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Discussion Starter #1
Hello from a land of ice, snow, and mountains. I'm looking to build a small 2 person electric car with 1 motor for each wheel direct drive. I know if I build a gasoline car that delivers 1 ft lb of peak torque to the tire surface per 5 lbs of weight I will have a pretty snappy car. I found this place trying to find out if the same numbers work with direct drive electric cars or if I need different ones.
 

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I'm looking to build a small 2 person electric car with 1 motor for each wheel direct drive. I know if I build a gasoline car that delivers 1 ft lb of peak torque to the tire surface per 5 lbs of weight I will have a pretty snappy car.
That doesn't make any sense, unless you assume a specific overall tire diameter. It is the force at the tire contact with the road, not the torque on the wheel, which matters. What tire diameter are you assuming?

For instance, if you put 1 lb-ft of torque into each of two wheels with a diameter of two feet, you get 1 pound of force at the tire contact (2 pounds of force in total) to drive the vehicle; if you put the same 1 lb-ft of torque into each of two wheels with a diameter of only one foot, you get 2 pounds per tire (4 pounds in total) to drive the vehicle. Of course with the smaller tire diameter the wheels turn twice as fast for the same road speed.
 

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Well, lets say you V8 has 300 ft lbs of peak torque. You run that through a 3:1 diff and you now have 900 lbs of torque. For a 28 inch tire, you would then need to multiply that by 12/14. That gives you 771 lbs of force at the tread. If the vehicle weighs 4,000 lbs, they you have about 5.2 pounds of vehicle per pound of force at the tire tread.
 

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Well, lets say you V8 has 300 ft lbs of peak torque. You run that through a 3:1 diff and you now have 900 lbs of torque.
That's 900 lb-ft (or ft lbs) of torque... you just missed the "foot" part.

For a 28 inch tire, you would then need to multiply that by 12/14. That gives you 771 lbs of force at the tread. If the vehicle weighs 4,000 lbs, they you have about 5.2 pounds of vehicle per pound of force at the tire tread.
So you do understand that torque by itself is useless, and you need to divide the torque by the torque arm (which is the tire outside radius, 14 inches in your example). That suggests that when you said this...
I know if I build a gasoline car that delivers 1 ft lb of peak torque to the tire surface per 5 lbs of weight I will have a pretty snappy car.
... you really meant
I know if I build a gasoline car that delivers 1 lb of peak force to the tire surface per 5 lbs of weight I will have a pretty snappy car.
Okay, minor miscommunication cleared up :)


That means 0.2 g of acceleration. If you could maintain that from a standstill to 60 miles per hour, and there was no rolling drag and no aerodynamic drag, that would mean a 14 second 0-60 time. That's not very quick - almost every available production car is quicker - and actual performance will be much slower, because
  • you won't have peak motor torque at all speeds, and
  • much of the force will be used to overcome drag, not accelerate the car
An actual 4000 pound car with a V8 engine capable of 300 lb-ft of torque would be much quicker, because it wouldn't be driven in one gear all of the time; it would go through a few gears with the engine torque multiplied by the transmission (a different multiplication for each gear ratio). The same would be true of an EV with a multi-speed transmission, but if you connect the motor directly to the final drive you only get the final drive torque multiplication (3:1 in your example).

I suppose this is sufficient performance to drive on public roads, but I wouldn't call it "pretty snappy". I've never owned a car that slow (and my first car had only a 78 horsepower engine), although my ten-ton motorhome might be similar.
 

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I'm looking to build a small 2 person electric car with 1 motor for each wheel direct drive.
When you say "direct drive", what do you mean? It should mean that the motor is connected directly to the hub, with a shaft and joints but no gearing; that's generally a terrible idea. What most people mean by "direct drive" in this forum is that the motor is connected directly to the final drive unit, with a shaft and joints but no additional gearing; there's nothing very direct about that. It often seems to be intended to mean a connection with a single gear reduction ratio, so it should be called "fixed ratio" or "single ratio"; nearly all production EVs have a single-ratio drive system.

Every production EV uses at least two stages of spur gears or a planetary gear set, with an overall drive ratio (motor speed to wheel speed) ranging from a low of less than 4:1 (the oddball Chevrolet Spark) to a high of more than 12:1 (Tesla Model S/X). Some conversions have only a single stage of reduction, typically using the ring-and-pinion gears of a final drive unit, which suits their low-speed motors.
 

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Discussion Starter #6
It would have sprockets to serve as a differential. By direct drive I mean no transmission. An yes, the particular car I am using as an example had a 3:1 first gear ratio, so it was actually getting up to 2700 ft lbs of torque to the axle.

I'm seeing people getting away with 100 lbs per hp ratio in this forum with electric motors hooked to transmission. I understand this is due to the different torque curve of the electric motors, but many people have also said if you don't use a transmission, you need a bigger motor. Thus my question of what ratios people look for when going this route?
 

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It would have sprockets to serve as a differential.
With a motor per wheel, there is no differential, so it looks like you mean a chain and sprockets to provide speed reduction (torque multiplication).

By direct drive I mean no transmission.
Okay, so fixed-ratio.

... many people have also said if you don't use a transmission, you need a bigger motor. Thus my question of what ratios people look for when going this route?
Traditional brushed DC motors have a narrow powerband, so with a fixed drive ratio they are usually far off their peak power, and so need to be relatively large compared to a motor driving through a multi-speed transmission.

Most people who do a conversion without keeping the original transmission have only a common final drive unit for reduction; it usually doesn't have a high enough reduction ratio so the motor isn't used effectively if it is capable of high speed. The motor size ends up being determined by required torque to the wheels, which in combination with the gear reduction determines required motor torque.

The ratio determination with a fixed-ratio system like this can be quite simple: the ratio is a high (as much reduction) as possible while keeping the motor speed safe and effective (within the motor's maximum speed, and at a speed where the motor can produce close to full power) at the planned maximum speed of the car. For instance, an early Nissan Leaf motor produces full power to about 9800 rpm, and the car has a 93 mph top speed and 24" tall tires... all of which lead to a ratio of almost 8:1.
 
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