Jim has this as 3.160493827.

In this post i want to link Jim's finding to Hugh's calculation of the base perimeter of the GP.

3.160493827 / 1.333r = 2.37037037 / 1.333r = 2.777r / 1.333r = 1.333r.

The final figure is 4/3.

Over to Hugh.

He takes the square root of this number and mutiplies it by real pi to get to the actual perimeter as follows.

root 4/3 = 1.154700538 x 3.141592654 = 3.6276 ( rounded to 4 sig figures in excel )

Multiply this by 100 x 100 to get 36276 inches this is 755.75 imperial feet per the Encycopedia Brittanica.

[www.britannica.com]

This gives a side of the GP as 9069 inches and a cubit of 9060 / 440 = 20.61136364 inches.

So if the cubit is 20.61818181818r the side is 9072 so 12 inches different to Petrie's perimeter. This seems to be a huge error when Petrie's meticulous approach is considered.

and if the cubit is 20.625 the side is going to be 9075 inches so 24 inches different as a perimeter.

It is possible to interpret Hugh's finding very simply.

One solar day is represented by 25 imperial inches.

9069 / 25 = 362.76

But if the perimeter is used one solar day is represented by 100 imperial inches. This is 8.333r feet.

If the cubit is of variable length the actual meaning of the perimeter can be worked out in solar days and obviously for me it will be a side of 366.666r units giving a perimeter of 1466.666r units.

the conversion factor to produce this is 326484 / 33000

326484/12 = 27207 this is Hugh's calculation of Thom's megalithic yard.



Edited 3 time(s). Last edit at 02-Feb-18 00:27 by DavidK.