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
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.
Public-domain text, read in full here on John Shaqi.
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