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54. The Megalithic Yard and a Soli-Lunar Mile

Updated: Dec 13, 2022


St Michael's Mount, Cornwall, Wikimedia Commons

A 'Soli-lunar' mile


There are 200 'soli-lunar' miles between Saint Michael's Mount and the Mont Saint-Michel, a soli-lunar mile being a mile divided by the ratio between a solar year and a lunar year. The distance in regular miles is 206.14 miles, which divided by 365.242199 / 354.36708 is 200.002. The ratio between the solar year and a lunar month, multiplied by 20 gives a number close to Phi, 1.617042. There is an approximate connection between Phi and the 1 lunation / solar year, as 20 x 29.53059 / 365.242199 = 1.6170415183 The distance between Stonehenge and Lundy, famed for being a side of Robin Heath's lunation triangle, can also be interpreted in the light of a soli-lunar mile. This distance can be measured as 123.69 miles, which divided by (365.242199 / 354.36708) gives 120 'soli-lunar' miles. (Thanks to Jim Alison for his help working on that.) 123.69 miles are close to 200 / 1.61803 miles.

If you divide 123.69 miles by 2,500, you get a number that's quite close to the measure given between station stones 92 and 93, and between station stones 91 and 94, as Robin Heath pointed out in the Lost Science of Measuring the Earth.

Mont Saint Michel, Wikimedia Commons

What is a Megalithic Yard?


A 'soli-lunar' mile is very close to 2,000 of Thom's Megalithic Yards, which he valued at close to 2.72 feet. A mile of 5280 x 365.242199 / (354.367 x 2,000) feet, or 2.64 x 365.242199 / 354.367 feet, which would give a value for the MY of 2.7210186 feet. 2.72102 feet multiplied by 2,000 x 354.36708 / 365.242199 is 5280, the total number of feet in a regular mile. So a mile multiplied by this soli-lunar ratio (not divided, as above) , then divided by 2,000, is the Megalithic Yard.


What is a mile?


In the Lost Science of Measuring the Earth, Robin Heath begins by fleetingly connecting the English foot to the equatorial circumference of the earth (equatorial circumference of the earth is 365.242199 x 360,000 =131,487,191.64 feet, or 24,902.877 miles). The main focus however is on the English foot connecting to a measure not of space, but of time. The foot represents "the difference in time between the lunar and solar year (10.875119 days)", according to Heath.


Heath and Michell go to a lot of trouble to include the Astronomical Megalithic Yard (AMY) / Druisian step of 2.71542857 feet into all this. In fact, there is no real need for the AMY to prove the connection between the English foot and the cycles of the sun and the moon, or to the Stonehenge landscape. One lunation in days, 29.53059, divided by the average difference in days between the lunar year and the solar year, 10.87512 days, multiplied by 7,000/36, is 5,279.996. There are of course 5,280 feet in a mile, so if you take this to be a value in feet, it’s very close to a mile.


In fact:

5,280 x 10.87512 x 36 / 70,000 = 29.5306

A mile, in feet, can therefore be thought of as a lunation in days multiplied by 70,000 / 36 and then by the difference in days between the lunar year and the solar year, 10.87512.


What is a foot?


The foot is a precise division of the Earth's circumference which is roughly equivalent to the size of a human foot - though of course a human foot can vary widely in size. The English foot divides reasonably nicely into the 24,883.2 mile polar circumference. (11 x 3⁴ x 2¹²). However, the foot (and the mile) divide even better into the equatorial circumference, as Hugh Franklin has pointed out. Taking 24,902.31984 miles as the value:


one mile = earth’s equatorial circumference /( √(π3 x 20,000,000)

= √(10,000,000 π) x √2 x π /earth’s equatorial circumference


So if you have a circle with a circumference of 24,902.31984, and the diameter of this circle is also the diagonal of a square. The square will have a side of 24,902.31984 /(√2x π), and the area of this square will be 10,000,000 π. This square would be equal in area to a circle with a diameter of 10,000,000. This would suggest that the equatorial circumference of our planet, via the geometries of the circle and the square, gave rise to the mile itself, as a unit of measure. The circumference of the earth as a value in miles is equated to a square with an area of 10,000,000 π.


