Updated: Sep 16
Euan MacKie published this table in his article "A New Look at the Astronomy and Geometry at Stonehenge":
He observed that a Druisian Foot of 0.333 metres fits remarkably well into the design of the Station Stone Rectangle and Heel Stone Triangle at Stonehenge. He also made the very interesting observation that the short side of the Station Stone Rectangle doubles up as the side of an octogon which fits perfectly in the Aubrey Circle. The Station Stone Rectangle is a 5:12 rectangle, and the large triangle with the apex at the point to the side of the Heel Stone and the base as the furthest Station Stone Rectangle length is a 10:13:13 triangle, which makes sense of the site. The proportions of a 10:13:13 triangle are quite interesting in themselves. Area = 60, perimeter = 36, height = 12, inradius = 10/3, apex to centroid = 8. But the way this goes together with the 5:12 rectangle is surprising. The line drawn from the base angle of the triangle to intersect the opposite side of the triangle at a right angle also passes through a station stone. For example, a line from Station Stone 93 to Station Stone 91 meets the side of the triangle at a right angle.
So can we go with a Station Stone Rectangle that measures 80 x 33.3 metres? As Jim Alison pointed out on GHMB:
MacKie gave 3 measures for the short sides of the station stone rectangle, and 4 measures for the long sides. Assuming that the station stone rectangle was intended to be rectangular, with the short side - long side proportion intended to be 5/12, we can take all seven of these measures to obtain an averaged length based on MacKie's survey data. The three lengths given for the short sides are 34.17 m, 32.70 m, and 33.23 m. The four lengths given for the long sides are 79.93 m, 79.78 m, 80.26 m, and 79.75 m. The four long side lengths add up to 319.76 m. The three short side lengths add up to 100.1 m. 100.1 m x 12/5 = 240.24 m, and 240.24 m plus 319.76 m = 560 m, divided by 7 = 80 m, exactly. Perhaps this is why MacKie says the Northern foot of .333...m is a better match for Stonehenge than the MY of 2.72 English feet. On the other hand, it is exactly the same perfect match for a megalithic yard of .333... m x 2.5, or 45 digits.
and Jim also pointed out:
MacKie gives the surveyed lengths of the sides of the heel stone triangle as: 39.97 m (Atkinson) - 95.97 m (Ranieri) - 104.00 m (Thom). In the case of the station stone triangle, all three sides add up to 240 MY, while in the case of the heel stone triangle, all three sides add up to 240 m. Argh.
80 x 100/3 (ancient) metres or 4320 x 1800 digits of 0.729166667". This gives an area of 7,776,000 square digits, or 8,000/3 square metres.
A circle with a diameter of 80 metres has, with pi as 22/7, a circumference of 9,900", or 750 Saxon feet of 13.2", which is also 7 x 9,900/7, so 7,000 x approximately the square root of 2 inches in the circumference.
The 104.00 metre side between the Heel Stone and Stone 92 is very intriguing...
If the Heel Stone to Stone 93 were the same length, also 104.00 metres, then the Heel Stone - Stone 92 - Stone 93 triangle would definitely be isosceles. Can it still be considered isoceles? It's not perfect, but almost. Is this because the axis of the triangle isn't supposed to go through the Heel Stone itself, but between it and a missing companion stone?
This is what the rectangle and triangle above might look like on a plan of the site:
I had a go at trying out the length and width of the Station Stone Rectangle to see if the dimensions linked up to Phi. In digits of 0.729166667", if you take the width and length to be 1,800 and 4,320 respectively, and use Phi as 55/34:
(1,800 + 4,320) x (55/34) = 9,900, echoing the circumference of a circle with a diameter of 80 metres, with pi as 22/7.
In metres, the width and length are 100/3 and 80.
((100/3) + 80) / 70 = 34/21 = 1.6190476
Is the length divided by Phi squared the diameter of the circle that passes through the centre of the Sarsen Stones?
Using Phi squared as 55/21 and pi as 22/7, that gives a diameter of 315 imp. feet exactly, or 324 x 11.6666", or 259.2 x 14.583333" or 96 ancient metres.
3150/12 x 21/55 x 22/7 = 315
Harry Sivertsen gives 316.751 feet for the circle passing through the centre of the Sarsen stones, Robin Heath 316.8 feet.
Another 288 connection: The Heel stone - 93-92 triangle, with sides 104, 104 and and 80 metres, would have a perimeter 288 metres, and a height of 96. The width of the Station Stone rectangle taken as 100/3 metres goes into the 96 metre height 2.88 times. (and the Aubrey Circle is 288 feet in diameter according to Hawkins.)
