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\begin{center}
On the Chronometric Determination of Longitudes
\ \\
(\textit{Astronomische Nachrichten}, Volume V, page 227)
\end{center}

Let $\theta$, $\theta'$, $\theta''$ etc. be the periods (n in number),
at which a chronometer has determined the differences $a$, $a'$, $a''$,
etc. with the times of places whose longitudes are $x$, $x'$, $x''$,
etc.\ $\theta$, $\theta'$, $\theta''$, being supposed reduced to the time
of a single place and u denoting the daily advances of the chronometer;
one would have, if the instrument were perfectly regular, the equations
$a - u - x = a' - 'u - x' = a'' - ''u - x'' = \dots .$
In order that these equations suffice for the determination of the
unknowns $x$, $x'$, $x''$, \dots, u, it is necessary, for one thing, to consider
one of the longitudes as given, and for another, it is necessary that
at least two observations have been made in the same place, so that at
least two of the unknowns $x$, $x'$, $x''$, etc. are equal to each other.  If
among these quantities there are only two which are identical, the
problem is completely determined; in the contrary case it becomes
indeterminate, and one should proceed to satisfy the equations
$\begin{array}{lllllllllllll} 0 &= &a &- &a' &+ &(\theta' &- &\theta)u &- &x &+ &x' \\ 0 &= &a' &- &a'' &+ &(\theta'' &-&\theta')u &- &x' &+ &x'' \\ 0 &= &a'' &- &a'''&+ &(\theta'''&-&\theta'')u &- &x'' &+ &x'''\\ \multicolumn{13}{c}{\dotfill} \end{array}$
as exactly as possible, from the inevitable imperfections of the
chronometer will never permit all of them to be satisfied rigorously.
However, one should not assign to these equations equal weight, for the
quantities
$\begin{array}{lllllllllll} a &- &a' &+ &(\theta' &- &\theta)u &- &x &+ &x' \\ a' &- &a'' &+ &(\theta'' &- &\theta')u &- &x'&+ &x'' \\ \multicolumn{11}{c}{\dotfill} \end{array}$
represent the accumulations of all the variations in the
motion of the chronometer in the intervals $\theta'-\theta$,  $\theta'' - \theta'$, etc. and if a good chronometer is involved to which one can
truly attribute an average motion without a variation which keeps
increasing in one directions, the average value to be expected for such
a sum can be considered as proportional to the square root of the
elapsed time.

Thus one should, in the application of the method of least squares,
consider the preceding equations as having weights inversely
proportional to the differences $\theta' - \theta$, $\theta'' - \theta'$,
$\theta''' - \theta''$, etc.

The solution then offers no difficulty, and furnishes the most
likely values of $x$, $x'$, $x''$, etc. as well as the weight of each
determination.

However I shall add several remarks.

\noindent
I:  If the first and last observation have been made at the same place,
the most probable value of u is that which results from comparison
of these extreme observations.

The calculations then become very simple, for by virtue of a theorem
which is very easy to demonstrate, one may replace u in the equations
by its most likely value, or, what comes to the same thing, on may use
this value as if it were exact to correct the observations and to reduce
them to those which would be made with a fictitious chronometer whose
rate of gain was zero.

\medskip

\noindent
II.  If one simply attributes to the various equations weights equal to
$\frac{1}{\theta''-\theta}\qquad\frac{1}{\theta''-\theta'}\qquad \frac{1}{\theta'''-\theta''}$
the unit of precision for the weights obtained will be the exactitude of
that which one would obtain by the aid of the same chronometer observed
only two times, and at one day's interval; but in order to compare the
results obtained by the aid of various chronometers of unequal precision,
on the greater or less perfection of such chronometer used.

