432 
Proceedings of the Royal Society 
Required its radius : 
Now V. of sphere = 4 7 tR 3 
o 
7125000 - 4 wR 3 
O 
.-. R 3 - 7125000 
4 7 r 
3 
.-. R = V 7125000 
4-1887902 
= 119-371 
log. 7425-000= 6-8527849 
log. =0-6220886 
O 
3)6-2306963 
119-371 = 2-0768987 
But we are getting on too fast. Now in spite of the presence of 
7 r are we to suppose the circle squared practically, as we have 
imagined, when suggesting that the area of the base of a square 
pyramid might be represented by 7 tR 2 ? 
To seek an answer to that question we must go back to that part 
of our investigation, where we had reason to believe that the con¬ 
nection between 116-3 and 365-24 was intentionally introduced 
as an exponent of the relation between diameter and circumference, 
and we may not unreasonably test the accuracy of our deductions 
by finding the area of the circle there expressed, trusting that if 
we are working in the right direction this step may lead to some 
further proof of its being so. 
But in so doing we should use the figures only as a guide to the 
intentions of the Great Architect, and having as we believe learnt 
that the “ year” of inches symbolises a circle of 365-256, &c., we 
may take as our starting-point the more accurate diameter repre- 
n1 356-256 .... , 
sented bv -or 116-264 pvramid inches. 
J t r 1 * 
To proceed. 
The area of a circle whose diameter is 116-264 is 10,616*65. 
This number in itself does not seem peculiarly suggestive, but 
when we recollect how remarkably both the east wainscot and 
granite floor* point to an accurately marked square of 103 Pyramid 
* Viz. the east wainscot, a vertical line 108 inches high, and of the floor, 
a special portion constructed in granite showing a horizontal line 108 inches 
long. 
