Archimedes - The Sand Reckoner

Translation by Thomas L. Heath (1897) (source)

There are some, king Gelon, who think that the number of the sand is infinite in multitude ; and I mean by the sand not only that which exists about Syracuse and the rest of Sicily but also that which is found in every region whether inhabited or uninhabited. Again there are some who, without regarding it as infinite, yet think that no number has been named which is great enough to exceed its multitude. And it is clear that they who hold this view, if they imagined a mass made up of sand in other respects as large as the mass of the earth, including in it all the seas and the hollows of the earth filled up to a height equal to that of the highest of the mountains, would be many times further still from recognising that any number could be expressed which exceeded the multitude of the sand so taken. But I will try to show you by means of geometrical proofs, which you will be able to follow, that, of the numbers named by me and given in the work which I sent to Zeuxippus, some exceed not only the number of the mass of sand equal in magnitude to the earth filled up in the way described, but also that of a mass equal in magnitude to the universe. Now you are aware that 'universe' is the name given by most astronomers to the sphere whose centre is the centre of the earth and whose radius is equal to the straight line between the centre of the sun and the centre of the earth. This is the common account (τἁ γραφόμενα), as you have heard from astronomers. But Aristarchus of Samos brought out a book consisting of some hypotheses, in which the premisses lead to the result that the universe is many times greater than that now so called. His hypotheses are that the fixed stars and the sun remain unmoved, that the earth revolves about the sun in the circumference of a circle, the sun lying in the middle of the orbit, and that the sphere of the fixed stars, situated about the same centre as the sun, is so great that the circle in which he supposes the earth to revolve bears such a proportion to the distance of the fixed stars as the centre of the sphere bears to its surface. Now it is easy to see that this is impossible; for, since the centre of the sphere has no magnitude, we cannot conceive it to bear any ratio whatever to the surface of the sphere. We must however take Aristarchus to mean this : since we conceive the earth to be, as it were, the centre of the universe, the ratio which the earth bears to what we describe as the 'universe' is the same as the ratio which the sphere containing the circle in which he supposes the earth to revolve bears to the sphere of the fixed stars. For he adapts the proofs of his results to a hypothesis of this kind, and in particular he appears to suppose the magnitude of the sphere in which he represents the earth as moving to be equal to what we call the 'universe'.

I say then that, even if a sphere were made up of the sand, as great as Aristarchus supposes the sphere of the fixed stars to be, I shall still prove that, of the numbers named in the Principles [Lost work of Archimedes], some exceed in multitude the number of the sand which is equal in magnitude to the sphere referred to, provided that the following assumptions be made.

1. The perimeter of the earth is about 3,000,000 stadia and not greater.

It is true that some have tried, as you are of course aware, to prove that the said perimeter is about 300,000 stadia. But I go further and, putting the magnitude of the earth at ten times the size that my predecessors thought it, I suppose its perimeter to be about 3,000,000 stadia and not greater.

2. The diameter of the earth is greater than the diameter of the moon, and the diameter of the sun is greater than the diameter of the earth.

In this assumption I follow most of the earlier astronomers.

3. The diameter of the sun is about 30 times the diameter of the moon and not greater.

It is true that, of the earlier astronomers, Eudoxus declared it to be about nine times as great, and Pheidias my father twelve times, while Aristarchus tried to prove that the diameter of the sun is greater than 18 times but less than 20 times the diameter of the moon. But I go even further than Aristarchus, in order that the truth of my proposition may be established beyond dispute, and I suppose the diameter of the sun to be about 30 times that of the moon and not greater.

4. The diameter of the sun is greater than the side of the chiliagon inscribed in the greatest circle in the (sphere of the) universe.

I make this assumption because Aristarchus discovered that the sun appeared to be about $n^{th}$ part of the circle of the zodiac, and I myself tried, by a method which I will now describe, to find experimentally (ὀργανικῶς) the angle subtended by the sun and having its vertex at the eye.