To arrive as it I suppose that the expressions
$\begin{array}{lllllllllll} a &- &a' &+ &(\theta' &- &\theta)u &- &x &+ &x' \\ a' &- &a'' &+ &(\theta'' &- &\theta')u &- &x' &+ &x'' \\ \multicolumn{10}{c}{\dotfill} \end{array}$ become $\lambda$, $\lambda'$, $\lambda''$, etc.
respectively, when one substitutes for the unknowns their most probable
values.  Let
$\frac{\lambda^2}{\theta'-\theta}\quad+\quad \frac{\lambda{\prime2}}{\theta''-\theta'}\quad+\quad \frac{\lambda^{\prime\prime2}}{\theta'''-\theta''}\quad+\dots=S$ if
$\nu$ is the number of unknowns and one puts
$m=\sqrt{\frac{S}{n-\nu-1}}$
the specific factor relating to each chronometer is proportional to
$\frac{1}{m^2}$ or to $\frac{n-\nu-1}{S}$ and one can consider $m$ as
the deviation of the average motion which is to be expected during a
day.

\medskip

\noindent
III:  The preceding rules are relevant to a chronometer whose motion is
not subject to any noticeable irregularity which increases with time.
If this hypothesis were untenable one might assume, when the
observations do not include an excessively long period, a variation in
the daily gain of the instrument, proportional to the time thus

The equations would then take the following form:
$\begin{array}{llllllllllllllll} 0 &= &a &- &a' &+ &(\theta' - &\theta)u &+ &(\theta^{\prime2} - &\theta^2)v &- &x &+ &x'\\ \multicolumn{15}{c}{\dotfill} \end{array}$

\medskip

\noindent
IV:  Concerning the solution of the equations according to the method of
least squares, it is perhaps not unuseful to recollect that one should
begin in most cases by calculating an approximate value for the unknowns,
and the apply the method to the determination of the small corrections
to which the values should be subjected.

It seemed useful to recall the general advice, because many calculators
seem to have forgotten it and been led to calculations which were more
laborious and perhaps less exact.

I have determined the behaviour of the following chronometers
{\small
\begin{center}
\begin{tabular}{l@{\,}r@{\ }l@{\,}r@{\,}r@{\,}r@{\,}r@{\,}r@{\,}
r@{\,}r@{\,}r@{\,}r@{\,}r@{\,}r@{\,}r@{\,}r}
&&&&&\multicolumn{2}{c}{1}&\multicolumn{2}{c}{4}
&\multicolumn{2}{c}{Breguet}&\multicolumn{2}{c}{Barraud}
&\multicolumn{2}{c}{Kassel} \\
&&&h&$'$&&&&&\multicolumn{2}{c}{3056}&\multicolumn{2}{c}{904}
&\multicolumn{2}{c}{1252} \\
Greenwich&30&June& 3&22&$-8'$&$17.14''$&$+1'$&$2.37''$ \\
&25 &July & 2  &15  &10  &44.39   &1 &32.15  &$+30'$&$59.75''$&$+48'$&$29.20''$
&$+50'$&$29.31''$ \\
&28 &$''$  & 3 &13 &11 & 0.69 &1 &36.96 &30 &50.07 &48 &40.24 &50 &39.69 \\
& 2 &Aug. & 1 &15 &11 &28.48 &1 &44.44 &30 &31.78 &48 &58.87 &50 &52.14 \\
&17 &Aug. &10 &18 &12 &59.40 &2 & 6.24 &29 &35.69 &49 &57.83 &51 &38.66 \\
&25 &$''$  &7  &27 &13 &47.98 &2 &15.84 &29 &10.48 &50 &27.15 &52 & 2.45 \\
&10 &Sep. &7  &40 &15 &24.47 &2 &40.36 \\
\ \\
Helgoland& 3&July& 3&40&$-40$&8.00&$-30$&26.84 \\
&22 &$''$  &12 &40 &42 & 2.02 &30 &3.89 &$-0$ &20.34 &$+16$ &47.39 &$+18$
&48.39 \\
& 5 &Aug. & 1 &48 &43 &18.11 &29 &43.35 & 1 &10.24 &17 &37.51 &19 &26.77 \\
&11 &$''$  &13 & 9 &43 &35.77 &29 &33.43 & 1 &32.75 &18 & 1.30 &19 &47.22 \\
&30 &$''$  &19 &30 &45 &53.08 &29 & 7.96 & 2 &40.67 &19 &17.03 &20 &47.68 \\
& 6 &Sep. & 3 & 6 &46 &51.56 &28 &58.94 & 3 & 4.55 &19 &43.80 &21 & 6.56 \\
& 7 &$''$  & 8 &42 &46 &38.72 &28 &56.71 \\
\ \\
Altona   & 6 &Aug.&5&55&$-51$&38.95&$-37$&55.76&$-9$&28.50&$+\ 9$&28.48&$+11$
&16.25 \\
& 9 &$''$  &12 &35 &51 &57.35 &37 &50.03 & 9 &38.81 & 9 &40.30 &11 &27.76 \\
&31 &$''$  & 9 &57 &54 &10.33 &37 &21.30 &10 &56.68 &11 & 5.92 &12 &25.96 \\
& 4 &Sept.&22 &12 &54 &39.16 &37 &15.21 &11 &15.36 &11 &24.49 &12 &48.10 \\
\ \\
Bremen&13&Aug.& 0& 2&$-47$&50.65&$-33$&16.49&$-5$&23.37&$+14$&21.86&$+16$&5.83
\end{tabular}
\end{center}
}

