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The Pamphlet Collection of Sir Robert Stout: Volume 54

[introduction]

Professor Piazzi Smyth tells us that the lower course of masonry is equal to 50 inches, then the thickness of each course as they ascend decreases till the 35th course is only equal to 27 inches in depth, while the 36th course suddenly increases to 50 inches, and that this thick 36th course of masonry is conspicuous on every side of the Great Pyramid. He then asks, what then was the extraordinarily important thing completed in these first 35 courses that the builders crowned them so majestically, honoured them, in fact, with a diadem of stone, whose 50-inch escarpment shines afar on every side, and marked them to all future time by the weight and size of the 36th, 37th, and other higher courses of extra thick masonry immediately above them?

I Will Answer This Important Question.

In the first place, there can be no doubt that this difference in the thickness of the masonry courses was continued in the outer casing stones, because the 50-inch white escarpment of the 36th course of masonry, which is visible to the naked eye so far off all round was Designed and left as a "Hailing sign," calling the attention of passersby to a Continually Increasing Deficiency during a certain period of time.

This period of time is equal to 10,482 years, equal to the Period of a "Zone of Water," which travels from one hemisphere into the other periodically, crossing the Equator once every 10,482 years, equal to half the period of the perihelion, which is equal to 20,964 solar years, each year composed of 365-24 days; then, 365-24 × 10 ÷ π = 1162′6, equal to the height of the 35th course above the basal plane.

The "Problem" of the Great Pyramid is built up in the terms of the inch—in the terms of, as the diameter is to the circumference of the circle, then 25827-20964 = 4863, and 36 ÷ (4863 ÷ 1000) × π ÷ 2X10 = 1162.6, equal to the height of the 35th course of masonry above the basal plane.

And 1162.6 × 10 equals EE, equals the diameter, of the circle, as shown in the Figure, because, the vertical height of the Great Pyramid, is as the radius of that circle—equal to 11626 ÷ 2 = 5813 inches.

Then, the height of the 35th course of masonry above the basal plane, is in the terms of the circle, then it is in the terms of page 4 the "Zone of water," whose period is equal to half that circle—equal to 10,482 years.

It was as far back as 1873, when I had roughly traced out the astronomical and physical cause of the existence of this "Zone of water," and a certain Deficiency, which was created during that period by the continued action of the Centrifugal force.

Therefore this remarkable difference in the thickness of the masonry courses was the first symbol that attracted my attention when I set myself to work to try to read the meaning of the Symbology of the Great Pyramid in 1879.

What made this symbol particularly attract my attention was the fact that, I was aware of a cause for a Deficiency which went on increasing during a certain period of time in one Hemisphere, while this "Zone of water" was retained in the opposite during 10,482 years by the Sun's force of attraction.

The position in latitude of the apex of this "Zone of water" governs the obliquity of the ecliptic; but the position in latitude of the apex is governed by the position of the perihelion point of the earth's orbit, because the sun will exert its maximum of force when, the earth is nearest the sun, i.e., in the perihelion of its orbit.

Consequently, the vertical height of the Great Pyramid is built up in the terms of the Lesser distance of the Sun. Therefore, twice the vertical height of the Great Pyramid, represents the Lesser distance of the sun from the earth, in the terms of the breadth of the earth from pole to pole—not the mean distance as supposed. Therefore, 91,840,000 miles will represent the Lesser distance of the Sun ±.

I will now test the 35th and 36th courses of masonry in the terms of the deficiency, in the terms of the inch, in the terms of π.

The deficiency caused by the action of the centrifugal force, according to the Theory, is equal to 0 5003 inches annually at the pole of that hemisphere which is free from the zone of water during 10,482 years.

0.5003 inches × 10,482 ÷ 12 = 437 feet 0.5003 × 10 × 1000 = 5003 inches. Then— 5003 = 3.699231 437 ÷ 10 = 43.7 = 1.640481 2.058750 π = 0.497149 3) 359.66 = 2.555899 119.886 = 2.078760 π = 0.497149 376.61 = 2.575909 and 376.61 + 1369.5 = 1746.1

page 5

equals the arc Aa + arc AE (as shown in the Figure) = 1744.5 ± equals 652 + 1092.5 ... ... ... ... ... ... ... = 1744.5 ± 5003 = 3.699231 35th Course = 27 = 1.431364 2.267867 π = 0.497149 582.18 = 2.765016 3 1746.54 = 1746.54

and 27 inches is equal to the thickness of the 35th course of masonry, and the decreasing thickness of the courses symbolises this deficiency at the end of the period; and the end of the period was to happen 1746 years ± after the foundation was laid, because the perihelion would then have arrived in the equinox and the apex at the equator.

