Low KV vs High KV motors

[Linkman81]

I have done a good amount of research and I feel that I am starting to understand the relationship of how the power systems fit together from motor to ESC to battery.

My question: Everything being equal, why would a higher RPM motor be qualified as a "speed" motor and a low RPM as a "trainer" type motor?

From watching videos, I've noticed that David would always put a high RPM/KV motor in his planes when wanting to go fast... here's my problem. From what I understand the higher the pitch of your prop the more current (amps) are drawn through your ESC from your battery. So if I had two 25mm diameter motors... one of them was 900KV and the other was 1800KV, both have a max amp draw rating of 30amps. From my understanding the lower KV motors have lower RPMs but higher torque and the higher KV motors have more RPM but less torque. I've noticed that it is customary to put the lower KV motor on a trainer with a big aggressively pitched prop, and a for a speed plane you would put a smaller prop with less pitch. My question is... wouldn't you have the same results (air speed) if you had a low rpm / high torque motor with a high pitch prop or a fast rmp / low torque motor with a low pitched prop. If you found a prop for each style but each pulled the same AMPs on their respective motors wouldn't the speed of the plane be the same?

I'm just trying to wrap my head around the physics of it... and I think I must not be getting something. I've only messed with motor so far, but if I had a low KV motor and a high KV motor, I would hook up my watt meter and switch out props until I got up near the max amps that motor was rated for... and do the same for the low KV one. In my mind if the motor specs were all the same between the two other than the KV and I found a prop that pulled near the max amp draw for both... they should perform the same given the same work being done in relation to the prop. Perhaps I'm way off. Give me a hint please. Thank you in advance.

Brandon

 

[IFlyRCstuff]

Higher rpm=faster
The second number in a prop rating is how many inches it moves forward per rotation (in a perfect world), so more rotation, more inches forward/faster.
and for the best setup you want low kv, with higher voltage. kV means thousand turns per volt × k/the number

 

[IFlyRCstuff]

The ESC supplies the current required to fuel what throttle setting it is at.
Imagine a dam. Voltage=how big a door is, and in turn the power. Current=amount of flow, which can differ based upon how open the door is. Watt=force of the water Amp=measure of force.

 

[joshuabardwell]

An airplane's nominal speed, known as pitch speed, is equal to the prop pitch times the prop RPM. Therefore, all else being equal, a higher kv motor will result in a faster-flying plane. Other things that make a plane fly faster are a higher-pitch prop and a higher voltage (3S, 4S, 6S) battery.

Increasing pitch has some limitations because a prop is actually a small airfoil, not just a flat fan blade. So in the same way that an airplane wing can stall if its angle of attack gets too high, a prop can stall if its pitch gets too high. Some planes that are designed for high speed will actually have their prop partially stalled when they are standing still. As the plane gains forward speed the angle of attack and the prop un-stalls and develops proper thrust.

Increasing voltage can have limitations in that the bearings of the motor may not be rated for the higher speed. Additionally, the tip of the prop must not exceed the speed of sound, both because the prop may be stressed and break, and because turbulent air flow will make the prop very inefficient. So for any given prop size, there is a certain maximum RPM above which the prop cannot be allowed to go.

 

[PHugger]

Imagine a dam. Voltage=how big a door in it is, or potential power. Current=flow, which can differ based upon how open the door is. Watt=force of the water Amp=measure of force.

I'm not sure this is correct.
To continue with your Dam (the watery kind) analogy -

Current = how big the door is
Voltage = how much pressure
Watt = Voltage x Current (work)



Best regards,
PCH

 

[Linkman81]

Higher rpm=faster
The second number in a prop rating is how many inches it moves forward per rotation (in a perfect world), so more rotation, more inches forward/faster.
and for the best setup you want low kv, with higher voltage. kV means thousand turns per volt × k/the number

Hi there IFlyRCstuff. I understand fully the numbers of a prop being the diameter and the pitch that would carry the prop forward. So a greater pitch would make something faster... BUT... the greater the pitch, the more current a motor will draw and you always will run into the max amps that a motor is rated for. So the more stress or resistance you put on a motor, the more current it will draw and you will have to stop increasing the pitch once it gets to the max amps of that motor unless you want to burn it out or damage. The point of my post is that the lower KV motor would be able to endure a LOT more stress or resistance... thus a much higher pitched prop before reaching it's max AMP draw since it is lower RPM and higher torque. I don't see how high rpm would mean you get higher speed if you have to use a prop with less pitch... it seems the advantages of either motor balance out if all other aspects of the motor are equal.

