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