% This paper has been transcribed in Plain TeX by
% David R. Wilkins
% School of Mathematics, Trinity College, Dublin 2, Ireland
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% Trinity College, 2000.
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\centerline{\Largebf ON R\"{O}BER'S CONSTRUCTION OF THE}
\vskip12pt
\centerline{\Largebf HEPTAGON}
\vskip24pt
\centerline{\Largebf By}
\vskip24pt
\centerline{\Largebf William Rowan Hamilton}
\vskip24pt
\centerline{\largerm (Philosophical Magazine, 27 (1864), pp.\ 124--132.)}
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\vfill
\centerline{\largerm Edited by David R. Wilkins}
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\centerline{\largerm 2000}
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\noindent
{\largeit On\/} {\largerm R\"{o}ber's} {\largeit Construction of
the Heptagon. By\/} {\largerm Sir}
{\largesc William Rowan Hamilton}, {\largeit LL.D., M.R.I.A.,
F.R.A.S., \&c., Andrews Professor of Astronomy in the University
of Dublin, and Royal Astronomer of Ireland\/}\footnote*{Communicated
by the Author.}.
\bigbreak
\vskip 12pt
\centerline{[{\it The London, Edinburgh and Dublin Philosophical
Magazine and Journal of Science,}}
\centerline{4th series, vol.~xxvii (1864), pp. 124--132.]}
\bigskip
1.
In a recent Number of the Philosophical Magazine, observations
were made on some approximate constructions of the regular
heptagon, which have recalled my attention to a very remarkable
construction of that kind, invented by a deceased professor of
architecture at Dresden, Friedrich Gottlob
R\"{o}ber\footnote\dag{The construction appears to have been
first given in pages 15, 16 of a quarto work by his son,
Friedrich R\"{o}ber, published at Dresden in 1854, and entitled
{\it Beitr\"{a}ge zur Erforschung der geometrischen Grundformen
in den alten Tempeln Aegyptens, und deren Beziehung zur alten
Naturerkenntniss}. It is repeated in page~20 of a posthumous
work, or collection of papers, edited by the younger R\"{o}ber,
and published at Leipzig in 1861, entitled {\it
Elementar-Beitr\"{a}ge zur Bestimmung des Naturgesetzes der
Gestaltung und des Widerstandes, und Anwendung dieser
Beitr\"{a}ge auf Natur und alte Kunstgestaltung}, von Friedrich
Gottlob R\"{o}ber, ehemaligen K\"{o}niglich-S\"{a}chsischen
Professor der Baukunst und Land-Baumeister. Both works, and a
third upon the pyramids, to which I cannot at present refer, are
replete with the most curious speculations, into which however I
have above declined to enter.},
who came to conceive, however, that it had been known to the
Egyptians, and employed by them in the building of the temple at
Edfu. R\"{o}ber, indeed, was of opinion that the connected
triangle, in which each angle of the base is triple of the angle
at the vertex, bears very important relations to the plan of the
human skeleton, and to other parts of nature. But without
pretending to follow him in such speculations, attractive as they
may be to many readers, I may be permitted to examine here the
accuracy of the proposed geometrical construction, of such an
isosceles triangle, or of the heptagon which depends upon it.
The closeness of the approximation, although short of
mathematical rigour, will be found very surprising.
\bigbreak
2.
R\"{o}ber's diagram is not very complex, and may even be
considered to be elegant; but the essential parts of the
construction are sufficiently expressed by the following
formul{\ae}: in which $p$ denotes a side of a regular pentagon;
$r$,~$r'$ the radii of its inscribed and circumscribed circles;
$r''$ the radius of a third circle, concentric with but exterior
to both; $p'$ a segment of the side~$p$; and $q$, $s$, $t$, $u$,
$v$ five other derived lines. The result is, that in the
right-angled triangle of which the inner diameter $2r$ is the
hypotenuse, and $u$,~$v$ supplementary chords, the former
chord~($u$) is {\it very nearly\/} equal to a side of a regular
heptagon, inscribed in the interior circle; while the latter
chord~($v$) makes with the diameter~($2r$) an angle~$\phi$,
which is {\it very nearly\/} equal to the vertical angle of an
isosceles triangle, whereof each angle at the base is triple of
the angle at the vertex. In symbols, if we write
$$u = 2r \sin \phi,\quad v = 2r \cos \phi,$$
then $\phi$ is found to be very nearly
$\displaystyle = {\pi \over 7}$. It will be seen that the
equations can all be easily constructed by right lines and
circles alone, having in fact been formed as the expression of
such a construction; and that the numerical ratios of the lines,
including the numerical values of the sine and cosine of $\phi$,
can all be arithmetrically computed\footnote*{The computations
have all been carried out to several decimal places beyond what
are here set down. Results of analogous calculations have been
given by R\"{o}ber, and are found in page~16 of the first-cited
publication of his son, with the assumption $p = \surd 3$, and
with one place fewer of decimals.},
with a few extractions of square roots.
