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\DOI{10.5802/crphys.246}
\datereceived{2025-02-26}
\daterevised{2025-03-13}
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\dateposted{2025-05-14}
\begin{document}

\begin{noXML}

\CDRsetmeta{articletype}{history-of-sciences}

\title{An explanation for Aristarchus' measurement of the sun's
distance}

\alttitle{Une explication \`{a} la mesure de la distance du soleil par
Aristarque}

\author{\firstname{Siebren} \lastname{van der Werf}\IsCorresp}
\address{University of Groningen, The Netherlands}
\email[S. van der Werf]{syvanderwerf@kpnmail.nl}
%\urladdr{https://siebrenvanderwerf.nl/index.php}

\author{\firstname{Tjeerd} \lastname{van Albada}}
\address{Kapteyn Astronomical Institute, University of Groningen, The
Netherlands}
\email[T. van Albada]{albada@astro.rug.nl}

\begin{abstract}
Aristarchus of Samos determined the angular distance between moon and
sun at half-moon as 87\textdegree, from which he deduced that the sun's
distance is ``between 18 and 20 times that of the moon''. In this
study, we show  that his result can be reproduced and understood when
it is assumed that the observation was made when, apart from being
perceived as straight, the shadow edge over the moon's surface was
vertical. A measurement could then have been done with instruments that
were available at the time: the gnomon and the quadrant. An example
that might have been observed  by Aristarchus himself is presented
and\break analyzed.
\end{abstract}

\begin{altabstract}
Aristarque de Samos a d\'{e}termin\'{e} que la distance angulaire entre
la lune et le soleil \`{a} la demi-lune \'{e}tait de 87\textdegree, ce
qui lui a permis de d\'{e}duire que la distance du soleil est \og entre 18 et 20 fois celle de la lune \fg. Son r\'{e}sultat peut
\^{e}tre reproduit et compris si l'on suppose que l'observation a
\'{e}t\'{e} faite lorsque le bord de l'ombre sur la surface de la lune,
en plus d'\^{e}tre per\c{c}u comme rectiligne, \'{e}tait vertical. Une
mesure aurait alors pu \^{e}tre effectu\'{e}e avec les instruments
disponibles \`{a} l'\'{e}poque : le gnomon et le quadrant. Un exemple
que Aristarque aurait pu observer lui-m\^{e}me est pr\'{e}sent\'{e} et
analys\'{e}.
\end{altabstract}

\keywords{\kwd{Aristarchus}\kwd{Ancient astronomy}\kwd{Distance of the sun}}

\altkeywords{\kwd{Aristarque}\kwd{Astronomie ancienne}\kwd{Distance du soleil}}

\maketitle

\end{noXML}

\section{Introduction}\label{sec1}

Aristarchus of Samos (circa 310--230 BC) is known as the first
astronomer to propose a heliocentric model of the universe. His own
written work about his ideas on the universe has not survived.
Fortunately, his book 
{${\Pi}$}{\textipa{E}}${\rmrho}\grave{\rmiota}$
{${\rmmu}$}{\textipa{E}}${\upgamma}${\textipa{E}}\textipa{T}$\tilde{\rmomega}{\rmnu}$
{${\rmkappa}{\rmalpha}\grave{\rmiota}$}
{\aunient{1F00}}${\uppi}\mathrm{o}{\rmsigma}{\rmtau}{\rmeta}{\rmmu}\acute{\rmalpha}{\rmtau}{\rmomega}{\rmnu}$
{\bunient{1F21}}${\rmlambda}\acute{\rmiota}\mathrm{o}{\rmupsilon}$ 
${\rmkappa}{\rmalpha}\grave{\rmiota}$
${\rmsigma}{\rmvarepsilon}{\rmlambda}\acute{\rmeta}{\rmnu}{\rmeta}{\varsigma}$
(\textit{On the Sizes and Distances of the Sun and the Moon}) has been
preserved and it has been translated into many languages. The list of
references~\cite{1,2,3,4,5,6,7} is far from complete, but it contains
sources that are readily accessible. The first edition to include the
Greek original is that of 1688 by John Wallis~\cite{1}.

The oldest preserved copy of Aristarchus' treatise is the
10{th}-century manuscript with catalog number Vat.gr.204 in the Vatican
Apostolic Library. It has been digitized and is easily accessible. This
seems to be the version that most other surviving copies are based
upon, and it is the primary source that Thomas Heath used in his
\textit{Aristarchus of Samos, the ancient Copernicus}~\cite{2}. In the
present work, we rely on the Greek text as we find it in Heath's work.

