Is there any advantage to running a lower kv motors with a lower gear ratio (larger motor pulley) vs higher kv/higher gear ratio? I read that kv is also the motor torque constant (via some unit conversions, kv = N*m/A), and ran some numbers, and it looks like for a given board speed, it doesnt matter what kv motor you use, you will draw the same amperage from the batteries.
KV vs Gearing?
- Thread starter ProxRB
- Start date
psychotiller
10 kW
The higher motor KV you choose the lower Torque at start up you'll receive. The best thing you could do is choose a motor lower than 300kv. Otherwise you will have to push yourself up to speed before your motor will be very effective. Also, you'll most likely encounter some hills on a ride. Low KV motors wont have the tendency to bog down while climbing. Of course, gearing is a factor. I have an 8s board, 290kv motor, geared @ 1.77 which loves to go fast (upper 40's) But it requires a generous push to get going.
I have another board, 245kv motor, geared @2.0, or 2 to 1, 22v, 30mph, barely push it at all to get going.
If you tried to use an 800kv motor on the latter board I mentioned it would want to go 100mph, but it wouldn't really have the torque to get there and breaking would probably suck.
On a urethane board 2-1 to 3-1 is a good range to shoot for.
I have another board, 245kv motor, geared @2.0, or 2 to 1, 22v, 30mph, barely push it at all to get going.
If you tried to use an 800kv motor on the latter board I mentioned it would want to go 100mph, but it wouldn't really have the torque to get there and breaking would probably suck.
On a urethane board 2-1 to 3-1 is a good range to shoot for.
torqueboards
1 MW
Yeah, I'd say 170kv to 280/290kv.
Lower KV is better but also depends on your gearing ratio.
I wouldn't go higher than 300kv as there is not much torque at all.
290kv works fine on a dual motor but torque is a bit lacking on a single motor setup.
Higher voltage + more torque is ideal.
Lower KV is better but also depends on your gearing ratio.
I wouldn't go higher than 300kv as there is not much torque at all.
290kv works fine on a dual motor but torque is a bit lacking on a single motor setup.
Higher voltage + more torque is ideal.
that0n3guy
1 mW
- Joined
- May 2, 2015
- Messages
- 16
I'd say just design your gearing and KV based on your requirements. See this: http://vedder.se/2014/10/chosing-the-right-bldc-motor-and-battery-setup-for-an-electric-skateboard/
For example, more efficient KV based on battery:
12s:
3.8*12 = 45.6v
8600 / 45.6 = 189 KV
or
10s:
3.8*10 = 38v
8600 / 38v = 226 KV
Then for gearing:
Gear Ratio:
http://www.advanced-ev.com/Calculators/TireSize/
RPM: 8600
Tire Diameter: 8.75 (8.5" to 9" depending on air pressure... I have pneumatic tires)
MPH: 25
Differential Ratio: ~1:9 (this seems high... I don't see this type of ratio anywhere on this forum)
For example, more efficient KV based on battery:
12s:
3.8*12 = 45.6v
8600 / 45.6 = 189 KV
or
10s:
3.8*10 = 38v
8600 / 38v = 226 KV
Then for gearing:
Gear Ratio:
http://www.advanced-ev.com/Calculators/TireSize/
RPM: 8600
Tire Diameter: 8.75 (8.5" to 9" depending on air pressure... I have pneumatic tires)
MPH: 25
Differential Ratio: ~1:9 (this seems high... I don't see this type of ratio anywhere on this forum)
parajared
10 kW
You can get to a point where the shaft diameter cannot withstand the torque of being overgeared down but I don't understand the difference between a direct drive 50kv motor and a 100kv motor with a 1:2 reduction. Both arrangements turn at the exact same RPM: wouldn't they be almost exactly the same performance?
parajared said:
I'm not sure about performance-wise but I know gearing down the output of a motor is less efficient. Also, using current to control the motor's output is far better than using a fixed gear ratio which doesn't always work in all situations. It would be nice if E-boards can have variable-gear ratios.
that0n3guy
1 mW
- Joined
- May 2, 2015
- Messages
- 16
Pediglide said:
From a mechanical engineering standpoint, I think the only thing that would make one less efficient would be gear/belt friction.
According to Vedder's article, the 14 pole motor that is at 8600 RPM would be the "most" efficient. At 50kv motor at 10s would turn at 1900rpm, a 100kv motor at 10s would turn at 3800rpm. Since the efficiency curves are ...well...curves, the number closest to 8600 would be more efficient. So the 100kv motor would be more efficient at 10s b/c its closer to 8600 rpm.
