The Architect's Plan
The Direct Measurement of Sekeds
Khafre Pyramid Computations from Part 1 - (Refer to both Figures 5 and 6 in the text, and to the diagrams below.) :
-- As explained in the text, Line OD (the extension of line OP) is at a 27°16' angle with the horizontal. The Tangent of this angle is .5154, the Sine is .45813.
-- As also explained in the text, Line OE (the extension of line OL) is at a 21°22.6' angle with the horizontal. The Tangent of this angle is .39142. The Sine is .36450.
OP is 1.272727 units in length (by construction). Angle POY is 27°16'. Therefore, PY/OP = the Sine of 27°16' = .45813. And so, PY = .583074. This is the elevation of the shifted pyramid above its original base level in the diagram.
PY is the same as gi. The Tangent of the angle at K is gi/Ki = 1.3333, so Ki = .583074/1.3333 = .4373065. Oi = 1 - Ki = .562694. This is the same length as line gmm which sits directly above line Oi.
We know that OY = 1.131313 because it is half the side of the square having the same area as the circle of radius 1.2727. Therefore, iY = gP = .562694 + 1.131313 = 1.694007. PB is then 2 - 1.694007 = .305993. (Point B denotes the pyramid's north base edge. See next diagram).
In triangle PDB above, we've just found that PB = .305993. We know that angle DPB is 27°16' (Tangent .5154) and that angle PBD is 53°8' (Tangent 1.3333). Therefore, DJ/PJ = .5154; DJ/JB = 1.3333; and PJ + JB = .305993. The combined formula therefore becomes: .5154(.305993) = (1.3333+.5154)JB. JB is then .0853065 (=9.18 m), and PJ will be .2206865 (= 23.75m). PJ/DP = Cosine of 27°16' = .88888. Therefore, DP = .248273 (=26.72m). DJ = .5154xPJ = .1137418 (=12.24m). This is the elevation of the entrance to the Upper Descending Passage. (Note that half the base length of the Khafre Pyramid = 107.6 m.)
VX/OX also equals .5154. (Point O is at the pyramid's center. See earlier diagram). OX = 1.0476, and so VX = .539933. Since NX = PY, NV = NX - VX = PY - VX = .583074 - .53993 = NV = .043141(= 4.64m). Next, NV/PV = Sine 27°16' = .45813. PV = .043141/.45813 = .0941675 (=10.13m). The full length of DV is then DP + PV = DV = 26.7m + 10.13m = 36.85m. This is the length of the Upper Descending Passage.
Line OW = OY + YW, but YW is the same as PB. Therefore, OW = 1.1313+.305993 = 1.437306. Angle HOW is 21°22.6', and the Tangent of this angle is .39142. So, HW/OW = .39142 and so HW = 1.437306x.39142 = .56259. With BW equal to PY, BH then equals BW - HW = .583074 - .56259 = .020484. Angle BEH is also 21°22.6' as a result of parallel lines, therefore BH/BE = .39142, and so BE = .020484/.39142 = .0523325(= 5.63 meters). This is the distance from the pyramid's edge to the lower passageway's entrance.
Line OZ = 1.185185, and so LZ/OZ equals the Tangent of 21°22.6' = .39142. Therefore, LZ = .463905. Moving right along, we have RZ - LZ = RL. RZ is the same as PY, and so RL = .583074 - .463905 = .119169 (= 12.82m). This is the depth below grade of the bottom horizontal section of the Lower Passage.
RL/ER = .39142 (from Tangent of 21°22.6'), so ER = .119169 ÷ .39132 = .304453. The Sine of 21° 22.6' = .36450 = RL÷EL, so EL = .119169/.36450 = .32694 (= 35.18m). This is the length of the descending section of the Lower Entrance Passage.
The distance between Point L and Point T equals 1.185185 - 1.0476 = .137585 (= 14.8m). This is the horizontal section of the Lower Passage.
