THE WILSFORD BARROWS
Wilsford Barrows are situated just over 3^{1}/2 miles SSW of Durrington Walls Henge. The location seems to demonstrate all of the attributes of a major resolving area for incoming vectors and a very interesting classroom offering multiple tutorials. Here is a sampling from Wilsford Barrows:
1. The distance is 20000' and the angle is 20.25-degrees back to Durrington Walls.
The distance aspect is self-explanatory. Remember how the distance from the centre of Durrington Walls Henge to the southeastern extremity of Berwick St. James Henge was 40,000', well this is exactly half of this distance or ^{1}/6534th the distance around the equator. According to the "11" method of navigation the Earth was configured to have a circumference of 24750-miles (of 5280-feet each) or 130680000'. The sum of 13068-feet equals 6534 X 2. It appears obvious that this marker-mound was used mnemonically for relationships to the size of the Earth under, probably, two different navigational systems in use (the "11" and the "6&7" methods of positional plotting at sea).
The angle value is part of a mathematical progression that is especially useful in lunar calculations. For example: There would be 17.5 intervals of 20.25-days in a lunar year of 354.375-days; there would be 126 intervals in the 7.2 lunar year period monitored within the lunisolar Sabbatical Calendar (2551.5-days). There would be 336 intervals of 20.25-days in the in the 6804-day lunar nutation cycle. The 20.25, 40.5*, 60.75, 81 ...162 ....1215 ...etc., progression will provide all of the essential lunar cycle numbers.
2. The a double set of lines traverse a mound, giving distance codes of 19200' & 19250' respectively, then resolve upon a further mound at 19440'. There are two marginally different return degree angle codes, 20.61818182 & 20.625.
The 19200' distance is navigational and a mathematical progression based upon this value will provide many essential navigational numbers. The circle diameter that best encompasses the Stonehenge site (with exception to the Avenue) is 384' (North to South), which is 192' X 2.
GREEK VOLUMES…DRY.
1 Medimnus…3110.4 cubic inches, equals:
6 Hecteus…518.40 cubic inches, or
48 Choenix…64.80 cubic inches, or
192 Coytle…16.20 cubic inches, or
1152 Cyathus…2.70 cubic inches.
2b. A distance of 19250' would equate to 3500 ancient fathoms of 5.5' each (the 6' fathom is relatively modern, but the older 5.5', or merchant fathom, survived intact certainly through the 19th century in Britain and was, possibly, still in use into the 20th century).
2c. The distance of 19440' is a dynamic navigational code and part of a navigational mathematical progression. For example: the sum of 194.4-miles would be ^{1}/128th of the 24883.2-mile equatorial circumference. It's important to remember that the compass reduces by ^{1}/2, ^{1}/4, ^{1}/8, ^{1}/16, ^{1}/32, ^{1}/64, ^{1}/128, etc.
GREEK VOLUMES…LIQUID.
1 Metretes…2332.8 cubic inches equals:
12 Chous…..194.4 cubic inches, or
144 Cotyle…16.2 cubic inches, or
576 Oxybaphhon…4.05 cubic inches, or
864 Cyathus…2.70 cubic inches.
3. A yellow line brushing the western edge of mound 2 is on an angle of 20.736-degrees (the largest Egyptian Royal Cubit was 20.736" or 1.728') return and would also equate to 19440'.
?. That a position would have been situated here in the middle of what is now a cultivation, is hypothetical but still very likely. Although not marked by an individual mound, a line coming into the centre of a garden cultivation, shows a distance of 19008' at a return azimuth angle of 20.833333-degrees.
The distance would have been 19008', a dynamic navigational code, with a return angle back to Durrington Walls Henge of 20.8333333-degrees (20^{5}/6ths ... a navigational code).
4. Again, the distance is 19200' and the return angle back to Durrington Walls Henge is 21-degrees.
The Hebrew Royal Cubit was 21", as was the Celtic Royal Cubit. Both groups used a variety of shorter common cubits as well.
5. The distance is 20160' and the angle is 21.3333333-degrees back to Durrington Walls Henge.
The length is navigational coding and the sum of 2016' equates to ^{1}/3rd of 1-minute of arc for a world configured to be 24883.2 Greek miles in circumference. This distance across the Salisbury Plains would therefore equate to 3^{1}/3rd minutes of arc.