Let us for example take Breguet's chronometer 3056.  Let zero be
the longitude of Helogoland, -x that of Greenwich, y that of Altona.
I do not take account here of that of Bremen, since having only one
observation for this town, it is impossible to control it.  I count
the time from the first comparison of the Chronometer No. 1 (Greenwich
June 30 3h 22m).  Substituting for Breguet's chronometer a fictitious
instrument with daily advance zero, we find
\begin{center}
\begin{tabular}{l@{\ }l@{\ }l@{\ }l@{\ }ll@{\ }l@{\ }l@{\ }
l@{\ }ll@{\ }l@{\ }l@{\ }l@{\ }l@{\ }l}
\multicolumn{1}{c}{$\theta$} &&&&&
\multicolumn{1}{c}{$\theta$} &&&&&
\multicolumn{1}{c}{$\theta$} \\
22.4 &+ &$60.20''$&  &   &37.1 &$-$& 434.98&$+$&$y$&61.6&$+$& 61.32\\
25.0 &+ &1949.60&$-$&$x$&40.4 &$-$& 433.49&$+$&$y$&62.2&$-$&432.53&$+$&$y$\\
28.0 &+ &1950.87&$-$&$x$&42.4 &$+$&  59.88&&&66.8&$-$&434.98&$+$&$y$\\
32.9 &+ &1950.29&$-$&$x$&48.3 &$+$&1949.60&$-$&$x$&68.0&$+$&60.19\\
35.9 &+ &  59.08&   &   &56.2 &$+$&1952.74&$-$&$x$
\end{tabular}
\end{center}
In the equations above, the unknowns $x$ and $y$ are separated, which
facilitates their determination; we find for $x$ four determinations
$\begin{array}{cc} &\text{Weight}\\ 1889.40'' & \frac{1}{2.6} = 0.38 \\ 1891.21 & \frac{1}{3.0} = 0.33 \\ 1889.78 & \frac{1}{5.9} = 0.17 \\ 1891.42 & \frac{1}{3.4} = 0.19 \end{array}$
from which one obtains
$x = 1890.36''\qquad1.07$
and similarly one finds
$y = 494.12''\qquad3.83$
According to these values, the fictitious chronometer would indicate,
in Helgoland time
{\footnotesize
\begin{center}
\begin{tabular}{rrrrrrrrrrrr}
$\theta$&&\multicolumn{2}{c}{$\lambda$}
&$\theta$&&\multicolumn{2}{c}{$\lambda$}
&$\theta$&&\multicolumn{2}{c}{$\lambda$} \\
22.4  &60.20''&    &     &37.1  &59.14 &$+$ &0.06''&61.6  &61.32 &$-$ &1.06 \\
25.0  &59.24 &$-$ &0.96 &40.4  &60.63 &$+$ &1.47 &62.2  &61.59 &$-$ &0.27 \\
28.0  &60.51 &$+$ &1.27 &42.4  &59.88 &$-$ &0.75 &66.8  &59.14 &$-$ &2.45 \\
32.9  &59.93 &$-$ &0.58 &48.3  &59.24 &$-$ &0.62 &68.0  &60.19 &$+$ &1.05 \\
35.9  &59.08 &$-$ &0.85 &56.2  &62.38 &$+$ &3.14
\end{tabular}
\end{center}
}
from which one obtains
\begin{align*}
S &= 6.00 \\
m &= \sqrt{\frac{6}{13-3}}
\end{align*}
and the standard error to be expected is
$\text{for\ }x\ 0.75'',\qquad\text{for\ }y\ 0.40''$
The results furnished by the five chronometers give
\begin{center}
\begin{tabular}{lccccc}
&&&&\text{Standard error} \\
&&&&\text{to be expected}&\text{Weight} \\
Breguet &$x$ &= &1890.36''          &0.75            &\phantom{0}1.78\\
Kassel  &    &  &1893.39           &0.67            &\phantom{0}2.23\\
Barraud &    &  &1892.32           &0.49            &\phantom{0}4.16\\
\multicolumn{1}{c}{1}&&&1892.39           &0.43            &\phantom{0}5.41\\
\multicolumn{1}{c}{4}&&&1892.52           &0.35            &\phantom{0}8.16\\
\cline{6-6}
Average &$x$ &= &1892.35           &                &21.74
\end{tabular}
\end{center}
Similarly one finds for $y${lccccc}
\begin{center}
\begin{tabular}{lcccccc}
&&&&\text{Standard error} \\
&&&&\text{to be expected}&\text{Weight} \\
Breguet &$y$ &= &494.12           &0.10  &\phantom{0}6.25 \\
Kassel  &    &  &493.89           &0.36  &\phantom{0}7.72 \\
Barraud &    &  &493.67           &0.21            &14.79 \\
\multicolumn{1}{c}{1}&&&493.98           &0.29            &11.89 \\
\multicolumn{1}{c}{4}&&&494.16           &0.24            &17.36 \\
\cline{6-6}
&&&                 &                &58.01 \\
\end{tabular}
\end{center}

