The Architect's Plan
The Computations
1) To Find the
.618 Division on the Pyramid's
Base
2) To
Find the Perpendicular from Line
OM
3) To
Find The Descending Passage Length in the Pre-Shift
Diagram
4) To
Find The Length of The (Upper) Descending Passage in the Post-Shift
Diagram
5) To
Find The Pre-shift Descending Passage Endpoint Below the Pyramid's
Base
6) The
Pre-Shift Ascending Passage
Location
7) The
Post-Shift Ascending Passage
Location
8)
Ascending/Descending Passage Intersection
Point
9) The
Pre-Shift Queen's Passage Level and
Length
10)
The Derivation of the .4472
Line
11)
The Anticipated King's Chamber
Location
12)
Computing the Entrance Passage's East/West
Location
13)
The Queen's Chamber Location South of the Central Axis in the
Pre-shift Diagram
14)
Computing the Queen's Shaft Angles in the Pre-Shift
Diagram
15)
Computing the Distance To the Queen's Chamber Shaft
"Doors"
16)
Computing the Queen's Chamber Shaft Distance to the Vesica
Piscis
17)
357.8" Marker Location in The Ascending
Passage
18)
Location of a Possible West Descending
Passage
1) To Find The .618 Division On The Pyramid's Base.
(This is to find the Point P of Figure 4 in the text - resulting from Line OL.The diagram below is an enlargement from Figure 4. Note that described here is a derivation resulting from the point where the .5 circle meets the pyramid's side. If the location of Point P is to be derived from the point where Line OL crosses the pyramid's side, then the Point P location computes to be at .617789. At the scale of the pyramid, this is within 2 inches of what is given below. - Further note that the 'pyramid side' derivation will follow from the finding in Figure 4 of the main text that the cosine of the angle that line OL makes with the horizontal is 1/1.2727)
Start with, H / Y = 1.2727 and X + Y = .5, so X = .5 - Y
Next, from the right triangle of sides X, .5, H we have: X(sqrd) + H(sqrd) = .5(sqrd)
Squaring H / Y = 1.2727, we can get H(sqrd) = 1.619765xY(sqrd)
And from squaring X = .5 - Y, we can get: X(sqrd) = .25 - Y - Y(sqrd)
And so, X(sqrd) + H(sqrd) = .5(sqrd) = .25 - Y - Y(sqrd) + 1.619765xY(sqrd) = .25
Therefore, 2.619765 x Y(sqrd) - Y = 0
So, 2.619765 x Y(sqrd) = Y, and so Y = 1/ 2.619765 = .3817136
Since .5 + X + Y = 1, then .5 + X = 1 - .3817136 = .6182864 = .5 + X.
2) Locating The Perpendicular Dropped Fom the Line OM / Pyramid Side Intersection:
Again from Figure 4 in the text: Point G below depicts where line OM crosses the pyramid's side. Point M stands 1 unit above the pyramid's base, and line OM is 1.2727 units long. (as explained in the text, the radius of the initial circle equals 1 unit and so the half-side of the circumscribed square will also be 1 unit. Line OM meets the circle whose radius equals the diameter of the outer circle, this diameter having been shown to equal 1.2727 units.)
The angle at O is therefore derived from the angle whose sine is 1/1.2727 = .7857 = 51°47'. The tangent of this angle is 1.2700
Therefore, A = OV, and B = NV. Point Z is the actual mid-point of Line ON, the half-side of the pyramid's base.
And so, H/A = 1.2700 and H/B = 1.2727. As a result, A = 1.0021259xB.
Since A + B = 4,534.4 inches, we have A = 4,534.4" - B, and so substituting This "A" into the formula above we get:
1.0021259xB + B = 4,534.4". Therefore, A = 2,269.6" and B = 2,264.8". The actual mid-point of Line ON will be at 4534.4"/2 = 2,267.2", only 2.4 inches from Point V.
3) To Find The Descending Passage Length in the Pre-Shift Diagram
In the Pre-Shift Diagram, the Descending Passage is at a 26°33.9' angle, and passes
through the .618 point.