[Up to this point the treatise has been literally translated because of the historical interest attaching to the ipsissima verba of Archimedes on such a subject. The rest of the work can now be more freely reproduced, and, before proceeding to the mathematical contents of it, it is only necessary to remark that Archimedes next describes how he arrived at a higher and a lower limit for the angle subtended by the sun. This he did by taking a long rod or ruler, fastening on the end of it a small cylinder or disc, pointing the rod in the direction of the sun just after its rising (so that it was possible to look directly at it), then putting the cylinder at such a distance that it just concealed, and just failed to conceal, the sun, and lastly measuring the angles subtended by the cylinder. He explains also the correction which he thought it necessary to make because "the eye does not see from one point but from a certain area."]

The result of the experiment was to show that the angle subtended by the diameter of the sun was less than $1/164$-th part, and greater than $1/200$-th part, of a right angle.

To prove that (on this assumption) the diameter of the sun is greater than the side of a chiliagon, or figure with 1000 equal sides, inscribed in a great circle of the "universe."

Suppose the plane of the paper to be the plane passing through the centre of the sun, the centre of the earth and the eye, at the time when the sun has just risen above the horizon. Let the plane cut the earth in the circle $EHL$ and the sun in the circle $FKG$, the centres of the earth and sun being $C$, $O$ respectively, and $E$ being the position of the eye.

Further, let the plane cut the sphere of the "universe" (i.e. the sphere whose centre is $C$ and radius $CO$) in the great circle $AOB$.

Draw from $E$ two tangents to the circle $FKG$ touching it at $P$, $Q$, and from $C$ draw two other tangents to the same circle touching it in $F$, $G$ respectively.

Let $CO$ meet the sections of the earth and sun in $H$, $K$ respectively; and let $CF$, $CG$ produced meet the great circle $AOB$ in $A$, $B$.

Join $EO$, $OF$, $OG$, $OP$, $OQ$, $AB$, and let $AB$ meet $CO$ in $M$.

Now $CO > EO$, since the sun is just above the horizon.

Therefore $\angle PEQ > \angle FCG$.

And $\angle PEQ > 1/200 R$ but $< 1/164 R$, where $R$ represents a right angle.

[diagram]

Thus $\angle FCG < R$, a fortiori,

and the chord $AB$ subtends an arc of the great circle which is less than $1/656$-th of the circumference of that circle, i.e.

$$AB < (\text{side of 656-sided polygon inscribed in the circle})$$

Now the perimeter of any polygon inscribed in the great circle is less than $44/7 CO$. [Cf. Measurement of a circle, Prop. 3.]

Therefore $AB : CO < 11 : 1148$,

and, a fortiori, $AB < 1/100 CO$$. ($\alpha$)

Again, since $CA = CO$, and $AM$ is perpendicular to $CO$, while $OF$ is perpendicular to $CA$,

$$AM = OF$$

Therefore $AB = 2 AM = (\text{diameter of sun})$.

Thus $(\text{diameter of sun}) < 1/100 CO$, by ($\alpha$),

and, a fortiori, $(\text{diameter of earth}) < 1/100 CO$. [Assumption 2]

Hence $CH + OK < 1/100 CO$,

so that $HK > 99/100 CO$,

or $CO : HK < 100 : 99$.

And $CO > CF$, while $HK < EQ$.

Therefore $CF : EQ < 100 : 99$. (β)

Now in the right-angled triangles $CFO$, $EQO$, of the sides about the right angles,

$OF = OQ$, but $EQ < CF$ (since $EO < CO$).

Therefore $∠ OEQ : \angle OCF > CO : EO$,

but $< CF : EQ$. [The proposition here assumed is of course equivalent to the trigonometrical formula which states that, if $\alpha$, $\beta$ are the circular measures of two angles, each less than a right angle, of which $\alpha$ is the greater, then $$\frac{\tan(\alpha)}{\tan(\beta)} > \frac{\alpha}{\beta} > \frac{\sin(\alpha)}{\sin(\beta)}$$]

Doubling the angles,

$\angle PEQ : \angle ACB < CF : EQ$

$< 100 : 99$, by (β) above.

