THE UNKNOWN GREAT PYRAMID



THE GREAT PYRAMID MAY 2002

THE MATHEMATICS OF THE FOUR SHAFTS

By Jeremy Potter

Preliminary Information

It is well known that there are two shafts rising upwards from the King’s Chamber and a further two rising up from the Queen’s Chamber. The purpose of these shafts has always been a matter of conjecture, especially since it was discovered that the two lower shafts were built closed at both ends. The answer was that the shafts were actually diagonals on squares and rectangles seen on a pyramid section east to west and viewed from a distance. None of the slope angles at the Great Pyramid were measured in degrees, minutes, and seconds, but in tangents, and they were taken from a 90-degree triangle formed by half a pyramid on its base. The tangent of the pyramid base angle was found by dividing the height by half the base. The distances and angles on the pyramid, mainly in whole number cubits, could then be obtained by multiplying or dividing on the number given by the tangent. It was very simple but very profound. The tangential value of its base angles will then have been 280 / half-base 220 = Tan 1.272727 the original design slope angle of the Great Pyramid and given in number form. The equivalent value in degrees can then be found from the Standard Four Figure Mathematical tables issued by Cambridge University to give a slope of 51 degrees 50 minutes 30 seconds, or just less than 52 degrees, as commonly assumed.

Nothing more could be found at the Great Pyramid without an accurate survey given in architectural format so as to avoid ambiguity. The Italian engineers Rinaldi and Maragioglio had carried out such a survey in 1965 and they had given the existing distances and levels of the Great Pyramid on a set of large-scale drawings. Without recourse to those drawings it would be impossible to draw any reliable conclusions. They were used as a reference during the investigations that follow but the survey was done in metres and that made it necessary to convert metres to cubits, the unit used by the original builders.

The question was, how many cubits were there to the metre. It was an absurdity because the metre did not exist at the time of the pyramids whereas the unit foot might have done. The Cole Borchardt Survey of 1925 had found the original corner stones buried in the sand and they found that the distances between the four corners were very close to 756 feet. The builders had been trying to build a square of sides 756 feet and the differences were due to simple construction error. The pyramid might have been 280 cubits high and 440 cubits on its base, before its casing stones were taken, and if that were true then the correct conversion was 756 / 440 = 1.718181 feet per cubit. The British Museum says that it was 1.7183 feet per cubit. The difference is small indeed.

The geometrical principles that had been adopted for the planning of this unique pyramid were pyramids inside pyramid squares. The first of these was the Great Pyramid Square whose sides were 440 cubits inside which was the Great Pyramid on its base and 280 cubits high. The pyramid (or isosceles triangle) inside such a square will carry certain properties. One of these is that if the two slopes are extended, and reflected about the square until they exit the square, they will create 11 layers of 40 whole number cubits, and 14 divisions, making 154 sub-rectangles, all found by calculation using Tan 1.2727 recurring. The numbers were dedicated.

Piazzi Smythe, Astronomer Royal of Scotland had in 1868 published his Life and Work at the Great Pyramid in which he had advocated that the area of one sloping side was equal to the height squared, but he did not know then that the cubit was equal to 1.718181 feet because the corners stones had not then been found. Nor did he know about 280 and 440 cubits. He thought that the cubit was 25 inches long. He also claimed that the value for PIE could be found at the Great Pyramid and for that Sir Henry James, Director of the Ordnance Survey, derided him and he lost his credibility. In fact, he was right and he can now be vindicated.

280

356.0898

220 440

The value of П = 3.1416, the circumference of a circle divided by the diameter. If twice the pyramid height becomes the diameter of that circle and the pyramid perimeter is the circumference then Pyramid П is 1760 / 560 = 3.1428571. The number is close but not quite close enough to the correct value, and that was the problem. Very strangely, the identical number also resides in 440 cubits / 14 and to six decimal places but 10 times larger. This was not the true value for Pie at the Great Pyramid but it was there nonetheless. It was based Smythe’s other theory that the area of one of the sloping sides was equal to the height squared.

Height squared = 280 x 280 = 78400 sq cubits

Half base squared = 220 x 220 = 48400 sq cubits

78400 + 48400 = 126800 sq cubits

Square root 126800 = 356.0898 cubits

Area sloping side = base x half height

Area sloping side = 178.0449 x 440 sq cubits

Area sloping side = 78339.756 sq cubits

Square root 78339.769 = 279.8924 cubits

If the height of the square, derived from the area of one sloping side, is seen as a pyramid height, then it falls short of 280 cubits by a distance of 280 – 279.8924 = 0.1076 rather oddly non-recurring cubits. If then the distance of 0.1028 cubits is added to a pyramid 280 cubits high, that gives 280.1076 cubits for a new conceptual pyramid height. Now if the theory that the pyramid height was the radius of a circle, and the perimeter was the circumference, was correct, then the exact value for Pie will appear.

Pie = Perimeter / New pyramid height

Pie = (440 x 4) / (280.1076 x 2)

Pie = 1760 / 560.2152

Pie = 3.1416

Such an extraordinary result, on a pyramid that archaeologists claim was never designed at all, must prove that the pyramid was 280 cubits high and 440 cubits on its base when its original casing stones were in place. The long hidden meaning behind the four shafts can now be revealed as they had originally related with a pyramid 280 cubits high, and 440 cubits on its base.

THE SOUTH SHAFT, KING’S CHAMBER.

The South Shaft runs a short distance horizontally from the south wall of the King’s Chamber before rising out of the pyramid at an inclination of exactly 45 degrees until it passes out of the south slope, but because the original casing stones are missing at the point of exit, it is impossible to know the where exit level had occurred by physical measurement.

The Rinaldi and Maragioglio survey of 1965 had climbed the pyramid and taken levels off the existing pyramids stones around the shaft exit point and they found levels of 79.24 metres (150.30 cubits) on course number 102, and 79.88 metres (152.53 cubits) on course number 103, but they had not measured on course 104. They had then shown a drawn section through the shaft exit point and had extended the line of shaft at 153.31 cubits above base where they thought it had originally passed out of the pyramid when the casing stones were in place.

