# Torque required for light electric tractor



## Brendan B (Apr 30, 2012)

Hi,
I am looking at building a tractor that will be 1200kgs and needs to go up a 35% slope at a working speed of 5km/h. What would be the torque required at the wheel to get it up there?


----------



## PStechPaul (May 1, 2012)

There are some good threads about HP and torque in another forum where I am active:
http://www.mytractorforum.com/showthread.php?t=234238
http://www.mytractorforum.com/showthread.php?t=117929
http://www.mytractorforum.com/showthread.php?t=224315

Based on your information of 1200 kg 35% slope 5 km/h, you can figure it out like this:

The slope is what determines the force required along the axis of the tractor, so disregarding friction and other losses, you need about 33% of the 1200 kg, or 396 kg. If the drive wheel (tire) diameter is 0.5 meters, the radius is 0.25 m, so the torque is 99 kg-m or 971 N-m. The circumference of the tire is 1.57 meters, so 5 km/hr is 3183 rev/hr or 53 RPM, which works out to 5388 watts, or 7.2 HP. I used this on-line calculator: http://www.magtrol.com/support/motorpower_calc.html

I have an electric vehicle power calculator on my website that might be helpful. I made it quite a while ago and it may not be totally accurate, but it might be a good starting point: http://pstech-inc.com/VehiclePower.xls

I'm not used to metric calculations, but this sounds about right.


----------



## Brendan B (Apr 30, 2012)

Thanks Paul.
The calculations have been helpful and shows that I have more than enough power and torque from what I plan to put on the tractor to move it around.
Now the fun begins.


----------