In his treatise, Aristarchus first presents a number of working
hypotheses. Numbers~1--4 pertain to the distances of the sun and the
moon, while 5 and 6 deal with their sizes. In this work, we will only
consider distances.

{\noindent}
Hypothesis~1: The moon receives its light from the sun.

{\noindent}
Hypothesis~2: The earth may be considered as a point and also as the
center of the moon's sphere.

{\noindent}
Hypothesis~3: When we see the moon as half, the great circle that
separates the dark side from the illuminated one is oriented along our
line of vision.

{\noindent}
Hypothesis~4: When we see the moon as half, its [angular] distance
from the sun is less than a quadrant by one thirtieth of a quadrant.

Having presented his hypotheses, Aristarchus then proceeds by
clarifying them and discussing their consequences in a number of
propositions, eighteen in total. He is careful to remark in
proposition~2 that the sun is bigger than the moon and that the part of
the moon's surface that is illuminated must therefore be larger than a
hemisphere. But in proposition~4, he states that the effect is small:
at half-moon, the terminator circle and the ``maximum circle'' are less
than 1/3960 of a quadrant (1\textdegree/44) apart. The circle over the
moon's surface that separates the dark and the illuminated parts does
not sufficiently differ from the maximum circle to be perceptible to
our eyes. This is the {a posteriori} justification for
neglecting, in hypothesis~3, the subtle difference between the shadow
edge over the moon's spherical surface and the great circle (or
``maximum circle'') parallel to it. In addition, the statement that
this dividing circle is oriented along our line of vision is more
precisely phrased in proposition~5, where it is stated  that this
circle and the connecting line between its center and our eye are in
one and the same plane.

It is important to note that Aristarchus uses here the visual acuity of
the human eye: lines apart by 1\textdegree/44 $=$\ 1.36$^{\prime}$ cannot
be distinguished. In today's optometry ``normal'' 20/20 vision (10/10
vision in France) is defined just a bit sharper as an angular
perception limit of 1$^{\prime}$. In his hypothesis~6, Aristarchus had
originally adopted a semidiameter of 1\textdegree{} for the moon, which
is far too big. Archimedes~\cite{8,9,10} tells us in his
\textit{Psammites}, also known as \textit{Arenarius}, or \textit{The
Sand Reckoner}, that Aristarchus corrected this in hindsight to
15$^{\prime}$.  Since we see the moon's semidiameter as 15$^{\prime}$,
a 1$'$ displacement of the middle of the terminator from the half-moon
situation, corresponds to a change of the phase angle, from
90\textdegree{}  to [90\textdegree{} ${\pm}$ sin$^{-1}$(1/15)]\ $=$\ 
93.8\textdegree{} or 86.2\textdegree{} and hence a phase of  [1\ ${+}$\
cos(93.8\textdegree)]/2\ ${=}$\  0.467 or [1\ ${+}$\
cos(86.2\textdegree)]/2\ ${=}$\ 0.533.  For all phases closer to 0.5,
the moon will be judged as half.

Hypothesis~4 still puzzles us today. It states that the angular
distance of the moon from the sun, its \textit{elongation}, would
measure 87\textdegree{} when the moon is seen as half. The situation is
illustrated in Figure~\ref{fig1}. The ratio of the sun's distance to
that of the moon would be 1/cos(87\textdegree) $=$\ 19.1. In
Aristarchus' time, trigonometric tables were not yet available.
Instead, he established an upper limit of 20 and a lower limit of 18.
Our today's knowledge is that this sun/moon distance ratio is much
larger: 389.18, based on time-averaged distances. The correct value for
the elongation of the moon is therefore 89.85\textdegree. Atmospheric
refraction makes  us see the sun higher than its true altitude. For the
moon, the combined effect of parallax and refraction is to lower its
apparent altitude. As a result, the apparent elongation may be smaller
than the true one, but this effect is not enough to bring it below
89.5\textdegree. The puzzling question remains: how could Aristarchus
have been so far off? 

\begin{figure}
{\vspace*{4pt}}
\includegraphics{fig01}
{\vspace*{4pt}}
\caption{\label{fig1}The Earth--Moon--Sun plane. Seen from the earth,
the elongation would be ``a~quadrant minus one thirtieth part of it'',
or 90\textdegree{}\ $-$\ 3\textdegree{}\ $=$\ 87\textdegree; see
hypothesis~4.}
{\vspace*{4pt}}
\end{figure}

