The Architect's Plan
The Direct Measurement of Sekeds
Bent Pyramid Computations from Part 1 - Refer to Figure 4 in the text - Note that "Z" and "Y" are again relative to each situation.
1) Elevation of the Bend: The Cosine of the angle of Line OB = .7857/1 = 1.1/1.4 = .7857
This is the Cosine of the angle 38°13', whose tangent is .78739. (Note that this point can be determined on the basis of the 1.4 design pyramid alone, and so the elevation of the bend can be determined before any design choice is made as to what angle the pyramid will take above the bend.)
Following the same procedure as before we have:
.78739 Z = 1.4 Y, and Z = 1 - Y -------------Therefore, .78739 = 2.18739 Y
Y = .3599678
Z = .6400322
H = .5039548
With the half-base of this pyramid (i.e., Z+Y) equaling 94.8 meters, the elevation of Point N (given as "H", above) is 47.78 meters.
2) Height of the Bent Pyramid: The pyramid's exterior angle at Point N changes to 43°21.8'. The tangent of this angle is .9444. The height of AM in Figure 4 (reproduced below) is then found from: AM/Z = AM/.6400322 = .9444. AM works out to .604472 units = 57.3 meters. The height of the entire pyramid is then H + AM = .5039548 + .604472 = 1.10843 units = 105.08 meters.
3) Elevation of the Lower (Northern) Passage Entrance: The Cosine of line OE in Figure 4 in the text is 1.1/1.11 = .9910. The angle here is then 7°41.5', and the Tangent of this angle is .13506.
Using the same method as has been used above, one gets: .13506 = 1.53506 Y
The elevation of Point K is then .12318. This times the 94.8 meter half-base of this pyramid = 11.68 meters.
Again, from Figure 4 (reproduced above) the angle of line OD is 27°16', and so .5154 = 1.9154 Y. Therefore, Y computes out to be .26908. Z is then .730918, and the elevation for Point H in this case will be .376715. The length of line OH (= KZ) can then be computed via either the Pythagorean Theorem or from the Sine of 27°16' (= .45813). OH works out to be 77.95 meters.
4) Elevation of Ehe Upper (Western) Entrance Passage - (Refer to Figure 5 in the text - reproduced below, and note that the side of the pyramid is shown on the left and not on the right): The Cosine of Line OY is 1/1.1 = .90909. The angle is therefore 24°38', and the Tangent of this angle is .45854.
So, in the same way as before, H/Z = .45854; H/Y = 1.4; and Z+Y = 1 (and so Z = 1-Y)
This leads to the formula of .45854 = (1.4 + .45854) Y. H, in this case, then works out to be .3454087 = 32.74 meters = the western entrance elevation.
The Cosine of line OJ is .86428, and this is the cosine of the angle 30°12'. The Tangent of 30°12' is .58201. Using the same techniques, the elevation of Point G computes to .4111048, and the length of OG to .8172732 (= 77.47 m).
To find the length of OX in Figure 5, two changes need to be made - both of these on the basis of the findings for the precursor pyramid as referenced in the text: 1) the Z+Y length needs to be made equal to the 78.6 meter half-base amount of the precursor pyramid; and, 2) the exterior angle changes to the near 60° exterior angle of that earlier design stage. The Tangent of 60° is 1.732.
Therefore, the formulae become: H/Z = .58201 (the Tangent of 30°12'); H/Y = 1.732 (the Tangent of 60°); and Z+Y = 78.6 meters. Therefore, .58201 Z = 1.732 Y, and Z = 78.6m - Y. And so, .58201(78.6m - Y) = 1.732Y. Then, 2.314Y = 36.04 meters, resulting in Y = 19.76 meters. Z is therefore 78.6 - 19.76 = 58.84 meters.
This Z of 58.84 meters is then the base of the right triangle of which OX is the hypotenuse. Since OX is at an angle of 30°12', its Cosine (i.e., adjacent/hypotenuse) = .86428 = 58.84m/OX. OX then works out to be 68 meters.
Bent Pyramid Computations from part 2 - (Refer to Figure 6 in the text).
As seen in Figure 6 in the text, the Bent Pyramid is basically designed to incorporate two pyramid shapes - one with an exterior angle having a tangent equal to 1.4, and the other with an exterior angle having a tangent of .9444. The visible manifestation of this is, of course, the bend in the pyramid's side - which according to the derivation diagram for this pyramid occurs at an elevation of 47.48 m, and results from the radial line to Point 8 (which also goes to Points 9 and 10.) The mathematical explanation for all of this can be found in the Computation section to Part 1 of this essay.
