%%\label{ch.air2}
\myquote{What we need to do is to look at how you make air travel more energy efficient, how you develop the new fuels that will allow us to burn less energy and emit less.}{Tony Blair}
%% http://www.metro.co.uk/news/article.html?in_article_id=32127&in_page_id=34
%% 9 Jan 2007
%%% _flight.tex has further notes
\myquote{Hoping for the best is not a policy, it is a delusion.}{Emily Armistead, \ind{Greenpeace}}
\marginfig{
%\begin{figure}
%\figuremargin{
\begin{center}
\begin{tabular}{@{}c@{}}
\lowres%
{\mbox{\epsfxsize=53mm\epsfbox{../../images/PUBLICDOMAIN/Arctic_ternsSmall.jpg.eps}}}%
{\mbox{\epsfxsize=53mm\epsfbox{../../images/PUBLICDOMAIN/Arctic_terns.jpg.eps}}}\\
\lowres%
{\mbox{\epsfxsize=53mm\epsfbox{../../images/PUBLICDOMAIN/BartailedGodwitSmall.jpg.eps}}}%
{\mbox{\epsfxsize=53mm\epsfbox{../../images/PUBLICDOMAIN/BartailedGodwit.jpg.eps}}}\\
\lowres%
{\mbox{\epsfxsize=53mm\epsfbox{../../images/PUBLICDOMAIN/AirForceOneSmall.jpg.eps}}}%
{\mbox{\epsfxsize=53mm\epsfbox{../../images/PUBLICDOMAIN/AirForceOne.jpg.eps}}}\\
\end{tabular}
\end{center}
\caption[a]{
Birds:\index{bird}
two Arctic terns, a bar-tailed godwit, and a Boeing \ind{747}.
}
}%
\noindent
What are the fundamental limits of travel by flying?
Does the physics of flight require an unavoidable
use of a certain amount of energy, per \tonne, per kilometre flown?
What's the maximum distance a 300-\tonne\ Boeing 747 can fly?
What about a 1-kg
bar-tailed godwit or a
100-gram Arctic tern?
Just as \chref{ch.car}, in which we estimated consumption by cars,
% invalid carriages,
was followed by \chref{ch.car2}, offering a model of where the energy
goes in cars,
% invalid carriages,
this chapter fills out \chref{ch.air},
discussing where the energy goes in planes.
%%%% This discussion may interest you, because
% You'll be able to answer the question
% `if I take 30\,\kg\ less luggage with me on my trip,
% does that make any difference to the energy consumed by the
% plane?'
The only physics required
% you'll need to understand this
is \ind{Newton's laws}\index{physics!Newton's laws}
of motion, which I'll describe when they're needed.
This discussion will allow us to answer questions such as
``would air travel consume much less energy if
we travelled in slower propellor-driven planes?''\index{myth!planes should fly slower}
There's a lot of equations ahead: I hope you enjoy them!\index{flight!myth about going slower}\index{plane!myth about flying slower}
% But if not, feel free to skip to the next chapter (\pref{chsolar}).
\section{How to fly}
Planes (and birds) move through air, so, just like cars and trains,
they experience a drag force, and much of the energy
guzzled by a plane goes into pushing the plane along
against this force.\index{bird!theory of} Additionally, unlike cars and trains,
planes have to
expend energy {\em{in order to stay up}}.
Planes stay up by throwing air down.
When the plane pushes down on air, the air pushes up on
the plane (because Newton's third law tells it to).
As long as this upward
push, which is called lift,
is big enough to balance the downward weight of the plane, the plane
avoids plummeting downwards.
When the plane throws air down, it gives that air kinetic energy.
So creating lift requires energy.
The total power required by the plane is the sum of the
power required to create lift and the power required to
overcome ordinary drag. (The power required to create lift is usually
called ``induced drag,'' by the way. But I'll call it the lift power, $P_{\rm lift}$.)
% \marginpar{*** \ \ \ Warning \ \ \ ***}
% \marginpar{*** Equations ahead ***}
The two equations we'll need, in order to work out a theory of flight,
are Newton's second law:
\begin{equation}
\mbox{force} = \mbox{rate of change of momentum},
\end{equation}
and Newton's third law, which I just mentioned:
\begin{equation}
\mbox{force exerted on A by B} = -\,
\mbox{force exerted on B by A}.
\end{equation}
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
{\em{equal}\/} to the power required to overcome drag.
So the requirement to ``stay up'' {\em{doubles}\/} the power required.
\begin{figure}
\figuremargin{
\begin{center}
\begin{tabular}{ll}
\makebox[0in][l]{\sf Before} &
{\mbox{\epsfbox{crosspad/plane2c.1}}} \\
\makebox[0in][l]{\sf After} &
\hspace*{1.01in}{\mbox{\epsfbox{crosspad/plane2c.2}}}\\
\end{tabular}
\end{center}
}{
\caption[a]{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.
}
\label{fig.sausage}
}
\end{figure}
\begin{figure}
\figuremargin{\small
% mpost earth.mp ; tex wrapper ; dvips wrapper ; gv wrapper
\begin{center}
\begin{tabular}{cc}
\raisebox{3mm}{\mbox{\epsfbox{crosspad/plane3a.1}}} &
{\mbox{\epsfbox{crosspad/plane3a.2}}}\\
{\sf Cartoon} &
{\sf A little closer to reality} \\
\end{tabular}
\end{center}
}{
\caption[a]{Our cartoon assumes that
the plane leaves
a \ind{sausage} of air moving down in its wake.
% The diameter of the sausage is roughly equal to the wingspan
% of the plane.
% cross-sectional area $\As$ is r
A realistic picture involves a more complex
swirling flow. For the real thing, see \protect\figref{fig.planeflow}.
}
\label{fig.swirl}
}
\end{figure}
%\subsection{The physics of flight}
Let's make a cartoon of the lift force on
a plane moving at speed $v$.\index{cartoon!flight}
In a time $t$ the plane moves a distance $vt$ and leaves
behind it a sausage of downward-moving air (\figref{fig.sausage}).
We'll call the cross-sectional area of this sausage \As.
This sausage's diameter is roughly equal to the \ind{wing}span $w$
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 interesting
than this sausage picture: each wing tip leaves behind it a \ind{vortex},
with the air between the wingtips moving down fast, and the
air beyond (outside)
the wingtips moving up (figures \ref{fig.swirl} \& \ref{fig.planeflow}).
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.\index{bird!formation flying}\index{formation flying}\index{flight!formation}
Anyway, let's get back to our \ind{sausage}.
\marginfig{
\begin{center}
\begin{tabular}{c}
\lowres%
{\mbox{\epsfxsize=50mm\epsfbox{../../images/PUBLICDOMAIN/Airplane_vortex_tiny.jpg.eps}}}%
{\mbox{\epsfxsize=50mm\epsfbox{../../images/PUBLICDOMAIN/Airplane_vortex_small.jpg.eps}}} \\
\end{tabular}
\end{center}
\caption[a]{
Air flow behind a plane.
