Kanga Wrote:
-------------------------------------------------------
> "Which, in your opinion, came first:
> a) the height of the Chamber from floor to ceiling
> is to be half the diagonal length of the
> floorplan
> or
> b) the height of the Chamber walls is to be half
> the diagonal length of the 10rc x 20rc floorplan @
> 11.18rc plus one-quarter of a royal cubit @ 0.25rc
> = 11.43rc?"
>
> In my opinion, the height of the chamber was
> chosen so that the re would be a 3-4-5 triangle
> across the chamber. This set the height as 313
> fingers. Then the floor slabs were chosen as 7
> fingers so as to bring the wall height up to 80
> palms. This was then divided into 5 lots of 16
> palms.
>
> In my opinion, the primary reason for the
> dimensions of the chamber is to create the 3-4-5
> triangle. The fact that there are two 2 x 1
> rectangles in the chamber is incidental.

Hello Kanga,

Thank you for answering my questions.

Another question, if I may.
why was the height of the KC walls increased to 80 palms?

The usual and most obvious possible answer to this is because it then makes the perimeter of the Chamber's side (north and south) walls equal* to the length of the walls multiplied by 3 1/7 (height = length 20rc x 3.143 ÷ 2 - length 20rc = 11.429rc).

Interestingly, if the walls of the Chamber were designed to be 11.429rc and the vertical distance between the surface of the raised floor and the Chamber ceiling was intended to be half the diagonal length of the Chamber's floor plan (i.e. 11.18rc), then the appearance of Phi and the 3:4:5 triangle (or ratios very, very close to them) in this Chamber could be fairly argued to be purely unintentional.

Regards,

MJ

*there is a difference of about 0.1"/2.5mm per wall course between the required theoretical height and the actual height of the wall courses (2.286rc @ 20.632"/524mms = 47.16"/1197.9mms against Petrie's mean height 47.04"/1194.82mms)