Because of the bend, computations related to radial lines which cross the pyramid's side above this point become a little more involved than those which occur below the bend. The next diagram explains this dilemma.
A radial line from Point O which crosses the pyramid's side below the bend (Point B to Point N) is crossing the side a right triangle which has a tangent of 1.4, and a base length of OB.
A radial line from Point O which crosses the pyramid's side above the bend is crossing the side a right triangle which has a tangent of .9444 and a base length of OD. To find where a perpendicular dropped from this crossing point would meet the base (Point S in the example above), one has to take into account that what you want to know is where Point S sits relative to the midpoint of OB (the length of the as-built pyramid) - and not relative to the midpoint of OD.
To begin, we first need to find the length of OD. We know from previous discussions that NT is 47.78 m. The tangent of the angle at D is .9444, and so TD = 47.78m/.9444 = 50.59 m. The tangent of the angle at B is 1.4, and similarly TB = 47.78m/1.4 = 34.13 m. Therefore, BD is (50.59 m - 34.13m =) 16.46 m. We know from survey information that OB equals 94.8 m in the actual pyramid. Therefore, OD = 94.8m + 16.46m = 111.26 m.
The first derivation in the text is in regard to Angle # 13. Since this occurs above the bend, we will employ the above in showing it's derivation. (I will be presenting these derivations more or less in the order in which they are called for in the text.)
Intersection Point # 13 at tangent = .9444 side - In Figure 6 we see that the radial line to this intersection point also goes through points 14 and 15. The Sine of this angle is .7857, and so the angle is 51°47'. The tangent of 51°47' is 1.27. Therefore,
H÷OS = 1.27, and H÷SD = .9444, and OS + SD = 111.26m = OD
Therefore, OS = (111.26m - SD). Therefore,
1.27(OS) = .9444xSD Plugging in for OS from above
1.27 (111.26m - SD) = .9444xSD, and so
141.3 m = 2.2144xSD
SD = 63.8m
From OS = (111.26m - SD), OS = 47.46m
Since surveys show OB to be equal to about 94.8m, the midpoint of the half-side is 47.4 m. The Lower Entrance to the Bent Pyramid was apparently centered on the perpendicular to the 47.46 m point..
Intersection Point # 11 at tangent = .9444 side - The cosine of the angle created by this radial line is .7857÷1.11 = .70784, and this is the cosine of Cos 44°56.4'. The tangent of 44°56.4' is .99792. This line lies above the exterior bend, and so the tangent of the exterior angle for the actual pyramid' side will be .9444, and the base length will be 111.26 m. In the same way as with the Point # 13 diagram above, H/OS = .99792; and H/SD = .9444. Also, OS + SD = OD = 111.26m And so, OS = 111.26m - SD
.99792(OS) = .9444xSD
.99792(111.26m - SD) = .9444xSD
111.029m = 1.9423xSD
SD = 57.162m
From OS = 111.26m - SD, OS = 54.098m
From H = .9444xSD, H = 53.99 m
Intersection Point # 18 at tangent = .9444 side - The sine of the angle created by this radial line is 8888/1, and therefore the angle is 62°44' . The tangent of 62°44' is 1.9402 This line crosses the pyramid's side above the bend, and so the tangent of the exterior angle will be .9444, and the base length will be 111.26 m. In the same way as before,
H/OS = 1.9402; and H/SD = .9444. Also, OS + SD = 111.26m = OD
Therefore, OS = 111.26m - SD
1.9402(OS) = .9444xSD
1.9402(111.26m - SD) = .9444xSD
215.87 = 2.8846xSD
SD = 74.834 m
From OS = 111.26m - SD, OS = 36.43 m. Note that 94.8m/2 -36.43m = 10.97m
From H = .9444xSD, H = 70.67 m.
The difference between the 70.67 m and the 53.99 m elevations (i.e., H) is 16.68 m. By moving the perpendicular associated with Intersection Point # 18 to the point at which the radial line to Intersection Point # 11 crosses the pyramid's side, it will identify the depth below grade of this pyramid's Lower Chamber.
Intersection Point # 16 at tangent = .9444 side - The sine of the angle created by this radial line is .8888/1.11 = .8008, and so the angle is 53°12.5' . The tangent of 53°12.5' is 1.3371.This line crosses the pyramid's side above the bend, and so the tangent of the exterior angle will be .9444, and the base length will be 111.26 m. In the same way as before,
H/OS = 1.3371; and H/SD = .9444. Also, OS + SD = 111.26m = OD
Therefore, OS = 111.26m - SD
1.3371(OS) = .9444xSD
1.3371(111.26m - SD) = .9444xSD
148.766 m = 2.2815xSD
SD = 65.205 m
From OS = 111.26m - SD, OS = 46.05 m
From H = .9444xSD, H = 61.58 m m.