The South Shaft Exit Point

104 153.31

Level 154

103

102

Original slope

Existing Stones

The engineers had assumed that the line was on the shaft floor whereas it was actually on the upper surfaces. The reason for that was because the horizontal parts of the shafts align with the entrance corridor ceiling at the King’s Chamber below, and at a particular level that had formed a datum above base. If the upper surfaces are used then the level where the shaft had passed out of the original pyramid slope was very slightly higher, and it was almost certainly at 154 cubits above base, the same in cubits as the number of rectangles on the pyramid square. If the level of the exit point is 154 cubits above base then another pyramid will exist above that level and it will be 280 – 154 = 126 cubits high.

280

South

154

Shaft

Tan 1.2727

220 99 121

With the tangent on the pyramid slopes known, the length of the half-base of that pyramid can be calculated at 126 / Tan 1.2727 = 99 cubits to give a full base of 99 x 2= 198 cubits for a pyramid 126 cubits high. This will mean that the exit point of the South Shaft, King’s Chamber, is not only 154 cubits above base but also 99 cubits south of pyramid centre.

The relationship between the base of the upper pyramid and that of the Great Pyramid now produces another revelation. A pyramid square can now be added, so that its corners abut the pyramid slopes.

The First Pyramid Square at level 154 cubits

280

154

82

44

99 121

A notional square exists of sides 198 cubits and its upper corners abut the original slopes of the Great Pyramid at an assumed level of 154 cubits above base. The square will therefore extend below the base by 198 – 154 = 44 cubits and a second pyramid 126 cubits high can be constructed on the base of the square. Its tip will extend above the Great Pyramid base by 126 – 44 = 82 cubits, and on the pyramid centreline.

The Great Step at the Grand Gallery

Pyramid centre

Grand Gallery

To King’s Chamber 82 cubits above base

Great Step

Pyramid Tip

The level on the top of the Great Step at the Grand Gallery leading into the entrance corridor of the King’s Chamber is known because the RM Survey had taken the level at 42.96 metres above base, and that would convert to 42.96 x 1.0936 x 3 / 1.71818 = 82.03 cubits. But for the smallest of errors, that is 82.00 cubits above base, and it has been found according to the pyramid inside the pyramid square.

The pyramid geometry is proving the reason for the Step. In that case the builders were able to place a stone at a level high up inside the pyramid with incredible accuracy, and only tangents and pyramid squares would show where this was on the pyramid section.

The Second Pyramid Square at level 126 cubits.

280

154

126

116

121 99

Because there exists a level of 154 cubits on the shaft exit point there can also exist a pyramid 154 cubits high at that level with a half-base of 154 / Tan 1.2727 = 121 cubits. A second notional square therefore exists of sides 121 x 2 = 242 cubits and its topside will abut the pyramid slopes at a level of 99 x Tan 1.2727 = 126 cubits above base.

The base of the second pyramid square will extend below the base of the Great Pyramid by 242 – 126 = 116 cubits. The base dimensions are now reversed from 99 and 121 cubits for the first square to 121 and 99 cubits for the second square. The geometry is revolving about a centreline.

The Second Height on the Second Pyramid Square.

280

242

154

126

38

242

116

Because the square extends below base by 116 cubits and the pyramid is 154 cubits high, its tip must extend above the base by 154 – 116 = 38 cubits. The Great Pyramid is 280 cubits high and so the tip intrudes into the height by 280 – 38 = 242 cubits. But 242 cubits is also the length of side of the second pyramid square. The distance of 242 cubits now exists on the shaft exit point and the second pyramid tip.

The difference between the levels of 154 cubits, and 126 cubits, is 28 cubits, and that is 10 times smaller than 280 cubits, height of the Great Pyramid. The mean level between the levels of 154 cubits and 126 cubits will then be (28 / 2) + 126 = 140 cubits above base and that is half the height of 280 cubits. The pyramid has been bisected.

The Slope on the Diagonal

280

275

275

North

154

Line of Shaft

154

121 99 55 165

165

99 + 55 = 154 cubits

220 – 55 = 165 cubits

121 + 99 + 55 = 275 cubits

According to Petrie, the shaft slope angle was something slightly greater than 45 degrees, but the Gantenbrink survey of 1993 using a robotic crawling device, decided that the angle was at 45 degrees. It was actually the diagonal on a square at 45 degrees. In fact it was the diagonal of three consecutive squares at the pyramid.

If the shaft is a geometrical line on section it is also acting as a diagonal on squares of sides 154 cubits, 165 cubits, and 275 cubits crossing the pyramid base at 154 cubits due north of the exit point. Because the exit point is 99 cubits due south of pyramid centre, the point of crossing on the base must then be 154 – 99 = 55 cubits north of pyramid centre, and it divides the Great Pyramid base of 440 cubits into 8 sections exactly.

The Shafts at the King’s Chamber

South North

King’s Chamber

Datum 84

X

3 10

The Rinaldi and Maragioglio survey gives a vertical distance of 1.14 metres from the King’s Chamber floor to the upper reveal of the south shaft opening on the south wall, converting to 2.17 cubits. If a level of 82 cubits is taken on the floor from the Great Step then the level of point X at the change of direction of the shaft and on its upper surfaces is about 84.17 cubits above base giving a probable design level of 84 cubits above base for the datum at the King’s Chamber.

The distance between the north wall of the King’s Chamber and pyramid centre is 330.6 inches or (330.6 / 12) / 1.7183 = 16.03 cubits, as given by the Petrie Survey of 1880 and the RM survey had found that the chamber is 5.24 metres wide, or 10 cubits wide exactly. They also found that the distance between the point change of direction for the south shaft and the face of the south wall is 1.5 metres, or 2.86 cubits. The distance from point X to pyramid centre will then have been somewhere very close to 16.03 + 10 + 2.86 = 28.89 cubits. The design distance was probably 29 cubits to pyramid centre.

The Original Position of the South Shaft

154

70 Centre

South

Shaft Chamber

84

55

55

Tan 1.2727

220 55

The coordinates on point X on the shaft as seen on the east-west pyramid section will then be 84 cubits above base and 29 cubits south of pyramid centre. The true position of the South Shaft can now be shown relative to the pyramid geometry. The shaft rises at 45 degrees and is a diagonal on three squares. The lower square is of sides 55 cubits giving a level of 55 cubits on pyramid centre. The second square on the same diagonal is of sides 29 cubits, creating the next level of 55 + 29 = 84 cubits. The third square on the diagonal must then be of sides 154 – 84 = 70 cubits and its south side must be 70 + 29 = 99 cubits south of pyramid centre and that was the half base distance on the first pyramid square.

