RLH Wrote:
-------------------------------------------------------
> Sirfiroth Wrote:
> --------------------------------------------------
> -----
> >
> > Problems 48 and 50 of the RMP demonstrate the
> area
> > of a circle, #48 with a octogon, and #50 with
> the
> > (8/9d)^2, but there is nothing in any of the
> > papyri that even comes close to pi. One
> would
> > think if pi was as important to them, as it
> is to
> > us, they would have had a hieroglyph to
> represent
> > it. One can be reasonably sure the Ancient
> > Egyptians did not use 3 1/7 (pi) nor any
> value
> > relating to pi in any manner! The 11/2 seked
> of G1
> > explains this phenomena quite easily.
> >
> > Regards,
> > Jacob
>
>
> Hello Sirfiroth
>
> Do any of the MP’s have a problem to find the
> circumference of a circle when the diameter is
> known?

Not a one, pi was unnecessary as they seem to have calculated everything from the side of the square.

> Would you agree that if the AE did not know the
> formula (A = pi r squared) then there would not
> be any reason for them to use pi(22/7)?

The method demonstrated by reverse engineering G1 is more accurate than the RMP #50's (8/9d)^2 for the area of a circle. The square's perimeter 1760 / 4 = 440
The Circle of 1760 / (44/7) = 280 (this is not necessary by their methods just used for demonstration) This demonstrates the circle's circumference is equal to the perimeter of the square as found in G1.
Therefore:
440 / 280 = 11/7 as the ratio between one side of the square and radius of the circle.

> Could they know that C = D * pi (22/7) and yet
> not know that radius needed to be squared for
> areas and volumes?

Pi is not necessary for any of these values.
The 5 1/2 seked has a base to height ratio of 5 1/2 palms to 7 palms (one cubit), this creates a triangle with a base of 5 1/2 palm with a height of 7 palms providing a foumula for creating the angle of G1. For simplicity of concept remove the fractional portions, that gives a rise of 14 palms to a run of 11 palms. Giving a whole number ratio of 14:11 between the height and base respectively, which is our tan 14/11 for an angle of 51º 50' 33.98”.
What is the area of the circle with a perimeter of 1760 units?
Their method was simple 440 squared equals 193600 cubits squared
193600 * 14/11 = 246400 as the area of a circle.
Using this method the area of the circle always has a 14:11 ratio to the area of a square.

246400 area of a circle times 4 = 9856000 the surface area of a sphere with a circumference of 1760, for any sphere the circle area times four equals surface area.

Also it should be noted that the area of a circle times 4/3 * r (which is 7/11 side) equals the volume of the sphere which in the case of G1 is 246400 * 4/3 * 280 = 91989333 1/3.

>
> I’m just asking is it fair to judge what the AE
> knew about 3 1/7 from the problems in the MP’s.
>
> Regards
> RLH
>
We see pi because that is what we were trained to do, never dreaming of a world without pi, or any other possibility.

Regards
Jacob