The Cosine of angle YOF = 1.131313/1.3333 = .84848, and so this angle is 31°57.2'. The Tangent of 31°57.2' = .62374. OM equals .8888 units, and so, SM/OM = SM/.88888 = .62374. Therefore, SM = .554435. UM is the same as PY, so US = UM - SM = .583074 - .554435 = .028639 (= 3.08m). This is the depth of the horizontal section leading to the main chamber. Note that this correlation was not presented in the text.
Point Q is located at the same elevation as Point S (That is, at SM's elevation) = .554435, and is located on the edge of the R =1 circle. Therefore, a line to Point Q from Point O will create an angle whose Sine is .554435/1, and so this angle is found to be 33°40.3'. The Tangent of 33°40.3 is .66621. To get Q's horizontal distance from Point O, Tan 33°40.3' = .554435/Horiz. Dist. = .66621. The horizontal distance computes to be .83222. A horizontal line from Point Q will meet line NX at a Point S1, with this point being the same distance as US is (=.028639) down from Point N. Line QS1 will be 1.0476 - .83222 = .2153775 units long, and NT = RL = .119169, so TS1 = NT - US = .119169 - .028639 = .09053. Therefore, using the Pythagorean Theorem, TQ can be found: (.09053)sqrd + (.2153775)sqrd = TQsqrd. TQ = .23363 (= 25.14m). This is the length of the ascending section of the Lower Passage. (Note that the tangent line TQ makes with the horizontal is then .09053/.2152775 = .42033 = tangent of 22°48').
Point Q is .83222 units horizontally from Point O. It is therefore .83222 -.4373065 = .3949132 units (= 42.49m) from Point O2 (refer to diagram on preceding page for the .4373065 figure). This is the length of CQ. Line QS1 was given above and is .2153775 (= 23.17m). Therefore, Point S1 (or Point V) is 42.49m+23.17m = 65.66m from Point G.
The length PN equals 1.131313 - 1.0476 = .083713. NV/PN = .5154, and so NV = .0431456 ( = 4.64m). This then places Point V at an elevation that is 4.64 - 3.08 = 1.56m lower than Point S1 (or Points Q and S). This raises the interesting issue of whether this difference in elevation between Points V and Q actually exists in the pyramid. Legon (1989, 31-32) has both of these points at the same depth of 3.55 meters (he labels them as Points K and Point G). M&R (1966, 54) say that Point V (which they label Point L) "is about 3 meters below the paving of the courtyard"). In order to conclusively resolve this relatively small disparity, and the few other minor disparities between the theory presented here and the as-built pyramid, it would be extremely helpful to have a definitive survey for the entire Khafre Pyramid - inside and out - one that would be of the same level of detail as Petrie's survey of the Khufu structure.
Lastly, Legon (1989,31) lists the bottom end (which he labels Point K) of the upper descending passage as being 42.77 meters from the pyramid's nortth base edge. This then implies that it is 107.6 - 42.77 = 64.83m from the pyramid's center. This is about .8 meters less that the 65.66m length for this distance (line VG) given by the diagram. (See above).
Khafre Pyramid Computations from Part 2 - (Refer to Figure 8 in the text):
Derivation of The Upper Descending Passage Lateral Location
The figure below is taken directly from Figure 8 in the text. It shows the square whose 1/2 side is .7857 units. This square is drawn symmetrically around the diagram's horizontal midline, Line LC The top of the square intersects the side of the 1.2727 pyramid (the inner pyramid) at Point A. AB is the perpendicular dropped to Line LC, meeting it at Point B.
Therefore, AB/BC = 1.2727. Since we know that AB = .7857, BC = .7857/1.2727 = .61735. With OC equal to 1 unit, OB = 1 - .61735 = .38265 units. The midpoint of OC will obviously be at .5 units, and so Point B is .5 - .38265 = .11735 units to the left of this midpoint. At the scale of the actual pyramid, 1 unit = 107.6 m, and so .11735 = 12.627 m (= 497 inches). This is then the proposed intended lateral location for the axis of the Upper Descending Passage.
Main Chamber's West Wall Location -
The diagram above will serve as a generic representation for the following Khafre Pyramid radial line intersection computations. OD = A , and DC = B.