The square of sides 70 cubits will have a perimeter of 70 x 4 = 280 cubits and that is the Great Pyramid height. The length of the shaft from exit point to change of direction will be the hypotenuse of a square of sides 70 cubits, or the square root of 70 squared + 70 squared = 98.99 cubits, or as close to 99 cubits as it could get. The perimeter and the diagonal of a square of sides 70 cubits are mathematically meaningful. That must show that the line on the shaft as existing is the same line at that determined by the geometry of pyramids and pyramid squares.

THE NORTH SHAFT, KING’S CHAMBER

The first problem associated with this shaft is that it is not straight and has been found to divert laterally as it progresses upwards, during which time it also changes direction vertically as well. As a result there can be no specific diagonal on a rectangle or a square, and the tangent of the shaft slope is impossible to find, but that did not mean that the shaft was useless. The RM Survey had also taken the levels on the stone courses on the existing the North Slope where the shaft exits the pyramid.

The North Shaft Exit Point

151.31 cubits above base

151.25

103

102

Tan 1.2727

Tan 1.2500

Existing Stones

The level on course number 103 was given on the engineer’s drawing at 79.24 metres above base, converting to 151.31 cubits above base. They had extrapolated the line where they had thought the original shaft had passed out of the original pyramid slope when the casing stones were in place. They had naturally assumed a pyramid 280 cubits high, but they could have been wrong. No one knows if this shaft had passed out of the pyramid slope because the casing stones are missing. The level that they gave is suggestive that something quite different had occurred in this location, and it had involved another pyramid of another height.

The opposite shaft, when it was acting as a diagonal on a square, had produced a square of sides 275 cubits, falling short of the built pyramid height by 5 cubits. Perhaps that had been intended as a means of linking the north shaft with the south shaft. If that were so then the pyramid that had related with the north shaft had never been built in stone but was conceptual. It might have been 275 cubits high, and one way of finding out is to relate the two pyramid heights with shaft exit points where Z is the unknown level for the north shaft exit point.

154 x 275 = Z x 280

Z = (154 x 275) / 280 = 151.25 cubits

That looks just about right considering that 151.31 cubits had been found by survey on the floor of the shaft for a pyramid 280 cubits high, whereas if the pyramid was less high, giving a less steep base angle, and a lower shaft exit point, and on the upper surfaces, the exit point would be at 151.25 cubits above base, and must be correct.

The First Pyramid Square

275

151.25

Shaft

46.75

99 121

That would give a slope angle of 275 / 220 = Tan 1.2500. There would then be another pyramid above the level of 151.25 cubits and it would be 275 – 151.25 = 123.75 cubits high, and its half base would be 123.75 / Tan 1.2500 = 99 cubits exactly!

But that had been the half-base distance for a pyramid 280 cubits high with a shaft exit point at 154 cubits above base. The pyramid squares are of sides 198 cubits for both of these pyramids and they are common to both. That must prove a pyramid 275 cubits high with a shaft exit point at 151.25 cubits above base, was correct, and it did relate with the shaft opposite.

The Pyramid on the Pyramid Square

275

151.25

77

46.75

99 121

The pyramid square will extend below the base of the Great Pyramid by 198 – 151.25 = 46.75 cubits. Another pyramid inside that square will be 99 x Tan 1.2500 = 123.75 cubits high and its tip will extend above the base by 123.75 – 46.75 = 77 cubits exactly. That level is 5 cubits below 82 cubits, the level on the top of the Great Step, and the pyramid to which it relates at 275 cubits high is 5 cubits below the tip level of the pyramid 280 cubits high. These pyramids are inter-related.

Two Pyramids and two Shaft Exit Points

280 275

2.75

154 151.25

82 77

Tan 1.2727 Tan 1.2500

When the upper parts of the two pyramids are shown together it quickly emerges that the two pyramid squares are linked. The difference between the two exit point levels is 2.75 cubits coming off a pyramid 275 cubits high and 100 times greater. The difference between the two pyramid tips is 5 cubits and at the chamber level it is another 5 cubits. A pyramid 5 cubits high will have a half base of 5 / Tan 1.2727 = 3.928571 cubits divisible into the half base of 220 cubits by 56. There are 56 layers of 5 cubits on a pyramid 280 cubits high, and another 55 layers of 5 cubits on a pyramid 275 cubits high.

The heights 280 – 82 = 198 cubits and 275 – 77 = 198 cubits are the same, the length of side of the first pyramid square for both pyramids. There is no doubt then that an invisible pyramid 275 cubits high, and sloping at Tan 1.2500, was a geometrical reality.

The Second Pyramid Square

275

151.25

123.75

118.25

121 99

A pyramid 151.25 cubits high will carry a half base of 151.25 / Tan 1.2500 = 121 cubits, and a second pyramid square is created of sides 121 x 2 = 242 cubits. The square will abut the pyramid slopes at a level of 99 x Tan = 123.75 cubits above base and extend below the pyramid base by 242 – 123.75 = 118.25 cubits.

The Pyramid on the Second Pyramid Square

275

123.75

33

26.4

118.25

121

A pyramid drawn on the second pyramid square will be 121 x Tan 1.2500 = 151.25 cubits high but because its base is 118.25 cubits below Great Pyramid base, its tip will be 151.25 – 118.25 = 33 cubits above base. A pyramid 33 cubits high will have a half base of 33 / Tan 1.2500 = 26.4 cubits long. The probability is that there exists another pyramid on the next stage and it is 10 times larger and 264 cubits high.

Before that stage is reached however, there is another aspect of a pyramid 275 cubits high because a line drawn on one base position is able to pass out of its opposite slope at 90 degrees to create another important shaft slope angle.

The Diagonal at 90 degrees on the Slope

275

90°

Tan 1.2500 Tan 0.8000

220

If the diagonal passes out of the opposite slope at 90 degrees it will form a right-angled triangle where one side is known, at 440 cubits, and one base angle is known, at Tan 1.2500 together with 90 degrees. That makes it possible to calculate the value of the second base angle because there are always 180 degrees inside any triangle. From the Standard Tables…

Tan 1.2497 = 51 degrees 20 minutes 00 seconds

Tan 0.0003 = 00 degrees 00 minutes 22.5 seconds

Tan 1.2500 = 51 degrees 20 minutes 22.5 seconds

That angle, given degrees, can now be deducted from 90 degrees and the number of degrees remaining will then be able to give the missing base angle in degrees and then in tangential form. It is the number in tangent that is being sought.

