and Newton’s third law, which I just mentioned:

force exerted on A by B = − force exerted on B by A

(C.2)

If you don’t like equations, I can tell you the punchline now: we’re going

to find that the power required to create lift turns out to be *equal* to the

power required to overcome drag. So the requirement to “stay up” *doubles*

the power required.

Let’s make a cartoon of the lift force on a plane moving at speed *v*. In

a time *t* the plane moves a distance *vt* and leaves behind it a sausage of

downward-moving air (figure C.2). We’ll call the cross-sectional area of

this sausage *A*_{s}. This sausage’s diameter is roughly equal to the wingspan

*w* of the plane. (Within this large sausage is a smaller sausage of swirling

turbulent air with cross-sectional area similar to the frontal area of the

plane’s body.) Actually, the details of the air flow are much more interest-

ing than this sausage picture: each wing tip leaves behind it a vortex, with

the air between the wingtips moving down fast, and the air beyond (outside)

the wingtips moving up (figures C.3 & C.4). This upward-moving

air is exploited by birds flying in formation: just behind the tip of a bird’s

wing is a sweet little updraft. Anyway, let’s get back to our sausage.

The sausage’s mass is

*m*_{sausage} = density × volume = *ρvtA*_{s}

(C.3)

Let’s say the whole sausage is moving down with speed *u*, and figure out

what *u* needs to be in order for the plane to experience a lift force equal to