Photo by NASA Langley Research Center.
}\label{fig.planeflow}
}
The sausage's mass is
\beq
m_{\rm sausage} = \mbox{density} \times \mbox{volume}
= \rho vt \As.
\eeq
%
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 its weight $mg$.
The downward momentum of the sausage created in time $t$ is
\beq
\mbox{mass} \times
\mbox{velocity}
=
m_{\rm sausage} u = \rho vt \As u.
\eeq
And by Newton's laws this must equal the momentum
delivered by the plane's weight in time $t$, namely,
\beq
m g t.
\eeq
Rearranging this equation,
\beq
\rho vt \As u = m g t ,
\eeq
we can solve for the
required downward sausage speed,
\[
u = \frac{m g }{ \rho v \As } .
\]
Interesting! The sausage speed is {\em{inversely}\/} related
to the plane's speed $v$. A slow-moving plane has to throw
down air
% the air that it encounters
harder than a fast-moving plane,
because it encounters less air per unit time.
That's why landing planes, travelling slowly,
have to extend their flaps: so as to create a larger and steeper
wing that deflects air more.
What's the energetic cost of pushing the sausage down at the required speed $u$?
The power required is
\beqan
P_{\rm lift} &=& \frac{ \mbox{kinetic energy of sausage} }
{ \mbox{time} }
\\
&=& \frac{1}{t} \frac{1}{2} m_{\rm sausage} u^2 \\
&=& \frac{1}{2t} \rho vt \As \left( \frac{m g }{ \rho v \As } \right)^2
\\
&=&
\frac{1}{2} \frac{(mg)^2}{\rho v \As}
.
\eeqan
% Notice that this lift-related power scales as the weight of the
% plane {\em{squared}}.
The total power required to keep the plane going is the sum of
the drag power and the lift power:
% and the power required to overcome drag:
\beqan
P_{\rm total} &=& P_{\rm drag} + P_{\rm lift}\\
&=&
\frac{1}{2} c_{\rm d} \rho \Ap v^3
+ \frac{1}{2} \frac{(mg)^2}{\rho v \As}
,
\eeqan
where $\Ap$ is the frontal area of the plane and $c_{\rm d}$
is its drag coefficient (as in \chref{ch.cars2}).
The fuel-efficiency of the plane, expressed as the
energy per distance travelled, would be
\beq
\left. \frac{ \mbox{energy} }{ \mbox{distance} }\right|_{\rm ideal} =
\frac{ P_{\rm total} }{ v } =
\frac{1}{2} c_{\rm d} \rho \Ap v^2
+ \frac{1}{2} \frac{(mg)^2}{\rho v^2 \As}
,
\label{eq.forceplane}
\eeq
if the plane turned its fuel's power into drag power
and lift power perfectly efficiently. (Incidentally,
another name for ``energy per distance travelled'' is ``force,''
%
and we can recognize the two terms above as
the drag force $ \frac{1}{2} c_{\rm d} \rho \Ap v^2$
and the lift-related force
%% (also known as the induced drag)
% [CHECK]
$\frac{1}{2} \frac{(mg)^2}{\rho v^2 \As}$.
The sum is the force, or ``thrust,'' that specifies exactly
how hard the engines have to push.)
\amarginfig{t}{% produced by gnuplot < gnu.flight
\small \begin{center}
\sf thrust (kN) \hfill \,\hspace*{1cm} \\[1.5mm]
\mono%%% \raisebox{-2mm}{$v$}
{\epsfxsize=53mm\epsfbox{figs/flightv.eps}}%
{\epsfxsize=53mm\epsfbox{figs/flightvC.eps}}%
\\
\hfill speed (m/s)
\end{center}
\caption[a]{The force required to keep a plane moving,
as a function of its speed $v$,
is the sum of an ordinary
drag force $ \frac{1}{2} c_{\rm d} \rho \Ap v^2$ -- which increases
with speed --
and the lift-related force (also known as the induced drag)
$\frac{1}{2} \frac{(mg)^2}{\rho v^2 \As}$ -- which decreases
with speed.
There is an ideal speed, $v_{\rm optimal}$, at which the
force required is minimized. The force is an energy per distance,
so minimizing the force also minimizes the fuel per distance.
To optimize the
fuel efficiency, fly at $v_{\rm optimal}$.
This graph shows our cartoon's estimate of the
thrust required, in kilonewtons, for a
Boeing \ind{747}\nlabel{p747da} of mass 319\,t, wingspan 64.4\,m, drag coefficient 0.03,
and frontal area 180\,m$^2$, travelling
in air of density $\rho = 0.41\,$kg/m$^3$
% .4135
(the density at a height of 10\,km),
as a function of its speed $v$ in m/s.
Our model has an optimal speed $v_{\rm optimal} = 220\,\m/\s$ (540\,mph).
%% sausages
For a cartoon based on \ind{sausage}s, this is a good match to real life!}
\label{fig.forcesum}
}%
Real jet engines have an efficiency of about $\epsilon = 1/3$,\nlabel{peffEng}
so the energy-per-distance of a plane travelling at speed $v$ is
\beq
\frac{ \mbox{energy} }{ \mbox{distance} } =
\frac{1}{\epsilon} \left(
\frac{1}{2} c_{\rm d} \rho \Ap v^2
+ \frac{1}{2} \frac{(mg)^2}{\rho v^2 \As}
\right).
\eeq
This energy-per-distance is fairly complicated; but
it simplifies greatly if we assume that the plane is {\em{designed}\/}
to fly
at the speed that {\em{minimizes}\/} the energy-per-distance.
The energy-per-distance, you see,
has got a sweet-spot as a function of $v$ (figure \ref{fig.forcesum}).
The sum of the two quantities
$ \frac{1}{2} c_{\rm d} \rho \Ap v^2$ and
$ \frac{1}{2} \frac{(mg)^2}{\rho v^2 \As} $
is smallest when the two quantities are equal.
This phenomenon is delightfully common in physics and engineering:
two things that don't obviously {\em{have}\/} to be equal {\em{are}\/}
actually equal, or equal within a factor of 2.
% For example, in
% Newtonian mechanics,
% the average kinetic energy of a pendulum,
% and the average of its potential energy (relative to its energy
% when hanging vertically) are equal;
% in thermodynamics, the
% average kinetic energy of a molecule in the air is exactly
% half the average of its potential energy (relative to zero defined by
% the earth's surface); and
% another engineering example ....
So, this equality principle tells us that the optimum speed
for the plane is such that
\beq
c_{\rm d} \rho \Ap v^2 = \frac{(mg)^2}{\rho v^2 \As} ,
\eeq
\ie,
%\beq
% v_{\rm opt}^4 = \frac{(mg)^2}{\rho^2 \As c_{\rm d} \Ap } ,
%\eeq
\beq
\rho v_{\rm opt}^2 = \frac{mg}{ \sqrt{c_{\rm d} \Ap \As } } ,
\label{eq.optV}
\eeq
%\beq
% v_{\rm opt} = \left( \frac{(mg)^2}{\rho^2 \As c_{\rm d} \Ap }
%\right)^{1/4}.
