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Megalithic Yard Surprises
If two equilateral triangles are formed into a 'Lozenge' inside the Vesica, (often regarded as a depiction of female genitalia), is it entirely a coincidence that this Lozenge shape was chosen in Heraldry (albeit in Medieval times) to show a woman's coat of arms? (There is a precedent for the Lozenge shape, in the golden plaques from the Bush Barrow near Stonehenge, and as Norman Lockyer pointed out there is a great equilateral triangle with 6 mile sides, linking Stonehenge, Old Sarum and Grovely Castle).

1. A circumference of 22 MY... (diameter = 7.00281+ MY, 22/7 = approximation for p)
....Has an SS of 5 1/2 ft., 3 ´ SS = 16.5 ft = 1 Rod, Pole or Perch
....Has an LS of 11 ft., 6 ´ LS = 66 ft., = 22 yds. = 1 cricket pitch,
one chain or one acre width.
NOTE : 5 1/2 ft is equal to 180 Barleycorns*
11 ft. is thus equal to 360 Barleycorns, and the
area of a square with sides of 360 Bc's is
1/360 th of an ACRE
ANY circumference in Feet, divided by 11 will give the number of Barleycorns per degree (360 degs in a circle).
NB Theoretically, a Barleycorm ought to be 1/M inches (.36755.. in), but custom has it as being .3666.. in, 1/36 th of a Northern foot of 13.2 in.
An SS of .36755" gives a circumference of 4"
2. A circumference of 1760 MY.....
.....Has an SS of 440 ft., 3 ´ SS = 1/4 MILE.
.....Has an LS of 880 ft., 6 ´ LS = 1 MILE
3. Radius of above circle is 280.112+ MY , which is the number of units in the height
of a Pi  type pyramid which has a base side of 440 units. (i.e. the Great Pyramid in cubits).
4. 250 MY is 9/10 ths the base of the Great Pyramid (755.749735ft., or 440 cubits.)***
Thus 360 ´ Gt. Pyr. base = 100000 MY ( Pi pyramid diagram bottom of page).
5. If the radius of the circle is 1 MY or 1M, then the SS of it's triangle is p/2, and the
LS is p. This Long Side of p when formed into a circle , gives a DIAMETER of 1
unit  the same 1 unit used in the SS which produced the quadrant MY (or M) in
the first place.
6. MY is thus equal to p/2 ´ Ö3, or 2.720699046....... , which is extremely close to Thoms median of 2.72 +/ .003 ft.
EVER SINCE THERE HAS BEEN A 'FOOT', THERE HAS BEEN A MEGALITHIC YARD. (and vice versa?)
If one understands this 30/60 triangle method, then the formula to play around with using any units is :
Circumference / M = 4 SS
( e.g. 1/4 metre in SS will produce ONE unit of 2.7206+ metres in Circumference).
7. He also found that 2.5 MY lengths were used, which he called a Megalithic Rod. My
bicycle has 26 inch diameter wheels, so at each rotation I am pleased to know that I am covering (very nearly!) one Meg Rod. (diameter = 25.98.. inches)
Decimal division apparently did not come on the scene until the Middle Ages, so simple fractional maths, multiplying and dividing, halving and quartering etc., must have been well within the capabilities of the Stoneagers. If 120 MY were required in a circumference, using p to find a diameter needing 38.197.. MY, might have been beyond them, but an SS of 30 ft is rather ingenious.
It is not necessary to know the length of the MY. It is rather like saying that if my diameter is 5 units, then the circumference is 5 lengths of "3 + something" units, but we don't actually have a measured length of 3.141592.. units. (see additions page for Arabian measures.)
[ I might wager that if during the course of measuring artefacts or stones, measures of 3.1415+ feet had been found over a wide area, it would be assumed that 1 foot (12 Imperial inches) must be the base unit.
(A diameter of one foot having a circumference of 3.1415+ feet)..
Thus if any multiples of p feet were to be found, the base unit would still be 1 foot. Thoms 2.720699046 is a multiple of p (being the quadrant p/2 ´ Ö3), and the 1 foot part of the circle is to be found not in the diameter of course, but in the Short Side of the 30/60 triangle.]
If Thom's MY surfaces from his accurate measuring of dimensions in FEET, does it mean that Stoneage man measured in Feet? Would the quantum of 2.72 ft. have appeared if Thom had been measuring in Jewish cubits or any other unit?
To construct an Equilateral triangle from a Pitype triangle :
Divide the Base side by M/2 (1.360349...) and retain the same Vertical Height.
To construct a Pitype triangle from an Equilateral triangle :
Multiply the Base side by M/2 (1.360349...) and retain the sameVertical Height
Alternatively, the Vertical Heights can be similarly manipulated.
The latitude of the Great Pyramid is close to 30 deg North. Tangent of this latitude is .577350269.
5773.50269 inches = 481.1252243 feet = Pyramid Height
Pyramid Base = 755.7497351 feet = 9068.99682 inches. and 3 x 9068.9+ = 27206.99046
Did you realise that the square root of 20 million Pi equals Equator Diameter?
* Barleycorns. 1 Mile = 172800 33 ft. = 1080
(.3666") 1/2 mile = 86400 22 ft. = 720
1/4 mile = 43200 1 Pole = 540
1/8 mile = 21600
Similarly, the number of Barleycorns (.3675525" each) in the SS is the number of INCHES in the Quadrant.
Number of INCHES in the SS is the number of Meg Feet in the circle circumference.
(! MY = 3 Meg Ft., 1 Meg Ft = 10.8827962.. inches).
BUT ALSO: It is remarkably easy to work in Fractions of an Inch in the SS, and to have Fractions of a Megalithic Yard in the circumference.
For example : 25 7/16 MY in circumference would require an SS of (25 7/16) / 4 = 6 +1 7/16 ft.
= 6 ft + 3" + 21/16ths" (7/16 *3)
= 6 ft + 3" + 1" + 5/16"
SS = 6 ft 4" 5/16"
An SS of 91' 5 3/8" = (91 * 4 = 364 MY) in circumference
5" = 1 + 2/3 MY in circumference
3/8" = 3/24 MY
3/24 + 2/3 = 19/24 MY, total = 365 19/24 MY.
So one can never be sure that a nonintegral MY circumference was NOT intended, for the SS may well have been designed in Feet, Inches and Fractions of an Inch.
(Although such accuracy of fractional inches would be most difficult to achieve in the field).
Thus it is probable that the Megalithic Yard may never be proved....... but ..
THERE IS A "LINEAR" MEGALITHIC YARD
by using 22 / 7 as a substitute for Pi as in the diagram below.
( 11/7 * sqrt 3 = 2.72179... so 11* sqrt 3 = 7 * 2.72179...)
***
W.M.F.Petrie, 'Pyramids and Temples of Gizeh' 1883, gives the 1/2 base of the Great Pyramid as 115.117 mtrs ± 6.35 mm. which is 755.7501219 English feet for the base side at the upper limit.
A difference of .004628 inches from 100,000 MY / 360
(which base side if converted into a 60deg Equilateral triangle, would be 555.555.. ft.)