A widespread opinion is that Aristarchus and his Alexandrian
contemporaries did consider the argumentation and the mathematics as
more important than the actual measurement. In his monumental work
\textit{A History of Ancient Mathematical Astronomy}, Otto Neugebauer
writes \cite[p.~271]{11}: ``\ldots{} there is much in the astronomy
of Eudoxus, Aristarchus, and Archimedes (i.e. in the period just
preceding Apollonius) that shows a lack of interest in empirical
numerical data in contrast to the emphasis on the purely mathematical
structure''. About Aristarchus, he writes ``I think this [i.e.
Neugebauer's] analysis leads to the conclusion that Aristarchus'
treatise on the sizes and distances is a purely mathematical exercise
which has little to do with practical astronomy'' \cite[p.~643]{11}.

Earlier, Paul Tannery in \textit{Recherches sur l'histoire de
l'astronomie ancienne} was of the same opinion: ``il semble que tous
les auteurs aient volontairement m\'{e}connu que la science des astres
repose essentiellement sur l'observation'' \cite[p. 43--44]{12}.

Different yet is the critique of Reviel Netz in his \textit{Ludic
Proof}: Alexandrian scientists would primarily have been interested in
storytelling and Aristarchus and Archimedes would have made ``the
deliberate choice to mystify their audiences'' \cite[p. 42]{13}.

It has been suggested, by Evans \cite[p.~72]{14} that, instead of
measuring the elongation, Aristarchus could have estimated it from the
time it takes the moon to go from its third quarter to its first
quarter. Assuming that the moon moves at a constant velocity along its
orbit, the finite distance to the sun would cause the path from third
quarter to first quarter via new moon to be shorter than the remaining
path from first quarter back to third quarter via full moon. This
argument is also given by Carman and Buz\'{o}n \cite[p.~191--192]{7}
and by Papathomas~\cite{15}. The idea is, however, easily rebutted: the
time from third to first quarter is not constant at all, as Aristarchus
must undoubtedly have been aware of. Instead, it oscillates between
13.5 and 16~days, around half a synodic month, which is 14 days, 18 h
and 22 min on average. This is illustrated in Figure~\ref{fig2}, where
this difference is plotted for the years~2025 and 2026.

\begin{figure}
{\vspace*{4pt}}
\includegraphics{fig02}
{\vspace*{4pt}}
\caption{\label{fig2}Time differences of first and third quarter moon
for 2025--2026. Red dots: the times from third quarter to first quarter
via new moon. Blue triangles: half of the
synodic month, evaluated as the time between two successive first
quarters.}
{\vspace*{4pt}}
\end{figure}

Below, we will analyze a first quarter observation that Aristarchus
might have witnessed on 16~June 255 BC. The preceding third quarter
fell 15 days and 2~h and 40~min earlier, on 1 June 255 BC, when both
sun and moon were below the horizon. But even reckoned from the time of
moonrise, just over four hours later, there still remained 14 days and
22~h and 30~min to the first quarter on 16 June, \textit{longer---not
shorter}---than half a synodic month. 

It has also been suggested that Aristarchus may not have performed any
measurement at all~\cite{14,16}. 

In our view, however, there is a sufficient number of indications in
Aristarchus' text to suggest that he does refer to an actual
measurement of the distance to the sun. For example, this is the case
in the wording of propositions~2 and 4 above and in the missing part of
a quadrant in Hypothesis~4. Therefore, in this article we wish to
assume that he did intend to perform a measurement as precisely as
possible. We show that his result can be reproduced and understood if
we extend his hypothesis~4 with the further specification that the
observation was made at an instant that the shadow edge over the moon's
surface was seen as vertical. An accurate measurement would then have
been possible by using the gnomon and the quadrant. Both instruments
did exist in Aristarchus' time and must have been available to him.

In fact, it is hard to imagine that he could have made a measurement
with a backward tilting half-moon, and when the half-moon is tilting
forward, the sun is already below the horizon.

\section{Elongation and tilt angle}

As an example, Figure~\ref{fig3} illustrates the situation as it
occurred in Alexandria on 16 June 255 BC at 16:43:25 UT, about a
quarter of an hour before sunset. In reproducing the ephemerides, we
use the planetary program SkyMap Pro~9, of Chris Marriott~\cite{17}.
Other professional planetary programs~\cite{18,19,20} use the same
routines and give identical results.

\begin{figure}
\includegraphics{fig03}
\caption{\label{fig3}Celestial sphere, illustrating the situation as
seen from Alexandria (in P) on 16 June 255 BC, just before sunset.}
\end{figure}

