1. The document discusses how Archimedes may have calculated the area under a parabola and the centroid of a parabolic region without calculus, using only geometry. It estimates that Archimedes' results came close to matching what is obtained through calculus. 2. It suggests experiments Archimedes could have performed to estimate the centroid of a half-parabola by constructing cardboard models and balancing them. The results of these experiments closely matched the calculus solution. 3. It examines how Archimedes may have estimated the arc length of a parabola and spiral using measurement tools available to him like dividers, and how his work laid the foundations for later mathematicians like Pappus

1. Noname manuscript No.(will be inserted by the editor)
Pretending to Be Archimedes
Properties of the parabola and spiral he could have learned
without calculus
Douglas Leadenham
Received: date / Accepted: date
Abstract The 1998 purchase of the Archimedes Palimpsest allowing technically
advanced imaging of the underlying Greek text along with similarly advanced
analysis of the Antikythera Mechanism, found in 1900 and with recently dated
construction at between 150 and 100 BCE, that now permits construction of a
working replica, has renewed interest in the mathematical works of Archimedes.
Archimedes was the only genius of antiquity capable of designing the Antikythera
Mechanism, and its dating to the 2
nd
century BCE places him at the time of the
design. It is desired to compare some of Archimedes mathematical results with
those same ones now taught as examples in calculus classes to see how close he
may have been to the development of the calculus.
Keywords Archimedes · Pappus's Centroid Theorem
PACS 01.65.+g
Mathematics Subject Classication (2010) 01A20 · 97-03
1What calculus taught us
Every beginning student learns how to nd the area under a parabola and from
that the area between a chord of the parabola and the curve y = x2
. In the simple
case of a parabola bounded by the origin and a point P(xp, yp), the area under
the parabola is
1
3 x3
p or
1
3 xpyp, where the rectangle bounding the parabola has area
Ar = xpyp. Thus the area of the parabola is
2
3 Ar.
Archimedes (287-212 BCE) was proud of his result that the area of a parabola
is
4
3 of the area of the inscribed triangle. Draw a diagonal in the rectangle from
O to P and see that Archimedes was right. His result is Proposition 17 of tetrag-
wnisvmåc parabol¨c, Quadrature of the Parabola. He developed his result by the
D. Leadenham
675 Sharon Park Drive, No. 247, Menlo Park, California 94025
Tel.: 650-233-9859
E-mail: douglasleadenham@gmail.com

2. 2 Douglas Leadenham
Fig. 1 Constructed half-parabola
method of exhaustion whereby he constructed a sequence of triangles inside and
circumscribing the parabola, subdividing them into trapezia, or what we call trape-
zoids, using the lever principle to add up their areas as he went along, until he
could no longer construct a smaller triangle with his straightedge and compass.
The end result is obtained when the areas inside and outside the parabola con-
verge. This result is also Proposition 1 of Method, that Archimedes gave the title,
perÈ tÀn mhqanikÀn jewrhmˆtwn präc Eratosvjènhn; èfodoc, to Eratosthenes, con-
cerning Mechanical Theorems; Method.[Palimpsest I][Palimpsest II] His method
approached what we today call innitesimals, at the end of a sequence. This is
the foundation of integral calculus, and we justiably admire Archimedes for this
work.[Works]
2Pretend to be Archimedes the experimenter
Archimedes, besides being a mathematician, was also a physicist and inventor. He
rst dened the principle of the lever, which had been used for centuries before
him as the balance beam of scales in the marketplace. We can suppose that the
idea of dividing the parabola into narrow strips parallel to its axis had occurred to
him, and that each strip would balance at its midpoint. Divide the parabola again
into narrow strips parallel to the chord from the axis to xp, and these too would
balance at their midpoints. The intersection of both series of midpoints should tell

