Well i'm building something on my free time and I want to buy an engine and I would like to know if it is strong enough to carry me uphill. Is there a formula i can use?

This is what is know to me (added an engine i have my eyes on): Engine (hub engine would be inside wheel): Volts - 36v Watts - 361w Torque - 22.83 N.m RPM - 151 n

body: Wheel radius - 15.3 cm Body weight (with batteries, chassis, my weight and motor weight included) - 75 kg

Transmission: Hub engine - not transmission. the engine itself spins and is located at the center of the wheel (i think this means that the differential is 1). environment: expected to go ~35 degrees uphill

At what speed am i expected to go uphill? At what acceleration? How do i calculate this?

Further more, i found lots of engines that only state watt, RPM and volt. is this enough to calculate torque?

thank you very much in advance!


closed as unclear what you're asking by Chetan Bhargava, Daniel Grillo, Leon Heller, Matt Young, Olin Lathrop Jul 6 '14 at 22:20

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    \$\begingroup\$ I'd say this was more physics than electronics... \$\endgroup\$ – Majenko Jul 6 '14 at 18:04
  • 3
    \$\begingroup\$ Ben, please use proper English capitalization. \$\endgroup\$ – Nick Alexeev Jul 6 '14 at 18:32
  • \$\begingroup\$ Any motor can carry you uphill, in theory, if geared down far enough. \$\endgroup\$ – Spehro Pefhany Jul 6 '14 at 21:27
  • \$\begingroup\$ @Nick: Yup. After reading the title and the first line, I had enough of the OP thumbing his nose at us. Downvoting, closing, and moving on. \$\endgroup\$ – Olin Lathrop Jul 6 '14 at 22:20

You haven't said anything about gearing so I'll assume direct drive.

Force = 22 Nm / radius 0.153m = 143N.

Gradient : 35 degrees.

Lifting force = Driving force / sin(35 degrees) = 249N.

So it should be able to power nearly 25kg up that hill if it doesn't have to accelerate.


Assuming no losses, it takes 750W to raise 75kg one meter/s (75*9.8~=750). So, with a motor about half that power your vertical speed will be 0.48m/s.

At 35 degrees incline, your ground speed will 0.48/sin(35) = 0.84m/s.

0.84m/s divided by 2*pi*0.153 gives 0.87 rev/s or 52 rpm, so you need a reduction ratio of 3:1, which agrees with Brian's answer that direct drive raises 25kg.

  • 1
    \$\begingroup\$ Worth pointing out that both answers make no allowance for vehicle or battery weight : if you can keep the weight inc.motor and battery down to 50kg then 5:1 gearing would just do it. 6:1 or more would be required for some acceleration and friction, and even that might be optimistic. \$\endgroup\$ – Brian Drummond Jul 6 '14 at 20:30

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