The Solution of the Pyramid Problem; or, Pyramid Discoveries: With a New Theory as to their Ancient Use — John Shaqi
The Solution of the Pyramid Problem; or, Pyramid Discoveries: With a New Theory as to their Ancient UseBallard, Robert
History
The Solution of the Pyramid Problem; or, Pyramid Discoveries: With a New Theory as to their Ancient Use
Ballard, Robert
Pyramids
I thus make Cheops, Cephren, and Mycerinus, respectively, 81·56,
101·93, and 106·93 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 138½ ft. below base of Cheops (or 82·19 R.B.
cubits).
Piazzi Smyth makes the pavement of Cheops 1752 British inches (or 86·64
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 101·93 cubits, and Cheops' base of
452 cubits being 81·56 cubits above Wild's datum, the difference in
level of their bases is, 20·37 cubits.
The ratio of base to altitude of Cheops being 330 to 210, therefore
20·37 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 altitude 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 295·98
Apex of Cheops above Base Cheops 287·77
Apex of Mycerinus above Base Cheops 156·51
Base Cephren above base of Cheops 20·37
Base Mycerinus above base of Cheops 25·37
Piazzi Smyth makes the top of the tenth course of Cheops 414 pyramid
inches above the pavement; and 414 divided by 20·2006 equals 20·49 R.B.
cubits.
But I have already proved that Cheops' 420 cubit base measure occurs at
a level of 20·37 cubits above pavement; therefore is this level the
level of the top of the tenth course, for the difference is only 0·12
R.B. cubits, or 2½ 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· = 84·25
Cheops' Base, as before stated 9131·05 = 452·01
King's Chamber, "True Length," 412·132 = 20·40
" " "True First Height," 230·389 = 11·40
" " "True Breadth," 206·066 = 10·20
He makes the present summit platform of Cheops 5445 pyramid inches above
pavement. My calculation of 269·80 R.B. cub. (See Fig. 21) is equal to
5450 pyramid inches--this is about 18 cubits below the theoretical apex.
Public-domain text, read in full here on John Shaqi.
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