Professor Ahmes (5)
cir...@access.ch
perhaps the most complex problem of the Rhind Mathematical Papyrus
is no. 7 which I shall discuss in several posts.
Ahmes multiplies 1/4 + 1/28 by 1/2 + 1/4 and obtains 1/2:
'4 '28 x 1 '2 '4 = '2
This calculation might concern the area of a board measuring
'4 '28 royal cubits x 1 '2 '4 royal cubits = '2 square cubit
or
8 fingers x 49 fingers = 392 square fingers
Let us transform the area of the panel into a square. Hwo long are
the diagonals of the square? Exactly 1 royal cubit or 28 fingers.
b
a m c
d
The square abcd has an area of 392 square fingers. The diagonals amc
and bmd measure 28 fingers each. Now let us imagine a circle around m
through the four corners of the square. We obtain a figure consisting
of four triangles (amb bmc cmd dma) and four arcs (ab bc cd da).
If the arcs and triangles are made of stone (sawn out of stone disks),
7 arcs will weigh as much as 4 triangles. This leads to the following
numbers:
area of one arc ---------------- 4 units
area of one triangle ----------- 7 units
area of the square ------------ 28 units = 392 square fingers
area of the circle ------------ 44 parts = 616 square fingers
If the diameter of a circle measures 28 fingers, the area measures 616
square fingers. The same value is obtained via the following equation:
'4 x 28 fingers x 28 fingers x 3 '7 = 616 fingers
If we replace the 28 fingers by the word 'diameter' and the number 3 '7
by the name 're' (number of the circle, name inspired by the sun god Re)
we obtain the following formula:
area of the circle = '4 x diameter x diameter x re
Regards Franz Gnaedinger Zurich cir...@access.ch
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cir...@access.ch
cir...@access.ch wrote:
part 1 of an interpretation of problem no. 7 of the Rhind Mathematical
Papyrus. Here follows part 2.
A board measures 8 fingers x 49 fingers, its area 392 square fingers
or half a square cubit.
Let us draw a circle around it and calculate its area by means of the
formula
'4 x diameter x diameter x 3 '7
The square diameter x diameter is given by the sum
8f x 8f plus 49 f x 49 f = 2465 square fingers
We put this number in the above formula and obtain
'4 x 2465 square fingers x 3 '7 = 1936 '2 '7 '7 square fingers
The number of the circle is slightly smaller than 3 '7. Therefore we
may neglect the fractions '2 '7 '7 and keep 1936 square fingers only.
A circle of this area is easily squared, for 1936 = 44 x 44.
A circle around a panel measuring 8 fingers x 49 fingers and a square
measuring 44 fingers x 44 fingers have about the same area.
Exact square: 44.000073 f x 44.000073 f
The same circle circumscribes a panel measuring 23 fingers x 44 fingers,
for
23f x 23f plus 44f x 44f = 2465 square fingers
A panel measuring 8 fingers x 49 fingers and another one measuring
23 fingers x 44 fingers are circumscribed by the same circle. From
this can be won a simple method for the transformation of a square
into a circle of the same area.
We divide a square as follows:
a f b ab = cd = ac = bd = 44 parts
m af = cg = 23 parts
c g d fb = gd = 21 parts
Square abdc measures 44 parts x 44 parts while rectangle afgc measures
23 parts x 44 parts. We draw the diagonals ag and cf. They cross each
other in point m. We draw a circle around m through a, f, g and c. The
circle has the same area as the square abcd (with a tiny mistake only).
cir...@access.ch
cir...@access.ch wrote:
part 2 of an interpretation of problem no. 7 of the Rhind Mathematical
Papyrus. Here follows part 3.
Problem no. 7 is based on the key number 44. This number allowed us
to transform a square into a circle of the same area. The same number
can be used for the squaring of the circle:
Let us imagine a circle of the diameter 44 fingers and calculate its
area by means of the following formula:
'4 x diameter x diameter x 3 '7
We obtain
'4 x 44 fingers x 44 fingers x 3 '7 = 1521 '7 square fingers
The number of the circle is slightly smaller than 3 '7. Therefore we
let go the fraction '7 and keep 1521 square fingers. A circle of this
area is easily squared, for 1521 = 39 x 39.
Exact square: 38.9939... f x 38.9939... f
A circle of the diameter 44 fingers and a square of the side length
39 fingers have about the same area.
The numbers 44 and 39 stay in a similar ratio as the numbers 9 and 8,
what is shown by a pair of products:
9 x 39 = 351 8 x 44 = 352
This leads to a simpler formula: a circle of the diameter 9 units
and a square of the side length 8 units have about the same area.
This simple formula, useful for beginners and for practical reasons,
is found in problem no. 50 and bases on the value '81 of 256 (265/81)
for re or pi.
cir...@access.ch
cir...@access.ch wrote:
part 3 of an interpretation of problem no. 7 of the Rhind Mathematical
Papyrus. Here follows part 4.
A board measures 8 fingers x 28 fingers plus 14 fingers plus 7 fingers.
It may be divided in 3 panels A B C like this:
A 8f x 28f B 8f x 14f C 8f x 7f
Let us imagine a semicircle around rectangle A, a circle around B and
a sphere around C. The area of semicircle A, of circle B and the surface
of the sphere are given by the formulas
A '8 x diameter x diameter x 3 '7
B '4 x diameter x diameter x 3 '7
C diameter x diameter x 3 '7
The squares diameter x diameter are given by the sums
8f x 8f plus 28f x 28f = 848 square fingers
8f x 8f plus 14f x 14f = 260 square fingers
8f x 8f plus 7f x 7f = 113 square fingers
The areas measure then
'8 x 848 square fingers x 3 '7 = 333 '7 square fingers
'4 x 260 square fingers x 3 '7 = 204 '7 '7 square fingers
113 square fingers x 3 '7 = 355 '7 square fingers
The number of the circle is slightly smaller than 3 '7. Therefore we
let go the fractions and keep 333, 204 and 355 square fingers. Thus
we obtain the following equations:
'8 x 848 square fingers x re = 333 square fingers
106 square fingers x re = 333 square fingers
'4 x 260 square fingers x re = 204 square fingers
113 square fingers x re = 355 square fingers
For the number of the circle 're' (a name inspired by the sun god Re)
we find the approximate values
'106 of 333 '65 of 204 '113 of 355
cir...@access.ch
cir...@access.ch wrote:
part 4 of an interpretation of problem no. 7 of the Rhind Mathematical
Papyrus (RMP 7). Here follows a simple method for obtaining approximate
values or re or pi. The name 're' is inspired by the sun god Re whose
hieroglyph was a small circle and who had many names while no one knew
his real name ...
The number of the circle is smaller than 4 and a little greater than 3.
When we know such a pair of values we can get more values by adding the
numbers like this:
4 (plus 3) 7 10 13 16 19 22 25 28
1 (plus 1) 2 3 4 5 6 7 8 9
By carefully measuring out a circle or by rolling off stone disks on
a carefully prepared ground one may find that the number of the circle
is greater than 3 and a little smaller than '6 of 19.
By writing 3 above 1, and 6 above 2, and 9 above 3, and by adding
19 above 6 we obtain the following numbers:
3 22 6 25 44 9 28 47 66 ... 256
1 7 2 8 14 3 9 15 21 ... 81
The value '81 of 256 is implied in RMP 50 while the values '7 of 22
and '14 of 44 and '21 of 66 equal 3 '7 - a number found in RMP 38.
Please imagine a stone disk of the diameter 1 royal cubit. Roll it off
7 times on a carefully prepared course. How far will you get? On the
average 22 royal cubits minus about 1 Shu mark ('4 finger). This means
that the number of the circle is slightly smaller than '7 of 22 or 3 '7.
By writing 3 above 1 and by adding several times 22 above 7 we obtain
the following sequence:
3 (plus 22) 25 47 69 91 113 135 157 179 201 223
1 (plus 7) 8 15 22 29 36 43 50 57 64 71
245 267 289 311 333 355 377
78 85 92 99 106 113 120
The values '106 of 33 and '113 of 355 found via RMP 7 appear in this
sequence. More handy values are
'99 of 311 = 3 '9 '33 and '120 of 377 = 3 '10 '24
By writing 6 above 2 and by adding 22 above 7 we obtain the following
sequence:
6 (+22) 28 50 72 ... 204 ... 336 ... 424 ... 600
2 (+ 7) 9 16 23 ... 65 ... 107 ... 135 ... 191
In this sequence appears the value '65 of 204 found via RMP 7. We
solved RMP 36 by means of the better value '135 of 424 = 3 '9 '45 '135
(see: Professor Ahmes 4). Inverse values:
'336 of 107 = '4 '21 '48 '600 of 191 = '4 '24 '60 '100
Regards FRanz Gnaedinger Zurich cir...@access.ch
Steve Whittet
In article <6rj6g3$h7l$1...@nnrp1.dejanews.com>, cir...@access.ch says...
>
>In article <6rggrc$hqh$1...@nnrp1.dejanews.com>,
> cir...@access.ch wrote:
>
>part 4 of an interpretation of problem no. 7 of the Rhind Mathematical
>Papyrus (RMP 7). Here follows a simple method for obtaining approximate
>values or re or pi. The name 're' is inspired by the sun god Re whose
>hieroglyph was a small circle and who had many names while no one knew
>his real name ...
