Tuesday, 17 June 2008

Appendix: Calculation of Pyramid-Related Positions

“Right then,” I said. “So the area of a square is just the side length squared, unsurprisingly enough. So side 2 gives area 2 x 2 = 4. So if the outer square had a side 2, then the inner one, if it had an area half that of the outer one, would have an area of 2, i.e. half of 4. So its side length squared would equal 2, in other words it would have a side length of the square root of 2, which is [tap, tap, tap] around 1.414.
“So we want to check if the diagram shows two such squares?” said Santina. “Where the lengths are in the ration of 2 : 1.414, yes?”
‘That’s it. So, what is the size of the smaller square? What we have is a half the diagonal of the smaller square, that is to say from the centre to the corner, being equal to the radius of the circle, do you agree?”
They looked at the diagram again, saw how the radius of the circle was indeed half the diagonal of the smaller square, and then agreed.
“And you agree that the radius of the circle is also half of the side of the larger square?”
Again, after looking at the diagram for a moment, and as you can see for yourself, they agreed that yes, the radius of the circle was indeed equal to half the side of the larger square.
“So we can work out the relative sizes of the two squares. Half the diagonal of the smaller square is half the side length of the larger square. So before we imagined that the larger had the side length of 2, so that the smaller would have the side length of 1.414. So if this circle radius is half the length of the large square, and the larger square is side 2, then this radius length will have the unit value of 1. With me so far.”
They nodded.
“Right so the full diameter of the circle is 2, and the area of the larger square is 4. What then is the side of the smaller square?”
“Um…” said Laura.
“Well,” I said, putting on a somewhat teacherly tone, “what do we know about it?”
“We know that half of its diagonal is 1 unit long,” said Santina.
“Right, so how can we work out the length of the side?”
They weren’t too sure.
“We can use Pythagoras’ theorem for right angle triangles,” I said. “So if half the diagonal is 1, then the full diagonal is 2. So we have a right angle triangle with two sides of the length of the side of the smaller square, and the hypotenuse – the longest side of the diagonal – is length 2. Clear?”

“Perfectly.”
“Good. But let’s half the size of that so we can work with our unit length, and then we can double it up again later. So, call half the side of the smaller square s, and then we know by Pythagoras’ theorem for right angle triangles that s squared plus s squared is equal to the square of 1. 1 squared is just 1 x 1 = 1. So 2 x s squared = 1.
Divide both sides of the equation by two, and then we have s squared = ½.
So s = the square root of 0.5, which is [tap, tap, tap, tap] 0.7071.
But remember we halved the size of the triangle, so let’s double it up again. s is half of the side of the square, so the side is that times 2, and 0.7071 x 2 is 1.414. Good. This is what we were looking for. To find the area of this smaller square, we square its side length, and squaring 1.414, as we have already seen in reverse, gives the value of 2. So the area of the inner circle is 2, and the area of the outer circle is 2 x 2 = 4, so, yes, the inner one does have an area that is half of the outer one, and our original theory was surely correct.”



Laura said “And what was that theory again, just to remind me?”
Santina said “I said that I was sure that the 50th course, referred to in the clue, was the height on which the King’s Chamber within the pyramid was placed. And then Luca said that in plan view the area above the height of the King’s Chamber and that below are equal. And along with the clue was a diagram, which Luca has now shown by maths to be a diagrammatic statement of the same thing.”

“Cool!” said Laura. “I got all that. I’m not sure I could repeat it all back to you, but I got it. Which I’m quite surprised about as I was never much of a master at maths at school.”
“The link between the mind and motivation is amazing,” said Santina. “It’s surprising what can be achieved when you are genuinely interested in something.”
“Only one thing left to do then,” I said, “We have to find the latitude of the top and bottom of the 2 by 1 rectangle of southern Britain, and then check that the Long Man – which must be the giant with the poles in the poem – is at this height up the pyramid plan.”
“Let’s do that then,” said Laura.

This I did, as I will detail a little further down, and it turned out to be true to a very precise degree – as precise as it is possible to be, as I then informed the other two. So we had made definite progress, and basically solved this current clue, but as for what this was all about in more general terms, well it all still seemed quite obscure to us, but little did we didn’t realize then how the next stages of unraveling would both explain and confirm the pattern.

Said Laura “But what about these official’s poles that he has in his hands? And what is this business of He-who-is-in-his-Ka?”

Laura’s question was not something we were able to answer at that time, and it wasn’t essential to finding the next clue, but at a later time we did find the solution to this, as you shall see.