The zenith for the observer is in P and he sees the sun along PS at a
zenith angle of b $=$ ($90\text{\textdegree} - H_{\mathrm{S}}$), or, at height
$H_{\mathrm{S}}$ above the horizon. Similarly, the moon is seen along
PM at a zenith angle of a $=$ ($90\text{\textdegree} - H_{\mathrm{M}}$), or, at
height $H_{\mathrm{M}}$. The angle $\gamma = \Delta AZ =
AZ_{\mathrm{S}} - AZ_{\mathrm{M}}$ is the difference in azimuth between
sun and moon. The moon is illuminated along SM. Since here $\beta=
 90\text{\textdegree}$, the moon is seen upright. The side c $=$\ SM represents
the elongation $\uppsi$, or \textit{lunar distance} as it is known in
nautical astronomy. The well-known cosine rule~\cite{21} gives:
{\begin{equation}\label{eq1}
\cos({\uppsi}) = \sin(H_{\mathrm{S}})\sin(H_{\mathrm{M}}) + 
\cos(H_{\mathrm{S}})\cos(H_{\mathrm{M}})\cos({\Delta}AZ).
\end{equation}}\unskip
The equation applies equally to the apparent elongation
${\uppsi}^{\prime}$, which is expressed in terms of the apparent heights
$H^{\prime}_{\mathrm{S}}$ and $H^{\prime}_{\mathrm{M}}$, wherein the effects
of refraction and parallax have been 
included~\cite[and older references quoted therein]{22}. Both are
vertical corrections and they do not affect the azimuths. Therefore:
{\begin{equation}\label{eq2}
\cos(\uppsi^{\prime}) = \sin(H^{\prime}_{\mathrm{S}})
\sin(H^{\prime}_{\mathrm{M}}) + \cos(H^{\prime}_{\mathrm{S}})
\cos(H^{\prime}_{\mathrm{M}}) \cos({\Delta}AZ).
\end{equation}}\unskip
The moon's tilt angle, $T$, is found from~\cite{23,23a,24}:
{\begin{equation}\label{eq3}
\tan (T) = [\sin(H_{\mathrm{S}})\cos(H_{\mathrm{M}}) -
\cos(H_{\mathrm{S}}) \sin(H_{\mathrm{M}}) \cos({\Delta}AZ)]/
[\cos(H_{\mathrm{S}}) \sin({\Delta}AZ)].
\end{equation}}\unskip
This equation is less well known and not easily searchable. We
therefore repeat its derivation~\cite{24} in the Appendix of the
present work, together with the slightly different form given by Myers
and Myers~\cite{25}.

Figure~\ref{fig4}a presents a panoramic view of the moon over the
day, in steps of one hour. Horizontally, the azimuth angles are shown
as compass directions. Vertically, the moon's height is plotted as its
tangent, with tilt angles calculated from Equation~(\ref{eq3}). The
dashed curve represents  the celestial equator. In a panoramic view,
half of it is above the horizon, stretching from east to west, via
south. The part from west via north back to east  is below the horizon.

\begin{figure}
\includegraphics{fig04}
\caption{\label{fig4}Panoramic views of: (a) The moon, as observed on
16 June 255 BC from Alexandria, in steps of 1~h. At meridian transit
(15:52:47~UT), the moon's declination is 
6\textdegree{}30$^{\prime}$8$^{\prime\prime}$ and its altitude 
64\textdegree{}53$^{\prime}$4$^{\prime\prime}$. The dashed curve
indicates the celestial equator. (b) Moon and sun at 16:43:25~UT, when
the moon is upright. The sinusoidal curve is the great circle through
the positions of the moon and the sun.}
\end{figure}

The phase of the moon changes from 44.3\% at its rise to 50.5\% at moon
set. Just under 0.5\% in an hour. For an observer without modern
equipment to zoom in, and relying only on his naked eye vision, it will
be impossible to tell whether the moon's phase is precisely half or
whether it is off by a few percent~\cite{14}. Also, the elongation
changes slowly, by about half a degree in one hour. By the time the
moon's phase reaches precisely 50\%, the sun is already below the
horizon. 

In contrast to the phase and the elongation, the moon's tilt angle
changes rapidly. One hour before its shadow edge is upright (which
happens at 16:43:25 UT), it is tilted backwards by 27.7\textdegree. One
hour after, its forward tilts is ${+}$18.3, hence a change of
46\textdegree{} in two hours. This represents a change of a full degree
in just over two and a half minutes. 