The radial line to Intersection Point # 12 (this radial line also goes to #13 and #14) has a sine of .7857, and so the angle is 51°47'. The tangent of 51°47' is 1.27. The wall location is determined by the perpendicular dropped from the point where the radial line intersects with the side of the exterior (tan = 1.3333) pyramid. And so, H/A = 1.27; and H/B = 1.3333. A + B = 1, and so A = 1 - B. Therefore,
H = 1.27xA = 1.3333xB
1.27 (1 - B) = 1.333xB
1.27 = 2.60333xB
B = .4878367
A = .5121633 (x 107.6 m = 55.11 m)
H = .6504473 (x 107.6 m = 69.99 m)
Line OC equals 1 unit, and is half the full base length of the pyramid. It's midpoint (m) is at .5 units, and so Line OD (= A) extends (.5121633 - .5 =) .0121633 units beyond it. With 1 unit equaling 107.6 m at the pyramid's actual scale, Point D will be 1.31 m to the right of this .5 midpoint.
Main Chamber's North Wall Location -
The radial line to Intersection Point # 16 has a sine of .84848, and so the angle is = 58°2.8'. The tangent of 58°2.8' is 1.6032. The wall location is determined by the perpendicular dropped from the point where the radial line intersects with the side of the interior (tan = 1.2727) pyramid. And so, H/A = 1.6032; and H/B = 1.2727. A + B = 1, and so A = 1 - B. Therefore,
1.6032 = 2.8759xB
B = .55746
A = .44254 (x 107.6 m = 47.62 m)
With half of the pyramid's base length equaling 107.6m, the midpoint of the half-side will be 53.8 m north of the pyramid's center. 53.8m - 47.62m = 6.18m. Superimpose the midpoint of the half-side over the midpoint of the pyramid's full western side to derive the north wall location.
Lower Chamber's North/South Lateral Location - Intersection Point # 2.
The radial line to Intersection Point # 2 (which also goes to #3 and #4) has a cosine of .5154, and so the angle is 27°16'. The tangent of 27°16' is .5154. The Lower Chamber's North/South Lateral Location is determined by the perpendicular dropped from the point where the radial line intersects with the side of the interior (tan = 1.2727) pyramid. And so, H/A = .5154; and H/B = 1.2727. A + B = 1, and so A = 1 - B. Therefore,
5154 = 1.7881xB
B = .28824 (x107.6 m =31.01 m )
A = .71176 (x 107.6 = 76.58 m)
H = .36684 (x107.6 m = 39.47 m)
Lower Chamber's West Wall Location - Intersection Point # 16
The radial line to Intersection Point # 16 has a sine of .84848, and so the angle is = 58°2.8'. The tangent of 58°2.8' is 1.6032. The wall location is determined by the perpendicular dropped from the point where the radial line intersects with the side of the exterior (tan = 1.3333) pyramid. And so, H/A = 1.6032; and H/B = 1.3333. A + B = 1, and so A = 1 - B. Therefore,
1.6032 = 2.9365 B
B = .54595
A = .45405 (x 107.6 m = 48.86 m)
H = .727933 (x 107.6 m = 78.33 m)
The midpoint of the half-side is 107.6÷2 = 53.8 m. This minus 48.86 m = 4.94 m.
Lower Chamber's Floor Level - Intersection Points # 20 and # 16
Just as the Lower Chamber's north to south location was based on the perpendicular dropped from the side of the inner (tan = 1.2727) pyramid, so too are the two determinate derivations for the chamber's floor elevation.
The radial line to Intersection Point # 20 has a sine of .931217, and so the angle is = 68°37.3' . The tangent of 68°37.3' is 2.5546. The wall location is determined by the perpendicular dropped from the point where the radial line intersects with the side of the interior (tan = 1.2727) pyramid. And so, H/A = 2.5546; and H/B = 1.2727. A + B = 1, and so A = 1 - B. Therefore,
2.5546 = 3.88273 B
B =.66747
A = .33253 (x 107.6 m = 35.78)
H = .849486 (x 107.6 m = 91.4 m)
The radial line to Intersection Point # 16 has a sine of .84848, and so the angle is = 58°2.8'. The tangent of 58°2.8' is 1.6032. The wall location is determined by the perpendicular dropped from the point where the radial line intersects with the side of the interior (tan = 1.2727) pyramid. And so, H/A = 1.6032; and H/B = 1.2727. A + B = 1, and so A = 1 - B. Therefore,
1.6032 = 2.8759xB
B = .55746
A = .44254 (x 107.6 m = 47.62 m)
H = .70948 (x 107.6 m = 76.34 m)
Subtracting the "H" value for # 16 from # 20 = 91.4 m - 76.34 m = 15.06 m.
