How Much Power Does It Take To Move A Person Around On A Scooter Or Cycle?
By: Karl T. Ulrich
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:
- The power required to overcome the rolling resistance of the wheels on the pavement.
- The power required to overcome the wind resistance associated with moving the vehicle/passenger through the air.
- 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:
- Total power required is strongly influenced by speed.
- At high speeds, the effect of wind resistance will be very large (because it depends on S cubed).
- 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 (This includes air density factor for sea-level air. Picky engineers: see note below.)
- 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
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
How much power is consumed in the same situation except traveling up a 2% grade at 12 mph?
- now G = 0.02
Total-Power = (a+G)WS + bFS^3 = (0.04+0.02)734 x 5.4 + 0.4 x 0.8 x 5.4 x 5.4 x 5.4 = 238 + 49 = 287 watts
What happens if the scooter is going 20 mph on the flat?
- now S = 20 mph = 9 meters/second
Total-Power = (a+G)WS + bFS^3 = (0.04+0)734 x 9 + 0.4 x 0.8 x 9 x 9 x 9 = 264 + 233 = 497 watts
Note that climbing a 2% grade (G=0.02) consumes the same power as rolling on tires with a rolling resistance of 2% (a=0.02).
Note that the power required to propel a vehicle does not depend on the power source. In other words, a scooter powered by kicking requires the same power for the same speed, weight, etc. as a scooter powered by an electric motor.
Finally, let me note that the vast majority of small electric vehicle manufacturers do not appear to know these basic laws of physics. I see a lot of scooters with advertised top speeds of 15-17 mph, yet with 9 inch pneumatic tires, and with motors and transmissions that can deliver only about 150 watts to the wheels. You can calculate for yourself that the specs must be highly exaggerated (or the manufacturers must assume that a small child is riding the scooter down a big hill...). Caveat emptor.
NOTE TO THE PICKY ENGINEERS (you know who you are): The expression for wind resistance is actually rho/2 * Cd * F * S^3, where rho is the density of air and Cd is a non-dimensional drag coefficient. Since rho is around 1.2 kg/m^3, the rho/2 term is around 0.6. To simplify these calculations, I included this factor of 0.6 in computing the "coefficient" b.