 

[Linkman81]

To add on to my previous post... imagine this example

Two power systems where everything is the same but the motor. Same battery, same esc, same motor diameter, same max amp rating for the motor. The only thing that is different between the two setups is the KV or RPM of the motors.

1000kv motor with a 8x10 prop pulls the max amp rating of a motor at 30 amps

2000kv motor with a 8x5 prop pulls the max amp rating of a motor at 30 amps

The first (lower KV) motor can use a higher pitched prop before maxing out the amp rating of it's motor because it has more torque but less revolutions per minute. The second (higher KV) motor can spin twice as fast, but can only accommodate a prop that has a pitch half as high as the first before maxing out the amp draw of the motor. You can see that even though one can spin twice as fast, the prop is only carrying it half as far.

This is what I'm asking about. If all other thing being equal, why would a lower KV motor be any less speedy than a higher KV motor when it can accommodate a prop with more pitch before reaching it's max amps. Does this example hold up in practice? Is there something else that I'm not anticipating? My argument assumes that the scale at which motors' RPMs goes up, the torque equally and linearly goes down. If this is not the case... let me know.

 

[IFlyRCstuff]

I'm not sure this is correct.
To continue with your Dam (the watery kind) analogy -

Current = how big the door is
Voltage = how much pressure
Watt = Voltage x Current (work)



Best regards,
PCH

Yes, I mean, think, bigger door=more power over an area smaller door=less power over an area VOLTAGE
CURRENT=amount of flow (which could be interpreted as door size)

 

[IFlyRCstuff]

To add on to my previous post... imagine this example

Two power systems where everything is the same but the motor. Same battery, same esc, same motor diameter, same max amp rating for the motor. The only thing that is different between the two setups is the KV or RPM of the motors.

1000kv motor with a 8x10 prop pulls the max amp rating of a motor at 30 amps

2000kv motor with a 8x5 prop pulls the max amp rating of a motor at 30 amps

The first (lower KV) motor can use a higher pitched prop before maxing out the amp rating of it's motor because it has more torque but less revolutions per minute. The second (higher KV) motor can spin twice as fast, but can only accommodate a prop that has a pitch half as high as the first before maxing out the amp draw of the motor. You can see that even though one can spin twice as fast, the prop is only carrying it half as far.

This is what I'm asking about. If all other thing being equal, why would a lower KV motor be any less speedy than a higher KV motor when it can accommodate a prop with more pitch before reaching it's max amps.

Good question hmm...
Think about it, you don't fly at full throttle if you are a beginner, but you can punch out, so in all reality, it's mainly that it's less wear and tear. That is how interpret it, but nothing else seems to trigger my mind.
Back to my dam example.
Now imagine throttle as door being open or closed, or in between. The current affects the amps directly. More throttle, higher current, dam erodes away (hopefully not but...)
So, larger door, more current and erosion.

 

[joshuabardwell]

Really, I think the analogy should be:

Volts (Pressure) = The height of the water column (feet) which determines pressure (PSI).
Ohms (Resistance) = The size of the opening (square area).
Amps (Current) = The flow rate of the water (gallons per minute).
Watts (Power) = Flow rate * pressure. You wouldn't typically encounter this in a water system, but in a hydraulic system like on a tractor, you'd see it.

 

[IFlyRCstuff]

*vin diesel voice* We think differently, So here is my final interpretation of the analogy.

Door size-potential power or Voltage
Openness of door-current, dependent on Size of door
Amps-relative to door height (irrelevant to voltage directly)
Watts-column of water/fforce.

This is just how I think of it, but any way you think of it, it holds true that you can alter the current through voltage, which alters amps through throttle.

Sorry for the confusion

Hehehe...

 

[Linkman81]

I've seen argument about comparing water and electricity on every forum known to man. I think that's why I tend to not try and think of them in that way, although it is helpful in learning the basics, I just don't know how'd you'd apply it in this situation... because what analogy would the prop be in a water system if a higher pitched prop draws more current from the power source?

From my limited understanding, a motor with no prop will only pull a given amount of current, usually much lower than you'd think. The more resistance you put on that motor (progressively higher pitched props) given then same throttle setting, the more current it will try to pull to achieve your selected throttle setting.