The issue is first whether the Egyptians realized that the ratio of
a circles diameter to its radius was a constant and secondly if they
had one value only for this or more than one.
We both agree that the Egyptians used the hieroglyphic re to refer
to any and all of the properties of the sun including its property
of being round, and that they observed the sun in the process of
which they learned a good deal about it.
They were also interested in divisions of time and measured degrees
which implies some interest in dividing up a circle and in measuring
cycles and parts of cycles.
>The number of the circle is smaller than 4 and a little greater than 3.
>When we know such a pair of values we can get more values by adding the
>numbers like this:
>
> 4 (plus 3) 7 10 13 16 19 22 25 28
> 1 (plus 1) 2 3 4 5 6 7 8 9
>
>By carefully measuring out a circle or by rolling off stone disks on
>a carefully prepared ground one may find that the number of the circle
>is greater than 3 and a little smaller than '6 of 19.
The Egyptians measured with knotted cords. An Egyptian might
put a string across the diameter of a pot and around a pot its
circumference in attempting to see how much it would hold.
He might reason that a circle is sort of like a square with
the corners knocked off
>
>By writing 3 above 1, and 6 above 2, and 9 above 3, and by adding
>19 above 6 we obtain the following numbers:
>
> 3 22 6 25 44 9 28 47 66 ... 256
> 1 7 2 8 14 3 9 15 21 ... 81
This is a really nice expansion. You have sort of simultaneously
addressed aproximations to pi and the tabular use of unit fractions
taking a somewhat different tangent ...
Imagine that the top row is the knotted cord wrapped around the
circumference and the bottom row the knotted cord held across
the diameter and that they represent measurements taken of a
pot whose circumference and diameter were checked at intervals
from top to bottom.
It would be evident that he ratio of circumference to
diameter was a constant just inspecting the measures.
3, 3 '7, 3, 3 '8, 3 '7, 3, 3 '9, 3 "15, 3 '7, ...3 '8 '61...
could be taken to have a value somewhere around 3 '7 or
any of the other aproximations could be used as well if
that made a calculation more convenient
>The value '81 of 256 is implied in RMP 50 while the values '7 of 22
>and '14 of 44 and '21 of 66 equal 3 '7 - a number found in RMP 38.
256/81 is what we derive as the ratio used in the problems but
as you point out the Egyptians who used septenary measurements
probably found 3 '7 a comfortable practical aproximation for
everyday calculations and their problems do get set up using
those ratios even if not explicitly mentioning a situation
which requires pi to be used in the calculation.
>
>Please imagine a stone disk of the diameter 1 royal cubit. Roll it off
>7 times on a carefully prepared course. How far will you get? On the
>average 22 royal cubits minus about 1 Shu mark ('4 finger). This means
>that the number of the circle is slightly smaller than '7 of 22 or 3 '7.
The Isis glyph on their rulers can be used so that measuring from
each end to the mark which follows it gives 22 and 7. This is at
the very least convenient even if not indended.
>
>By writing 3 above 1 and by adding several times 22 above 7 we obtain
>the following sequence:
>
> 3 (plus 22) 25 47 69 91 113 135 157 179 201 223
> 1 (plus 7) 8 15 22 29 36 43 50 57 64 71
>
> 245 267 289 311 333 355 377
> 78 85 92 99 106 113 120
>
>The values '106 of 33 and '113 of 355 found via RMP 7 appear in this
>sequence.
I had never before seen anyone demonstrate how the Egyptians might
have found the ratio of 113 to 355 for the diameter of a circle to
its circumference.Its interesting that many of these ratios are
aproximated by using dual systems of unit measures
5 hands: 1 link
4 pes: 1 remen
6 pes: 1 long remen
7 long greek feet(13"):"5 orguia
10 pes:1 yard
12 pes:1 ell
13 feet(12 long greek feet):1 short pace
15 feet(12 remen):5 pes
4 sp: 5 lgf
18 greek feet:6 pes
etc;
More handy values are
>
> '99 of 311 = 3 '9 '33 and '120 of 377 = 3 '10 '24
>
>By writing 6 above 2 and by adding 22 above 7 we obtain the following
>sequence:
>
> 6 (+22) 28 50 72 ... 204 ... 336 ... 424 ... 600
> 2 (+ 7) 9 16 23 ... 65 ... 107 ... 135 ... 191
I know the Egyptians made a lot of tables and some of the mathematical
papyri are quite similar comparing two lists of numbers. Is there any
evidence that they hit on your method or used a similar method for a
different purpose?
>
>In this sequence appears the value '65 of 204 found via RMP 7. We
>solved RMP 36 by means of the better value '135 of 424 = 3 '9 '45 '135
>(see: Professor Ahmes 4). Inverse values:
>
> '336 of 107 = '4 '21 '48 '600 of 191 = '4 '24 '60 '100
A very interesting post Franz, its a simple method. If the Egyptians
used such tables you could make a case that they had the method and
simply applied it.
The one I would like to see if they hit on would be 3 '8 '64...
using the horus eye divisions, or better 3 '8 '61 ...
>
>Regards FRanz Gnaedinger
regards
steve
cir...@access.ch
whi...@shore.net (Steve Whittet) wrote:
> I know the Egyptians made a lot of tables and some of the mathematical
> papyri are quite similar comparing two lists of numbers. Is there any
> evidence that they hit on your method or used a similar method for a
> different purpose?
Thank you, Steve. At least someone reads my messages :)
Let me answer your question by means of an interpretation of the problems
no. 21-23 of the Rhind Mathematical Papyrus (RMP 21-23).
Ahmes finds the following equations:
RMP 21 '5 '15 plus "3 '15 equals 1
RMP 22 '5 '10 plus "3 '30 equals 1
RMP 23 '4 '8 '10 '30 '45 plus '9 '40 equals "3
I multiply the first equation by a factoer of 15 and again by 9 palms,
the second one by a factor of 10 and again by 5 palms, and the third
one by a factor of 360 and again by 1 palm. Thus I obtain
4 + 11 = 15 --- 36 palms + 99 palms = 135 palms
3 + 7 = 10 --- 15 palms + 35 palms = 240 palms
49 + 191 = 240 --- 49 palms + 191 palms = 240 palms
Let us use the measurements 36, 99, 135, 15, 30, 50, 191, 49, 240 palms
as diameters of nine circles and calculate their circumferences by means
of the formula circumference = diameter x 3 '7 :
36 palms x 3 '7 = 113 '7 palms
99 palms x 3 '7 = 311 '7 palms
135 palms x 3 '7 = 424 '7 '7 palms
15 palms x 3 '7 = 47 '7 palms
35 palms x 3 '7 = 110 palms
50 palms x 3 '7 = 157 '7 palms
191 palms x 3 '7 = 600 '7 '7 palms
49 palms x 3 '7 = 154 palms
240 palms x 3 '7 = 754 '7 '7 palms
The number of the circle is slightly smaller than 3 '7. Therefore we
let go the fractions and keep the whole numbers. Thus we obtain the
following diameters, circumferences and ratios:
36 palms and 113 palms '36 of 113 = 3 '9 '36
99 palms and 311 palms '99 of 311 = 3 '9 '33
135 palms and 424 palms '135 of 424 = 3 '9 '45 '135
15 palms and 47 palms '15 of 47 = 3 '9 '45
35 palms and 110 palms '7 of 22 = 3 '7
50 palms and 157 palms '50 of 157 = 3 '10 '25
191 palms and 600 palms '600 of 191 = '4 '24 '60 '100
49 palms and 154 palms '7 of 22 = 3 '7
240 palms and 754 palms '120 of 377 = 3 '10 '24
All these values are found in the sequences I discussed in my previous
message. Especially fine values are 3 '9 '33 and 3 '10 '24. I found
evidence for their use in several problems of the RMP.
Regards Franz Gnaedinger Zurich cir...@access.ch
cir...@access.ch
cir...@access.ch wrote:
> Thank you, Steve. At least someone reads my messages :)
A welcome for John, my second reader.
He uses the same notations as I do (for example 3 '7 instead of 3 1/7)
and he finds my work interesting although rather speculative.
John, I don't really speculate, but ... You'll see.
You certainly know the begin of Papyrus Rhind. Here is a french version
taken from LA GEOMETRIE EGYPTIENNE, Contribution de l'Afrique antique
a la Mathematique mondiale, by Theophile Obenga (L'Harmattan, Paris 1995),
a book I like very much:
'Method correct (tp-hsb) d'investigation (n h3t) dans (m) la nature (ht)
pour connaitre (rh) tout ce qui existe (ntt nbt), chaque mystere (snkt nbt),
tous les secrets (st3t nbt) ...'
Ahmes makes big promises about telling us all secrets and mysteries,
then he carries out rather simple calculations. Was he an impostor?
A case for Alan Dunsmuir, our well-known professor of kookology?
I went to my strologer and asked her if she please may get in contact
with Ahmes, whereupon she lit a candle, drew the curtains and had a long
look into her crystal sphere, whispering her magic syllables. Finally,
she succeeded. We saw only flickering shadows, yet we could her a clear
voice of a man asking: Hello, who calls me?
My astrologer: Good morning, professor Ahmes, can you hear me? We read
your famous papyrus and wonder about your secrets and mysteries. Can
you tell us more please?
