\documentclass[12pt]{article}% fleqn
\usepackage[latin1]{inputenc}
\usepackage{natbib}%{mybibliog}
\usepackage{mycenter}
\usepackage[allowmove]{url}
%\mathindent0cm
%\parindent0cm
\textheight=10.73in
\topmargin=-0.75in
\oddsidemargin=-0.5in
\textwidth=7in
\pagestyle{empty}
\usepackage{epsf}
\usepackage{natbib}%{lsalike}% defines citation commands
% Pick a bibliography-style. I recommend abbrvnat.
% The natbib package supports many styles.
% Change this style to change the appearance of the bibliography.
%
%\bibliographystyle{unsrtnat}
%\bibliographystyle{plainnat}
\bibliographystyle{abbrvnat}
\usepackage{graphicx}
\usepackage{cft} % provides header and footer defns and page sizes
\usepackage{cents} % provides \cents
\usepackage{ourbox2} %
\usepackage{textcomp} % provides textdegree
\usepackage{fancybox}% Provides ability to put verbatim text inside boxes
\usepackage{boxedminipage}
\usepackage{myalgorithArticle}% provides ``floatseq'' for labelling figs and tables
\usepackage{mycaption}% defines ``\indented''and \@makecaption; and the notindented style used in figure captions
\usepackage{marginfig}% Defines many macros for making various styles of figure with captions, and framedalgorithms
\usepackage{booktabs}% makes nice quality tables
\usepackage{ragged2e}% provides \justifying
\usepackage{mycenter}% modifies center to reduce vertical space waste - useful for figures, etc.
\begin{document}
%\bibliographystyle{lsalikedjcmsc}%.bst
\pagestyle{empty}
\renewcommand{\textfraction}{0.01}
% \input{psfig.tex}
\input{/home/mackay/tex/newcommands1}
\input{/home/mackay/tex/newcommands2}
% \input{/home/mackay/tex/newcommands2e}
% textcomp provides textdegree
% Alternatively, you could use: \newcommand{\degrees}{\ensuremath{^{\circ}}}%
\newcommand{\europe}{125}
\newcommand{\Wmm}{\ensuremath{\rm{W}\!/{\rm{m}}^2}}
\newcommand{\mWmm}{\ensuremath{\rm{mW}\!/{\rm{m}}^2}}
\newcommand{\Wm}{\ensuremath{\rm{W}\!/{\rm{m}}}}
\newcommand{\degree}{\textdegree}
\newcommand{\degrees}{\textdegree}
\renewcommand{\unit}{{\sf{unit}}}
\newcommand{\units}{{\sf{units}}}
\newcommand{\COO}{\mbox{CO$_2$}}
\newcommand{\per}{{\rm per}}
\newcommand{\litre}{{\rm litre}}
\newcommand{\mile}{{\rm mile}}
\newcommand{\gallon}{{\rm gallon}}
\newcommand{\km}{{\rm{km}}}
\renewcommand{\kg}{{\rm{kg}}}
\newcommand{\kJ}{{\rm{kJ}}}
\newcommand{\GW}{{\rm{GW}}}
\newcommand{\MJ}{{\rm{MJ}}}
\newcommand{\J}{{\rm{J}}}
\newcommand{\kwh}{{\rm{kWh}}}
\newcommand{\kWh}{{\rm{kWh}}}
\newcommand{\kwhe}{{\rm{kWh}\ensuremath{^{\rm(e)}}}}
\newcommand{\kWhe}{{\rm{kWh}\ensuremath{^{\rm(e)}}}}
\newcommand{\GWhe}{{\rm{GWh}\ensuremath{^{\rm(e)}}}}
\newcommand{\kw}{{\rm{kW}}}
\renewcommand{\m}{{\rm{m}}}
\newcommand{\kW}{{\rm{kW}}}
\newcommand{\mW}{{\rm{mW}}}
\newcommand{\W}{{\rm{W}}}
\newcommand{\MW}{{\rm{MW}}}
\newcommand{\uday}{{\rm{day}}}
\newcommand{\days}{{\rm{days}}}
\newcommand{\hour}{{\rm{hour}}}
\newcommand{\mph}{\,{\rm{mph}}}
\newcommand{\s}{{\rm{s}}}
\renewcommand{\N}{{\rm{N}}}
\newcommand{\person}{{\rm{person}}}
\newcommand{\miles}{{\rm{miles}}}
% \newcommand{\mile}{{\rm{mile}}}
\newcommand{\years}{{\rm{years}}}
\newcommand{\uyear}{{\rm{year}}}
\newcommand{\Lomborg}{Lomborg}
%\newcommand{\url}[1]{{\tt{#1}}}
\thispagestyle{empty}
%\section*
\title{Under-estimation of the UK Tidal Resource}
\author{ {\large David J.C. MacKay} \\
Cavendish Laboratory,
% Department of Physics\\
University of Cambridge\\
{\tt{mackay@mrao.cam.ac.uk}}}
\date{January 20, 2007 -- Draft 2.6 (some typos corrected and references added \today)}
