Table Showing the Comparative Measures of Lines.

(Fig. 67.)

ME	= 2000000	=	diameter.
AB	= 1902113	=	AD ÷ DB
MB	= 1618034	=	MC + MH	=	MP + PB
AS	= 1538841·5
EP	= 1453086	=	AG + FB
AF	= 1175570	=	AE = GB
MC	= 1000000	=	radius = CD + DX	=	CH + CX
AD	= 951056·5	=	DB = DS
PB	= 854102
QS	= 812298·5
MP	= 763932	=	CH × 2	=	base of Cheops.
AG	= 726543	=	GH = XH	=	HN = PF = FB = Slant edge of Cheops. = slant edge of Pent. Pyr.
DE	= 690983	=	DH = XD	=	apothem of Pentagonal Pyramid.
MH	= 618034	=	MN = XC =	{	apothem of Cheops. altitude of Pentagonal Pyramid. side of decagon inscr'd in circle.
MS	= 500000
	485868	=	{	mean proportional between MH and HC altitude of Cheops.
OP	= 449027	=	GF = GD + DF
HC	= 381966	=	half base of Cheops.
SO	= 363271·5	=	HS
CD	= 309017	=	half MH
PR	= 277516
GD	= 224513·5
SP	= 263932

The triangle DXH represents a vertical section of the pentagonal pyramid; the edge HX is equal to HN, and the apothem DX is equal to DE. Let DH be a hinge attaching the plane DXH to the base, now lift the plane DXH until the point X is vertical above the centre C. Then the points A, E, B, O, N of the five slant slides, when closed up, will all meet at the point X over the centre C.

We have now built a pyramid out of the pentangle, whose slope is 2 to 1, altitude CX being to CD as 2 to 1.

Apothem	DX = DE
Altitude	CX = HM or MN
Altitude	CX + CH = CM radius.
Apothem	DX + CD = CM radius.
Edge	HX = HN or PF

Note also that

(MP)	= CH
2	= CH
OP	= HR

Let us now consider the Pentangle as the symbol of the Great Pyramid Cheops.

The line MP = the base of Cheops.

The line CH = half base of Cheops.

The line HM = apothem of Cheops.

The line HN = slant edge of Cheops.

Thus: Apothem of Cheops = side of decagon.

Apothem of Cheops = altitude of pentagonal pyramid.

Slant edge of Cheops = slant edge of pentagonal pyramid.

Now since apothem of Cheops = MH

and half base of Cheops = HC

then do apothem and half base represent, when taken together, extreme and mean ratio, and altitude is a mean proportional between them: it having already been stated, which also is proved by the figures in the table, that MC : MH :: MH : HC and apoth: alt :: alt : half base.

Thus is the four pointed star Cheops evolved from the five pointed star Pentalpha. This is shown clearly by Fig. 68, thus:—

Fig. 68.

Within a circle describe a pentangle, around the interior pentagon of the star describe a circle, around the circle describe a square; then will the square represent the base of Cheops.

Draw two diameters of the outer circle passing through the centre square at right angles to each other, and each diameter parallel to sides of the square; then will the parts of these diameters between the square and the outer circle represent the four apothems of the four slant sides of the pyramid. Connect the angles of the square with the circumference of the outer circle by lines at the four points indicated by the diameters, and the star of the pyramid is formed, which, when closed as a solid, will be a correct model of Cheops.

Calling apothem of Cheops,	MH = 34
and half base,	HC = 21
as per Figure 6. Then— MH +	MC = 55

and 55 : 34 :: 34 : 21·018, being only in error a few inches in the pyramid itself, if carried into actual measures.

The ratio, therefore, of apothem to half-base, 34 to 21, which I ascribe to Cheops, is as near as stone and mortar can be got to illustrate the above proportions.

Correctly stated arithmetically let MH = 2.

Then		HC	= √5 - 1
		MC	= √5 + 1
	and altitude of Cheops		= √(MH × MC)

Let us now compare the construction of the two stars:—

Fig. 69.	Fig. 70.
TO CONSTRUCT THE STAR PENTALPHA FIG. 69.	TO CONSTRUCT THE STAR CHEOPS, FIG. 70.
Describe a circle. Draw diameter MCE. Divide MC in mean and extreme ratio at H. Lay off half MH from C, to D. Draw chord ADB, at right angles to diameter ECM. Draw chord BHN, through H. Draw chord AHO, through H. Connect NE. Connect EO.	Describe a circle. Draw diameter MCE. Divide MC in mean and extreme ratio, at H. Describe an inner circle with radius CH, and around it describe the square a, b, c, d. Draw diameter ACB, at right angles to diameter ECM. Draw Aa, aE, Eb, bB, Bd, dM, Mc, and cA.