Upper Horizontal Passage Correlation Points (Refer to Figure 12 in the Text)
Intersection Point # 10 - The radial line to Intersection Point # 10 is determined by a base of 1 unit and a hypotenuse of 1.333 units, and so 1/1.333 = .75 = Cos 41°24.5' . The tangent of 41°24.5' is .88188. A perpendicular dropped from the point where this line crosses the outer (tan = 1.333) pyramid's side will lead to the following: H/A = .88188; and H/B = 1.3333. A + B = 1, and so A = 1 - B. Therefore,
.88188 (1 - B) = 1.333xB
88188 = 2.2152 B
B = .39810 (x107.6 m = 42.84 m)
A = .601897 (x107.6 m = 64.76 m)
H = .5308016 (x107.6 m = 57.11 m)
Intersection Point # 11 - The radial line to Intersection Point # 11 is determined by a height of 1 unit and a hypotenuse of 1.333 units, and so 1/1.333 = .75 = Sin 48°35.4' . The tangent of 48°35.4' is 1.1339. A perpendicular dropped from the point where this line crosses the outer (tan = 1.333) pyramid's side will lead to the following: H/A = 1.1339; and H/B = 1.3333. A + B = 1, and so A = 1 - B. Therefore,
1.1339(1 - B) = 1.333xB
1.1339 = 2.4672xB
B = .459583
A = .540417 (x107.6 m = 58.15 m)
H = .612779 (x107.6 m = 65.935 m)
Intersection Point # 12 - The perpendicular associated with this radial line's intersection with the outer (Tan = 1.333) pyramid's side was given above in regard to the Main Chamber's West Wall Location.
The Computation of the Two Possible Upper Entrance Elevations - (Refer to Figure 13 in the text)
With the end of the Descending Passage positioned at the perpendicular associated with the radial line to Intersection Point # 10 (which see above), this perpendicular will be 64.76 m north of the pyramid's center (and therefore it will be 42.84 m south of the pyramid's north edge.) If the depth of the end of this passage is considered to be at the Point S of Figure 11 in the text, it will be 3.082 m below the pyramid's base. These are the parameters needed to compute the two possible entrance locations at the pyramid's side.
Points P and E are the two ends of the Upper Descending Passage as seen in Figure 13 in the text. It is here given the diagram's predicted length of 36.85 m. Point P is located at the perpendicular associated with Intersection Point # 10, and so sits 42.84 m south of the pyramid's north base edge. It is also located the derivation diagram's 3.082 m below the pyramid's base level. Point T is the line of the pyramid's side extended down to the level of Point P.
