Calculating torque needed to climb a graded hill

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
• 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.

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

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

• 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