# Selecting electric motors once power requirements are calculated

• Automotive

## Main Question or Discussion Point

So for a school project I gotta design an electric car that can meet minimum range and top speed requirements for how heavy it is. I have calculated the amount of power needed(kW) for its heaviest load situation to reach the min speed and range, along with the necessary energy(kWh) required for this. Now when it comes to selecting a capable motor for this what kind of factor of safety should be used in terms of power?

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If you've factored in wind and rolling resistance, you shouldn't need much of a safety factor... wind resistance is the harder to calculate since it's hard to know the frontal area of the car and the coefficient of resistance without wind tunnel testing, and it is increases with the square of velocity... If you're not going very fast, I'd go with a 20% safety factor and select then next larger motor.. IE.. you calculate your requirement as 10kw, add 20% = 12kw, next readily available size is probably 15kw.. that would give a 50% safety factor, which I would think is plenty... Of course the motor controller has to be matched to that power rating too.

Nidum
Gold Member
If you've factored in wind and rolling resistance, you shouldn't need much of a safety factor... wind resistance is the harder to calculate since it's hard to know the frontal area of the car and the coefficient of resistance without wind tunnel testing, and it is increases with the square of velocity... If you're not going very fast, I'd go with a 20% safety factor and select then next larger motor.. IE.. you calculate your requirement as 10kw, add 20% = 12kw, next readily available size is probably 15kw.. that would give a 50% safety factor, which I would think is plenty... Of course the motor controller has to be matched to that power rating too.
Only calculated rolling resistance since minimum top speed needed is only 4 mph. Which the power req. at the wheels to do this is only 6.5 kW and the energy req. for that speed and required range was 10.4 kWh. Thanks for the helpful reply. My next question is how would you determine the gear ratio needed to operate around this speed? A lot of the motors with the power req. i needed have much higher rated top speeds than needed and currently somewhat lost on how to determine necessary gear ratios for the speeds of operation. Thanks

OK, lets say you have a 12" tire, or about 36" circumference, and since your minimum top speed is 4mph, I'll round it up to 5 mph.

5280 feet per mile, or 26400 feet per hour or 8800 tire revolutions per hour, or ~150 RPM.
If your motor speed is 2000 RPM, you need a 2000/150 reduction, or 13.3:1.
Most differentials have a reduction ratio of about 3.5:1 (though you can get others), so you'd need a primary reduction of 13.3/3.5 or 3.8:1

How heavy do you figure this vehicle will be?, what is the maximum steepness of the slope it must climb? 6.5kW @ 4mph is a lot of pulling power
6.5kW = 8 3/4 hp
4mph = 440ft/min or 7.3 ft/s
8 3/4 Hp = 4812 ft*lb/s
4812/7.3 = 650 lbs of pulling power @ 100% efficiency and 4mph.

Is this a tractor?

CWatters
Homework Helper
Gold Member
6.5kW is about 8HP. I might be wrong but that seems a lot for 4mph? Is it an electric car or more like a go-cart?

On this web site it details a 10HP electric bike that can do 65mph (see #4).
https://www.electricbike.com/10-fastest-electric-bikes/

I think the Wright Brothers calculated that around 8HP was enough for them to fly although their motor actually produced nearer 12HP if I remember correctly.

OK, lets say you have a 12" tire, or about 36" circumference, and since your minimum top speed is 4mph, I'll round it up to 5 mph.

5280 feet per mile, or 26400 feet per hour or 8800 tire revolutions per hour, or ~150 RPM.
If your motor speed is 2000 RPM, you need a 2000/150 reduction, or 13.3:1.
Most differentials have a reduction ratio of about 3.5:1 (though you can get others), so you'd need a primary reduction of 13.3/3.5 or 3.8:1

How heavy do you figure this vehicle will be?, what is the maximum steepness of the slope it must climb? 6.5kW @ 4mph is a lot of pulling power
6.5kW = 8 3/4 hp
4mph = 440ft/min or 7.3 ft/s
8 3/4 Hp = 4812 ft*lb/s
4812/7.3 = 650 lbs of pulling power @ 100% efficiency and 4mph.

Is this a tractor?
its a low speed transportation/cargo car

Does it need to go up any inclines? I think if you could sacrifice some acceleration you may increase battery life significantly.

Pulling a number out of my arse for the weigh of this thing, I'll switch to metric measure, and say 300kg (660 lb)... and I already figured the pulling power at 650 lbs... pretty much 300kg or 3000N
F = MA, so A = F/M
A = 3000/300 = 10m/s2

This means that it can climb vertical walls given the traction

So even at 650 lbs this thing will accelerate at a rate equivalent to gravity...
So how long does it take to get to top speed?
Vf = Vi + AT
We know the final velocity to be 4 MPH, or 1.8m/s, and acceleration to be 10m/s2

so T = 1.8/10, or .18 seconds to go from standstill to top speed...

if the thing is 10 times heavier (6500 lbs, 3 metric tonnes) we have a 1/10th the acceleration (1m/s2), meaning it would achieve top speed in 1.8 seconds... still not bad!

x
Does it need to go up any inclines? I think if you could sacrifice some acceleration you may increase battery life significantly.

Pulling a number out of my arse for the weigh of this thing, I'll switch to metric measure, and say 300kg (660 lb)... and I already figured the pulling power at 650 lbs... pretty much 300kg or 3000N
F = MA, so A = F/M
A = 3000/300 = 10m/s2

This means that it can climb vertical walls given the traction

So even at 650 lbs this thing will accelerate at a rate equivalent to gravity...
So how long does it take to get to top speed?
Vf = Vi + AT
We know the final velocity to be 4 MPH, or 1.8m/s, and acceleration to be 10m/s2

so T = 1.8/10, or .18 seconds to go from standstill to top speed...

if the thing is 10 times heavier (6500 lbs, 3 metric tonnes) we have a 1/10th the acceleration (1m/s2), meaning it would achieve top speed in 1.8 seconds... still not bad!
Aight so I started from square 1 again and heres all the needed requirements-
-Transport car
-low speed
-minimum range 5 miles
-minimum top speed 5mph
-must be able to climb 30 degree gradient

-So I re did all my calculations and found that the power required for this was 3.96kw, this was calculated by using the force necessary to move the fully loaded weight of 1500 lb up a 30 degree incline at a top speed of 3.6 mph(F.S. = 1.2)

-Next I calculated the rpm of the tire needed for 3.6, which turned out to be about 67.4 rpm with an 18" diameter tire.

-So now I am back to selecting a motor. When selecting motors, would I select a motor based on peak hp numbers?

-Then once I select a motor, my gear reduction/ratio needed would the rpm of the motor divided by the rpm of the tire.

Selecting the motor will depend a little on the duty rating.. how often it will need to be climbing that slope, and how often it's starting and stopping.. For the details of that you will have to look at the data sheet for each model of motor.
I run some 3ph AC motors at above their rated power by running them at a higher frequency... this can mean I can gear them lower (thus requiring less startup amperage) and revving them higher can also improve fan cooling efficiency, though they usually lose some electrical efficiency if you go much above their rated speed...

Yes, you would need to select your motor, and then calculate the reduction ratio from it's max speed.. you should be pretty close and can add a fudge factor one way or another.. more gear ratio for more power or less gear ratio for higher speed...