$$\left\{ \vcenter{\halign{$\displaystyle #$\hfil&\quad\hfil
$\displaystyle # \, $&$\displaystyle #$\hfil\cr
(r + r')^2 = 5 r^2
& {r' \over r} = & 1.2360680\cr
\noalign{\vskip3pt}
p^2 = 4 (r'^2 - r^2)
& {p \over r} = & 1.4530851\cr
\noalign{\vskip3pt}
{p' \over p} = {r + {1 \over 2} r' \over r + r'}
& {p' \over r} = & 1.0514622\cr
\noalign{\vskip3pt}
q^2 = p^2 - p'^2
& {q \over r} = & 1.0029374\cr
\noalign{\vskip3pt}
s^2 + ps = (p - q + r)^2
& {s \over r} = & 0.8954292\cr
\noalign{\vskip3pt}
r''^2 = r^2 + s^2
& {r'' \over r} = & 1.3423090\cr
\noalign{\vskip3pt}
t^2 = \left( {r' r''\over r} \right)^2 - (r'' - r)^2
& {t \over r} = & 1.6234901\cr
\noalign{\vskip3pt}
u^2 = 2r (2r - t)
& {u \over r} = & 0.8677672\cr
\noalign{\vskip3pt}
v^2 = 2rt
& {v \over r} = & 1.8019379\cr
\noalign{\vskip3pt}
u = 2r \sin \phi
& \sin \phi = & 0.4338836\cr
\noalign{\vskip3pt}
v = 2r \cos \phi
& \cos \phi = & 0.9009689\cr}}
\right.
\leqno {\rm (A)}$$
\bigbreak
3.
On the other hand, the true septisection of the circle may be
made to depend on the solution of the cubic equation,
$$8 x^3 + 4 x^2 - 4x - 1 = 0,$$
of which the roots are
$\displaystyle \cos {2\pi \over 7}$,
$\displaystyle \cos {4\pi \over 7}$,
$\displaystyle \cos {6\pi \over 7}$.
Calculating then, by known methods\footnote\dag{Among these the best
by far appears to be Horner's method,---for practically arranging
the figures in the use of which method, a very compact and
convenient form or scheme was obligingly communicated to me by
Professor De Morgan, some time ago. We arrived independently at
the following value, to 22 decimals, of the positive root of the
cubic mentioned above:
$$\cos {2\pi \over 7}
= 0 \cdot 62348 \,\, 98018 \,\, 58733 \,\, 53052 \,\, 50.$$
I had however found, by trials, before using Horner's method, the
following approximate value:
$$\cos {2\pi \over 7} = 0 \cdot 62348 \,\, 98018 \,\, 587;$$
which was more than sufficiently exact for comparison with
R\"{o}ber's construction.},
to eight decimals, the positive root of this equation, and thence
deducing to seven decimals, by square roots, the sine and cosine
of
$\displaystyle {\pi \over 7}$,
we find, without tables, the values
$$\vcenter{\halign{\hfil $\displaystyle # \,$&\hfil
$\displaystyle #$&$\displaystyle \, #$\hfil\cr
\cos {2 \pi \over 7} =
&x &= 0.62348980;\cr
\noalign{\vskip3pt}
\sin {\pi \over 7} =
&\sqrt{\vphantom{\biggl(}} {1 - x \over 2}
&= 0.4338837;\cr
\noalign{\vskip3pt}
\cos {\pi \over 7} =
&\sqrt{\vphantom{\biggl(}} {1 + x \over 2}
&= 0.9009689;\cr}}$$
and these last agree so nearly with the values (A) of $\sin \phi$
and $\cos \phi$, that at this stage a doubt may be felt,
{\it in which direction does the construction err}. In fact,
R\"{o}ber appears to have believed that the construction
described above was geometrically rigorous, and had been known
and prized as such from a very remote antiquity, although
preserved as a secret doctrine, entrusted only to the initiated,
and recorded only in stone.
