voltage. The example of the scorpion 2520 highlights this as the 1360kv motor and the 1880kv motor both run on 6s. If you change voltage than all things are not equal. The OPs original question was assuming both motors are running on 12s

But what you are missing is that you can gear up the lower Kv motor to produce the same output power as the high Kv motor. There really is no difference here, unless you fix the pinion size. In that case you can get more power out of a high Kv motor using the same battery pack, but of course use more input power too.

You never want to use peak power by the way--it occurs at 1/2 the no-load rpm--more or less what you were saying. At that rpm, the motor is turning half the input power into heat, and half into shaft power. Continuously that would most likely burn up the motor (unless the motor was way oversized for the particular application)/

[Jmann841]

Originally Posted by ahahn

But what you are missing is that you can gear up the lower Kv motor to produce the same output power as the high Kv motor. There really is no difference here, unless you fix the pinion size. In that case you can get more power out of a high Kv motor using the same battery pack, but of course use more input power too.

You never want to use peak power by the way--it occurs at 1/2 the no-load rpm--more or less what you were saying. At that rpm, the motor is turning half the input power into heat, and half into shaft power. Continuously that would most likely burn up the motor (unless the motor was way oversized for the particular application)/

No, that is incorrect and what you are missing is it does matter.

You can up the gearing to produce the same headspeed but the max power is higher on a higher KV motor, assuming all things equal including voltage. Max power is rated around 50% of no load rpm, what mechanical gearing is irrelevant as it is the load applied to get the motor down to its 50% of no load rpm that will be the peak power the motor can produce. Power is power, it does not care about how it is applied as it is just work being done.

Changing the mechanical gearing does not change the max power rating (Wattage) of the motor.

The mathmatical equations in the link below can explain it better than us arguing back and forth. Plug some numbers in.

but the bottom line up front is this: "For the same motor, ideally if you apply double the voltage you'll double the no-load speed, double the torque, and quadruple the power. This is assuming of course the DC motor doesn't burn up, reach a state which violates this simplistic ideal motor model, etc."

http://electronics.stackexchange.com...rushless-motor

For a real world reference look at the max watts/power of these motors.

http://www.scorpionsystem.com/catalo.../HK_2520_1360/

http://www.scorpionsystem.com/catalo.../HK_2520_1580/

http://www.scorpionsystem.com/catalo.../HK_2520_1880/

[simplechamp]

In the scenario of same voltage same gearing the higher kv is going to be spinning the head faster. I think we all agree spinning a head at 2500RPM is going to take more power than spinning at 2000RPM. So Jmann I do see the point you are making, but it is not that relevant to the initial proposed scenario. I think this is the point ahahn is trying to make to you as well. We want to talk about a constant headspeed, and a constant power to make that headspeed by using different combos of motor kv and gearing.

It sounds like the consensus is that there are just too many moving parts (both literally and figuratively) to say one is better than the other.

[Jmann841]

Originally Posted by simplechamp

In the scenario of same voltage same gearing the higher kv is going to be spinning the head faster. I think we all agree spinning a head at 2500RPM is going to take more power than spinning at 2000RPM. So Jmann I do see the point you are making, but it is not that relevant to the initial proposed scenario. I think this is the point ahahn is trying to make to you as well. We want to talk about a constant headspeed, and a constant power to make that headspeed by using different combos of motor kv and gearing.

It sounds like the consensus is that there are just too many moving parts (both literally and figuratively) to say one is better than the other.

I understand this but that is not what I am saying.

You can change the gearing so it's the same headspeed but max power does not change as it is the max the motor can produce.

To help everyone understand i'll try to explain it better.

Take two of the same exact motors, same battery, same voltage, same everything except KV (so winding will be a bit different but same copper fill).

Put the motor on a Dyno (NOT on a Heli) and test the max power. The higher KV motor should have a higher max power.

Now if you put it on the heli and gear them to have the same headspeed (different mechanical gearing) the max power is still more on the higher KV motor because the motor can produce more power... In a hover they will produce the same power because the load is the same, but, in a situation where you need max power they will be different. Say full pitch hurricanes. The higher KV motor will have a higher max power.

