**Pappus on Archimedes' spiral**©- trans. by Henry Mendell, Cal. State U., L.A.

Return to Vignettes of Ancient Mathematics

Pappus of Alexandria, __Mathematical Collection__ IV §§21-25,
pp. 234.1-242.12:

Prop. 21: Construction of Archimedes spiral and basic symptom

Prop. 22: spiral area is 1/3 the circle enclosing it. Solid proof by infinitary compression

Prop. 23.: generalization of Prop. 22 to include arcs of spiral.

Prop. 24: A spiral area formed by a 'radius' of the spiral to the whole spiral area is as the cube of the bounding straight lines.

Prop. 25: Ratios of the areas of the quadrant spiral areas.

Prop. 21 (p. 234.1-234.2). Conon, the Samian geometer, proposed the theorem of the spiral inscribed on a plane, but Archimedes, using wonderful intuition demonstrated it. Now the line gets this sort of generation.

(diagram 1: moving or still) Let there be a circle whose center is B, but whose line from the center (radius) is BA. Let straight-line BA move so that B remains while A moves regularly along the circular-arc of the circle, and at the same time let some point on it start out from B and move down it regularly towards A, and in equal time let the point from B traverse BA and A the circular-arc of the circle. The point moving down BA describes a line in rotation, such as BEZA, and its beginning is point B, and the beginning of the rotation is BA. And this line is called 'spiral'. And here is the principal symptom of it.

(diagram 2) Whatever line, such as BZ, is drawn through it and is extended, as the whole circular-arc of the circle is to circular-arc ADG, so is straight-line AB to BZ.

And this is easy to see from the generation. For while point A traverses the whole circular-arc of the circle, in this time the point from B traverses BA, and while A traverses circular-arc ADG, in this time the point from B traverses straight-line BZ. And these motions have equal speeds with themselves (i.e., they move with uniform speed), so that they are also proportional.

Corollary:
(diagram 3)
This is also obvious: that any straight-lines drawn from B to
the line that enclose equal angles exceed one another by an equal
amount.

Prop. 22 (p. 236.3-238.19). (diagram
1) The figure enclosed by the spiral and the straight line
at the beginning of the rotation is proved a third part of the
circle enclosing it.

(general diagram)

(diagram
2) For let there be the circle and the mentioned line, and
let a right-angled parallelogram (rectangle)
be displayed KNLP, (diagram
3) and let circular-arc AG, a part of the circular-arc of
the circle, be taken out, and straight-line KR be the same part
of KP, and let GB as well as BA be joined, (diagram
4) and let RT be be parallel to KN, join
KL^{1} and WM to KP,
and let circular-arc ZH be about center B. (diagram
5) And so since as straight-line AB is to AH, i.e., BG to
GZ, so is the whole circular-arc of the circle to GA (for this
is the principal symptom of the spiral), (diagram
6) but as the circular-arc of the circle is to GA, PK is to
KR but as PK is to KR, LK is to KW, (diagram
7) i.e., RT to RW, and, therefore, as BG is to GZ, TR is to
RW. (diagram 8)
And by ratio conversion BG : BZ = TR : TW.

(diagram 9) And, therefore, as the square on BG is to the square on BZ, so is the square on RT to the square on TW. But as the square on BG is to the square on BZ, so is section ABG to section ZBH. (diagram 10 2D version or 3D version) But as the square on RT is to the square on TW, so is the cylinder on parallelogram KT about axis NT to the cylinder on parallelogram MT about the same axis. (diagram 11: 2D version or 3D version) And, therefore, as section GBA is to section ZBH, so is the cylinder on parallelogram KT about axis NT to the cylinder on parallelogram MT about the same axis.

(diagram 12: 2D version or 3D version) Similarly, if we posit GD equal to AG and RC equal to KR, and we construct the same things, as section DBG will be to EBQ, so will the cylinder on parallelogram RF about axis TF to the cylinder on parallelogram XF about the same axis.

(diagram 13: 2D version or 3D version) And proceeding in the same manner we will prove that as the whole circle is to all the figures inscribed from sections in the spiral, so is the cylinder on parallelogram NP about axis NL to all the figures composed of cylinders inscribed in the cone generated from triangle KNL about axis LN.

(diagram 14: 2D version or 3D version) And again as the circle is to all the figures composed from sections circumscribed about the spiral, so is the cylinder to all the figures composed of cylinders and circumscribing the same cone. From this it is obvious that as the circle is to the figure between the spiral and straight-line AB, so is the cylinder to the cone. But the cylinder is three times the cone. Therefore, the circle is also three times the mentioned figure.

^{1}
KL is needed to determine W as the intersection of RT and KL,
while KL is never explicitly constructed in the text.

Prop. 23 (p. 238.20-25).

(diagram 1) In
the same way we will prove that if some line as BZ is drawn through
to the spiral, and a circle is inscribed through Z about center
B, the figure enclosed by spiral ZEB and straight-line ZB is a
third part of the figure enclosed by circular-arc ZHQ of the circle
and by straight-lines ZBQ.

The proof will follow the proof of Prop.
23. (diagram 2: 2D
version or 3D
version, 3.1). As before, given any
circular section GBE which touches the spiral and a similar ciruclar
section ZBH, ZB : BE = arc ZH : GE, so that section ZBH : section
GBE = ZB^{2} : BE^{2}. Let arc HZ be a part of
arc ZHQ (i.e., for some n, n*HZ = ZHQ). Take a rectangle RMLT
(for convenience, TR : CT = BZ : AB, to facilitate comparison
with the diagrams of Prop. 22) with its diagonal RL, and divide
it up so that:

RW is the same part of RM as GZ is of BZ, i.e., arc ZH of arc
ZHQ.

