Kick Scooter Power: How Much Do You Need?

I’ve seen scooters and cycles claiming 200 Watts and 18 mph top speed and others claiming 400 Watts and 12 mph top speed. How much power does it actually take to move a person around on a scooter or cycle? Bicycles, scooters, and even automobiles are all governed by the same fundamental power requirements. At constant speed, the power required to move the vehicle and the passenger goes to three places:

  1. The power required to overcome the rolling resistance of the wheels on the pavement.
  2. The power required to overcome the wind resistance associated with moving the vehicle/passenger through the air.
  3. The power required/provided to move the vehicle and passenger up/down any incline (if not traveling on flat pavement).

We can write this as an equation:

Total-Power = Power-rolling-resistance + Power-wind-resistance + Power-hill-climbing

(Note that Total-power is the power delivered to the driving wheel of the vehicle net of any friction in the transmission and inefficiencies in the power system.)

To a first approximation, power-rolling-resistance is in turn determined by the weight of the vehicle/passenger (W), the speed of the vehicle (S), and a coefficient that characterizes the rolling resistance of the wheel (a). Power-rolling-resistance = aWS

To a first approximation, Power-wind-resistance is determined by the “frontal area” (F) of the vehicle/passenger (the area of the outline of the vehicle/passenger when viewed from the front), a coefficient (b) that characterizes the shape of the vehicle/passenger, and the CUBE of the speed (S x S x S).

Power-wind-resistance = bFS^3

Power-hill-climbing is determined by the grade of the hill (G), the weight of the vehicle/passenger (W), the speed (S) of the vehicle/passenger. Power-hill-climbing = GWS

So, the entire equation is: Total-Power = aWS + bFS^3 + GWS = (a+G)WS + bFS^3

Before we do some calculations, we can make some interesting observations:

  1. Total power required is strongly influenced by speed.
  2. At high speeds, the effect of wind resistance will be very large (because it depends on S cubed).
  3. Light vehicles/passengers have an overall advantage. In fact, although W does not appear in the expression for wind resistance, frontal area (F) is highly correlated with W, so overall size/weight pretty much influences all three categories of power consumption.

Now, some approximate numbers. (I use metric units, but provide some examples and conversion factors for those of you who think in English units.)

  • a = coefficient of rolling resistance
    • 0.008 for high-pressure 700mm road bike tire
    • 0.020 for a mountain bike tire
    • 0.040 for a typical (e.g., 9 inch) pneumatic scooter tire
  • W is weight in Newtons (1 pound = 4.45 Newtons)
  • S is speed in Meters/Second (1 mph = 0.45 meters/second)
  • b = drag factor in kg/m^3 
    • 0.6 for a square-edged box
    • 0.4 for most human-like shapes
    • 0.2 for a egg-shaped object
  • F = frontal area in square meters
    • 0.4 for a crouched racing cyclist and bicycle
    • 0.6 for an upright cyclist and bicycle
    • 0.8 for a standing scooter rider
  • G = height of climb/distance of climb (e.g., % grade)
    • Typical maximum railroad grade = 0.02
    • Typical maximum bike path grade = 0.05
    • Typical maximum overpass grade = 0.08
    • Maximum grade on Pike’s Peak mountain road = 0.10
    • Powell St. in San Francisco (cable cars) = 0.17

Example

  1. How much power is consumed to propel a medium-sized (165 lb.) adult standing on a scooter with 9 inch pneumatic tires traveling at 12 mph?
    • W = 165 lb. = 734 Newtons
    • S = 12 mph = 5.4 Meters/second
    • a = 0.040
    • b = 0.4
    • F = 0.8 square meters
    • G = 0

    Total-Power = (a+G)WS + bFS^3 = (0.04+0)734 x 5.4 + 0.4 x 0.8 x 5.4 x 5.4 x 5.4 = 159 + 49 = 208 watts

By: Karl Ulrich

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