The following letter was sent from Br. Morton Edgar to Mr. Bell, who, it is believed, was Br. Richard Hill’s grandfather.

This letter was found in a copy of David Davidson and H. Aldersmith’s book entitled "The Great Pyramid-It’s Divine Message", published in 1924 by Williams and Norgate, Ltd, London.

Br. Rodney Hugelman owns this book. He obtained it from a library that was willing to sell it. He was particularly interested in this book because it has several notes in the margins by Morton Edgar. Apparently, this book belonged to him and was sent to another (perhaps to Mr. Bell?) with these rather critical notes in the margins. Most of these marginal notes bear Br. Morton’s initials or his name, to indicate their authorship.

27 JUL 1933 Morton Edgar 27 Aytoun Road, S.1.

Glasgow, Scotland Dear Mr. Bell, I have been going over some of Mr. Davidson’s figures in his diagrams on his page 198, Plate XXXVIII, particularly Fig. B, the centre one of that Plate.

I do not question the proportions of that diagram, for it is merely one of many such proportionate agreements to be found in the Great Pyramid.

What I want to know is how he gets his figures for determining the angle of the passages in the Pyramid. He says that this angle is 26º 18’ 9".63, whereas to demonstrate the correct proportions of this diagram the angle should be 26º 18’ 9".725+ Can you determine which of these is correct?

Because Davidson takes a slightly greater day-value for the Solar Tropical Year than I do, and also a very slightly greater inch length for the "Pyramid Inch", his dimensions for the various parts of the Pyramid (which are largely related to these two factors) are necessarily greater than the dimensions that I understand to be correct, theoretically considered. But the proportions are the same, no matter what the actual dimensions.

I here forward you a diagram that demonstrates the same method of determining the passage-angle, using another proportion, namely, a Square and a Circle of the same area, -in this case 144,000 multiplied by 10. This is really the same method as used by Davidson in his diagram B, only instead of taking the hypotenuse (A-T) of his right-angled triangle as representing the diameter of a circle (the circle necessarily being of the same area as the square PQRS), he makes it the Radius of the Year-Circle.

In other words, the Square with a side-length equal to 10 times the solid cubic diagonal of the King’s Chamber (P-R), has an area that is equal to the circle of which the vertical height of the whole Pyramid is the diameter.

The same thing is found by another proportion: If the side-length of the Square is the same number of inches as there are days in the year, and we find the area of this square, and draw a Circle of the same area, then the diameter of this circle is equal to the length of the King’s Chamber.

In all cases, if the diameter of the equal-area Circle is used as the hypotenuse of a right-angled triangle, and half the side length of the Square as the Perpendicular, the resultant angle from the base-line is the same as the Passage-angle, which I believe is correctly stated as 26º 18’ 9".72515+. Can you verify this?

The formula for the Sine of the Passage-angle is: The square-root of Pi, divided by 4. While the formula for the Sine of the Casing-stone angle (called the "Pi" angle) is: 4, divided by the square-root of (the square of Pi, plus 16).

These two angles are thus directly connected (The Tangent of the Casing-stone angle is: 4 divided by Pi. While the Tangent of the Passage-angle is: The square-root of Pi, divided by the square-root of (16 minus Pi).

On page 308 of my Vol. II "Great Pyramid Passages" I have presented the trigonometrical values *{of}* the Passage-angle, 26º 18’ 9".725+.

With regards, Yours sincerely,

Morton Edgar