Note that the symptom of the spiral is
that

ZB : EB = arc ZHQ : arc HQ,

while because of the diagonal RL,

RT : FS = RM : WM, or RW is the same part of RM as RO is of RT
as arc ZH is of arc ZHQ.

Hence, RT : OT (=FS) = ZB : EB

and RT^{2} : OT^{2} = ZB^{2} : EB^{2}
= cylinder on RT with height RW : cylinder on OT with height OT
(=RW).

(Diagram 3D 3.2) Hence,

the circular section ZHQB : the (yellow) circular sections within the spiral = the whole
cylinder RH : the (yellow) cylinders under diagonal RL.

(diagram 3: 2D version or 3D version) As we take smaller parts, the yellow sections approach the area bounded by the spiral ZEB and the cylinders approach a cone with surface RL. This cone is 1/3 the cylinder, so that the spiral area is 1/3 the circular segment ZHQB.

We would expect a version of this argument in Pappus to say that the circular section ZHQB is to all the circular sections in ZEB as the cylinder to the cone. Clearly, the circular sections in ZEB are infinite in number.

And so, the demonstration is something of this sort, but we
next inscribe a theorem concerning the same line that is well
worth relating.

Prop. 24 (p. 238.29-240.32). (general
diagram)

(diagram
1) For let there be the circle mentioned in the generation
and the spiral itself AZEB. I say that any line drawn such as
BZ, the figure enclosed by the whole spiral and straight-line
AB is to the figure enclosed by spiral ZEB and straight-line BZ,
so is the cube on AB to the cube on ZB.

(diagram 2 = general diagram) For let there be inscribed a circle ZHQ through Z and about center B. (diagram 3) And so, (3) since as the figure enclosed by line AZEB and straight-line AB is to the figure enclosed by line ZEB and straight-line ZB, so is circle AGD to the figure enclosed by circular-arc ZHQ and straight-lines ZBQ, (1-2) since each was proved to be a third part. (6) (diagram 4) But circle AGD is to the area taken out by straight-lines ZBQ and circular-arc ZHQ has the ratio composed from the ratio which circle AGD has to circle ZHQ and the ratio which circle ZHQ has to the area taken out by straight-lines ZBQ and circular-arc ZHQ. (4) But as circle AGD is to circle ZHQ, so is the square on AB to tthe square on BZ, (5) but as circle ZHQ is to the mentioned area, its whole circular-arc is to ZHQ, i.e., the circular-arc of circle AGD to GDA, i.e., because of the symptom of the line, straight-line AB to BZ, and, (6) therefore, the figure between the spiral and straight-line AB has to the figure between the spiral and BZ the ratio composed from the ratio of AB to the square on ZB and the ratio of AB to BZ. This ratio is the same as the ratio of the cube on AB to the cube on BZ.

Basically,

1. Spiral
area AZEBxAB
= 1/3 Circle AGD

2. Spiral
area ZEBxZB
= 1/3 Circular section ZHQ

3. Spiral
area AZEBA
: Spiral area ZEBZ
= Circle AGD
: Circular section ZHQ

4. Circle
AGD
: Circle ZHQ
= AB^{2}
: ZB^{2}

5. Circle
ZHQ : Circular section ZHQ = Circumference QZHQ : Circular-arc ZHQ = Circumference AGDA : Circular-arc GDA = AB : ZB

6. Circle
AGD
: Circular section ZHQ = Compound of Circle AGD : Circle ZHQ by Circle ZHQ : Circular section ZHQ = Compound of
AB^{2}
: ZB^{2}
by AB
: ZB
= AB^{3}
: ZB^{3}

7. Hence,
Spiral area AZEBxAB
: Spiral area ZEBxZB = AB^{3} : ZB^{3}

Prop. 25 (p 242.1-12)

From this it is obvious that if
the spiral is supposed as well as the circle about it, and AB
is extended to D and GZEK is drawn at right angles to it, for
which the area between line BLE and straight-line BE is of one
figure, then in the case of these
sorts, the area between line NME and straight-lines NBE is seven,
and that between line ZQN and straight-lines ZBN is 19, and that
between line AXZ and straight-lines ABZ is 37 (since these are
clear from the previously proved theorem), and that where AB is
4, ZB is three, BN is 2, and BE is 1. For this is clear from the
symptom of the line and the fact that circular-arcs AG, GD, DK,
KA are equal.

The ratios of the bounding lines is obvious.

Also obviously, 1 + 7 + 19 + 37 = 64
= 8^{3}.

Spiral area ELB
: The whole spiral area AZNEB = BE^{3} : AB^{3}
= 1^{3} : 4^{3} = 1 : 64

Spiral area NEB : The whole spiral area AZHEB = BN^{3}
: AB^{3} = 2^{3} : 4^{3} = 8 : 64

Hence, Spiral area NME : The whole spiral area AZHEB = (8-1):64
= 7 : 64

Spiral area ZNEB : The whole spiral area AZHEB = BZ^{3}
: AB^{3} = 3^{3} : 4^{3} = 27 : 64

Hence, Spiral area ZQN : The whole spiral area AZHEB = (27-8) : 64
= 19 : 64

Spiral area AXZ
: The whole spiral area AZHEB = (64-27) : 64 = 37 : 64