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

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


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 As. 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

msausage = density × volume = ρvtAs


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

Figure C.2. A plane encounters a stationary tube of air. Once the plane has passed by, the air has been thrown downwards by the plane. The force exerted by the plane on the air to accelerate it downwards is equal and opposite to the upwards force exerted on the plane by the air.
Figure C.3. Our cartoon assumes that the plane leaves a sausage of air moving down in its wake. A realistic picture involves a more complex swirling flow. For the real thing, see figure C.4.
Figure C.4. Air flow behind a plane. Photo by NASA Langley Research Center.