Therefore, A + B = 1 - .618 = .382. (and so A = .382 - B)
H/A = .5, and H/B = 1.2727
Therefore, .5A = 1.2727B
Substituting in for A = .382 - B
.5 (.382 - B) = 1.2727B, so .191 = 1.7727B
B = .1077452 x 4534.4" = 488.56" = Distance of Entrance from the north edge.
A = .2742548 x 4534.4" = 1243.58"
H = .1371274 x 4534.4" = 621.8" = Elevation above the base of the Entrance Passage.
H/C = Sine of 26°33.9' = .4472 , so C = H/.4472 = 1390.43" = Length of Entrance Passage.
4) The Length of The (Upper) Descending Passage in the Post-Shift Diagram
Due to both components of the shift, the intersection point of the Descending Passage with the Pyramid's base is moved 96" + 37.8" = 133.8" to the south. 133.8" divided by by 4534.4" = .0295077.
Therefore, .618 - .0295077 = .5884923. And so, 1 - .58849 = .4115077 = A + B
Therefore, A = .4115077 - B
Again, H/A = .5, H = .5A and H/B = 1.2727, H = 1.2727B
And so, .5 (.4115077 - B) = 1.2727B
And then, .2057538 = 1.7727B
B = .116068 x 4534.4 = 526.3" = distance south from northern base edge of desc. passage entrance
A = .2954397 x 4534.4 = 1339.64"
H = .1477196 x 4534.4 = 669.82" = elevation of post-shift desc. passage entrance
H/C = sin 26°33.9' = .4472, and so C = .3303635 = 1497.8" = length of post-shift desc. passage
5) The Pre-shift Descending Passage Endpoint Below the Pyramid's Base
PH is a continuation of line "C" in #'s 3 and 4 above. Refer to Diagram 7 in the text. Z is the mid-point of the right-side radius, and P is the .618 location on this radius.
Therefore, P - Z = .618 - .5 = .118
Angle ZPT is 26°33.9', and so ZT/ZP = .4472
Therefore ZT = .118 x .4472 = .05277
PT = 2 x ZT = .10554
ZH = OZ = .5
TH(sqrd) + ZT(sqrd) = .5(sqrd),
so TH(sqrd) = .25 - .0027846 = .2472
Therefore, TH = .4972, and so PT + TH = PH = .10554 + .4972 = .60274 (= 2733") = the pre-shift length of the Desc. Passage below the pyramid's base
PV ÷ PH is the cosine 26°33.9' = .8944. From above, PH = .60274
And so, PV/.60274 = .8944, and therefore .60274 x .8944 = .53909 (= 2444.5")
Because the angle is 26°33.9', VH = PV/2 = .53909/2 = .269545 (= 1222.25") = The pre-shift diagram's depth of the descending passage.
OV = .618 - PV = .618 - .53909 = .07891 = 357.8". This is the pre-shift diagram's location for the end of the Descending Passage north of the pyramid's center.
6) The Pre-Shift Ascending Passage Location
7236 - .618 = .1056. This divided by 2 = .0528. This times 4534.5" = 239.4" Therefore, H/239.4" = sine of 26°33.9' = .5. And so, H = 119.7" This is the pre-shift elevation for the intersection of the Ascending Passage with the Descending Passage.
239.4/C = cosine of 26°33.9' = .8944 , and so C = 267.7" This is the pre-shift length of that portion of the Descending Passage between the pyramid's base and the Ascending Passage.
7) The Post-Shift Ascending Passage Location
In the Pre-Shift Diagram the Descending Passage intersects the pyramid's base at the .618 point. In the Post-Shift: from .618 subtract 96" + 37.8" = 134.06", ( 134.06" ÷ by 4534.4" = .029565) note that the Descending Passage does not undergo the 25.6" extra shift. Therefore, the shifted Descending Passage now intersects at .618 - .029565 = .58844 north of the pyramid's center.
