Part 6
(_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.
{apothem of Cheops. MH = 618034 = MN = XC = {altitude of Pentagonal Pyramid. {side of decagon inscr'd in circle.
MS = 500000
{mean proportional between MH and HC 485868 = { {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
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.
TO CONSTRUCT THE STAR PENTALPHA FIG. 69.
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.
Fig. 70.
TO CONSTRUCT THE STAR CHEOPS, FIG. 70.
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 [Pi] 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 420 Base angle 51° 51′ 14·3″ {Altitude 267·380304 &c. {Apothem 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 420 Base angle 51° 49′ 37-42/471″ {Altitude 267·1239849 &c. {Apothem 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 420 Base angle 51° 51′ 20″ {Altitude 267·394839 &c. {Apothem 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.).
§ 17. THE MANNER IN WHICH THE SLOPE RATIOS OF THE PYRAMIDS WERE ARRIVED AT.
The manner in which I arrived at the Slope Ratios of the Pyramids, viz., 32 _to_ 20, 33 _to_ 20, and 34 _to_ 21, for _Mycerinus_, _Cephren_, and _Cheops_, respectively (_see Figures_ 8, 7 _and_ 6), was as follows:--
First, believing in the connection between the relative positions of the Pyramids on plan (_see Fig_. 3, 4 _or_ 5), and their slopes, I viewed their positions thus:--
Mycerinus, situate at the angle of the 3, 4, 5 triangle ADC, is likely to be connected with that "primary" in his slopes.
Cephren, situate at the angle of the 20, 21, 29 triangle FAB, and strung, as it were, on the hypotenuse of the 3, 4, 5 triangle DAC, is likely to be connected with _both_ primaries in his slopes.
Cheops, situate at the point A, common to both main triangles, governing the position of the other pyramids, is likely to be a sort of mean between these two pyramids in his slope ratios.
Reasoning thus, with the addition of the knowledge I possessed of the angular estimates of these slopes made by those who had visited the ground, and a useful start for my ratios gained by the reduction of base measures already known into R.B. cubits, giving 420 as a general base for Cheops and Cephren at one level, and taking 210 cubits as the base of Mycerinus (half the base of Cephren, as generally admitted), I had something solid and substantial to go upon. I commenced with Mycerinus. (_See Fig_. 71.)
_Fig. 71. (Mycerinus)_
LHNM represents the base of the pyramid. On the half-base AC I described a 3, 4, 5 triangle ABC. I then projected the line CF = BC to be the altitude of the pyramid. Thus I erected the triangle BFC, ratio of BC to CF being 1 to 1. From this datum I arrived at the triangles BEA, ADC, and GKH. GK, EA, and AD, each represent apothem of pyramid; CF, and CD, altitude; and HK, edge.
The length of the line AD being √(AC² + CD²), the length of the line HK being √(HG² + GK²), and line CH (half diagonal of base) being √(CG² + GH²). These measures reduced to R.B. cubits, calling the line AC = ratio 4 = 105 cubits, half-base of pyramid, give the following results:--
R. B. BRITISH CUBITS. FEET. Half-base LA = 105·000 = 176·925 Apothem EA = 168·082 = 283·218 Edge HK = 198·183 = 333·937 Altitude CD = 131·250 = 221·156 Half diag. of base CH = 148·4924 = 250·209
and thus I acquired the ratios:--
Half-base : Altitude :: Apothem : Base. = 20 : 25 :: 32 : 40 nearly.
To place the lines of the diagram in their actual solid position--Let AB, BC, CA and HG be hinges attaching the planes AEB, BFC, CDA and HKG to the base LHNM. Lift the plane BCF on its hinge till the point F is vertical over the centre C. Lift plane CDA on its hinge, till point D is vertical over the centre C; then will line CD touch CF, and become one line. Now lift the plane AEB on its hinge, until point E is vertical over the centre C, and plane HKG on its hinge till point K is vertical over the centre C; then will points E, F, D and K, all meet at one point above the centre C, and all the lines will be in their proper places.
The angle at the base of Mycerinus, if built to a ratio of 4 to 5 (half-base to altitude), and not to the more practical but nearly perfect ratio of 32 to 20 (apothem to half-base) would be the complement of angle ADC, thus--
4 165″ --- = ·8 = Tan. < ADC = 38° 39′ 35--- 5 477
312″ ∴ < DAC = 51° 20′ 24--- 477
but as it is probable that the pyramid was built to the ratio of 32 to 20, I have shown its base angle in Figure 19, as 51° 19′ 4″.
Figure 72 shows how the slopes of _Cephren_ were arrived at.
_Fig. 72. (Cephren)_
LHNM represents the base of the pyramid. On the half-base AC, I described a 3, 4, 5 triangle ABC. I then projected the line CF (ratio 21 to BC 20), thus erecting the 20, 21, 29 triangle BCF. From this datum, I arrived at the triangles BEA, ADC, and GKH; GK, EA and AD each representing apothem; CF and CD, altitude; and HK, edge. The lengths of the lines AD, HK and CH being got at as in the pyramid Mycerinus. These measures reduced to cubits, calling AC = ratio 16 = 210 cubits (half-base of pyramid) give the following result.
R. B. BRITISH CUBITS. FEET. Half-base 210·00 353·85 = LA Apothem 346·50 583·85 = EA Edge 405·16 682·69 = HK Altitude 275·625 464·43 = CD Half-diag. of base 296·985 500·42 = CH
thus I get the ratios of--Apothem : Half-Base :: 33 : 20, &c. The planes in the diagram are placed in their correct positions, as directed for Figure 71.
The angle at the base of Cephren, if built to the ratio of 16 to 21 (half-base to altitude), and not to the practical ratio of 33 to 20 (apothem to half-base), would be the complement of < ADC, thus--
16 16″ -- = ·761904 = Tan. < ADC = 37° 18′ 14-- 21 46 30″ ∴ < DAC = 52° 41′ 45-- 46
but as it is probable that the pyramid was built to the ratio of 33 to 20, I have marked the base angle in Fig. 17, as 52° 41′ 41″.
I took _Cheops_ out, first as a [Pi] pyramid, and made his lines to a base of 420 cubits, as follows--
Half-base 210 Altitude 267·380304 Apothem 339·988573 (_See Fig_. 73.)
_Fig. 73. (Cheops) _
But to produce the building ratio of 34 to 21, as per diagram Figure 6 or 9, I had to alter it to--
Half-base 210 Altitude 267·394839 Apothem 340°
Thus the theoretical angle of Cheops is 51° 51′ 14·3″, and the probable angle at which it was built, is 51° 51′ 20″, as per figure 15.
Cheops is therefore the mean or centre of a system--the slopes of Mycerinus being a little flatter, and those of Cephren a little steeper, Cheops coming fairly between the two, within about 10 minutes; and thus the connection between the ground plan of the group and the slopes of the three pyramids is exactly as one might expect after examination of Figure 3, 4 or 5.