[Elvin]

How to clcuate the required torque for a car to climb hill using electric moter? Car weight is 190kg, speed is around 25-35km/h. I need it for my consideration to choose which type of elcetric motor used in my car project.

[jcaron2]

What's the approximate diameter of the wheels? Larger diameter wills require more torque but lower RPMs for a given speed. Also, are there any gears, sprockets/chains, belt/pulley drives between the motor and the wheels, or is it a direct connection? If not direct, the driving mechanism will affect the required torque.

[ebaines]

Couple of other points to consider:

1. How steep is the hill? For practical purposes can we assume no greater than a 15% grade?

2. How much weight will the car carry? You say the car's mass is 190 Kg, but what about the driver? Are there passengers? Cargo?

3. I think you'll find that a surprising small motor is all that's required to maintain a constant speed up a hill. More important to the calculation of motor size is how quickly do you want it to accelerate up to to 35 Km/h? And will that acceleration occur on the hill?

[ebaines]

Some (very) rough calculations:

The force that needs to be applied between wheel and ground to (a) accelerate to 35 Km/h (10 m/s) in 10 seconds while (b) traveling up a 15% grade, and (c) to overcome wind resistance is calculated as follows:



Note that for the wind resistance calculation I have assumed that the car has a frontal area of 1 m^2 and a drag coefficient of 0.7.

Now add it up:



If the drive wheels are 0.7 m in diameter, that means you need about 158 N-m of torque to be applied to the wheel by the drive train. Typical drive trains are perhaps 75% efficient, so that means the motor must produce about 210 N-m of torque.

The power required is:



Again - this is a very rough calculation.

[jcaron2]

Just a technicality, but it looks like you're saying 15% grade is 15% of the angle of a 100% grade (pi/4). That's different than my definition. Isn't 15% grade another way of saying 15% slope (i.e. rise/run = .15; i.e. theta = atan(.15); i.e. sin(theta)=.15/sqrt(1^2 + .15^2))?

[ebaines]

Just a technicality, but it looks like you're saying 15% grade is 15% of the angle of a 100% grade (pi/4). That's different than my definition. Isn't 15% grade another way of saying 15% slope (i.e. rise/run = .15; i.e. theta = atan(.15)

Yep, you're right. A 15% grade = slope of 0.15 = arctan(0.15) = 8.5 degrees. I had it as 0.15 x 45 degrees = 6.75 degrees. So yes, a 15% grade is steeper than what I used. Which means a bit more power is needed.

By the way - I was on a road last week with a 16% grade, and it was really steep!

[jcaron2]

Coincidentally I was in San Francisco last week, and I walked up Filbert Street. According to Wikipedia, it's one of the steepest navigable streets in the Western Hemisphere, with a slope of 31.5%. I have to say, I was pretty tired by the time I got to the top and turned around.

[Elvin]

Details about the car.
The desired acceleration is 1.2 to 1.5m/s2.

The uphill is around 2.83 degree or can be say as 5% inclined hill.

The car is 120kg and the driver is 70kg. So the total weight is 190kg.

The overall effective diameter of the wheel is 597mm.

[ebaines]

Elvin - use this data with equations I gave earlier and see what you get for the required torque. Post back with your answer and we'll check it for you.

[jcaron2]

One other thing, EB: In your calculation for required power, did you miss a factor of 2 in the value of omega (radial velocity)? I think it should be .

On the other hand, presumably the motor doesn't require this same amount of torque when it's already moving at full speed (~10 m/s), since no further acceleration is required at that point. It only has to overcome wind, friction, and gravity. So maybe 4.5kW is still pretty reasonable.

Either way, hopefully you've given then OP enough for a decent ballpark estimate.

[ebaines]

I did make a mistake - actually two, but not what you suggest. Despite what I wrote, what I did was multiply the torque value of 158 N-m by omega of 28.57/s to get 4500 watts. What I should have done is multiply 210 Nm by 28.57/s, which yields 6000 watts.

Of course the OP's data is different from what I assumed - he needs slighter quicker acceleration, but on a more gentle hill and using a smaller diameter wheel/tire. His torque requirement turns out to be a bit less than what I calculated.

[Elvin]

Force acceleration = 285N
Force gradient = 92N
Force wind = 40N
Force total = 417N
r = 0.2985 m
V=rw where w=v/r=9.72/0.2985=32.56rad/s
T = 124.5 Nm
Power required,P = Tw = 4052.9W

But what power train system should be used if the motor used is 24V 800W to run the vehicle?

[jcaron2]

Elvin, do you have the motor's specification for max RPMs? Or, better yet, a plot of the torque curve? As I'm sure you know, both the power draw and the torque change as a function of rotational speed. The torque starts at its maximum when the motor is not spinning, then decreases linearly as the motor's speed goes up. Meanwhile, the power delivered by the motor starts at zero (even though the torque is maximum at zero RPMs, no power is actually being generated unless the motor turns). As rotational speed increases, the power increases rapidly at first, following a parabolic trajectory, until it reaches its maximum at 1/2 the max RPMs (the max being the speed at which the motor's torque falls to zero). Beyond 1/2 speed, the power output drops again until it gets back to zero at max speed. There's a picture of a typical torque curve below, which I got from this site, which explains this a lot better than I just did. :)



Anyway, once we know the max rated RPMs of the motor, we can figure out what gear ratio is required to minimize the time required to reach 25-35 km/h.