But $\angle PEQ > 1/200 R$, by hypothesis.

Therefore $\angle ACB > 99/20000\ R > 1/203\ R$.

It follows that the arc AB is greater than $1/812$-th of the circumference of the great circle $AOB$.

Hence, a fortiori, $AB > (\text{side of chiliagon inscribed in great circle})$, and $AB$ is equal to the diameter of the sun, as proved above.

The following results can now be proved :

$$(\text{diameter of "universe"}) < 10,000\ (\text{diameter of earth})$$

and $(diameter of "universe") < 10,000,000,000\ \text{stadia}$.

(1) Suppose, for brevity, that $d_u$ represents the diameter of the "universe," $d_s$ that of the sun, $d_e$ that of the earth, and $d_m$ that of the moon.

By hypothesis, $d_s \ngtr 30\ d_m$, [Assumption 3]

and $d_e > d_m$; [Assumption 2]

therefore $d_s < 30\ d_e$.

Now, by the last proposition,

$d_s > (\text{side of chiliagon inscribed in great circle})$,

so that $(\text{perimeter of chiliagon}) < 1000 d_e$. $< 30,000\ d_e$.

But the perimeter of any regular polygon with more sides than 6 inscribed in a circle is greater than that of the inscribed regular hexagon, and therefore greater than three times the diameter. Hence

$$(\text{perimeter of chiliagon}) > 3 d_u$$

It follows that $d_u < 10,000\ d_e$.

(2) $(\text{Perimeter of earth}) \ngtr 3,000,000\ \text{stadia}$. [Assumption 1]

and $(\text{perimeter of earth}) > 3\ de$.

Therefore $d_e < 1,000,000\ \text{stadia}$,

whence $d_u < 10,000,000,000\ \text{stadia}$.

Assumption 5

Suppose a quantity of sand taken not greater than a poppy-seed, and suppose that it contains not more than $10,000$ grains.

Next suppose the diameter of the poppy-seed to be not less than $1/40$-th of a finger-breadth.

Orders and periods of numbers

The traditional greek numerals are given by greek letters as :

• ${\rm A},{\rm B},\mathrm{\Gamma },\mathrm{\Delta },{\rm E},Ϝ,{\rm Z},{\rm H},\mathrm{\Theta }$ for numbers $1$ through $9$
• $\mathrm{\text{Ι}},\mathrm{\text{Κ}},\mathrm{\Lambda },\mathrm{\text{Μ}},\mathrm{\text{Ν}},\mathrm{\Xi },\mathrm{\text{Ο}},\mathrm{\Pi },\mathrm{Ϙ}$ for numbers $10$ through $90$,
• $\mathrm{\text{Ρ}},\mathrm{\text{Σ}},\mathrm{{\rm T}},\mathrm{\text{Υ}},\mathrm{\text{Φ}},\mathrm{{\rm X}},\mathrm{\text{Ψ}},\mathrm{\Omega },\mathrm{ϡ}$ for numbers $100$ through $900$

with larger numbers given by a factor of $1000$ indicated with an additional sign, typically the symbol ͵. For instance, Α is $1$, but ͵Α is $1000$. The myriad here being denoted by a iota with that sign, ͵Ι, $10,000$.

1. We have traditional names for numbers up to a myriad ($10,000$); we can therefore express numbers up to a myriad myriads ($100,000,000$). Let these numbers be called numbers of the first order.

   Suppose the $100,000,000$ to be the unit of the second order, and let the second order consist of the numbers from that unit up to $(100,000,000)^2$.

   Let this again be the unit of the third order of numbers ending with $(100,000,000)^3$; and so on, until we reach the $100,000,000$-th order of numbers ending with $(100,000,000)^{100,000,000}$, which we will call $P$.