%\eeq
%\beq
% v_{\rm opt} =
% \frac{(mg)^{1/2}}{\rho^{1/2} w^{1/2}
% ( c_{\rm d} \Ap )^{1/4} }.
%\eeq
This defines the optimum speed if our \ind{cartoon of flight} is accurate; the
cartoon breaks down if the engine efficiency $\epsilon$ depends
significantly on speed, or if the speed of the plane exceeds
the speed of sound (330\,m/s); above the speed of sound, we would need
a different model of drag and lift.
Let's check our model by seeing what it predicts is the
optimum speed for
a 747 and for an \ind{albatross}.
%% nonstop flight record march 23-24 Washington to Cape City??
%% 16560km. range
%% cabin width 6.1m, vertical height 10m
We must take care to use the correct air-density: if
we want to estimate the optimum cruising speed for a 747 at
$30\,000$ feet, we must remember that air density drops with
increasing \ind{altitude} $z$
as $\exp ( - m g z / kT )$, where $m$ is the mass of \ind{nitrogen} or \ind{oxygen}
molecules, and $kT$ is the thermal energy (\ind{Boltzmann}'s constant
times absolute temperature).
The density is about $3$ times smaller at that
altitude.%
%% I get exp(-1.007) when I use T=300, m=28 amu, g=9.81... factor of 2.737
%% but it varies with temperature...
%% The
\begin{table}
\figuremargin{
\begin{center}\begin{tabular}{lccc}\toprule
{\sc{Bird}} & & 747 & Albatross \\ \midrule
Designer & & Boeing & natural selection\\
% Mass &$m$ & 181,000\,\kg & 8\,\kg \\ % without fuel
% 170,000 without fuel
% Fuel volume 240,000\,\litre
% Fuel mass 240,000\,\litre * 0.7 kg/l = 168\,000 kg
% passenger: 80kg. luggage: 40. food containers, water, drinks? 10 kg each?
% Mass of passengers and luggage 416 * ( 80 + 40 +10) = 54080
%% * For a typical international flight, one 747 operator uses about 5.5 tons (5,000 kg) of food supplies and more than 50,000 in-flight service items.
%% http://www.boeing.com/commercial/747family/pf/pf_facts.html
%%
Mass (fully-laden) &$m$ & 363\,000\,\kg & 8\,\kg \\
%% including fuel
%%% from www.boeing.com/
% % http://www.boeing.com/commercial/747family/pf/pf_classics.html
%%% max takeoff weight for a 747-100 was 333400kg. fuel capacity 183380l , range 9,800km. (452 passengers max)
%%% TYPICAL cruise speed = mach 0.84 = 555mph = 895 km/h
%%% max takeoff weight for a 747-200 was 374850kg. fuel capacity 199000l , range 12,700km. (452 passengers max)
%%% TYPICAL cruise speed = mach 0.84 = 555mph = 895 km/h
%%% max takeoff weight for a 747-300 was 374850kg. fuel capacity 199000l , range 12,400km. (496 passengers max)
%%% TYPICAL cruise speed = mach 0.85 = 565mph = 910 km/h
%%% http://www.boeing.com/commercial/747family/pf/pf_400er_prod.html
%%% max takeoff weight for a 747-400ER was 412775kg. fuel capacity 241000l , range 14,200km. (524 passengers max, 416 also typical) (has 2 extra fuel tanks in a cargo hold)
%%% 416 seems to be the standard number
%%% TYPICAL cruise speed = mach 0.855 = 567mph = 912 km/h
%%
%%
%% according to narita airport site,
%% maximum takeof weights of 7474s are
%% 335000, and 322000, for two different 747s
%% I estimate Mass at takeoff is 403000
%% I think a reasonable est for the halfway mass is 319000.
%% pr 181000 + 0.5*168000 + 54080
Wingspan&$w$& 64.4\,\m & 3.3\,\m \\
Area$^\star$ &$\Ap$& 180\,$\m^2$ & $0.09\,\m^2$ \\
%% filling factor is 0.0434
Density&$\rho$& $0.4\,\kg/\m^3$ & $1.2\,\kg/\m^3$ \\
Drag coefficient & $c_{\rm d}$& 0.03 & 0.1 \\
\midrule
Optimum speed & $v_{\rm opt}$ & 220\,\m/\s & 14\,\m/\s \\
& & = 540\,\mph & = 32\,\mph \\
\bottomrule
\end{tabular}
\end{center}
}{
\caption[a]{Estimating the optimal speeds for a \ind{jumbo jet} and
an \ind{albatross}.
%% CHECK whether the mass includes passengers and cargo and fuel.
%% Find out the actual air density at 30000 feet and recompute.
% http://www.pdas.com/m1.htm
%% see pdas.tex
% Density at 10km height is 0.4135 kg/m$^3$
% v_{\rm opt} =
% \frac{(mg)^{1/2}}{\rho^{1/2} w^{1/2}
% ( c_{\rm d} A )^{1/4} }.
% pr ( 363000.0 * 9.81 )**0.5 / ( 1.3 *exp(-1.007) * 64.4 )**0.5 / (0.03 * 180)**0.25
% 223 m/s
% * 3600 / 1500
% REDO THIS EXACTLY
$\star$ Frontal area estimated for 747 by taking cabin width (6.1\,m)
times estimated height of body (10\,m) and
adding double to allow for
the frontal area of
engines, wings, and tail;
for albatross, frontal area of 1 square foot estimated from a
photograph.
% Drag coefficient for 747 from
% www.aerospaceweb.org
% Cessna is same. Learjet is 0.02.
% Albatross assumed to be same as 747, though J Exp Biol paper by Hedenstrom and Liechti vol 204 (6) 1167-1175
% says ``large birds have cd=0.2''
%% C M Bishop paper says cd=0.1 is conceivable.
%% What does the nice MIT book say?
}
\label{tab747}
}
\end{table}%
The predicted optimal speeds (\tabref{tab747})
are more accurate than we have a right
to expect!
The 747's optimal speed is predicted to be 540\,\mph,
and the albatross's, 32\,\mph\ -- both very close to the true cruising
speeds of the two birds\index{bird!speed}\index{flight!optimal speed}
(560\,mph and 30--55\,mph respectively).
%% animals.
%% cruising speed of 747 is cruising speed 907km/h
%% mach 0.85
% Economical cruising speed 747: 900km/h.
% http://www.boeing.com/commercial/747family/pf/pf_400_prod.html
%% \section*{Notes}
Let's explore a few more predictions of our cartoon.