Professor Ahmes: I hear you well. You are interested in my secrets?
You wish to hear about mysteries? My good people: you must know that
a secret is a secret only as long as I keep quiet whereas when I tell
you my secret it ain't a secret anymore. I mean I would love to tell
you my secrets but it's a logical impossibility. Ever so sorry ...
My astrologer: Come on, professor Ahmes, you are joking! We like your
papyrus very much. The handwriting is beautiful, having an almost musical
qualitiy. I am sure you will tell us more.
Professor Ahmes: You are kind. So let me say this. You must know that
we teach all our pupils by the same numbers, may they be five or twenty
years old. The young ones learn how to calculate with unit fractions
while the older ones learn how to solve geometric problems. I give you
an example:
A plus '7 x A equals 19 (A + A/7 = 19)
A plus '2 x A equals 16 (A + A/2 = 16)
A plus '4 x A equals 15 (A + A/4 = 15)
A plus '5 x A equals 21 (A + A/5 = 21)
How much is A? When you know how to solve these problems, I come up
with the same numbers and give you a geometric task by saying that
my numbers 19, 16, 15 and 21 are the peripheries of four rectangular
triangles while the numbers '7 x A and '2 x A and '4 x A and '5 x A
represent a side of the respective triangle. Now tell me please: how
long are the other sides? Their sum is given by number A, but me likes
to know how long ea
My astrologer: Professor Ahmes? Do you hear me? We lost you. Please
come back. Hello? --- Nothing. Sorry, seems we really lost him.
Now you see how I get my informations ...
By the way, can you solve the above problems?
cir...@access.ch
cir...@access.ch wrote:
a fancy; here follows a serious interpretation of the problems
no. 24-27 of the Rhind Mathematical Papyrus:
RMP 24 A + A/7 = 19 16 '2 '8 plus 2 '4 '8 equals 19
RMP 25 A + A/2 = 16 10 "3 plus 5 '3 equals 16
RMP 26 A + A/4 = 15 12 plus 3 equals 15
RMP 27 A + A/5 = 21 17 '2 plus 3 '2 equals 21
The formula
A plus A/n equals P
can be interpreted as a rectangular triangle of a special kind:
P = side a + side b + side c = periphery
A/n = side a
A = side b + side c (c being the hypothenuse)
How long are the sides b and c?
* The difference c - b is easily found: divide a by n, or A by nxn,
or P by n(n+1)
** subtract c - b from A, thus you obtain 2b
*** from this you get b
**** add c - b to b, thus you obtain c
Let me solve RMP 24 in this way. I hope you can follow my steps:
A plus '7 x A equals P
16 '2 '8 plus 2 '4 '8 equals 19
* side a = 2 '4 '8 = '8 x 19
* n = 7
* side c - side b = side a divided by n = 'n x side a
* side c - side b = '7 x '8 x 19 = '56 x 19
** side c + side b = A - (c - b) = '8 x 133 - '56 x 19
** = '56 x 931 - '56 x 19 = '56 x 912 = '7 x 114 = 2b
*** b = '2 x '7 x 114 = '7 x 57 = 8 '7
**** c - b = '56 x 19 = '168 x 57 = '3 '168
**** c = b + (c - b) = 8 '7 + '3 '168 = 8 '3 '7 '168
RMP 24 2 '4 '8 8 '7 8 '3 '7 '168 (7 - 24 - 25)
RMP 25 5 '3 4 6 "3 (4 - 3 - 5)
RMP 26 3 5 '2 '8 6 '4 '8 (8 - 15 - 17)
RMP 27 3 '2 8 '3 '15 9 '10 (5 - 12 - 13)
The first triangles are based on the triples 7-24-25 and 3-4-5 which
are part of a method of calculating the circle (see: School of Imhotep
1, 2 and 3, to be found via http://www.dejanews.com and Power Search).
cir...@access.ch
cir...@access.ch wrote:
an interpretation of RMP 24-27. RMP 24-30 are written in a column of
the papyrus and treat the same kind of problems. Here follows RMP 28.
Given in the time of Pharaoh Bill Clinton and Vizier Tim Burners-Lee.
A x "3 equals B
B + A equals C
C - '3 x C equals 10
9 x "3 equals 6
6 + 9 equals 15
15 - '3 x 15 equals 10
6 plus 9 equals 15 ------ 9 plus 6 equals 15
15 minus 5 equals 10 ------ 10 plus 5 equals 15
Let us suppose that the periphery of a rectangular triangle measures
15 royal cubits while one side measures 6 or 5 royal cubits. How long
are the three sides? If we apply the method found via RMP 24-27 we
find the following measurements:
2 '2 6 6 '2 royal cubits or 70-168-182 fingers
3 '2 '4 5 6 '4 royal cubits or 105-140-175 fingers
How long are the diameters of the circumscribed circles? They are
given by the hypothenuses: 182 and 175 fingers.
How long are the diameters of the inscribed circles? They are found
by means of a simple formula: sum of cathetes minus hypothenuse
70 fingers + 168 fingers - 182 fingers = 56 fingers
105 fingers + 140 fingers - 175 fingers = 70 fingers
One diameter measures exactly 56 fingers = 2 royal cubits while
the other one measures exactly 70 fingers.
Now let me ask for the equilateral triangle whose inscribed circles
have the same diameter.
The equilateral triangle can be calculated by means of a number
column:
1 1 3
2 4 6
1 2 3 a b 3a
3 5 9 a + b b + 3a 3(a + b)
8 14 24
4 7 12
11 19 33
30 52 90
15 26 45
41 71 123
112 194 336
56 97 168 and so on
If the side of an equilateral triangle measures 97 fingers, the
diameter of the inscribed circle measures 56 fingers = 2 royal cubits
(mistake 0.06 millimeters) while the height measures '2 x 168 fingers
= 84 fingers = 3 royal cubits (mistake 0.08 millimeters).
Multiply every measurement by a factor of 1 '4, thus you obtain
a height of 105 fingers and a dimaeter of 70 fingers.
cir...@access.ch
cir...@access.ch wrote:
an interpretation of RMP 28. Here follows RMP 29, part 1
A plus "3 x A = B
B plus '3 x B = C
C divided by 3 = 10
13 '2 + 9 = 22 '2
22 '2 + 7 '2 = 30
30 divided by 3 = 10
13 '2 + 9 = 22 '2 4 '2 plus 3 equals 7 '2
22 '2 + 7 '2 = 30 7 '2 plus 2 '2 equals 10
Please imagine two rectangular triangles whose peripheries measure
7 '2 and 10 royal cubits while one side measures 3 or 2 '2 royal
cubits. By applying the method found via RMP 24-27 and by using
Nut marks instead of royal cubits (1 royal cubit = 168 Nut marks)
we obtain the following triangles:
210 -- 504 -- 546 Nut marks periphery 1260 Nut marks
560 -- 420 -- 700 Nut marks periphery 1680 Nut marks
The diameters of the inscribed circles are found as follows:
210 Nm plus 504 Nm minus 546 Nm equals 168 Nut marks
560 Nm plus 420 Nm minus 700 Nm equals 280 Nut marks
Now let me ask for the equilateral triangles whose inscribed circles
have the same diameters.
I multiply 168 and 280 Nut marks by a factor of 1 '2 and obtain
252 and 420 Nut marks. These are the heights of our triangles.
How long are the sides? I draw up a number column:
1 1 3
2 4 6
1 2 3
3 5 9
8 14 24
4 7 12
11 19 33
30 52 90
15 26 45
41 71 123
112 194 336
56 97 168
...............
If the side of an equilateral triangle measures 97 units, the height
measures '2 x 168 or 84 units. When we know the height and wish to
know the side we may divide the height by 84 and multiply the result
by 97:
252 Nm : 84 = 3 Nm 3 Nm x 97 = 291 Nut marks
420 Nm : 84 = 5 Nm 5 Nm x 97 = 485 Nut marks
sides ................................ 291 and 485 Nut marks
heights .............................. 252 and 420 Nut marks
radii of the inscribed circles ....... 84 and 140 Nut marks
radii of the circumscribed circles ... 168 and 280 Nut marks
cir...@access.ch
cir...@access.ch wrote:
an interpretation of RMP 29, part 1. Here follows part 2
We found an equilateral triangle of these measurements:
side 291 Nut marks = 48 '2 fingers = 1 '2 '7 '14 '56 royal cubits
height 252 Nut marks = 42 fingers = 1 '2 royal cubits
radius of the inscribed circle 84 Nm = 14 f = '2 royal cubit
radius of the circumscribed circle 168 Nm = 28 f = 1 royal cubit
What is the area of the circumscribed circle? radius x radius =
28 fingers x 28 fingers = 784 square fingers. I multiply 784 by
3 '9 '33 and 3 '10 '24 and round the results:
784 x 3 2352 784 x 3 2352
784 x '9 87 784 x '10 78
784 x '33 24 784 x '24 33
------------------------------------------
sum 2463 sum 2463
The circle of the diameter 1 royal cubit or 28 fingers has an area
of 2463 square fingers. This number yields a new value for re or pi:
'784 x 2463
This value belongs to a sequence:
2177 (311) 2199 2221 2243 2265 2287 2309 2331 (333)
693 ( 99) 700 707 714 721 728 735 742 (106)
2353 2375 2397 2419 2441 2463 2485 (355) 2507 2529
749 756 763 770 777 *784 791 (113) 798 805
2551 2573 2595 2617 2639 (377)
812 819 826 833 840 (120)
You may remember the rectangular triangle measuring 560-420-700
Nut marks. The diameter of the circumscribing circle is given
by the hypothenuse 700 Nut marks. How long is the circumference?
