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The Pamphlet Collection of Sir Robert Stout: Volume 54

An answer to Professor Piazzi Smyth's questions as to the meaning of the symbols of the Great Pyramid of Egypt

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Why is the Bulk of the Ocean Retained in the Southern Hemisphere?

Printed by Wilsons and Horton, New Zealand Queen Street, Auckland. 1885

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An Answer Professor Piazzi Smyth's Questions as to the Meaning of the Symbols of the Great Pyramid of Egypt.

Printed by Wilsons and Horton, New Zealand Queen Street, Auckland. 1885

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The Thickness of the Masonry Courses of the Great Pyramid of Egypt.

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

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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

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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,

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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,

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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.

Meaning of the Symbols.

In the first place, the Great Pyramid of Egypt is a Prehistoric Monument, designed and erected by the people of a former period, and left by them as a "Scientific" record, or "Book of Stone," to make known to us (a people of a distant posterity, of another and later period), their knowledge of the existence of a "Zone of water," its period, arcs of time, etc. The Problem is written in the language of symbols and numbers, in the terms of the inch, in the terms of π, so that the people of the next period should be able to read it, as soon as they understood the "reason why" the bulk of the ocean was retained in the Southern Hemisphere at the present day, and they knew the value of

The builders knew there was no problem in "physical" science so important for us to know as the problem built up there, because it gives the astronomical and physical forces which govern the obliquity of the ecliptic, and reveals to us some of the lost pages of the true history of the past, prior to the Deluge of our history.

This ancient monument, then, is erected on a parallel of latitude which marks the boundary of the "Zone of water," during their period of repose, when this "Zone of water" was in the Northern Hemisphere, as it is now in the South. For we are passing through Our Period of Repose at this time. In fact, we are more than half-way through our period of repose "without knowing anything about it." * [If we did there would be less talk of the end of the world every few years.]

The period of this "Zone of water" is equal to 10,482 years from the time it crosses the equator, till, its return to the equator again.

During each period it is in motion 2739 years, and stationary during 7743 years, its apex is equal to 2185 feet, and it extends on either side over 1304 ± miles of latitude. Then, it follows that as the Great Pyramid marks the parallel of latitude, of the boundary, of this "Zone of water" during their period of repose, when it (the "zone of water") was in the Northern Hemisphere, it becomes evident that it was physically necessary that the foundation should be laid, and the building complete before the "zone of water" began its return journey to the south, because if it was not, the site would be under water.

Now, I have abundant evidence to prove that the foundation

* Note.—The perihelion point of the earth's orbit is now about 10 degrees past the solstice ± on its way towards the opposite equinoctial point.

page 9 was laid 375 years ± before the end of that former period of repose.—See figure.

A a = 375 ± and A a + A E = 1744.5 = date = 375 years + 1369.5 years = 1744.5 years, then, by the symbology. It symbolises that, 652 years after the foundation was laid, the apex of the "Zone of water" would be symbolically at the entrance.

You notice those Inclined Stones above the entrance, revealed now the outer stones are removed.* They were placed within the interior of the building for preservation, so that in future time, when we should be able to read this Book of Stone, we would understand that these Inclined Stones symbolised the apex, as being then symbolically at the entrance 652 years after the foundation was laid.

One of the acute angles of these stones is equal to 51°, then in the terms of the inch, in the terms of π

51 ÷ (652 ÷ 100) × π x7 ÷ l00 = 1.72018 1) I-1.7203 × (365.24 ÷ ?) }=652ins. II-One of the obtuse angles of these stones is equal to 128°, Then, 1284 ÷ (652 ÷ 10) × π ÷ 7 × 10 = 11.92143) 11.9223 × 7 ÷ 2 ÷ π × (5276 ÷ 100) } = 652in.

And by the symbology 1,092.5 years after the apex has arrived at the entrance it was to be at the equator, symbolised by the junction of the passages, then 652 + the whole length of the entrance passage is equal to 1,744.5 ±, because the whole length of the entrance passage is equal to (2,185 ÷ 2 = 1,092.5 inches ±) equal the half of the height, of the apex of the "zone of water" in the terms of the inch (See figure). If we call the Circle a representation of the Earth, then E E E represents the Equator and E will represent the equinoctial point, and the perihelion point of the earth's orbit was to be in the equinox 1,744.5 years after the foundation was laid.

Then, the Ascending and Descending passages symbolise the two hemispheres.

The apex is symbolically at the equator, then its arm (as I call it) would reach 1,302 miles across the equator into the opposite hemisphere, and the arm would then come under the influence of the Deficiency.

This is clearly symbolised by the symbology of the Horizontal Passage.

I will again refer you to the figure. You see those two square blocks of stone above the entrance passage; they are 100 inches by 100. Now, the line of intersection of these two stones, if extended

* Note.—See plates in Piazzi Smyth's work.