We can check whether the force (\ref{eq.forceplane}) is compatible with the
known thrust of the 747. Remembering that at the optimal speed, the two
forces are equal, we just need to pick one of them and double it:
\beqan
\mbox{force} &=&
\left. \frac{ \mbox{energy} }{ \mbox{distance} }\right|_{\rm ideal} =
\frac{1}{2} c_{\rm d} \rho \Ap v^2
+ \frac{1}{2} \frac{(mg)^2}{\rho v^2 \As}
\\
&=&
c_{\rm d} \rho \Ap v_{\rm opt}^2
%\\
%&=&
% c_{\rm d} \rho \Ap \left( \frac{(mg)^{1/2}}{\rho^{1/2} w^{1/2}
% ( c_{\rm d} \Ap )^{1/4} }\right)^2
\\
&=&
c_{\rm d} \rho \Ap \frac{mg}{\rho ( c_{\rm d} \Ap \As )^{1/2} }
\\
&=&
\left(\frac{c_{\rm d} \Ap}{\As}\right)^{1/2} {mg} .
\label{eq.forceplane2}
\eeqan
%% WHICH IS INDEPENDENT OF RHO
Let's define the filling factor $f_A$ to be the area ratio:
\beq
f_A = \frac{\Ap}{\As} .
\eeq
\marginfig{
% \begin{figure}
\begin{center}
\begin{tabular}{@{}c@{}}
{\mbox{\epsfxsize=53mm\epsfbox{figs/plane3.eps}} }\\
\end{tabular}
\end{center}
% }{
\caption[a]{Frontal view of a Boeing 747, used to estimate the
frontal area $\Ap$ of the plane.
% This area is probably an
% underestimate of the effective frontal area, since the air doesn't flow
% straight past the wing.
The square has area $\As$ (the square of the wingspan). }
\label{fig.plane3}
}%
%% plane3.eps Frontal view of a Boeing 747. This area is probably an
%% underestimate of the effective frontal area, since the air doesn't flow
%% straight past the wing. ?
(Think of $f_A$ as the fraction of the square occupied by
the plane in figure \ref{fig.plane3}.)
%% -- perhaps $f\simeq 0.07$.) I am using 0.04.
Then
\beqan
\mbox{force} &=&
(c_{\rm d} f_A)^{1/2} {(mg)} .
\label{eq.forceplane3}
\eeqan
Interesting!
Independent of the density of the fluid through which
the plane flies, the required \ind{thrust}
(for a plane travelling at the optimal speed)
is just a dimensionless
constant $(c_{\rm d} f_A)^{1/2}$ times the weight of the plane.
This constant, by the way, is known as the \ind{drag-to-lift ratio}
of\index{lift-to-drag ratio} the plane.
(The lift-to-drag ratio has a few other names:
the \ind{glide number}, \ind{glide ratio},
\ind{aerodynamic efficiency},
\amargintab{t}{
\begin{tabular}{lr} \toprule
Airbus A320 & 17 \\
Boeing 767-200 & 19 \\
Boeing 747-100 & 18\\ % 17.7\\
Common Tern
% ({\em{sterna hirundo}})}
& 12\\
Albatross
% ({\em{diomeda exulans}})
& 20\\
\bottomrule
\end{tabular}
\caption[a]{Lift-to-drag ratios.
% Bigger is better.
% http://aerodyn.org/HighLift/ld-tables.html
% standard glideslope for approaching heathrow is 3 degrees
% That corresponds to 19.
}\label{tabLtD}
}%
or \ind{finesse}; typical values are shown in \tabref{tabLtD}.)
% Does my $f_A$ agree with the value obtained by computing $f_A$!?
Taking the jumbo jet's figures, $c_{\rm d} \simeq 0.03$ and
%% $f_A \simeq 0.07$,
%% the above is a guess from the picture; I don't think I ever used 0.07 in calculations
$f_A \simeq 0.04$,
%% 0.0434
%% 0.029 from using the figs I assumed before
we find the required \ind{thrust} is
\beq
(c_{\rm d} f_A)^{1/2} \, {mg} = 0.036\, mg = 130\,\kN .
\eeq
%% pr 363000.0*9.81 * (0.03 * 0.0434)**0.5
%% 128493.48982922
%% m = 363000 Kg fully laden << check whether it was including cargo etc
How does this agree with the 747's spec sheets?
In fact each of the 4 engines has a maximum thrust of about
250\,\kN, but this maximum thrust is used only during take-off.
%% http://www.boeing.com/commercial/747family/pf/pf_400er_prod.html
%% For a 747-400, 276,000N. (EACH according to wikipedia)
During cruise, the thrust is much smaller:
% http://www.anirudh.net/seminar/html/
% A 389kN engine will cruise at 70kN.
% A 276kN engine at 50.4kN.
the thrust of a cruising 747 is 200\,kN, just 50\% more than
our cartoon suggested.
%%
%% I wonder if the engines operate at maximum all the way, or just at takeoff?
%% When a 747 pilot lands, he aims to have
%% less than 40,000 kg of fuel on board, but is obliged to have...?
%% http://www.hoppie.nl/flightline/old/fl/logs.php/logs.php?forg=KBFI
%% see also notes/flight.tex
Our cartoon is a little bit off because our estimate of the
drag-to-lift ratio was a little bit low.
\amarginfig{t}{
\begin{center}
\begin{tabular}{@{}c@{}}
\lowres{\mbox{\epsfxsize=53mm\epsfbox{../../images/Cessna.310Small.jpg.eps}}}%
{\mbox{\epsfxsize=53mm\epsfbox{../../images/Cessna.310.jpg.eps}}}
\\
\end{tabular} \\
\end{center}
\caption[a]{
\ind{Cessna} 310N:
{\eccol{60\,kWh per 100 passenger-km}}.
A Cessna 310 Turbo carries
6 passengers (including 1 pilot)
at a speed of
370\,km/h.
Photograph by Adrian Pingstone.
}
}This thrust can be used directly to deduce the
transport efficiency\index{transport!efficiency!plane}
achieved by any plane.
We can work out two sorts of transport efficiency:
the energy cost of moving {\em{weight}\/} around, measured in
% passenger--miles per gallon
kWh per {\tonne}-kilometre; and the energy cost of
moving people, measured in kWh per 100 passenger-kilometres.
\subsection{Efficiency in weight terms}
Thrust is a force, and a force is an
% useful
energy per unit distance.
The total energy used per unit distance is bigger by a factor
($1/\epsilon$), where $\epsilon$ is the efficiency of the engine, which we'll
take to be $1/3$.
Here's the {gross transport cost}, defined to be the
{energy per unit weight (of the entire craft) per unit distance}:
\beqan
\mbox{transport cost}
% ({kWh per {\tonne}--kilometre})
% \nonumber \\
&=&
\frac{ 1 }
{\epsilon}
\frac{ \mbox{force} }
{\mbox{mass}}
\\ & = & \frac{ 1 }
{\epsilon}
\frac{ (c_{\rm d} f_A)^{1/2} {mg} }
{ m}
\\ & = &
\frac{ (c_{\rm d} f_A)^{1/2} }
{\epsilon}
g .
