Optimal gear ratio for electric engine

15416 Views 16 Replies 8 Participants Last post by brian_, May 2, 2018

electric_driver

Discussion Starter · Feb 4, 2017 (Edited)

I wanted to ask what the gear ratio should be of my electric vehicle to allow it to have the fastest speed possible, while not blowing up my engine.

The vehicle I'm building has a Servovision 10 kW electric motor. I'll be running it at 48 volts. Further details of the motor are at servovision.

Clear from the table is the fact that the engine would deliver 3500 to 4000 RPM.
The diameter of my wheels is 24 cm (it's a kart). The drive shaft is 4 cm in diameter.

Is there any formula to calculate the optimal gear ratio ? And am I missing one or more variables to calculate this ? If so, what ?
Also, does resistance increase if the terrain is sloped (as I assume), and if so, is this calculated in in the formula ?

What I found so far is the following pages:
http://simplemotor.com/calculations/
http://www.bmikarts.com/Gear-Ratio-Speed-Calculator-MPH_ep_84-1.html
http://www.hotrod.com/articles/ctrp-0007-maximum-gear-ratio-equation/
http://www.instructables.com/id/Understanding-Motor-and-Gearbox-Design/

None of these however answers my question, although they do contain formula's. The instructables post also mentions there's something like a "Stall Torque" and a "Stall current", but I doubt any motor manufacturer will provide any such specifics, so I guess I best just calculate it out using other data as the basis.

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dcb

· #2 · Feb 4, 2017

in order to pick the best gear ratio for top speed, you have to know the power vs mph for your vehicle. This is a function of rolling resistance, CDA (coefficient of drag * frontal area), weight, slope. weight is straight forward. Rolling resistance can be done at very low speed (measure force @ 1mph), CDA can be done with a "coastdown" test (google it), and slope is a sine function of weight.

Once you can figure out power required vs speed, you figure out where that graph crosses your motor output power. Cross reference to the motor torque/rpm graph/data to figure out the motor rpm for that power, compare it to your wheel rpm at the computed top speed, and that is your ratio.

Typically though something that sized is probably sprocket-ed, and sprockets are cheap, so just experiment.

piotrsko

· #3 · Feb 4, 2017

10 k motor is 13.4 hp. not a lot to play with there.At 48 volts that will be 208.3 amps. Without looking at motor specs, I suspect you'll be voltage limited based on experience.

Now you need to find out how much torque /hp you are making at say 3500 motor rpm from the published motor specifications. You should notice the curve is kind of flat, but generally you want the peak power rpm at what speed you want to run. Motor rpm at peak power times gear ratio equals tire rpm. Solve for ratio.

electric_driver

Discussion Starter · #4 · Feb 4, 2017 (Edited)

I think I found the answer:
the data needed are the nominal and stall torque (when the vehicle is being driven, so not taken when not under load). The ratio between these is probably also the ratio the engine can take (as the engine will about get it a same amount of times harder then the ratio used).

For simple electric DC motors, that ratio is about 1:5 (see http://www.me.umn.edu/courses/me2011/arduino/technotes/dcmotors/motor-tutorial/ ), and it is also about the same as what most people use in their karts (see http://www.diygokarts.com/vb/showthread.php?t=25255 )

For more powerful electric engines like mine (10 kW), the ratio is probably much higher. There's no data on it (since I have not data when being driven -under load-), but when spinning without load, the nominal vs stall torque is about 1:51 (29000/565, see http://www.servovision.com/hub motor/Motor 10 KW/6-26-2013 2-45-09 PM.jpg ).

As for the torque the engine provides: that's about 14,45 Nm @ 3618 rpm
Calculation:
engine provides 5470 watt-hours @ 3618 rpm
Pout = τ * ω
ω = rpm * 2π / 60
so ω = 3618 * 0,1046
ω = 378,44 rad/s
so τ = Pout / 378,44
so τ = 5470 / 378,44
so τ = 14,45 Nm

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dcb

· #5 · Feb 4, 2017

ok, You aren't wrong, 14.4nm @ 3618rpm = 5.47kw.
(hint, there are also calculators for that http://www.wentec.com/unipower/calculators/power_torque.asp )

but it doesn't answer your question. top speed is a function of power. The motor makes 5.47kw at 3618rpm, you still need to figure out how fast to turn the wheels.