Note.—5,276 is equal to the first four figures of the mass of the moon as used in the calculation, and 652 inches is equal to the height of the entrance.

page 10 each way, would connect the Horizontal Passage and the outer face of the Great Pyramid; and the distance from the horizontal passage to the outer face is equal to 2185 ±, and from the point of intersection of this extended line with the outer face, down along the inclined face, is equal to half, equal 1092.5 ±. Then these two blocks of stone symbolise the two hemispheres, the line of intersection the equator, and they symbolise to us that in 1092.5 years ± the apex of the "Zone of water" would be at the equator, and that the Horizontal Passage symbolises the Arm of the "Zone of water," reaching the Deficiency. This Deficiency is symbolised by the floor, by that portion which you see is deficient in height, and the length of this low portion deducted from the whole length of the floor is equal to 1302± equals the length of the arm of the "Zone of water."

We have arrived then, symbolically, at the end of that former period, or any period, because we have arrived at the end of the 1744.5 years ±, and, after the 375 years had elapsed, the perihelion would then be within 23½° of the equinox E' or E, and it is during those years when the perihelion point of the earth's orbit is within 23½° of the equinox, that this "Zone of water" is in motion. Then, it follows that, during the 1369.5 years of the last arc of time of that former period, as this "Zone of water" advances from the north to the south, as it approaches the equator, the change of position of this body of water must effect the "centre of gravity of the earth," consequently the obliquity of the ecliptic would decrease till, at the end of the period, it would be reduced to "zero," because the apex, being then at the equator, neither pole would gravitate towards the sun when the earth arrives in the perihelion. This is symbolised in the horizontal floor of the King and Queen's Chambers. This would be the period known in geology as the carboniferous period, when ice and snow have been comparatively unknown at the poles.

The apex is now, symbolically, at the equator. The arm has reached the deficiency, consequently, currents would be set in motion (for this force would be stronger than the force of the sun for the time being), and these currents would increase in volume and velocity, and carry this "Zone of water" into the opposite hemisphere and submerge it, as is symbolised by the subterranean chamber, with its up and down and apparently unfinished floor, which symbolises to us the ceasing of all human undertakings, when the ocean breaks its bounds and causes universal destruction, and the death of all science for ages and ages after.

Here, we arrive at the cause of the drift period, and the knowledge of this will open to our contemplation some of the lost pages of the true history of the past—prior, to the deluge of our history—and enable us to trace out some near approach to the page 11 history of the many ruins of antiquity, of which, at present, we know nothing.

Then, when we arrive at the junction of the passages, we have arrived at the end of the one period and at the beginning of the new.

The Ascending Passage symbolises the first arc of time of a new period; its Portcullis entrance symbolises the serious difficulties and obstructions which will check the advance of the descendants of the remnants of the people during the first centuries of the new period, for no sooner do they overcome one difficulty than another presents itself equally formidable, and the more hard to bear because "Hope deferred makes the heart grow sick." But necessity has no law, and the force of circumstances compel them to proceed so that as time rolls on, obstacle after obstacle is removed till at length the end of this first arc of time is reached—the first 1369.5 years has passed.

The height of the south end of this passage is = 43.6 inches ±, and 1369.5 ÷ π = 435.934 ÷ 10 = 43.6 = the height of the south end of this passage and 1369.5 ÷ 435.936 = π.

The length, of this Ascending Passage along the floor, is equal to 1,542 inches, the π angle at the base of the Great Pyramid is equal to 51° 51′ 14″ = 51.8504; the height of the apex of the "zone of water" equals 2,185, then,

2185 ÷ 51.8504 × π ÷ 2 ÷ 6 = 11.0525} ... = 1542in. 11.0525 × 8 ÷ 100 × 4 ÷ π × 1369.5} equals the length of this passage. 1302.983 × 3 × 2 ÷ 100 ÷ π × 61.83 ... ... = 1542in. and 2739 ÷ 1542 × π × 7÷3 × 100 ... ... ... = 1302.07 equals the length of the arm ±, as is symbolised by the Horizontal Passage floor.

The Transverse Plates.

These plates have the passage cut through them—they form roof, sides, and floor—they symbolise the "Zone of water." they surround the passage as this "Zone of water" surrounds the earth, and the position of each is in the terms of the Problem, in the terms of the inch, etc., which will be too long to treat of here.

Then, when we have arrived at the south end of this passage, we have arrived, symbolically, at the commencement of the Period of Repose, when the "Zone of water" will remain at rest during all these 7743 years. I will ask you to refer to the figure. You see the arc EB equals 1369.5 years, equals half of AB, and when the perihelion is distant more than 23½° from the equinox, the obliquity of the ecliptic has reached its maximum of inclination ±, and during the 7743 years the perihelion is travelling from B round past the solstice S, to the point C, or within 23½° of the opposite equinox, the "Zone of water" remains stationary.

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This is symbolised by the Ramps, because the Ramps symbolise the Periods of Repose which occur in each hemisphere, and the space between them symbolises the arc AB = 2,739 years the space is equal to 42 inches, then, 42 × (1304 ÷ 10 ÷ 2) = 2739; this, ÷ 2 = 1369.5 = the arc EA × 2 = the arc AB, etc.

