- The tube of air created in a time
*t*has a volume*Avt*, where*A*is

the cross-sectional area of the tube, which is similar to the area of

the front view of the car. (For a streamlined car,*A*is usually a little

smaller than the frontal area*A*_{car}, and the ratio of the tube’s effective

cross-sectional area to the car area is called the drag coefficient*c*_{d}.

Throughout the following equations,*A*means the effective area of

the car,*c*_{d}*A*_{car}.) The tube has mass*m*_{air}=*ρAvt*(where*ρ*is the

density of air) and swirls at speed*v*, so its kinetic energy is:

and the rate of generation of kinetic energy in swirling air is:

So the total rate of energy production by the car is:

power going into brakes | + | power going into swirling air | (A.2) |

= ^{1}⁄_{2}m_{c}v^{3}/d |
+ | ^{1}⁄_{2}ρAv^{3}. |

Both forms of energy dissipation scale as *v*^{3}. So this cartoon predicts that

a driver who halves his speed *v* makes his power consumption 8 times

smaller. If he ends up driving the same total distance, his journey will

take twice as long, but the total energy consumed by his journey will be

four times smaller.

Which of the two forms of energy dissipation – brakes or air-swirling –

is the bigger? It depends on the ratio of

(*m*_{c}/*d*)/(*ρA*) .

If this ratio is much bigger than 1, then more power is going into brakes; if

it is smaller, more power is going into swirling air. Rearranging this ratio,

it is bigger than 1 if

*m*_{c} > *ρAd*.

Now, *Ad* is the volume of the tube of air swept out from one stop sign

to the next. And *ρAd* is the mass of that tube of air. So we have a very

simple situation: energy dissipation is dominated by kinetic-energy-being-

dumped-into-the-brakes if the mass of the car is *bigger* than the mass of

the tube of air from one stop sign to the next; and energy dissipation is

dominated by making-air-swirl if the mass of the car is *smaller* (figure A.4).

Let’s work out the special distance *d** between stop signs, below which

the dissipation is braking-dominated and above which it is air-swirling

dominated (also known as drag-dominated). If the frontal area of the car

is:

*A*_{car} = 2 m wide × 1.5 m high = 3 m^{2}