Jim Alison came up with the idea of a rhombus to try to explain the discrepancy between the diameter of the lintel circle of 100 Roman feet, or 1600 digits, versus the short sides of the station stone rectangle, of 40 MY, or 100 Northern feet, or 1800 digits., as per below:
So if the '1' side of the rhombus triangle is 40 metres, the two other sides are 120 and 126.4111 metres, the area of the triangle is 2400 square metres, and of the rhombus made up of four of these triangles 96 square metres. The hypothenuse of this triangle (in orange in the diagram) intersects the width of the Station Stone Rectangle at the 2/5th part. This matches Robin Heath's Lunation Triangle, the line drawn from the apex of where the 12 and 13 sides meet, down to the base, at the 2/5th or 3:2 point, creates a line 12.369 units long, corresponding to the average number of lunations in a solar year, 12.3683.
(The rhombus triangle has a smaller angle of 18.435 degrees, and if you look at the triangle made up then by a section of the hypothenuse of the rhombus triangle, half the length of the station stone rectangle, and a section of the width of the rectangle, using this same angle, the shorter side is 13.33337 metres, which is 80 /6). Curiously, the smaller triangle, in turquoise in the diagram, is also a 1:3:√10 triangle. Four of these turquoise triangles together have a combined area of 1,066.666 square metres, or 7,776 square remens. (compare with the 7,776,000 square digits for the area of the Station Stone Rectangle)
It's worth noting briefly that the famous Station Stone Rectangle is not a perfect rectangle in fact, or at least, it depends on the way you measure it out. There is scope, with these large stones, to allow for a perfect rectangle, but also to dispute the idea. Here is a diagram I made based on the angles given by Gerald Hawkins in his addition to Flinders Petrie's book on Stonehenge. You can see the angles of the rectangle are not, according to Hawkins, right angles. This does not mean that they weren't set out to be however, and the geometry that arises from the presence of a 5:12 rectangle is amazing. The line going from the Heel Stone to the mid point between Stones 92 and 93 isn't quite at a right angle with the Station Stone Rectangle either. The centre to the Heel Stone is azimuth 51.13° degrees, according to Hawkins, and the azimuth from Stone 91 to Stone 94 is given as 319.83°.
319.83 - 180 - 51.13 = 88.7, not quite 90.
Or the same goes for Stones 92 to 93, given as 320.19°. 320.19 -180 - 51.13 = 89.7, almost but not quite 90.
But maybe this is splitting hairs, as the rectangle itself isn't perfectly rectangular if the azimuths are anything to go by. (The azimuth for Stones 92 to 93 minus the azimuth for Stones 94 to 93 gives: 320.19 - 230.63 = 89.56)
If we just go with the hypothetical 13:13:10 isosceles triangle for the large triangle that links up the Hell Stone (or the point between it and a missing cmpanion stone) and the two corners of the Station Stone rectangle that are furthest from it, with sides 104, 104 and and 80 metres, then there are some interesting results, like a perimeter of 288, and a height of 96. The Aubrey Circle is 288 feet in diameter according to Hawkins.
Gerald S. Hawkins notes that:
"The Aubrey Holes vary from 2.5 to almost 6 feet in width and between 2 to 5 feet in depth and were steep sided and flat bottomed. Although irregular in shape, there was little irregularity in their spacing. They formed a very accurately measured circle 288 feet in diameter with a 16 foot interval between their centre points. The greatest radial error was 19 inches and greatest circumferential or interval spacing error was 21 inches. Let it be noted that such accurate spacing of 56 holes around the circumference of so large a circle was no mean engineering feat."
Stonehenge Decoded (1965)
Thom I think gives 285.60 feet for the diameter, though.
For Jim Alison, the Megalithic Yard is 45 digits of 0.729166667", or 5/6 of the 54 digits that make up an ancient metre of 39.375". Jim noted that the perimeter of the station stone triangle is contained 200,000 times in the circumference.
(using 24,857.954545 miles and 54 x 0.729166667")
Jim Alison also had the idea to draw four circles, each centered on a corner of the Station Stone Rectangle, with a radius equal to the width of the rectangle, so according to Mac Kie, 100/3 metres.
He wrote in his post:
It occurred to me that a vesica pisces with short sides of the station stone rectangle being a common radius may explain the somewhat horseshoe instead of circular layout of the trilithons. In the drawing below, the two trilithons closest to the Heel stone are centered on the diagonals from the station stone rectangle, perpendicular to diagonals, at the point where the station stone vesica pisces circle intersects the diagonal. The next two trilithons from the Heel stone are also perpendicular to the diagonals, but the edge of the trilithons are on the diagonal, rather than the center. The back Trilithon is centered on and perpendicular to the site axis, and the back of the trilithon is on the same circle drawn on the intersections of the vesica pisces station stone circles and the diagonals of the rectangle. This has the effect of equalizing the distance from each trilithon to the next, and also creating the horseshoe effect, because of the two trilithons that are adjacent to the diagonals, rather than being centered on them
These circles are very interesting in themselves too. The area of each circle, with pi = 22/7, is 3,492.06349 sq metres, or 5,414,062.5 sq inches.
