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So I'm new to this forum. My name is Roel and I recently bought an old 1951 Francis-Barnett motor frame without the engine and I am looking into turning it into electric drive.


There is one question that I am stuck on at the moment and that is motor voltage. Almost everywhere I look I read "more voltage is more speed", which is true. But doesn't required power also come in to play?


For example, according to some calculations and reference builds on the web I came to the conclusion that I need about of 6.2 kW to get my bike to 90km/h. With the circumference of my wheels of 2 meters this comes to 750 wheel rpm. Let's also say I use a 50t rear sprocket.


Now I've been looking at the ME-1004 motor, and it has a nominal rpm of about 3100 (loaded) at it's rated voltage of 48 Volts I assume. 3100/750= 4.13, so if I use a 12t-13t front sprocket, won't this motor give me the required speed? If so, what would be the advantage of getting a motor with a higher voltage rating? Will it be more efficient because it runs at a lower current for the given output power?

I may be overlooking something here since this is all new to all of this, so any input is appreciated.
 

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I think what you are missing is the relationship of RPM and Torque to Horse Power. HP = RPM x T / 5252. You are looking at a 6.2 Kwh motor or roughly 6 to 7 HP at some RPM and Torque value. You have to sacrifice Torque to get higher RPM's.

Example you have two 48 volt motors:

1. Generates 6 HP @ 3200 RPM with 9.8 ft-lbs of torque.
2. Generates 6 Hp @ 10,000 RPM with 3.2 ft-pounds of torque.

So which motor do you use. Well you use the one that generates the most power at the maximum engine RPM required for top speed. So if your target speed is 30 mph. you would need to know the motor RPM, differential ratios, and tire diameters to work that out.

You are correct in thinking higher voltage = higher RPM's as a general rule. HP is just like electrical power watts. Watts and Horse Power are products of Voltage and Current for watts, and RPM and Torque for HP.

I can get say 6000 watts with:

6000 Watts = 1 Volts x 6000 Amps turning a motor at 1 RPM with 31,512 ft-pounds of torque. You could pull a train a few inches per minute

or

6000 watts = 6000 volts x 1 Amp. Turning a motor at 6000 RPM with 5.3 ft-lbs of torque. making your bike go say 30 mph with the right transmission system. That is assuming that is enough power to overcome weight, rolling resistance, and COD.


So you could pull a train with one motor, o rmake your bike go 30 mph with a speed motor, but forget about pulling a train with it.
 

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HP = RPM x T / 5252...
All good, but the torque in that formula must be in pound-feet (or foot-pounds... no need to argue that here), so
Power {in horsepower} = Speed {in RPM} x Torque {in lb-ft} / 5252​
The factor of 5252 is just all of the conversion factors for units of measure piled together, and is specifically for these particular units.


If you would rather use metric system units:
Power = Speed x Torque​
or in SI units
Power {in watts} = Speed {in radians/s} x Torque {in Nm}​
or with a more convenient measure of speed
Power {in watts} = Speed {in rev/min} x Torque {in Nm} x 2xPi/60
Power {in watts} = Speed {in rev/min} x Torque {in Nm} x 0.105​
 

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... what would be the advantage of getting a motor with a higher voltage rating? Will it be more efficient because it runs at a lower current for the given output power?
A motor designed for higher voltage at the same shaft speed and power will use less current (in proportion to the voltage increase), but that doesn't make it any more (or less) efficient.

I may be overlooking something here since this is all new to all of this, so any input is appreciated.
You are actually in an excellent position for someone who is new to this. You appear to have a better understanding of at least this aspect of design (power requirement and gearing) than some people who have actually built a vehicle.

One issue is that the motor will reach peak power at the maximum vehicle speed, which is the way to get the highest possible road speed from the motor, but it will take a very long time to get there, because there is less power available at lower speed.

Also, if 6.2 kW is just enough to keep it moving at 90 km/h, it will never get there because as the bike gets close to that speed there is almost no extra power to accelerate it. You need enough to keep the bike moving, plus some extra to account for headwinds and uphill grades, plus more power to accelerate it.
 

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Also, if 6.2 kW is just enough to keep it moving at 90 km/h, it will never get there because at the bike gets close to that speed there is almost no extra power to accelerate it. You need enough to keep the bike moving, plus some extra to account for headwinds and uphill grades, plus more power to accelerate it.
This is an excellent point, and one often missed by DIY. If you look at any commercial passenger vehicle like a car and how the motor or engine is sized if it takes say 26 hp to make your Honda Civic cruise at 70 mph, you want a motor with 3 to 4 times more peak Horse Power 75 to 100 HP for passing and acceleration.

So if you calculate it will take the full 6 HP of your motor to make the bike go say 60 mph top speed means it is a 30 mph cursing vehicle using 1.5 to 2 HP. Enough left to get up to 40 mph to pass, or accelerate 0 to 30 mph just fine but getting to 50 will take a while, and 60 will only happen down hill.


Another thing you have to look at is you have a fixed Differential Ratio transmission just like a golf cart with Direct Drive. You can get a golf cart to go 60 mph and faster, but not with a brushed DC Motor. You need a high speed motor with flat torque up to 5000 RPM, and flat HP up to 10,000 RPM because even with 22 inch tires with a fixed ratio of 12:1 up to 17:1 requires a very high PPM using Direct Drive. You would have to use an AC Induction Motor or BLDC in a golf cart. A DC motor operates at much lower range of RPM and the peak power bandwidth is in the lower half of RPM range. To get both Torque and Speed is with a Transmission to change differential ratios to get higher wheel rpms. HPEV makes one heck of a 48 to 96 volt golf cart motor called the AC9. 6 HP continuous at 8000 RPM, and 40 HP peak @ 3700 RPM.
 

