The Lintel Circle

Professor Thom has given the outer circumference of the Sarsen lintels as 120 Meg. Yards, and the Inner circumference as 112.5 Meg. Yards.

Gerald Hawkins accurate survey, has 10.826 feet as the average length of the outer edge of a lintel stone, 3.28 feet as the average width, and 2.62 feet as the average height.

Thus Thom's measure would give a lintel stone as being 4 Meg yds. outer length, and 3.75 Meg Yds. as the inner arc. The 4 MY fits in with Hawkins, for it is 10.88279+ ft., the width, as determined from the figure below, is 3.24759+ ft.

The lintel stones are in perfect proportion to the inner and outer diameters, and once again the Ö3 ft. and MY measures fall into place, due to the inbuilt width of the lintel stones. This width, 3.24759+ ft., is the largest diameter circle which can be constructed within the outer edge of the lintel stone, and have the same circumference as the lintel stone inner arc.

First decide how many times the circumference is to be divided. Add 2 to this number and divide it into the outside diameter. And there you have it!

Try it, any circle, any number of segments. The result will be the diameter of the circle standing on an arc equal to it's cirumference.

Another 'quirk' of the geometry. If an SS is constructed in the lintel width circle, it will be 11.25 inches, reflecting the 112.5 MY in the inner Circumference. Indeed, any circle drawn with the same circumference as the arc it stands on is long, will have an SS in inches which reflects the number of MY to be found when that arc is multiplied by 30.

I have just acquired W.M.F. Petrie's book "Stonehenge: Plans, Description, and Theories.", where in Chapter 9, 'Sequence shown by Measures', he discusses a unit of 224.8 +/- .1 inches which he found mainly in the outer bank and ditch. Also, in his "Inductive Metrology", he has shown a prehistoric mean unit of 22.51 +/- .02 inches.

I need hardly point out that the sides of an equilateral triangle which can form the Lintel width circle diameter, is 22.5 inches., and thus the circumference is (22.5 * Ö3 * p) = 122.431... inches, = 10.202...ft., = 3.75 Meg Yds.

Whenever a diameter is divided by 30, the resulting SS of the circle within the lintel width will have an SS of the original diameter divided by 60Ö3, (my Stonehenge diameter), and this Lintel width SS * 120 * 2.720699046... is the original circumference.

If the main circle SS and the diameter divisor are equal, then the Lintel width will always be 2Ö3, and Lintel SS = 1

In 'The Measure of Albion' by Robin Heath and John Michell, p.78., the Sarsen diameter is given as 104.272 ft, and the lintel width as 3.47574857 ft. (one Royal Yard, diameter / 30), so Diameter / 60Ö3 = 1.003357788 ft., and this SS multiplied by 120 * 2.720699046... is the circumference of 327.58.... ( Incidentally, 1.00335778ft is 12.042+ inches, and there are 120.429.. MY in 327..58+ ft.)

Example:- Sector angle (above diagram) = 12 degrees, Area = 282.7433 sq.ft. = 10 Pi sq.yds

The geometry of the Lintel Circle is quite intruiging, and the interior geometry of the Sarsen circle can be drawn with just a straight edge and compasses.