The Ascending Passage: starts at .7236. To this then add 96" and subtract 37.8" - due to the two components of the shift. Next add back the 25.6" which results from the second shift (for a total add of 96"-37.8"+25.6" = 83.8" = .01848). And so, the beginning point of the Ascending Passage at the Pyramid's base level =.7236 + .01848 = .74208 (this x 4534.4" = 3365")
To determine all the lengths in the diagram directly above, we have:
.74208 - .58844 = .15364 = A + B, so A = .15364 - B
H/A = .5 = tangent 26°33.9' angle. The tangent of 26°14' = .49278, and this = H/B. So H = .49278xB = .5A. Substituting in for A from above, .5 (.15364 - B) = .49278B
Therefore, .07682 = .99278xB
B = .07738 (x 4534.4" = 350.9")
A = .07626 (x 4534.4" = 345.8")
H = .03813 (x 4534.4" = 172.9")
D ---- H/D = .4472, D = .085264 = 386.6") - from sine of 26°33.9'
U ---- H/U = .44203, U = .08626 = 391") - from sine of 26°14'
8) Ascending/Descending Passage Intersection Point: (Refer to the Diagram and findings in Demonstration # 7 directly above)
From: .74208 - B = .74208 -.07738 = .6647 = 3014". This is passageway intersection point's distance from pyramid center, and 1 - .6647 = .3353 = 1520.4" equals its distance from the north edge. Petrie gives this distance as 1517.8" (pg. 65 and chart on 95).
From The Descending Passage in the Post-Shift Diagram (see # 4 above), the sloping length of the Descending Passage as it extends above the pyramid's base = .3303635 = 1497.8". The part below the intersection with the Ascending Passage was just shown in # 7 to be 386.6", and so the distance from the Descending Passage's Entrance to this intersection = 1497.8" - 386.6" = 1111.2". This is exactly the same distance as given by Petrie (pp. 61, 62).
9) The Pre-Shift Queen's Passage Level and Length (the diagram here is an expansion from Figure 7 in the text)
Line PL is the base (Line DF) of the inverted interior triangle DFB in Figure 5 in the text, and so it sits .4472 above the horizontal diameter (see explanation # 10 below). Therefore, AP = 1 - .4472 = .5528.
Since angle PAT is 26°33.9', PT/AP = .4472. (See next derivation below) And so, PT = .247212. GK is the Ascending Passage, and so it is at the 26°33.9' angle. PC is the sloping roof of the Grand Gallery at this same angle. Angle KBC is a right angle, so angle BCA is 63°26.1' (from 90° - angle PAT) Therefore, angle PCT is 63°26.1' minus 26°33.9' = 36°52.2' The sine of this angle is .6. So, .6 = PT/PC and so PC = .247212/.6 = .41202. PB is then found from PB/PC = sine 26°33.9' , and so PB = PC x .4472 = .184255.
Therefore, AB = AP + PB = .5528 + .184255 = .737055.
Point P is located at .4472 in elevation above the pyramid's base. K is located at .3618. Therefore PK = CQ = BS = .4472 - .3618 = .0854 = BS.
So, to find the elevation of line SQ, which is the elevation of the Queen's Passage level, AS = AB + BS = .822455, and so the elevation equals
1 - .822455 = .177545. This x 4534.5" = 805.06"
The length of SQ = BC = AB/2 = .3685275. This x 4534.5" = 1671", and is the length of the initial diagram's Queen's Passage. This actually remains about the same after both shifts because the lowering of the angle of the Ascending Passage to 26°14' cuts this length by 23.5" while the 25.6" northward shift pushes the length back to 1673.2". (Note that BC/AB = tangent of 26°33.9' = .5)
10) The Derivation of the .4472 line (line DF's elevation above the pyramid's base in Figures 2 and 5 in the text, and of the .7236 length
Lines BD and AE intersect at the midpoint of the horizontal radius, a point we'll call Point N.
Line OB equals 1, ON = .5. Using the Pythagorean Theorem, Line BN computes to be 1.118034. Since Line DN = .5, Line BD = 1.618034.
Let us call the point of intersection of Line BA with Line DF Point P.
