The Solution of the Pyramid Problem Part 13

R. B. CUBITS.

{Base 420 Base angle 51 49' 37-42/471? {Alt.i.tude 2671239849 &c.

{Apothem 3397875153 &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? {Alt.i.tude 267394839 &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 Gzeh 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 Gzeh, we will summarise the proportions of the three main pyramids as shewn by my dimensions and ratios, very nearly, viz.:--

_Mycerinus. Base : Apothem :: Alt.i.tude : Half-Base._ as shown by the ratios, (_Fig_. 13), 40 : 32 :: 25 : 20.

_Cephren. Diagonal of Base : Edge :: Edge : Alt.i.tude._ 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 : Alt.i.tude :: Alt.i.tude : 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 alt.i.tude of the pyramid. The angle NHM in the heptangle is, 38 34' 17142?, 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 alt.i.tude (apothem being 283 ft. 1 inch, and alt.i.tude 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 alt.i.tude 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, alt.i.tude; and HK, edge.

The length of the line AD being v(AC + CD), the length of the line HK being v(HG + GK), and line CH (half diagonal of base) being v(CG + GH). 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 = 105000 = 176925 Apothem EA = 168082 = 283218 Edge HK = 198183 = 333937 Alt.i.tude CD = 131250 = 221156 Half diag. of base CH = 1484924 = 250209

and thus I acquired the ratios:--

Half-base : Alt.i.tude :: 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 alt.i.tude), 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, alt.i.tude; 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 21000 35385 = LA Apothem 34650 58385 = EA Edge 40516 68269 = HK Alt.i.tude 275625 46443 = CD Half-diag. of base 296985 50042 = 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 alt.i.tude), 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="">

? < 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 Alt.i.tude 267380304 Apothem 339988573 (_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--