\maketitle
\begin{abstract}
A widely-quoted estimate of the practical
UK tidal resource is
12 TWh/y \citep{BlackVeatch}.
I believe this is an underestimate, because
it is based on an incorrect physical model of
the flow of energy in a tidal wave.
In a shallow-water-wave model of tide, the true flow
of energy is greater than the Black-and-Veatch
flow by a factor of $d/h$, where
$d$ is the water depth and $h$ is the tide's vertical amplitude.
The tidal resource may therefore have been underestimated by
a factor of about 10.
\end{abstract}
The widely-quoted
estimate of the practical
UK tidal resource is
12\,TWh/y
(equivalent to
an average production of 1.4\,GW, or
0.5\,kWh per person per day) \citep{BlackVeatch}.
In a two-page comment on the DTI Energy Review,
% S.H.\ Salter
\citet{Salter2} suggests that this standard figure may well be an
under-estimate (see also \citet{SalterTaylor}).
Salter estimates that the dissipation by friction on
the sea bed of the Pentland Firth alone is 100\,GW (peak). He argues
that turbines could be inserted as a sea-bed substitute
there and would deliver up to 40\,GW (peak).
In this note, I present back-of-envelope models of tidal power
that concur with and amplify Salter's view. In sum, the method
used by Black and Veatch to estimate the tidal resource, namely
estimating the kinetic energy flux across a plane, is flawed because
(except in certain special cases)
{\em the power in tidal waves is not equal to the kinetic energy flux
across a plane}.
These back-of-envelope models are not new.
Essentially identical models are analysed in
greater detail by \citet{Taylor1920}.
\section{Tides as tidal waves}
Follow a high tide as it rolls in from the Atlantic.
The time of high tide becomes progressively later as we move east up the
English channel from the Scillys to Portsmouth and on to Dover.
Similarly, a high tide moves clockwise round Scotland, rolling
down the North Sea from Wick to Berwick and on to Hull.
These two high tides converge on the Thames Estuary. By coincidence, the
Scottish wave arrives nearly 12 hours later than the one that came via
Dover, so it arrives in near-synchrony with the next high tide via Dover,
and London receives the normal two high tides per day.
\begin{figure}
% \figuremargin{
\figuredangle{
\begin{center}
\mbox{\epsfbox{metapost/tide.2}}
\end{center}
}{
\caption[a]{A shallow-water wave.
The wave has energy in two forms:
potential energy associated with raising water out of the
light-shaded troughs into the heavy-shaded crests;
and kinetic energy of all the water moving around
as indicated by the small arrows.
The speed of the wave, travelling from left to right, is indicated by
the much bigger arrow at the top.
For tidal waves,
a typical depth might be 100\,m,
%% v = sqrt(g d) = sqrt( 1000 )
the crest velocity 30\,m/s,
the vertical amplitude at the surface 1 or 2\,m,
and the water velocity amplitude
%% U = v h/d = 30 * 2/100 = 0.6
0.3 or 0.6\,m/s.
}
\label{fig.shallowwave}
}
\end{figure}
\Figref{fig.shallowwave} shows a model for a tidal wave travelling
across relatively shallow water.
This model is intended as a cartoon, for example, of tidal
crests moving up the English channel,
towards the outer Hebrides, or down the North Sea.
The model neglects Coriolis forces.
[The Coriolis force causes tidal crests and troughs to
tend to drive on the right -- for example,
going up the English Channel, the high tides are higher
and the low tides are lower on the French side of the
channel. By neglecting this effect I may have introduced
some error into the estimates.
The analysis of \citet{Taylor1920} includes the Coriolis effect,
and includes the possibility that there is a second tidal wave running
in the opposite direction.]
The water has depth $d$.
Crests and troughs of water are injected from the left hand side
by the 12-hourly ocean tides.