2. Suppose the numbers from $1$ to $P$ just described to form the first period.

   Let $P$ be the unit of the first order of the second period, and let this consist of the numbers from $P$ up to $100,000,000\ P$.

   Let the last number be the unit of the second order of the second period, and let this end with $(100,000,000)^2\ P$.

   We can go on in this way till we reach the $100,000,000$-th order of the second period ending with $(100,000,000)^{100,000,000}\ P$, or $P^22$.

3. Taking $P^2$ as the unit of the first order of the third period, we proceed in the same way till we reach the $100,000,000$-th order of the third period ending with $P^3$.
4. Taking $P^3$ as the unit of the first order of the fourth period, we continue the same process until we arrive at the $100,000,000$-th order of the $100,000,000$-th period ending with $P^{100,000,000}$. This last number is expressed by Archimedes as "a myriad-myriad units of the myriad-myriad-th order of the myriad-myriad-th period (αἱ μυριακισμυριοστᾶσ περιόδου μυριακισμυριοστῶν ἀριθμῶν μυρίαι μυριάδες)", which is easily seen to be $100,000,000$ times the product of $(100,000,000)^{99,999,999}$ and $P^{99,999,999}$ i.e. $P^{100,000,000}$.

Octads

Consider the series of terms in continued proportion of which the first is $1$ and the second $10$ [i.e. the geometrical progression $10^0, 10^1, 10^2, 10^3, \ldots$]. The first octad of these terms [i.e. $10^0, 10^1, 10^2, \ldots, 10^7$] fall accordingly under the first order of the first period above described, the second octad [i.e. $10^8, 10^9, \ldots, 10^{15}$] under the second order of the first period, the first term of the octad being the unit of the corresponding order in each case. Similarly for the third octad, and so on. We can, in the same way, place any number of octads.

Theorem

If there be any number of terms of a series in continued proportion, say $A_1, A_2, A_3, \ldots, A_m, \ldots, A_n, \ldots, A_{m+n-l}, \ldots$ of which $A_l = 1$, $A_2 = 10$ [so that the series forms the geometrical progression $10^0, 10^1, 10^2, \ldots, 10^{m-1}, \ldots, 10^{n-1}, \ldots, 10^{m+n-2}, \ldots$], and if any two terms as $A_m$, $A_n$ be taken and multiplied, the product $A_m A_n$ will be a term in the same series and will be as many terms distant from $A_n$, as $A_m$ is distant from $A_1$; also it will be distant from $A_l$ by a number of terms less by one than the sum of the numbers of terms by which $A_m$ and $A_n$, respectively are distant from $A_1$.

Take the term which is distant from $A_n$, by the same number of terms as $A_m$ is distant from $A_1$. This number of terms is $m$ (the first and last being both counted). Thus the term to be taken is $m$ terms distant from $A_n$, and is therefore the term $A_{m+n-1}$.

We have therefore to prove that

\begin{equation} A_m A_n = A_{m+n-1} \end{equation}

Now terms equally distant from other terms in the continued proportion are proportional.

Thus $A_m / A_1 = A_{m+n-1} / {A_n}$.

But $A_m= A_m A_1$, since $A1 = 1$.

Therefore $A_{m+n-1} = A_m A_n$. (1)

The second result is now obvious, since $A_m$ is $m$ terms distant from $A_1$, $A_n$ is $n$ terms distant from $A_1$, and $A_{m+n-1}$ is $(m+n-1)$ terms distant from $A_1$.