3. Archimedes 3
Fig. 2 Cardboard half-parabola
where the parabola's centroid is. One of the lost works of Archimedes is perÐzugÀn,
On balances or On levers, so we can only infer his thinking on this.[Works]
2.1 The centroid of a right-angled, planar gure with a parabolic boundary, from
calculus
Denition 1 x−coordinate of the centroid
Given that Ap is the area of the parabola, and x is the x−coordinate of the
centroid,
x =
1
Ap
Pˆ
O
x(yp − x2
)dx
The parabola has been turned upside down in order to calculate areas under
the curve.
x =
3
2x3
p


Pˆ
O
ypxdx −
Pˆ
O
x3
dx

 =
3
2x3
p
yp
x2
p
2
−
x4
p
4
x =
3
2x3
p
x4
p
4
=
3
8
xp

4. 4 Douglas Leadenham
Denition 2 y−coordinate of the centroid
y =
1
Ap
Pˆ
O
y
√
ydy
y =
3
2xpyp
Pˆ
O
y
3
2 dy =
3
2xpyp
y
5
2
p
5
2
y =
3
5y
3
2
p
y
5
2
p =
3
5
yp
2.2 The centroid from geometrical construction
Construct a parabola with compass and straightedge using the denition as the set
of points equidistant from a focus and a straight line, called the directrix. Locate
by construction the midpoint of the axis at
1
2 yp, and the midpoints of the chord
segments on both sides of the axis at ±1
2 xp. Extend these to the parabola and
then locate the four midpoints of those lines. Extend these until they intersect
at two points symmetric about the axis. Archimedes could easily have done this,
and he had dividers to measure their lengths. Doing this results in a construction
centroid C(xc,yc) = (0.358xp, 0.625yp).
If we can assume that he had the equivalent of a French curve or else was very
good at eyeball-tting, we can get another estimate. Draw a smooth curve from
the vertex through the midpoints of the lines bisecting the axis and ending at the
midpoints of the two half-chords. Then draw smooth curves from the midpoint
of the axis through the two midpoints of the lines from the midpoints of the two
half-chords that end on the parabola. The intersections of these four curves give
another estimate of the centroid. Call this estimate C(xf ,yf ), with the subscript
f referring to tted or French, as you desire. This result is, as best as can
be drawn on a sheet of notebook paper, C(xf , yf ) = (0.388xp, 0.584yp). If we
assign the calculus result the notation with subscript t, to signify theory, we
have C(xt, yt) = (0.375xp, 0.600yp). So the two estimates Archimedes could have
made bracket the calculus result. So, more applause for Archimedes.
There is more. Archimedes actually did prove this result with geometry. This is
in àpipèdwn ÐsvorropiÀn b, On the Equilibrium of Planes II, Proposition 10.[Works]
Archimedes himself understood and expressed the dierence between a workable
mathematical model that gives useful results and a mathematical proof. The model
justies further investigations, but it is not a proof.
2.3 The centroid from a physics experiment
Simply construct a half-parabola on a sheet of paper and cut out the same g-
ure from a piece of a cardboard box. Balance the cardboard half-parabola on a
Sharpie
® point that leaves a mark. The mark will be at the centroid. Calling

5. Archimedes 5
this point C(xe, ye), we obtainC(xe, ye) = (0.375xp, 0.608yp). The precision of the
black mark is about the same as the thickness of the cardboard. See Figure 1.
Coincidentally, this result is even better than the construction estimates. Experi-
menter Archimedes would certainly have done this, either with a sheet of copper
or a thin ceramic plate. He is also likely to have marked the centroid thus obtained
and tossed the half-parabola into the air, giving it some rotation to see that the
object rotates about its center of gravity. A thin plate of uniform thickness and
composition will do that, conrming his intuition that the centroid is the center
of gravity of such objects.
3Arc length
Archimedes could have measured the parabolic arc length S with dividers. With
the cardboard object just described we have xp = 6.4, yp = 12.0 and S = 14.
It's hard to know what he would have done with this measurement, but we do
know what Pappus of Alexandria (c.290-c.350 CE) could have done with it 550
years later, along with the centroids of triangles and semicircular arcs. He de-
veloped formulas for the volume of solids of revolution about an external axis,
and for the surface area of surfaces of revolution about an external axis. See
http://mathworld.wolfram.com/PappussCentroidTheorem.html.
3.1 Arc length of a parabolic arc from calculus
Denition 3 Arc length S
S =
Pˆ
O
ds =
Pˆ
O
dx2 + dy2 =
Pˆ
O
dx2 + (2xdx)2
S =
Pˆ
O
1 + 4x2dx
The WolframAlpha
® online integrator does this in a second or so, giving the
indenite integral
S =
1
4
2x 4x2 + 1 + sinh−1
(2x) + constant
Archimedes had no knowledge of hyperbolic functions, and this analytic solu-
tion is often used in textbooks to show methods used to obtain the anti-derivative.
In old-fashioned notation this is equivalent to
S =
1
4
2x 1 + 4x2 + ln 2x + 1 + 4x2
P
O
Here the unit of variable x corresponds to
1
3.5 xP = xE cm, with a precision
of about ±0.1, so the working unit variable x = 1.83 in the formula for S. The