The Grand Gallery

Symbolises the two periods, equal to 20,964 years, and the high steps at the south end symbolise the "Zone of water" passing from one hemisphere into the other, as from A towards B, or vice versa, as the case may be.

The vertical height of the Grand Gallery varies from 339.05 ± to 339.8±; the mean is given as equal to 339.3 ±.

The distance of the moon, as used in the Theory, is equal to 237,000 miles.

237,000in. ÷ 365.24 × π 6 × 100 ... ... ... ... = 339.758 Then, the greater and lesser distances of the sun are equal to 95,000,000 and 92,000,000 miles ±, and 95000000 ÷ 92,000,000 × π ÷ 3 × 100 × π = 339.71 and 92000000 ÷ 882000 ÷ 3 × 10 ... ... ... ... ... = 339.7761 The cubic diagonal of the coffer is equal to 87.13in. ± 87.13 ÷ (7743 ÷ 1000) × π × 3 × π ... = 339.36 and 2160 ÷ 2 ÷ 10 × π ... ... ... ... ... ... = 339.3 The length of the Grand Gallery above the Ramps is equal to 1881.6 inches, width at ceiling is equal to 43.7 inches; then, 5003 ÷ 43.7 × π ÷ 3 × π = 376.65) 376.62 ÷ 3 ÷ π × 47.067} = 1881.6 equals the length of the Grand Gallery above the Ramps, and 47.067 is equal to 47° 4′, equals the angle A EE, etc., and 26° 18′ = 26.3 ÷ 4.863 × π × 8 ÷ 4x l0 = 339.8 then 10482-7743 = 2739, this ÷ (1304 ÷ 10 × 2) =42 equals the spaces between the Ramps; then, 92000000 ÷ 339.3 × π × 2 ÷ 1000 = 1701.66 = 1702 equals the height of the floor of the King's Chamber above the basal plane; and 93500000 ÷ 2739 ÷ 2 ... ... ... ... ... ... ... ... = 1702.95 95000000 ÷ 132.934 × 3 ... ... ... ... ... ... = 1703.01 237000 ÷ 8687.87 × π ÷ 4 × 100 = 1701.85 = 1702.0 The difference of the periods equals 4863. 4863 ÷ 339.3 × π 3 ÷ 3 × 10 = 50.0026 4863 ÷ 304.4 × π ... ... ... = 50.189 = 50.2 25327-20964 = 4863 1701.72 ÷ 9 ÷ π × 83.13 = 5003 ÷ 10 ÷ 100 = 0.5003 equals the annual deficiency. Then, 5003 ÷ 1702 × π × 5 × 10 ... ... ... ... = 461.17 in. and the diagonal of the floor of the King's chamber = 461.19 in.

page 13

The Grand Gallery symbolises the whole period of the perihelion, and the two hemispheres in the terms of the figure. Let A D and B C represent the Ramps on either side; make the spaces DC, A B = 42; then you have the floor of the Grand Gallery. Call the length equal to 1881.6 ±; then,

20964 ÷ 1881.6 × π × 2 ÷ 10 = 7.0026 = 7= the number of overlaping stones on either side; then, one of the heights of the Ramps equals 20.07, and

7×3×10x5 ÷ π × 9 = 3008 } 3007.8 ÷ 9 ÷ π × 20.07*} ... ... ... ... = 2185 equals the height of the apex of the "Zone of water" and the Ramps symbolises the "Zone of water," and 1744.5 + 1263.3 years ... ... ... ... ... = 3007.8 years and 3007.8 years × 61.83″ ÷ 60 ÷ 60 ... ... ... ... =51° 39′ 48″ equals the latitude of the apex of the "Zone of water" during the period of repose ±. Then, 2739 ÷ 20.07 × π =428.4 } 428.4 × 7 ÷ 100 = 29.938 } ... ... ... ... =29? 59′ 16″ equals the latitude of the boundary; if you add the length of the arm, equal 1304 miles of latitude ... ... ... ... ... = 21° 42′ 00″ then, 29° 59′ 16″+ 21° 42′ 00″ ... ... ... ... ... =51° 41′ 16″ ± The other height of the Ramps is equal to 20.96in.; then, 92000000 ÷ 7391.5 × π ÷ 3 ÷ 10= 1309.3} 1309.26 ÷ 10 × 3 × 8 ÷ π × 20.96 } ... = 20964 20964 ÷ 5813 × π × 10 = 113.3 × 3 = 339.9 } 339.91 ÷ × 2 ÷ π × 304.4 } ... = 20964 equals the whole period of the perihelion, and 304.4 is equal to the transverse height of the Grand Gallery; then, 20964 ÷ 339.8 × π × 3 × 3 ... ... ... ... = 1744.38 2185 ÷ 111.8 × π × 3 × 10 ... ... ... ... = 1745.34 is equal to the date of laying the foundation, equal to the Arc, aA + AE = 375 ÷ 1369.5 years = 1744.5 ±, as shown in the figure. The latitude by obs. =29° 58′ 51″= 29 981 × 10 = 299.81. 299.81 = 2.476846 7 = 0.845098 1.631748 π = 0.497149 1.134599 The obtuse angle of the Inclined} Stones = 128° } = 128 = 2.107210 1744.28 = 3.241809 = 1744.28

