The Architect's Plan
The Direct Measurement of Sekeds
Note that the below derivations for the entrance passage elevation and length refer to these positions in the original design, prior to the secondary implementation of a steeper exterior angle for the pyramid's side. See the analysis of the Perring and Dorner surveys for the changes in entrance elevation and passage length due to the steepened angle. See the essay on Apparent Triples for an explanation of the change in exterior angle from 43°22' to about 44°42'.
Entrance Passage Elevation - (Refer to Figure 7 in the text).
The triangle above is a not-to-scale representation giving the diagrametric for computing the Entrance Passage's height and length.
1) In the case of Point K in Figure 7 in the text: The length of line OE is .94444 units, and line OP is .88888 units.The Cosine of angle POE is therefore .88888/.94444 = .94117. This is the Cosine of an angle of 19°45'. The Tangent of 19°45' is .35904.
The angle of the pyramid's exterior is given as 43°21.8', and so the tangent there is .94444. This leads to the two formulae: H/Z = .35904, and H/Y = .94444
Combining these: .35904Z = .94444Y. In addition, Z + Y = 1 unit. And so, Z = 1 - Y. Substituting this in, .35904(1-Y) = .94444Y which leads to: .35904 = .94444Y + .35904Y = 1.30348 Y
Y = .35904/1.30348 = .275447 (x 359.625 feet = 99.06')
Z = .724553 (x 359.625 feet = 260.57')
H = .260144 (x 359.625 feet = 93.55')
Since the full base length of the pyramid can be taken as being very near 719.25 feet, the length of OB in the diagram (that is, 1 unit) will equal 359.625 feet at the pyramid's scale. H will therefore be 93.55 feet. This is the height of entrance passage. (A slight digression here: Only the Perring and Dorner surveys venture credible readings for the Red Pyramid's base length. Perring found 719.42 feet, Dorner found 219.08 meteres (= 718.76 feet). I have chosen the 719.25' figure as something of a compromise.
2) In the case of Point L in Figure 7 in the text, the Cosine of the line OD is .8888/1, and so the angle here is 27°16' . The Tangent of 27°16' is .5154, and so following the same procedure as used above, the "Y", "Z", and "H" values will follow from: .5154 = 1.45984 Y1 Z1 = 1 - Y1
Y1 = .3530514
Z1 = .6469486
H1 = .3334373
The length of line OL of Figure 7 ("C" in the triangle above) can be found either via the Pythagorean Theorem or by using the Cosine, where Z1/C = .8888. With Z1 being the above .6469486, OL will then be .7278244 = 261.74 feet
Entrance Passage Length and Location of the 1st Chamber's South Wall - (refer to Figure 8 in the text):
3) With the shift, line OL moves down to Point K to become line KN - which descends at the 27°16' angle of line OL. The triangle relationship shown on the previous page can now be used to compute the length of line KG.
Since H is known (it is 93.55 feet - see #1 above), H/W = .5154, and so W = 181.5 feet.
W/KG = .8888, and so KG = 204.2 feet.
Therefore, GN = OL - KG = KN - KG = 261.74' - 204.2' = 57.54 feet.
TG/GN = Cos 27°16' = .8888, and so TG = 51.15 feet (and TN = this x .5154 = 26.36 feet).
Knowing both "W" and "Y" (that is, length Y from the 19°45' diagram),
GB = W + Y = 181,5 + 99.06 = 280.6 feet.
Therefore, Point G is 359.625' - 280.56' = 79.06 feet from the pyramid's center.
What follows are the derivations referred to in the text, but which have not been derived there. The diagram below will serve as a template. "H" is the perpendicular dropped from the point at which the radial line crosses the pyramid's side.
The Intersection Point Perpendiculars - (Refer to Figure 9 in the text).
Intersection Point # 4 creates a right angle with the pyramid's base that has a base of .8395 units and a hypotenuse of 1 unit, and so it has a cosine of .8395. Therefore, the angle is 32°54.8', and the tangent of this angle is .64726. The tangent of 43°21.8', the pyramid's exterior angle, is .9444. Half the length of the pyramid's side is 359.625 feet. Half the length of the half-side is 179.8 feet.