\bigbreak
4.
The following is an easier way, for a reader who may not like so
much arithmetic, to satisfy himself of the extreme closeness of
the approximation, by formul{\ae} adapted to logarithms, but
rigorously derived from the construction. It being evident that
$$r' = r \sec {\pi \over 5},
\quad\hbox{and}\enspace
p = 2r \tan {\pi \over 5},$$
let $\phi_1 \, \ldots \, \phi_6$ be six auxiliary angles, such
that
$$r' = 2r \tan \phi_1,\quad
p' = p \sin 2 \phi_2,\quad
p - q = r \tan^2 \phi_3,$$
$$p - q + r = {\textstyle {1 \over 2}} p \tan 2 \phi_4,\quad
s = r \tan 2 \phi_5,\quad
r(r'' - r) = r' r'' \sin \phi_6;$$
we shall then have the following system of equations, to which
are annexed the angular values, deduced by interpolation from
Taylor's seven-figure logarithms, only eleven openings of which
are required, if the logarithms of two and four be remembered, as
they cannot fail to be by every calculator.
$$\left\{ \vcenter{\halign{$\displaystyle #$\hfil&\quad\hfil
$\displaystyle #\, $&&\hfil$\displaystyle #$ \cr
& &\phantom{=}\circ&{}'&{}''\hskip1.5em\cr
\cot \phi_1 = 2 \cos {\pi \over 5}
& \phi_1 =& 31 & 43 & 2 \cdot 91 \cr
\noalign{\vskip3pt}
\sin 2 \phi_2 = \cos^2 \phi_1
& \phi_2 =& 23 & 10 & 35 \cdot 52 \cr
\noalign{\vskip3pt}
\tan^2 \phi_3 = 4 \sin^2 \phi_2 \, \tan {\pi \over 5}
& \phi_3 =& 33 & 51 & 31 \cdot 90 \cr
\noalign{\vskip3pt}
\cot 2 \phi_4 = \cos^2 \phi_3 \, \tan {\pi \over 5}
& \phi_4 =& 31 & 41 & 39 \cdot 37 \cr
\noalign{\vskip3pt}
\tan 2 \phi_5 = 2 \sin^2 \phi_4 \, \sec 2 \phi_4 \, \tan {\pi \over 5}
& \phi_5 =& 20 & 55 & 15 \cdot 93 \cr
\noalign{\vskip3pt}
\sin \phi_6 = \sin^2 \phi_5 \, \cot \phi_1
& \phi_6 =& 11 & 54 & 22 \cdot 60 \cr
\noalign{\vskip3pt}
\cos^2 \phi = \cos \phi_6 \, \sec 2 \phi_5 \, \tan \phi_1
& \phi =& 25 & 42 & 51 \cdot 4\, .\cr}}
\right.
\leqno {\rm (B)}$$
It is useless to attempt to estimate hundredths of seconds in
this last value, because the difference for a second, in the last
logarithmic cosine, amounts only to ten units in the seventh
place of decimals, or to one in the sixth place. But if we thus
confine ourselves to tenths of seconds, a simple division gives
immediately that final value, under the form
$${\pi \over 7} = {180^\circ \over 7}
= 25^\circ \,\, 42' \,\, 51'' \cdot 4;$$
it appears therefore to be difficult, if it be possible, to
decide by Taylor's tables, whether the equations (B), deduced
from R\"{o}ber's construction, give a value of the angle~$\phi$,
which is {\it greater\/} or {\it less\/} than the {\it seventh
part of two right angles}. (It may be noted that
$2 \tan \phi_1 = 2$; but that to take out $\phi_1$ by this
equation would require another opening of the tables.)
\bigbreak
5.