The crests and troughs move with velocity
\beq
v= \sqrt{ g d }.
\label{eqv}
\eeq
We assume that the wavelength is much bigger than the depth.
Call the vertical amplitude of the tide $h$.
For the standard assumption of nearly-vorticity-free flow,
the horizontal velocity is near-constant with depth.
The velocity is proportional to the surface displacement
and has amplitude $U$, which can be found by conservation of mass:
\beq
U = v h/d.
\label{eqU}
\eeq
If the depth decreases gradually, the wave velocity $v$ reduces.
% and the amplitude of tidal wave increases (as $1/d^{1/4}$, as we'll see).
For the present discussion we'll assume the depth is constant.
Energy flows from left to right at some rate.
How should this total tidal power be estimated?
And what's the maximum power that could be extracted?
One suggestion is to choose a cross-section and estimate the average
flux of kinetic energy across that plane. This kinetic-energy-flux
method is used by Black and Veatch to estimate the UK
resource. In this toy model, we can also compute the total power by
other means.
We'll see that the kinetic-energy-flux answer is incorrect by a significant
factor.
The peak kinetic-energy flux at any section
is
\beq
K_{\rm BV} = \frac{1}{2} \rho A U^3 ,
\eeq
where $A$ is the cross-sectional area.
The true total incident power is a standard textbook calculation;
one way to get it is to find the total energy present in one wavelength
and divide by the period; another option is to imagine replacing a vertical
section by an appropriately compliant
piston and computing the average work done on the
piston. I'll do the calculation both ways.
The potential energy of a wave (per wavelength and per unit width
of wavefront) is
\beq
\frac{1}{4} \rho g h^2 \lambda
\eeq
The kinetic energy (per wavelength and per unit width
of wavefront) is identical to the potential energy.
So the true power of this model shallow-water tidal wave is
\beq
\mbox{Power} = \frac{1}{2} ( \rho g h^2 \lambda) \times w / T
= \frac{1}{2} \rho g h^2 v \times w ,
\eeq
where $w$ is the width of the wavefront.
Substituting
% $A = w d$ and
$v=\sqrt{ g d }$,
\beq
\mbox{Power}
= \rho g h^2 \sqrt{ g d } \times w /2.
= \rho g^{3/2} \sqrt{ d } h^2 \times w /2.
% = \rho g h^2 \sqrt{ g } \times A / \sqrt{d} ,
\label{eqP}
\eeq
% cf Taylor
% \frac{1}{2} g \rho U w d h \cos \frac{2 \pi}{T}( T_1- T_0 ) ,
% Use U = h sqrt(g/d)
% \rho g^{3/2} \sqrt{ d } h^2 \times w .
% \frac{1}{2} g \rho g**1/2 d**1/2 h**2 w
Let's compare this power with
the kinetic-energy flux $K_{\rm BV}$.
Strikingly, the two expressions scale differently with amplitude.
Using the amplitude conversion relation (\ref{eqU}),
the crest velocity (\ref{eqv}), and $A = w d$,
we can re-express
the kinetic-energy flux
as
\beq
K_{\rm BV} = \frac{1}{2} \rho A U^3
= \frac{1}{2} \rho w d (v h/d)^3
% = \frac{1}{2} \rho w (\sqrt{ g d })^3 h^3 / d^2 .
= \frac{1}{2} \rho w \left( g^{3/2} / \sqrt{ d } \right) h^3 .
\eeq
Thus the kinetic-energy-flux method suggests that the total
power of a shallow-water wave scales as amplitude cubed;
but the correct formula shows that the power goes as amplitude squared.
The ratio is
\beq
\frac{ K_{\rm BV} }
{ \mbox{Power} }
= \frac{ \rho w \left( g^{3/2} / \sqrt{ d } \right) h^3
}{ \rho g^{3/2} h^2 \sqrt{ d } w }
= \frac{ h
}{ d}
\eeq
Thus estimates based on the kinetic-energy-flux method may be
too small by a significant factor, at least in cases where this
shallow-water cartoon of tidal waves is appropriate.
Moreover, estimates based on the kinetic-energy-flux method
incorrectly assert that the total available power at springs is
greater than at neaps by a factor of eight (assuming an amplitude
ratio, springs to neaps, of two); but the correct answer is that the
total available power of a travelling wave scales as its amplitude
squared, so the springs-to-neaps total-incoming-power ratio is four.
