## Geometry 3: Making a circle from a counted length

The number of days in four years is a whole number of 1461 days if one approximates the solar year to 365¼ days. This number is found across the Le Manio Quadrilateral (point N to J) using a small counting unit, the “day-inch”, exactly the same length as the present day inch. It is an important reuse of a four-year count to be able to draw a circle of 1461 days so that this period of four years can become a ouroboros snake that eats its own tale because then, counting can be continuous beyond 1461 days. This number also permits the solar year to be counted in quarter days; modelling the sun’s motion within the Zodiac by shifting a sun marker four inches every day. Figure 1 How a square of side length 11 will equal the perimeter of a circle of diameter 14

Our goal then is to draw a circle that is 1461 day-inches in perimeter. From Diagram 1 we know that a rope of 1461 inches could be divided into 4 equal parts to form a square and from that, an in-circle to that square has a diameter equal to a solar year of 365¼ days. Also, with reference to Figure 1, we know that the out-circle will have a diameter of 14 units long relative to the in-circle diameter being 11 units long, and this out-circle will have the perimeter of 1461 inches that we seek. Figure 3 A general method, using the equal perimeters model, applied to a 4 solar year day count of 1461 day-inches, found as a linear count at the Manio Quadrilateral. A square, formed from this linear count, can be transformed into an outer circle of equal perimeter using the simple geometry of π as 22/7.

For this, the solar year rope (the in-circle diameter) needs to be divided into 11 parts. Start by choosing a number that, when multiplied by 11, is less that 365 (and a 1/4). For instance, 33. A new rope will be formed, 11 x 33 = 363 inches, marked every 33 inches to provide 11 divisions. Through experience, we discover we need 2 identical ropes so as to make practical use of the properties of symmetry through attaching ropes to both ends of the solar diameter rope.

Place one rope at the West side of the in-circle diameter and swing it up until it touches the in-circle. Place the other rope at the East side of the in-circle diameter and swing it down until it touches the edge of the in-circle. Now connect the 33 inch marks between the 2 ropes. This will divide the 365 1/4 diameter into 11 segments.

Seven of those segments are the new radius to create the 1461 inch outer-circle. Figure 3 Division of the in-circle into eleven equal parts so as to select 7 units as a radius rope to then form the circle of diameter 14 units and perimeter 1461 inches.

This novel application of the equal perimeters model, rescued from Victorian textbooks by John Michell and applied by him most memorably perhaps to Stonehenge and the Great Pyramid (in Dimensions of Paradise) is a general method for taking a counted length and reliably forming a radius rope able to transform that counted length into a circle of the same perimeter as the square, easily formed by four sides ¼ of the desired length.

The site survey at the start, drawn by Robin Heath, appeared in our survey of Le Manio.

## A Pyramidion for the Great Pyramid

image: By 1200 BC, the end of the Bronze Age, the Egyptian map of the world (above) showed nine bows or latitudes, numbers 4 to 9 including the Nile Delta, Delphi, Southern Britain and Iceland, a map based on an ancient geodetic survey.

This post explores a pyramidion, now lost, which exceeded the apex height of the pyramid, so as to model the different reference latitudes established by geodetic surveys and encoded within their metrology and the Great Pyramid (by 2500 BC). This pyramidion would have sat on the flat top of the pyramid, 480 feet above the base of the pyramid.

In All Done With Mirrors, John Neal described how the full height of the pyramid, reaching to its natural apex, would have been just over 481 feet. Most pyramids probably had a pyramidion since a number have been found elsewhere that repeat aspects of or have a name carved on them, of a specific pyramid. Sitting on their apex, they often repeat the form of the larger pyramid, and are scale models of a specific pyramid. In the case of the Great Pyramid, exactly 441th of its natural apex is missing, and this is likely to be because a pyramidion once stood on the flat top the actual pyramid.

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## Recalibrating the Pyramid of Giza

Once the actual height (480 feet) and actual southern base length (756 feet) are multiplied, the length of the 11th degree of latitude (Ethiopia) emerges, in English feet, as 362880 feet. However, in the numeracy of the 3rd millennium BC, a regular number would be used. In the last post, it was noted that John Neal’s discovery of such rectangular numbers to define degrees of latitude, multiplied the pyramid’s pointed height (481.09 feet) by the southern base length (756 feet) to achieve the length of the Nile Delta degree of latitude and, repeating Neal’s diagram relating the key latitudinal degrees of the ancient Model as figure 1, the Ethiopian degree is 440/441 of the Nile Delta degree. As shown above, the length of the 756 foot southern base is changed, when re-measured in the latitudinal feet for Ethiopia; it becomes the harmonic limit of 720 feet of 1.05 feet – normally called the root Persian foot.

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## paper: The Origins of Day-Inch Counting

ABSTRACT
This paper presents the theory that in the Megalithic period, around 4500-4000 BCE, astronomical time periods were counted as one day to one inch to form primitive metrological lengths that could then be compared, to reveal the fundamental ratios between the solar year, lunar year, and lunar month and hence define a solar-lunar calendar. The means for comparison used was to place lengths as the longer sides of right angled triangles, leading to a unique slope angle. Our March 2010 survey of Le Manio supports this theory.

## Units within the Great Pyramid of Giza

There is a great way to express pi of 22/7 using two concentric circles of diameter 11 and 14 (in any units). Normally, a diameter of 7 gives rise to a circumference of 22, when pi is being approximated as 22/7 (3.142587) rather than being the irrational number 3.141592654 … for then, the 14 diameter should have a circumference of 44, which is also the perimeter of the square which encloses a circle of diameter 11.

The square of side 11 and
the circle of diameter 14
will both have the same perimeter. Figure 1 The Equal Perimeter model of two circles, the smaller of which has an out-square of equal perimeter to the greater circle
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