Figure~\ref{fig4}b shows the situation when the shadow edge of the
moon is precisely vertical. This happens at 16:43:25 UT, when the
moon's phase is 47.9\%. We propose that this is the configuration
described by Aristarchus. Table~\ref{tab1} lists the heights of the
moon and the sun, their azimuths, and the elongation for this instant
in time. In addition to the true values, the apparent values for the
heights and for the elongation are also shown. The apparent elongation
is almost exactly equal to Aristarchus' value of 87\textdegree, in
agreement with hypothesis~4.

%tab1
\begin{table}
\caption{\label{tab1}Alexandria, 31\textdegree{}12$^{\prime}$N,
29\textdegree{}55$^{\prime}$E, anno 255~BC, at UT $=$\ 16~h 43~m
25~s}
\begin{tabular}{cccc}
\thead
&amp; True height &amp; Apparent height &amp; Azimuth \\
\endthead
Sun &amp; 2\textdegree{}18.7$^{\prime}$ &amp; 2\textdegree{}34.4$^{\prime}$ &amp;
296\textdegree{}2.8$^{\prime}$ \\
Moon &amp; 62\textdegree{}38.2$^{\prime}$ &amp; 62\textdegree{}11.1$^{\prime}$ &amp; 
207\textdegree{}14.6$^{\prime}$ \\
Elongation &amp; True 87\textdegree{}23.8$^{\prime}$ &amp; Apparent
87\textdegree{}9.9$^{\prime}$ &amp; Phase $=$\ 0.4786 
\botline
\end{tabular}
\end{table}

\section{How Aristarchus could have measured the moon's elongation from
the sun}

Aristarchus is said to have invented the scaph\`{e}
(${\rmsigma}{\rmkappa}{\acute{\rmalpha}}{\rmvarphi}{\rmeta}$), a hollow half or quarter
sphere, equipped with a shadow pin. For measuring the moon--sun
distance it would, however, not have been practical, since the moon
does not cast a shadow, at least not one that is perceptible in
daylight. For Aristarchus, the zero-tilt moment would have been an
attractive criterion to base his measurement upon, since, in addition,
the moon was already indistinguishably close to being half full. 

A gnomon, a vertical stick in its simplest form, on a horizontal
sundial would then have been very much suited. The moon's azimuth can
be found with great precision by aligning the straight shadow edge over
its surface with the gnomon. In the horizontal plane, the direction
that is at right angles with this moon azimuth gives a point on the
horizon, which we denote by $Q$ (of ``quadrant''). This point marks
precisely the 90\textdegree{} difference from the moon's position in
both azimuth and elongation. This observation may also be understood
from Figure~\ref{fig4}b, where this great circle is shown in a
panoramic view. Half of it is above the horizon while  the other half
is below, as is the case  for any great circle. Its maximum is halfway
the ``above horizon'' part. When in this point, the moon is seen
upright. It is illuminated by the sun (hypothesis~1) and the connecting
path to the sun is locally horizontal in our 2D panoramic rendering. 

Figure~\ref{fig5} shows a detail of Figure~\ref{fig4}b, as seen in the
direction of the sun.

\begin{figure}
\includegraphics{fig05}
\caption{\label{fig5}A close-up of Figure~\ref{fig4}b in the
direction at right angles with the azimuth of the moon. The ``triangle
of missing fractions'', with $Q$-Sun as its hypotenuse, is really a
spherical triangle with the foot of the gnomon as its center, but
because of it being very small, it is a flat triangle to good
approximation. The sides and angles are related as in
formula~(\ref{eq4}).}
\end{figure}

For the elongation and the sun--moon azimuth difference, their
``missing parts'' of filling a full quadrant are ($90\text{\textdegree}
-\uppsi$) and ($90\text{\textdegree} -\Delta AZ$), and these angles are
related by:
{\begin{equation}\label{eq4}
\cos(H_{\mathrm{M}}) = \cot(\Delta AZ)/\cot( \uppsi) \approx
(90\text{\textdegree} - \Delta AZ)/(90\text{\textdegree} - \uppsi).
\end{equation}}\unskip
Equation~(\ref{eq4}) follows from Napier's mnemonic rules~\cite{21} by
taking $\beta= 90\text{\textdegree}$ in the spherical triangle PSM of
Figure~\ref{fig3}. 