*

* Note.—One of the measures of the Transverse Plates is equal to 1212; then, 7743 ÷ 1212 × π =20.07.

page 14

2185 ÷ 128 × π =53.628 × 7 = 375.396 + 1369.5 = 1744.896 5003 ÷ 9 × π ... ... ... ... ... ... ... ... =1746.4 2185 ÷ lll.8 × π × 3 × l0 ... ... ... ... ... =1745.34 and 11.63 + 50.2 = 61.83″. Then, 1744.5 years × 61.83″ ÷ 60 ÷ 60 is = 29° 58′ 9″.

When we reach the South end of the Grand Gallery we have symbolically reached the end of the Period of Repose, and are about entering upon the last 1369.5 years of the period when the "Zone of water" will be approaching the equator and the obliquity of the ecliptic decreasing, till, at the end of the period, the angle will be reduced to zero, symbolised by the horizontal stone (which is called the High Step), because the motion of this "Zone of water" north or south must affect the centre of gravity of the earth. This centre is symbolised by the Boss on the Granite Leaf, and motion is symbolised by all the grooves. For instance, the distance of the centre of the Boss from the west wall face is equal to 19.5 inches, then the groove in the wall of the Grand Gallery is= 1878.24

5003 ÷ 1878.24 × π × 7 ÷ 3 ... ... ... ... ... =19.502in.

This directs your attention to the Boss by the measure. There are 28 square holes in the Ramps, then,

2185 ÷ 28 × π X7 ÷ 100 ... ... ... ... ... ... ... ... = 17.1612 and the grooves of the granite leaf is ... ... ... ... = 17.l6in. and 12913.26 ÷ 237000 × π =17.153 ... ... ... ... = 17.16in. ± and the diagonal of the Base is ... ... ... =12913.26in.

This High Step extends back equal to 61 inches, and

2185 ÷ 9 × π =762.7 × 8 ... ... ... ... ... ... =61.016

equals the measure of this stone which symbolises the "Zone of water." The distance from the groove end of the Granite Leaf in the Ante-chamber is equal to 3.25 inches,

2185 ÷ 3.25 × π ÷ 4 ÷ 10 ... ... ... ... =51.6625 =51° 39′ 45″ 12913.26 × 4 ÷ 1000 ... ... ... ... =51.65304 = 51° 39′ 10.8″ and the latitude 29° 58′ 51″ + 21° 42′ 0″ ... ... ... ... = 51° 40′ 51″

equals the latitude of the apex ± and 1302 miles ÷ 60 = 21° 42′ as used above. The Granite Leaf symbolises the Flood Gates of the oceans. There is a stone on which it rests; this stone is equal to 43.7; this × 10 × 5 = 2185 equals the height of the "Zone of water," which is symbolically passing the Flood Gates, at the end of the period.

The height of the floor of the King's Chamber above the basal plane is 1702 inches, and the diagonal of the floor is 461.19; then,

5003 ÷ 1702 × π x5xl0 ... ... ... ... =461.17 and 29.9826 x2 × 100 ÷ 9 ÷ 4 ÷ π × 412.13 ... ... ... =2185 then 2185 ÷ 235.5 × π × 7 ÷ 2 ÷ 2 ÷ 10 = 51.00375 = 51° 00′ 13″

page 15

equals the acute angles of the Inclined Stone above the Entrance Passage } = 51?00 00 5003 ÷ 230.47 × π ÷ 3 ÷ 10 = 2.27433) 13221 ÷ 2.2744 } ... = 5813

The Deficiency, so distinctly pointed out by the Difference in the thickness of the masonry courses, is also symbolised by the difference in the height of the Two Wainscots in the Ante-chamber; they also symbolise the "Zone of water" in one hemisphere. Then,

5003 = 3.699231 West Wainscot = 111.8 = 2.071882 1.627349 π = 0.497149 2)133.2 = 2.124490 66.6 = 1.823472 π = 0.497149 1.326323 East Wainscot = 103.1 = 2.013259 2185.4 = 3.339582 = 2185.4

equals the Height of the Apex of the "Zone of water"; and 2185.4 ÷ 5 = 437, equals the deficiency at the end of the 10,842 years (in inch feet); and 437 ÷ 10 = 43.7, equals the width of the stones in the ceiling of the Grand Gallery, equals the depth of the stone under the Granite Leaf, and 43.7 × 10 × 5 = 2185, and the stone under the Granite Leaf symbolises the "Zone of water" passing the Flood Gates of the Ocean, which open at the end of the period; for then the Ocean breaks its bounds, and causes universal destruction and the "Death of all Science "for ages after.

The three cylindrical hollows on the top of the West Wainscot are cut down equal to 8.5 inches.