6. 6 Douglas Leadenham
absolute value can be dropped for calculation in the rst quadrant. We compute
the denite integral to be 3.97, and after scaling the result up to the measured
unit length, the adjusted arc length from the formula is 13.9 cm. The measured
arc length by rotating the arc of the cardboard along a ruler is 14 cm.
The reason the arc length scale has to be adjusted is that all parabolas are sim-
ilar. The unit formula variable is determined by the point on the parabola equidis-
tant from the x− and y−axes, besides being equidistant from the focus and direc-
trix. This illustrates a dierence between physics and mathematics. Mathematics
works with pure numbers, whereas physics works with observables, mathematical
objects that can be measured. Measurements require an anthropic scale with stan-
dard units of measurement. Archimedes could have done this and his compass and
straightedge drawings show that he knew where the unique equidistant point was
on the parabola.
1
Archimedes is known to have designed the Antikythera Mechanism; Cicero
saw it and gave a description.[Works] It was able over short time frames to predict
eclipses, so Archimedes certainly had precision measuring instruments.[Mechanism]
Archimedes could have measured arc length with dividers, and he could have
determined likewise the point on the parabola equidistant from the symmetry axis
and the line through the vertex parallel to the directrix. However, he could not
have set the scale without some understanding of Cartesian coordinates. Without
the scale factor it would have been hard for him to see the connection of xp and
yp to arc length or to the relation of y as the square of x. For that matter, it's
hard for us to make the connection of xp and yp to arc length.
It is quite likely that he could have made a parabolic arc of copper wire and
tried tossing it into the air the same way as for the planar parabolic object. He
would have seen it rotate about its center of gravity, but the mathematical centroid
of a wire is dened relative to its external axis of revolution.
3.2 Arc length of a spiral arc
Besides parabolas, Archimedes worked extensively with spirals. Those he could
construct are the eponymous Archimedean spirals in perÈ èlÐkwn, On Spirals. These
spirals have a radius that grows in direct proportion to the orthogonal angle tra-
versed. The polar expression is
r = aθ
1 The important real-life observation of this fact appears in the relation of the orbit of the
planet Jupiter with Earth's orbit. Comets have parabolic orbits, and a hypothetical comet in
the ecliptic plane of the planets that is on a collision trajectory with Earth, reaching perihe-
lion at Earth's orbit, will have its point of equal distance from its x− and y− axes at 4.94
astronomical units from the Sun, point E in Figure 1, just less than the distance of Jupiter's
orbit from the Sun at 5.2 AU. (An astronomical unit, abbreviated AU, is the Earth-Sun dis-
tance.) Jupiter's orbit is shown in Figure 1 as the large circular arc, and Earth's orbit is the
small circle. A situation in which a hypothetical such comet got past point E, headed toward
Earth, was portrayed in the movie, Deep Impact, (1998)©Paramount Pictures and Dream-
Works Pictures. What happens in three-body dynamical systems with two masses, Sun's and
Jupiter's, much larger than the third mass, the comet's, is that the small mass, approaching
point E, is perturbed out of its original orbit into a dierent one that would not impact Earth,
thereby avoiding an extinction level event. The planet Jupiter is Earth's guardian.