90 degrees 00 minutes 00 seconds

-51 degrees 20 minutes 22.5 seconds

38 degrees 39 minutes 37.5 seconds

From the Standard Tables…

38 degrees 40 minutes 00 seconds = Tan 0.8002

- 00 degrees 00 minutes 22.5 seconds = Tan 0.0001.8

38 degrees 39 minutes 37.5 seconds = Tan 0.8000

Flinders Petrie had also measured the inclination of the South Shaft of the Queen’s Chamber during his famous survey of 1880-1882. He found that the shaft slopes at 38 degrees 28 minutes 00 seconds. The two findings are very close together with one calculated from the hidden geometry and the other measured by physical means. Petrie was a mere 11 minutes 22.5 seconds short, and since a calculated angle is always going to be much more accurate than one measured physically, it looks as if Petrie was slightly wrong. That would say that the intended design slope angle was precisely at Tan 0.8000. If that applied to the South Shaft of the Queen’s Chamber for another pyramid not yet found, then Tan 0.8000 had belonged to the Queen and was appropriate.

THE SOUTH SHAFT, QUEEN’S CHAMBER

The line of the South Shaft of the King’s Chamber above had passed out of the Great Pyramid base at 154 – 99 = 55 cubits north of pyramid centre. The South Shaft of the Queen’s Chamber is directly below and this slope too might have passed out of the base of the Great Pyramid at 55 cubits north of centre. If it had, then this can be shown in section.

The Slope on the Diagonal

The Slope on Diagonal

Pyramid Centre

132

88

Slope on Shaft

44

Tan 1.2000 Tan 0.8000

110 55 55 55

If the line carrying the same slope angle of Tan 0.8000 as found on the pyramid 275 cubits high, is transferred to a section through the South Shaft of the Queen’s Chamber, and if it also begins at 55 cubits north of pyramid centre on pyramid base, as the shaft above had done, then the line would rise in three stages of 44 cubits, and on three distances of 55 cubits, to give a slope angle of 44 / 55 = Tan 0.8000. On the third stage the line would pass through a pyramid slope at 44 x 3 = 132 cubits above base. But the distance between the starting point on the pyramid base and the south base pyramid corner is 220 + 55 = 275 cubits, where the shaft has traversed horizontally by 55 x 3 = 165 cubits. The distance that had not been traversed on the pyramid base is 275 – 165 = 110 cubits. That creates a right-angled triangle 132 cubits high and 110 cubits on its base to give a slope angle of 132 / 110 = Tan 1.2000.

A pyramid with base angles of Tan 1.2000 and a half base of 220 cubits would be 220 x Tan 1.2000 = 264 cubits high. That was the number that had come from the half-base of the earlier pyramid 33 cubits high and sloping at Tan 1.2500, to give 26.4 cubits and 10 times smaller. If that were true then a third pyramid has now appeared for the South Shaft of the Queen’s Chamber and it is 264 cubits high.

The Slope on the Diagonal for the shaft, rising at Tan 0.8000, would not however be the line of the shaft that was built in stone because the shaft changes direction in the wrong place some distance south of the Queen’s Chamber wall. There were then two slopes for this shaft.

The Slope of Diagonal to pyramid centre, and notional

The Slope of Shaft offset from pyramid centre, and built in stone

There was an important geometrical reason why there were two slopes emanating from the chamber and it would only be fully known when the exit point level for the shaft in stone is found. That point will probably be at the location of the small door that Rudolf Gantenbrink had discovered to worldwide acclaim in 1993.

The two pyramids that were 280 cubits high and 264 cubits high can now be shown together. It is apparent that the exit point at 132 cubits above base is half the height of a pyramid 264 cubits high. That would indicate that the exit point is correct. A pyramid above that level would be 132 cubits high and its half base would be 132 / Tan 1.2000 = 110 cubits long. A square would then exist of sides 110 x 2 =220 cubits and that is the half base distance of the Great Pyramid base, which is 440 cubits and another confirmation that the slope angle is correct!

The Square, the Rectangle, and the Diagonals

264 280

275

220

Slopes on Diagonal

132

55 55

275

220

88

220 110 110

Because the Slope on Diagonal begins its journey at 55 cubits north of pyramid centre, it forms the diagonal of a rectangle 220 + 55 = 275 cubits long, and 275 x Tan 0.8000 = 220 cubits high. But the diagonal of the South Shaft, King’s Chamber above had also passed through a square of sides 275 cubits in the same location, for it too began its journey at 55 cubits north of pyramid centre. The result is that the inclinations begin at the same point, the upper slope rises at Tan 1.0000, and the lower slope rises at Tan 0.8000. These were the two slope angles that Bauval and Gilbert had claimed in their book The Orion Mystery were aimed at the stars Orion for Osiris, and Sirius for Isis, in times long ago. If that was so then Tan 1.0000 and Tan 0.8000 determined the alignment on the stars giving rise to the number 18, the active principle for Isis.

The Optimum Pyramid Square.

264

165

Optimum Square

Tan 1.2000

82.5 137.5

There is only one square that will fit into an isosceles triangle in the form of a pyramid without leaving a gap or a surplus and that is the Optimum Pyramid Square. It carries a special significance at the Great Pyramid but only in whole number cubits on a pyramid 264 cubits high. The Square is proven because its half base is 82.5 cubits, the difference on the pyramid half base at 220 cubits, is 137.5 cubits, and the height of the pyramid square within a pyramid 264 cubits high with base angles Tan 1.2000 will be 137.5 x Tan 1.2000 = 165 cubits. The Optimum Square is of sides 165 whole number cubits or twice 82.5 cubits.

The remainder distances on the pyramid base at 137.5 cubits each side together add up to 275 cubits, the height of the first notional pyramid for the North Shaft of the King’s Chamber. By this means, the Optimum Square brings the two pyramids together.

The Triangle at the Queen’s Chamber

Slope on Diagonal

Slope on Shaft

Datum 44

7.3333

4.2 ? 5

The Rinaldi and Maragioglio Survey shows that the level of the floor of the entrance corridor to the Queen’s Chamber is 21.74 metres above base and the corridor is 1.29 metres high on the roughly finished floor. That will give a level of 23.03 metres above base for the upper parts of the horizontal shaft, and that converts to 43.98 cubits, or 0.02 cubits short of a datum level of 44 cubits above base.