The angle at OBN = 26°33.9', so BP/BD = cosine 26°33.9' = .8944
Therefore, BP = 1.618034 x .8944 = 1.4472. Since OB = 1, OP = .4472.
Furthermore, if BP = 1.4472, then because angle OBN = 26°33.9', the length of DP will be 1.4472 ÷ 2 = .7236.
11) The Anticipated King's Chamber Location
In the Pre-shift Diagram, the King's Chamber elevation is .3618 units (i.e., 1/2 of .7236).
Had the King's Chamber been designed to sit against the edge of the .5 circle, this distance from the center line would compute by finding X, and then subtracting X from .5
From Pythagorean Theorem, X = .3451086 .5 - .3451086 = .1549 (x4534.4" )= 702.38"
12) Computing the Entrance Passage's East/West Location - Refer to Figure 12 in the text.
A line from the diagram's center (Point O) is drawn to the point where the top horizontal line of the Pi rectangle intersects the pyramid's side. This point is then .7236 units above the pyramid's base.
Therefore, .7236/Y = 1.2727, and so Y = .568555.
Therefore, X = 1 - .568555 = .431445.
Point E is then .5 - .431445 = .068555 units to the left of the mid-point of the half side (Point Z - as shown in Fig. 12 of the text, but not shown here). .068555 x4534.4" = 310.86"
13) The Queen's Chamber Location South of the Central Axis in the Pre-shift Diagram - Refer to Figure 7 in the text.
This computation is similar to that just done in Demonstration # 11 for the King's Chamber. The Queen's Chamber pre-shift elevation was earlier shown in Computation 9 to be at .177545 units above the pyramid's base.
From the Pythagorean Theorem, (.177545)sqrd + x sqrd = .5 sqrd
And so, X = .467416
Therefore, .5 - X = .5 - .467416 = .032584. This times 4534.4" = 147.75"
14) Computing the Queen's Shaft Angles in the Pre-Shift Diagram
At the Queen's Passage elevation of .177545 units, the height therefore gained by each shaft will equal .7236 - .177545 = .546055. (See Figure 7 in the text). Note that this derivation occurs prior to the implementation of the drop in elevation caused by the Step in the Queen's Passage - a drop of 20 inches. It would appear that this drop did not effect the dynamics f the derivation.)
The base of the right triangle created by the south shaft equals .7236 - .0156139 = .707986. (This being because the exterior of the chamber's south wall sits 147.8" south of the center axis - as shown above in Computation 13. The wall is 77 inches thick, and so the interior side of the south wall sits 70.8" south of the center axis. 70.8"/4534.4" = .0156139)
Therefore, the tangent of the south shaft's angle will be = .546055/.707986 = .77128. This is the tangent of the angle 37°38.5'.
See the note at the end of Demonstration # 15 below regarding the alternative instance in which the design of the Queen's Chamber shafts was based on the initial decision to simply place the east/west mid-line Queen's Chamber directly on the east/west axis of the whole pyramid.(as a result of the decision to align it with the perpendicular dropped from line OM as has been explained in the text in conjunction with Diagram 12).
Since the interior of the south wall is 70.8" south of the center axis, and since the Queen's Chamber is 206" north to south, the interior of the north wall must be 206" - 70.8" = 135.2" north of the center axis. And so, 135.2"/4534.4" = .0298165
Therefore, the base of the right triangle formed by the north shaft will be .7236 - .0298165 = .6937835.
Therefore, the tangent of the north shaft's angle will be = .546055/.69378 = .78707. This is the tangent of the angle 38°12.3'.
--- However, If the
Queen's Passage elevation in an earlier design phase of the project
had been intended to be in line with the entrance to the descending
passage, and so be at .13713 units elevation (this being the
pre-shift elevation of the entrance to the Descending Passage - see
above Demonstration # 3 "To Find The Descending Passage Length in the
Pre-Shift Diagram"),
the height gained by the shafts would have then equaled .7236 -
.13713 = .58647.
