The Solution of the Pyramid Problem Part 2

Figures 13 and 14 show the ratios of Mycerinus, viz., half base to apothem, 20 _to_ 32 _exactly_, and half base, alt.i.tude, and apothem respectively, as 20, 25, and 32 very nearly.

Also full diagonal to edge as 297 to 198, nearly. A peculiarity of this pyramid is, that base is to alt.i.tude as apothem is to half base. Thus, 40 : 25 :: 32 : 20; that is, half base is a fourth proportional to base, apothem, and alt.i.tude.

-- 5. THE EXACT DIMENSIONS OF THE PYRAMIDS.

R.B. Cub. Brit. Ft.

Fig. 15. Cheops. 452 = 76162 287767 = 484887 Fig. 16. Cheops. 3659047 = 616549 430058 = 724647 6392244 = 1077093

Figures 15 to 20 inclusive, show the linear dimensions of the three pyramids, also their angles. The base angles are, Cheops, 51 51' 20"; Cephren, 52 41' 41?; and Mycerinus, 51 19' 4?.

R.B. Cub. Brit. Ft.

Fig. 17. Cephren. 420 = 70770 27561 = 46440 Fig. 18. Cephren. 34650 = 58385 40516 = 68269 59397 = 100084

R.B. Cub. Brit. Ft.

Fig. 19. Mycerinus. 210 = 353.85 168 = 283.08 Fig. 20. Mycerinus: 131.14 = 220.97 198.10 = 3337985 2969848 = 50042

In Cheops, my dimensions agree with Piazzi Smyth--in the base of Cephren, with Vyse and Perring--in the height of Cephren, with Sir Gardner Wilkinson, nearly--in the base of Mycerinus, they agree with the usually accepted measures, and in the height of Mycerinus, they exceed Jas. J. Wild's measure, by not quite one of my cubits.

In my angles I agree very nearly with Piazzi Smyth, for Cheops, and with Agnew, for Cephren, differing about half a degree from Agnew, for Mycerinus, who took this pyramid to represent the same relation of [Pi]

that P. Smyth ascribes to Cheops (viz.: 51 51' 14?3), while he gave Cheops about the same angle which I ascribe to Mycerinus.

I shall now show how I make Cephren and Cheops of equal bases of 420 R.B. cubits at the same level, viz.--that of Cephren's base.

John James Wild made the bases of Cheops, Cephren, and Mycerinus, respectively, 80, 100, and 10490 cubits above some point that he called Nile Level.

His cubit was, I believe, the Memphis, or Nilometric cubit--but at any rate, he made the base of Cephren 412 of them.

I therefore divided the recognized base of Cephren--viz., 70775 British feet--by 412, and got a result of 17178 British feet for his cubit.

Therefore, his measures multiplied by 17178 and divided by 1685 will turn his cubits into R.B. cubits.

I thus make Cheops, Cephren, and Mycerinus, respectively, 8156, 10193, and 10693 R.B. cubits above the datum that J. J. Wild calls Nile Level. According to Bonwick's "Facts and Fancies," p. 31, high water Nile would be 138 ft. below base of Cheops (or 8219 R.B.

cubits).

Piazzi Smyth makes the pavement of Cheops 1752 British inches (or 8664 R.B. cubits) above _average Nile Level_, but, by scaling his map, his _high Nile Level_ appears to agree nearly with Wild.

It is the _relative levels_ of the Pyramids, however, that I require, no matter how much above Nile Level.

Cephren's base of 420 cubits being 10193 cubits, and Cheops' base of 452 cubits being 8156 cubits above Wild's datum, the difference in level of their bases is, 2037 cubits.

The ratio of base to alt.i.tude of Cheops being 330 to 210, therefore 2037 cubits divided by 210 and multiplied by 330 equals 32 cubits; and 452 cubits minus 32 cubits, equals 420.

Similarly, the base of Mycerinus is 5 cubits _above_ the base of Cephren, and the ratio of base to alt.i.tude 32 to 20; therefore, 5 cubits divided by 20 and multiplied by 32 equals 8 cubits to be _added_ to the 210 cubit base of Mycerinus, making it 218 cubits in breadth at the level of Cephren's base.

Thus, a horizontal section or plan at the level of Cephren's base would meet the slopes of the Pyramids so that they would on plan appear as squares with sides equal to 218, 420, and 420 R.B. cubits, for Mycerinus, Cephren, and Cheops, respectively.

