Perfect Pyramid

The following calculations show that the Perfect Pyramid is a function of the Golden Ratio. 

The Perfect Pyramid built as the Great Pyramid of Giza is special because of a particular constraint. In history, only Flerodotus has alluded to it. " The key is that the area of a side face is equal to the square of the height of the pyramid. Herodotus statement turns on the translation of the Greek word epipoles, which means either in elevation, hy the lateral surface or simply in surface... 

So the unique formula for a Perfect Pyramid is that the square of the height divided by the square of the side length is equal to the Golden Ratio divided by four ... 

The inverse of h/s is 1.5723. This approximates to O (1.618) with an error of only 2.8%. Thus the ratio of the side to the height of the Perfect pyramid is also the Golden Ratio O. 

The Golden Ratio is not the only primary constant inherent in the perfect Pyramid. While the Perfect Pyramid has no need for the knowledge of tt, nor precisely demonstrates tt, it is the case that 2/tt is an arbitrary value very close to the perfect value for h/s in (6) above of... 

Again, let the Perfect Pyramid be laid flat so the sides splay into a Cross Patte. The area of a circle inscribed within the Cross Patte is ... 

Since the Area of the Base of the Perfect Pyramid is the side length squared, s ... 

Pyramid, make it into a Cross Patte and inscribe a circle. The area of the circle is twice that of the square base of the pyramid. Using a square made from the diagonal of the base of the pyramid rather than the base of the pyramid itself, the twice factor disappears. This is because the diagonal measures sV2. The area of such a square is 2.s. Thus the esoteric solution from the Perfect Pyramid is ... 

With a pair of Compasses centered in the middle of the Chessboard, draw a Circle so it passes through points on each side, each two squares in from each corner. This circle has an area that equals that of the square with an error of 1.8%, less than the 2.8% inherent in the Perfect Pyramid method. Schoolchildren sometimes prove the areas are equal using paper. They cut away the corners of the square and rearrange these trimmings to fill the segments of the circle. 

As expected, the N3P3 ring is found to be perfectly planar. On the other hand, exocyclic nitrogen atoms have a highly pyramidal character, in contrast with the... 

Carbanions of the type [HjCR ], [HCR R ] and [CR R R ] (R = NO and R R = CN, NO, NO2) can be considered to be resonance-stabilized, nonlinear pseudohalides. All experimentally known resonance-stabilized methanides are reported to be planar or nearly planar (Table 1). While the parent ion, the methanide anion HsC, adopts a pyramidal structure [Afipianar-pyramidai = 9.8 kJmol rf(CH) = 1.099 A, <(HCH) = 109.7° cf. rf(CH) = 1.093 A, <(HCH) = 109.6°] due to the lack of delocalization (no resonance for the p-AO-type lone pair possible) , substitution of one H atom by NO results in a planar anion since the empty jr -orbitals of the NO group are perfectly suitable to delocalize the carbon lone pair. Further substitution of the second H atom again results in planar anions, and the same holds for the third substitution in case of R = CN. In case of R = NO and NO2, the third substitution leads either to a propeller-type structure with only a small distortion from planarity or one NO2 group is twisted by 90°, nevertheless leaving the central carbon in an almost trigonal planar environment . ... 

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