The engineers also gave a distance of 2.29 metres between the change of direction of the shaft, measured on the floor, and the face of the south wall. That converts to 4.37 cubits on the floor of the horizontal section of shaft but the exit points of the upper two shafts of the King’s Chamber were on the upper surfaces. That would apply for this shaft too because the uppers surfaces align with the entrance corridor ceiling just as they do for the shaft above. The distance measured by the engineers from the wall to the change of direction on the shaft, and on the floor, would then be very slightly too long. The intended design distance was therefore something around 4.20 cubits.

The Datum and the Double Square

264

132

36.6666

Datum 44

88

110

The pyramid square of sides 220 cubits will extend below the pyramid base by 220 – 132 = 88 cubits and the pyramid inside the square will be 132 cubits high. That gives rise to another revelation. The pyramid base inside the square will then be 132 – 88 = 44 cubits above the base.

A pyramid 44 cubits high exists and it bears on the datum at 44 cubits above base, as found by the RM survey confirming geometrically that 44 cubits above base on the stone ceiling of the entrance corridor was a fact and that the upper surfaces of the two horizontal sections of shaft had also marked that datum level. But the half base of a pyramid 44 cubits high is 44 / Tan 1.2000 = 36.6666 cubits long, creating a double square that is 36.6666 cubits high.

The Triangle and the Double Square

Slope on Shaft Pyramid centre

Slope on Diagonal

Datum 44

36.6666

X

9.1666 x 7.3333

DOUBLE SQUARE

Pyramid Base

36.6666

What the Double Square reveals

The upper surface is 44 – 36.6666 = 7.3333 cubits below the datum of 44 cubits above base and 10 times smaller than the length at 73.3333 cubits, where the factor of 10 appears again. There are 6 layers of 7.3333 cubits below the datum and 5 layers of 7.3333 cubits below the top surface of the Double Square. That would create a triangle 7.3333 cubits high on which would ride the Slope on Shaft on its hypotenuse with the Slope on Diagonal on its corner. The triangle would then be 7.3333 / Tan 0.8000 = 9.1666 cubits long and is now able to show if the change of direction on the upper surfaces of the shaft that was built in stone was correct.

The distance that had been found by measurement was approximately 9.20 cubits, but the correct design distance was 9.1666 cubits just found. The horizontal upper surfaces of the shaft that was built in stone can now be verified on the triangle.

The Slope on Diagonal enters the Double Square at a point 9.1666 cubits north of pyramid centre and creates a triangle X. Since the half width of the Double Square is 36.6666 cubits, it is 36.6666 – 9.1666 = 27.5 cubits long and 27.5 x Tan 0.8000 = 22 cubits high. Once again the factor of 10 appears because 27.5 cubits are 10 times smaller than 275 cubits, the height of the second pyramid, and 22 cubits are 10 times smaller than 220 cubits, the pyramid half-base. There must then exist a direct geometrical connection between the North Shaft, King’s Chamber, for a pyramid 275 cubits high, and the South Shaft, Queen’s Chamber, for a pyramid 264 cubits high, on opposite faces of the Great Pyramid slopes.

The Double Square can also be seen as a half-cube resting on the base of the Great Pyramid and 36.6666 cubits high. The sum of its perimeter, as seen from above, would then be (36.6666 x 2) x 4 = 293.3333 cubits, and the sum of its vertical corners would be 36.6666 x 4 = 146.6666 cubits. That gives rise to another surprising result because 293.3333 added to 146.6666 gives 440 cubits exactly, the Great Pyramid base. One full side of the pyramid base exists on the edges of the notional Double Square, derived from a notional pyramid 264 cubits high, with a notional exit point at 132 cubits above base, originating on a shaft angle sloping at Tan 0.8000. The hidden geometry must then be correct for such a result could not possibly occur by chance.

The Position of the Gantenbrink Door

Slope on Diagonal

90°

132

Slope on Tan 1.2000

Chamber? 7.3333

127.6

A B

Slope on Shaft

6.1111

Enough of the hidden information has now been found to see how this affects the shaft, built in stone, where it terminates at the upper levels, the place where Rudolf Gantenbrink had discovered his mysterious door.

The slopes for pyramids 280 cubits high, 275 cubits high, and 264 cubits high, can now be shown at the level of the exit points for the Slope on Diagonal, and the Slope on Shaft. The Slope on Shaft must be 7.3333 cubits below the exit point for the Slope on Diagonal because it is already that distance below at the chamber position, and the two diagonals are parallel. A triangle therefore exists at the exit position, 7.3333 cubits high and 7.3333 / Tan 1.2000 = 6.1111 cubits long. The shaft in stone will pass through its corner (on its upper surfaces) and it might then be expected to terminate on the face of the slope for a pyramid 264 cubits high sloping at Tan 1.2000.

If the Gantenbrink Door does occur on this triangle it will probably do so on the slope for a pyramid 264 cubits high, and in doing so it would then create the triangles A and B. Triangle A would carry a slope of Tan 1.2000 and triangle B would carry a slope of Tan 0.8000, both on a common base 6.1111 cubits long. The height of these triangles will then make it possible to determine the level of the Gantenbrink Door.

2.4444 x Tan 1.2000 = 2.9333 cubits

3.6666 x Tan 0.8000 = 2.9333 cubits

The heights on triangles A and B are the same at 2.9333 cubits when on a common base of 2.4444 + 3.6666 = 6.1111 cubits. The exit point is at 7.3333 –2.9333 = 4.4 cubits below 132 cubits above base, the exit point for the Slope on Diagonal, and that puts the exit point of the Gantenbrink Door at 132 – 4.4 = 127.6 cubits above the base of the Great Pyramid. That is a very important level and it has never before been found.

Once again the factor of 10 appears, because the exit point is 4.4 cubits below the level of 132 cubits, and 10 times smaller than the datum at 44 cubits above base, both points of which are attached to the shaft on its upper surfaces. This height will also be found on the north slopes.

The length of the shaft

Now that the exit point of the South Shaft of the Queen’s Chamber has been found, its length can also be found for it will be the hypotenuse of a right-angled triangle. The triangle will be 127.6 – 44 = 83.6 cubits high but the base is rather more difficult to find. A pyramid above the level of 127.6 cubits will be 264 – 127.6 = 136.4 cubits high and its half base will be 136.4 / Tan 1.2000 = 113.6666 cubits. From that must be deducted 9.1666 cubits for the change of shaft direction at the chamber level on pyramid centre, giving a base distance of 113.6666 – 9.1666 = 104.5 cubits, the length of the triangle. The length of shaft on the hypotenuse of the triangle will then be the square root of 83.6 squared + 104.5 squared = 133.82 cubits = 70.08 metres.