I will now divide 5003 by the 35th and 36th courses— 5003 = 3.699231 35th Course = 35 = 1.544068 2.142928 π = 0.497149 449.07 = 2.652312 3592.56 × 10 = 35926 ± The first five figures of the mass of the sun as used in the Theory is ... ... ... ... ... ... = 35926 5003 = 3.699231 36th Course = 36 = 1.556303 2.142928 π=0.497149 436.59 = 2.640077 and 436.59 ÷ 10 = 43.7 (inch feet)

Equal to the width of the stones in the ceiling of the Grand Gallery. The difference in the width of this chamber as it ascends symbolises this increasing deficiency, and 43.7 × 10 = 437, and 437 × 5 = 2185 equal to the height of the apex of the "zone of water," as required by the Theory. Then, 2185 2 = 1092.5 equals the whole length of the entrance passage; then, the height of the entrance above the pavement equals 652 and 652 + 1092.5 = 1744.5 equals the date in inch years.

5003 ÷ (5813 ÷ 10) × π = 27.031

18225 ÷ π ÷ 3 × ÷ × 4 ÷ 3 = 27.00

page 6

Equal to the thickness of the 35th course of masonry, 20964 years-2739 = 18225 years.

35926 is equal to the first five figures of the mass of the sun, as used in the calculation of the sun's force.

35926 = 4.555407 20964 ÷ (1322.1 × ?) = 4.98107 = 0.697323 3.858084 π = 0.497149 22658 = 4.355233 9 2)203922 3)101961 33987 ÷ 100 ... ... ... ... ... ... = 339.87

And 95000000 ÷ 92000000 × π ÷ 3 × 100 = 108.133 × π = 339.71 These are the greater and the lesser distances or the sun, and they have given the same, in the terms of π.

339.3 ÷ 47.067 × π X9 ÷ 2 ÷ 3X10 ... = 339 666 And 47.067 = the angle in the figure. Then,

5276 ÷ (9200.0000 ÷ ?) × π × 6 × 10 ... = 339.71 And 5276 is equal to the first four figures of the mass of the moon as used in the theory.

Then, 5276 = 3.722305 20964 ÷ (1322.1 ÷ ?) = 0.697323 3.024982 π = 0.497149 2249.3 = 3.522131 4 3)8997.2 2999.066 ÷ 100 = 29.99066 = 29° 59′ 26.4″

*

equals the latitude or the Great Pyramid ± in the terms of the inch in the terms of π,

and the latitude, by observation, is equal to ... ... 29° 58′ 51″

The next Symbol which attracted my attention was the Displacement of the Entrance and the Difference in the Height of the two Wainscots in the Ante-chamber.

The Displacement of the Entrance is equal to 295.7, or 296± inches.

The annual motion of the perihelion is equal to 11″.63 + 50".2 = 61″.83. Then,

page 7

5813 ÷ 61.83 × π ... ... ... ... ... ... ... = 295 36in. 7391.5 ÷ (20964 ÷ 100) × 26.302 ... ... ... = 295.26in. 8687.87 ÷ 9 × π ÷ 10 ... ... ... ... ... ... ... = 295.3in. 237000 ÷ 4 ÷ 2 ÷ 1000 ... ... ... ... ... ... ... = 295.25in. 882000 ÷ (7743 ÷ 27.5) × π × 3 ÷ 10 ... = 295.257in.

The length of the floor of the entrance passage is equal to 985 inches.

985 = 2.993436 25827 ÷ (7743 ÷ ?) = 10.4711 = 1.020314 1.973122 π = 0.497149 295.31 = 2.470271 ... = 295.31 in. The Height of the Entrance ... ... =652in.± 26° 18′ 43.2″ = 26.3125 × 4 ÷ 100 × 4 ÷ π × 486.3 =652 and the difference of the periods equals 4863 years 25827-20964 = 4863 years 296 ÷ 273.9 × π × 3 ÷ π × 15 ... ... ... ... ... ... = 4863 4863 ÷ 15 × π ÷ 3 ÷ π × 273.9 ... ... ... ... ... = 296

And the thickness of the granite leaf, north and south, is equal to 15 inches.

Having given you proof that my reading of the meaning of the difference in the thickness of the masonry courses was the correct one, you must bear in mind that I am writing this in the fourth month of the year 1885, and that it was in 1879 when I first read that symbol. These proofs are the result of six years' close study and testing of all the measures of this ancient monument. It was not till after more than four years' work that it occurred to me to set out the accompanying figure; the doing so necessitated the putting on one side all former work and commencing upon one uniform method—using the log. 0.497149 as a constant log. in all cases—and as I had to test the measures of the Great Pyramid inside and out, great and small, coffer included, not only by all the astronomical quantities used in the Theory; but by the arcs of time as shown in the figure, and the angles, which are subtended by those arcs of time and the complements of those angles and the complements of the arcs of time; also by the periods of the equinox, the period of the perihelion, and by the difference of these two periods; by the height of the "apex" of the "zone of water," and its period, etc., and by the annual motions of these two periods, and by the sum of these two annual motions, etc.

This will give you some idea of the labour I have bestowed upon the reading of the symbols and measures of this Prehistoric Monument,

page 8

In my earlier attempts at reading the meaning of the symbols I could not grasp the meaning of all, but since then more of them have revealed themselves as 1 proceeded with the investigation.

I will now give you a brief account of the

* Note.—20964 years less 7743 = 13221 years.