based on other karts, my swag is that you can expect maybe 80kph for 5.47kw, assuming you set up the gearing so the wheels are turning 80kph when the motor is at 3618rpm. Which depends on wheel circumference.

https://www.google.com/url?sa=t&rct...SvAJ8ERWh4DQQ7HzQvgbnQ&bvm=bv.146094739,d.eWE

electric_driver

Discussion Starter · #6 · Feb 5, 2017 (Edited)

Hi dcb,

I don't target to reach 80 kph, rather I want the kart to go as fast/powerful as it is capable of going (so highest torque & highest speed, and engine running at say 75% of its full capability). In addition, the data available is for the situation in which the engine isn't mounted to the kart (kart isn't driving). If its connected to the kart, the speed (rpm) will be lower, and the torque (Nm) will be higher. These figures can be considerably different. If you look at http://www.me.umn.edu/courses/me2011/arduino/technotes/dcmotors/motor-tutorial/, you'll see that the difference between the nominal torque and stall torque is 1200% .

The only thing that's useful to calculate from the "14,45 Nm @ 3618 rpm" is that it's only 5,47 kw so 1/1,8 of the full power (10 kw) -when the engine is running free-. So, for this situation, it could easily handle 1,8x more resistance, or hence a gear ratio of 1:1,8
But as said, that's just for when the engine is running freely (not in the kart), and for giving the engine no stress/resistance at all. My guess is that it could handle up to a ratio of 1:51 (see previous posts), at least in an ideal situation (flat terrain, little or no road resistance, ...). So, it's probably going to be less than that, but more than 1:5 (which is the ratio that is normal in electric karts).

The only option I see to make the kart run more efficient would be to put in a gear box, which has say 10 gears, allowing it to shift up from 1:5 to 1:51, and just try the gears out (or use some sort of instrument that can calculate the above and store the data -not sure whether such a tool exists-). And even then, if a gear is found that still wouldn't be ideal since it would only run perfect on flat terrain.

Another possibilility is to use such a calculation tool, and have a 1:5 to 1:51 gearbox, and have the system change gears depending on the calculations it makes. Again, this probably doesn't exist yet, and I doubt electric cars today hence run very efficient.

As for your speed calculation, that isn't possible with the engine speed of the free-running engine data. Here's the silly high speed it gave:
vehicle speed per minute = engine rpm * gear ratio * 3,14 * wheel diameter
@ 3500 engine rpm and no gears, vehicle speed would be 3500 * 1 * 3,14 * 24 cm
@ 3500 engine rpm and no gears, vehicle speed per minute would be 263760 cm
or hence 2,63760 km per minute or 158,25 km/h;
with a 1:5 gearing, it would be 791,25 km/h

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dcb

· #7 · Feb 5, 2017

You are not understanding my point, and you have some misconceptions as well.

Do you understand my comments about drag (CDA) and rolling resistance and weight and slope. Do you understand that those are the main determining factors in determining the POWER required to maintain a given speed? For a given power (5kw, 10kw, you aren't saying what power level you are using for top speed gearing), and a given CDA/RR/weight/slope there is a 1:1 correspondence with a given top speed, but you still have to do "impedance matching" by getting the gearing right. You have not demonstrated an understanding of this fundamental relationship.

nobody is using 10 speeds, mostly 1 or 2. Your thinking is off there. You can get away with a smaller motor with a multi-speed transmission, but the trade-off is that smaller motors are less efficient (and there are more losses in a typical multi-speed transmission).

electric_driver

Discussion Starter · #8 · Feb 5, 2017

dcb said:
Do you understand my comments about drag (CDA) and rolling resistance and weight and slope. Do you understand that those are the main determining factors in determining the POWER required to maintain a given speed?

Sure I do. But I don't know any of these data, the kart hasn't even been completed yet. There's a tool for calculating this at http://ecomodder.com/forum/tool-aero-rolling-resistance.php but I can't fill in any data since I don't have that data. I just wanted to figure it out some other way -mathematically-, and didn't care it would not be extremely precise. Rather, I just wanted to get some idea of what gearing I'd need to use, before starting to experiment with gears in real life.

dcb

· #9 · Feb 5, 2017

Are you trying to figure out how hard you can push that motor?