7. Archimedes 7
Fig. 3 Constructed Archimedean unit spiral
with coecient a the proportionality constant. Small values of a yield tight
spirals, and large values give open spirals that grow quickly. These spirals are
constructable. The unit spiral, with a = 1 is the easiest, and also illustrates the
point made in the case of the parabola that the mathematics requires pure numbers
without measurement units. Expressed this way,
r
a
= θ
the factor a is a scale factor that shows that all Archimedean spirals are geo-
metrically similar. Figure 3 shows the construction. It is rather busy because the
denition is one that equates the linear variable r with the angular variable θ.
The rst thing geometry students learn is that the measure of an angle is quite
dierent from linear measure. For this reason the increments of radius must be
small enough to permit the approximation of linearity. The angular increments
used in construction are a little less than 4º. Physicists often make this approxi-
mation for angles less than 5º. On the completed diagram, AOB is equilateral,
so m∠AOB = π
3 = 60º. Point E is the intersection of the unit spiral with the
circle, taken as the unit circle, so
r
a = 1. Thus, m∠EOB = 1. Archimedes is thus
the originator of the dimensionless angle measure commonly known as the radian.
Generations of physicists owe him a yet another debt of gratitude. The diagram
shows the radian to be a little less than
π
3 . It is precisely
360º
2π .
Spiral arc length from origin O to any spiral point P is calculated from
S =
Pˆ
O
ds =
Pˆ
O
r2dθ2 + a2dθ2
This simplies to a single variable integral:

8. 8 Douglas Leadenham
S =
Pˆ
O
a θ2 + 1dθ
Thus,
S =
a
2
θ θ2 + 1 + ln θ + θ2 + 1
P
O
Again, Archimedes could only estimate or measure this physically.
The point Q is the rst point at which the vertical tangent diverges. The
vertical tangent slope is y(x) = dy
dx , which requires transforming the polar equation
r = aθ to Cartesian form in order to nd the angle of point Q. The Cartesian
equivalent is
x2 + y2 = a arctan(
y
x
)
Use implicit dierentiation.
1
2
2x + 2yy
x2 + y2
= a
xy − y
x2 + y2
Solve for y (x).
y (x) = −
ay + x x2 + y2
y x2 + y2 − ax
This diverges if the denominator is zero.
y x2 + y2 − ax = 0
x2 + y2 = a
x
y
r
a
= cot θ =
aθ
a
After a simple last step we have an expression for the angle θq of point Q:
θq tan θq = 1
A quick visit to WolframAlpha yields the solution θq = 0.8603336 = 49.3º.

9. Archimedes 9
3.3 Area under the spiral arc from O to Q
Finding this area is an interesting exercise that allows calculation of the volume
of a bowl from volume of a solid of revolution of a spiral-bounded area about an
axis dened as the x−coordinate of the point Q. In addition to the area, we will
need the y−centroid of the area measured from the axis of revolution.
Area under the spiral arc requires Cartesian coordinates: x = r cos θ, y =
r sin θ. On the spiral, x = aθ cos θ, y = aθ sin θ. First we need the area between the
spiral and the x−axis that we will subtract from the area of the rectangle dened
by the origin at point O and the point Q. Archimedes could have done this the
way he did for the parabola, but there is no record that he actually did.
A =
ˆ
ydx =
Qˆ
O
r sin θ (cos θdr − r sin θdθ)
A =
Qˆ
O
aθ sin θ (cos θadθ − aθ sin θdθ) = a2
Qˆ
O
θ sin θ (cos θdθ − θ sin θdθ)
Here it can be seen how the scale factor a sets the physical measurement scale.
In Figure 3 the large circular arc is the unit circle, so, as drawn, r = a = 13.35cm.
We have two integrals to evaluate with help from WolframAlpha
®:
A = a2