I look up the above sequence and find 2199 Nut marks = 13 royal
cubits 2 fingers 1 Re mark.
If the diameter of a circle measures 25, 26, 27, 28, 29 or 30 royal
cubits = 700, 728, 756, 784, 812 or 840 fingers, the circumference
measures 2199, 2287, 2375, 2463, 2551 or 2639 fingers.
cir...@access.ch
cir...@access.ch wrote:
an interpretation or RMP 29. Here follows RMP 30
A x "3 '10 = 10 A = 13 '23 = '23 x 300
The multiplication may be given as an addition:
A x "3 plus A x '10 equals 10
A equals '23 x 300, therefore we obtain
'23 x 200 plus '23 x 30 equals '23 x 230
I multiply each term by 23 palms or 92 fingers and obtain
200 palms plus 30 palms equal 230 palms
800 fingers plus 120 fingers equal 920 fingers
Let us imagine a rectangular triangle. Its periphery measures 920
fingers while the short side measures 120 fingers. By applying the
method found via RMP 24-27 we obtain the following sides:
120 + 391 - 409 fingers = 102 fingers = 306 Maat marks
Now let me ask for the equilateral triangle whose inscribed circle
has the same diameter.
We obtain the height of an equilateral triangle by multiplying the
diameter of the inscribed circle by a factor of 1 '2. As the diameter
measures 102 fingers = 306 Maat marks, the height measures 153 fingers
= 459 Maat marks.
How long are the sides? I draw up my number column:
1 1 3
2 4 6
1 2 3
3 5 9
8 14 24
4 7 12
11 19 33
30 52 90
15 26 45
41 71 123
112 194 336
56 97 168 and so on
An example. If the height of an equilateral triangle measures
97 fingers, the side measures 2 x 56 fingers = 4 royal cubits.
The height of our triangle measures 465 Maat marks, therefore the
side measures 2 x 265 = 530 Maat marks (mistake 0.05 millimeters).
A task for you:
The circumscribed and inscribed circles (radii 306 and 153 Maat marks)
form a ring. Please calculate its area and compare it with the area of
a circle drawn around a side of the triangle (radius 265 Maat marks).
You may do the same with a square, regular pentagon, hexagon ...
RMP 31 is a difficult one. I will ask my astrologer to get in contact
with Ahmes again.
cir...@access.ch
cir...@access.ch wrote:
an interpretation of RMP 30 (sorry for having omitted the line
153 265 459 of my number column). Here follows RMP 31
My astrologer (softly touching her crystal sphere): Professor Ahmes?
Me: I wonder about your funny division 33 divided by 1 "3 '2 '7 equals
14 '4 '56 '97 '194 '338 '679 '776
Ahmes: An impressive number, eh? Goes about a granary. I will explain
it for you.
The side of a regular hexagon measures 66 fingers. The circumscribed
circle and the inscribed circle form a ring. Now build a cylindrical
granary on this ring. The height shall measure 5 '2 royal cubits or
154 fingers. What will the capacity of my granary be?
Me: Area of the inscribedd circle x height
Ahmes: So let me first calculate the radius of the inscribed circle.
Please have a look at one of my number patterns:
1 1 3
2 4 6
1 2 3
3 5 9
8 14 24
4 7 12
11 19 33
30 52 90
15 26 45
41 71 123
112 194 336
56 97 - - 168
If the side of a regular hexagon measures 97 fingers, the diameter
of the inscribed circle measures about 168 fingers or 42 palms.
'42 x 97 equals 1 "3 '2 '7
The side of my hexagon measures 66 fingers. 66 divided by 2 equals 33.
33 divided by 1 "3 '2 '7 equals 14 '4 '56 '97 '194 '388 '679 '776.
Hence the *radius* of my circle measures 14 ... palms.
Now please calculate the area of my circle.
Me: A hopeless case!
Ahmes: Not really. We can solve this problem in another way. I tell
you one of my theorems. Here I go. The area of the ring of a regular
polygon equals the area of a circle around one side of the polygon.
The side of my hexagon measures 66 fingers. Hence the area of the
ring measures
'4 x 66 fingers x 66 fingers x 3 '9 '33 = 3421 square fingers
The area of the circumscribed circle measures
66 fingers x 66 fingers x 3 '9 '33 = 13684 square fingers
Now the area of the inscribed circle measures
13684 sf - 3421 sf = 13684 sf x '2 '4 = 10263 square fingers
The volume of my granary measures practically 72 cubic cubits:
154 fingers x 10263 square fingers = 1,580,502 cubic fingers
72 x 28 fingers x 28 fingers x 28 fingers = 1,580,544 cubic fingers
Now please fill my granary up to a height of 1 royal cubit and trans-
form the volume into a cube. The edge of this cube will measure about
66 fingers:
28 fingers x 10263 square fingers = 287,364 cubic fingers
66 fingers x 66 fingers x 66 fingers = 287,496 cubic fingers
(Continuation follows)
cir...@access.ch
cir...@access.ch wrote:
an interpretation of RMP 31. Here follows an interview with Ahmes
My astrologer: Thank you, professor Ahmes, for your kind explanations.
Me: I checked your numbers on my pocket calculator and got a volume
of 72.0021... cubic cubits and an edge of 65.991... fingers.
Ahmes: Funny numbers. Please, what is a pokit calculator?
Me: A calculating device, some kind of a machine.
Ahmes: And what is a masheen?
Me: How shall I explain this? You may think of an ushebti who works
for me. Some machines calculate. Other ones carry us around. Still
other ones copy newspapers and books. Papyri, I mean. Hundreds of
copies a day ...
Ahmes: Are you making fun of me? I need weeks for a single copy.
My astrologer: Dear professor Ahmes, you must know that we call you
from another time --- some 3,600 years ahead of your days ...
Ahmes: Are you kidding?
Me: Our Pharaoh is called Bill Clinton, our Memphis Washington.
Our viziers don't build pyramids any longer. They build other
things like for example the *world wide web*
Ahmes: ???
Me: Imhotep and his successors piled up millions of blocks while
one Tim Burners-Lee succeeded in connecting millions of human minds
in a so-called web, allowing everyone to talk with everyone else
and exchange ideas all ove
...: Doug the weasel
Ahmes: What?
Me: Oh, this is Marianne. Or Saida? No one knows her name. Anyway,
she is in love with one Doug.
...: weaselissimo
Me: If only Doug would marry her! But he is not willing to do so.
Therefore she is flaming him. All the time and everywhere. She will
never give up on him.
Ahmes: Wait a minute. You say you live in my future. You command
ushebtis who work inbelievably fast. You can even connect your
minds in a so-called web - whatever this may be. But you are not
able to find a cure for a suffering heart?
Bobby: Life is sad, life is a bust
Ahmes: All these voices ...
Bobby: all you can do is do what you must
Ahmes: ... do they mean ...
Bobby: I do it for you, honey baby can't you tell?
Ahmes: ... I am in this - web?
Bobby: I'm no monkey but I know what I like
Alan: Oh my, the kooks are loose
Bobby: ... buckets of moonbeeeeams ...
(Continuation follows)
cir...@access.ch
cir...@access.ch
began an interview with Ahmes
John: Good morning, professor Ahmes. My name is John.
Ahmes: Good morning, John.
John: I worked on the Kahun Papyrus and proposed to interprete some
problems in a geometric way. Now you tell us that we can solve many
of your problems in such a manner. You came up with the formula
A plus 'n x A equals B
Ahmes: B may be seen as the periphery of a rectangular triangle,
A as the sum of two sides. Divide A by n, thus you obtain the third
side. Divide the number of this side by n again, thus you will obtain
the difference of the two other sides.
John: I take A as periphery - and I find interesting numbers too.
What intrigues me somehow.
Ahmes: John, come over and join my class! I always tell my pupils:
play with my numbers, vary my problems, invent a new problem on the
base of my equation. Many problems proposed by our pupils challenge
us and sometimes even provide new insights ...
Here is my first equation of the above kind:
A plus '7 x A equals 19
When I take 19 as periphery I obtain the following sides:
8 '3 '7 '168 8 '7 2 '4 '8 units
My triangle is based on the triple 7-24-25.
Now please show me your solution derived from periphery A.
John: 7 '4 '16 '96 6 "3 '4 '96 2 '4 '8 units
basic triple 12-35-37
Ahmes: Fine. A nice solution.
Now for my second equation:
A plus '2 x A equals 16 6 "3 4 5 '3 units
John: I got ------------------------- 5 '3 - 5 '3 units
Ahmes: My numbers are based on the triple 3-4-5 whle you found a
special case of a triangle: one side disappears. In fact you found
a line measuring 5 '3 units. Does a line have a periphery? You may
say yes when you think of a race: run from A to B, around the pole
and back to A again.
John: My solution is based on the triple 0-1-1, zero one one.
Ahmes What does 'zero' mean?