\label{eqTe}
\eeqan
So the transport cost is just a dimensionless quantity (related
to a plane's shape and its engine's efficiency), multiplied by
$g$, the acceleration due to gravity.
Notice that this gross transport cost applies to all planes, but depends only
on three simple properties of the plane: its drag coefficient, the shape of the plane,
and its engine efficiency.
It doesn't depend on the size of the plane, nor on its weight, nor on the density
of air.
If we plug in $\epsilon = 1/3$ and assume a lift-to-drag ratio of 20
% , $c_{\rm d} = 0.03$, and $f_A \simeq 0.04$,
we find the gross transport cost of {\em{any}\/} plane, according to our cartoon, is
%%pr 3.0 * sqrt( 0.1*0.05 )
%% 0.212132034355964
%%pr 3.0 * sqrt( 0.03*0.04 )
%% 0.1
%%pr 3.0 / 20
%% 0.1
\[
0.15 \, g
\]
% WAS 0.212132034355 * 9.81 ((m / s) / s) = 0.578059794 (kWh per {\tonne}) per km
% IS 0.2832
% IS NOW: 1.47 J/kg/m -> 0.41
or
\[
0.4 \, \kWh / \mbox{\tonne-km}.
\]
\subsection{Can planes be improved?}
If engine efficiency can be boosted only a tiny bit by technological
progress, and if the shape of the plane has already been essentially
perfected, then there is little that can be done about the dimensionless quantity.
The transport efficiency is close to its physical limit.
The aerodynamics community say that the shape of planes could
be improved a little by a switch to blended-wing bodies, and that the drag coefficient
could be reduced a little by
% a technology called
\ind{laminar flow control}, a technology that reduces the growth of turbulence
over a wing by sucking a little air through small perforations in
the surface
\citep{BraslowBook}.
Adding laminar flow control to existing planes
would deliver a 15\% improvement in drag coefficient,
and the change of shape to blended-wing bodies is predicted to improve the
drag coefficient by about 18\%
\citep{GreenAviation}.
And \eqref{eqTe} says that the transport cost is proportional to the
square root of the drag coefficient, so improvements of $\cd$ by
15\% or 18\% would improve transport cost by 7.5\% and 9\% respectively.
\marginfig{
\begin{center}
\begin{tabular}{@{}c@{}}
\lowres{\mbox{\epsfxsize=53mm\epsfbox{../../images/Learjet.60Small.jpg.eps}}}%
{\mbox{\epsfxsize=53mm\epsfbox{../../images/Learjet.60.eps}}}% was jpg and crashed
\\
\end{tabular} \\
\end{center}
\caption[a]{
``Fasten your cufflinks.''
% A Learjet 60
A Bombardier \ind{Learjet} 60XR
carrying 8 passengers at 780\,km/h
% \myurl{http://www.bombardier.com/}
has a transport cost of
{\eccol{150\,kWh per 100 passenger-km}}.
Photograph by Adrian Pingstone.
}
}
%Learjet 60 parked at Filton Airfield, Bristol, England.
%Taken by Adrian Pingstone in September 2004 and released to the public domain.
% Cessna 310N (UK registration G-YHPV, year of build 1968) at Kemble Airfield, Gloucestershire, England.
% Photographed by Adrian Pingstone in July 2005 and released to the public domain.
% \subsection{}
This gross transport cost is the energy cost of moving weight around, {\em{including the
weight of the plane itself}}.
To estimate the energy required to move freight by plane,\index{air freight!energy consumption}
per unit weight of freight,
we need to divide by the fraction that is cargo. For example, if a full
\ind{747} freighter is about $1/3$ cargo, then its transport cost
is
\[
0.45 \, g,
\]
or roughly 1.2\,kWh/\tonne-km.
%
% I watched a UK freight train, it has 22 full size containers and four half containers
This is just a little bigger than the transport cost of a truck, which is 1\,kWh/{{\tonne}-km}.
\subsection{Transport efficiency in terms of bodies}
Similarly, we can estimate a passenger transport-efficiency for a
\ind{747}.
\beqan
\lefteqn{\mbox{transport efficiency
({passenger--km per litre of fuel})
} } \nonumber \\
&=&
\mbox{number of passengers} \times \frac{ \mbox{energy per litre} }{
\frac{ \mbox{thrust} }
{\epsilon} }
\\ & = &
% jet fuel (135 000 btu) / US gal = 37 626 696.3 J / litre
%%% roughly 24\,000\,\kJ \, \per\, \litre
%%% 9 kWh/l is 32.4 MJ/l or 162 MJ/gal<> should use 37.6 MJ/l
\mbox{number of passengers} \times
\frac{ {\epsilon} \times \mbox{energy per litre} }
{ \mbox{thrust} }
\\
&=& 400 \times \frac{1}{3} \frac{ 38 \, \mbox{MJ} / \mbox{litre} }
{ 200\,000 \, \N }
\\
&=& 25 \,\mbox{passenger--km per litre}
\eeqan
This is a bit more efficient than a typical single-occupant car
% (40\,mpg)
(12\,\km \,\per\, \litre).
% 1 / (33 (miles per Imperial gallon)) = 11.6821994 km per litre
% We can find the actual efficiency of a 747
% by looking up its specifications:
%%% 747-???
% 10.45 kWh/l
%%% pr 100.0 * 199000 / 12700 / 416.0
%%% 3.46 l/100km each
%%% 747-400
%%% pr 100.0 * 240000 * 10.45 / 14205 / 416.0
%%% 42.44 kWh per 100 km
%%% pr 416 * 14205.0 / 240000
%%% 24.62 km per litre
% it is indeed 25\,passenger-km per litre.
So travelling by plane is more energy-efficient than car if
there are only one or two people in the car;
and cars are more efficient if there are three
or more passengers in the vehicle.
%%% _flight.tex further notes removed from here
\subsection{Key points}
We've covered quite a lot of ground! Let's recap the key ideas.
Half of the work done by a plane goes into {\em{staying up}};
the other half goes into {\em{keeping going}}.
The fuel efficiency at the optimal speed,
expressed as an energy-per-distance-travelled,
was found in the force (\ref{eq.forceplane3}), and
it was simply proportional to the weight of the plane; the constant
of proportionality is the drag-to-lift ratio, which is determined by
the shape of the plane. So whereas lowering speed-limits
for cars would reduce the energy consumed per distance travelled,
there is no point in considering speed-limits for planes. Planes that are up in the
air have optimal speeds, different for each plane, depending
on its weight, and they already go at their optimal speeds.
If you ordered a plane to go slower, its energy consumption would {\em{increase}}.
The only way to make a plane consume
fuel more efficiently is to put it on the ground and stop it.
Planes have been fantastically optimized, and there is no prospect
of significant improvements in plane efficiency.
(See pages \pageref{pPlaneTargets} and \pageref{pAirbus}
for further discussion of the notion
that new superjumbos are ``far more efficient'' than old jumbos;
and \pref{pTurboprop} for discussion of the notion that turboprops
are ``far more efficient'' than jets.)