 

[joshuabardwell]

1000kv motor with a 8x10 prop pulls the max amp rating of a motor at 30 amps
2000kv motor with a 8x5 prop pulls the max amp rating of a motor at 30 amps

This is what I'm asking about. If all other thing being equal, why would a lower KV motor be any less speedy than a higher KV motor when it can accommodate a prop with more pitch before reaching it's max amps.

One answer to your question is that manufacturers don't make an infinite variety of motors. If you need a motor that puts out a certain number of watts, there may only be two or three different options in any manufacturer's range, and they will be within a certain range of kv values, and that will be that. So a lot of the time, you start by finding a motor in the appropriate size, and then work from there with kv rating and prop size.

Let's see what eCalc says about your example, though. Start with the SunnySky X2212 kv980 running on 3S. Put an 8x10 prop on it. eCalc estimates 8646 rpm with an estimated pitch speed of 82 mph, pulling 12.3 amps. Now let's take the same motor and double its kv and put an 8x5 prop on it, while leaving all other characteristics the same. eCalc estimates 13,247 rpm, with a pitch speed of 63 mph, drawing 26.5 amps, which is over the motor's rated limit.

The short version is that there is more to the equation than kv, voltage, and pitch. It takes much more power to spin the half-pitch prop twice as fast, because air resistance rises non-linearly. This is the same reason why you need a LOT more horsepower to push the top speed of a car up just a little bit, and why aerodynamics become much more important for high speed cars than for low-speed ones. It is basically always more efficient to spin a larger, higher-pitch prop slower, compared to spinning a smaller, lower-pitch prop faster.

Compare the graphs below for more insight (apologize that the X and Y axes are not the same, but that's how eCalc spits them out).

980 kv:


1960 kv:

 

[joshuabardwell]

Ibecause what analogy would the prop be in a water system if a higher pitched prop draws more current from the power source?

I agree that the analogy is probably more confusing than just learning about electricity in the first place, but just because I love pedantics, I think the answer to your question is the pitch of the blades on an impeller, or something like that.

 

[Robyle3]

I know this thread is really old, but i just wanted to throw my 2 cents into the “understanding electric power” analogies ring.
When it comes to electric motors, imagine a huge muscle-man pulling a rope attached to a propeller shaft. The more muscular, the more power he can initially start with (voltage). It also makes sense that the prop will spin faster the more power he starts with (kv), and a shaft with no propeller would turn a lot faster. The longer he can can pull it, the longer the propeller will keep turning (amps). And the speed (current) at which he pulls will tell how much thrust the prop makes. If different men try to make it turn, each will have a unique overall ability (watts).

Volts= Initial power
Kv= Number of rotations given consistent power
Amps= a combo stat of how fast that initial power goes
Current= How fast you can go
Watts= overall ability

All the stats depend on the voltage of the battery (or muscle strength), and the resistance (pitch) of the prop. Thats why most builds aleays start with motor/prop selection, as that tells you what size esc to use, and what battery you want for a specific flight style (i.e. a 4s will give you a rocket while a 3s gives you a slow trainer).

A prop with lots of resistance (pitch) will require more power to turn, but at any given rpm will give more thrust than a low resistance (pitch) prop of the same diameter. High kv motors are paired with low pitch props because they dont need as much tourque to cut through the air, so therefore can actually reach higher rpms to produce more thrust. Low kv motors are paired with big props to let the prop push more air back with less rpm, making for a more gentle flying style. Think of indoor rubber models that use giant 2’ props turning at MAYBEE 23 rpm, vs 100mph foamies that use 8x4 props with ridiculously high kv motors.

Thats my understanding, anyway.

 

[d8veh]

In simple terms kV doesn't directly affect torque nor power. It affects speed and efficiency.

A 1000kV motor with a 3S battery will have the same maximum speed as a 1500kV motor with 2S, but the 1000kV one will make nearly 50% more power with that battery throughout the rpm range, but not at maximum rpm. That would show up as it being more punchier.

There are a lot of different things going on at the same time, so it's difficult to make simple rules, but, basically, you need to have a propeller, motor, battery and ESC that all have matching parameters. Here are some things to help. You have to have all of them figured out in your head at the same time when choosing stuff:

Before we start, what's written below is on the basis of maximum throttle. When you reduce the throttle, it has the same effect as lowering the battery voltage, i.e. the speed reduces and the whole efficiency curve moves down the rpm scale with it. In actual fact, the ESC reduces the current by pulsing it with shorter pulses as the throttle reduces until you get no pulses at zero throttle.