The question now arises, does this pyramid Cheops set forth by the relations of its altitude to perimeter of base the ratio of diameter to circumference; or, does it set forth mean proportional, and extreme and mean ratio, by the proportions of its apothem, altitude, and half-base? The answer is—from the practical impossibility of such extreme accuracy in such a mass of masonry, that it points alike to all, and may as fairly be considered the exponent of the one as of the others. Piazzi Smyth makes Cheops 761·65 feet base, and 484·91 feet altitude, which is very nearly what he calls a Π pyramid, for which I reckon the altitude would be about 484·87 feet with the same base: and for a pyramid of extreme and mean ratio the altitude would be 484·34 feet.

The whole difference, therefore, is only about six inches in a height of nearly five hundred feet. This difference, evidently beyond the power of man to discover, now that the pyramid is a ruin, would even in its perfect state have been inappreciable.

It appears most probable that the star Pentalpha led to the star Cheops, and that the star Cheops (Fig. 70) was the plan used by the ancient architect, and the ratio of 34 to 21, hypotenuse to base, the template used by the ancient builders.

Suppose some king said to his architect, "Make me a plan of a pyramid, of which the base shall be 420 cubits square, and altitude shall be to the perimeter of the base as the radius of a circle to the circumference."—Then might the architect prepare an elaborate plan in which the relative dimensions would be about—

			R. B. CUBITS
Base angle 51° 51′ 14·3″	}	Base Altitude Apothem	420 267·380304 &c. 339·988573 &c.

The king then orders another pyramid, of the same base, of which altitude is to be a mean proportional between apothem and half-base—and apothem and half-base taken as one line are to be in mean and extreme ratio.

The architect's plan of this pyramid will be the simple figure illustrated by me (Fig. 70), and the dimensions about—

			R. B. CUBITS
Base angle 51° 49′ 37	42″ 471	}	Base Altitude Apothem	420 267·1239849 &c. 339·7875153 &c.

But the builder practically carries out both plans when he builds to my templates of 34 to 21 with—

			R. B. CUBITS
Base angle 51° 51′ 20″	}	Base Altitude Apothem	420 267·394839 &c. 340

and neither king nor architect could detect error in the work.

The reader will remember that I have previously advanced that the level of Cephren's base was the plan level of the Gïzeh pyramids, and that at this level the base of Cheops measures 420 R.B. cubits—same as the base of Cephren.

This hypothesis is supported by the revelations of the pentangle, in which the ratio of 34 to 21 = apothem 340 to half-base 210 R.B. cubits, is so nearly approached.

Showing how proportional lines were the order of the pyramids of Gïzeh, we will summarise the proportions of the three main pyramids as shewn by my dimensions and ratios, very nearly, viz.:—

Mycerinus. Base : Apothem :: Altitude : Half-Base.
as shown by the ratios, (Fig. 13), 40 : 32 :: 25 : 20.
Cephren. Diagonal of Base : Edge :: Edge : Altitude.
as shown by ratios, (Fig. 12), 862 : 588 :: 588 : 400.
Cheops. (Apothem + Half Base): Apoth. :: Apoth. : Half Base.
as shown by the ratios, (Fig. 9), 55 : 34 :: 34 : 21.
and—Apothem : Altitude :: Altitude : Half-B.

Similar close relations to other stars may be found in other pyramids. Thus:—Suppose NHO of figure 69 to be the NHO of a heptangle instead of a pentangle, then does NH represent apothem, and NO represent base of the pyramid Mycerinus, while the co-sine of the angle NHM (being MH minus versed sine) will be equal to the altitude of the pyramid. The angle NHM in the heptangle is, 38° 34′ 17·142″, and according to my plan of the pyramid Mycerinus, the corresponding angle is 38° 40′ 56″. (See Fig. 19.) This angular difference of 0° 6′ 39″ would only make a difference in the apothem of the pyramid of eight inches, and of ten inches in its altitude (apothem being 283 ft. 1 inch, and altitude 221 ft.).