Does that make more sense now? This is why I stated the mechanical gearing is irrelevant, we are talking about max mechanical power to the shaft of the motor, what happens after that is not important for this discussion.

If that still doesn't make sense maybe this will help:

Take a motor, doesn't matter the motor Let's say it's a 6s motor for simplicity sake. If you get a max power of 2,000 watts on this motor on 6s then the same exact motor, except run at 8s, will produce more than 2000 watts. This is why everyone claims 4s is so powerful compared to 3s. They are running 4s batteries (higher voltage) on a 3s motor and seeing a lot more power.

This time we are changing the voltage part of the equation instead of the kv of the motor, But, it is the same effect as keeping the voltage the same and changing the KV of the motor.

[simplechamp]

Originally Posted by Jmann841

Now if you put it on the heli and gear them to have the same headspeed (different mechanical gearing) the max power is still more on the higher KV motor because the motor can produce more power... In a hover they will produce the same power because the load is the same, but, in a situation where you need max power they will be different. Say full pitch hurricanes. The higher KV motor will have a higher max power

I'm sorry but I don't agree. If the higher kv, higher reduction motor pulls 2000W to hold a governed 2000RPM headspeed in full pitch hurricanes then the lower kv, lower reduction will be about 2000W to do the same thing.

You say yourself in a hover the load is the same for both. That is true. But you imply in a full pitch situation the load between the two is now different, and that is not true. If they are governed and actually holding the headspeed the load is the same.

[Jmann841]

Originally Posted by simplechamp

I'm sorry but I don't agree. If the higher kv, higher reduction motor pulls 2000W to hold a governed 2000RPM headspeed in full pitch hurricanes then the lower kv, lower reduction will be about 2000W to do the same thing.

You say yourself in a hover the load is the same for both. That is true. But you imply in a full pitch situation the load between the two is now different, and that is not true. If they are governed and actually holding the headspeed the load is the same.

Gah, the example is irrelevent as max power is achieved with what the max the motor can produce. I used the full pitch hurricanes as an example if one motor is more powerful than the other then say it takes 3000 watts to do full pitch hurricanes sustained but the less powerful motor can only do 2000 watts then it will blow out the heli aka, headspeed will drop to it's max power out. It was just an example of when you might find the max power of the motor being used, it was not very clear and I should not have used that, what I was trying to portray is in an example where max power is needed you will have more power with the higher KV motor so it can handle the move better (assuming it requires more power than the lower KV motor can put out, aka more bogging)

think about it like it is on a dyno, we are talking MAX power, so whatever move requires more power than the lower KV motor can handle.

The application is irrelevent for this discussion as the point is a high KV motor, or same kv but higher voltage, assuming all things equal, will produce more max power a the motor shaft. You can apply it to the application however you want, but at some point, or in some manuever, there will be a move that requires more power than what the less powerful motor can handle. Or maybe not and you are well over powered either way. but at the end of the day, the higher KV motor is capable of more power. Go to the extremes... Can a 450 class motor power a 700 sized heli? If it takes 400 watts of power to hover a 700 sized heli then yes it can. But can it do full pitch pumps that required 2000 watts of power? No it can not. That is what I was trying to say because one is more powerful than the other. I don't know what the watt requirements for full pitch hurricanes are or if it is at that limit. For simplicity's sake let's just talk about the motor being directly connected to a Dyno machine that can measure peak power.

[1BOHO]

I disagree with these ideas for a few reasons:

The power is the same as soon as you are able to pull enuf amperage out of the lower kv which would require more load. This is why companies give an entire series in their line the same wattage output and the per voltage output(KV) and amperage follow suit dictated by ohms law.The higher kv will pull more amps with the same load as the lower but it cannot handle the same voltage as the lower kv with less amp draw using higher voltage.
It comes out to be the same. You cant compare them at the same voltage really because the higher kv will not be able to withstand the same voltage as the lower kv at the end of the day.You also need to account for the mechanical maxes of the motors rpm in your comparisons.