There is a comparison between the distance as found on the pyramid geometry and the distance travelled by Gantenbrink’s robot when it had found the door high up in the shaft. According to Bauval and Gilbert, the machine had travelled some 65 metres before coming to a stop in front of the door. That would be just about right for an actual shaft length of 70 metres, less say 3 metres between the camera and the door, and say another 2 metres between the chamber and the change of shaft direction to give a distance very close to 65 metres, indicating that the door is located within the shaft where the geometry says that it should be. The geometry also shows that there was enough stone beyond the exit point at 127.6 cubits above base for a small, unknown, chamber to exist, that was around 2 cubits high.

The Original Pyramid Tip

A section through the summit of the Great Pyramid can now be shown as once it might have looked with the tip of a pyramid 264 cubits high for the South Shaft of the Queen’s Chamber, and a pyramid 275 cubits high for the North Shaft of the King’s Chamber. The mast is shown as it had been erected in the 19th century and as it exists today. Petrie had been to the summit and taken levels on the existing stones in 1880, and he said that the NE corner was 5407.9 inches above base and the SW corner was 5409.2 inches above base. The average height was 450.71 feet or 262.31 cubits. The height, when seen in cubits, was revealing

The Original Tip Pyramids.

280

Crowning Pyramid

275

Tip Pyramid

Mast

Existing stones 262

264

The two levels found on the existing summit stones are just over 262 cubits above base. Had Petrie given any existing levels at the summit that were above 264 cubits, above pyramid base, then the hidden geometry might well have been invalidated. That says that there was probably a truncated Tip Pyramid 11 cubits high, and sloping at Tan 1.2727, at the first level, followed by a second smaller Crowning Pyramid 5 cubit high at the second level, also sloping at Tan 1.2727. The two pyramids, when combined, gave a height of 16 cubits and there were 16 layers, 16.5 cubits high, on a pyramid 264 cubits high. Each of those layers would then have been 10 times smaller than 165 cubits, and that was the length of side for the Optimum Square within the pyramid 264 cubits high. The Optimum Square had related with the original stones at the pyramid tip.

The Transit of Venus in Egypt

This was the part of the pyramid that had inspired the Transit of Venus Party, mentioned by Petrie, to erect a mast in 1874 to restore the original 280 cubits above base, and with it, the tip stones that had existed before the summit was vandalised. They were the English astronomers who had been sent to the Great Pyramid by Sir George Airy, Astronomer Royal at the Greenwich Observatory, led by William de Wiveleslie Abney, an engineer and photographer who might have invented the Abney Level, an instrument for measuring heights. He had used a map that showed that the astronomers had drawn a line of longitude and latitude across the pyramid tip. They had also known the meaning behind 280 cubits, for the height to the top of the mast was 5776 inches, according to Petrie, and that converts to 5776 / 12 / 1.718181 = 280.14 cubits, which was very close indeed to the correct height if they were not aware of the original design height of the Great Pyramid. Not only had these astronomers and mathematicians known about 280 cubits, they had recognised it as being important enough to record with their mast, itself a difficult operation that would have involved some advance planning.

THE NORTH SHAFT, QUEENS CHAMBER

The first thing to notice from the survey of the north shaft of the Queen’s Chamber is that its outlet point is directly opposite the south shaft by line and level so that the west reveal of the north shaft is opposite the east reveal of the south shaft. By this means, a north-south centreline across the chamber is created and because the chamber is 11 cubits long, it is also divided into two distances of 5.5 cubits each side. Since the north shaft is opposite its partner, there is a probability that it was integrated in the same way as the upper shafts belonging to the King’s Chamber were integrated, and they would therefore use a common pyramid square.

Section through the Queen’s Chamber

Slope on Diagonal

North Shaft

Datum 44

5 4.20 ? 6.80

The square for south shaft was sides of 220 cubits at a level of 132 cubits for a pyramid 264 cubits high, and the same logic would say that the north shaft was also carrying a square of 220 cubits on its sides for a pyramid height as yet unknown. There had already been a clue as to what that pyramid height might be because the number 242 appeared as the length of side for the common pyramid square for pyramids 280 and 275 cubits high. Looking at this in respect of the earlier 264 cubits, it seemed to be correct and it might then simply be a matter of drawing it out and seeing if it made sense.

The Pyramids for the Queen’s Chamber

264 242

132 121

Shafts

99

110 110

The pyramid square would then be of sides 220 cubits and if the missing pyramid was 242 cubits high, and the shaft exit point was at half that level, it would be 121 cubits above base. The difference in shaft exit levels would then be 11 cubits, and that would certainly make sense. The square would extend below the base by 220 – 121 = 99 cubits and the pyramid base angles would be 242 / 220 = Tan 1.1000.

The First Slope on the Diagonal

242

121

Datum 44

62.85714

110 110

The line of shaft was probably the hypotenuse of a right-angled triangle coming off the common datum level of 44 cubits above base. That would give a triangle 121 – 44 = 77 cubits high and its base would be half of half the pyramid half-base, or 121 / Tan 1.1000 = 110 cubits. That would form the Slope on Diagonal because the inclination begins at pyramid centre whereas the shaft that was built in stone begins at about 9.20 cubits north of pyramid centre. The inclination of the Slope of the Diagonal would then be at 77 / 110 = Tan 0.7000, a very interesting number in sequence with Tan 0.8000 opposite.

The Slope on the Diagonal passes through the pyramid base on a distance of 44 / Tan 0.7000 = 62.85714 cubits south of centre and, improbable as it might seem, it divides into the pyramid base of 440 cubits by 7 exactly. What now exist is a height of 77 cubits, a slope of Tan 0.7000, and 7 divisions on the pyramid base. The number 7 is clearly being promoted on this shaft, but there was a problem.

The opposite shaft had begun its inclination at 55 cubits north of pyramid centre on base and by the same logic, the north shaft should also be able to achieve this by the same amount south of pyramid centre. It was not doing so and that was because there was a second Slope on the Diagonal.

