§ 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.)

(Mycerīnus) Fig. 71.

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.CUBITS. BRITISH 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= ·8 = Tan. < ADC= 38° 39′ 35165″
5 477
∴ < DAC= 51° 20′ 24312″
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.

(Cephren) Fig. 72.

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. CUBITS.BRITISH FEET.
Half-base 210·00353·85 = LA
Apothem 346·50583·85 = EA
Edge 405·16682·69 = HK
Altitude 275·625464·43 = CD
Half-diag. of base 296·985500·42 = 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= ·761904 = Tan. < ADC= 37° 18′ 1416″
21 46
∴ < DAC= 52° 41′ 4530″
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 Π pyramid, and made his lines to a base of 420 cubits, as follows—

Half-base210
Altitude267·380304
Apothem339·988573 (See Fig. 73.)

(Cheops) Fig. 73.

But to produce the building ratio of 34 to 21, as per diagram Figure 6 or 9, I had to alter it to—

Half-base210
Altitude267·394839
Apothem340

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.