50030 ÷ 7391.5 × π × 4 ÷ 10 ... ... ... ... ... ... ... =8.50568 882000 ÷ (412.13 × ?) × π ÷ 2 ÷ 100 ... ... ... =8.5 4863 ÷ 359.26 × π × 2 ÷ 10—... ... ... ... ... =8.505 237000 ÷ 7743 ÷ 6 ÷ 3 × 10 ÷ 2 ... ... ... ... ... =8.50005 111.8-103.1 = 8.7. The whole length of the Horizontal Passage is equal to 1517.9 inches, this ÷ 273.9 × π ÷ 2 ... ... ... ... ... ... ... =8.7 3393 ÷ (50.034 ÷ ?)X π × 4 ÷ 3 ÷ 10 ... ... ... ... ... ... ... =8.7215 2739 ÷ 989 × π ... ... ... ... ... ... =8.7005 5276 ÷ [20964 ÷ (7743 ÷ ?)] × π ÷ 2 ÷ 10 ... ... ... ... ... ... =8.72

page 16

and 92000000 ÷ (109.25X ?)x π ÷ 10 ... ... ... ... ... ... ... =8.5 20964 ÷ 12913.26x π ÷ 3 ÷ 2 × 10 ... ... ... ... ... =8.5003 then 2185 ÷ 8.7 × π ÷ 3 ÷ 10 = 26.3063 ... ... ... ... ... =26° 18′ 22″ equals the angle of the passages ±. 35926 ÷ 412.13 × π =273.86 × 10 ... ... ... ... =2739 ± 35926 ÷ 206.06 × π ÷ 4=136.9325 × 10 ... ... =1369.5 5276 ÷ 10 ÷ π x(61.83x4 ÷ ?) ... ... ... ... ... ... ... ... =13221 equals 20964 years, less 7743 years... ... ... = 13221 years 237000 ÷ 48.9 × π =15.22x2x 10 ... ... ... ... ...=304.4 equals the transverse height of the Grand Gallery. 2739 ÷ 9=304.333

The King's Chamber symbolises the end of the period. The apex of the "Zone of water" has then arrived back at the equator; this is symbolised by the floor rising up on the lower course of stones equal to 5 03, because 235.5-230.47 = 5.03. Then the Arm is equal to 1303.2±; this ÷ 3 = 434.4 and 434.4 × 5.03 = 2185 equals the height of the apex; and 1302.l ÷ 100 ÷ π × 515.16 equals 2185, and 515.16 is equal to the cubic or solid diagonal of the King's Chamber; and,

5003 ÷ 412.13 × π × 7 ÷ 4 = 66.7397 } 66.73 × 100 × π =20964 ÷ 2 } ... ... ... ... = 10,482

equals the period of the "Zone of water," and 1369.5 years + 7743 + 1369.5 = 10,482 years, equal to the Half of the Circle as shown in the figure; then,

2185 ÷ 412.13x π × 4 × 9 ÷ 100 ÷ 2=29.9826 =29° 58′ 57″ 2185 ÷ 309.15X π × 3 x9 ÷ 10 ÷ 2 =29.97945=29° 58′46″

and 412.13 is equal to the length, while 309.15 is equal to the diagonal of one of the walls of the King's Chamber; and,

5003 ÷ 18 × π × 2 ... ... ... ... ... = 1746.36equals the Date 20964 ÷ 339.3 × π x3x3 ... ... ... =1746.9 equals the Date and the Foundation was laid ... ... 1744.5 or 1746.5 BF and aA + AE=375 + 1369.5 (in inch years) =1744.5 ± BF (in inch years).

When we arrived at the Junction of the Passages, we symbolically commenced a new period. We have now arrived in the King's Chamber, and at the end of the second period, and 10,482 × 2=20964 years equals the whole circle.

I have told you that the tranverse plates symbolise the "Zone of water."

One of the distances of these plates is ... =799in. The length of the King's Chamber is ... =412.13in. 412.13 ÷ 2=206 065, this × 4 ... ... ... ... =82.428in.

page 17

Then, 82.428= 1.916075 π =0.497149 1.418926 Measure of Transverse Plate = 799 = 2.902547 2)20964 = 4.321473 10482 ... ... ... ... ... = 10,482 equals Half the Circle as shown in the Figure; and 206.06 ÷ 10 ÷ π × (132.934 ÷ 10) ... ... ... ... =87.19in. equals the greater cubic diagonal of the Coffer in the King's Chamber.

And the angle BEC is equal to 132° 56′ = 132.934 as used above. The annual Deficiency is equal to 0.5003in. at the pole.

.5003 × 10 × 1000 = 5003in. ÷ 1702 × π × 5 ... =461.17

and the diagonal of the floor of the King's Chamber = 461.19 ±

Are these proofs? I ask; or, can former writers prove their assertions by the same scientific method? It is Truth we want, not assertion.

