Energy and Human Ambitions on a Finite Planet, 2021a

16 Small Players 283
executing circular paths as they gain <strong>and</strong> lose gravitational potential
energy. Meanwhile, the energy in a packet of waves travels efficiently
onward, until reaching a barrier like a shoreline, where they break in a
display of kinetic energy. It is hard to stare at a coastline being pounded
by surf without admiring the power of nature. What if we harnessed
that power?
As a means of estimation, let’s imagine sinusoidal 25 waves1min
amplitude (peak-to-trough) arriving every 6 seconds, traveling at about
3 m/s. Crests must then be 18 m apart, since the distance between crests
is their rate (3 m/s) times time (6 seconds). We call this the wavelength,
symbolized by λ (Greek lambda). See Figure 16.2 for the layout.
25: . . . shaped like a sine wave
energetically equivalent blocky wave
A
shave off
flip <strong>and</strong> place
Area=
Figure 16.2: Sinusoidal waves have amplitude
A, <strong>and</strong> wavelength λ. In terms of harvesting
the gravitational potential energy,
we can think of it as lopping off the crest <strong>and</strong>
flipping it over onto the trough to level the
water surface. In doing so, we move some
mass, m, down a height h to get mgh of energy.
The block-equivalent is shown below,
where the area <strong>and</strong> average height of the
sinusoidal trough/crest has been faithfully
captured by rectangles of height ∼0.39A
<strong>and</strong> length ∼0.41λ. From these, it is possible
to figure out the potential energy associated
with the wave.
In order to figure out the energy involved, we need the mass of water
raised <strong>and</strong> a height to which it is raised. Notice that in Figure 16.2,
the potential energy in the wave can be extracted by making the water
flat again, which is equivalent to taking all the water from the crest
<strong>and</strong> putting it into the trough. We just need to know how much water
we’re moving, <strong>and</strong> through what height. Figure 16.2 has done the fancy
math already <strong>and</strong> redrawn the wave as rectangular chunks that have
equivalent area as the sinusoidal crest <strong>and</strong> trough <strong>and</strong> also the same
average (midpoint) height relative to the average surface height (dashed
line). From this, we learn that the wave crest has area Aλ/2π <strong>and</strong> the
height of the displacement is πA/8, where A is the wave amplitude from
the top of the crest to the bottom of the trough.
Author’s note: we’ll be going through a
bit of a derivation here—not because wave
power is particularly important, but because
physicists sometimes can’t help it. Feel free
to skip past all the equations, as this is
meant as a sort of entertainment. What can
I say: the introduction had a warning about
skipping the whole chapter, yet here you
are.
To assess potential energy as mgh, we need three pieces. We already
know g ≈ 10 m/s 2 , <strong>and</strong> we now know that h πA/8. The mass is a
density, ρ, times a volume. We already have the area of the crest cross
section as Aλ/2π. To get a volume, we need a length along the wave,
which we are free to make up as a variable we’ll call l. The mass, m, of our
block of water is then ρAλl/2π. Putting this together, the gravitational
potential energy (GPE) associated with putting the water back to a flat
state along a length, l, of one wave is
E GPE mgh ρAλl
2π g πA 8 ρλlgA2 . (16.4)
16
© 2021 T. W. Murphy, Jr.; Creative Commons Attribution-NonCommercial 4.0 International Lic.;
Freely available at: https://escholarship.org/uc/energy_ambitions.