Therefore, if the south shaft angle is refigured based on this
height, the tangent becomes .58647/.70799 = .82836. This is the
tangent of the angle 39°38.2' . Gantenbrink found the angle of
the south shaft to be 39°36.5', and so it is possible that this
angle is a vestige of an earlier design rationale.
15) Computing the Distance To the Queen's Chamber Shaft "Doors" - Refer to Figures 15 and 16 in the text, and also see Demonstration #14 above for computation of the shaft angles.
The text explains that the Pre-shift Diagram's angle of the north shaft works out to be determined from .54605/.69378 = .78707 = the tangent of 38°12.3', and that the diagram's south shaft works out to be = .54605/.70799 = .77127 = the tangent of 37°38.5'.
We start with FE = FO = .5 (Both are radii of the left-hand side .5 circle. Also, as shown in #14 above, the distance of B (the inside surface of the chamber's south wall) from the pyramid's central axis is 70.8".
AB therefore equals FO - 70.8" = .5 - .0156135 = .4843865
CA/AB = Tan 37°38.5' = .77127, so CA = .3735927
AF = Queen's Passage elevation = .177545 (see #9 above)
CF = CA + AF, therefore CF = .3735927 + .177545 = .5511377
CE = CF - FE , so CE = .5511377 - .5 = .0511377
CA/BC = Sin 37°38.5' = .61073, therefore BC = .6117149
Line EV is drawn tangent to the circle, meeting Line BC at Point V - a point which is not the same as Point P.
CE/CV = Sin 37°38.5' = .61073, therefore CV = .083732
CE/EV = Tan 37°38.5' = .77127, so EV = .0663032
FEsqrd + EVsqrd = FVsqrd so, FV = .5043769
VT = FV - .5 = .0043769
Tan of angle FVE = FE/EV = 7.5411, so angle = 82°26.2'
Angle TVP is therefore 180° minus 37°38.5' and minus the 82°26.2', and so TVP = 59°55.3'
Cosine of angle TVP is therefore .50118 and this = VT/ VP
And so, VP = .0043769/.50118 = VP = .0087331
BP = BC - CV - VP , and so BP = .6117149 - .083732 - .0087331 = .5192498 = 2354.5 inches
Subtract out the 98" for the diagonal through the chamber wall (refer to Figure 15 in the text) and you have 2354.5 - 98 = 2256.5" = 188' = 57.3 meters = distance to the south shaft "door" (from the exterior of the south wall). The north shaft is calculated exactly the same way with plugging in its angle and location parameters - and gives a distance of about 56.5 meters. Note that this is a 'line of sight' calculation, and that the actual north shaft does not travel a straight route, but instead angles around the Grand Gallery. Also note that these findings are based on Gantenbrink's findings for the thickness of the Queens Chamber's two walls.
(The average cosine value for the angles of the two shafts is about .78. Since the wall thickness is about 77 inches, the diagonal length of the line of each shaft continued through the wall will be about 77 inches divided by .78 = 98 inches.)
A Further Note: As is mentioned above in Demonstration # 14, it is also possible that the decision was made very early in the pyramid's design process to have the east/west axis of the Queen's Chamber lie on the east/west axis of the entire pyramid. If this is in fact the case, then the design angle for both the north and south shafts would have wound up being 37°55.3'
To derive this angle: In triangle HGB above, side HG will remain .7236-.177545 = .546055. However, side GB will now be .7236 - 103" (103" being half the north/south length of the Queen's Chamber). 103" divided by 4534.4" (i.e., the half-length of the entire pyramid) yields .0227152. Therefore, GB = .7236 - .0227152 = .7008848. Therefore, HG/GB = .546055/.7008848 = .7790937, and this is the tangent of 37°55.3'.
Using this angle to now compute the distance to the "door" will yield a distance from the exterior of the south wall of 2245.5" = 57.036 meters.