The angle of 21.3333333-degrees is simply part of a mathematical progression resolving on whole number expressions of 64. Such a progression would have been very useful for measuring lunar time intervals within the Sabbatical Calendar and also for angle determinations in navigation and the 360-degree compass.
6. The distance is 19687.5' and the angle is 21.6-degrees.
The distance code relates to the lunar cycle and the sum of 19.6875-days (19^{11}/16ths) equates to ^{1}/18th of the lunar year. This value was adopted very anciently by the Egyptians and then by the latter Greeks - Romans and used widely in "gold standards" (grain weight).
GREEK COMMERCIAL WEIGHT. 1 Talent………….590625.0
grains equals: |
EGYPTIAN GOLD STANDARD. 1 Egyptian Talent………393750
grains, equals: |
ROMAN DOUBLE-BEQA SYSTEM. 1 Libra………
4725 grains, equals: |
The angle value of 21.6-degrees relates to the Precession of the Equinoxes, where the Sun spends 2160-years in each House of the Zodiac during the cycle of Precession (25920-years). The ancient British Bushel was, quite apparently, 2160 cubic inches (^{1}/10th of an ancient Babylonian Homer).
BABYLONIAN CAPACITY.
1 Archane…… 129600 cubic inches,
equals:
6 Homer…… @ 21600 cubic inches, or
36 Artaba….. @ 3600 cubic inches, or
216 Sutu…… @ 600 cubic inches, or
2160 Qa……. @ 60 cubic inches.
6b. The intended distance is 20160' (3^{1}/3rd minutes of equatorial arc) and the angle to Durrington Walls is 21.6-degrees.
The line coming from Durrington Walls centre overshoots the small marker by a few feet.
7. The distance is 19200' and the angle is 21.78-degrees to Durrington Walls.
The sum of 19200' would equate to 1/6804th of the equatorial circumference of 24883.2 Greek miles. Remember: the lunar nutation cycle endures for 6804-days.
The angle of 21.78-degrees contains coding related to the equatorial circumference of the Earth under the "11" family of numbers geodetic system and 21780' would equate to ^{1}/6000th of the equatorial circumference under the 24750-mile assignment.
7b. The distance is 19440' and the angle remains 21.78-degrees.
The distance code relates to the 24883.2 mile equatorial circumference.
The sum of 1944-miles X 2 X 2 X 2 X 2 X 2 X 2 X 2 = 24883.2-miles. The code and related progressions thereof was of such importance in the ancient world that it was incorporated into Weights, Measures and Volume standards.
GREEK VOLUMES…LIQUID.
1 Metretes…2332.8 cubic inches equals:
12 Chous…..194.4 cubic inches, or
144 Cotyle…16.2 cubic inches, or
576 Oxybaphhon…4.05 cubic inches, or
864 Cyathus…2.70 cubic inches.
8. The distance is 19635' and the angle is 22-degrees.
The distance relates to a fraction of PI, rendered as an expanded value. The sum of 1963.5 would be ^{1}/16th of 31416. It would appear that ancient initiates learnt whole numbers values before reducing them to a lesser whole number and fraction. Learning aids, to demonstrate ratios like PI (3.1416) & PHI (1.6180339), could be set out as circles, where the full outer circumference was reduced into divisions.
The 22-degree return angle is simply a part of the "11" family of numbers and 22' would equate to ^{1}/240th of a mile.
9. The distance is 19687.5' and the angle is 22.222222-degrees.
The angle is simply 22^{2}/9ths and the sum of 22.222222 is simply 200 ÷ 9.
10. The distance is 19440' and the angle is 22.5-degrees.
The degree angle is ^{1}/16th of 360-degrees.
11. The distance is 19313.75' and the angle is 22.9166666-degrees.
The distance relates to the PHI reciprocal of 6.1804*. The sum of 1931.375 would be ^{1}/32nd of 61804. Amongst the "times-tables" that students had to memorise it would have been of great importance to commit to memory progressions based upon the PHI reciprocal, as round tubs and vessels used in the market places to dispense very precise quantities of grain or liquid goods were constructed according to this value.
The angle aspect relates to a division of the equatorial circumference of the Earth under the 24750-mile assignment, wherein a degree of arc was 68.75-miles (22.9166666-miles X 3). This value rendered as a fraction is 22^{11}/12ths.
