## Posts Tagged ‘ellipse’

### Saqqara Ostrakon

December 5, 2012

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

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

### Quantum Arithmetic

November 22, 2012

This is another brief expert from my book “Talking to the Birds”, I hope you enjoy it. I have included some notes from Ben Iverson as well.

Only a trained mathematician/geometer can recognize the missing elements in the geometric canon, that which has been ignored for the last 60 years, you cannot accidentally or casually find this most unique arithmetic, it is not based just on porisms. It is not number theory or invented mathematics, it does not need calculus, there is no approximations, it does begin its understanding as the trivial terms that most assume to be understood. Let us begin with the trivial terms so we can dispense with all the mystery.

Quantum Arithmetic uses a natural number system which has a base in of all of the prime numbers which occur in the problem you need to solve. All its numbers are interlocked in a geometric arrangement. All numbers are positive integers, no decimal points or irrational numbers. Addition, subtraction, multiplication and division is all that is mainly required, the use of square roots and diadic fractions are also used, they are mainly used to verify the results and find the whole number ratio’s. The ordering into Par numbers can also verify and help understand the internal arrangements.

There are sixteen primary identities. The first four are given the identity of “a”, “b” “e” and “d”. These, are the roots of its given problem, and is the base numbers. The next twelve identities are the upper case letters A through L. They are combinations, of the first four base numbers, and are usually considered as one dimension, higher than the base identities. They will denote linear dimension, surface areas, or volumes, for the three standard dimensions above the roots.

There is also one dimension below the “roots” b, e, d, and a. They are called “quaternions”, and are the square roots of the root numbers. In conventional mathematics, these are called “Gaussian Integers”. They are integers, only when their base number is a perfect square. A problem in QA is well defined and problems solvable when only one of the upper case identities is assigned a value and the name of that identity. All of the values within a problem in QA are intertwined this gives you hundreds of ways to solve any given problem.

The Quantum Number for one figure is the same for all geometric figures. A single quantum number defines the magnitude of the measurements, as they are related between themselves. Different geometric figures can be connected and their dimensions are calculated once for all of the various shapes. The shapes used are:

1 Right triangles

2 Equilateral triangle

3 Isosceles Triangles

4 Triplet circles (Koenig Series)

5 Triplet Squares (Koenig Series)

6 Ellipses

Each shape is calculated from a single Quantum Number. The one requirement is that the second and third integers of the Quantum Number must be prime to each other. These are in the Fibonacci sequence. They derive from Euclid VII, Proposition 28. The first and fourth integers become, “Sum and Difference” numbers.

All Quantum ellipses will give the value one when you apply the Steiner Inellispe porism. The numeric laws it contain can be geometrically reconstructed with s0me of the most excellent programs such as GeoGebra, First you need to understand the construction of the “Quantum Ellipse”. The usual 2 foci and semi-major length/third point will not suffice only Conic section ellipse will create the correct elliptical form, if you have 5 points that are aligned to the quantum arithmetic standard. I hope some of these diagrams will help clarify all the parameters. The best way to construct an ellipse of this sort is to start with a string with length 3+4+5=12 a Pythagorean triangle. This method was re-discovered by James Clerk Maxwell in the mid 1800’s.

This type of geometry had been developed by Ben Iverson from the 1950’s through to to 1990’s extended by Dale Pond and myself over the last 20 years.

After many calculations and with the application of Quantum Arithmetic, I have created a beginning for the foundation of the Quantum 120degree Isoceles. The smallest integer units begin at the 4 digit root numbers, which means some calculations are up to 20 numbers long This study has revealed the Quantum Equilateral also within its perimeter, having now multiple Z line integers.

Listed below is the some of the raw data for one of the smallest Quantum Isosceles 120 degree Triangle, the added Var1,Var2,Var3,Var4,Var5,Var6 are part of an incomplete data list, you can ignore them or just make each equal zero.

b = 2646 = d – e

e = 4343 = a – d

d = 6989 = b + e

a = 11332 = d + a

B = 7001316 = b x b

E = 18861649 = e x e

D = 48846121 = d x d

A = 128414224 = a x a

C = 60706454 = 2 x d x e

Fe = 29984472 = d x d – e x e + Var1

G = 67707770 = d x d + e x e

L = 151687580848524 = ( d x d x d x e – e x e x e x d ) / 6

H = 90690926 = ( a x a + 2 x b x a – a x a ) / 2

I = 30721982 = ( e x e + 2 x d x e ) – d x d

J = 18492894 = d x d – d x e

K = 79199348 = d x d + d x e

We = 109552575 = d x e + d x a

X = 30353227 = d x e

Ye = 79568103 = 2 x d x e + d x d + Var2

Ze = 98060997 = e x e + d x d + d x e + Var3

Wib = 328657725 = 3 x ( d x e + d x a )

Fi = 109552575 = d x e + d x a + Var4

Yi = 219105150 = 2 x ( d x e + d x a ) + Var5

Zi = 109552575 = d x e + d x a + Var6

Wis = 189750626 = ( d x e + d x a ) x sq 3

Wh = 94875313 = ( ( d x e + d x a ) / 2 ) x sq 3

Thank you

Regards Arto