\subsection{Intuition}
Why is
the kinetic-energy-flux method wrong for tidal waves in open shallow
water?
One way to think about it is to make an analogy with other
processes where a moving body delivers energy to another body.
If I grab someone with one hand and shake them around, how much power
am I delivering? Can we find the power by putting
a section through my arm and working out the kinetic-energy-flux of my arm?
No! My arm might be made of balsa wood -- that would completely change the
kinetic-energy flux, but would not change the effect on the receiving body.
One reason people get confused about the power in a wave is because
they think that the power moves at the same speed as the water.
There are a few ways to see that this is not generally true:
note the speed at which high tides move up the English channel
or down the North Sea -- they move at hundreds of miles per hour,
while the water itself moves only at one or two miles per hour.
Another thought experiment is to imagine a travelling transverse wave,
where there is no motion at all in the direction
of travel; in this case it is particularly clear that the
kinetic energy method gives an incorrect answer.
The tidal wave conveys energy not because a piece of water moves
along, carrying
that energy with it as kinetic energy, but because the weight of water
in a tidal peak exerts a pressure on neighbouring water, and that
pressure does work as the water moves.
We can compute the power using this idea of the body of water on the
left doing work on the body of water on the right.
\subsection{Power flux using forces}
Let's repeat the power calculation using forces.
Consider a piston pressing against a wall of water, behaving just
as an adjacent body of water would.
During the half period when
the piston moves to the right, while a crest
is present, the work done on the piston at depth $z$ is
of order $P^{+}(z) U T$ per unit area, where
$P^{+}(z)$ is the pressure at depth $z$, $U$ is the velocity and $T$
is the period. As the crest passes, the peak pressure is:
\beq
P^{+}(z) = \rho g ( z+h ).
\eeq
When the piston moves to the left, the pressure is lower:
\beq
P^{-}(z) = \rho g ( z-h );
\eeq
and the work done on the piston is
$- P^{-}(z) U T$ per unit area (neglecting constants
of order 1, as before).
The net work done on the piston (per unit area) is
\beq
P^{+}(z) U T
- P^{-}(z) U T
= UT ( \rho g ( z+h ) - \rho g ( z-h ) )
= UT ( \rho g ( 2 h ) ) ,
\eeq
independent of $z$.
Integrating up over the area $A=wd$,
the average power delivered to the piston is
\beq
\mbox{Power}
= w d UT ( \rho g ( 2 h ) ) / T
= 2 w v h \rho g h
= 2 w \rho g^{3/2} \sqrt{d} h^2 .
\eeq
In this expression, the factor of 2 is bogus:
I should have done the integral.
This answer agrees with the other outcome (\ref{eqP}).
\section{Reconciliation in a special case}
The kinetic-energy-flux method is not always wrong.
In the special case of tidal flow through a narrow cleft
connecting two immense reservoirs, one of which is tidal
and one of which is scarcely so, it gives the right answer.
An example of such a cleft might
by the Strait of Gibraltar, connecting
the tidal Atlantic with the not-very-tidal Mediterranean.\footnote{
In fact, the Strait of Gibraltar
is much more complicated, with density
differences simultaneously driving a surface inflow and
a deeper outflow
% upper 0.6m/s to the east
% lower 0.4m/s to the west
% tidal current max = 5 knots. strongest currents are
% midway between tides.
\cite{Gibral99,Gibral02}.
% gibraltar's camarinal
% sill is ten minutes of lat wide and roughly 300 m deep
}
Imagine that at high water there is a height drop of $h$
between stationary waters on the two sides.
Assuming vorticity-free flow from the high side up to the outlet,
the velocity $U$ of water at the outlet of the cleft (at any depth)
can be estimated by applying Bernoulli's formula
along a streamline connecting that water to a virtually-stationary
upstream origin:
\beq
\frac{1}{2} \rho U^2 = \rho g h
\eeq
\beq
U = \sqrt{2 g h }
\eeq
The kinetic-energy flux
% as before.
is
\beq
\frac{1}{2} \rho A U^3 =
\frac{1}{2} \rho A \sqrt{2 g h }^3 =
\rho A \sqrt{2} g^{3/2} h^{3/2}
\label{crackK}
\eeq
The total power arriving can also be written down in terms
of potential energy drop:
\beq
\rho g h U A = \rho g h \sqrt{2 g h } A .
