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{\bf Question}
On board the Cutty Sark at Greenwich one can see the tonnage
calculation done in 1869 by the ship's designer Hercules Linton
(not Brunel!). This involves working out the areas of eleven
equally spaced cross sections of the hull. For each one, Linton
divided half the cross section unto six horizontal strips of equal
width, and then used Simpson's rule. At the largest cross section
the depth of the hull is 20.4 feet, and its half-widths at the
seven levels are 17.15, 17.40, 17.45, 17.30, 16.10, 10.65, 0.60
feet.
Show that the area of the cross section is just over 600 square
feet. If you knew all eleven cross-sectional areas, how would you
calculate the volume if the hull, given that the length of the
Cutty Sark's hull is 213 feet?
PICTURE \vspace{2in}
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{\bf Answer}
For those of you who don't know, the Cutty Sark is a ship! It was
a Tea Cutter and was one of the fastest ships of its day (19th
century). It's also the name of a good pub on the waterfront at
Greenwich, but that's another story...
We must use Simpson's rule with 7 ordinates (with 6 strips). Look
at the diagram on the sheet and we have $h=\ds\frac{204}{6}$ with
$y_1=17.15,\ y_2=17.40,\ y_3=17.45,\ y_4=17.30,\ y_5=16.10,\
y_6=10.65,\ y_7=0.60$
Simpson with 7 ordinates gives
$\begin{array} {rcl} \rm{Area} & = & \ds\frac{20.4}{6 \times
3}(\underbrace{17.15+0.60}+4 \times
(\underbrace{17.40+17.30+10.65})\\ & & \ \ \ \ \ \ \ \ \ \ \ \
y_1+y_7 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ y_2+y_4+y_6
\\ & & +2\times(\underbrace{17.45+16.10}))\\ & & \hspace{.8in} y_3+y_5\\ & = & 301.75\ \rm{square\ feet}
\end{array}$
This is only $\ds\frac{1}{2}$ the area of cross-section (see
diagram on sheet) so total cross-sectional area=\un{$603.5 \approx
600$ square feet}.
The volume of the hull would be given by
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volume=$\ds\int_0^L$(area of elemental cross section)$\times \,dx$
where $\,dx=$width of elemental cross-section, $L$=length of
strip.
With only 11 cross-sectional areas we approximate this by
Simpson's rule with 11 ordinates, 10 strips. Hence here
$h=\ds\frac{213}{10}=21.3$, $y_i=$area of section $i$
volume $\approx
\ds\frac{h}{3}(y_1+4y_2+2y_3+4y_4+2y_5+4y_6+2y_7+4y_8+2y_9+4y_{10}+y_{11})$
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