# Electric scooter acceleration

I tested two electric scooters with almost same power (1500 W), but very different performances: A "Zem Star 45" (1500 W, 60 Nm) can reach 50 km/h in "a few seconds" (not measured actually), An "Ingaeta Ingo" (1500 W, ??? Nm) took more than 10 seconds to reach 50 km/h.

How much does torque affect acceleration performance?

Given two identical scooters (same weight, same engine power, same wheels size), rgardless of wheels and air friction, how can I calculate acceleration given the torque on a flat street?

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I add: Star 45 has a 60V motor, Ingo has a 48V one. How does this relate to speed and acceleration? –  jumpjack Sep 12 '12 at 19:00

A typical electric motor has its torque varying linearly with speed. You can translate this through the gearbox into traction $F$ and speed $v$ to arrive at the acceleration function ignoring air resistance

$$a(v) = \frac{F_0}{m} \left( 1- \frac{v}{v_f} \right)$$

where $F_0$ is the initial (peak) traction (relating to the peak torque of motor), and $v_f$ the top speed of vehicle (relating to zero load speed).

To get the time to reach a speed $v$ you do

$$t(v) = \int_0^v \frac{m/F_0}{ \left( 1- \frac{v}{v_f} \right) }\;{\rm d}v = \frac{m\,v}{F_0} \left(1-\frac{v}{2\,v_f} \right)$$

inverting yields

$$v(t) = v_f \left( 1-\sqrt{1-\frac{2 F_0}{m v_f} t} \right)$$

In addition, to get the distance traveled you do

$$x(v) = \int_0^v \frac{m v/F_0}{ \left( 1- \frac{v}{v_f} \right) }\;{\rm d}v = \frac{m v^2 (3 v_f-2 v)}{6 F_0 v_f}$$

inverting yields

$$v(x) = \frac{v_f}{2} + v_f \sin\left( \frac{1}{3} \sin^{-1}\left( \frac{12 F_0}{m v_f^2} x -1 \right) \right)$$

Edit 1

The power produced by the motor is $$P = T(\omega)\, \omega = m\, v\, a(v) = F_0 v \left( 1- \frac{v}{v_f} \right)$$ peak power occurs at $v=v_f$ with value $P_{peak} = F_0 v_f / 4$

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A typical electric motor has constant torque. –  jumpjack Sep 12 '12 at 18:56
If it had constant torque then it would produce infinite power (as the speed increases). It has limited power as derived by $P = T(\omega) \omega = F(v) v$ –  ja72 Sep 13 '12 at 11:54

The force between the driven wheel and the road is given by:

$$F = ma = \frac{\tau}{r}$$

where $\tau$ is the torque and $r$ is the driven wheel radius, so the acceleration is:

$$a = \frac{\tau}{m \space r}$$

NB this is the torque at the wheel, while the torque the manufacturer quotes is normally the torque at the engine crankshaft. The torque at the wheel will depend on how much the engine speed is geared up or down.

At any speed greater than zero you need to subtract the aerodynamic drag off the force, but the simple expression for the acceleration - torque relation will be valid at low speeds.

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What about hub motors? –  jumpjack Sep 12 '12 at 18:58
It doesn't matter what type of motor it is. All that matters is the torque it produces at the wheel. –  John Rennie Sep 13 '12 at 6:13