To fix then decisively the {\it direction\/} of the {\it error\/}
of the approximation, and to form with any exactness an estimate
of its {\it amount}, or even to prove quite satisfactorily by
{\it calculation\/} that any such error {\it exists}, it becomes
necessary to fall back on arithmetic; and to carry at least the
first extractions to several more places of decimals,---although
fewer than those which have been used in the resumed computation
might have sufficed, except for the extreme accuracy aimed at in
the resulting values. For this purpose, it has been thought
convenient to introduce eight auxiliary numbers
$a \, \ldots \, h$, which can all be calculated by square roots,
and are defined with reference to the recent equations (B), as
follows:
$$a = 1 + 2 \tan \phi_1;\quad
b = 4 \cos 2 \phi_2 \cot {\pi \over 5};\quad
c = 2 \cos 2 \phi_2 - \cot {\pi \over 5};$$
$$d = \sec 2 \phi_4;\quad
e = \sec 2 \phi_5;\quad
f = 2 \cos^2 \phi;\quad
g = 2 \cos \phi;\quad
h = 2 \cos {\phi \over 2};$$
or thus with reference to the earlier equations (A):
$$a = {r + r' \over r};\quad
b = {8 q r \over p^2};\quad
c = {2(q - r) \over p};\quad
d = {2s + p \over p};\quad
e = {r'' \over r};$$
$$f = {t \over r};\quad
f = {v \over r};\quad
h^2 = {2r + v \over r};$$
and respecting which it is to be observed that $c$, like the rest,
is positive, because it may be put under the form
$$c = \sqrt{\vphantom{\biggl(}} {14 - 2 \surd 5 \over 5}
- \sqrt{\vphantom{\biggl(}} {5 + 2 \surd 5 \over 5},$$
and $14 - 2 \surd 5 > 5 + 2 \surd 5$, because $9 > 4 \surd 5$,
or $9^2 > 4^2 \,\, 5$. With these definitions, then, of the numbers
$a \, \ldots \, h$, and with the help of the following among
other identities,
$$\eqalign{
\cos {7 \phi \over 2} \, \sec {\phi \over 2}
&= 2 \cos 3 \phi - 2 \cos 2 \phi + 2 \cos \phi - 1 \cr
&= 2 (2 \cos \phi - 1) \cos 2 \phi - 1,\cr}$$
I form {\it without tables\/} a system of values below, the early
numbers of which have been computed to several decimals more than
are set down.
$$\left\{ \vcenter{\halign{$\displaystyle #$\hfil&\quad\hfil
$\displaystyle # \, $&$\displaystyle #$\hfil\cr
a^2 = 5
& a = & 2 \cdot 23606 \,\, 79774 \,\, 99789 \,\, 6964\cr
\noalign{\vskip3pt}
b^2 = 8 + {72 a \over 25}
& b = & 3 \cdot 79998 \,\, 36545 \,\, 96345 \,\, 0138\cr
\noalign{\vskip3pt}
c^2 = {19 \over 5} - b
& c = & 0 \cdot 00404 \,\, 29449 \,\, 23565 \,\, 7641\cr
\noalign{\vskip3pt}
d^2 = 1 + (2 - c)^2
& d = & 2 \cdot 23245 \,\, 25898 \,\, 01044 \,\, 7849\cr
\noalign{\vskip3pt}
e^2 = 1 + (5 - 2a) (d - 1)^2
& e = & 1 \cdot 34230 \,\, 90137 \,\, 74792 \,\, 5831\cr
\noalign{\vskip3pt}
f^2 = (5 - 2a) e^2 + 2e - 1
& f = & 1 \cdot 62349 \,\, 00759 \,\, 24105 \,\, 2470\cr
\noalign{\vskip3pt}
g^2 = 2f
& g = & 1 \cdot 80193 \,\, 78878 \,\, 99638 \,\, 5912\cr
\noalign{\vskip3pt}
h^2 = 2 + g
& h = & 1 \cdot 94985 \,\, 58633 \,\, 65197 \,\, 2049\cr
\noalign{\vskip3pt}
\sin {\pi - 7 \phi \over 2}
= h \left( (f - 1)(g - 1) - {1 \over 2} \right)
& \phantom{h} = & + 0 \cdot 00000 \,\, 06134 \,\, 49929 \,\, 1683.\cr}}
\right.
\leqno {\rm (C)}$$
Admitting then the known value,
$$\pi = 3 \cdot 14159 \,\, 26535 \,\, 89793 \ldots,$$
or the deduced expression,
$$1'' = {\pi \over 648000}
= 0 \cdot 00000 \,\, 48481 \,\, 36811 \,\, 095 \ldots,$$
I infer as follows:
$$\left\{ \eqalign{
{\pi - 7 \phi \over 2}
&= + 0'' \cdot 12653 \,\, 31307 \,\, 822,\cr
{\pi \over 7} - \phi
&= + 0'' \cdot 03615 \,\, 23230 \,\, 806,\cr
\phi
&= 25^\circ \,\, 42' \,\, 51'' \cdot 39241 \,\, 91054 \,\, 91,\cr}
\right.