Application to the number of the sand

By Assumption 5,

$$(\text{diam. of poppy-seed}) \ngtr (\text{finger-breadth})$$

and, since spheres are to one another in the triplicate ratio of their diameters, it follows that

\begin{eqnarray} (\text{sphere of diam. 1 finger-breadth}) &\ngtr& 64,000\ \text{poppy-seeds}\\ &\ngtr& 64,000 \times 10,000\\ &\ngtr& 640,000,000\\ &\ngtr& \ 6 \text{units of second order} + 40,000,000 \text{units of first order grains of sand}\\ (\text{a fortiori})&\ngtr& \ 10 \text{units of second order of numbers} \end{eqnarray}

We now gradually increase the diameter of the supposed sphere, multiplying it by $100$ each time. Thus, remembering that the sphere is thereby multiplied by $100^3$ or $1,000,000$, the number of grains of sand which would be contained in a sphere with each successive diameter may be arrived at as follows.

Diameter of sphere.	Corresponding number of grains of sand.
(1) $100$ finger-breadths	$< 1,000,000 \times 10$ units of second order $< (7th term of series) $\times$ (10th term of series) $<$ 16th term of series [i.e. $10^{15}$] $< 10^7$ or $10,000,000$ units of the second order.
(2) $10,000$ finger-breadths	$< 1,000,000 \times $ (last number) $< (7th term of series) $\times$ (16th term of series) $<$ 22nd term of series [i.e. $10^{21}$] $< 10^5$ or $100,000$ units of the third order.
(3) $1$ 1 stadium (< $10,000$ finger-breadths)	$< 100,000$ units of the third order.
(4) $100$ stadia	$< 1,000,000 \times $ (last number) $< (7th term of series) $\times$ (22nd term of series) $<$ 28nd term of series [i.e. $10^{27}$] $< 10^3$ or $1000$ units of the fourth order.
(4) $10,000$ stadia	$< 1,000,000 \times $ (last number) $< (7th term of series) $\times$ (28nd term of series) $<$ 34th term of series [i.e. $10^{33}$] $< 10$ units of the fifth order.
(4) $1000,000$ stadia	$< (7th term of series) $\times$ (34th term) $<$ 40th term of series [i.e. $10^{39}$] $< 10^7$ or $10,000,000$ units of the fifth order.
(4) $10,000,000,000$ stadia	$< (7th term of series) $\times$ (46th term) $<$ 52nd term of series [i.e. $10^{51}$] $< 10^3$ or $1,000$ units of the seventh order.

But, by the proposition above,

$$(\text{diameter of "universe"})<10,000,000,000\text{stadia}$$

Hence the number of grains of sand which could be contained in a sphere of the size of our "universe" is less than $1,000$ units of the seventh order of numbers [or $10^{51}$].

From this we can prove further that a sphere of the size attributed by Aristarchus to the sphere of the fixed stars would contain a number of grains of sand less than $10,000,000$ units of the eighth order of numbers [or $10^{56+7}= 10^{63}$].

For, by hypothesis,

$$(\text{earth}) : (\text{"universe"}) = (\text{"universe"}) : (\text{sphere of fixed stars})$$

And

$$(\text{diameter of "universe"}) < 10,000\ (\text{diam. of earth})$$

whence

$$(\text{diam. of sphere of fixed stars}) < 10,000\ (\text{diam. of "universe"})$$

Therefore

$$(\text{sphere of fixed stars}) < (10,000)^3\ (\text{"universe"})$

It follows that the number of grains of sand which would be contained in a sphere equal to the sphere of the fixed stars

\begin{eqnarray} &<& (10,000)^3 \times 1,000\ \text{units of seventh order}\\ &<& (\text{13th term of series}) × (\text{52nd term of series})\\ &<& \text{64th term of series [i.e. $10^{63}$]}\\ &<& 10^7\ \text{or}\ 10,000,000\ \text{units of eighth order of numbers.} \end{eqnarray}

Conclusion

I conceive that these things, King Gelon, will appear incredible to the great majority of people who have not studied mathematics, but that to those who are conversant therewith and have given thought to the question of the distances and sizes of the earth, the sun and moon and the whole universe, the proof will carry conviction. And it was for this reason that I thought the subject would be not inappropriate for your consideration.