The number placed under the heading of weight in the last column
is the reciprocal of the square of the standard error to be expected,
taking as unit weight that which corresponds to observations giving a
standard error to be expected of 1'', so that, for Altona, the standard
error to be expected is $\sqrt{\frac{1}{58.01}}=0.13''$; but it is preferable
to consider the numbers in the last column as indicating merely ratios,
and to deduce the absolute precision from the difference between the
values of these final results found for x and y by means of each chronometer.
The precision found in this way will be a little too large, since the
determinations of time at Greenwich, at Helgoland and at Altona do not
have an absolute precision, so that consequently whatever the number of
chronometers, the errors arising from this source will always have some
effect in each final result.

One may similarly, in the following way, obtain the longitude of
Bremen.

Let $z$ be this longitude to the east of Helgoland; the comparison of
the Bremen chronometer gives the position of the fictitious chronometer as
$- 164.52'' + z$
and one deduces from comparison with previous results
$\begin{array}{cc} & \text{Weight} \\ z = 225.40'' & \frac{1}{1.4} = 0.7; \end{array}$
the others give
$\begin{array}{rr} z = 224.76 & \frac{1}{4.5} = 0.2 \\ 225.24 & 0.9 \end{array}$
The weight 0.9 should be multiplied by $\frac{10}{6.000}$; the five
chronometers give
\begin{center}
\begin{tabular}{lcc}
Breguet         &225.24      &\phantom{0}1.5 \\
Kassel          &225.84      &\phantom{0}1.9 \\
Barraud         &225.39      &\phantom{0}3.6 \\
\multicolumn{1}{c}{1}   &226.04      &\phantom{0}2.9 \\
\multicolumn{1}{c}{4}   &224.86      &\phantom{0}4.3 \\
\cline{3-3}
&&           14.2
\end{tabular}
\end{center}

The longitude of Bremen, which according to his would be $268.54''$
to the west of Altona, is naturally affected by errors in the determination
of the time at Bremen, and this difference appears to be too small by
several seconds.  According to my triangulations, the tower of Anagarius
is $273.51''$ of time to the west of Gottingen, and the observatory of Olbers
$271.19''$.

\bigskip\bigskip\bigskip

\noindent
Taken from \textit{Work (1803--1826) on the Theory of Least Squares},
trans. H F Trotter, Technical Report No.5, Statistical Techniques Research
Group, Princeton, NJ: Princeton University 1957.

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