
Re: Ariadne's thread
Posted:
Jan 19, 2012 2:35 AM


RMP 46 and 47
Two square containers measure
10 by 10 by 3 '3 royal cubits 10 by 10 by 5 royal cubits
I am assuming that the floor of each container measures 10 by 10 royal cubits. Now let me inscribe an octagon in that square:
(ASCII drawing omitted, can't be rendered in proportional font)
Square ABCD  AB = BC = CD = DA = 10 royal cubits
inscribed octagon abcdefgh  ab=bc=cd=de=ef=fg=gh=ha
A side of the square measures 10 royal cubits or 70 palms or 280 fingers. How long is the side of the regular octagon within the frame of the square?
1 1 2
2 3 4
5 7 10
12 17 24
29 41 58
70 99 140
169 ... ...
side of octagon  side of circumscribed square
7  5 + 7 + 5 = 12
10  7 + 10 + 7 = 24
17  12 + 17 + 12 = 41
24  17 + 12 + 17 = 58
41  29 + 41 + 29 = 99
58  41 + 58 + 41 = 140
By doubling the numbers of the last line we find:
side of octagon 116  side of square 82+116+82 = 280
The side of the square measures 10 royal cubits or 70 palms or 280 fingers, hence the side of the inscribed octagon measures practically 116 fingers or 4 cubits 1 palm.
Now let me go a step further and transform the square containers into granaries of the same volume but in the shape of prisms. The floor of these containers is given by the above octagon. The area of an octagon is smaller than that of the square. Accordingly, the granaries will be taller. How do we calculate their heights? Again by means of the above number pattern:
height of granary based on square 10 24 58 140 ...
height of granary based on octagon 12 29 70 169 ...
The former square containers were 3 '3 and 5 royal cubits tall. The new granaries on the base of the octagon are taller. By using the first numbers 10 and 12 we find the heights:
3 '3 royal cubits x '10 x 12 = 4 royal cubits
5 royal cubits x '10 x 12 = 6 royal cubits
Very simple numbers.
Now let us consider only the second granary. Wishing to get a more accurate volume I use the numbers 140 and 169:
former height 5 royal cubits or 140 fingers
new height 140f x '140 x 169 = 169f = 6 cubits 1 finger
Calculating the area of the inner wall (right parallelepiped and prism):
base 10 c x 10 c height 5 royal cubits periphery 4 x 10 c = 40 royal cubits area of wall 40 c x 5 c = 200 square cubits
side of octagon 116 fingers or 4 cubits 1 palm height 169 fingers or 6 cubits 1 finger periphery 8 x 116 fingers = 928 fingers or 33 cubits 1 palm area of wall 33 '7 c x 6 '28 c = 200 '28 '196 cc
The walls of the two granaries have practically the same area. Even better: they have exactly the same area (the minor error in the above result is due to the margins of error in the values '41 of 58 and '140 of 169 used for the calculation of the octagon).
Imagine a pair of ideal granaries with vertical walls. One granary is based on a square. The other granary is based on the regular octagon inscribed in the square. If the granaries have the same capacity, their walls have the same area. Or if the inner walls have the same area, the two granaries have the same volume.