\leqno {\rm (D)}$$
and think that these twelve decimals of a second, in the value of
the angle~$\phi$, may all be relied on, from the care which has
been taken in the calculations.
\bigbreak
6.
The following is a quite different way, as regards the few last
steps, of deducing the same ultimate numerical results.
Admitting (comp.\ Art.~3) the value\footnote*{Compare a preceding
note.},
$$2 \cos {2\pi \over 7} = z
= 1 \cdot 24697 \,\, 96037 \,\, 17467 \,\, 06105,$$
as the positive root, computed by Horner's method, of the cubic
equation
$$z^3 + z^2 - 2z - 1 = 0,$$
and employing the lately calculated value~$f$ of $1 + \cos
2\phi$, I find by square roots the following sines and cosines,
with the same resulting error of the angle~$\phi$ as before:
$$\left\{ \eqalign{
\sin {\pi \over 7} = {1 \over 2} \sqrt{2 - z}
&= 0 \cdot 43388 \,\, 37391 \,\, 17558 \,\, 1205;\cr
\cos {\pi \over 7} = {1 \over 2} \sqrt{2 + z}
&= 0 \cdot 90096 \,\, 88679 \,\, 02419 \,\, 1262;\cr
\sin \phi = {1 \over 2} \sqrt{4 - 2f}
&= 0 \cdot 43388 \,\, 35812 \,\, 03469 \,\, 1138;\cr
\cos \phi = {1 \over 2} \sqrt{2f} = {1 \over 2} g
&= 0 \cdot 90096 \,\, 89439 \,\, 49819 \,\, 2956;\cr
\sin \left( {\pi \over 7} - \phi \right)
&= + 0 \cdot 00000 \,\, 01752 \,\, 71408 \,\, 3339;\cr
{\pi \over 7} - \phi
&= + 0'' \cdot 03615 \,\, 23230 \,\, 806.\cr}
\right.
\leqno {\rm (E)}$$
\bigbreak
7.
If we continue the construction, as R\"{o}ber did, so as to form
an isosceles triangle, say ${\rm A} {\rm B} {\rm C}$, with $\phi$
for its vertical angle, and if we content ourselves with
thousandths of seconds, the angles of this triangle will be as
follows:
$$\left\{ \eqalign{
{\rm A} = \phi \hskip2em
&= 25^\circ \,\, 42' \,\, 51'' \cdot 392;\cr
{\rm B} = {\pi - \phi \over 2}
&= 77^\circ \,\, 8' \,\, 34'' \cdot 304;\cr
{\rm C} = {\rm B} \hskip2em
&= 77^\circ \,\, 8' \,\, 34'' \cdot 304;\cr}
\right.
\leqno {\rm (F)}$$
and we see that each base-angle exceeds the triple of the
vertical by only about an eighth part of a second, namely by
that small angle which occurs first in the system (D), and of
which the sine is the last number in the preceding system (C).
And if we compare a base-angle of the triangle thus constructed,
with the base-angle
$\displaystyle {3 \pi \over 7}
= 77^\circ \,\, 8' \,\, 34'' \cdot 2857 \ldots$
of the true triangle, in which each angle of the base is triple
of the angle at the vertex, we find an error in excess equal
nearly to $0'' \cdot 018$, or, more exactly,
$${\rm B} - {3\pi \over 7}
= + 0'' \cdot 01807 \,\, 61615 \,\, 403,$$
which amounts to {\it less\/} than a {\it fifth-fifth part\/} of
a {\it second}, but of which I conceive that all the thirteen
decimals here assigned are correct. And I suppose that no artist
would undertake to construct a triangle which should more
perfectly, or so perfectly, fulfil the conditions proposed. The
{\it problem}, therefore, of constructing {\it such\/} a
{\it triangle}, and with it the {\it regular heptagon}, by
{\it right lines and circles only}, has been
{\it practically solved\/} by that process which R\"{o}ber
believed to have been {\it known\/} to the ancient Egyptians, and
to have been {\it employed\/} by them in the architecture of some
of their temples---some {\it hints}, as he judged, being
intentionally preserved in the details of the workmanship, for
the purpose of being {\it recognized}, by the initiated of the
time, or by men of a later age.
\bigbreak
8.