Not having recourse to spherical trigonometry, yet Aristarchus could
have come to the same result by reasoning as follows: it is of course
trivially true that 90\textdegree{} in azimuth is a quadrant of a full
circle along the horizon. He would likely have found the missing
fraction of the elongation by noting that from the upright moon to the
point $Q$ spans exactly one quarter of the moon--sun great circle. The
sun's azimuth difference from the moon falls short of the
90\textdegree{} point $Q$ by 1.2\textdegree{} (see Table~\ref{tab1}),
and it would have been quite easy to measure this. For Aristarchus, the
procedure would then have been to evaluate the corresponding missing
fraction in elongation by producing a line from $Q$ via the sun in the
direction of the moon and measuring it against the azimuth deficit that
he had found, or directly, by using a handheld quadrant. In doing so,
he would, of course, have used the sun's elevation as he saw it.
Refraction had lifted the sun by an amount of 15.7$^{\prime}$, about
equal to its semi-diameter, and for Aristarchus the missing fraction
would have been 2\textdegree{}50$^{\prime}$, almost precisely ``one
thirtieth of a quadrant''. The resulting  apparent elongation of
87\textdegree{}10$^{\prime}$ is the same as that obtained from a
direct calculation, using Equation~(\ref{eq2}), see Table~\ref{tab1}.

\section{A \textit{Gedankenexperiment}: the moon's elongation and
longitude}

One might ask if the result of 87\textdegree{} is an artefact of the
method that we describe: finding the elongation at a time when the
half-moon's terminator is seen vertical. Let us therefore imagine that
a similar observation was not only conducted in Alexandria, but also in
other important centra of the ancient world, Basrah in Mesopotamia, the
Greek and Phoenician settlements in the Mediterranean, Syracuse,
Carthage, Massilia and Cadiz. And, just for the sake of argument, let
us include the most westerly island of the Azores, Ilha do Faial.

The results are given in Table~\ref{tab2}. The third column gives the
time when the moon's shadow edge is locally seen as vertical. Over a
time of five and a half hours, the true elongation increases from
86\textdegree{}42.1$^{\prime}$ for the Basrah observation to
89\textdegree{}49.4$^{\prime}$ on the Azores. For each later
observation, the height of the sun is lower, almost 3\textdegree{} for
Basrah and only just above the horizon for the Ilha do Faial. At the
same time, the phase gets closer and closer to 0.5000. According to
Aristarchus' criterion for visual acuity, formulated above, all phases
are close enough to 0.5000 to make the moon look like half.  

%tab2
\begin{table}
\caption{\label{tab2}Elongation and phase at other places at the local
time of upright moon, 16 June 255 BC}
\begin{tabular}{ccccccc}
\thead
Place &amp; Coordinates &amp; Time (UT) &amp; True elong. &amp; App. elong. &amp;
$H_{\mathrm{Sun}}$ &amp; Phase \\
\endthead
Basrah        &amp; 30\textdegree{}31$^{\prime}$N, 47\textdegree{}46$^{\prime}$E &amp; 15:27:08 &amp; 86\textdegree{}42.1$^{\prime}$ &amp; 86\textdegree{}29.8$^{\prime}$ &amp; 2\textdegree{}56.4$^{\prime}$ &amp; 0.4726 \\
Alexandria    &amp; 31\textdegree{}12$^{\prime}$N, 29\textdegree{}55$^{\prime}$E &amp; 16:43:25 &amp; 87\textdegree{}23.8$^{\prime}$ &amp; 87\textdegree{}09.9$^{\prime}$ &amp; 2\textdegree{}18.7$^{\prime}$ &amp; 0.4786 \\
Syracuse      &amp; 37\textdegree{}04$^{\prime}$N, 15\textdegree{}11$^{\prime}$E &amp; 18:01:01 &amp; 88\textdegree{}06.3$^{\prime}$ &amp; 87\textdegree{}50.9$^{\prime}$ &amp; 1\textdegree{}33.9$^{\prime}$ &amp; 0.4848 \\ 
Carthage      &amp; 36\textdegree{}47$^{\prime}$N, 10\textdegree{}12$^{\prime}$E &amp; 18:21:01 &amp; 88\textdegree{}17.3$^{\prime}$ &amp; 88\textdegree{}01.2$^{\prime}$ &amp; 1\textdegree{}25.1$^{\prime}$ &amp; 0.4864 \\ 
Massilia      &amp; 43\textdegree{}16$^{\prime}$N, 05\textdegree{}24$^{\prime}$E &amp; 19:02:26 &amp; 88\textdegree{}39.9$^{\prime}$ &amp; 88\textdegree{}23.9$^{\prime}$ &amp; 1\textdegree{}00.0$^{\prime}$ &amp; 0.4897 \\ 
Cadiz         &amp; 36\textdegree{}29$^{\prime}$N, 06\textdegree{}15$^{\prime}$W &amp; 19:28:56 &amp; 88\textdegree{}54.4$^{\prime}$ &amp; 88\textdegree{}36.1$^{\prime}$ &amp; 0\textdegree{}54.5$^{\prime}$ &amp; 0.4918 \\
Ilha do Faial &amp; 38\textdegree{}34$^{\prime}$N, 28\textdegree{}42$^{\prime}$W &amp; 21:09:09 &amp; 89\textdegree{}49.4$^{\prime}$ &amp; 89\textdegree{}27.4$^{\prime}$ &amp; 0\textdegree{}08.6$^{\prime}$ &amp; 0.4998
\botline
\end{tabular}
\end{table}

