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The Pamphlet Collection of Sir Robert Stout: Volume 54

The Grand Gallery

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.