1. Higher voltage means more RPM and more current. Too much current will burn your motor.
2. Peak power comes at the RPM about 0.75 x kV x battery voltage and peak efficiency a bit lower.
3. Motor torque is directly proportional to current.
4. The current is astronomically high at low RPM (full throttle) and ramps down to zero at maximum RPM, where max RPM is kV x battery voltage, i.e.the faster the motor goes, the less current it draws. Too slow will burn the motor.
5. Larger diameter or higher pitch propeller makes a bigger load, so it slows down the motor. The aim is to use a propeller that gets the motor up to a high enough speed that the current is low enough not to burn it. That's if you want maximum power.
6. The battery C-rating has to be enough to provide the current that the motor will take. My guideline is C x MAh x 0.5 should be higher than the motor current.
7. The ESC has to be rated higher than the motor current. I use a safety factor of 0.75, so a 40A ESC for a 30 amp motor.
8. Any watts that you measure cannot be taken as power output, neither can current be taken as a measure of power nor can it be used to judge likeliness of burning because the efficiency varies so much throughout the RPM range. It's not unknown to have motors running with only 50% efficiency, which means that a 240w reading would be 120w of output power and 120w of burning power, while as a 200w reading with the motor running at peak efficiency (80%) would be 160w output power and only 40w burning power.

Can anybody add to this or is there anything that needs changing?

I've left out propeller efficiency because I'm not sure of rules and relationships. The best is to actually measure thrust with different propeller and battery options. I think that the best measure is thrust per watt consumed, which will give some idea of how efficient your system is.

 

[DamoRC]

In simple terms kV doesn't directly affect torque nor power. It affects speed and efficiency.

A 1000kV motor with a 3S battery will have the same maximum speed as a 1500kV motor with 2S, but the 1000kV one will make nearly 50% more power with that battery throughout the rpm range, but not at maximum rpm. That would show up as it being more punchier.

There are a lot of different things going on at the same time, so it's difficult to make simple rules, but, basically, you need to have a propeller, motor, battery and ESC that all have matching parameters. Here are some things to help. You have to have all of them figured out in your head at the same time when choosing stuff:

1. Higher voltage means more RPM and more current. Too much current will burn your motor.
2. Peak power comes at the RPM about 0.75 x kV x battery voltage and peak efficiency a bit lower.
3. Motor torque is directly proportional to current.
4. The current is astronomically high at low RPM (full throttle) and ramps down to zero at maximum RPM, where max RPM is kV x battery voltage, i.e.the faster the motor goes, the less current it draws. Too slow will burn the motor.
5. Larger diameter or higher pitch propeller makes a bigger load, so it slows down the motor. The aim is to use a propeller that gets the motor up to a high enough speed that the current is low enough not to burn it. That's if you want maximum power.
6. The battery C-rating has to be enough to provide the current that the motor will take. My guideline is C x MAh x 0.5 should be higher than the motor current.
7. The ESC has to be rated higher than the motor current. I use a safety factor of 0.75, so a 40A ESC for a 30 amp motor.
8. Any watts that you measure cannot be taken as power output, neither can current be taken as a measure of power nor can it be used to judge likeliness of burning because the efficiency varies so much throughout the RPM range. It's not unknown to have motors running with only 50% efficiency, which means that a 240w reading would be 120w of output power and 120w of burning power, while as a 200w reading with the motor running at peak efficiency (80%) would be 160w output power and only 40w burning power.

Can anybody add to this or is there anything that needs changing?

I've left out propeller efficiency because I'm not sure of rules and relationships. The best is to actually measure thrust with different propeller and battery options. I think that the best measure is thrust per watt consumed, which will give some idea of how efficient your system is.

I think this is an excellent summary but there are a couple of items I would change or re-write for clarity.