That is more of a black art. Basically though you are limited by heat buildup and the laws of thermodynamics, and validated through testing.

http://www.servovision.com/hub motor/Motor 10 KW/BLDC Motor 48VCdata.pdf

it says the rated power is 6.5kw @ 3500 rpm, which is 17nm. and 150 amps (48v). It should be able to do that all day.

torque is related to current, so that little motor makes ~ 0.113 nm per amp, regardless of the rpm (for the most part). No load it makes ~82 rpm/volt. Amps = heat.

You can see that efficiency drops off a cliff at very low load, but from 5nm up it stays above 80% (so don't drive it at low load).

There is no picture of the motor proper, so I don't know if it is internally cooled with a fan, or if you use an external radiator, but an internal fan is rpm sensitive, so don't expect to run 150 amps continuous below 3500 rpm.

You can usually throw more volts at a motor (>48v) and get it to make the same torque at a higher rpm (thus more power). The time limitations are determined by the current (heat) and cooling capacitiy.

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electric_driver

Discussion Starter · #10 · Feb 6, 2017 (Edited)

dcb said:
Are you trying to figure out how hard you can push that motor?

Yup. That's exactly what I like to do.

My first idea was that there was going to be some formula to calculate this out. I think that the nominal and stall torque would probably be an indicator to work out the gear ratio's, and I still think that's the case. I even think that the nominal and stall torque ratio difference of electric engines that are just running free could be used to figure out the gear ratio of the completed vehicle, even if it is being driven. This, because the ratio's between those 2 ratio's (non-driven and driven ratio of nominal and stall torques) would probably remain the same. The only variables that would need to be integrated in a formula to figure it out could be
* weight of the vehicle
* speed at which it is being driven (because the higher the speed, the more air resistance, and the air resistance tends to be a significant value -see http://www.lowtechmagazine.com/2008/09/speed-energy.html -
I think that for this latter, you need:
1 kwh /50 kmph for a vehicle driven at 45 kmph
and
1 kwh /15 kmph for a vehicle driven at 90 kmph
). Sadly, such a formula doesn't yet exist since no one has ever collected data to calculate such a formula.

But, since you said that "

dcb said:
you are limited by heat buildup and the laws of thermodynamics, and validated through testing.

", there's another way !

Some quick googling revealed that 2 units are used for heat: the joule and the watt.
So, it stands to reason that my engine is able to withstand "10 kw of heat" continuously.
So, a 1:1,8 gear could definitely be used (6,5 kw * 1,8 = 10 kw), but as most people use a 1:5 on their electric kart, and as nominal vs stall torque indicates that up to 1:51 can be used, it's going to be able to handle much more, continuously.

To calculate this better, a (kitchen) thermometer (measuring range 10-250°C) could probably be used, when driving the vehicle, and using a known gear ratio. What I don't know however, is where and how I best measure it (i.e. what heated part of the engine do I need to measure) and would a kitchen thermometer do, or do I need say an infrared thermometer ?
Another thing I don't know is how hot the engine can safely get (in °C, not in watt). I'm assuming I could just measure the operating temperature (°C) when operating at 10 kW, but it could probably take more heat than that. According to a quick search (https://www.rcgroups.com/forums/showthread.php?1329288-Normal-brushless-motor-temperature-range ) , I assume I can get to between 71°C-80°C with my engine (which is a BLDC aka brushless direct current engine).

Regarding the cooling: there are 2 versions of the engine; one being cooled with a fan, the other being oil-cooled. I got the version that is oil-cooled.

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dcb

· #11 · Feb 6, 2017 (Edited)

"So, it stands to reason that my engine is able to withstand "10 kw of heat" continuously."

Good luck with that... (p.s. it is really (1-efficiency) * power rating that it needs to withstand)

edit 2, here is a thermocouple installation in a stator fyi (the red lead)

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Duncan

· #12 · Feb 6, 2017

Hi Electric

Your 10Kw motor - do you know the duty cycle?

