Presented by Bruce C. Friedman
Page images created: January 2002
Last modified: June 29, 2003
To any visitors:
Please note this part [most parts] of the site is in progress and you may question me further about this analysis at the following Email link. brucefriedmanDCG@aol.com
The data, as it appears in, columns 11 and 12 of the Kahun IV, 3 "fragment".
I believe the analysis by R. Gillings correctly identifies the numbers on the page but is otherwise in error.
|Line #||SUM/Series||Multiplier||Checked: Yes or No|
|V||1/12, 2/3, 3||9||Yes|
The first numbers in Column 12 (twelve)
are 10, 100.
Although this is often read as one number, 110, my analysis is predicated on this as 10 (ten portions) and 100 (a quantity of one hundred).
The list of ten portions from column 12:
(Hi)=13.75 shown as 13, 2/3, 1/12
The other portions shown as the Hi portion less 5/6 in successive steps which ends with the Lo portion.
(Lo)=6.25 shown as 6, 1/6, 1/12
This 5/6 (or 10/12) is the typical gap of 2/3, 1/6.
1. Now find the midpoint of ab.
2. Name this midpoint e.
3. Continue line segment cb to point f, such that bf = the mean variant = 3.75
4. Draw triangle bef.
5. Continue fe to line segment da at point g.
Note the following:
Given: bf = MV = 3.75
We now find, be = P/2 = 5
We now find, cf = 13.75 = Hi
We now find, dg = 6.25 = Lo
We now find that:
bf: be: fe = 3:4:5
*Unlike the previously discussed relationship [ HILO+MV^2=Q], we only have one way to distribute this quantity using the 3,4,5 triangle!
Now lets look at the actual fragment!
Column 11 Column 12
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