Chris Tedder Wrote:
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> Thank you Alex for clearing up perceived
> discrepancies between Petrie’s and Dorner’s
> surveys. It's reassuring to know that Petrie’s
> survey is reliable as its consistent with Dorner’s
> modern survey. Petrie has an average 189.459m, and
> Dorner, 189.43m for the length of a side - a mere
> 29mm difference, which means they are virtually
> identical given the ruinous state of the pyramid.
>
> However, there is still the problem of converting
> their figures to cubits.
>
> A 362 cubit square base requires a cubit length of
> 523.3mm
>
> A 360 cubit square base requires a cubit length of
> 526.2mm.
>
> So which one was used?
Neither. The cubit they used was 523.8mm, the same as that used in Khufu's Pyramid. 189.459m divided by 0.5238m is 361.70 cubits, or 361c 5p. This is the length of the bend-line, 235c 5p, plus 2 x 63c.
> Other calendrical connections are also possible.
> The summer solstice has 14 hours from sunrise to
> sunset, and 10 hours from sunset to sunrise, a
> ratio of 7:5. The winter solstice has 14 hours
> from sunset to sunrise, and 10 hours from sunrise
> to sunset, also a ratio of 7:5 - based on this
> ratio, the angle of incline is 54.462°.
The slope of the lower part is 10:7 (55.00°), not 7:5.
> At the equinoxes, the day/night relationship is
> 1:1 that gives an angle of incline of 45° (sqd 7).
>
> A pyramid that has equilateral triangular faces,
> has an angle of incline of 54.74°, and a corner
> edge angle of 45°. The closest whole number seqed
> to the 'perfect' pyramid form is sqd 5, with an
> angle of incline of 54.462°...
Incorrect. Root 2 lies midway between 7/5 and 10/7. In other words,
r2 is just as close to 10/7 as it is to 7/5. In any case, the slope is an approximation to
r2:1. This gives it faces that are equilateral triangles (or close to it) - if the faces were projected upwards to a point.
The profile triangle for the lower portion is 63-90-110. This is a proxy for a 1:
r2:
r3 right triangle. In my opinion, this pyramid (lower portion) is meant to represent an octahedron (or the half of it which is above ground).
> Leaving aside all these musings, do you know how
> Dorner determined the base level of the so-called
> “inner pyramid” that is hidden deep inside the pyramid?
He took the face of the inner pyramid as indicated by the disjunction in the western passage. The disjunction is at right angles to the passage, which has a slope of 4:7, so he took the slope of the inner pyramid as 7:4, and projected the line of disjunction down to the platform, making a base length of about 300 cubits.