I've been designing an electric skateboard which is driven by only one of the rear wheels (for the moment). As I was picking out motors for the board, I soon realized I would need to find a balance between torque needed to climb hills, and RPM to get a good top speed.

So I'm searching for the best KV (RPM/v) motor/gearing ratio combination.

For all calculations I've assumed the following:

  • Total mass (board + rider) = 85 kg
  • Voltage = 12 Cells * 3.2V (LiFePo4) = 38.4 V
  • Max Speed = 40 km/h = 11.11 m/s
  • Hill Grade = 15%
  • Coefficient of friction between rubber and asphalt: 0.65
  • Diameter of wheel = 80 mm

I tried assuming maximum kinetic energy as being equal to maximum gravitational potential energy at the top of the hill.

$$E_{k,max} = \frac{1}{2}mv_{max}^2 = E_g = mgh_{max}$$

Then solving for $h$.

I'd take that and use $y/x = 0.2$, solve for $x$, then use some trigonometry to setup a Newton's Second Law problem and solve for force needed to overcome friction and gravity (Equilibrium problem).

Then using $T = F \times r$, find torque needed at the wheels, assume some gearing ratio that keeps that and the $V_{max}$. Then solve for RPM and divide by total voltage for motor KV.

My mechanics are rather rusty and I could use some help.

up vote 9 down vote accepted

I think you're overcomplicating things. To push 85 kg up a 15% slope against gravity of 9.8 m/s2 requires a force of

$$ \sin (\arctan (0.15) ) = 0.1483 \approx 0.15 $$

$$F = 85 kg \cdot 9.8 \frac{m}{s^2} \cdot 0.15 = 125 N$$

With an 80 mm wheel, this requires a torque of

$$T = 125 N \cdot 0.04 m = 5 Nm$$

To do this at a forward speed of 11.11 m/s requires a fair amount of power:

$$P = Force \cdot Velocity = 125 N \cdot 11.11 \frac{m}{s} = 1400 W$$

which is about 36 A from your 38.4 V battery pack, and your wheels will be turning at

$$\frac{11.11 \frac{m}{s}}{\pi\cdot 0.08 \frac{m}{rev}} \cdot 60 \frac{sec}{min} = 2600 RPM$$

Does any of this help you move forward?

  • Thanks a bunch, I think I got on one pattern of thinking and found it hard to go back! This is what I'm looking for. – Satchel Jan 4 '16 at 5:39
  • 2
    The calculation of force should be $F=m \, g\, \sin{\theta}$. In this case, $\sin(\tan^{-1}0.15) \approx 0.15$, so it works, but that's not the general case. – Carlton Jan 4 '16 at 13:18
  • @Carlton: Do you understand that we're just doing some ballpark estimation here? Small-angle approximations are perfectly acceptable. – Dave Tweed Jan 4 '16 at 13:46
  • 1
    @DaveTweed - While I would agree that small-angle approximations are acceptable, OP's "mechanics are rather rusty", and you didn't state/show that you were using any small-angle approximations. – Chuck Jan 4 '16 at 13:48
  • @Chuck thanks for the concern, I did pick up on that however I just assumed that he did that calculation and rounded off to 0.15 – Satchel Jan 4 '16 at 19:24

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