And so H/B = .94444; H/A = .64726; and A + B = 1 and so A = 1 - B
.64726xA = .94444xB, and so .64726 (1-B) = .94444xB and
.64726 = .64726xB + .94444xB = 1.5917xB
Therefore, B = .64726 ÷ 1.5917 = .406647
A = .59335 (x 359.625 feet = 213.385 feet)
H = .35274 (x 359.625 feet = 138.115 feet)
Intersection Point # 6 creates a right triangle with a base of .7857 units and a hypotenuse of 1 unit, and so it has a cosine of .7857. The angle is therefore 38°13', the tangent of which is .78739. Plugging this into the same formula as above yields:
.78739 = 1.73183 B
B = .454656
A = .545343 (x 359.625 feet = 195.43 feet)
H = .429398 (x 359.625 feet = 154.42 feet)
Intersection Point # 10 has a sine of .74205/.9444 = .7857 (Intersection Point
# 11 has a sine of .7857/1, and so results from the same angle, which is 51°47'. The tangent of 51°47' is 1.27. Therefore,
1.27 = 2.21444 B
B = .57351
A = .42649 (x 359.625 feet = 153.376 feet)
H = .5416423 (x 359.625 feet = 194.79 feet)
Intersection Point # 14 has a sine of .8395 ÷ .94444 = .88888. Intersection Point # 15 has a sine of .88888÷ 1, and so results from the same angle. .8888 is the sine of 62°44', and the tangent of 62°44' is 1.9402. Therefore,
1.9402 = 2.88464 B
B = .6726
A = .3274 (x 359.625 feet = 117.74 feet)
H = .63522 (x 359.625 feet = 228.44 feet)
Intersection Point # 16 has a sine of 8888 ÷ .9444 = .9411. This is the sine of 70°15', and the tangent of 70°15' is 2.7852.Therefore,
2.7852 = 3.729644 B
B = .7467736
A = .2532264 (x 359.625 feet = 91.07 feet)
H = .7052861 (x 359.625 feet = 253.64 feet )
By moving the perpendicular of Intersection Point # 14 up to the location of Intersection Point # 16, the # 14 perpendicular will end 253.64' - 228.44' = 25.2 feet above the pyramid's base. This is the elevation of the floor of the Red Pyramid's 3rd Chamber.
The Proposed Descending Passage to the Red Pyramid's 3rd Chamber
The derivation of the 25.2 foot elevation of the 3rd Chamber above the pyramid's base has just been given in regard to Intersection Points # 14 and # 16. The proposed possible descending passage to this chamber from the pyramid's west face utilizes Intersection Points # 1 and # 2, and so the particulars of these points will now be derived..
Intersection Point # 1 has a cosine of .8888/.9444 = .941176, and so the angle is 19°45'. The tangent of 19°45' is .35904. Therefore,
.35904 = 1.30348 B
B = .2754464
A = .7245536 (x 359.625 feet = 260.57 feet)
H = .2601437 (x 359.625 feet = 93.55 feet)
Intersection Point # 2 has a cosine of .8395/.9444 = .8888/1, therefore the angle is 27°16'' . The tangent of 27°16' is .5154, and so:
.5154 = 1.45984 B
B = .3530514
A = .6469486 (x 359.625 feet = 232.66 feet)
H = .3334373 (x 359.625 feet = 119.91 feet)
The above diagram is similar to that of Figure 13 in the text. Here, however, we have the radial lines to Intersection Points # 1 and # 2 crossing the pyramid's west side (we are looking from the south). The section of the radial line to Point 1 which goes from Point O to the pyramid's side is shifted up to the point where the radial line to Point 2 crosses the pyramid's side. The other end of this shifted line can therefore be seen extending to the right of the pyramid's vertical axis (line OC), and has been extended (by the dotted line) to meet the pyramid's base at a Point J. Also shown is the position of the 3rd Chamber relative to this shifted line.
Since the Line MJ begins at the pyramid's side at the elevation shown for Angle 2, we can see from the above calculations for this angle that it (Point M) is 119.91 feet above the pyramid's base. The angle taken by this shifted line is that of Angle # 1, or 19°45'. The tangent of this angle is .35904. Note that this shift is the reverse of that done to establish the line of this Pyramid's descending Passage. (See Part 1 of this essay).