The Second Slope on Diagonal

242

North

121

5

Datum 44

55 110 110

There exists a second Slope on the Diagonal and it begins its rise at 55 cubits south of pyramid centre on the pyramid base. This can be shown where the slope angle on the triangle is now 121 / 165 = Tan 0.7333 and that was the height in linear cubits x 10 of the triangle below the level of 44 cubits for the shaft opposite. The two shafts had connected through that number and the pyramid height of 242 cubits must be correct.

The hypotenuse of the triangle passes through the pyramid at centre, and at just below 44 cubits. Why was it just below? The triangle south of pyramid centre must be 55 x Tan 0.7333 = 40.3333 cubits high and it falls short of 44 cubits by 3.6666 cubits, and that is 10 times smaller than the height of the double square for the south shaft opposite.

A small triangle in the vicinity of the chamber now exists 3.6666 cubits high and 3.6666 / Tan 0.7333 = 5 cubits long. But the RM survey says that the chamber is 5.23 metres wide converting to 10 cubits, and half of that is the same 5 cubits. That explains why the chamber is 10 cubits wide and bisected east to west on pyramid centre, as found by Petrie.

The Slope on Shaft

242

121

9.20

Datum 44

110

The two Slopes on the Diagonal do not however give the Slope on Shaft that was built in stone and was offset to the north. That will carry another angle in tangent and it too will be numerically meaningful. Petrie had measured the shaft inclination at 37 degrees 28 minutes, and from the Standard Tables that would give an approximate tangential value of…

37 degrees 20 minutes 00 seconds = Tan 0.7627

00 degrees 08 minutes 00 seconds = Tan 0.0037

37 degrees 28 minutes 00 seconds = Tan 0.7664

The clue had rested in the triangle 7.3333 cubits high, with a base of 9.1666 cubits, for the south shaft opposite, where 9.1666 – 5 = 4.1666 cubits was the correct distance between the north wall of the chamber and the change of direction of the shaft. That had in turn created a triangle 110 – 9.1666 = 100.8333 cubits long instead of the 100.8 cubits long as had been found by measurement on the floor of the shaft when it should have been on the ceiling. The angle of slope for the north shaft was therefore 77 / 100.8333 = Tan 0 7636 recurring cubits measured on the upper shaft surfaces. From the Standard Tables that gives an angle of…

Tan 0.7627 = 37 degrees 20 minutes 00 seconds

Tan 0.0009 = 00 degrees 02 minutes 00 seconds

Tan 0.7636 = 37 degrees 22 minutes 00 seconds

The geometry says that the design angle for this shaft was 6 minutes of arc less than the angle that Petrie had measured. The survey information had said that shaft begins to rise at approximately 4.20 cubits from the north wall, or about 9.20 cubits from pyramid centre, at 44 cubits above base. The measured distance had been very slightly in error.

The Three Slopes through the Queen’s Chamber

First Slope on Diagonal

Second Slope on Diagonal

47.50

Converging

A

44.00

B

40.3333 7

C

5

37.00

Slope on Shaft

9.1666

The three slopes of inclination can be shown at the Queen’s Chamber in the form of triangles with their respective slopes on the hypotenuse.

Triangle A. Slopes at Tan 0.7000 on a base of 5 cubits and is 5 x Tan 0.7000 = 3.5 cubits high. That is also half of 7.

Triangle B. Slopes at Tan 0.7333 on a base of 5 cubits and is 5 x 7.3333 = 3.6666 cubits high. That is 10 times smaller than 36.6666 cubits.

Triangle C. Slopes at Tan 0.7636 recurring on a base of 9.1666 cubits, and it is 9.1666 x Tan 0.763636 = 7 cubits high.

It has been commonly stated that the Queen’s Chamber was intended as the burial place for the un-named pharaoh, but was then the subject of a change of mind in favour of the chamber above. The geometry shows that it could not possibly have been a mistake, and was instead, very cleverly planned, whether or not it was also a tomb for a pharaoh.

The three slope angles with their respective triangles can now be seen in relation with the cross section through the Queen’s Chamber, and what they show is that they had much to do with why the chamber is 10 cubits wide, and why the datum was 44 cubits above base.

The link between the slopes and the shafts

The next stage in the pyramid investigation can now centre on what had actually existed at the levels around the shaft exit points when the original casing stones were in place. Notional rectangles north and south are the means of doing this, the positions determined by the levels and distances relevant to the shaft exit points. The results demonstrate beyond doubt that the exit points were valid entities and were in existence when the pyramid was first built. They will show that the three invisible pyramids and the one built in stone, were the basis for the pyramid. The numbers, areas, and levels, will tell the story of the pyramid plan.

THE RECTANGLE ON THE SOUTH SLOPES

Tan 1.2727

154

A

King

145.2

B

99

113.6666

132

C

127.6

Queen

Slope on Diagonal

Tan 1.2500

A vertical rectangle has been created on the two shaft exit points and its width is determined by the difference between the two half base distances on pyramid squares. In the case of the upper exit point it was 99 cubits from pyramid centre. In the case of the lower exit point, it was the half base distance of a pyramid 264 – 127.6 = 136.4 cubits high, sloping at Tan 1.2000, or 136.4 / Tan 1.2000 = 113.6666 cubits. The rectangle was 113.6666 – 99 = 14.6666 cubits wide. The height was the difference between the exit point levels of 154 cubits and 127.6 cubits to give 26.4 cubits and that was significant because it was also 264 / 26.4 = 10 times smaller than the height of its enclosing pyramid. The height and the width will then be found to relate directly with the Great Pyramid base area at 440 x 440 cubits = 193600 square cubits.

Total Area of the Rectangle

The height = 154 – 127.6 = 26.4 cubits

The width = 113.6666 – 99 = 14.6666 cubits

The area = 26.4 x 14.6666 = 387.1999 square cubits

193600 / 387.1999 = 500 times

Area of Rectangles B + C

The height - 14.6666 x Tan 1.2000 = 17.6 cubits

The width = 14.6666 cubits

The area = 17.6 x 14.6666 = 258.1333 square cubits

193600 / 258.1333 = 750 times

Area of Rectangle B

The height = 145.2 – 132 = 13.2 cubits

The width = 14.6666 cubits

The area = 193.5999 square cubits

193600 / 193.5999 = 1000 times

Area of Rectangle A

The height = 26.4 – (14.6666 x Tan 1.2000) = 8.8 cubits

The width = 14.6666 cubits

The area = 8.8 x 14.66666 = 129.0666 square cubits

193600 / 129.0666 = 1500 times

Area of Rectangle C

The height = 132 – 127.6 = 4.4 cubits

The width = 14.6666 cubits

The area = 4.4 x 14.6666 = 64.5333 square cubits

193600 / 64.5333 = 3000 times

Rectangle A is twice area Rectangle C

Rectangle B is three times area Rectangle C

Rectangles A, B, C, are six times area C

Rectangles B and C are twice area Rectangle A

There is no doubt then that the rectangle on the two exit points on the two pyramid slopes to the south was a viable, geometrical, entity that was put there to relate mathematically with the pyramid base area. The results are all in whole numbers in half hundreds, hundreds, and thousands, as areas related with the Great Pyramid base area, and that could not be random.