In the calculation of the Sun's force you see I use the Sun's diameter as equal to 882,000 miles, the Sun's lesser distance as equal to 92,000,000 miles. Then, in the terms of the Inch, etc.,

92000000 ÷ 882000 ÷ 5 = 39.97766 1} ... ... =26? 18′ 00″ 39.98 × 4 ÷ 8 ÷ π × 4.863 = 26.3} equals the mean angle of the Passages ±. 26.3 ÷ 4.863 × π × 8 ÷ 4 ÷ π ÷ π × } ... ... = 132? 56′ 05″ 38.61 = 132.934} equals the angle B E C, as shown in the Figure; and 132.934 ÷ 60 ÷ 60x 61.83″ ... ... ... ... ... ... ... = 7743 years

equals the are B s C, which subtends the above angle, as shown in the Figure: and

10482 years-7743 = 2739. This divided by 2 = 1369.5 years. 10482 × 2 = 20964 = the Circle

and the 38.61 used above is equal to the mean external breadth of the Coffer in the King's Chamber; and

26.301 ÷ 2 × 10x 3 × 237000 ... ... ... ... =93500000*

equals the mean distance of the Sun (in inch miles), and 237000 is equal to the distance of the Moon as used in the calculation.

You will notice above that I have found the angle B E C equal 132° 56′ and the arc subtending it. I will now find the angle A E B and are subtending it, as follows:—

The inner depth of the Coffer is equal to 34.31in. ±, and the width of the Grand Gallery above the Ramps is equal to 82.2in. ±,

* Note.—(11025.3 ÷ 10 ÷ 3=387.51. This × 237,000=91,840,000).

page 18 then 34.305 × 10 ÷ 3 ÷ π × 82.2 = 2992 2992 ÷ 34.31 × π = 27.396 × 100 = 2739.6

equals the are AB = AE + EB = 1369.5 × 2 = 2739.0

as shown in the Figure; then,

2739 years × 61.83″ ÷ 60 ÷ 60 = 47.067 ± = 47° 4′ 0″ equals the angle AEB, as shown in the Figure.

The thickness of the walls of the Coffer is equal to 5.99; then again,

2992 ÷ 5.99 × π × 3 = 4707 ÷ 100 = 47.7 = 47° 4′ 12″ and 2739 + 7743 = 10482 years equals the period of the "Zone of water" equals Half the Circle.

The mean inner length of the Coffer is equal to 77.85in.

2992 ÷ 77.85 × π × 4 = 483.0 ÷ 100 = 4.830

equals the mean rise of the daily tide, equals 4.837 ± as used in the calculation. Then the cubic diagonal of the Coffer is equal to 87.15 ±, and,

2992 ÷ 87.15 × π = 109.25. This × 10 × 2 = 2185

equals the height of the Apex of the "Zone of water," equals the Force of the Sun as compared to that of the Moon, according to their mass and distance equals as 2185 : : 4.837.

5003 ÷ 461.19 × π × 2 × 9 ÷ 10 ÷ 2 = 30-030 30.033 × 100 × 6 ÷ π × 4 = 18225

equals 20964 years-2739 years = 18225 years

In the following calculation of the Sun's force, etc., as you see, I use the following numbers, and I will test each as I give them to you. Then, in the terms of the Inch, in the terms of π—

92000000 ÷ π × 7 × 94 ÷ 4 4 ÷ 1000 = 461.2275in.

the diagonal of the floor of the King's Chamber = 461.19in.

then, 237.000in. × π × 7 ÷ 4 = 1302.73in.

the low portion of Horizontal Passage floor deducted from the whole length = 1302.0in.

and 882.000 × π × 3 ÷ 100 = 83.1255in.

the diagonal of inner west wall of the Coffer = 83.13in.

then 2160 × π ÷ 2 ÷ 10 = 339.29in.

the mean vertical height of the Grand Gallery = 339.3in.

and 359.26 × π × 7 ÷ 3 ÷ 10 = 26.3293 = 26° 19′ 47.48″

equals the angle of the Passages ± = 26° 18′ 00″

then 5276 × π × 7 ÷ 3 ÷ 1000 = 38.677 in.

the mean outside breadth of the Coffer = 38.61 in.

4.837 × π × 7 ÷ 2 = 53.186 = 53.2 in.

the height of the north end of the Ascending Passage = 53.2 in.

2185 × π = 6834 ÷ 2 ÷ 100 = 34.32in.

the mean internal depth of the Coffer = 34.31 in.

If a sphere of the mass of the Moon at her distance has a force of attraction acting upon the waters of the ocean sufficient to cause a mean rise of tide equal to 4.837ft., what should be the force of page 19 attraction of a sphere of the mass of the Sun at its distance (when the Earth is in the perihelion of its orbit) in those parallels of latitude where it exerts its greatest force throughout the period of 10,482 years?

The Moon's diameter ... = 2160 = 10.003362 .5236 = -1.719000 5276, etc., etc. ... = 9.722362 The Sun's distance ... =92000000 = 15.927576 25.649938 The Sun's diameter ... = 882000′ = 17.836407 .5236 = -1.719000 35926. etc., etc. ... = 17.555407 The Moon's distance ... = 2370002 = 10.749400 28.304807 Mean rise of daily tide... = 4.837 = 0.684576 28.989383 25.649938 2185ft ... ... ... ... ... ... = 3.339445 The Sun's force: the Moon's force: 2185ft. : : 4 837ft.