\marginfig{
\begin{center}
\begin{tabular}{@{}c@{}}
\lowres{\mbox{\epsfxsize=53mm\epsfbox{../../images/Easyjet737SmallC.jpg.eps}}}%
{\mbox{\epsfxsize=53mm\epsfbox{../../images/Easyjet737LargeC.jpg.eps}}}
\\
\end{tabular} \\
\end{center}
\caption[a]{
Boeing 737-700:
{\eccol{30\,kWh per 100 passenger-km}}.
Photograph \copyright\ Tom Collins.
}
}
% Boeing 737-700:
% where is the calculation?
%% easyJet Boeing 737-700 G-EZJC departing Bristol Airport. Photo copyright Tom Collins
% Boeing 737-700,
% photo copyright Tom Collins.
% http://links.jstor.org/sici?sici=0908-8857(199606)27%3A2%3C118%3AMWSCCS%3E2.0.CO%3B2-R
% article saying that swas crossing iceland to scotland have to use
% fat weighing 25% of their lean body mass
\subsection{Range}
Another prediction we can make is, what's the range of a plane or bird
-- the biggest distance it can go without refuelling?\index{bird!range}\index{range!of bird}
You might think that bigger planes have a bigger range, but the prediction of
our model is startlingly simple.
The range of the plane, the maximum distance it can go before refuelling, is
proportional to its velocity and to the total energy of the fuel,
and inversely proportional to the rate at which it guzzles fuel:
\beq
\mbox{range} =
v_{\rm opt} \frac{\rm energy}{\rm power}
= \frac{\mbox{energy}\times \epsilon}{\rm force} .
\eeq
Now, the total energy of fuel is the calorific value of the fuel, $C$ (in joules per kilogram),
times its mass; and the mass of fuel is some fraction $f_{\rm fuel}$
of the total mass of the plane.
So
\beq
\mbox{range} = \frac{\mbox{energy} \, \epsilon}{\rm force}
= \frac{ C m \epsilon f_{\rm fuel}}{ (c_{\rm d} f_A)^{1/2} {(mg)} }
% = \frac{ C \epsilon f_{\rm fuel}}{ (c_{\rm d} f_A)^{1/2} {g} }
= \frac{ \epsilon f_{\rm fuel}}{ (c_{\rm d} f_A)^{1/2} }
\frac{C}{g} .
\eeq
It's hard to imagine a simpler prediction:
the range of any bird or plane is the product of a dimensionless
factor
$\left(\frac{ \epsilon f_{\rm fuel}}{ (c_{\rm d} f_A)^{1/2} }\right)$
which takes into account the engine efficiency, the drag coefficient, and
the bird's geometry, with
a fundamental distance,
\[
\frac{C}{g},
\]
which is a property of the fuel and gravity, and nothing else.
No bird size, no bird mass, no bird length, no bird width;
no dependence on the fluid density.
So what is this magic length?
It's the same distance whether the fuel
is goose fat or jet fuel: both these fuels are essentially
hydrocarbons (CH$_2$)$_n$.
Jet fuel has a calorific value of $C = 40$\,MJ per kg. The distance
associated with jet fuel is
\beq
d_{\rm Fuel} =
\frac{C}{g}
= {4000\,\km} .
\eeq
% Back to the birds.
The range of the bird is the intrinsic range of the fuel,
\amarginfignocaption{t}{
\small
You can think of $d_{\rm Fuel}$
as the distance that the fuel could throw itself if it
suddenly converted all its chemical energy to kinetic energy and launched
itself on a parabolic trajectory with no air resistance.
[To be precise, the distance achieved by the optimal parabola is
twice $C/g$.] This distance is also the {\em{vertical}\/} height to which the fuel could throw
itself if there were no air resistance.
Another amusing thing to notice is that
the calorific value of a fuel $C$, which I gave in joules
per kilogram, is also a squared-velocity (just as the energy-to-mass
ratio $E/m$ in \ind{Einstein}'s $E=mc^2$ is a squared-velocity, $c^2$):
$40\times 10^6$\,J per kg is $(6000\,\m/\s)^2$.
So one way to think about fat is ``\ind{fat is 6000 metres per second}.''
% 12\,000 miles per hour}''.
If you want to lose weight by going jogging, 6000 m/s
(12\,000 mph) is the speed you should aim for in order
to lose it all in one giant leap.
}%
4000\,km,
times a factor
$\left(\frac{ \epsilon f_{\rm fuel}}{ (c_{\rm d} f_A)^{1/2} }\right)$.
If our bird has engine efficiency $\epsilon=1/3$ and
% I switched to the
drag-to-lift ratio
$(c_{\rm d} f_A)^{1/2} \simeq 1/20$, and if nearly half of the bird is fuel
(a fully-laden \ind{747} is 46\% fuel),
we find that all birds and planes, of whatever size,
% (as long as they have roughly the same geometry),
% 1.0/6.0 / 0.03
% was 0.03 , changed to 1/20
% 1.0/6.0 / (1/20) * 4000
have the same range:
about three times the fuel's distance -- roughly 13\,000\,km.
%% actual mass of plane is 170 without fuel, and fuel is same again; (168)
%% takeoff weight fully laden is 363
This figure is again close to the true answer:
the nonstop \ind{flight record}\index{record, flight} for a \index{747!flight record}747
(set on March 23--24, 1989)
% 10500 miles is 16898 km
was a distance of 16\,560\,km.
And the claim that the range is independent of bird size is
supported by the observation that birds of all sizes, from great geese
down to dainty swallows and arctic tern, migrate
intercontinental distances.
%
The longest recorded non-stop flight by a bird was\index{bird!longest flight}
a distance of 11\,000\,km, by a \ind{bar-tailed godwit}\index{godwit}.\nlabel{pgodwit}
How far did \index{Fossett, Steve}{Steve Fossett} go in
the specially-designed
\ind{Scaled Composites} Model 311 \ind{Virgin} Atlantic \ind{GlobalFlyer}?
% 25766 feb 2006.
41\,467\,km.
\tinyurl{33ptcg}{http://www.stevefossett.com/html/main_pages/records.html}
% 6389.3 miles 42\,469\,km.
% The Longest Distance Aircraft Flight - The Absolute Non-Stop Distance Record
%% OK, how about the passenger-miles per gallon?
An unusual plane:
83\% of its take-off weight was fuel; the flight made careful use of the
jet-stream to boost its distance.
Fragile, the plane had several failures along the way.
One interesting point brought out by this cartoon:
if we ask ``what's the optimum air-density to fly in?'',
we find that the {\em{thrust}\/} required (\ref{eq.forceplane2}) at the optimum speed is independent
of the density. So our cartoon plane would be equally happy
to fly at any height; there isn't an optimum density;
the plane could achieve the same miles-per-gallon
in any density; but the optimum {\em{speed}\/} does depend on the
density ($v^2 \sim 1/\rho$, \eqref{eq.optV}).