John: Nothing.
Ahmes: How strange.
Me: While you write n n n l l
n n l l
we simply write 504 - five zero four.
Ahmes: I will have a sleepless night ...
John: My best wishes, professor.
Ahmes: Thank you, John, you are very kind.
John Legon
>In article <6sqp8f$e5p$1...@nnrp1.dejanews.com>,
> cir...@access.ch
>began an interview with Ahmes
>
>John: Good morning, professor Ahmes. My name is John.
>
>Ahmes: Good morning, John.
Thank you, professor. It's nice to be here.
>
>John: I worked on the Kahun Papyrus and proposed to interprete some
>problems in a geometric way. Now you tell us that we can solve many
>of your problems in such a manner. You came up with the formula
>
> A plus 'n x A equals B
>
Yes, I wrote an article about a fragment of papyrus from Kahun. Sadly,
the scribe had bad hand-writing (not like yours, professor!), and the
sage Griffith made an error in his transcription. It was only a tiny
error, but it changed the interpretation considerably. And the problem?
To divide a quantity of 100 into 10 shares in arithmetic progression
such that the largest share is equal to twice the smallest share.
Interesting, eh?
And yes, professor, I have interpreted the 'aha' problems geometrically,
but not as you have. Please go on...
>Ahmes: B may be seen as the periphery of a rectangular triangle,
>A as the sum of two sides. Divide A by n, thus you obtain the third
>side. Divide the number of this side by n again, thus you will obtain
>the difference of the two other sides.
>
>John: I take A as periphery - and I find interesting numbers too.
>What intrigues me somehow.
>
To put it mildly.
>Ahmes: John, come over and join my class! I always tell my pupils:
>play with my numbers, vary my problems, invent a new problem on the
>base of my equation. Many problems proposed by our pupils challenge
>us and sometimes even provide new insights ...
>
>Here is my first equation of the above kind:
>
> A plus '7 x A equals 19
>
>When I take 19 as periphery I obtain the following sides:
>
> 8 '3 '7 '168 8 '7 2 '4 '8 units
>
>My triangle is based on the triple 7-24-25.
>
>Now please show me your solution derived from periphery A.
>
It gave me a headache to do this, but...
>John: 7 '4 '16 '96 6 "3 '4 '96 2 '4 '8 units
>
> basic triple 12-35-37
>
>Ahmes: Fine. A nice solution.
It's kind of you to say so.
>
>Now for my second equation:
>
> A plus '2 x A equals 16 6 "3 4 5 '3 units
>
>John: I got ------------------------- 5 '3 - 5 '3 units
>
>Ahmes: My numbers are based on the triple 3-4-5 whle you found a
>special case of a triangle: one side disappears. In fact you found
>a line measuring 5 '3 units. Does a line have a periphery? You may
>say yes when you think of a race: run from A to B, around the pole
>and back to A again.
>
>John: My solution is based on the triple 0-1-1, zero one one.
>
>Ahmes What does 'zero' mean?
>
>John: Nothing.
>
>Ahmes: How strange.
>
>Me: While you write n n n l l
> n n l l
>
>we simply write 504 - five zero four.
>
Well, if the n's are tens and the l's are ones then I would have read
this as fifty-four. But I get your (decimal) point. A question of
the place-value system. And the fact that the scribes had nothing to
express nothing with. Except nothing...
>Ahmes: I will have a sleepless night ...
>
>John: My best wishes, professor.
>
>Ahmes: Thank you, John, you are very kind.
>
>(Continuation follows)
I look forward to it.
Thank you, Franz.
--
John Legon
cir...@access.ch
John Legon <jo...@nospam.legon.demon.co.uk> wrote:
> >Me: While you write n n n l l
> > n n l l
> >
> >we simply write 504 - five zero four.
> >
> Well, if the n's are tens and the l's are ones then I would have read
> this as fifty-four. But I get your (decimal) point. A question of
> the place-value system. And the fact that the scribes had nothing to
> express nothing with. Except nothing...
Me: Thank you, John, for correcting my silly mistake. By the way:
I find it amazing how f e w mistakes professor Ahmes made on his
long papyrus.
Ahmes (yawning): I had a sleepless night, thinking about your number
'zero' ...
John: Sorry, professor.
My astrologer: Please go on in your own ways, for my crystal sphere
translates every language and number system.
Me: Fine.
John: Here is a problem of a papyrus from Kahun:
To divide a quantity of 100 into 10 shares in arithmetic progression
such that the largest share is equal to twice the smallest share ...
How do you solve such a problem?
Ahmes: Well, our young pupils lay out beans:
A
A b
A b b
A b b b
A b b b b
A b b b b b
A b b b b b b
A b b b b b b b
A b b b b b b b b
A b b b b b b b b b
Replace every 'A' by nine 'b's:
b b b b b b b b b
b b b b b b b b b b
b b b b b b b b b b b
b b b b b b b b b b b b
b b b b b b b b b b b b b
b b b b b b b b b b b b b b
b b b b b b b b b b b b b b b
b b b b b b b b b b b b b b b b
b b b b b b b b b b b b b b b b b
b b b b b b b b b b b b b b b b b b
Now the first line contains 9 beans and the last one 18 beans.
Count all the beans. You will obtain 135 beans.
135 beans equal 100. Hence a single bean equals 100 divided by 135
= 20 divided by 27 = '3 '3 '27 '27 = "3 '18 '54.
Bean 'A' equals 9 x 20 divided by 27 = 20 divided by 3 = 6 "3.
From this you will obtain the whole progression.
Later on, our pupils learn how to calculate the beans instead of
counting them:
b ....................... 1 x 2 divided by 2 = 1
b
b b ..................... 2 x 3 divided by 2 = 3
b
b b
b b b ................... 3 x 4 divided by 2 = 6
b
b b
b b b
b b b b ................. 4 x 5 divided by 2 = 10
b
b b
b b b
b b b b
b b b b b ............... 5 x 6 divided by 2 = 15
and so on
On a third level of teaching and learning I will use my numbers
1 x 2 and 2 x 3 and 3 x 4 and 4 x 5 and 5 x 6 ... for developing
a most amazing series:
1 = '1
1 = '1x2 '2
1 = '1x2 '2x3 '3
1 = '1x2 '2x3 '3x4 '4
1 = '1x2 '2x3 '3x4 '4x5 '5
1 = '1x2 '2x3 '3x4 '4x5 '5x6 '6 and so on
You see, John: we go from simple tasks to more and more demanding
ones.
May I ask for your geometric interpretation of the 'aha' problems?
My astrologer: Sorry, professor, my crystal sphere is heating up.
Translating hieroglyphs over so many years and miles uses quite
some astral energy, you know.
Ahmes: Why, so long everybody.
John Legon
A fascinating solution for a problem in a papyrus from Kahun: to divide
a quantity of 100 into 10 shares in arithmetic progression such that the
largest share is equal to twice the smallest share. Thanks, Franz.
The scribe's reasoning? In my interpretation, he knew that the average
share should be 10, and that the sum of the smallest and largest shares
should be 20. And since the largest share had to be twice the smallest,
these shares should be 13 '3 and 6 "3 respectively, with a difference
between them of 6 "3. With 10 shares there are 9 common differences, so
each difference should equal:
'9 x 6 "3 = "3 '18 '54
which is a single 'bean' using your method. However, the scribe didn't
much like this awkward number, and actually used a 'rounded up' common
difference of "3 '6.
As his working shows, he then multiplied half this number by the number
of common differences, and added the result of 3 "3 '12 to the average
share, to obtain a largest share of 13 "3 '12. From this he subtracted
"3 '6 repeatedly, to give each of the smaller shares as listed on the
papyrus. The 10 shares still added up to exactly 100, but since the
smallest share was now 6 '6 '12, the 1:2 relation with the largest share
was only approximately achieved.
I must add that the nature of the problem was not stated on the papyrus,
only the numbers themselves. The revered Griffith (no mathematician)
misread the numbers 100 and 10 at the head of the working as 110, and
thus caused great confusion to scribe Gillings (q.v.).
[Snip]
>
>You see, John: we go from simple tasks to more and more demanding
>ones.
>
I can believe it, professor...
>May I ask for your geometric interpretation of the 'aha' problems?
>
By all means. For me, these problems can have a practical application.
Although usually translated as 'quantity' or 'heap', the Egyptian term
'aha' has the literal meaning of 'a rising up' or 'height'. It might
therefore be seen as the vertical side of a right-angled triangle, or as
the vertical rise of a sloping passage or shaft.
Suppose we begin to build a pyramid entrance passage, starting from
ground level and rising at a constant inclination, to meet the side of
the pyramid also starting from ground level, and rising at a constant
angle of slope. At what height above the ground will the passage meet
the sloping side of the pyramid? Let's take a simple example...
According to robotics expert Gantenbrink, the southern 'air-shaft' from
the King's Chamber has an inclination of 45* 0' 0", or exactly the
diagonal angle of a square: a slope of 1 rise on 1 base. The pyramid
profile is 14 rise on 11 base, or 1 rise on '2 '4 '28 base. We equate
the 1 rise of the shaft with the 1 rise of the pyramid slope and call
this the 'rising up' or 'aha'. The base-line, or horizontal distance
between the lower end of the shaft and the pyramid slope at the same
level, will be the sum of the horizontal subtentions of both slopes.