Thats my logic, I'm not sure if its entirely true, but it plays out.
But you can use 172V for the 50kv if you want to and get 8600rpm. You can also use half of that voltage for 100kv to get the same rpm, but you would need higher amps to get the same watts, which will produce more heat, thus less efficiency from an electrical standpoint.
I don't totally get it, while ratios make the motor's life easier, there are limits to the gearing benefits (or else nobody would go hub motors?)
Correct me if I'm wrong : since efficiency curves are curves, for a fixed Voltage a 100Kv 1:2 ratio motor will be advantaged against a 50Kv direct drive motor only when he goes past the RPM limit of the direct drive? Yet it will need more RPM to achieve the same power = less powerful acceleration caused by longer momentum of power.
Plus you need to take in account the throttling of the motor which have to vary twice its number of revolutions to either brake or accelerate, thus fight its own weight and inertial energy twice more.
Then you add belt friction and everything else.
So I think that at low speed and from standstill the direct drive has the advantage both in control and sheer torque, the geared motor will get advantageous once launched and raising from a fixed RPM. Since it is closer to 8600 magic numbers, it will hold better in this "mid to high" RPM range.
So it depends on the use, for a picture drag race would give an edge to the hub from the starting line then when you reach the 1/3 of the mile the geared motor will catch up or power up better till the finish line? Most people would only need this 1/3 of the mile power range in urban environment. The rest of the power curve is for racing.
Correct me if I'm wrong : since efficiency curves are curves, for a fixed Voltage a 100Kv 1:2 ratio motor will be advantaged against a 50Kv direct drive motor only when he goes past the RPM limit of the direct drive? Yet it will need more RPM to achieve the same power = less powerful acceleration caused by longer momentum of power.
Plus you need to take in account the throttling of the motor which have to vary twice its number of revolutions to either brake or accelerate, thus fight its own weight and inertial energy twice more.
Then you add belt friction and everything else.
So I think that at low speed and from standstill the direct drive has the advantage both in control and sheer torque, the geared motor will get advantageous once launched and raising from a fixed RPM. Since it is closer to 8600 magic numbers, it will hold better in this "mid to high" RPM range.
So it depends on the use, for a picture drag race would give an edge to the hub from the starting line then when you reach the 1/3 of the mile the geared motor will catch up or power up better till the finish line? Most people would only need this 1/3 of the mile power range in urban environment. The rest of the power curve is for racing.
that0n3guy
1 mW
- Joined
- May 2, 2015
- Messages
- 16
There are a number of things getting mixed up here (though I'm not sure if it effects anything). Efficiency is are not the same as torque. Max efficiency does not equal max torque. So just because you are more efficient, doesn't mean that you have more torque, so I'm not sure the drag racing example is true.
I've not seen any documentation on torque vs KV. Is torque linear as KV changes (like RPM is)? Ie...does 100KV motor have half the torque at 1900RPM as a 50KV motor at 1900rpm? Does torque even have anything to do with RPM? I asked a question on vedders article.
Also, some quotes from vedders article:
From Vedders article: "For the same motor, the low KV versions have more windings with thinner wire, while the high KV versions have less windings with thicker wire. As long as they have the same mass of copper, they are exactly the same in regard to max power output, torque, efficiency, max RPM; but at different currents/voltages"
Also: "Because of the square relation, it is desirable to run at as high speed and low torque as possible as long as we stay below 8.6k RPM. To express the square relation in some numbers, having double the RPM and half the torque at a certain power output will cause four times less losses. "
So 100kv at 1:2 ratio is exactly the quote above... This setup would require double the RPM and half the torque on the motor for the same wheel speed/torque.
I've not seen any documentation on torque vs KV. Is torque linear as KV changes (like RPM is)? Ie...does 100KV motor have half the torque at 1900RPM as a 50KV motor at 1900rpm? Does torque even have anything to do with RPM? I asked a question on vedders article.