THE RECTANGLE ON THE NORTH SLOPES

Tan 1.2500

151.25

A

King

110

B

137.5

133.1

99

C

121

Queen Tan 1.1000

The rectangle for the exit points on the north slopes is rather different in that it does not give areas related with the pyramid base area except that in the middle on area B. There would, however, be many more rectangles if all four slopes were shown and they would give further results.

Total Area of the Rectangle

The height = 151.25 – 121 = 30.25 cubits

The width = 110 – 99 = 11 cubits

The area = 30.25 x 11 = 332.75 square cubits

193600 / 332.75 = 581.8181

Area of Rectangle A

The height = 11 x Tan 1.2500 = 13.75 cubits

The lower level = 137.5 cubits above base

The lower level is in the same number as the height

The width = 110 – 99 = 11 cubits

The area = 13.75 x 11 = 151.25 square cubits

The upper level = 151.25 cubits above base

The upper level is in the same number as the area

193600 / 151.25 = 1280

Area of Rectangle C

The height = 11 x Tan 1.1000 = 12.1 cubits

The lower level = 121 cubits above base

The lower level is in the same number as the height

The width = 110 – 99 = 11 cubits

The area = 12.1 x 11 = 133.1 square cubits

The upper level = 133.1 cubits above base

The upper level is in the same number as the area

193600 / 133.1 = 1454.5454

Area of Rectangle B

The height = 137.5 – 133.1 = 4.4 cubits

The width = 110 – 99 = 11 cubits

The area = 4.4 x 11 = 48.4 square cubits

193600 / 48.4 = 4000 times

The Total Rectangle is 2.5 times Rectangle C

The Total Rectangle is 2.2 times Rectangle A

THE RECTANGLE ON FOUR PYRAMID SLOPES

Tan 1.2727

Tan 1.2500

154

151.25

145.2

99 140

137.5

133.1 132

110

121

Tan 1.2000

Tan 1.1000

Lower level = 121 cubits above base

Height = 11 x Tan 1.1000 = 12.1 cubits

Lower level is in the same number as the height

Upper level = 121 + 12.1 = 133.1 cubits above base

Area covered = 11 x 12.1 = 133.1 sq cubits

Upper level is in the same number as the area

Lower level = 132 cubits above base

Height = 11 x Tan 1.2000 = 13.2 cubits

Lower level is in the same number as the height

Upper level = 132 + 13.2 = 145.2 cubits above base

Area covered = 11 x 13.2 = 145.2 square cubits

Upper level is in the same number as the area

Lower level = 137.5 cubits above base

Height = 11 x Tan 1.2500 = 13.75 cubits

Lower level is in the same number as the height

Upper level = 151.25 cubits above base

Area covered = 11 x 13.75 = 151.25 square cubits

Upper level is in the same number as the area

Lower level = 140 cubits

Height = 11 x Tan 1.2727 = 14 cubits

Lower level is in the same number as the height

Upper level = 154 cubits above base

Area covered = 11 x 14 = 154 square cubits

Upper level is in the same number as the height

THE FOUR GREAT PYRAMIDS

SOUTH NORTH

Exit Level Line Angle Pyramid Height Base Angle Entity

154 Tan 1.0000 280 cubits Tan 1.2727 Osiris

151.25 Unknown 275 cubits Tan 1.2500 Unknown

132 Tan 0.8000 264 cubits Tan 1.2000 Isis

121 Tan 0.7000 242 cubits Tan 1.1000 Set

Exit Level Shaft Angle

154 Tan 1.0000 = 45 degrees 00 minutes 00 seconds

151.25 Varies throughout

127.6 Tan 0.8000 = 38 degrees 39 minutes 37 seconds

121 Tan 0.7636 = 37 degrees 22 minutes 00 seconds

CONCLUSION

The four shafts rising from the two chambers of the Great Pyramid were intended to act as diagonals on squares and rectangles, and the exit points were measured on their upper surfaces, at pre-planned levels. The levels were in turn dependant on the correct pyramid slopes angles measured in tangent, and the slope angles came about because they were derived from four pyramid heights on a common base, with three of them invisible and notional. The dimensions and levels were given in cubits, and metres had played no part in the pyramid construction.

The discovery that the shafts were geometrical and had carried a message will open the door to many more secrets hidden away inside this pyramid and the hunt has already begun. The reason why the chambers were of a certain size, and were located where they were inside the pyramid, can now be found, for they too were mathematically based, as were the five strange voids that were built directly above the King’s Chamber. The Queen’s Chamber had not been a mistake, as currently advocated, and the levels that had once existed on the floor can now be found.

The results coming from the rectangles on the pyramid slopes prove that they were valid and real, and could not have occurred by chance, or by any fantastically difficult manufacturing process, for the findings were based on a factual survey taken directly from the stones. That says that whoever it was who had conceived this plan, some 5000 years ago, and as current theory would state, they were technically advanced to a level far in excess of anything thought possible at the present time.

REFERENCES

Bauval R and Gilbert G. The Orion Mystery. Heinemann 1994

Bauval R and Hancock G. Keeper of Genesis. Heinemann 1996

Bauval R. Secret Chamber. Century (Random). 1999

Edwards I. E. S. The Pyramids of Egypt. Viking 1986

Lawton I and Ogilvie-Herald C. Giza The Truth. Element Books 1990

Maragioglio V and Rinaldi C A. L’Architectura delle Pirimidi Memphite. Rapallo. 1965

Petrie W M F. The Pyramids and Temples of Giza. London 1883

Tompkins P. The Secrets of the Great Pyramid. Penguin 1973

Wilson C. From Atlantis to the Sphinx. Virgin 1996

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