Motors are rated by duty cycle - it could be 10%
Which means 6 min at power then 54 min cooling
My 10kw motor is 100% rated - it weighs 102Kg

Joules and watts
1 watt is 1 Joule per second
So a 100 watts is 100 joules per second or 60,000 Joules in a minute

dh1

· #13 · Feb 6, 2017

I think your at the point where you need some practical testing, get the cart on the road and take some readings with an amp meter and spedo.
Are you using chain drive? Be easy to change gear ratios so you can load the motor as you want.
As for where to start take your best guess using the info you can find.

zaraf

· #14 · May 2, 2018

dcb said:
in order to pick the best gear ratio for top speed, you have to know the power vs mph for your vehicle. This is a function of rolling resistance, CDA (coefficient of drag * frontal area), weight, slope. weight is straight forward. Rolling resistance can be done at very low speed (measure force @ 1mph), CDA can be done with a "coastdown" test (google it), and slope is a sine function of weight.

Once you can figure out power required vs speed, you figure out where that graph crosses your motor output power. Cross reference to the motor torque/rpm graph/data to figure out the motor rpm for that power, compare it to your wheel rpm at the computed top speed, and that is your ratio.

Typically though something that sized is probably sprocket-ed, and sprockets are cheap, so just experiment.

Dear dcb,

I am undergoing the same Problem for determining the gear ratios. I have calculated the theoretical Power vs Velocity/RPM based on the losses incurred by the vehicle. Now what I dont understand is that this Power curve gives the power to sustain a particular speed. Where is the accelerating force and power. Do we need to take it into account for the gear ratios?
Additionally I have calculated the Power to accelerate from 0-13.8ms (50KMH) in 10 sec using KE gained/10sec = approx 10KW. Now my question is that is this 10KW needed on top of the Power vs Velocity/RPM calculated before, like simply adding? How can I understand this 10KW for acceleration from the Power vs RPM curves of my Motor?

Emyr

· #15 · May 2, 2018

zaraf said:
Now what I dont understand is that this Power curve gives the power to sustain a particular speed. Where is the accelerating force and power. Do we need to take it into account for the gear ratios?
Additionally I have calculated the Power to accelerate from 0-13.8ms (50KMH) in 10 sec using KE gained/10sec = approx 10KW. Now my question is that is this 10KW needed on top of the Power vs Velocity/RPM calculated before, like simply adding? How can I understand this 10KW for acceleration from the Power vs RPM curves of my Motor?

The acceleration calculations should already be accounting for losses at a given speed.

At constant speed, the driving force is equal to the total losses (aero drag, rolling resistance, drivetrain losses).

When accelerating, the rate of acceleration correlates to the driving force minus the total losses at the current speed.

For a constant power output, acceleration will trend towards 0 as the vehicle approaches theoretical max speed, unless the motor is rev-limited.

zaraf

· #16 · May 2, 2018

Emyr said:
The acceleration calculations should already be accounting for losses at a given speed.

At constant speed, the driving force is equal to the total losses (aero drag, rolling resistance, drivetrain losses).

When accelerating, the rate of acceleration correlates to the driving force minus the total losses at the current speed.

For a constant power output, acceleration will trend towards 0 as the vehicle approaches theoretical max speed, unless the motor is rev-limited.

Hello,

In my Power vs RPM curve, Power is calculated with Power = Force x Velocity. Now the Force contains all the losses but no accelerating force. Thats my question, how to incorporate the Accelerating Power into the curve. And secondly does it Impact the gear Ratio.

brian_

· #17 · May 2, 2018

Emyr said:
The acceleration calculations should already be accounting for losses at a given speed.

No, zaraf has not calculated what acceleration would result from applying power in excess of that used to overcome drag; he is calculating the power to overcome drag and the power to accelerate (increase kinetic energy) separately, which makes perfect sense:

zaraf said:
I have calculated the theoretical Power vs Velocity/RPM based on the losses incurred by the vehicle.
...
Additionally I have calculated the Power to accelerate from 0-13.8ms (50KMH) in 10 sec using KE gained/10sec = approx 10KW.

With these two power demands established, zaraf's question is then:

zaraf said:
Now my question is that is this 10KW needed on top of the Power vs Velocity/RPM calculated before, like simply adding?

Yes, if you want to maintain a constant acceleration.

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