Also, some quotes from vedders article:
From Vedders article: "For the same motor, the low KV versions have more windings with thinner wire, while the high KV versions have less windings with thicker wire. As long as they have the same mass of copper, they are exactly the same in regard to max power output, torque, efficiency, max RPM; but at different currents/voltages"
Also: "Because of the square relation, it is desirable to run at as high speed and low torque as possible as long as we stay below 8.6k RPM. To express the square relation in some numbers, having double the RPM and half the torque at a certain power output will cause four times less losses. "
So 100kv at 1:2 ratio is exactly the quote above... This setup would require double the RPM and half the torque on the motor for the same wheel speed/torque.
torqueboards
1 MW
While efficiency is great.. I don't think it's always ideal.
Just because something is efficient doesn't mean it's best to have.
Just because something is efficient doesn't mean it's best to have.
True (I get it better now, thank you for the clarifications 
), but you can also fight the square relations with higher voltage / lower amperage and you get the same benefits...
In the end it really depends on what power you need? You may want to get both of best worlds (gearing and voltage upping) past a certain point. Remember that the wheel size itself can transform into a kind of gearing for a hub motor.
We are a minority to go as high as 10S or even higher in the e-board/scooter/skating world, e-bikers have many more options (well, things are starting to evolve and we are catching up).
), but you can also fight the square relations with higher voltage / lower amperage and you get the same benefits... In the end it really depends on what power you need? You may want to get both of best worlds (gearing and voltage upping) past a certain point. Remember that the wheel size itself can transform into a kind of gearing for a hub motor.
We are a minority to go as high as 10S or even higher in the e-board/scooter/skating world, e-bikers have many more options (well, things are starting to evolve and we are catching up).
Ok, so I've done a lot of thinking about this and I'm pretty sure I got this figured out. Given two different setups:
1) 150 kv motor, geared 24:36 with 83mm wheels, 6s LiFePo, theoretical top speed of 19.2 mph
2) 300 kv motor, geared 12:36 with 83mm wheels, 6s LiFePo, theoreitcal top speed of 19.2 mph
Since motor torque constant = 1/Kv, if you do the math you find that:
1) motor torque constant = 0.0637 N*m/A
2) motor torque constant = 0.0318 N*m/A (half of motor 1)
However, your gearing ratio makes up for that, so torque at the wheels is just
1) torque at wheel = 0.0955 N*m/A
2) torque at wheel = 0.0955 N*m/A (the same)
Now, electrical losses are proportional to the electrical power put into the motor, which is given by P = I*I*R.
The way the manufacturers actually achieve the different kv ratings is by changing the number of turns in the motor. Motor 1 has twice the number of turns as motor 2. Obviously, this requires wires that are twice as long, but to fit the extra length they have to use wire that is exactly half the cross sectional area. Since the resistance of a wire is proportional to Length/Area, as long as the wires are made of the same material, the two motors have the exact same electrical resistance.
And since both motors draw the same current per unit torque at the wheels, they consume the same electrical power to move the same mechanical load and therefore identical electrical losses as well.
What does all this mean? To my eye, I would say that for a given line of motors (say the Turnigy SK3), and assuming you gear it properly:
1) Kv has no impact on wheel torque and speed assuming you gear it properly
2) Kv has no impact on electrical efficiency
3) Pick a Kv that stays under 8k rpm at max speed to avoid electrical inefficiencies not covered in this summary
4) Pick a lower Kv whenever possible since your mechanical drivetrain will likely be more efficient at lower speeds
5) Pick a motor whose Kv value works well with the commonly available pulleys/belts to get you the top speed you want.
6) If you're running an unsensored setup, dont pick a motor with a Kv that is too low or you'll have more issues with cogging when launching from a standstill (since our ESCs have trouble sensing the position of a VERY slowly rotating motor). Frankly though, I'm sure this is basically completely negligable.
This doesnt tell you anything about how to compare different lines of motors (i.e. SK3 vs NTM vs EMP motors) because across brands there may be other design (stator diameter, shaft supports etc) or quality differences (bearing quality, copper alloy) that will affect things. To make that decision, we'd really need to empirically measure the motor torque constant since and look for the motor that produces the highest torque per unit amperage. But at least if you've decided on the brand of motor you want to use, this should help you decide what version to buy.
1) 150 kv motor, geared 24:36 with 83mm wheels, 6s LiFePo, theoretical top speed of 19.2 mph
2) 300 kv motor, geared 12:36 with 83mm wheels, 6s LiFePo, theoreitcal top speed of 19.2 mph
Since motor torque constant = 1/Kv, if you do the math you find that:
1) motor torque constant = 0.0637 N*m/A
2) motor torque constant = 0.0318 N*m/A (half of motor 1)
However, your gearing ratio makes up for that, so torque at the wheels is just
1) torque at wheel = 0.0955 N*m/A
2) torque at wheel = 0.0955 N*m/A (the same)
Now, electrical losses are proportional to the electrical power put into the motor, which is given by P = I*I*R.