So all else being equal, our cartoon plane would have the shortest
journey time if it flew in the lowest-density air possible.
Now real engines' efficiencies aren't independent of speed and air density.
As a plane gets lighter by burning fuel, our cartoon says
its optimal speed at\index{flight!optimum height}
a given density would reduce
($v^2 \sim mg/(\rho (c_{\rm d} \Ap\As)^{1/2})$).
So a plane travelling in air of constant density should
slow down a little as it gets lighter.
But a plane can both keep going at a {\em{constant speed}\/}
and continue flying
at its {\em{optimal}\/} speed if it increases its altitude so as to reduce the
air density.
%% As a long-distance plane gets lighter,
%% the optimal altitude for flight increases.
% typically from 31,000 feet to 39,000 feet by the end of the flight.
Next time you're on a long-distance flight, you could check
whether the \ind{pilot} increases the cruising height\index{height, of flight}
from, say, 31\,000 feet to 39\,000 feet by the end of the flight.
\subsection{How would a hydrogen plane perform?}
We've already argued that the efficiency\index{hydrogen!plane}
of flight, in terms of energy per {\tonne}-km,
is just a simple dimensionless number times $g$.
Changing the fuel isn't going to change this
fundamental argument. Hydrogen-powered planes
are worth discussing if we're hoping to reduce
climate-changing emissions. They might also have better
range. But don't expect them to be radically more
energy-efficient.
\subsection{Possible areas for improvement of
plane efficiency}
Formation flying\index{formation flying}\index{bird!formation flying}
in the style of \ind{geese} could
give a 10\% improvement in fuel efficiency (because the \ind{lift-to-drag
ratio} of the formation is higher than
that of a single aircraft), but this trick relies, of course, on the
geese wanting to migrate\index{migration} to the same destination at the same time.
Optimizing the hop lengths:\nlabel{OptHop}
long-range planes (designed for a range of say 15\,000\,km)
are not quite as fuel-efficient as shorter-range planes, because
they have to carry extra fuel, which makes less space for
cargo and passengers. It would be more energy-efficient
to fly shorter hops in shorter-range planes.
% source page 4.2.1 (p474) of greens.pdf GreenAviation
The sweet spot is when the hops are about 5000\,km long, so
typical long-distance journeys would have one or
two refuelling stops \citep{GreenAviation}. Multi-stage
long-distance flying might be about 15\% more fuel-efficient;
but of course it would introduce other costs.
\subsection{Eco-friendly aeroplanes}
Occasionally you may hear about people making
eco-friendly aeroplanes.
Earlier in this chapter, however, our cartoon
made the assertion that the transport cost of {\em{any}\/} plane
is about
\[
0.4 \, \kWh / \mbox{{\tonne}-km}.
\]
According to the cartoon, the only ways in which a plane could
significantly improve on this figure are to reduce air resistance
(perhaps by some new-fangled vacuum-cleaners-in-the-wings trick)
or to change the geometry of the plane (making it look more like
a glider, with immensely wide wings compared to the fuselage,
or getting rid of the fuselage altogether).
So, let's look at the latest news story about ``eco-friendly aviation''
and see whether one of these planes can beat
the $0.4$\,kWh per {{\tonne}-km} benchmark.
% it's a load of hype
If a plane uses less than 0.4\,kWh per {{\tonne}-km}, we might conclude that the cartoon is defective.
%% http://news.bbc.co.uk/1/low/technology/7384788.stm
% another article about ^^ electric plane see taurusPlane image
\marginfig{
\begin{center}
\begin{tabular}{@{}c@{}}
{\mbox{\epsfxsize=53mm\epsfbox{../../images/ELECTRA3.eps}}}\\
\end{tabular} \\
\end{center}
\caption[a]{
The Electra F-WMDJ:
\eccol{11\,kWh per 100\,\pkm}.
Photo by Jean--Bernard Gache.
\myurlb{www.apame.eu}{http://www.apame.eu/}
}
}
% energy coefficent 2% of petrol
% APAME Electra December 23rd, 2007
% http://www.timesonline.co.uk/tol/news/world/europe/article3123681.ece
The \ind{Electra}, a wood-and-fabric single-seater,
flew for 48 minutes for 50\,km around the southern Alps
on 23rd December, 2007
\tinyurl{6r32hf}{http://www.theaustralian.news.com.au/story/0,25197,23003236-23349,00.html}.
The Electra has a 9-m wingspan and an 18-kW
% http://www.acv05.fr/News05.html
% a 25\,hp 18.64
electric motor powered by 48\,kg of lithium-polymer batteries.
% Ms Lavrand said the fuel cost per hour of the Electra was E1 compared with about E60 ($100) for an equivalent petrol-driven machine.
The aircraft's take-off weight is
265\,kg
(134\,kg of aircraft, 47\,kg of batteries, and
84\,kg of human cargo).
% On 23rd December, 2007 it flew a distance of 50\,km.
% at 75\,km/h.
% Finesse 13.
If we assume that the battery's energy density was 130\,Wh/kg,
and that the flight used 90\% of a full charge (5.5\,kWh),
% doublecheck: 18kW * 48/60 is 14.4 kWh (if the engine goes full power)
% but it doesn't. It has extra power for climb.
% 11 kWh per 100 pkm
the transport cost was roughly
% 1e-3*130*0.9*47 / (50*0.265)
\[
0.4\,\kWh/\mbox{\tonne-km},
\]
which exactly matches our cartoon.
This electrical plane is
not a lower-energy plane than a normal fossil-sucker.
% Electraflyer - C all electric airplane based on a Moni Motorglider
% 5.6kWh (35kg) battery. blogger said range 1.5-2 hours at 70mph.
% where that info from?
% The batteries for the plane came from Valence Technology.
Of course, this doesn't mean that electric planes are
not interesting.
% the energy consumption of a plane is not the only issue.
If one could replace traditional planes by alternatives with equal
energy consumption but no carbon emissions, that would certainly be a useful
technology.
And, as a person-transporter, the Electra delivers
a respectable \eccol{11\,kWh per 100\,\pkm}, similar to the electric
car in our transport diagram on \pref{passenger}.
But in this book the bottom line is
always: ``where is the energy to come from?''
\subsection{Many boats are birds too}
Some time after writing this cartoon of flight,
I realized that it applies to more than just the
birds of the air -- it applies to hydrofoils,\index{hydrofoil}
and to other high-speed \ind{watercraft}\index{boats as planes}\index{planes!boats as an example of}
too -- all those\index{flight!boats that fly}
that ride higher in the water when moving.
\begin{figure}
\figuremargin{\small
\begin{tabular}{@{}cc}
\mbox{\epsfbox{metapost/hydrof.155}}&
\mbox{\epsfbox{metapost/hydrof.156}}
\\
{\sf side view} & {\sf front view} \\
\end{tabular}
}{
{\mbox{\epsfxsize=53mm\epsfbox{../../images/HydrofoilC.eps}}}\\
\caption[a]{Hydrofoil.\index{hydrofoil}
Photograph by Georgios Pazios.