*Footnote: EACH COUSIN NATION CREATED PERFECT VOLUME VESSELS BY USING THE PHI FORMULA.
Although the cousin nations made their measures either the same or in easily calculable ratios to their trading neighbors, they also required a precise formula, standard for fashioning very individual circular jar or tub vessels for their own coded volumes of preference. The formula, used universally amongst these ancient trading nations, for "squaring the circle" appears strongly to be:
10 inches ÷ PHI (1.6180339) = 6.18034 inches.
For example:
* 1 Egyptian Theban tub @ 11664 cubic inches could have a circular base of 18.541 inches (3 X 6.18034") and sides 43.2 inches high. There would be 270 X 43.2 cubic inches in 11664 cubic inches The length of the Great Pyramid is 432 Hebrew/ Celtic Royal Cubits of 21-inches).
* 1 Greek Metretes vessel @ 2332.8 cubic inches (actually a liquid volume) could have a circular base diameter of 12.36068 inches (2 X 6.18034") and sides 19.44 inches high. The 19.44 number was used for lunar calculations and the Roman Pace @ 58.32 inches was 3 X 19.44 inches. There would be 120 X 19.44 cubic inches in 2332.8 cubic inches.
* The Hebrew Homer @ 28512 cubic inches could have a circular base diameter of 30.9017 (5 X 6.18034") and sides that were 38.016 inches high. The number 38.016 is a navigational use number and the Roman Amphora @ 1900.8 cubic inches was 50 X 38.016. Alternatively the Hebrew Homer @ 28512 cubic inches was 750 X 38.016 cubic inches.
* The Roman Amphora @ 1900.8 cubic inches could have a circular base diameter of 12.36068 inches (2 X 6.18034") and sides 15.84 inches high. The 15.84 number was used in navigation and there would be 120 X 15.84 cubic inches in 1900.8.
* The Babylonian Archane @ 129600 cubic inches could have a circular base diameter of 49.44272 inches (8 X 6.18034") and sides that extended above the base 67.5 inches. The number 129600 was used in navigation. The sum of 12960 years is half the cycle of the Precession of the Equinoxes. There would be 1920 X 67.5 cubic inches in 129600 cubic inches.
Any precise volume standard used by the cousin nations could be fashioned with tremendous precision as a circular vessel when the base diameter was in allotments of 6.18034 inches. The vessels could be more squat than tall or vice-versa ... it didn't matter, as long as the base retained the 6.18034 inch progression in it's diameter. The same formula, in lesser ratio, could be used to fabricate tumblers, jars or everything down to small cups for use by wine, beer or mead vendors within commercial premises.
The 6.18034 number could also be pressed into service if it was necessary to lay out circular land plots of precise square footage area. For example, an Egyptian Pyramid Acre of 28800 square feet would be a circle with a diameter of 31 X 6.18034 feet. An acre of 43560 square feet (1 furlong X 1 chain) would be a circle of 38.1 X 6.18034 feet.
It seems evident that there was one variation of the old Scottish ell (generally accepted as 37 inches, but 37.125" in remote antiquity) that was, quite simply, 6 X 6.18034 inches originally. This PHI based rendition of the Scottish ell would work very fluidly in laying out circles of desired square footage area with reasonable calculation ease. This is, undoubtedly, one of the surviving measurements carried from Egypt to France and Britain by about 5000 BC. Half a PHI-based Scottish ell could be used effectively to make old English bushel barrels or tubs of 2160 cubic inches (^{1}/10th of a Babylonian Homer).
In ancient England the tub created to measure out a correct bushel volume was described as, "any round measure with a plain and even bottom, being 18.5 inches wide throughout and 8 inches deep". This surviving volume from the Old English Winchester Standard, undoubtedly, had a pedigree founded upon yet older traditions passed down from distant ancestors, but had "drifted" by .041" (about ^{1}/25th of an inch) off the original standard.
Using the PHI based rendition listed above at 6.18034 inches the final capacity is: 6.18034 X 3 = 18.54102 inches ÷ 2 = 9.27051 (squared) = 85.94235566 X PI = 270 square inches X 8 = 2160 cubic inches...with a margin of error of .033 of a cubic inch.
11b. The distance is 19555.555555' and the angle is 22.9166666-degrees.
