Trilithon Ellipse page added June 2002
Hawkins Photogrammetry added 7 Sept 2002
Megalithic Inch added Jan 2003
Lintel Circle added Feb 2003, Additional Gem Feb 2004
Updated June 5 2003 with VARA note to Additions
Updated Feb 3 2005 (Interesting maths,Miles/Equator page).
* Feb 2006 . PETRIE'S 22.5 inch Measure, on LINTEL page
"...... It would be ironical if, after all, the linear measure used in stone circles was akin to the Imperial Foot."
Aubrey Burl. "Stone Circles of the British Isles" .1976.pp.72
If Professor Thom in his study and measurements of Stone Circles was using a tape measure marked in Imperial Feet, then it should be no surprise to find a unit equivalent to his stated 2.72 ± .003 feet.
This site aims to show that there is such a measure to be found naturally when one examines circles and their allied geometry. If the results of the following exercises can not be accepted as an origin for the Megalithic Yard (or for that matter Imperial measure as well) at least they can form a basis for practical geometry lessons in schools!.
First, draw a 30°/60°triangle. That is, one having a Short Side (SS) of one (1) unit, and a hypotenuse (Long Side LS) of two (2) units. Now use the third side as the radius and draw a circle.
The relationship of the length of a quadrant of that circle to the Short Side, is 2.72+ : 1
Alternatively, 4 x the units in the SS is the number of 2.72+ lengths in the circumference.
It is not necessary to know length of radius or to know p
Thus any circle or even horizon, can be divided into any number of degrees or equal divisions, by using if you wanted to, any short length of stick or chord as an SS. (Why not a Barleycorn?). Find it's quadrant equivalent and extrapolate by sighting. (see diagram below), but if your stick is one (1) ft. long, then magic happens!
e.g. SS = 1 ft. Circ ( in Meg Yards) = SS ´ 4 = 4 MY
SS = 3 inches. Circ = .25 ft ´ 4 = 1 MY. (or Inches/3 = MY in circumference).
Stonehenge Circumference = 120 MY, thus SS = 120/4 = 30 ft.
The Area of a square drawn on the SS, is 1/6th of the Area of a square inscribed in the circle.
M = 2.72+ MY = 2.72+ feet.
e.g. Say you have a huge circle of unknown circumference on which you would like to place 99 evenly placed stones.
1. First (after finding the centre of that circle) construct a circle of C = 99 MY., based on a triangle with an SS of 99 / 4 which thus must have an SS of 24 3/4 ft., and a hypotenuse (LS) of 49 1/2 ft..(Only a small arc is needed)
2. On this arc lay out your standard MY (derived from an SS of 1 ft. or that short length of stick!), and at each end of it place a sighting post.
3. From the centre of the circle, extend sight lines over the posts to the outer circle and insert two posts there.
4. Stepping the chord of these outer posts around the unknown circumference, will divide it into 99 equal divisions.
The above diagram shows larger and smaller circles divided into 360 divisions. If the circle to be divided is not too big, then string/rope is better than sighting posts.
How many Sun diameters fit into the Horizon circle?
Set two posts 1 Meg yard apart, facing toward say where the sun sets. View the sun between the posts and walk backwards until the edges of the sun disc exactly touch both posts. You are now at the centre of a circle , radius you to the posts. Draw out half of this circle and construct the 30°/60° triangle. Short side will be found to be180ft. and the Hypotenuse 360 ft. Thus the circumference of the circle will be 720 Meg yards. Answer 720 sun diameters.
Quadrant of 1 inch has an SS of 1 / 2.720699+ inches, or .3675525+ inches, which looks pretty much like a Megalithic Barleycorn. Three Barleycorns SS produces a circle of 12 Inches. (3 MegBc's = 1.102657791 Inches. Compare the inch of the Northern or Saxon Foot, 1.1 of the current inch.)
Barleycorns. A stock item in any Neolithic farmer's armoury. So there is no need for a Megalithic centre dispersing Megalithic Yard-sticks to all parts of the realm. Barley corns in the SS lead to the Inch and foot, which used in turn as an SS produce the Megalithic foot and Megalithic Yard.
Another surprising relationship, is that when a regular (60°) Tetrahedron is inscribed inside a Sphere, it's surface area to that of the surface area of the sphere, is also 1 : 2.72069904
(say the Tetrahedron surface area is 4 sq.units, then the sphere surface area is 10.8827+ sq.units).
Jan 2002, site updated April 2002 and June 2002, Feb 2004