Unbeknown to me at this time, Jenny had now arrived in Canterbury and was at this moment making her sweet way towards the Cross Keys Inn, occasionally trying her mobile to see if I had yet switched on my phone. We, however, proceded to check out, pick up our packs and walk off down the street. That I didn’t see her as we passed, which we must surely have done, can be put down to the fact that I had no reason to believe that she would be in the area. That she did not see me must have had some more elaborate reason – perhaps she stopped for a brief while to look into a shop window. Anyway, the upshot was that we passed by, traveling in opposite directions. She, upon arriving at the inn, learnt after some initial inquiries that we had checked out and left with our packs. We, arriving at the station, learnt that the train was cancelled and another not due for an hour. She, upon hearing the sad news of our departure, ordered herself a g’n’t from the bar and sat down to consider her options. We sat on our packs on the platform mulling over the next part of the clue about the giant.

“Luca,” said Santina, “you never did explain how you found that the Long Man is at the latitude of the King’s Chamber.”


“OK, so to show that the Long Man is at the height of the King’s Chamber up the pyramid plan, we find out the latitude of the point where the 2 by 1 diagonal reaches the sea in the far west of Cornwall, to give us the base latitude, 50’07’’. Next we do the same for the location where it meets the sea in the far east at Lowestoft, giving us the top of the rectangle, at 52’29’’. We convert these to decimal figures: 50.117 and 52.483, and then we find the difference, to give us the height of the rectangle in degrees of longitude. (The distance of a degree of latitude stays more or less the same as you get further North or South, unlike degrees of longitude.)”

“Sorry about this,” said Laura, “but what exactly is longitude and latitude?”
I said “Latitude rings go around the world parallel to each other, and measure how far north or south you are, and longitudes run perpendicular to these, to measure how far east or west you are.”
“Thanks.”
“So this gives us the height of the 2 by 1 rectangle: 2.367 degrees of longitude. The ratio of the sides of the outer and inner squares in the diagram is 1.414 / 2, which is 0.707. This measures off 1.673 degrees down the rectangle, which finds the latitude of 52.483 – 1.673 = 50.81.


Converting back to degrees and minutes this gives 50’49’’ degrees. And the latitude of Wilmington, according to the index of places in the Times Atlas of the World, is precisely this: 50’49’’, that height up the pyramid where the area of the faces above and below is equal.”
“I believe you on the maths,” said Laura, “but I don’t quite see how just by applying the same ratio you get the…oh hang on, yes, no it’s ok. I can see it now. The ratio of the height up a triangle at a place on its hypotenuse, or whatever you call it, is the same as the ratio of the horizontal distance along the…thingy.”

LUCA: Exactly.

Jenny sat for some time trying to reach me on the mobile and then phoned the boss who concurred with her feeling that it was time to give up and head for home. What else was there to do?

We boarded the train and sat waiting for it to depart. Shortly before it did so a somewhat dejected features photographer for the quarterly journal Wessex and Weald boarded a few carriages behind us, and plonked herself down in a seat.

Some twenty or so minutes into the journey I felt the call of nature, and doddered in lurching zigzags down the train towards its front, but reached a dead end before I reached a toilet. So I then zigzagged back in the other direction until I did find a toilet. I pressed a button the size of a puma’s paw and the door slid slowly open like something off Star Trek. In I went, pressed another large button to close the door, and in some haste lowered those parts of my clothing that would have hindered my purposes while sitting at stool. Looking up to the wall it then came to my attention that there were in fact two of these large buttons, one marked DOOR and the other LOCK. I realized that simply pressing DOOR to close it was not enough, even though I could think of no earthly reason why anyone would want to close the toilet door from the inside and not also lock it. I, personally, decided I would rather not be disturbed, so without further ado, I stood up, shuffled a couple of paces forward, and pressed the LOCK button. To my very great surprise, the result of this action was that the door began to open. With one hand I attempted to halt its relentless motion, (but I may as well have tried to stop the march of time and the tides themselves), as with equal ineffectiveness my other hand groped at the undergarments around my ankles in an attempt to lift them. And still the door kept opening.

Suddenly there in the open doorway was Jenny Love-Interest.

LUCA: Jenny!

JENNY: Luca!

LUCA: This is a turn-up, and no mistake!

JENNY: That’s no turnip!

Shielding my modesty with my hands, I stood up and attempted to explain.

LUCA: Hi, this was just…I didn’t mean to…I tried to lock it but the stupid button…been to Canterbury…looking for the pyramid and stuff…how have you been?

Jenny just smiled, pressed the outside button to close the door, and said, quietly and calmly: “I’ll speak to you when you’re finished.”

LUCA: Oh yes, good idea…speak in a mo.