Another way of rendering conceivable the extreme smallness of the
practical error of that process, is to imagine a series of seven
successive chords inscribed in a circle, according to the
construction in question, and to inquire {\it how near\/} to the
initial point the final point would be. The answer is, that the
{\it last\/} point would fall {\it behind\/} the {\it first}, but
only by about {\it half a second\/} (more exactly by
$0'' \cdot 506$). If then we suppose, for illustration, that
these chords are {\it sevven successive tunnels}, drawn
{\it eastward\/} from station to station of the {\it equator of
the earth}, the last tunnel would emerge to the {\it west\/} of
the first station, but only by about {\it fifty feet}.
\bigbreak
9.
My own studies have not been such as to entitle me to express an
opinion whether the architectural and geometrical drawings of
R\"{o}ber in connexion with the plan of the temple at Edfu, and
his comparisons of the numbers deduced from the {\it details\/}
of his construction with French measurements previously made, are
sufficient to bear out his conclusion, that the process had been
anciently used: but I wish that some qualified person would take
up the inquiry, which appears to me one of great interest,
especially as I see no antecedent improbability in the
supposition that the construction in question may have been
invented in a very distant age. The geometry which it employs is
in no degree more difficult than that of the Fourth Book of
Euclid\footnote*{The segment~$p'$ of the side~$p$ of the
pentagon, and the fourth proportional
$\displaystyle {r' r'' \over r}$
to the three radii, which enter into the equations (A), and of
which the latter is the greater segment of the third diameter,
$2r''$, if this last be cut in extreme and mean ratio, may at
first appear to depend on the Sixth Book of Euclid, but will be
found to be easily constructible without going beyond the Fourth
Book.};
and although I have no conjecture to offer as to what may have
{\it suggested\/} the particular process employed, yet it seems
to me quite as likely to have been discovered thousands of years
ago, perhaps after centuries of tentation, as to have been first
found in our own time, which does not generally attact so much
importance to the heptagon as a former age may have done, and
which perhaps enjoys no special facilities in the search after
such a {\it construction}, although it supplies means of proving,
as above, that the proposed solution of the problem is {\it not
mathematically perfect}.
\bigbreak
10.
If R\"{o}ber's professional skill as an architect, combined with
the circumstance stated of his having previously invented the
construction for himself did really lead him to a correct
interpretation\footnote*{It ought in fairness to be stated
that R\"{o}ber's {\it interpretation\/} of Egyptian antiquities
included a vast deal more than what is here described, and that
he probably considered the {\it geometrical part\/} of it to be
the {\it least interesting}, although still, in his view, an
essential and {\it primary element}.}
of the plan of the temple at Edfu, which he believed to embody a
tradition much older than itself, we are thus admitted to a very
curious glimpse, or even a partial view, of the nature and
extent, but at the same time the imperfection, of that knowledge
of geometry which was possessed, but kept secret, by the ancient
priests of Egypt. I say the {\it imperfection}, on the
supposition that the above described construction of the
{\it heptagon}, if known to them at all, was thought by them to
be {\it equal in rigour}, as the elder R\"{o}ber appears to have
thought it to be, to that construction of the {\it pentagon\/}
which Euclid {\it may\/} have learned from them,
{\it rejecting\/} perhaps, at the same time, the {\it other\/}
construction, as being not based on demonstration, and not by him
demonstrable, although Euclid may not have {\it known\/} it to
be, in its result, imperfect. The interest of the speculation
stretches indeed back to a still earlier age, and may be
connected in imagination with what we read of the ``wisdom of the
Egyptians.'' But I trust that I shall be found to have
abstained, as I was bound to do, from any expression which could
imply an acquaintance of my own with the arch{\ae}ology of Egypt,
and that I may at least be pardoned, if not thanked, for having
thus submitted, to those who may be disposed to study the
subject, a purely mathematical\footnote\dag{{\it Note added during
printing.}---Some friends of the writer may be glad to know
that these long arithmetical calculations have been to him rather
a relaxation than a distraction from his more habitual studies,
and that there are already in type 672 octavo pages of the
`Elements of Quaternions,' a work which (as he hopes) is rapidly
approaching to the stage at which it may be announced for
publication.}
discussion, although connected with a question of other than
mathematical interest.
\bigbreak
\line{\hfil W.~R.~H.}
Observatory of Trinity College, Dublin,
December~22, 1863.
\bye