The way the manufacturers actually achieve the different kv ratings is by changing the number of turns in the motor. Motor 1 has twice the number of turns as motor 2. Obviously, this requires wires that are twice as long, but to fit the extra length they have to use wire that is exactly half the cross sectional area. Since the resistance of a wire is proportional to Length/Area, as long as the wires are made of the same material, the two motors have the exact same electrical resistance.
And since both motors draw the same current per unit torque at the wheels, they consume the same electrical power to move the same mechanical load and therefore identical electrical losses as well.
What does all this mean? To my eye, I would say that for a given line of motors (say the Turnigy SK3), and assuming you gear it properly:
1) Kv has no impact on wheel torque and speed assuming you gear it properly
2) Kv has no impact on electrical efficiency
3) Pick a Kv that stays under 8k rpm at max speed to avoid electrical inefficiencies not covered in this summary
4) Pick a lower Kv whenever possible since your mechanical drivetrain will likely be more efficient at lower speeds
5) Pick a motor whose Kv value works well with the commonly available pulleys/belts to get you the top speed you want.
6) If you're running an unsensored setup, dont pick a motor with a Kv that is too low or you'll have more issues with cogging when launching from a standstill (since our ESCs have trouble sensing the position of a VERY slowly rotating motor). Frankly though, I'm sure this is basically completely negligable.
This doesnt tell you anything about how to compare different lines of motors (i.e. SK3 vs NTM vs EMP motors) because across brands there may be other design (stator diameter, shaft supports etc) or quality differences (bearing quality, copper alloy) that will affect things. To make that decision, we'd really need to empirically measure the motor torque constant since and look for the motor that produces the highest torque per unit amperage. But at least if you've decided on the brand of motor you want to use, this should help you decide what version to buy.
thepronghorn
1 kW
Edit: The first example is also wrong, you need to compensate for lower Kv by increasing voltage, not increasing gear ratio.
I think you got a little confused at the end. Kv is directly proportional to torque constant, so if you have one, then you have the other. Where motor quality comes into play is basically more copper fill equals more power handling in the same size, stronger magnets means lower Kv so you can put in more volts before you exceed the rpm limit, and thinner laminations mean lower eddy current losses due to high rpm's.
If you have two identical motors except that one is wound for 50Kv and the other 100Kv, they both have the same torque, rpm, and power capabilities, but the 50Kv motor will require twice the voltage and half the amps for the same output as the 100Kv motor.
In order to get the max power out of your motor, you want to gear as low as possible with the largest reasonable wheel pulley and smallest reasonable motor pulley, then up the motor voltage and/or Kv until you reach your desired top speed. Most of us are hitting the edge of our motor's torque capabilities while staying well under the rpm's where core losses start to come into play. By gearing as low as possible, we can maximize our thrust for a given motor torque and then use voltage and/or higher Kv's to get the speed we want.
Motors are dominated by copper losses (energy lost as heat due to winding resistance) at low rpm's and iron losses (energy lost as heat due to hysteresis and eddy currents in the stator) at high rpm's. Max power and efficiency for a motor is achieved when these losses are balanced. If you spin your motor slower than it was designed, you lose out on some of its available power.
I think you got a little confused at the end. Kv is directly proportional to torque constant, so if you have one, then you have the other. Where motor quality comes into play is basically more copper fill equals more power handling in the same size, stronger magnets means lower Kv so you can put in more volts before you exceed the rpm limit, and thinner laminations mean lower eddy current losses due to high rpm's.
If you have two identical motors except that one is wound for 50Kv and the other 100Kv, they both have the same torque, rpm, and power capabilities, but the 50Kv motor will require twice the voltage and half the amps for the same output as the 100Kv motor.
In order to get the max power out of your motor, you want to gear as low as possible with the largest reasonable wheel pulley and smallest reasonable motor pulley, then up the motor voltage and/or Kv until you reach your desired top speed. Most of us are hitting the edge of our motor's torque capabilities while staying well under the rpm's where core losses start to come into play. By gearing as low as possible, we can maximize our thrust for a given motor torque and then use voltage and/or higher Kv's to get the speed we want.