Qˆ
O
θ sin θ cos θdθ −
Qˆ
O
θ2
sin2
θdθ


The rst I1 is
I1 =
1
8
[sin 2θ − 2θ cos 2θ] + const
The second I2 is
I2 =
1
24
4θ3
+ 3 − 6θ2
sin 2θ − 6θ cos 2θ + const
These simplify somewhat:
A =
a2
24
6θ2
sin 2θ − 4θ3
Q
O
=
a2
24
6θ2
q sin 2θq − 4θ3
q
Subtract this area from the area of the bounding rectangle to get the area of
the spiral segment:
Aspiral = Arect − A =
a2
2
θ2
q sin 2θq − A
Aspiral =
a2
2
θ2
q sin 2θq −
a2
24
6θ2
q sin 2θq − 4θ3
q
Finally,

10. 10 Douglas Leadenham
Aspiral =
a2
12
θ2
q (3 sin 2θq + 2θq)
This area can be computed simply by using the polar form of the spiral and
polar coordinates to get the area of the wedge between the spiral and the radius
to θq and then adding the area of the triangle right above it.
3.4 Centroid of a spiral segment
This is a bit tricky, but Archimedes' Method, Arqim douc perÐ tÀn mhqanikÀn
jewrhmˆtwn pråc Eraosvjènhn; ëfodoc shows how he would have approached the
problem.[Palimpsest I] We have area of the bounding rectangle, area under the
spiral arc (now designated Au), and area of the spiral segment. The two smaller
areas would balance the rectangle when hung at opposite sides of a balance beam
or lever. Thus,
Aspiralyspiral + Auyu = Arectyrect
The rectangle's vertical centroid coordinate yrect = 1
2 aθq sin θq by symmetry.
That leaves yu to evaluate in order to obtain the desired yspiral result.
yu = yq −
1
Au
Qˆ
O
y · ydx = yq −
1
Au
Qˆ
O
(r sin θ)2
(cos θdr − r sin θdθ)
yu = yq −
1
Au
Qˆ
O
a2
θ2
sin2
θ cos θadθ − a3
θ3
sin3
θdθ
yu = yq −
a3
Au


Qˆ
O
θ2
sin2
θ cos θdθ −
Qˆ
O
θ3
sin3
θdθ


With much more help from WolframAlpha
® we have:
yu = yq −
a3
108Au
81θ3
− 432θ cos θ − 216θ2
− 432 sin θ − 9θ3
cos 3θ
Q
O
This simplies a little:
yu = yq−
a
3θ2
q sin 2θq − 2θ3
q
9θ3
q − 48θq cos θq − 24θ2
q − 48 sin θq − θ3
q cos 3θq
Aspiralyspiral = Arectyrect − Auyu

11. Archimedes 11
3.5 Centroid of a parabolic arc
If Archimedes stopped at this point, no one can blame him. Those with some
calculus knowledge can continue in the way that Pappus of Alexandria did without
calculus for the right circular cone and sphere. Let us determine the centroids of
a parabolic arc relative to radii of revolution about the x− and y−axes.
Denition 4 Centroid of an arc about a radius of revolution
xa =
1
S
Sˆ
xds =
1
S
Sˆ
x dx2 + dy2
For a parabolic arc rotated about the y−axis, this is
xa =
1
S
Sˆ
xds =
1
S
xpˆ
0
x 1 + 4x2dx
Following some instructive integral calculus methods, the result is
xa =
1
12S
1 + 4x2
p
3
2
− 1
Now rotate the parabolic arc about the x−axis. To simplify the calculation,
turn the parabola upside down and the coordinates -90
º so the half-parabola is
half the cross section of a bowl. For rotating about the now vertical x−axis, one
obtains
ya =
1
S
Sˆ
y dx2 + dy2 =
1
S
xpˆ
0
yp − x2
1 + 4x2dx
Following more instructive integral calculus methods with help from WolframAlpha
®,
the result is
ya =
1
S



yp
xpˆ
0
1 + 4x2dx −
xpˆ
0
x2
1 + 4x2dx



ya =
1
S
yp
1
4
2x 1 + 4x2 + ln 2x + 1 + 4x2
P
O
−
1
64
2x 8x2
+ 1 1 + 4x2 − ln 2x + 1 + 4x2
P
O
We must complete the calculation for the limits of integration.
ya =
1
S
yp
16
64
2xp 1 + 4x2
p + ln 2xp + 1 + 4x2
p −
1
64
2xp 8x2
p + 1 1 + 4x2
p − ln 2xp + 1 + 4x2
p
From these results it is instructive to apply Pappus's rules and obtain surface
areas of a vase by rotating about the y−axis, and of a bowl by rotating about the
x−axis.