The parallel of latitude where this force of the Sun will be exerted is governed by the position of the perihelion point—as it is near or distant from the equinoctial point on either side.

In the Figure, the Circle represents the period of the perihelion equal to 20,964 years from the time it leaves one equinoctial point till its return to the same again; its annual motion is equal to 11.63″ + 50.2″= 61.83″.

Because the motion of the perihelion and the precession move in opposite directions, therefore I use the sum of these two motions equal 61.83″.

And it is during those years when the perihelion point of the Earth's orbit is within 23½° of the Equinox on either side, or between A and B, or C and D, as shown in the Figure, that this "Zone of water" is in motion, moving to the north or south as the case may be.

In the calculation, the distance of the Sun from the Earth is equal to 92,000,000 miles equal the lesser distance, because it is when the Earth is in the perihelion of its orbit that the Sun will Note.—93500000 ÷ 35926 ÷ 2 = 1301.3. Then 92000000 ÷ 296 × π × 4 ÷ 3 × 10 = 1301.026} 1301.0 ÷ 100x3 ÷ π × 7391.5 } =91840.000 91840 × 1000=91840000 equals the lesser distance of the Sun ± in Inch miles. page 20 exert its greatest force upon the waters under its influence during the period.

Then the obliquity of the ecliptic will be governed by the position in latitude of the apex of this "Zone of water," and the position in latitude of the apex will be governed by the position of the perihelion point of the Earth's orbit.

Therefore the Sun exerts its greatest force upon the waters of the ocean, first in one hemisphere, then in the other, according to the position of the perihelion point. And, as the perihelion passes the equinox once every 10,482 years, the apex of this "Zone of water" will be at the equator once every 10,482 years.

The Deficiency is caused by the continued action of the centrifugal force (during the 10,482 years) where the waters of that hemisphere which is free from the "Zone of water" during the above number of years. It is well known that the tendency of the centrifugal force is to urge the waters to recede from the axis and flow towards the equator and there heap themselves up.

The effect of the centrifugal force upon the other hemisphere is counteracted by the presence of the "Zone of water" during the 10,482 years.

The existence of this "Zone of water" acts as a counteracting force to the continued action of the centrifugal force.

If this "Zone of water" did not exist, the bulk of the ocean would be heaped up at the equator at the present day.

It has been the motion of this "Zone of water" periodically from one hemisphere into the other which has caused all the great changes of climate, etc., which have occurred in remote periods.

And the deposition of the various layers of sedimentary strata geologically proves it, because the material of each successive layer has been brought from a distance.

All this evidence proves that the annual precession of the equinox goes on uninterruptedly independent of the change which takes place in the obliquity of the ecliptic. As this "Zone of water" moves from one hemisphere into the other periodically, the motion of the poles of the Earth in their curvilinear path will be found to be the result, and not the cause, of the precession.

This will make it almost certain that the motion of the pole will be caused by magnetic influences which will not allow the pole to remain outside the tangent of the radius when the Earth arrives in the equinox—the Earth and the Sun being each magnets.

Any of you who are acquainted with the use of the globes will find that the precession can be accounted for independent of the angle of the ecliptic (by the use of two globes).

If they will set out the Earth's orbit, place the one representing the Earth in its orbit, the other to represent the Sun, you will soon see that by the use of one globe only the cause of the precession is page 21 lost sight of, because you cannot move one ecquinoctial point without moving its opposite an equal arc, and with the globe (representing the Earth) in its orbit you will see that the Earth must arrive annually in the equinox before it has completed a sidereal revolution by an arc, equal the arc of precession; in fact, the Earth is annually behind an arc, equal the precession. This retreat of the Earth does not affect the sidereal year. But to arrive at how much the Earth is behind in any number of years, you have to multiply the annual precession by the number of years elapsed

That the Great Pyramid treats of the obliquity of the ecliptic I have hundreds of proofs. I can find you seven, or more, for every mile of change of angle, and each of these seven shall be within twenty seconds of arc of each other, and each a separate problem in the terms of the Inch, in the terms of ?, in the terms of the Figure. This prehistoric monument is one of the grandest studies on the face of the earth. It is a Book of Stone, written in the language of symbols and numbers, in the terms of the Inch, in the terms of ?. If you heed it not now, future ages will find Truth written there on every line.

