> It is a brilliant solution, so simple, is it
As far as I know, other than myself, no one has published anything like or even remotely similar.
> It is the missing piece in the jigsaw puzzle Beyond pi!!
> More complex yet simpler.
Just an accidentally rediscovered part of the Ancient Egyptian methods and for me it has been an indispensable key aiding in decoding how the interior dimensions were derived for the two upper chambers of G1.
> Thom said
> 'In terms of sheer brainpower they were my
They did have much simpler solutions for the more complex problems.
> An Oxford Professor of Engineering Science at the
> top of his game.
> this is not in our maths textbooks is it?
> is it in Egyptian solutions?
> not here
> Is it one of the qustions.
> It does prove they knew 22/7 because the solution
> is 2 Pi with radius 1. They can't not have known.
As I've stated many times before, I believe they were at least cognizant of pi but chose to use the 5 1/2 seked. Nor would I would say it constitutes any form of proof of their knowledge.
Looking at some of the problems shown (in the link you provided) in the papyri based solutions, G1 represents a much simpler and accurate method for finding the area of a circle, in the form of (d^2 / (14/11) = circle area): circle area x 2 = surface area of a hemisphere with diameter d: circle area x 4 = surface area of a sphere with a diameter equal to d. There is a certain elegance in the simplicity of their system.
i.e. G1 280 cubits radius x 2 = 560 cubits diameter: 560^^2 = 313600 cubits sq, 313600 / (14/11) = 246400 cubits sq area of circle, 246400 x 2 = 492800 cubit sq. surface area Hemisphere diameter 560: 2246400 x 4 = 985600 cubits sq. surface area of a sphere diameter 560:
This is really cool. If you have just the numbers and this time lets call them inches
560^2 x 4 = 985600 985600 / 27 slots on the GG ramp = 36503 19/27, 36503 19/27 / 4 = 9125 25/27 inches.
in decimals 36503.703703703..../ 4 = 9125.925925.... a figure within .03 inch of Petrie's mean socket base dimension of G1 perimeter. Or for the side length simply divide the area of the circle by 27. 560^2 / (14/11) / 27 = 9125 25/27 inches.
How is that for a coincidental anomaly?