A Further Further Note: All of the above has been derived using the elevation of the Queen's Passage level (as derived earlier in Section # 9). The Queen's Chamber as constructed, however, due to the lowering caused by the Queen's Passage Step sits a royal cubit (.524 m) lower. Therefore, if we accept that the decision was made to align the Queen's Chamber on the central axis of the pyramid (see note above), then the actual length of the shaft will be the 57.036m given above plus the .524m due to the lowering caused by the Step, yielding a total length of 57.56 meters. This exactly matches the 57.5 meters found by Gantenbrink for the shaft length from the exterior of the Chamber's south wall to the first "door" reached in the shaft.
16) Computing the Shaft Distance to the Vesica Piscis
Line NM is the same line as BD in Figure 2 in the text. It is therefore equal to 1.618034 units. Line BN we will call "A", Line TB we will call "P", and Line MT we will call "Z".
1) Therefore, (A+P)(sqrd) + Z(sqrd) = 1.618034(sqrd) in triangle NTM
2) We will call Line BM = X, and so, X(sqrd) = P(sqrd) + Z(sqrd), so Z(sqrd) = X(sqrd) - P(sqrd)
3) P/X = sin 37°38.5' = .61073, so P = .61073X (See Computation 14 above for the angle derivation for the south shaft).
4) Since ON = 1, A = BN = 1 + height of Queen's Chamber = 1.177545
From all of the above we get : A(sqrd) + 2AP + P(sqrd) + Z(sqrd) = 1.618034(sqrd)
And so, 1.38655 + 2.355(.61073X) + P(sqrd) + X(sqrd) - P(sqrd) = 2.168034
Therefore, X(sqrd) + 1.438269X - 1.231484 = 0
And solving this with the Quadratic equation yields X = .6032264 This times 4534.5" = 2,735.3 inches
Lastly, from this back out the 98 inches for the diagonal through chamber wall.
Therefore, the Shaft length to the Vesica Piscis = 2,637.3 inches = 219.8' = 66.987 meters
Note that this is for the south shaft, and is the measured from the exterior of the chamber's south wall. The corresponding 'line of sight' distance to the vesica piscis for the north shaft will be about 66.5 m. (As stated above at the end of Demonstration # 15, the actual north shaft does not travel a straight route, but instead angles around the Grand Gallery.)
A Further Note: As is mentioned in the text, it is also possible that the decision was made very early in the pyramid's design process to have the east/west axis of the Queen's Chamber lie on the east/west axis of the entire pyramid. If this is in fact the case, then the design angle for both the north and south shafts would have wound up being 37°55.3' (See further explanation at end of Section 15 above). If this is indeed the case, then recomputing the line of sight distance to the vesica piscis using the 37°55.3' angle will yield a distance of 2628.5" = 66.76 meters for both the north and south shafts. Again, this is a line of sight distance and as measured from the exterior side of each wall.
A Further Further Note: As detailed in the "Further Further Note" at the end of the previous section, if one takes into account the lowering of the Queen's Chamber elevation due to the Passage Step, then the derived shaft length from the Chamber's south wall to the "Gantenbrink Door" will be 57.56 meters. Assuming that the Queen's Chamber was aligned on the pyramid's central axis (as discussed in "A Further Note" in the previous section), then the proposed shaft length from the "Gantenbrink Door" to the Vesica Piscis location will be 66.76m minus 57.56m = 9.2 meters. As a result, the distance from the second "door" (discovered in the National Geographic robotic exploration of 2002) will be roughly 1/3 to 1/2 a meter or so less than this, depending on the actual distance between the two "doors" (reported to be about 7 inches) and the thicknesses of each "door".
17) 357.8" Marker Location in Ascending Passage
The derivation for the 357.8" location was given in Demonstration #5. The step in the Queen's Passage appears to be a marker for this meridian in the text. The following discusses where this same meridian would cross the Ascending Passage.
In the Pre-shift Diagram, the Ascending Passage has its point of beginning at the .7236 point. We are here trying to learn where along the Ascending Passage it will be a lateral 357.8 inches from the pyramid's central axis.