In support of what I have said respecting the obliquity of the ecliptic, notice the following evidence:—

20964. ÷ π ÷ 296.x π ÷ 3 ... =23.6066 =23° 36′24″ 8687.87 ÷ 7743 × ÷ π ÷ 2 ... =23.611 =23° 36′ 39.6″ 365.24 ÷ 182.25 × π × 3 ÷ 4 × 10 ÷ 2=23.609625=23° 36′ 34.6″ 51.8504* ÷ 9.2 × π × 4 ÷ 3 ... =23.606 =23° 36′ 21.6″ 132.934 ÷ (309.6 ÷ 10) × π × 7 ÷ 2 ÷ 2=23.605975=23° 36′ 21″ 7743 ÷ (10303.3 ÷ 10) × π ... =23.609 =23° 36′ 32″ 71317 ÷ π ÷ 3000 × π ÷ 4 ÷ 2 × 10 =23.6025 =23° 36′ 9″ then 23.6022 × 3 × 3 ÷ 2 = 107.71. This × 882000 = 95000000 equals the greater distance of the Sun (in inch miles); and 95000.000 ÷ 51.8504* × π × 4 ÷ 1000 =23.024 =23° 1′ 26″ 92000.000 ÷ 47.067 × π × 3 ÷ 4 ÷ 2 ÷ 100 = 23.028375=23° 1′ 42″ 2160 ÷ 442 × π ÷ 4 × 6 ... ... ... =23.0295 =23° 1′ 44″ 50.03 ÷ π ÷ 6.52 × π × 3 ... ... ... =23.0211 =23° 1′ 15″ 92000000 ÷ 91840000 × 7 × 10 ÷ 3 =23.03033 =23° 1′ 49.2″ then 5276 ÷ 51.8504 × π × 3 ÷ 4 ÷ 10 =23.97225 =23° 58′ 20″ 35926 ÷ [5813 ÷ (7743 ÷ ?)] × π ÷ 2 ÷ 100=23.975 =23° 58′ 30″ 20964 ÷ 10303.3 × π × 3 ÷ 4 ÷ 2 × 10 =23.9704 =23° 58′ 13″ etc., etc.

*

I could give you, as I said before, hundreds of proofs like these; in fact, I have in manuscript, proofs in the like manner for every answer I have given in these few pages.

What is the boasted date, given by the author of our inheritance, in comparison to the evidence of the date which is given

* Note.—51° 51′ 14″. equals the π angle of the casing stones.

page 22 here? I can give you a hundred proofs of the date, and it is an astronomical date, not a theological one.

The Great Pyramid is a scientific record, designed and erected by the people of a former period to symbolise to us a most interesting problem in physical science, which all people should know, for it explains the mode of operation of those Divine laws which govern the whole universe, and these laws are the laws ordained by the Great Architect of the universe.

But this prehistoric monument tells you nothing about the Christian dispensation; all that has been said to that effect is merely emotional assumption, arising from ignorance of the real meaning of the symbols and admiration of the grandeur of the whole structure. All people are apt to attribute to Divine agency great works which they cannot read the real meaning of.

Let us hope that those who have endeavoured to prove the truth of their assertions will, in honour of the builders, in honour of the Great Architect of the universe, now endeavour to make known the true reading of the symbols and measures of this ancient monument, free from all fallacy; that all nations may understand the astronomical and physical forces which govern the position of this "Zone of water," its period and its motions, and thus learn something of the true history of the past prior to the Deluge of our history.

John Leith,

Manukau South Head, Auckland, N.Z.

P.S.—To enable me to publish the Theory, and the meaning of all the Symbols in full, I must ask those who wish for more information to send in their names as willing to subscribe for it.

J. L.

Wilsons & Horton, Printers, Queen and Wyndham Streets.

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The Problem of the Great Pyramid of Egypt

Is built up in the terms of the Inch, in the terms of π. In the Figure, the Circle represents the Period of the Perihelion, equal to 20,964 solar years, each year equal to 365.24 days. Its annual motion is equal to 11.63″ + 50.2″ equals 61.83″.

365.24 +&c × 2 : : 1 : π.

Mathematical diagram

Arc AB is equal to 2739 years. Then, 2739 × 61.83″ ÷ 60 ÷ 60=47.067? equals Angle A E B=47° 4′ 0″.

Arc BC is equal to 7743 years. 7743 × 61.83″ ÷ 60 ÷ 60= 132.934° equals Angle B E C=132° 56′ 0″.

25,827 ÷ 7391.5x 1000=3494x6= 20,964 ÷ 2= 10,482 = the Period of the "Zone of water."

10,482-2739 = 7743. This, divided by 365.24 × π × 9 ÷ 2 ÷ 10=29.9706 equals 29° 58″ 14″ = the Latitude of the Great Pyramid in the terms of the Inch, in the terms of π.

π = Log. 0.497149, &c.

5813x2 = 11626 = the diameter of the Circle (as shown in the Figure) equals E E.

9131 in. × 4 ÷ 100 = 365.24.

In the Figure, aA = 375 years; add AE, 1369.5 years = 1744.5. This, multiplied by 61.83″ ÷ 60 ÷ 60 = 29° 58′ 9.44″. Latitude, by observation, equals 29° 58′ 51″ N.

The Foundation was laid 1744.5 years ± BF.

The Vertical height is in the terms of the Lesser distance of the Sun; therefore, 91,840,000 is equal to the Lesser distance ± (in inch miles).

51° 51′ 14″=51.8504+(61.83 x4) × π × 4 ÷ π × 625 = 20964 = the Circle.

Note.—The answers are to be read "symbolically" as equal to (in the terms of the Inch).

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Reprinted from the Proceedings of the Royal Society.