12. 12 Douglas Leadenham
Denition 5 Surface area of rotation of an arc about an external axis
A = 2πxaS = 2π
1
12S
1 + 4x2
p
3
2
− 1 S
Notice that the calculated S cancels. To conrm this, do the calculus problem for
a vase with a parabolic vertical midsection.
A =
ˆ
dA =
xpˆ
0
2πxds =
xpˆ
0
2πx 1 + 4x2dx
A =
π
6
1 + 4x2
p
3
2
− 1
This agrees, and the surface area of a vase of the size of the cardboard cut-out
is 343 cm
2
. In conrming the calculation please note that the working variable
for x = xp = 1.83, dimensionless. After applying the formula, it is necessary to
multiply A by the square of the scale factor, 3.52
. The bowl is harder to do, as
one can see.
A = 2πyaS =
2πS
1
S
yp
16
64
2xp 1 + 4x2
p + ln 2xp + 1 + 4x2
p −
1
64
2xp 8x2
p + 1 1 + 4x2
p − ln 2xp + 1 + 4x2
p
Again notice that the arc length S cancels. This too should be conrmed by
doing the calculus problem for a bowl with a parabolic horizontal midsection.
A =
ˆ
dA =
xpˆ
0
2π yp − x2
ds =
xpˆ
0
2π yp − x2
1 + 4x2dx
This is just the same integral already calculated, so agreement is assured. The
surface area of a bowl with vertical half-section equal to the cardboard cut-out is
587 cm
2
. In this calculation the working variable y = yp = 3.43, dimensionless.
Multiply by the scale factor 3.5 to get the physical model length of 12.
One may ask why the results with logarithms are so messy. That probably has
to do with the rst derivative discontinuity at the bottom of the bowl. The vase
is smooth at the bottom, but the bowl has a small peak there.
If one takes the cardboard half-parabola as an example, The vase is about the
size of a large champagne glass, and the bowl is about the size of a serving bowl.

13. Archimedes 13
4Volume of a paraboloid of revolution
Now apply Pappus's rule for the volume of a paraboloid of revolution from a
laminate.
Denition 6 Volume of a solid from revolution of a laminate about an external
axis
V = 2πrA
For the vase
V = 2πxA = 2π
3
8
xp
2
3
xpyp =
π
2
x2
pyp
For the bowl
V = 2πrA = 2π (yp − y) A = 2π
2
5
yp
2
3
xpyp =
8
15
πxpy2
p
Note please, that the radius of rotation is measured from the axis of rotation,
but the centroid y of the area A was calculated from the vertex of the parabola.
Now do the calculus problems for the two volumes. First the vase.
V =
ˆ
dV =
xpˆ
0
2πr yp − r2
dr
We obtain
V =
π
2
x4
p =
π
2
x2
pyp
and this agrees.
Archimedes obtained this result as the ratio of the volume of the paraboloidal
segment to the volume of the circumscribed cylinder.[Works]
Now for the bowl, again harder.
V =
ˆ
dV =
xpˆ
0
πy2
dx =
xpˆ
0
π yp − x2
2
dx
This results in
V =
8
15
πxpy2
p
This also agrees with the Pappus method.