Here an example. now the only thing that changes is the amperage and voltage LOW KV higher voltage less amperage high Kv lower voltage higher amperage equal the same power output whats different is at what speed is most efficient to do you work thats what the gears do....... sometimes you need an "armchair engineer" to think properly....



I think higher voltage with less amperage would always be a better way to go when possible on the electrical side. If the air frame cannot handle the torque then the motor isnt truly suitable for the vehicle.

I dont run helis but do you guys ever look at your blades tip speeds to decide whats most efficient??? Just wondering. Seems like a vehicle so dependent on it blades performance it might be worth a look see?????

http://mragheb.com/NPRE%20475%20Wind...ed%20Ratio.pdf

[Jmann841]

You are missing what I am saying as well, as I have explained a few times.

We are talking about using the SAME voltage. I don't know why this is so hard to get across.

Yes, when you change voltage your amps go up and power stays the same, however, with these motors, if you increase kv and keep voltage the same, or increase voltage and keep KV the same the power of the motor increases.

Look at the scorpion motors I posted in the previous example. Same votlage, different KV. The higher KV motor is rated at MORE power. It pulls a disproportional amount of amps for the increase in KV. If what you are saying is true, then increasing KV and keeping voltage the same would mean amp draw would decrease. This is NOT the case.

We are NOT talking about changing voltage. Read the OP's first post. He is asking about 2 different KV rated 44.4v motors.

I don't know how to make this any more simple to understand.

The motors you post are with different voltages. The OP is not asking about different voltages!

Scorpion 2520-1360kv Motor: Max Continuous Current 30 Amps
Max Continuous Power 630 Watts

Scorpion 2520-1880kv motor: Max Continuous Current 38 Amps
Max Continuous Power 800 Watt

They are both 6s motors with very similar wind, copper fill, and design. the Higher KV motor is MORE powerful. This all makes sense if you plug the math into the equations i put up earlier.

If you could pull 38 watts out of the 1360kv motor it would be the same power as the 1880kv motor, but you can not as it is less powerful on the same voltage. Now, if you increase the voltage of the 1360kv motor to say a 7s, you would see similar power out. But, again, we are NOT changing voltage here!

[1BOHO]

Because comparing two same series motors of different kv's at the same voltage makes no practical sense...... I dont know how much more simple I can state that.

Their output will be different at the same voltage hence KV........... how simple is that?

If the cost in your same voltage scenario is more amp draw with less TORQUE and you find that more efficient to TURN something have at it.....


TTYL
1BOHO

[Jmann841]

Originally Posted by 1BOHO

Because comparing two same series motors of different kv's at the same voltage makes no practical sense...... I dont know how much more simple I can state that.

Their output will be different at the same voltage hence KV........... how simple is that?

If the cost in your same voltage scenario is more amp draw with less TORQUE and you find that more efficient to TURN something have at it.....


TTYL
1BOHO

What do you mean it makes no sense!?

Is this not EXACTLY what the OP is asking?

Let me paraphrase the two motors he has for comparison:

"470kv at 44.4v, 8.9 gear reduction, approx. 2300 max calculated headspeed
560kv at 44.4v, 10.7 gear reduction, approx. 2300 max calculated headspeed"

Notice both are 44.4v volts? Am I missing something? Is this not exactly what the OP was talking about?

Please explain how that makes no sense?

Not only this, the example of scorpion motors is an exact practical application of this same thing. 3 separate motors, all same voltage, different KV. The KV differences is due to different mechanical gearing on different heli's, however, if brut power is what you want, which in many cases is what 3D heli's are all about, a higher gear reduction and higher KV motor is typically the higher power option. Seems to make perfect practical sense to me.

These are 3D helicopter applications, We are typically not concered about efficency but absolute power. This is exactly what is being discussed here and I have multiple examples stating along with the math.

We all understand Ohms law and it is at play here as well. Read my last few posts and maybe it will help you understand.