Intersection Point # 8 at tangent = 1.4 side - In Figure 6, the radial line to this intersection point also passes through Points # 9 and 10, and is below the bend. The cosine of the angle created by this line is .7857/1, and so the angle is 38°13'. The tangent of 38°13' equals .78739. Since this radial line crosses the pyramid's side below the bend, the opposite tangent will be 1.4, and the base length will be 94.8 m. Therefore, H/OS = .78739; H/SB = 1.4; and OS + SB = 94.8 m, which means OS = 94.8m - SB. Therefore,
.78739 (94.8 m - SB) = 1.4x SB
74.645m = 2.18739xSB
SB = 34.125 m
OS = 60.675 m so, 60.675 - 47.7 = 13.245 m
H = 47.775 m
Intersection Point # 11 at tangent = 1.4 side, Even though this radial line passes above the bend, this perpendicular falls from the intersection from the line with the line that is an upward extension of the lower side. H/OS = .99792, and H/SB = 1.4. OS now equals the actual base length of 94.8 m, and so OS + SB = OB = 94.8 m Therefore,
OS = 94.8m - SB and so,
.99792(94.8m - SB) = 1.4xSB
94.6 m = 2.398xSB
SB = 30.45 m
OS = 55.35 m
H 55.235 m
Intersection Point # 7 at tangent = 1.4 side - The radial line to this Intersection Point lies below the bend in the exterior side, and so the tangent of this exterior side will equal 1.4. The cosine of this radial line is .8888/1.11 = .8008, and so the angle is. 36°47.6'. the tangent of 36°47.6' is .74792, and so we have H/OS = .74792; H/SB = 1.4; and OS + SB = 94.8 m, which means that OS = 94.8 - SB. Therefore,
.74792 (94.8m - SB) = 1.4x SB
70.9 m = 2.14792xSB
SB = 33 m
OS = 61.79 m
H = 46.2 m
Intersection Point # 12 at tangent = .9444 side - This line is again above the bend, and so we return to the 'above the bend' strategy. The sine of the angle created by this radial line is .7857/1.11 = .70784, and so the angle is 45°3.6'. The tangent of 45°3.6' is 1.00206, and the tangent of the exterior angle will be .9444. The base length will be 111.26 m, and so H/OS = 1.00206; H/SD = .9444, and OS + SD = 111.26m. (This means that
OS = 111.26m - SD). Therefore,
1.00206(111.26 m- SD) = .9444xSD
111.489 m = 1.9465xSD
SD = 57.277 m
From OS = 111.26m - SD, OS = 53.983 m
From H = .9444xSD, H = 54.09 m.
Intersection Point # 21 at tangent = .9444 side - The sine of the angle created by this radial line is .98667/1, and so the angle is 80°38' . The tangent of 80°38' is 6.0624 This line crosses the pyramid's side above the bend, and so the tangent of the exterior angle will be .9444, and the base length will be 111.26 m. In the same way as before, H/OS = 6.0624; and H/SD = .9444. Also, OS + SD = 111.26m = OD And so,
OS = 111.26m - SD
6.0624(OS) = .9444xSD
6.0624(111.26m - SD) = .9444xSD
674.5 m = 7.0068xSD
SD = 96.26 m
From OS = 111.26m - SD, OS = 15 m
From H = .9444xSD, H =90.91 m.
Intersection Point # 22 at tangent = .9444 side - The sine of the angle created by this radial line is 1.1/1.11 = .991, and so the angle is 82°18.5' . The tangent of 82°18.5' is 7.3993. This line crosses the pyramid's side above the bend, and so the tangent of the exterior angle will be .9444, and the base length will be 111.26 m. In the same way as before, H/OS = 6.0624; and H/SD = .9444. Also, OS + SD = 111.26m = OD. And so,
OS = 111.26m - SD
7.3993(OS) = .9444xSD
7.3993(111.26m - SD) = .9444xSD
823.246 m = 8.34374xSD
SD = 98.666 m
From OS = 111.26m - SD, OS = 12.5937 m
From H = .9444xSD, H =93.184 m.
The difference between these last two elevations (H#22 - H#21) is (93.184m - 90.91m) = 2.274 m
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 Bent 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 24°38' can be closely approximated from a finding of 252/549 and so its seked will equal 2 royal cubits, 1 palm, 1 finger.
The angle of 30°12' can be closely approximated from a finding of 252/433 and so its seked will equal 1 royal cubit, 5palms, 1 finger.