14. 14 Douglas Leadenham
5Center of gravity
Archimedes worked extensively on centers of gravity of solid objects. Another of his
lost works is kentrobarikˆ, On centers of gravity. In the treatise, perÐ ìqoumènwn,
On oating bodies, he assumed or perhaps proved in the former work that the
center of gravity of a segment of a paraboloid of revolution is on the axis of the
segment at a distance from the vertex equal to
2
3 of its length.[Works] Any proof
by Archimedes has since been lost. At any rate, with Proposition 4 of àpipèdwn
ÐsvorropiÀn b, On the Equilibrium of Planes II, he proves that the center of gravity
of a planar parabolic segment cut o by a straight line lies on the diameter of the
segment.
Denition 7 Centroid of a solid body
Similar to the previous denitions for centroids, the centroid of a solid with
volume V is
zv =
1
V
Pˆ
O
zdV
We would formulate this calculation as a calculus problem in cylindrical coordi-
nates with the parabola opening upward in the z−direction:
zv =
1
V
Pˆ
O
zπr2
dz =
2
πr4
p
Pˆ
O
zπr2
dz
with z = r2
, r = x2 + y2. The serious student should conrm Archimedes' result
that he obtained without calculus. What can be said about the other two centroid
coordinates in this case? They are
xv =
1
V
Pˆ
O
xdV, yv =
1
V
Pˆ
O
ydV
The reader should explain this to himself.
I nd it humbling to learn that this 3-dimensional calculus problem yields a
result known to geometers 2,200 years ago. Modern scholars too often assume that,
when no written records survive from ancient times, the ancients were ignorant
and knew little. Sometimes we think too much of ourselves.
6Conclusion
Archimedes understood intuitively the role that the centroid plays in these calcu-
lations, but Pappus more than 5 centuries later was able to apply this formally to
the triangle, rectangle and semicircle to get his famous rules for the right circular
cone, right circular cylinder and sphere. What is clear is that Archimedes had not
discovered negative numbers, an essential part of algebra, and also had no equiv-
alent to Cartesian coordinates. Even if he had the concept of negative numbers,
he had no way to express them geometrically using only the rst quadrant. His

15. Archimedes 15
geometrical methods can be made to correspond to positive real numbers, but the
limitations of geometry are clear in his work. There is no documentary evidence
that he had attempted to work on arc length of a parabola, although he certainly
did much work with circular and spiral arcs, in perÈ èlÐkwn, On Spirals. A motiva-
tion of this paper was to see how far Archimedes had progressed toward calculus,
and it is clear that he had reached a dead end by the end of his life.
The recent high-tech imaging of the Archimedes Palimpsest, the same one
that Heiberg examined in 1907 that is the source material for The Method of
Archimedes in Heath's book, allowed double checking of the Greek headings listed
herein.[Palimpsest I][Palimpsest II] Another remarkable nd in the new imaging
is the rendering that shows Archimedes attributed to Euclid by name the earlier
work he needed. Heiberg had overlooked this attribution, and as a result Euclid's
existence had not been made absolutely certain up to 2008, at which time the
Archimedes Palimpsest Project data was released.[Palimpsest I] As Archimedes
lived rather soon after Euclid (. 300 BCE), we now know that Euclid was a real
individual and not a cadre of several astute geometers.[Palimpsest I]
tèloc
References
[Works] The Works of Archimedes. Edited in modern notation with introductory chap-
ters by T. L. Heath with a supplement, the Method of Archimedes, recently
discovered by Heiberg. Dover Publications, New York (Reissue of the Heath
Edition of 1897 and includes the Supplement of 1912). pp. xxi, xxxvii, cxlviii,
cliii-cliv, 218-220, 246.
[Mechanism] Freeth, T., Bitsakis, Y., Moussas, X., et al. Decoding the ancient Greek astro-
nomical calculator known as the Antikythera Mechanism. Nature 444 587-591
(30 Nov 2006)
[Palimpsest I] The Archimedes Palimpsest I: Catalogue and Commentary. Ed. Netz, R, Noel,
W., Tchernetska, N. and Wilson, N. Cambridge University Press, Cambridge,
UK (2011) pp. 11, 276, 296
[Palimpsest II] The Archimedes Palimpsest II: Images and Transcriptions. Ed. Netz, R, Noel,
W., Tchernetska, N. and Wilson, N. Cambridge University Press, Cambridge,
UK (2011) p. 69