And so one mile, in feet, can be defined as 70,000 lunations / (36 x the difference in days between solar and lunar years), which means earth’s equatorial circumference in feet

= 70,000 lunations x (√(π3 x 20,000,000) / ( 36 x the difference in days between solar and lunar years)

= √(10,000,000 π) x √2 x π x 5,280


And the equatorial circumference of the earth in miles = 365.242199 x 1,296 x the difference in days between solar and lunar years / 7 lunations.


John Neal's list of feet, as he wrote it in an article on the GHMB website, gives only a handful of feet which divide nicely into his, and Michell and Heath's, preferred 24,883.2 mile circumference for the earth. For example, Neal's Roman foot of 0.96 feet would give a digit (if there are 16 of them in the foot) of 0.72", and it goes 3⁵ x 4⁴ x 2,200 times into this circumference. His Assyrian foot of 0.9 feet goes 2¹⁴ x 3⁴ x 110 times into it, and would give a digit of 0.675", if divided by 16. But rather than defining all these units as specific divisions of the earth's circumference, or in terms of the Egyptian / Roman digit, Neal and Michell's system hinges on each unit's specific relationship to the English foot. For Neal, the English foot is "pivotal to the arrangement" (the worlwide metrological system), and "Root One" (see Ancient Metrology Vol II: The Geographic Correlation, p.18).


The authors of The Lost Science of Measuring the Earth are keen to connect the fact that one lunation in days divided by the difference in days between a solar and lunar year, 10.87512 days, is 2.715426. This is close to 19/7, and also to, 19.008/7, the number of feet in an AMY. This is then linked up to the closest match in Michell and Neal's impressive catalogue of units, the Druisian step, as it is defined by Michell and Neal. Perhaps all these links muddy the waters a little bit. Is there any real need for an AMY? Robin Heath says in the book he was keen to connect his 'silver fraction' 19/7, or 19.008/7, with Thom's Megalithic Yard, hence his creation of an alternative Megalithic Yard that just happens to fit in nicely with Neal's root canonical Belgic foot, multiplied by 2.5. As for the distances measured in AMY, they can also make sense without it.


Robin Heath brilliantly points out that a distance of 864/7 miles, which separates Lundy and Stonehenge, is simply 864/7 x 29.5306 x 70,000/(10.87512 x 36) feet, or 240,000 x 29.5306 / 10.87512 feet. So the Lundy - Stonehenge distance in feet = 240,000 lunations in days / the difference in days between a solar and lunar year. Robin Heath likes the 864/7 mile value for the Lundy - Stonehenge distance, but you could also think of it as 200/Phi miles. You can perfectly well measure 123.6 miles between Stonehenge and Lundy. As with the Aubrey circle, there is a little bit of give and take. I think that maybe from wanting to fit into both Thom's, and Michell and Neal's system, Robin Heath created a unit (the AMY / Druisian step) which was superfluous in the end, as he'd already proved the importance of the English foot in relation to time cycles. (As for the lunation triangle being a geometrical device for predicting the moon’s phases and eclipses, the unit of measurement it’s in is irrelevant. This is purely to do with proportion, in particular 5:12:13.) The English foot does indeed to be a lynch pin of historical metrology.


What is an inch?


There are 12 inches to the foot, 39.3700787102 to the metre... but what about in the circumference of the earth? Well, it could be measured in lots of ways, but I like this:


The meridian circumference of the earth = 1752 x 1,000,000 lunations / ( 3 x the difference in days between solar and lunar years x 176) inches


It seems the solar and lunar numbers, the dimensions of the earth, and a little pi and Phi thrown in, are all you need to create the most important units of measure.


