**Saqqara Ostrakon a Different and Exact Solution
**

**by Arto Heino 2012**

Having solved many numeric puzzles using Quantum Arithmetic, I decided to attack this interesting artifact with relish. It seems everybody had a go at this including Graham Hancock. I am glad to express my delight in the solution that I have found, revealing the true principles of Ancient Khemitian(Egyptian) geometry and part of the foundation of building and pyramid design and the reason for their construction principles. As I will show on a another blog entry, I have found one of the the final keys to knowledge of the ancient Pyramidic geometric cipher, no weird magic number or symbol required, this is truly exciting and I will notify all interested parties when my blog is published.

The Saqqra OstraKon was found at the excavations in Saqqara in 1925, the writing reveals this object has a geometric value. The inscription is a form of shorthand to a larger idea.

3 Cubits, 3 Palms, 2 Fingers = 98 Fingers = 1837.5 mm

3 Cubits, 2 Palms, 3 Fingers = 95 Fingers = 1781.25 mm

3 Cubits, 0 Palms, 0 Fingers = 84 Fingers = 1575 mm

2 Cubits, 3 Palms, 0 Fingers = 68 Fingers = 1275 mm

1 Cubit , 3 Palms, 1 Finger = 41 Fingers = 768.5 mm

1 Cubit = 7 Palms = 28 Fingers = 525 mm (using this conversion size in millimeters)

1 Palm = 4 Fingers = 75 mm

I Finger = 18.75 mm

The small curved line and a few strokes and the numbers above is all that was needed. The 98 Fingers was the first clue, this would be the Perihelion or in QA the J factor. The plane number 98 didn’t fit with any Quantum Ellipse so a multiplication or conversion number must have been used, after a few attempts it looks like the prime number 5 was the cipher.

G _ H = 490 = 5 x 98

N _ W = 475 = 5 x 95

D _ Z = 420 = 5 x 84

R _ B1 = 340 = 5 x 68

T _ C1 = 205 = 5 x 41

Now that I know J= 98 then as in all QA you can work backwoods to find the roots of the ellipse. These are the original root numbers that I found:

b = 2 e = 3 d = 5 a = 8 , these will give you J = 10 and the eccentricity of 0.6

b = 14 e 21 d = 35 a = 56 , these will give you J = 490 and the eccentricity of 0.6

The prime factor underlying the x 7 is a way of finding the other derived numbers in a simple translation as J=98 is not a Quantum root size it is a derivation from a physical measuring system from a Quantum number set. This diagram should explain the geometry.

The reason I exclaimed the true solution is because of the following hidden variables. The redrawn diagram everybody followed by Battiscombe Gunn is incorrect, looks like his pet theory of 28 fingers between lines was wrong, as you can see by the original carving the lines are not evenly spaced, it was only a crude diagram drawn without a rule, you cannot base the intention of the geometer on a rough sketch, only on the description he gives and the approximate position of all the lines. All QA requires is some real numbers and an approximate shape to define the correct form.

These are exact values with no decimal fractions:

H _ W = sqrt( 23376 )

H _ Z = sqrt( 106624 )

H _ B1 = sqrt( 220800 )

H _ C1 = sqrt( 394896 )

H _ I = Sqrt( 614656 )

These are the values when J=490, they reveal the correctness of the QA precision You will never arrive at this sort of numeric clarity using any mathematical tricks, the Quantum Ellipse is a unperturbed form and non empirical. The values when J=98 re listed below:

H _ W = sqrt( 935.04 )

H _ Z = sqrt( 4264.96 )

H _ B1 = sqrt( 8832 )

H _ C1 = sqrt( 15795.84 )

H _ I = sqrt( 24586.24 )

What we are looking for is the gap sizes, here is J=490

H _ W = sqrt( 23376 )

W _ Z = sqrt( 106624 ) – sqrt( 23376 )

Z _ B1 = sqrt( 220800 ) – sqrt( 106624 )

B1_ C1 = sqrt( 394896 ) – sqrt( 220800 )

C1_ I = Sqrt( 614656 ) – sqrt( 394896 )

Here are the final values decoded and back to empirical finger and metric measure

H _ W = sqrt( 935.04 ) = 30.5784237 Fingers = 573.3454456 mm

W _ Z = sqrt( 4264.96 ) – sqrt( 935.04 ) = 34.7282374 Fingers = 651.1544523 mm

Z _ B1 = sqrt( 8832 ) – sqrt( 4264.96 ) = 28.6720597 Fingers = 537.6011207 mm

B1_ C1 = sqrt(15795.84 ) – sqrt( 8832 ) = 31.7027812 Fingers = 594.4271479 mm

C1_ I = Sqrt(24586.24 ) – sqrt(15795.84 ) = 31.1184877 Fingers = 583.4718333 mm

These sizes were not needed by the Khemitian geometer to their requirements only the right angle edge and the curve was needed, only modern calculus require these values as to plot X * Y curves. The Khemitians had no need of such crude mathematical devices such as calculus, they treated the curve as a different type of measure than the straight line. Now we can extrapolate the Cubit/Finger and metric size of the whole ellipse.

Length of the ellipse is = 2450 = 490 x 5 in Fingers = 490 = 17 Cubits 3 Palms 2 Fingers = 9.1875 mtrs

Width of the ellipse is = 1960 = 392 x 5 in Fingers = 392 = 14 Cubits = 7.35 Mtrs

The values are correct as no extra bit of a finger is required, this one of the reasons the Khemitians used these types of measure. The translation into quantum units were simple and there were no fudge factors that required PI, PHI or Log, no fractional decimals, only whole numbers So the final outcome underlies how truly powerful Quantum Arithmetic is when fully understood and utilized in the right hands, when looking for a accurate solution without errors from empirical measurements and data from approximation using calculus.

These and many more are from my up and coming book “Talking to the Birds”, regards Arto

References

An article by Battiscombe Gunn

Published in Annales du Service des Antiquites de L’Egypte,

Volume 26, 1926, pages 197 – 202

Printed by the Institute of France, Oriental Archaeology,

Le Caire, France

Diagram was created by Arto Heino using GeoGebra

*** New Addition *** 2021

I can see some are not convinced of my odd methodology, here is ample proof that simplicity was the key.

As the diagram illustrates:

IH = 784

HF = 1470

IF = 1666

Add together to create the perimeter of the triangle, thus:

784+1470+1666 = 3920

Now return back to Finger measure:

3920 / 5 = 784

Now return back to Royal Cubits:

784/28 = 28

Thus if you have a string 28 RC long and joint it to make a loop, then you make two pegs on the ground at the distance of:

1470/5 = 294 fingers = 10 RC + 14 F

You can now trace this ellipse with a marker at a third point when you stretch the string. Refer to the diagram below using the 3-4-5 triangle giving 12 as the perimeter.

I hope that simplifies my reasoning why I use QA to solve these ancient problems. Regards Arto