Fig. 21.

R.B. Cub.

Apex of Cephren above Base Cheops 29598 Apex of Cheops above Base Cheops 28777 Apex of Mycerinus above Base Cheops 15651 Base Cephren above base of Cheops 2037 Base Mycerinus above base of Cheops 2537

Piazzi Smyth makes the top of the tenth course of Cheops 414 pyramid inches above the pavement; and 414 divided by 202006 equals 2049 R.B.

cubits.

But I have already proved that Cheops' 420 cubit base measure occurs at a level of 2037 cubits above pavement; therefore is this level the level of the top of the tenth course, for the difference is only 012 R.B. cubits, or 2 inches.

I wish here to note as a matter of interest, but not as affecting my theory, the following measures of Piazzi Smyth, turned into R.B. cubits, viz.:--

PYR. INCHES. R.B. CUBITS.

King's Chamber floor, above pavement 1702 = 8425 Cheops' Base, as before stated 913105 = 45201 King's Chamber, "True Length," 412132 = 2040 " " "True First Height," 230389 = 1140 " " "True Breadth," 206066 = 1020

He makes the present summit platform of Cheops 5445 pyramid inches above pavement. My calculation of 26980 R.B. cub. (See Fig. 21) is equal to 5450 pyramid inches--this is about 18 cubits below the theoretical apex.

Figure 21 represents the comparative levels and dimensions of Mycerinus, Cephren, and Cheops.

The following peculiarities are noticeable:--That Cheops and Cephren are of equal bases at the level of Cephren's base;--that, at the level of Cheops' base, the latter is only half a cubit larger;--that, from the level of Mycerinus' base, Cheops is just double the height of Mycerinus;--and that from the level of Cephren's base, Cephren is just double the height of Mycerinus; measuring in the latter case, however, only up to the level platform at the summit of Cephren, which is said to be about eight feet wide.

The present summit of Cephren is 2307 cubits above the present summit of Cheops, and the completed apex of Cephren would be 821 cubits above the completed apex of Cheops.

In the summit platforms I have been guided by P. Smyth's estimate of _height deficient_, 363 pyr. inches, for Cheops, and I have taken 8 feet base for Cephren's summit platform.

-- 6. GEOMETRICAL PECULIARITIES OF THE PYRAMIDS.

_In any pyramid, the apothem is to half the base as the area of the four sides is to the area of the base._

Thus--Ratio apothem to half base Mycerinus 32 to 20 " " " " Cephren 33 to 20 " " " " Cheops 34 to 21

AREA OF THE FOUR SIDES. AREA OF THE BASE.

Mycerinus 70560 44100 Cephren 291060 176400 Cheops 33077790 204304

All in R.B. cubits.

Therefore--32 : 20 : : 70560 : 44100 33 : 20 : : 291060 : 176400 34 : 21 : : 33077790 : 204304

[2]Herodotus states that "_the area of each of the four faces of Cheops was equal to the area of a square whose base was the alt.i.tude of a Pyramid;_" or, in other words, that alt.i.tude was a mean proportional to apothem and half base; thus--area of one face equals the fourth of 33077790 or 82694475 R.B. cubits, and the square root of 82694475 is 28756. But the correct alt.i.tude is 28777, so the error is 021, or 4 British inches. I have therefore the authority of Herodotus to support the theory which I shall subsequently set forth, that this pyramid was the exponent of lines divided in mean and extreme ratio.

By taking the dimensions of the Pyramid from what I may call its _working level_, that is, the level of the base of Cephren, this peculiarity shows more clearly, as also others to which I shall refer.

Thus--base of Cheops at working level, 420 cubits, and apothem 340 cubits; base area is, therefore, 176400 cubits, and area of one face is (420 cubits, multiplied by half apothem, or 170 cubits) 71400 cubits.

Now the square root of 71400 would give alt.i.tude, or side of square equal to alt.i.tude, 267207784 cubits: but the real alt.i.tude is v(340-210) = v71500 = 267394839. So that the error of Herodotus's proposition is the difference between v714 and v715.

Footnote 2: Proctor is responsible for this statement, as I am quoting from an essay of his in the _Gentleman's Magazine_. R. B.

This leads to a consideration of the properties of the angle formed by the ratio _apothem_ 34 to _half base_ 21, peculiar to the pyramid Cheops. (_See Figure 22._)