Motors are dominated by copper losses (energy lost as heat due to winding resistance) at low rpm's and iron losses (energy lost as heat due to hysteresis and eddy currents in the stator) at high rpm's. Max power and efficiency for a motor is achieved when these losses are balanced. If you spin your motor slower than it was designed, you lose out on some of its available power.
ProxRB said:
You got this wrong. If motor 1 has twice the number of turns (= twice as long wire) with half the copper thickness, it has four times higher electrical resistance. Therefore, it has four times more resistive losses at a given current. Therefore motor2 with 12:36 gearing is a MUCH better option if you stay below too high rpm.
torqueboards
1 MW
You guys so smart 
thepronghorn
1 kW
vedder said:
You're right I totally missed that. I didn't realize he compensated for lower Kv by decreasing the gear ratio. I assumed he had doubled the voltage and left the gearing the same. Since he instead geared higher, he did in fact make motor1 4x worse than motor2.
Ah, you're right, I messed up my math, the resistance for the low kv motor is 4x higher, not equal to that of the high kv motor. And since they both draw the same current per unit torque at the wheels, you want to use the highest kv motor practical.
ProxRB said:
In your example it is the highest kv motor, but the general conclusion is that you should have as much gearing as possible as long you stay below too high RPM on the motor. If you rewrite the equations, you always get the same amount of losses per torque unit for a given gearing, regardless of the motor kv if it is the same kind of motor. That is why KV does not matter for _motor_ losses. The kv value should be chosen to fit the motor to the electrical system, where the motor kv together with the battery voltage, wires and power stage makes a lot of difference.
By the way, comparing different motors is fairly easy. Take the winding resistance * kv * kv. This value should be as low as possible. Measuring the winding resistance (in case the manufacturer hasn't specified it or if the specification is wrong) can be done by using a lab power supply and run, say, 5A through two windings and measuring the voltage drop across them - then divide this voltage drop with the current.
This study is very useful if you are designing a motor and a drive system from scratch. We are lucky right now to have a much better selection of motor sizes, shapes, and Kv's, compared to what was available 5 years ago.
On a power board, I can see the size and selection of pulleys/sprockets being an issue. Larger diameter tires can help some, but selection is still limited. When presented with a choice, always defer to the choice that results in a higher magnet speed in the motor (up to a limit). I do not foresee a good future for hubmotor wheels in power boards, but I have been wrong before. A hubmotor by definition spins at the RPMs of the wheel. A 3:1 belted reduction (which allows the motor to spin 3 times faster) would be a major benefit to performance. Even 2:1 is better than nothing.
When presented between a choice of a higher voltage + lower amps, or lower voltage + higher amps, sometimes the problem is the amount of battery volume that can be attached to the vehicle. More volts means more cells in series. If you cut down on the Ah size of each cell to compensate (if battery pack volume remains constant), then it can have a big impact on range. So, as long as the board can hold enough cells to get the voltage and Ah you want/need...always lean towards more voltage (more amps to get the same power is possible, but creates MUCH more heat).
On a power board, I can see the size and selection of pulleys/sprockets being an issue. Larger diameter tires can help some, but selection is still limited. When presented with a choice, always defer to the choice that results in a higher magnet speed in the motor (up to a limit). I do not foresee a good future for hubmotor wheels in power boards, but I have been wrong before. A hubmotor by definition spins at the RPMs of the wheel. A 3:1 belted reduction (which allows the motor to spin 3 times faster) would be a major benefit to performance. Even 2:1 is better than nothing.
When presented between a choice of a higher voltage + lower amps, or lower voltage + higher amps, sometimes the problem is the amount of battery volume that can be attached to the vehicle. More volts means more cells in series. If you cut down on the Ah size of each cell to compensate (if battery pack volume remains constant), then it can have a big impact on range. So, as long as the board can hold enough cells to get the voltage and Ah you want/need...always lean towards more voltage (more amps to get the same power is possible, but creates MUCH more heat).
You are losing me a bit in the calculations :lol: Please clarify something for me :
-From Vedder's quote both motors have the same power with different V/A, I thought that for 2 identical motor one low wound and one high wound, the power output remained the same (V x A = W) but the RPM output was the variable. Where did the W go?
-From Vedder's quote both motors have the same power with different V/A, I thought that for 2 identical motor one low wound and one high wound, the power output remained the same (V x A = W) but the RPM output was the variable. Where did the W go?