The distance is 2.7-miles of 5280'.
12. The distance is 19313.75 (related to the reciprocal of PHI) and the angle is 23.04-degrees. An extension of this vector resolves in the centre of an arrowhead shaped mound @ 19687.5' from the centre of Durrington Walls Ring.
The angle is very important to navigation. The outer face measurement of the lintel stones on the Sarsen circle at Stonehenge was intended to convey 30 X 11.52' or 15 X 23.04' for an outer circuit sweep of 345.6-feet. Under the formula used anciently to describe the equatorial circumference as 24883.2-miles (either Greek miles of 5250' or British miles of 5280') the sum of 1-degree of arc was 69.12-miles or 23.04 X 3.
SAMPLES OF WILSFORD BARROWS COMPLEX, FURTHER WEST.
12. The distance is 19440' and the angle is 23.4375-degrees (23^{7}/16ths).
The distance code has already been discussed. The angle is part of a mathematical progression that relates to navigation. The value is ^{1}/32nd of 750.
13. The distance is 19800' and the angle is 23.76-degrees.
The distance value codes the diameter of the Earth, which is 7920-miles, wherein 1980-miles would be half the radius. This distance from Durrington Walls Henge would be 3.75-miles.
The degree angle relates to navigation and the sum of 237600' X 550 = 130680000 (the number of feet in the 24750-mile circumference).
The value of 2376 X 8 = 19008 and in each of 3 geodetic systems rendered in feet, i.e., 130636800', 130680000' or 131383296' , divided by 19008 the total given shows the value that would describe 1-degree of arc in miles (68.72727272; 68.75; 69.12).
14. The distance is 20000' and the angle is 24-degrees.
The distance is self-explanatory and the angle represents ^{1}/15th of a 360-degree circle.
14b. Again the distance is 20160' (3^{1}/3rd minutes of arc) and the angle remains 24-degrees.
15. The distance is 19687.5' and the angle is 24.3-degrees.
The distance code has been explained. The angle carried a very important lunar code and a mathematical progression, based upon 243, will deliver up many very useful lunar cycle numbers. For example: the lunar nutation cycle of 6804-days is 243-months of 28-days each. The 2551.5-day (7.2 lunar year period measured within the Sabbatical Calendar was 243 periods of 10.5-days (remember the ancient reed measurement was 10.5-feet and ancient savants probably carried half-reed, calibrated staffs upon which they could do their lunisolar calendar calculations).
15b. The distance is again 20000 and the angle remains 24.3-degrees.
15c. The distance is 21000' and the angle is 25-degrees. The western edge of the site probably complies to 25.2-degrees.
The distance out from the centre of Durrington Walls Henge is exactly 4 Greek miles of 5250' each. The angle is self explanatory. An angle of 25.2-degrees would also be very important to encode.
16. The distance is 20418.75' and the angle is 25.6-degrees.
The distance is ^{1}/6400th of the equatorial circumference of the Earth under the 24750-mile assignment. Alternatively, a position on the same mound could be read as 20412' and relate to ^{1}/6400th of the equatorial circumference of the Earth under the 24883.2 Greek mile assignment. The same mound, at its most extreme southerly point would achieve 20528.64' and relate to the "true" circumference of 24883.2-miles of 5280', but to reach that point of the mound the angle would have to swing to 25.78125-degrees (25^{25}/32nds), which is a navigational code (the sum of 25.78125-miles would equate to ^{1}/9600th of the 24750-mile circumference).
The angle of 25.6-degrees is navigational coding and a mathematical progression based upon 256 generates many very useful values, like 2304 (already discussed). In mnemonic usage, the sum of 25600 X 3 X 3 X 3 X 3 X 3 X 3 X 7 = 130636800' (the equatorial circumference under the 24883.2 Greek mile assignment).
17. The distance is 19687.5' and the angle is 26.25-degrees.
The angle is lunar and navigational coding simultaneously and the sum of 2625' equates to half a Greek mile. Lunar periods were easily broken down into increments of 5.25-days or 126-hours (twice 2.625-days). The value 5.25 was very useful for monitoring the lunar and solar periods within the Sabbatical calendar (Moon = 2551.5-days or 7.2 lunar years or 486 intervals of 5.25-days), (Sun = 2556.75-days or 7 solar years or 487 intervals of 5.25-days).