The area of two of these circles is exactly 55/21 (Phi squared) times the area of the rectangle.
(100/3)² x 22/7 x 2 = 55/21 x 80 x 100/3
I guess this might be a property of a 5:12 rectangle, though it depends on specific approximations of pi and Phi.
Is 100/3 metres divided by Phi relevant somehow?
100/3 x 34/55 = 20.6060606
100/3 x 55/89 = 20.599251
Maybe a link to the calendar?
365.242199 / (12 x 29.53059) x 200 x 55.34 = 333.4582, close to 1000/3
Or to Egypt? In the GP King's Chamber, the mean values of North length, east width and south length, added together and multiplied by 0.2 give the base width of the West width, 206.16". The width of the chamber is close to 1,000 / (3 x Phi) inches. To a couple of inches error, using the mean values, width + length ≈ 1,000 / φ
Using mean values for N and W, width + length + (width + length)/ Phi = 1,000.55
(width + length + (width + length)/ Phi)) / Phi ≈ width + length
Also, half of the floor width, 103.06 inches, plus the value of the height in inches, 230.09, makes 333.15, very close to one third of 1,000.
And, 206.12 x 1.618 = 333.50216
The combined values of the height and half the floor width make a figure close to the value of the width of the floor in inches multiplied by Phi, so if you were to draw a line of 333.15 inches along the longer wall, adjacent to the length of the chamber, from floor to ceiling, the remainder would come to precisely the middle of the floor. 333.50216 - 230.09 = 103.06, and 103.06 is exactly half of 206.12 inches, the width of the King's Chamber.
If the diagonal of the Station Stone rectangle is 13 x 80 / 12 = 260/3 metres, the area of a circle with the diagonal for its diameter is 5,901.587301. Dividing this by 13² and multiplying by 100 gives 3,492.06349 sq metres, the area of the smaller circle.
The circumradius for the smaller triangle created by the smaller section of the Heel Stone Triangle, to the left in the image above, is 40 metres, exactly half the length of the Station Stone Rectangle. This could reflect the diameter of the moon in a roundabout sort of way: 40 metres are 2,160 digits of 0.729166667" and the moon's diameter is approx. 2,160 miles. Or it could reflect the earth's circumference of 1,575,000,000 inches, as 2160 digits are 1,575 inches. (or 40 ancient metres are 40,000,000 metres divided by 1,000,000)
I was intrigued by the presence of the square root of 17 there: the line between the base of the Heel Stone triangle and the 2:3 point of the width of the rectangle (the Robin Heath line which expresses the average number of luantions ina solar year). Why is it 20 x √17 metres?
The triangle created by this line as hypotenuse, in purple above, has an area that is a third of the Station Stone Rectangle's.
The diagonal of the Station Stone Rectangle is central to Robin Heath's Lunation triangle idea, whereby a trianlgle made up of the legnth, width and diagonal of this rectangle is of special significance in terms of the moon and also the position of Stonehenge in the landscape. This triangle is a Ythagorean one: 5:2:13. The line drawn from one of the non right angles of the triangle towards the 2:3 point on the opposite side is, in its proportion to the rest of the triangle, exactly right to represent the average number of lunations in a year: between 12 and 13.
I keep wondering about the square root of 17 I came across in the length of the line drawn from a Statio nStone to the opposite side of the rectangle, at the 2:3 point, the line that Robin Heath has as his lunar year / solar year line. If you go with 100/3 and 80 metres for the size of the rectangle, this line is √17 x 20 metres. I came across something called a spiral of Theodorus, that Plato talks about. Like the metre, it fits surprisingly well. And √17 only became apparent to me once I looked at the site in metres (though of course these proportions don't depend on metres in any way). The base of the triangle, which is the length of the rectangle, is √16 x 20 metres. The 20 metres corresponds to the two thirds of the 33.3333 metre width of the rectangle. So this is taken as the base unit, and the spiral built around it, finishing up with its last hypothenuse precisely at the two thirds point of the width of the rectangle.
This spiral seems to fit with the long side of the Heel Stone - Station Stones triangle, and also with the Station Stone rectangle itself, not only with the lengths of the hypothenuses but also with the angles involved. What do you think?
Howard Crowhurst's triple square diagonal is the hypotenuse of a 1:3:√10 triangle, which is one of the triangles in the spiral. The same goes for his double square and quadruple squares, which appear as 1:√4:√5 and 1:√16:√17 triangles in the spiral. He also has a right-angled triangle with sides √10 x 100.0037 feet, 235 metres and 245 metres. ((Carnac the Alignments, Volume 1, page 51)
Below are two such spirals, centered on the two station stone pints furthest from the Heel Stone, and in the image below that, there are four spirals, each one centered on a corner of the Station Stone Rectangle.
So a few thoughts on the geometry at Stonehenge!