First, and I am just being an old pedantic bugger here, although there is an electrical term kV (kiloVolt or 1000 volts), when it comes to motors, there is no term kV or KV, only Kv, the motor velocity constant. Additionally, for practical purposes, although the Kv does relate the RPM of the motor and the voltage applied, the Kv is most accurately described as the voltage output from a brushless motor when it is run as a generator, i.e., if you turn a 1000Kv motor at 1000 rpm, it will generate 1V output across a pair of motor wires. It so happens that the converse is almost true, i.e., if you apply a properly commutated 1V signal to a 1000Kv motor it will spin at 1000 rpm (although in reality it is always a little less that this and only applies to an unloaded motor. I know you probably understand this but given the excellent summary and the fact that we could point new folks to these 8 points to understand their motors / props etc, I would be good to be accurate with the terminology

On point two - can you clarify what you mean here about "peak efficiency a bit lower"? I think peak electrical efficiency actually comes in at a higher RPM. Do you mean electrical efficiency or thrust out to power consumed?

I agree with point number 4 from a motor theory perspective, maximum current draw occurs at motor stall and decreases as the motor RPM increases. However, I am concerned that the way point number 4 is written would lead folks to think that they are less likely to burn out their motor at max throttle / max RPM (even though you stated low RPM at max throttle in your description). Is there are way to either remove point 4 (as point 5 gets to t meat of the matter), or to re-state to say that if you reduce the RPM of a motor by applying an increasing load you will increase the current the motor draws.

 

[Merv]

So a greater pitch would make something faster... BUT... the greater the pitch, the more current a motor will draw ....
I don't see how high rpm would mean you get higher speed if you have to use a prop with less pitch...

You are correct that increasing the pitch will make the plane go faster, up to a point. As @joshuabardwell stated. "Increasing the pitch of the prop has limitations". Imagine two extremes, one prop with no pitch, spinning it would not produce any thrust and little drag. Now imagine the other extreme, a prop with a pitch 90 degrees from the first. This prop would also produce no forward thrust, but it would produce a lot of drag or torque. Somewhere between these two extremes is the optimal point, the most forward thrust with the least amount of torque.

There is another relationship with Kv and props besides pitch. Generally low Kv motors take a larger diameter prop & high Kv take a smaller diameter prop. Also related to this drag or torque issue, larger diameter more drag. Lets do another thought experiment. Take two motors, both rated at 30 amps, one is 1000kv and uses 10x5 prop. The second is 2200kv & uses a 6x4 prop. Put both on a test stand, the low kv motor will produce more thrust and more torque. Now put both in a wind tunnel, set the wind at 20 mph. Now they will do about the same and both are less than on the bench. Crank the wind up to 40 mph. Now the high kv motor has some thrust but the low kv motor has none, too much energy producing torque.

So we do a dance with RPM's (voltage & kv), pitch & diameter. The question is at what air speed do we want thrust. At that air speed, what combination gives the most forward thrust with the least torque. Torque is just lost energy that not produce forward thrust.

Here is the bottom line, if you want to go fast, spin a smaller diameter prop faster. If you want to 3D, hang from the prop, spin a larger diameter prop slower.

 

[Merv]

imagine this example
1000kv motor with a 8x10 prop pulls the max amp rating of a motor at 30 amps
2000kv motor with a 8x5 prop pulls the max amp rating of a motor at 30 amps
Does this example hold up in practice? Is there something else that I'm not anticipating? My argument assumes that the scale at which motors' RPMs goes up, the torque equally and linearly goes down

Your example is correct in theory, if that were the case both would produce the same speed. But I do not think your amp draws are accurate. I don't believe it's a linear relationship, I believe its an exponential relationship between (pitch and amps) and (RPM's and amps). We would need to do the experiment to be sure.

In your example, your thinking about the motor is correct, more RPM = less torque (inch pounds), both produce the same horsepower but higher Kv has less torque. I think this relationship is linear. It's the prop that is the problem, more RPM's is going to need more torque, a lot more. Likewise more pitch is going to take more torque, a lot more. I don't believe these relationships are linear, I think they are exponential.

Think of how much energy the prop is absorbing to spin at a given pitch or RPM. If you go from 5,000 RPM's to 10,000 you will need more than 2X the energy to do it. And if you go to 15,000 RPM's you will need a lot more than 3X. Likewise for the pitch. The energy requirement will go up exponentially as RPM's and pitch increase.

 

[d8veh]

I think this is an excellent summary but there are a couple of items I would change or re-write for clarity.

First, and I am just being an old pedantic bugger here, although there is an electrical term kV (kiloVolt or 1000 volts), when it comes to motors, there is no term kV or KV, only Kv, the motor velocity constant.

i didn't know that. I've seen it written lots of ways. In maths, we usually use small letters for constants, which is why I wrote it that way. Thanks for the clarification.