and the drag coefficient is *c*_{d} = 1/3 and the mass is *m*_{c} = 1000 kg then the

special distance is:

So “city-driving” is dominated by kinetic energy and braking if the distance

between stops is less than 750 m. Under these conditions, it’s a good

idea, if you want to save energy:

- to reduce the mass of your car;
- to get a car with regenerative brakes (which roughly halve the energy

lost in braking – see Chapter 20); and - to drive more slowly.

When the stops are significantly more than 750 m apart, energy dissipation

is drag-dominated. Under these conditions, it doesn’t much matter

what your car weighs. Energy dissipation will be much the same whether

the car contains one person or six. Energy dissipation can be reduced:

- by reducing the car’s drag coefficient;
- by reducing its cross-sectional area; or
- by driving more slowly.

The actual energy consumption of the car will be the energy dissipation

in equation (A.2), cranked up by a factor related to the inefficiency of

the engine and the transmission. Typical petrol engines are about 25%

efficient, so of the chemical energy that a car guzzles, three quarters is

wasted in making the car’s engine and radiator hot, and just one quarter

goes into “useful” energy:

Let’s check this theory of cars by plugging in plausible numbers for motorway

driving. Let *v* = 70 miles per hour = 110 km/h = 31 m/s and

A = *c*_{d}*A*_{car} = 1 m^{2}. The power consumed by the engine is estimated to be

roughly

If you drive the car at this speed for one hour every day, then you travel

110 km and use 80 kWh of energy per day. If you drove at half this speed

for two hours per day instead, you would travel the same distance and

use up 20 kWh of energy. This simple theory seems consistent with the

ENERGY-PER-DISTANCE | ||
---|---|---|

Car at 110 km/h |
↔ | 80 kWh/(100 km) |

Bicycle at 21 km/h |
↔ | 2.4 kWh/(100 km) |

PLANES AT 900 KM/H | |
---|---|

A380 | 27 kWh/100 seat-km |