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Another thing you have to look at is you have a fixed Differential Ratio transmission just like a golf cart with Direct Drive. You can get a golf cart to go 60 mph and faster, but not with a brushed DC Motor. You need a high speed motor with flat torque up to 5000 RPM, and flat HP up to 10,000 RPM because even with 22 inch tires with a fixed ratio of 12:1 up to 17:1 requires a very high PPM using Direct Drive.
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A DC motor operates at much lower range of RPM and the peak power bandwidth is in the lower half of RPM range.
If you want to use a high reduction ratio, you need a high-speed motor, and vice versa.

jaaasgoed, I'm not saying whether or not the ME1004 is a suitable choice, but your gearing calculation is correct:
3100 RPM at the motor, divided by 50/12 by the chain drive, is 744 RPM at the wheel, which is 1488 metres per minute or 89 km/h.
The poor speed range of typical DC motors is an issue, as already discussed.

The broad power band of modern high-voltage AC motors is certainly preferable.
 

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HPEV makes one heck of a 48 to 96 volt golf cart motor called the AC9. 6 HP continuous at 8000 RPM, and 40 HP peak @ 3700 RPM.
That's certainly a more capable motor than an ME-1004... but it weighs 50 pounds (22.7 kg). That seems like too much for the light motorcycle, especially since that's just the motor, without a controller and (most importantly) the battery.
 

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Discussion Starter #8
Well, that's a lot of stuff to take in. Thank you guys for the quick responses.


I don't think the torque will be a problem. The motor will do ~300 Nm on the wheel I think. The motor has a torque of 24 Nm x 4.13 (gear ratio) / 0.31m (r of the wheel) =~ 300 N/m. If I take @brian_'s formula, I think I would only need 80 Nm to maintain the cruising speed?



@Sunking Agreed. The 6.2 kW would be the nominal output power for the given maximum cruise speed. The motor has a maximum power output of 16kW, so there is extra short-term power available to take it up to speed, climb small hills etc.



I've looked into AC motors, but the higher rpm almost always means you need an extra reduction to get the correct wheel speed. I would like to keep the design as simple as possible. This is also the reason I want to go for a 48 Volt setup. Building a 13S (18650 battery pack) is not really exotic in terms of BMS and charging. I'm looking at 13S20P of 2.6 AH cells to give me 52 Ah (~2400 Wh) for the range I'm going for, about 30 KM.


If it turns out this is not feasible, I will have to adjust my expectations for a lower voltage system. This is supposed to be a "cruise around the block" bike. I do not need highway speeds, but it's nice to have enough power/speed if need be.
 

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Discussion Starter #9
Just a test. My last reply message needed approval apparently, but has not yet come through, so just seeing if it was a fluke.
 

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Discussion Starter #10
So let's try again. If I'd only saved my text...


Thank you guys for the quick response, lots of info to take in.

I don't think the motor torque would be a problem. The motor does 24 Nm on the shaft. So this would be be 24*4.13 (gear ratio) / 0.31 (r of the wheel) =~ 319 Nm on the wheel. If I fill in brian's formula with the known parameters, I would only need ~80 Nm at cruising speed?


@Sunking. Agreed. The 6.2 kW is not the maximum motor power. Maximum short term power output is 16 kW, so there should be enough power available for getting up to speed, climbing small hills, etc.


I also looked into AC motors, but the main issue I have is that they need an extra reduction because of the higher RPM. I would like to keep the build as simple as possible. This is also one of the reasons I'm going for a lower voltage, if possible. Making a 13S pack is a lot less exotic then a 20S pack in terms of BMS and charging. This is going to be a "taking it for a spin around the block" build. I'm not looking for highway speeds, but a little extra power when needed would be nice. If this is not possible with the lower voltages, I will have to lower my expectations a little bit, which is not really a problem. That's why it's always nice to have some people with more experience to look and see if you have most of the basics right.


As a last note, this is the calculator I'm currently basing my choices on: http://www.electricbikesimulator.com/calculator.php?language=en&speed=90&range=30&weight=250&voltage=48
 

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Discussion Starter #11
And we try again. I hope these will not become double posts...maybe it's because of the link?


Thank you guys for the quick response, lots of info to take in.

I don't think the motor torque would be a problem. The motor does 24 Nm on the shaft. So this would be be 24*4.13 (gear ratio) / 0.31 (r of the wheel) =~ 319 Nm on the wheel. If I fill in brian's formula with the known parameters, I would only need ~80 Nm at cruising speed?


@Sunking. Agreed. The 6.2 kW is not the maximum motor power. Maximum short term power output is 16 kW, so there should be enough power available for getting up to speed, climbing small hills, etc.


I also looked into AC motors, but the main issue I have is that they need an extra reduction because of the higher RPM. I would like to keep the build as simple as possible. This is also one of the reasons I'm going for a lower voltage, if possible. Making a 13S pack is a lot less exotic then a 20S pack in terms of BMS and charging. This is going to be a "taking it for a spin around the block" build. I'm not looking for highway speeds, but a little extra power when needed would be nice. If this is not possible with the lower voltages, I will have to lower my expectations a little bit, which is not really a problem. That's why it's always nice to have some people with more experience to look and see if you have most of the basics right.
 

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The motor does 24 Nm on the shaft. So this would be be 24*4.13 (gear ratio) / 0.31 (r of the wheel) =~ 319 Nm on the wheel.
24 Nm x 4.13 is 99 Nm (the torque at the wheel)
then
24 Nm x 4.13 / 0.31 m is 319 N (the drive force at the tire)
That's 319 newtons of force, not 319 newton-metres of torque
 
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