The diagram above is an enlargement from the diagram previous. MU ÷ JU = .35904 (the tangent of 19°45'). Therefore, JU = 333.97 feet. From the computations regarding Angle # 2, we know that MU sits 232.66 feet from the central vertical axis of the pyramid. Therefore, Point J lies 333.97' - 232.66' = 101.31 feet on the other side of the central axis.
From the surveys we know that the west wall of the 3rd Chamber is 25.68 feet west of the central north to south axis, above shown as Point L. This means that the distance JL above will be 127 feet, and so the height of X (i.e., WL) can be found from the tangent of 19°45' with the formula X/JL = .35904. This leads to X/127' = .35904, and so X = 45.6 feet.
The floor of the 3rd Chamber sits about 25.6 feet above the pyramid's base, and so the here proposed descending passage (or some symbolic representation of it) would meet the 3rd Chamber's west wall at about an elevation of about 45.6' - 25.6' = 20 feet above the 3rd Chamber's floor level - unless, of course, the angle of this pproposed passage was steepened so as to allow it to meet the 3rd Chamber at its floor level.
Note that had this same maneuver been
implemented using the lines resulting from Angles # 4 and # 6, with
the Angle 4 line shifting up to the Angle 6 intercept at the
pyramid's side. This shifted line would compute to meet the 3rd
Chamber's west wall at an elevation of about 18.5 feet. The
passageway to this line would have then been at Angle 4's angle of
32°54.8', and it would have its point of beginning at an
elevation of about 154.42 feet above the pyramid's base on its
western flank. This is an alternate possibility for the proposed
descending passage line. (See computations for Angle # 6 above).
The Location of the Descending Passage Endpoint (Point "G" in diagram below) - (refer to footnote 16 in the text)
Maragioglio and Rinaldi (p. 128) quote Perring in saying that the original length of the descending passage was 62.63m (= 205.5 feet), and that the elevation of the original entrance above the pyramid's base was 28.65m (= 94 feet). They also add that they corroborated Perring's finding of 27°56' for the slope of the descending passage. These figures do not add up, since a right triangle with a height of 94 feet and a hypotenuse of 205.5' will have an angle nearer 27°14'. As I explain in a forthcoming essay, it is likely that Perring erred in compiling his results. His 94 foot elevation was the elevation of the existing entrance location as he then found it, and the existing passage length that he likely found was 196.5 feet. These figures are consistent with his stated passage angle of 27°56'.
From these results, and Perring's statement that the passage was then missing 4.5 feet of its length that it would need to meet a 43°36' exterior side, one can compute a passage length of 201 feet to a 43°36' side. From this one can then find that Perring would have found that an original entrance location at a 44°44' side would have been nearly 99 feet south of the pyramid's north base edge, and that the original passage length would have been nearly 204 feet.
Maragioglio and Rinaldi confirmed Perring's 27°56' measure for the slope of the descending passage, and so by using this figure and extrapolating Perring's data to a 44°44' north side, one will find that the 204 foot descending passage will end about 2 feet above the pyramid's base and about 4 feet to the north of where Dorner's survey has claimed it to be. This means that the center of the pyramid's Second Chamber will be about 4 feet to the north of the center of the pyramid, precisely as is predicted in the proposed derivation. (Refer to Figure 11 in the text).
Dorner (in "Neue Messungen") finds the elevation of the entrance to be at a significantly higher 101 feet, and that the entrance passage lies at the less steep angle of 26°30'. This leads him to the finding that the horizontal passage rests 2.93m (= 9.6') above the pyramid's base level and that the center of the Second Chamber aligns with the pyramid's center. I believe these findings may be incorrect, and that a new survey of this pyramid should be done to definitively resolve this issue. Further comparison and analysis of the Dorner and Perring surveys of the Red Pyramid will be soon posted at this website.
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.
The sekeds for the 27°16' and 43°21.8' angles have been given in the text.
Using the 1/9th sub-division, the seked of 19°45' can be derived from a finding of 702/252 units, which then results in a seked of 2 royal cubits + 5 and 1/2 palms.
Similarly, the seked for 7°42' works out to 7 royal cubits + 2 and 7/9th palms.
The seked for 30°12' could have been rounded off to 1 royal cubit + 5 palms.
For the 24°38' angle, it may have been given as 2 royal cubits + and 1 and 1/4th palms. And for 21°22.6', a seked works out to be 2 royal cubits + 3 and 3/4th palms.