In the present case, the horizontal subtention of the shaft is equal to
the rising up, and we can ask a question like those in the Rhind:
A rising up and '2 '4 '28 of it, added to it, equals the base-line.
What is the rising up?
In the general case, this method could have been used to solve other
passage and shaft intersections, by equating the sum of two fractions
with a given base-line. Evidence for this may be seen in the fact that
the northern shaft from the King's Chamber intersected the sloping side
of the pyramid at exactly the same level as the southern shaft, even
though the slope of the northern shaft is 7 rise on 11 base, or half the
pyramid profile of 14 rise on 11 base.
Gantenbrink states that this level is 80.63 ms +/- 4 cms over the base,
which is 154 equals 7 x 22 cubits. At this level, the diagonals of the
horizontal cross-section of the Great Pyramid measure 280 cubits, which
is also the height of the pyramid. At the floor level of the King's
Chamber, 82 cubits above the base, the diagonals of the horizontal
cross-section measure 440 cubits, which is also the side of the base.
[See J.A.R. Legon, _Discussions in Egyptology_ vol. 28 (1994) 29-34,
and vol. 33 (1995), 46-56].
>My astrologer: Sorry, professor, my crystal sphere is heating up.
>Translating hieroglyphs over so many years and miles uses quite
>some astral energy, you know.
But wait, I want to know what the professor thinks about the number
zero, and why do I get three (or four) basic triples from RMP 24-7 ?
>
>Ahmes: Why, so long everybody.
>
My best wishes, professor.
John
--
John Legon
Steve Whittet
jo...@nospam.legon.demon.co.uk says...
>
>In article <6t58ta$6em$1...@nnrp1.dejanews.com>, cir...@access.ch wrote:
>
>A fascinating solution for a problem in a papyrus from Kahun: to divide
>a quantity of 100 into 10 shares in arithmetic progression such that the
>largest share is equal to twice the smallest share. Thanks, Franz.
Seems like professor Ahmes! is starting a new class this fall...
Looking at what Franz is doing with the RMP I would be interested
in how this differs from the carpenters method of determining the
length of cripple jack rafters in framing a roof.
Todya the common difference in the length of rafters is often
taken off a table on the framing square. At some point it had
to occur to someone that there was a common difference and that
a table of its values would be useful.
My suspicion was that the unit spacing for rafters followed after
the use of unit spacing for the frames in boats and so was a rather
late development, but the dimensions given for some ancient buildings
make me wonder.
Anyway the Egyptians have a hieroglyphic for the framing square
and also used the plumb bob, level, and a number of different
measuring instruments, so I guess my question is to what extent
can you measure out an arithmentic solution to the problems in
the RMP using a graduated straightedge, compass and square...
>
>The scribe's reasoning? In my interpretation, he knew that the average
>share should be 10, and that the sum of the smallest and largest shares
>should be 20. And since the largest share had to be twice the smallest,
>these shares should be 13 '3 and 6 "3 respectively, with a difference
>between them of 6 "3. With 10 shares there are 9 common differences, so
>each difference should equal:
>
> '9 x 6 "3 = "3 '18 '54
>
>which is a single 'bean' using your method. However, the scribe didn't
>much like this awkward number, and actually used a 'rounded up' common
>difference of "3 '6.
I think the use of unit fractions, sequences and common differences
are all tied to the three clasic problems of antiquity, doubling
the cube, trisecting an angle and squaring a circle only without
the strict definition of what consitutes a straightedge.
Given that any carpenter can trisect a ninety degree angle,
what could a talented scribe do with a ruler.
Perhaps some of the properties of circles which have the same
circumference as the perimeter of squares or the development
of standards of measure where packing a number of unit values
into another unit value arrived in some fairly useful close
aproximations.
This all would seem to be implied by the septenary system of
measures used to lay out the unit rise and run.ie; the ruler
is 28 fingers long, one fourth that distance is seven fingers
if you measure seven fingers in from one end and 22 fingers in
from the other your ticks will enclose the Isis mark.
seven fingers is one hand plus one palm or 1 span.
14 fingers or '2 the royal cubit ruler is 14 fingers or 2 spans
The circuit of the wall of the kings chamber is to its length
as the circumference of a circle to its diameter because the wall
actually goes '14 its height below the level of the floor.
Its height is 10 royal cubits or 1 hayt and its span is 20
royal cubits or 2 hayt.
>
>[See J.A.R. Legon, _Discussions in Egyptology_ vol. 28 (1994) 29-34,
> and vol. 33 (1995), 46-56].
>
>>My astrologer: Sorry, professor, my crystal sphere is heating up.
>>Translating hieroglyphs over so many years and miles uses quite
>>some astral energy, you know.
>
>But wait, I want to know what the professor thinks about the number
>zero, and why do I get three (or four) basic triples from RMP 24-7 ?
I think those are what the scribe Gillings calls think of a number
problems. I am not sure how you would frame a geometric solution
for this but I guess it has something to do with the "3 table
and to me this suggests those pythagorean dotted number series...
>>Ahmes: Why, so long everybody.
>>
>My best wishes, professor.
>
>John
>
>--
>John Legon
regards,
steve
cir...@access.ch
John Legon <jo...@nospam.legon.demon.co.uk> wrote:
> But wait, I want to know what the professor thinks about the number
> zero, and why do I get three (or four) basic triples from RMP 24-7 ?
Ahmes: Well, John, solving my problems
A plus '7 x A equals 19 and A plus '2 x A equals 16
in your way is the same as solving the problems
A plus '6 x A equals 16 '2 '8 and A plus '1 x A equals 10 "3
in my way. When you do so you obtain a triple - if the divisor of
A is greater than 1, a whole number or a series of unit fractions.
When the divisor is 1 you get a solution based on the strange triple
zero one one.
I gave this number 'zero' a long and careful thought. Here are my
conclusions.
If you and your miraculous calculating ushebtis can handle this funny
number it's fine. I for myself am quite happy with our number system.
Please have a look at my personal calculating board. There are fields
from one to a million. A small bean 'o' means one or ten or a million,
depending on the field I lay the bean, while a large bean 'B' means
five or fifty or five millions. Here I carry out the multiplication
625 times 625. Being a learned man of the future you can certainly
follow my steps:
hundred ten
millions thousands thousends thousends hundreds tens ones
I--------I---------I---------I---------I----------I--------I--------I
I I I I I B o I o o I B I
I I I I I B o I o o I B I
I I I I I B o I o o I B I
I I I I I B o I o o I B I
I I I I I B o I o o I B I
I I I I B o I o o I B I I
I I I I B o I o o I B I I
I I I B o I o o I B I I I
I I I B o I o o I B I I I
I I I B o I o o I B I I I
I I I B o I o o I B I I I
I I I B o I o o I B I I I
I I I B o I o o I B I I I
I--------I---------I---------I---------I----------I--------I--------I
o o o B oooo B o o o B
Result: 3 hundred thousands 9 ten thousands 6 hundreds 2 tens 5 ones.
When I square the number 5 hundreds I obtain 5 B's which I lay on
the field hundred times hundred equals ten thousands. 4 B's on this
field equal 2 o's on the field hundred thousands. My result will
therefore be 2 hundred thousands 5 ten thousands.
These operations are a problem for beginners, but with a little
experience you will carry them out very quickly and without much
thinking anymore. And those pupils of mine who fail doing so will
leave my class anyway, sooner or later.
I imagine that you work with much bigger numbers than we do. In this
case it might be reasonable to mark the empty places by a number zero,
- e.g. 5 hundreds and zero tens and 4 ones - in order to render the
numbers in a more systematic way.
Used like this, the number zero would be a technical number while
our passion goes for a real number, namely for the number 1 and the
cosmos of numbers it contains. Every unit fraction is part of this
cosmos while it can evolve into growing series of unit fractions:
1 = '1
1 = '2 '2
1 = '2 '4 '4
1 = '2 '4 '8 '8
1 = '2 '4 '8 '16 '16
1 = '2 '4 '8 '16 '32 '32 and so on
1 = '1
1 = '1x2 '2
1 = '1x2 '2x3 '3
1 = '1x2 '2x3 '3x4 '4
1 = '1x2 '2x3 '3x4 '4x5 '5
1 = '1x2 '2x3 '3x4 '4x5 '5x6 '6 and so on
By the way, John, your funny number zero caused me a sleepless night.
If you ever come visit me in the Egypt of my days you owe me a jug of
beer!
Sincerly, Ahmes
cir...@access.ch
whi...@shore.net (Steve Whittet) wrote:
> Seems like professor Ahmes! is starting a new class this fall...
Ahmes: Welcome in my class :)
> Looking at what Franz is doing with the RMP I would be interested
> in how this differs from the carpenters method of determining the
> length of cripple jack rafters in framing a roof.
>
> Todya the common difference in the length of rafters is often
> taken off a table on the framing square. At some point it had
> to occur to someone that there was a common difference and that
> a table of its values would be useful.
>
> My suspicion was that the unit spacing for rafters followed after
> the use of unit spacing for the frames in boats and so was a rather
> late development, but the dimensions given for some ancient buildings
> make me wonder.