This \textit{Gedankenexperiment} illustrates that the elongation of
87\textdegree, observed by Aristarchus, is not an artefact of the
method. Had he done the experiment in a more westerly location, he
would have found a somewhat larger elongation.

\section{Discussion and conclusion}

As shown above, Aristarchus' fourth hypothesis that, at half-moon, its
angular distance to the sun is less than a full quadrant by one
thirtieth of it can be understood if we assume that the measurements
were made at a moment when the shadow edge over the moon's surface was
seen as vertical. This moment can be timed to within a few minutes. By 
contrast, the naked eye cannot pinpoint the precise first or last
quarter phase of the moon within several hours. 

Shadow-casting instruments available in Aristarchus' time, would not
have been practical, and the visor-equipped armillary sphere had not
yet been developed in his days. The additional requirement that the
moon should be seen as upright would, however, make it feasible to do
the observation using a gnomon. Indeed, it allows one to accurately
find a point on the horizon, which we name $Q$, that marks a full
quadrant's distance from the moon, both in azimuth and along the great
circle that connects moon and sun. The problem is then simplified from
finding the elongation itself to finding its missing fraction from
90\textdegree, and the gnomon would be an almost ideal instrument. 

As an example, we have analyzed the situation that Aristarchus may have
observed himself in Alexandria on 16 June 255 BC. Above, we have shown
how the observed elongation changes when measured on the same day, but
from different longitudes.

A similar experiment could also have been performed at the same
location, Alexandria, but on different dates. Table~\ref{tab3} gives
some examples of the resulting elongation, which ranges from
89\textdegree{}6.3$^{\prime}$ to 86\textdegree{}22.7$^{\prime}$. This
range is biased towards values smaller than 90\textdegree{} by the
trivial requirement that the sun must be above the horizon to be
visible. On the low side, it is restricted by the fact that for phases
smaller than about 0.467, the moon is perceived as not yet half, as
shown above in our discussion on visual acuity. The corresponding lower
limit on the apparent elongation is around 86\textdegree. Aristarchus'
value of 87\textdegree{} falls near the middle of this range.

%tab3
\begin{table}
\caption{\label{tab3}Elongation and phase seen from Alexandria for
upright moon on different dates}
\begin{tabular}{cccccc}
\thead
Date &amp; Time (UT) &amp; True elong. &amp; App. elong. &amp; $H_{\mathrm{Sun}}$ &amp;
Phase \\
\endthead
20 January 255 BC &amp; 15:17:42 &amp; 89\textdegree{}06.3$^{\prime}$ &amp; 88\textdegree{}45.8$^{\prime}$ &amp; 00\textdegree{}48.1$^{\prime}$ &amp; 0.4935 \\
25 October 267 BC &amp; 15:16:44 &amp; 88\textdegree{}23.8$^{\prime}$ &amp; 88\textdegree{}14.4$^{\prime}$ &amp; 00\textdegree{}50.5$^{\prime}$ &amp; 0.4873 \\
16 June 255 BC    &amp; 16:43:25 &amp; 87\textdegree{}23.4$^{\prime}$ &amp; 87\textdegree{}11.1$^{\prime}$ &amp; 2\textdegree{}18.7$^{\prime}$ &amp; 0.4786 \\
11 August 271 BC  &amp; 16:32:52 &amp; 86\textdegree{}22.7$^{\prime}$ &amp; 86\textdegree{}12.2$^{\prime}$ &amp; 2\textdegree{}35.8$^{\prime}$ &amp; 0.4697
\botline
\end{tabular}
\end{table}

From our analysis, one may conclude that Aristarchus' intention to
perform measurements as accurately as possible should not be
underestimated.

\section*{Declaration of interests}

The authors do not work for, advise, own shares in, or receive funds
from any organization that  could benefit from this article, and have
declared no affiliations other than their research organizations.

\section*{Acknowledgement}

Discussions with Luc Dettwiller are gratefully acknowledged.