So, we will say that line PF = A; line FT = B; and that line EF = H. We are given that line EP = 36.85 m, and that the tangent of the angle at T is 1.333. Therefore,
1) With Point T being 3.082 m below the pyramid's base level, it will then be 3.082 m / 1.333 = 2.31 m north of the pyramid's actual north base edge. Therefore,
PT = A + B. = 45.15 m. This means that A = 45.15m - B
As a result, A(sqrd) = 2038.5 - 90.3xB + B(sqrd) ------- I am dropping the 'meters squared' for now
2) H / B = 1.333, and so H = 1.333xB. Therefore H(sqrd) = 1.777xB(sqrd)
3) (36.85m)sqrd = 1,357.9 (meters squared)
4) From the diagram we can see that from the Pythagorean Theorem --- H(sqrd) + A(sqrd) = (36.85)sqrd
And so, plugging in the values for H(sqrd) and A(sqrd), we have:
36.85(sqrd) = 1.777xB(sqrd) + 2,038.5 - 90.3xB + B(sqrd) and this equals
1,357.9 = 2.777xB(sqrd) + 2,038.5 - 90.3xB; and so
2.777xB(sqrd) - 90.3xB + 680.6 = 0
Solving for B, the Quadratic Equation yields two results:
B1 = 12.11 m
B2 = 21.06 m
To find the height of H (which equals line EF) these results need to be multiplied by 1.333. And so,
H1 = 16.14 m
H2 = 28.07 m
Finally, to get the elevation of these entrance points relative to the pyramid's base level, 3.082 m needs to be backed out of the above results:
The elevation of E1 = 16.14 m - 3.082 m = 13.05 m
The elevation of E2 = 28.07 m - 3.082 m = 24.99 m
As stated, Legon found the end of the Upper Entrance Passage to be at a depth of 3.55 m. Recomputing the two possible Upper Entrance elevations elevations using Legon's depth of 3.55 m measurement will then yield:
The elevation of E1 = 17.577 m - 3.55 m = 14.03 m
The elevation of E2 = 26.99 m - 3.55 m = 23.44 m
It is therefore worth noting how sensitive the end results are to any minor change in any of the other parameters.
The Location of the Lower Ledge-Like Feature - Refer to Figures 8, 11, 12,and 13 in the text.
Point S in Figure 11 marks the intersection of the radial line to points # 2,3,4 with the S/2 = 1 unit line (which is the side of the square circumscribed around the 1 unit circle). The angle made by the radial line is 27°16', and so its tangent is .5154. Therefore, Point S sits .5154 units above the base of the pre-shifted pyramid. Since the base of the post-shifted pyramid is .583074 units above this pre-shifted base level, Point S will be .067674 units (x107.6m = 7.28m) below the base of the post-shifted pyramid. (See Computation section to Part 1 of this essay to see the derivation of the .583074 amount.)
The S/2 = 1 unit line is 1 unit north of the central axis of the pre-shifted pyramid. The shift deposits the post-shifted axis (.583074/1.3333 =) .437316 units to the north, and so the 1 unit line will be (1 - .437306 =) .56268 units from the post-shifted axis. At the pyramid's actual scale, .56268 units x 107.6m = 60.55 m. The north wall associated with the lower ledge measures to be 60.46m north of the pyramid's center, nearly spot on.
The Direct Measurement of Sekeds
Sekeds can be determined directly from the well drawn diagram. Needed will be an appropriately sized smooth flat surface upon which to work, a ruled straightedge which is accurately marked, a sharp stylus or writing tool, and a tool for scribing a circle. The author has found that by working carefully, and when starting with an initial radius equal in length to a royal cubit (524 mm), results within two minutes of arc of the ideal have been relatively easy to achieve. Note that one need not know ahead of time which subdivision of the finger (i.e., 1/9th, 1/16th. etc.) will work best for determining a length of run, although I have found the 1/9th unit to be most handy. The first measurement is to the nearest whole number of fingers, and this point is marked on the diagram. Trial and error with the variously sub-divided fingers on the cubit rod will suggest which division appears best for the measurement of the remaining part of a finger. The previous fingers are then simply multiplied by this sub-division amount and this total is added to the remaining amount that had been found for the tail-end.
Regarding the Khafre Pyramid:
With the Royal Cubit consisting of 7 palms of 4 fingers each, the full cubit contains 28 fingers. With each finger then further subdivided into 9ths, there are then 252 of these 9ths to a cubit. The seked of a 27°16' angle therefore can be found to have a horizontal run of 489 of these 1/9th units for each rise of 1 royal cubit (= 252 of the 1/9th units), and so equals 1 royal cubit, 6 palms, 2 1/3rd fingers (which is the same as 1 royal cubit, 6 and 7/12th palms).
The angle of 26°46' can be closely approximated from a finding of 252/500 and so its seked will equal 1 royal cubit, 1 palm and 8/9th palms.
The angle of 21°40' can be closely approximated from a finding of 252/634 and so its seked will equal 2 royal cubits, 3palms, and 2 + 1/2 fingers.