\label{crackP}
\eeq
These two equations (\ref{crackK},\ref{crackP}) agree.
The extractable power by stream-turbines in this situation
has been shown by \citet{Garrett05} to be
roughly 0.22 times the total power (\ref{crackP}).\nocite{Garrett07}
\section{Literature}
The fact that the power in a tidal wave
scales as amplitude-{\em squared\/} is present in
the detailed model of \citet{Taylor1920}.
Taylor's motivation in writing his paper, coincidentally,
was to correct an under-estimation of tidal dissipation!
He shows, assuming that the height and the current both vary sinusoidally,
that the flux of power passing into the Irish Sea
is
\beq
\frac{1}{2} g \rho U w d h \cos \frac{2 \pi}{T}( T_1- T_0 ) ,
\label{eqTay}
\eeq
where $U=1.63$\,m/s is the peak tidal flow,
$d=68$\,m is the depth of the channel,
$h=1.45$\,m is the average tidal height (the half-range)
along the line of width $w=80$\,km;
$T_1$ is the time of high water,
$T$ is the period of lunar tides, and
$T_0$ is the time of maximum current.
$U$ and $h$ are both proportional to amplitude, hence amplitude-squared.
Taylor concluded that a power of 64\,GW flowed into the
Irish Sea.
His formula (\ref{eqTay}) agrees with equation (\ref{eqP}).
He also estimated that three-quarters of this power was
dissipated in bottom-friction in the Irish Sea,
and one quarter re-emerged in a wave travelling in the
opposite direction.
This counter-travelling wave causes the rate of progress of high tides
along the coast to be different from the velocity of the tidal waves;
the relationship between the two velocities
depends on the phase difference between
the two waves.
Taylor also analysed the effect of
the moon itself on the energy
flow in the Irish Sea, finding its contribution
to be small (less than 10\% of the incoming flux above).
He estimated that the average dissipation rate in the Irish Sea
was 1.5\,\Wmm\ of sea floor.
% flow in 1640 , work done ON moon 110, dissipation 1530
Taylor's analysis includes several independent tests
and verifications of his model.
Taylor suggests that a reasonable model of frictional dissipation in
the Irish Sea (and for winds on Salisbury plain) is
\[
\mbox{power} = k \rho v^3 \ \mbox{(per unit area)},
\]
with $k=0.002$.
\citet{Flather1976} built a detailed numerical model of the
lunar tide, chopping the continental shelf around the British Isles into
roughly a thousand square cells.
Their friction model has mean dissipation
\[
\mbox{power} = k \rho v^3 \ \mbox{(per unit area)},
\]
with $k=0.0025--0.003$.
Flather estimates that the total average power entering this region
is 215\,GW.
According to his model,
180\,GW enters the gap between France and Ireland.
From Northern Ireland round to Shetland, the incoming power is 49\,GW.
Between Shetland and Norway there is a net loss of 5\,GW.
Measurements made over ten years near the edge of the continental shelf
by \citet{Cartwright1980}
verified and improved on Flather's estimates.
Their experiments indicate that the average M2
power transmission was 60\,GW between Malin Head (Ireland) and
\ind{Flor{\o}} (Norway)
and 190\,GW between Valentia (Ireland) and the Brittany coast near Ouessant.
The power entering the Irish Sea was found to be
45\,GW, and entering the North Sea
via the Dover Straits, 16.7\,GW.
Near the Orkneys the incoming powers
are 14\,GW and 12\,GW.
They try to estimate
the loss through bottom friction too
(using $k=0.0025$) and they
estimate that there is less dissipation
in the North Sea and Scottish waters
(40\,GW) than the incoming
power (77\,GW). They say they are not
sure exactly where the correction
to the loss arises.
On a later page they mention
finding that $k=0.005$
is sometimes a better model.
\section{Back to the shallow-water tidal wave model}
\subsection{Shelving}
If the depth $d$ decreases gradually and the width remains constant,
I'd guess there'll be minimal reflection
and the power of the wave will remain constant.
This means
$\sqrt{d} h^2$ is a constant, so we
deduce that the height of the tide scales with depth
as
$h \sim 1/d^{1/4}$.
\begin{figure}
\figuremargin{
\begin{center}
%% To get high resolution version, switch to the second line
{\mbox{\epsfxsize=4.995in\epsfbox{../../refs/DTIAtlas/TideLine.jpg.eps}}}
%{\mbox{\epsfxsize=4.995in\epsfbox{../../refs/DTIAtlas/TideLine.eps}}}
\end{center}
}{
\caption[a]{Two lines in the Atlantic.