.7236 x 4534.4" = 3281". Therefore, 3281" - 357.8" = 2923.3". The angle of the Ascending Passage in the Pre-shift Diagram is 26°33.9', and the cosine of this angle is .8944. This gives a passage length of 3268.4". If the passage angle is now lowered to 26°14' (as per the explanation in the text), the 3268.4" passage length will then extend to 349.3" north of the pyramid's center. (The cosine of 26°14' being .897)
Next, if we implement the 37.8" southward component of the first shift, and then the 25.6" northward second shift. This marker will then move to 337.1" north of the pyramid's center. This means that the marker will be 337.1"÷ .897 = 375.7" down the Ascending Passage from the north face of the Great Step (i.e., the pyramid's center).
Using Demonstration # 7, we can also compute that the marker is 3237.4" from the point at which the Ascending Passage meets the Descending Passage. Petrie (p. 65) says that the Queen's Passage intersects the Ascending Passage 1546.8" from this Ascending Passage beginning point. Therefore, the marker would, under this scenario, be about 2984.4" - 1546.8" = 1437.6" further up the Ascending Passage from the Queen's Passage entrance.
18) Location of a Possible West Descending Passage in The Khufu Pyramid
Following the same protocol as given in the Egyptian Pyramid Design paper, construct a pyramid with the same 14/22 proportions as are in the Khufu pyramid. Draw a circle centered on the midpoint of the pyramid's base using the height as a radius, and a circle using 1/2 the base as a radius.
Enter the area and circumference squares for each circle. Enter the lines from the diagram's center to the intersection points of these circles and squares.The lowest of these intersections will be at a 27°16' angle (cosine = 1.1313 / 1.2727 = .8888), the third of these will be at a 51°47' angle (sine = 1 / 1.2727 = .7857). Point W is the point at which this third line intersects with the pyramid's side.
The elevation of Point W (i.e., line WZ below) computes to be .635674 (See Demonstration #2 above for the method used for this calculation).
WT is at an angle of 27°16' - it is the lowest of the intersection lines slid upwards to Point W.
WZ is .635674 units. Since the angle at B is 51°50', and the angle of the third line intersection is 51°47', OZ will almost equal ZB. (Specifically, ZB = .49947 and so OZ = .50053).
OY is the distance that the interior side of the King's Chamber west wall extends west of the pyramid's center axis. Petrie says this distance is 107.7" (p. 83) - and so, 107.7/4534.4 = .0237517 units
Therefore, YZ = .50053 - .0237517 = .4767783, and this is then the same as TU (the other side of the rectangle)
The length of WU is found from WU/TU = Tan 27°16' = .5154, and so WU = .2457315
WZ - WU = .6356743 - .2457315 = UZ = .3899428 x 4534.4" = 1768". This is the elevation above the pyramid's base at which the descending line of WT - the line of the proposed West Descending Passage - intersects with the interior side of the King's Chamber west wall.
1768" is 79" higher than 1689" - the elevation level of the base of the King's Chamber's walls, and is therefore 74" higher than the elevation level of the floor in the King's Chamber. (It is possible, if not likely, that such a descending passage as described here would have been built at a slightly steeper angle so as to allow it to meet the King's Chamber west wall at the chamber's floor level.)
WT is found from TU/WT = .88888, and so WT = .536376 = 2,432"
The full length of the 27°16' intersection line from the center of the pyramid to the pyramid's side is .80074 (= 3630.9"). Shifted to Point W, this line could theoretically then continue descending downwards another 3630.9" - 2432" = 1199" beyond the King's Chamber wall in the diagram.
Petrie (p.38) implies that the width of the pyramid's casing (at the pyramid's base) is 33". Point W is at a .635674 = 2882.4" elevation, but this is at the theoretical casing exterior. Depending on the specific height and width of the original casing at this location, the elevation at the core masonry would be somewhat lower. Given the proposed 27°16' angle of the passage, and a 33' width for the casing block, the exit of the proposed passage at the core masonry would be anywhere from about 5" to about 45" lower than that given above.
The Derivation Diagram of the Khufu Pyramid
Copyright ©2004 L. Cooper (rc@atara.net) All Rights Reserved.