Ahmes: I must say I feel helpless when I hear your words. Please,
what is a 'cripple jack rafter'? You live some 3,600 years ahead
of my time. You are clever people and probably made inventions we
can't imagine. If you wish to get an answer from me please make it
simple. Or are you asking John? Well, he will certainly understand
your words and may reply by himself.
> Anyway the Egyptians have a hieroglyphic for the framing square
> and also used the plumb bob, level, and a number of different
> measuring instruments, so I guess my question is to what extent
> can you measure out an arithmentic solution to the problems in
> the RMP using a graduated straightedge, compass and square...
Ahmes: Feeling helpless again. However, I can tell you our principles.
We like to solve a mathematical problem in a convincing and pleasing
way. Then we use our solutions for deducing very simple and reliable
methods. I mean our scribes have a good life while our workers have
it rather hard. It's only fair if we do a lot of brainwork and equip
them with simple and reliable methods. In fact we have many solutions
for every kind of problem. Tell me a problem in a way I can follow,
and I may tell you our solutions - simple ones or more sophisticated
ones, as you like.
> I think the use of unit fractions, sequences and common differences
> are all tied to the three clasic problems of antiquity, doubling
> the cube, trisecting an angle and squaring a circle only without
> the strict definition of what consitutes a straightedge.
Ahmes: 'Dobling the cube' - this I understand. Please have a look
at one of my number patterns:
1 1 1 2
2 2 3 4
4 5 7 8
9 12 15 18
3 4 5 6
Go on like this. You will obtain the following number column:
1 1 1 2
2 2 3 4
4 5 7 8
9 12 15 18
3 4 5 6
7 9 11 14
16 20 25 32
36 45 57 72
12 15 19 24
27 34 43 54
61 77 97 122
138 174 219 276
46 58 73 92
104 131 165 208
235 296 373 470
531 669 843 1062
177 223 281 354
400 504 635 800 and so on
The edge of a cube measures 18 royal cubits or 504 fingers. Double
the volume and the edge will measure 635 fingers. Halve the volume
and the edge will measure 400 fingers.
504 divided by 400 equals 1 '4 '100
If you wish to double a volume, multiply every linear measurement
by a factor of 1 '4 '100.
635 divided by 800 equals '2 '4 '32 '80
If you wish to halve a volume, multiply every linear measurement
by a factor of '2 '4 '32 '80.
Steve Whittet
>
Finally! It's about time sci.archaeology began getting some class!
>
>> Looking at what Franz is doing with the RMP I would be interested
>> in how this differs from the carpenters method of determining the
>> length of cripple jack rafters in framing a roof.
>>
>> Today the common difference in the length of rafters is often
>> taken off a table on the framing square. At some point it had
>> to occur to someone that there was a common difference and that
>> a table of its values would be useful.
>>
>> My suspicion was that the unit spacing for rafters followed after
>> the use of unit spacing for the frames in boats and so was a rather
>> late development, but the dimensions given for some ancient buildings
>> make me wonder.
>
>Ahmes: I must say I feel helpless when I hear your words. Please,
>what is a 'cripple jack rafter'?
A person finds the length of a rafter by taking a unit rise and run.
where the rafters land on a hip or valley they are cut off and
called jack rafters. Where the pitch on the two sides of the hip
or valley is different the jacks are different lengths and so they
are sometimes called "cripples". the common difference in their
lengths depends on the pitch of the hip or valley and the pitch
of the roof.
How does a carpenter with little or no math skills know how long
and at what angle to make his cuts.
Although tables make the calculations easier the common difference
can also be made from manipulations of the framing square which is
simply a scale model of a right triangle. This is what I would call
a geometric solution and it raised the question in my mind how
much math you can do by mensuration or by constructing a table.
The important words are "table" and "common difference"; ie'
one of the pythagorean triangles is a table.
1 1
1 1 2 1
1 2 1 4 2
1 3 3 1 8 4
1 4 6 4 1 16 8
... each number is the sum of the two above it ...
the rows sum to powers of two and the numbers in the rows
give the coefficients of expansion ...
probabilities of heads or tails...etc;
or
1 1
1 3 4 3
1 3 5 9 5
1 3 5 7 16 7
...the sum of each row is the series of squares
the second difference is the series of odd numbers
etc;
In the RMP and EMLR there are a number of tables which seem
to lend themselves to being thought of as an extendable ruler
with 28 divisions which can be divided into '2, '4, '7, '14, '28
etc; and then these parts can be either further subdivided into
3 or 4 or 5 or more parts or added onto the base measure...
A royal cubit is divisible into fingers, inches, nails, palms,
hands, fists, spans, cubels, ...
In a way tables of unit fractions with their common differences
suggest continued fractions. A number of tables on different bases
placed next to each other can generates a pattern which provides
predictable results.
>You live some 3,600 years ahead
>of my time. You are clever people and probably made inventions we
>can't imagine. If you wish to get an answer from me please make it
>simple. Or are you asking John? Well, he will certainly understand
>your words and may reply by himself.
The example I gave is similar to what your weavers do in making
the design of Re on your fine linen. How does a weaver untrained
in mathematics calculate how many threads of warp and woof the
shuttle will cross above and below to lay out a circular pattern?
>
>> Anyway the Egyptians have a hieroglyphic for the framing square
>> and also used the plumb bob, level, and a number of different
>> measuring instruments, so I guess my question is to what extent
>> can you measure out an arithmentic solution to the problems in
>> the RMP using a graduated straightedge, compass and square...
>
>Ahmes: Feeling helpless again. However, I can tell you our principles.
>We like to solve a mathematical problem in a convincing and pleasing
>way. Then we use our solutions for deducing very simple and reliable
>methods. I mean our scribes have a good life while our workers have
>it rather hard. It's only fair if we do a lot of brainwork and equip
>them with simple and reliable methods. In fact we have many solutions
>for every kind of problem. Tell me a problem in a way I can follow,
>and I may tell you our solutions - simple ones or more sophisticated
>ones, as you like.
How would you instruct me to fabricate a cylindrical container
which will hold the same volume as a cubical one? Would it be
easier for me to get right if I made the circumference of the
circle the same as the perimeter of the square?
>
>> I think the use of unit fractions, sequences and common differences
>> are all tied to the three clasic problems of antiquity, doubling
>> the cube, trisecting an angle and squaring a circle only without
>> the strict definition of what consitutes a straightedge.
>
>Ahmes: 'Dobling the cube' - this I understand. Please have a look
>at one of my number patterns:
>
> 1 1 1 2
> 2 2 3 4
> 4 5 7 8
> 9 12 15 18
> 3 4 5 6
(curt 1)(curt 2) (cu rt4) (curt8)
(curt8) (curt 16) (cu rt32) (curt64)
(curt64) (curt 128) (cu rt256) (curt512 )
so far so good then it looks like you take numbers at random
9
3
7
etc
>
>Go on like this. You will obtain the following number column:
>
> 1 1 1 2
> 2 2 3 4
> 4 5 7 8
> 9 12 15 18
> 3 4 5 6
> 7 9 11 14
> 16 20 25 32
> 36 45 57 72
> 12 15 19 24
> 27 34 43 54
> 61 77 97 122
> 138 174 219 276
> 46 58 73 92
> 104 131 165 208
> 235 296 373 470
> 531 669 843 1062
> 177 223 281 354
> 400 504 635 800 and so on
>
>The edge of a cube measures 18 royal cubits or 504 fingers. Double
>the volume and the edge will measure 635 fingers. Halve the volume
>and the edge will measure 400 fingers.
That's a very interesting table. How could you arrive at it
in the form of a table of differences?
> 504 divided by 400 equals 1 '4 '100
>
>If you wish to double a volume, multiply every linear measurement
>by a factor of 1 '4 '100.
>
> 635 divided by 800 equals '2 '4 '32 '80
>
>If you wish to halve a volume, multiply every linear measurement
>by a factor of '2 '4 '32 '80.
How would it work if you wanted to pack cubes of unit sizes
in larger cubes so as to get aproximately double the volume
of a smaller cube?
How close could you come? Suppose you had two different standards
of measure, one for cubes containing unit cubes and one for
double cubes of unit cubes. What would the proportions of
the two measures be?
>
>Sincerly, Ahmes
Thank you Ahmes,
steve
John Legon
>
>Looking at what Franz is doing with the RMP I would be interested
>in how this differs from the carpenters method of determining the
>length of cripple jack rafters in framing a roof.
>
>Todya the common difference in the length of rafters is often
>taken off a table on the framing square. At some point it had
>to occur to someone that there was a common difference and that
>a table of its values would be useful.
I take it you mean the rafters that had to be cut to a progression of
different lengths to fit the sloping corner-edge of a roof? Like the
professor, I have not heard of the term 'cripple jack' before.
>
>My suspicion was that the unit spacing for rafters followed after
>the use of unit spacing for the frames in boats and so was a rather
>late development, but the dimensions given for some ancient buildings
>make me wonder.
I wonder about some of those dimensions myself. Perhaps there was some
use for arithmetic progressions in the Egyptian building trade, though I
can't think of one at the moment...
>
>Anyway the Egyptians have a hieroglyphic for the framing square
>and also used the plumb bob, level, and a number of different
>measuring instruments, so I guess my question is to what extent
>can you measure out an arithmentic solution to the problems in
>the RMP using a graduated straightedge, compass and square...