	Megalithic Yard Unearthed \| home Megalithic Yard Surprises \| Was Thom misled? \| Woodhenge \| Stonehenge and Sarsen circle \| The Lintel Circle \| Hawkins' Photogrammetry \| Trilithon ellipse \| Megalithic Foot and Megalithic Inch \| Arcs, chords & sagitta \| Pi-Pyramid and the Megalithic Yard \| Map Scales \| Miles and the Equator \| Additions, Vara etc. \| Contact Me \| CALCULATOR The Lintel Circle (additions March 2005) Professor Thom has given the outer circumference of the Sarsen lintels as 120 Meg. Yards, and the Inner circumference as 112.5 Meg. Yards. Gerald Hawkins accurate survey, has 10.826 feet as the average length of the outer edge of a lintel stone, 3.28 feet as the average width, and 2.62 feet as the average height. Thus Thom's measure would give a lintel stone as being 4 Meg yds. outer length, and 3.75 Meg Yds. as the inner arc. The 4 MY fits in with Hawkins, for it is 10.88279+ ft., the width, as determined from the figure below, is 3.24759+ ft. The lintel stones are in perfect proportion to the inner and outer diameters, and once again the Ö3 ft. and MY measures fall into place, due to the inbuilt width of the lintel stones. This width, 3.24759+ ft., is the largest diameter circle which can be constructed within the outer edge of the lintel stone, and have the same circumference as the lintel stone inner arc. To find the Diameter of this largest circle:- First decide how many times the circumference is to be divided. Add 2 to this number and divide it into the outside diameter. And there you have it! e.g. 30 sections in circumference Plus 2 = 32 Overall diameter 103.923. Divide by 32 = 3.247595+ Try it, any circle, any number of segments. The result will be the diameter of the circle standing on an arc equal to it's cirumference. Another 'quirk' of the geometry. If an SS is constructed in the lintel width circle, it will be 11.25 inches, reflecting the 112.5 MY in the inner Circumference. Indeed, any circle drawn with the same circumference as the arc it stands on is long, will have an SS in inches which reflects the number of MY to be found when that arc is multiplied by 30. **************************** Confirmation of PETRIE'S measure of 22.5 inches. (28 Feb 2006) I have just acquired W.M.F. Petrie's book "Stonehenge: Plans, Description, and Theories.", where in Chapter 9, 'Sequence shown by Measures', he discusses a unit of 224.8 +/- .1 inches which he found mainly in the outer bank and ditch. Also, in his "Inductive Metrology", he has shown a prehistoric mean unit of 22.51 +/- .02 inches. I need hardly point out that the sides of an equilateral triangle which can form the Lintel width circle diameter, is 22.5 inches.,** and thus the circumference is (22.5 * *Ö3 p) = 122.431... inches, = 10.202...ft., = 3.75 Meg Yds. **************************** Whenever a diameter is divided by 30, the resulting SS of the circle within the lintel width will have an SS of the original diameter divided by 60Ö3, (my Stonehenge diameter), and this Lintel width SS * 120 * 2.720699046... is the original circumference. If the main circle SS and the diameter divisor are equal, then the Lintel width will always be 2Ö3, and Lintel SS = 1 In 'The Measure of Albion' by Robin Heath and John Michell, p.78., the Sarsen diameter is given as 104.272 ft, and the lintel width as 3.47574857 ft. (one Royal Yard, diameter / 30), so Diameter / 60Ö3 = 1.003357788 ft., and this SS multiplied by 120 * 2.720699046... is the circumference of 327.58.... ( Incidentally, 1.00335778ft is 12.042+ inches, and there are 120.429.. MY in 327..58+ ft.) ADDITIONAL GEM (February 2004) IN THE OUTER SARSEN CIRCLE (radius 51.1962423) :- ANY SECTOR AREA in MEG.SQ.YARDS = SECTOR ANGLE x 10 / Pi Example:- Sector angle (above diagram) = 12 degrees, Area = 282.7433 sq.ft. = 10 Pi sq.yds Sector angle (12) x 10 = 120, / Pi = 38.1971863 sq.Meg.Yds. or Sector angle (1) x 10 = 10, / Pi = 3.18309 sq.Meg.Yds. Rem 1 sq.Meg. Ft, = 10.88278 x 10.88279 sq.inches = .822467 sq.ft 1 sq.Meg Yd,. = 2.72+ x 2.72+ sq.ft. = 7.402203 sq.ft. A Sector of 1 sq.Meg.Yd. has a sector angle of Pi / 10 degrees. (.3141592) The geometry of the Lintel Circle is quite intruiging, and the interior geometry of the Sarsen circle can be drawn with just a straight edge and compasses. Hugh Franklin February 2003 & Feb 2004 & Feb 2005 Home Back Next Top