}
\label{fig.hydrof}
}
\end{figure}
\Figref{fig.hydrof} shows the principle of the hydrofoil.
The weight of the craft is supported by a tilted
underwater wing, which may be quite tiny
compared with the craft.
The wing generates lift by throwing fluid down, just like the plane of
\figref{fig.sausage}.
If we assume that the drag is dominated by the drag on the wing, and that
the wing dimensions and vessel speed have been optimized to minimize the
energy expended per unit distance, then
the best possible
transport cost, in the sense of energy per {\tonne}-kilometre,
will be just the same as in
\eqref{eqTe}:
\beq
\frac{ (c_{\rm d} f_A)^{1/2} }
{\epsilon} g ,
\eeq
where $c_{\rm d}$ is the drag coefficient of the underwater wing, $f_A$ is
the dimensionless area ratio defined before, $\epsilon$ is the engine efficiency, and
$g$ is the acceleration due to gravity.
Perhaps $c_{\rm d}$ and $f_A$ are not quite
the same as those of an optimized aeroplane.
But the remarkable thing about this
theory is that it has no dependence on the density of
the fluid through which the wing is flying.
So our ballpark prediction is that the
transport cost (energy-per-distance-per-weight, including the vehicle weight)
of a hydrofoil is {\em the same\/} as the
transport cost of an aeroplane! Namely, roughly 0.4\,kWh per {\tonne-km}.
For vessels that skim the water surface, such as high-speed catamarans
and water-skiers, an accurate cartoon should also
include the energy going into making waves, but I'm tempted to guess that
this hydrofoil theory is still roughly right.
I've not yet found data on the transport-cost of a hydrofoil,
but some data for a passenger-carrying \ind{catamaran}
travelling at 41\,km/h seem to
agree pretty well:
% boat.tex ferry.tex
it consumes roughly 1\,kWh per {\tonne}-km.\nlabel{pfastcat}
It's quite a surprise to me to learn that
an island hopper who goes from island to island by plane
not only gets there faster than someone who hops by boat --
%% %%% Gender bias
he quite probably uses less energy too.
% new material BLIMPS? blimp0.tex
\input{blimp0.tex}
\section{Mythconceptions}
\beforeqa
\qa{The plane was going anyway, so my flying was energy-neutral.}{
This is false for two reasons. First, your extra weight on the plane
requires extra energy to be consumed in keeping you up.
Second, airlines respond to demand by flying more planes.
% example --
% United Airlines flights between Burbank and SFO --
% one departure every 20 mins? Each day, they would then cancel
% several of the flights and lump all the passengers into
% an appropriate number of planes.
}
\beginfullpagewidth
\small
\section{Notes and further reading}
\beforenotelist
\begin{notelist}
\item[page no.]
\item[\npageref{p747da}]
{\nqs{Boeing 747.}}
Drag coefficient for \ind{747} from
\myurl{www.aerospaceweb.org}.
Other 747 data from
\tinyurl{2af5gw}{http://www.airliners.net/info/stats.main?id=100}.
Albatross facts from
\tinyurl{32judd}{http://www.wildanimalsonline.com/birds/wanderingalbatross.php}.
% Albatross estimated drag coefficient (0.1)
%% assumed to be same as 747, though
% based on C M Bishop paper
% -- says $\cd=0.1$ is conceivable.
\item[\npageref{peffEng}]
{\nqs{Real jet engines have an efficiency of about $\epsilon = 1/3$.}}
% ***
% (``Specific fuel consumption.'')
% \myurl{http://adg.stanford.edu/aa241/propulsion/sfc.html}
% VV
Typical engine efficiencies are in the range 23\%--36\%
[\myurlb{adg.stanford.edu/aa241/propulsion/sfc.html}{http://adg.stanford.edu/aa241/propulsion/sfc.html}].
%% and especially http://www.grida.no/climate/ipcc/aviation/097.htm
%% VV
For typical aircraft, overall engine efficiency ranges between 20\% and 40\%,
with the best bypass engines delivering 30--37\% when cruising
[\myurlb{www.grida.no/climate/ipcc/aviation/097.htm}{http://www.grida.no/climate/ipcc/aviation/097.htm}].
% see also
%% http://www.grida.no/climate/ipcc/aviation/095.htm#741
%% and especially http://www.grida.no/climate/ipcc/aviation/097.htm
%% `For typical aircraft, overall efficiency ranges between 20 and 40\%'.
%% best bypass engines 30 to 37% at cruise.
You can't simply pick the most efficient engine however,
since it may be heavier (I mean, it may have bigger
mass per unit thrust), thus reducing
overall plane efficiency.
% Overall efficiency:
% ``The last piston-powered
% aircraft were as fuel-efficient as the current average jet.''
% \myurl{http://www.transportenvironment.org/Downloads-index-req-getit-lid-398.html}
% [Source missing
% \myurl{http://www.transportenvironment.org/}
\item[\npageref{pgodwit}]
{\nqs{The longest recorded non-stop flight by a bird\ldots}}
{\em New Scientist} 2492.
``Bar-tailed godwit is king of the skies.''
26 March, 2005.
% Magazine issue 2492.
% Tuesday,
11 September, 2007:
Godwit flies 11\,500\,km non-stop from Alaska to New Zealand.
\tinyurl{2qbquv}{http://news.bbc.co.uk/1/low/sci/tech/6988720.stm}
%% http://en.wikipedia.org/wiki/Bar-tailed_godwit
%% http://en.wikipedia.org/wiki/Image:BartailedGodwit24.jpg
% public domain image
\item[\npageref{OptHop}]
{\nqs{Optimizing hop lengths:
the sweet spot is when the hops are about 5000\,km long}}.
Source:
% page 4.2.1 (p474) of greens.pdf
\cite{GreenAviation}.
\item[\npageref{pfastcat}]
{\nqs{Data for a passenger-carrying \ind{catamaran}}.}
From \tinyurl{5h6xph}{http://www.goldcoastyachts.com/fastcat.htm}:
Displacement (full load) 26.3 tons.
On a 1050 nautical mile voyage she consumed just 4780 litres of fuel.
I reckon that's a weight-transport-cost of 0.93\,kWh per {\tonne}-km.
I'm counting the total weight of the vessel here, by the way.
The same vessel's {\em{passenger}}-transport-efficiency is roughly
35\,kWh per 100\,\pkm.\index{transport!efficiency!catamaran}
% added Sat 11/10/08
\item[\npageref{LunE}]
{\nqs{The Lun ekranoplan.}}
Sources: \url{www.fas.org}
\tinyurl{4p3yco}{http://www.fas.org/man/dod-101/sys/ship/row/rus/903.htm},
\citep{TaylorEkran}.
\item[Further reading:]
\cite{flight},
% %Tennekes
\cite{Wings99}.
\end{notelist}
\normalsize
\ENDfullpagewidth