Bathymetry data from DTI Atlas of Renewable Marine Resources.
\copyright\ Crown copyright.
}
\label{fig.TideLine}
}
\end{figure}
%% st kilda: springs = 2.8m, neaps = 1.3
%% shillay springs = 3.5m, neaps >= 1.5, maybe 1.8
%% /home/mackay/sustainable/refs/DTIAtlas> xv Z100.gif M100.gif
\subsection{Application to the UK}
Let's work out the power per unit length of wave crest
for some plausible figures.
If $d = 100\,m$,
and $h = 1$ or $2\,m$, the power per unit length of wave crest is
% in kW/m: p(h) = 0.5* 9.81**1.5 * 10 * h**2
\beq
\rho g^{3/2} \sqrt{ d } h^2/2
=
10000 \times 3 \times 10 \times (\mbox{1 or 4})/2
= \left\{ \begin{array}{ll} 150 \, \mbox{ kW/m } & (h = 1\,\m)\\
600 \, \mbox{ kW/m } & (h = 2\,\m) \end{array} \right. .
\eeq
These figures are impressive compared with the raw power per unit
length of Atlantic deep-water waves, 60--80\,kW/m gross
or 40--50\,kW/m net \citep{Mollison76}. Clearly,
%% 40 kW
the upper bound on tidal power is bigger by a factor of 10 or so
than that for waves.
\marginfig{
\begin{tabular}{cc}\toprule
$h$ & $ \rho g^{3/2} \sqrt{ d } h^2/2$ \\
(m) & (kW/m) \\ \midrule
0.9& 125 \\
1.0& 155 \\
1.2& 220 \\
1.5& 345 \\
1.75& 470 \\
2.0 & 600 \\ % 1230
2.25& 780 \\ % 1555
\bottomrule
\end{tabular}
\caption[a]{Power fluxes for depth $d=100\,m$. }
}
We can estimate the total incoming power from the Atlantic
by multiplying appropriate lengths by powers-per-unit-length.
My two lines A and B (\figref{fig.TideLine}) are both about 400\,km
long.
The tidal range (at springs) on line A at depth $d=100$\,m is $2h=3.5$\,m.
For neaps, I'll assume a range of $2h=1.8$\,m.
The tidal range (at springs) on line B at depth $d=100$\,m is $2h=4.5$\,m.
At neaps on line B, $2h=2.4$\,m seems a reasonable estimate.
Averaging the powers for springs and neaps,
the incoming tidal resource over line A is
120\,\GW,
and over line B, 195\,GW. A total of
%\beq
315\,\GW\
%\eeq
or
125\,kWh per person per day.
(Compare this with Salter's 100\,GW (peak) dissipation in Pentland Firth.)
In addition to lines A and B,
I should perhaps include a line joining Shetland to Norway:
much of the North Sea's tidal energy arrives across this line.\footnote{
See animation at
\url{http://www.math.uio.no/~bjorng/tidevannsmodeller/tidemod.html}
}
% Perhaps because of Coriolis force, the crests and troughs so generated
% tend to hug the British coast as they progress down the North Sea.
Let's throw in an extra 135\,GW for the North Sea (an overestimate
as I learned later from \citet{Cartwright1980}),
making a round total of
\beq
450\,\GW
\eeq
or 180\,kWh per person per day.
How much of this might conceivably be extracted?
If we say 5\%, and assume the conversion and transmission
steps are 50\% efficient, we arrive at
\beq
11\,\GW,
\eeq
or
4.5\,kWh per person per day.
\section*{Acknowledgements}
I thank Ted Evans,
Denis Mollison, Trevor Whittaker, Stephen Salter, and
Dan Kelley for helpful discussions.
\bibliography{cftbibs}%/home/mackay/bibs}
\marginfig{
\begin{center}
{\mbox{\epsfxsize=50mm\epsfbox{../../images/PUBLICDOMAIN/maps/northseaTID.eps}}}\\
\end{center}
\caption[a]{Average tidal powers measured by \citet{Cartwright1980}.
}
\label{fig.TideLine2}
}
\newpage
\thispagestyle{empty}
\input{PowerGrid.tex}
\end{document}
\noindent
\rule{\textwidth}{1pt}
%\ENDfullpagewidth
\input{PowerSheet1.tex}