In somewhat more refined circumstances, Jenny and I spent the rest of the return journey discussing excitedly the possibility of the trip to the sunny south to do the feature for the journal. I became entirely convinced of her happiness at the thought of spending this time in my company, which was a very pleasant affirmation of my hopes. I also attempted as best I could to explain the situation regarding the story about the inventor – that there was no such inventor nor invention and that I was in fact involved in research with a couple of colleagues – to whom I introduced her - research into a subject which at this stage it would be difficult to explain to the boss.

LUCA: So if you could, please, tell him that you didn’t manage to find me – I’m gonna a need a bit more time off work.

JENNY: No problem.


[And that was one of the good bits! But we’ll need to fast forward in my “novel” to the next bit of maths.]

“That’s not all,” Santina continued. “We did a bit of research. In Gods and Graven Images : The Chalk Hill Figures of Britain, by Paul Newman, is a great confirmation. Seen in the low winter sunshine of January ’69, Newman tells us, (and again in July ‘76 after a period of draught) was a 150 foot dog placed on the North side of the giant, that is to say to his right, or our left, as we look at him. In other words there was a dog placed just where Orion’s hunting dog, the constellation Canis Major, is located in the sky, to the left of Orion as we look at him (his right side as he faces us).”
“Crickey!”
“And get this,” Santina went on, “Rodney Castleden, whose work on the subject is referred to in Gods and Graven Images points out that in Petit Sainte Grail (c.1200) Peredur, the hero of the story, is sent to a mound beneath which is carved a figure of a man. And at Cerne Abbas we have a giant on the hillside, and above him on the summit of the hill is an ancient mound.”
“Lawks! Do you think the mound contains the Holy Grail?”



“It’s possible,” said Santina, “in the sense of something that represents it; represents the blood of the gods. But quite apart form anything else Castleden’s reference invalidates the oft’ repeated maxim that there are no references to the Cerne Abbas Giant from more than about 300 years ago. Added to that, an early account mentioned in John North’s book Stonehenge speaks of a cult of a god called Helith at Cerne Abbas, also at a much earlier time, which has been taken as meaning ‘man’ or ‘hero’.”
“…and the location is definitely correct?”
“Spot on,” I said. “Remember the latitude of Wilmington?”
“It’s on the tip of my tongue.”
“50’49’’. The latitude of Cerne Abbas – 50’49’’. Exactly the same. And we’ve verified the longitude co-ordinate as well. You can look at my working if you like:-

In an appendix in the back of Gilbert and Bauval’s Orion Mystery is a diagram showing measurements and a line extended from the Orion shaft so that it touches the tip of the top of the Queen’s Chamber at the exact central axis of the Pyramid at 53 royal cubits from the base, while the full height is 280. So the fraction of the full height is 53/280. Now we can work out how far this point is below the floor of the King’s Chamber. No figure is given for this in the Orion Mystery diagram, but we can work it out since we have calculated the latter as a fraction of 0.683/2.367 of the full height. In royal cubits this would be 80.79. So the distance between the tip of the roof of the Queens Chamber and this horizontal is 80.79-53, which is 27.79. Since the Orion Shaft is angled at 45 degrees, we can see that the distance horizontally from the vertical axis of the intersection point of the shaft and the King’s Chamber horizontal is the same as the vertical distance down to the top of the Queens Chamber. In a decimal longitude measurement this is 27.794/220x3.74=0.4725. The longitude of the central vertical in decimals is 1.99’W, and 0.4725’ west of here is 2.4625’. Converting to degrees and minutes this is 2’28’W. The Times Atlas gives the longitude of Cerne Abbas as 2’29W. Once again it is a mere 0’01’ away. The combination of such a close match both for the latitude and the longitude figures are pretty impressive, considering that not just any site is suitable for a large chalk hill figure.


“Splendid!” said Laura. “I can’t wait to go there!”
“And there is the name as well,” said Santina. “‘Cerne’ is very similar to ‘Herne’ as in Herne the Hunter, a hunter of the English oak forests who had two hunting dogs and would appear at night. In other words Herne was Orion, and again the two words are sonically very similar. ‘Herne’ / ‘Orion’.”
“So the line of the fiftieth course is actually the line of latitude that joins the two Albions,” said Laura, “the ancient chalk giants of the southern British downlands. I’m blown away, to be honest. Well I must say this is all really highly intriguing.”
“I know,” I said, “fascinating isn’t it? I’m rather concerned it may prove a distraction from less important things.”
“Don’t be daft! That reminds me: do you like my highlights?”
“Lovely.”
“It’s better shorter too,” added Santina.

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