\back{}

\appendix{}

\def\appendixlabel{Appendix \Alph{section}}

\def\theequation{\thesection.\arabic{equation}}
\setcounter{equation}{0}

\def\thefigure{\thesection\arabic{figure}}
\setcounter{figure}{0}

\section{The moon tilt}

In Figure~\ref{figA1}, the moon is in its last quarter and the observer
sees it west from the sun. The sun illuminates the moon along the
direction SM and the shadow line over the moon is perpendicular to it.
The observer in P sees the moon along the direction PM. If we denote
the angle between PM and SM as ${\alpha}$, then it follows immediately
that the inclination angle, which we will call $T$, is given by
($90\text{\textdegree} -\alpha$). Calculating ${\alpha}$ requires some
spherical trigonometry~\cite{21}. 

\begin{figure}
\includegraphics{fig06}
\caption{\label{figA1}The zenith for the observer is in P. Sun and moon
are shown as S and M, the foot points of their projections onto the
celestial sphere.}
\end{figure}

Use the following two relationships:
{\begin{eqnarray}
\sin({\alpha})/\sin(a) &amp;=&amp; \sin(\gamma)/\sin(c) \quad
(\mbox{sine rule}) \label{eqA1} \Seqnsplit
\sin(c)\cos({\alpha}) &amp;=&amp; \cos(a)\sin(b) - \sin(a)\cos(b)\cos(\gamma) 
\quad (\mbox{sine-cosine rule of the first kind}).\quad \label{eqA2}
\end{eqnarray}}\unskip
Eliminating $\sin(c)$ gives:
{\begin{equation}\label{eqA3}
\cot({\alpha}) = \cos({\alpha})/\sin({\alpha}) = [\cos({a})\sin({b}) -
\sin({a})\cos({b})\cos(\gamma)]/ [\sin({a})\sin(\gamma)].
\end{equation}}\unskip
The observer measures or calculates the moon's and the sun's altitudes
above the horizon, $H_{\mathrm{M}}$ and $H_{\mathrm{S}}$, and also
their azimuthal angles, $AZ_{\mathrm{M}}$ en $AZ_{\mathrm{S}}$. The
connection with the symbols, used above, is:
{\[
a = 90\text{\textdegree} - H_{\mathrm{S}}; \quad
b = 90\text{\textdegree} - H_{\mathrm{M}}; \quad
\gamma = \Delta AZ = AZ_{\mathrm{S}} - AZ_{\mathrm{M}}.
\]}\unskip
The moon's tilt angle is $T= 90\text{\textdegree} -\alpha$ and it is
found from:
{\begin{equation}\label{eqA4}
\tan(T) = [\sin(H_{\mathrm{S}})\cos(H_{\mathrm{M}}) -
\cos(H_{\mathrm{S}})\sin(H_{\mathrm{M}})\cos({\Delta} AZ)]/
[\cos(H_{\mathrm{S}})\sin({\Delta}AZ)].
\end{equation}}\unskip
This is the formula as we have given it in~\cite{23,23a} without proof and with
proof in~\cite{24}.

Its physical interpretation is that the nominator and the denominator
are the sideways and the normal projections, respectively, of the
vector product $\hat{\mathbi{r}}_{\mathrm{S}}\times
\hat{\mathbi{r}}_{\mathrm{M}}$, the cross product of the unit vectors
in the directions to the sun and the moon, as they are seen in a
cartesian frame with the $y$-axis pointing directly to the moon, the
$x$-axis in the observer's horizontal plane and the $z$-axis completing a
right-handed frame.

The observer may instead wish to define his coordinate frame as a
horizontal $x$--$y$ plane and his own zenith direction as the $z$-axis.
This involves a rotation about the $x$-axis over an angle
$H_{\mathrm{M}}$ which leaves the sideways $x$-component unchanged, but
multiplies the normal component by $\cos(H_{\mathrm{M}})$. The result
is then:
{\begin{equation}\label{eqA5}
\tan(T^{\prime}) = [\sin(H_{\mathrm{S}})\cos(H_{\mathrm{M}}) -
\cos(H_{\mathrm{S}})\sin(H_{\mathrm{M}})\cos({\Delta}
AZ)]/[\cos(H_{\mathrm{M}})\cos(H_{\mathrm{S}}) \sin({\Delta}AZ)].
\end{equation}}\unskip
In this form, the equation has recently also been derived by Myers and
Myers~\cite{25}.

\printbibliography
\refinput{crphys20250178-reference.tex}

\end{document}
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