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7 commentaires


D Kenworthy
D Kenworthy
25 nov. 2022

This is important to understand the distinction between Berriman and Neal and the use of the inch foot and yard in the calculation of Egyptian cubits.


https://www.youtube.com/watch?v=KfX2NNNx2_Q&t=5s


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D Kenworthy
D Kenworthy
25 nov. 2022

Thom Pi is the Thom mile 5440 representing the Nodal cycle x 0.8 divided by 10 eclipse danger periods = 1733.333 days


5440/1733.333 = Thom Pi ( base 100 pi)


https://www.academia.edu/41972558/The_ALOAM_Paper_13_Sacred_Base_10_Circles_and_Variable_Pi_Babylonian_Pi_base_10_circumference_C_100_D_32_Egyptian_Pi_base_11_C_22_D_7_Stonehenge_Pi_C_31416_D_10000_Stonehenge_Pi_Thom_C_408_D_130

extract from above

Stonehenge Thom Pi ( extended ) ( C 935 D 297 )

This is 408/130 x 325/324 = 935 / 297 Very important in linking the findings of Thom to those of Neal and Michell.1728 x 935/297 = 5440 the Thom mile in imperial feet.297 / 3 = 99 and this is the base of the imperial system

Thom’s units are designed to work with imperial ones as is evidenced by the

root 5 calculation discovered using the work of Crowhurst at Carnac.The value for root…


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D Kenworthy
D Kenworthy
25 nov. 2022

cubit x 1760 = giza base inches

cubit x 1200 = Earth circumference miles.


This is a metrological rule. It is not open for debate, it is a fact when there are 1760 cubits always in the Giza base.


Cubit x 1.584 = meg yard


meg yard x 1/2 = meter


more rules.


Thom meter = 39.168 inches (prime 17 always embedded in Thom's findings Nodal Cycle)


https://numbermatics.com/n/39168/


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D Kenworthy
D Kenworthy
25 nov. 2022

The Thom mile is 5440 feet and Bill Wilkinson discovered 15 of them between the Sanctuary and Woodhenge and this is exact measure.81600 feet = 15 x 5440 the Thom mile.


The Thom mile is the English mile x 34/33. The is absolutely no doubt about the integrity of Thom's discoveries. Just as the megalithic yard is 31.68 x 34/33 = 32.64. Not accepted by Neal and Sivertsen of course. We all see what we want to not what is there.


https://www.youtube.com/watch?v=7fq1Cbeeb_U


https://www.youtube.com/watch?v=iRgyiiuWrZg&t=2s


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Melissa Campbell
Melissa Campbell
25 nov. 2022
En réponse à

Could 81600 English feet also be 15 'soli-lunar' miles, ie. 15.4545 English miles divided by 365.242199 and multiplied by 12 x 29.53059? By the way 34/33 is very close to 365.242199 / 354.36708, the solar and lunar year ratio. Maybe it's meant to represent the sun and moon.

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D Kenworthy
D Kenworthy
24 nov. 2022

Robin Heaths AMY is 22/7 x 864/1000 = 2.71542857feet and 32.5851429 inches x 1111.111... gives a Giza base of 36205.7143 and dividing by 1760 a cubit of 20.571429 in x 1200 gives an earth circ in miles of 24685.714 miles


Take this from 25000 and get 314.28571 miles. I am not sure even Robin Heath knows this.


Also 24658.714 x 7 = 172800 I am sure he does not know this.

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Melissa Campbell
Melissa Campbell
25 nov. 2022
En réponse à

Hi David, thanks. The problem with the AMY is that it's too short, if it's going to fit into the Giza base 1111.111 times, and the earth circumference of 24685.714 is also much too small. But who knows what measure of the circumference may have been used at any given point? It could well have been much too long or too short. At the end of the day it's all conjecture, but this coincidence remains: the English foot and the modern metre both fit beautifully into estimates of the circumference that are close to each other and to current estimates, and both fit beautifully into many ancient site measures.

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