I guess the Egyptians could have used graphical methods to solve some of
these problems, or to provide a check on the results in some cases.
>I think the use of unit fractions, sequences and common differences
>are all tied to the three clasic problems of antiquity, doubling
>the cube, trisecting an angle and squaring a circle only without
>the strict definition of what consitutes a straightedge.
>
>Given that any carpenter can trisect a ninety degree angle,
>what could a talented scribe do with a ruler.
>
>Perhaps some of the properties of circles which have the same
>circumference as the perimeter of squares or the development
>of standards of measure where packing a number of unit values
>into another unit value arrived in some fairly useful close
>aproximations.
>
Well, I think there was always a problem translating the results of
calculations into dimensions. A lot of people don't realise, though,
that the Egyptians sometimes used natural fractions of the royal cubit
in their measurements, such as '2, '3, or '4, as well as the septenary
divisions found on the 'votive' cubit rods.
>>[Snip]
>
>>the northern shaft from the King's Chamber intersected the sloping side
>>of the pyramid at exactly the same level as the southern shaft, even
>>though the slope of the northern shaft is 7 rise on 11 base, or half the
>>pyramid profile of 14 rise on 11 base.
>>
>>Gantenbrink states that this level is 80.63 ms +/- 4 cms over the base,
>>which is 154 equals 7 x 22 cubits. At this level, the diagonals of the
>>horizontal cross-section of the Great Pyramid measure 280 cubits, which
>>is also the height of the pyramid. At the floor level of the King's
>>Chamber, 82 cubits above the base, the diagonals of the horizontal
>>cross-section measure 440 cubits, which is also the side of the base.
>
>This all would seem to be implied by the septenary system of
>measures used to lay out the unit rise and run.ie; the ruler
>is 28 fingers long, one fourth that distance is seven fingers
>if you measure seven fingers in from one end and 22 fingers in
>from the other your ticks will enclose the Isis mark.
>
>seven fingers is one hand plus one palm or 1 span.
>14 fingers or '2 the royal cubit ruler is 14 fingers or 2 spans
Certainly in my view, the septenary division of the cubit had a bearing
on the design factors in pyramids, etc. 28 fingers in the cubit and 280
cubits in the height of the Great Pyramid, for example. Perhaps the
professor knows how the division of the royal cubit into seven palms
came about.
>
>The circuit of the wall of the kings chamber is to its length
>as the circumference of a circle to its diameter because the wall
>actually goes '14 its height below the level of the floor.
>Its height is 10 royal cubits or 1 hayt and its span is 20
>royal cubits or 2 hayt.
According to this theory, and taking pi as 3 '7, the wall-height should
be 11 cubits 3 palms. The theory that the cubic diagonal of the chamber
is just 25 cubits gives a floor-to-ceiling height of 5*sqr(5) = 11.18..
cubits. The difference from the wall-height is then is about '4 cubit,
which is rather less than '14 of the height.
Petrie, who discussed these theories at some length, suggested that the
external proportions of the Great Pyramid were adjusted so as to express
the pi-relation more accurately than the value of 22/7.
Regarding RMP 24-7:
>
>I think those are what the scribe Gillings calls think of a number
>problems. I am not sure how you would frame a geometric solution
>for this but I guess it has something to do with the "3 table
>and to me this suggests those pythagorean dotted number series...
>
I understand, professor, that your Mesopotamian brethren devised a
method for evolving Pythagorean triples using the system you describe,
and recorded the results on one of their clay tablets (Plimpton 322).
I tend to think of Egyptian maths in terms of practical real-world
applications, but I don't doubt that their thinking went far beyond the
methods preserved in the papyri.
>
>>>Ahmes: Why, so long everybody.
>>>
>>My best wishes, professor.
>>
>>John
>>
>regards,
>
>steve
>
I'm sorry if I gave you a sleepless night, Franz. And by the way, I've
been taking a taking a look at your personal calculating board...
An offering of @ loaves of bread and @ jugs of beer for the professor!
--
John Legon
cir...@access.ch
John Legon <jo...@nospam.legon.demon.co.uk> wrote:
> An offering of @ loaves of bread and @ jugs of beer for the professor!
Ahmes: Thank you, John, you are so kind :)
(Ahmes and I go bubblefish again)
Ahmes: I am still wondering about this 'web' of your's. How does
it work?
Me: Well, we lay out a web of wires - fine threads made of copper.
Then we send through electricity - you may imagine a tame lightning
- and make it vibrate according to our voices.
Ahmes: Amazing. Some people here tame animals. I shure enough never
heard that someone tries to tame a lightning. You must be courageous
and very clever people.
Me: Clever and silly. For example many of our professors believe
that you have a rather poor idea of mathematics ...
Ahmes: Really?
Me: ... while others believe that the pyramids were built by giants
who came from the middle of the ocean, or even by lion people who
came from the heavens.
Ahmes: How strange. Don't you folks read our papyri? Don't you have
a look at the reliefs ornating the valley temples, causeways, pyramid
temples and temenos walls?
Me: Um ...
Ahmes: What?
Me: No, I can't tell you.
Ahmes: Please do. Go on.
Me: These buildings disappeared a long time ago. And so did many
papyri.
Ahmes: Our temples shall vanish? our memories go lost?
Me: The River Nile sinks and raises again ... Many of your works
will disappear in the sand --- and be found again in our days.
Even a mathematical papyrus by one Ahmes ...
Ahmes: In a good condition?
Me: In a very fine condition. It was acquired 140 years ago - I mean
it shall be acquired in some 3,500 years - by the British Museum.
Ahmes: ... by the pretty museum?
Me: British. Great Britain is an island. So to speak the motherland
of America. While America may be seen as a modern version of Egypt.
Ahmes: I get me a funny picture of your time.
Me: As we got a funny one of your's, I guess.
Ahmes: Yeah, really.
Me: Anyway, your ideas live on. The seed you plant will become
a strong tree.
Ahmes: What do you mean?
Me: Many people shall visit Egypt and learn most eagerly all they
can learn from your successors ...
Ahmes: Are you sincere?
Me: ... while your creation myth comes to life again in the physics
of our time -- only that we speak of a 'vacuum sea' instead of the
Primeval Water Nun, of 'virtual particles' instead of mud particles,
and of a ' singularity' instead of the Primeval Water Nun.
Ahmes: Single means one, alone. May it be that your word 'single...'
stays for our Primordial One?
Me: You may say so.
Ahmes: You mean you are still struggling with our main question:
how did the many come forth from the One?
Me: We do. Our professors propose many theories. One is called TOE
= theory of everything. Many theories are strange. Really strange.
A professor says: more crazy than all the ancient mythologies ...
Ahmes: Well, WE are not crazy. We experience life and render our
fine reasoning in the form of theological stories.
Me: I know, professor. We don't understand your symbols any more,
therefore we miss a lot you say. But we work on it. May I mention
professor Erik Hornung? a scribe whose papyri I estimate highly.
Ahmes: Very pleased
Me: Other ones explore your mathematical work. Milo Gardner found
the general formulas of your conversions "a equals 'b 'c 'd ...
Steve Whittet likes your system of measures and scales. John Legon
wrote two articles on your 'aha' problems. Unfortunately I was not
able to find his papyri. Anyway, I find his calculations interesting.
He tries to apply them to the shafts and slopes of Khufu's Horizon.
(Continuation follows)
cir...@access.ch
cir...@access.ch wrote:
> Steve Whittet likes your system of measures and scales. John Legon
> wrote two articles on your 'aha' problems. Unfortunately I was not
> able to find his papyri. Anyway, I find his calculations interesting.
> He tries to apply them to the shafts and slopes of Khufu's Horizon.
Ahmes: How can you know about these shafts??
Me: People broke an entrance into Khufu's pyramid a long time ago.
Nowadays we try to understand, preserve and restaure this magnificient
building. As well as we can. A pity we have no plans and papyri at
hand. Would be a great help.
Ahmes: Well, if you preserve and restaure the pyramids I will help
you. Here in my house I keep a precious papyrus from Khufu's time,
showing his horizon. I will tell you some numbers, then you draw
a plan for John. Fine?
Me: Very fine, thank you so much. (...)
77 121 99 99 121 77
o------o-----------o--------C--------o-----------o------H
I I
I I
I I 126
I I
o o B o D o o
I I 34 "3
o o F o G o o
I I
I I
I I 119 '3
I I
o------A o o o E------o
I I 44
o------o-----------o--------o--------J-----------o------o
Length of the grid 77+121+99+99+121+77 = 594 royal cubits
Height of the grid 126 + 34 "3 + 119 '3 + 44 = 324 royal cubits
Pyramid A-B-C-D-E-A
base A-E southern slope A-B-C northern slope C-D-E
height : base = 280 : 440 = 7 : 11
Main measure royal cubit (52.36 cm)
complementary measure magic rod (33.32 cm)
magic rod : royal cubit = 7 : 11
Northern shaft of the upper chamber D-H
tangent 126 : 121+77 = 7 : 11
ideal angle 32 degrees 28 minutes 16 seconds ...
actual angle 32 degrees 38 minutes (Gantenbrink)
32 degrees 28 minutes (Bauval)
Northern shaft of the lower chamber G-H
tangent 126+34"3 : 121+77 = 241 : 297
ideal angle 39 degrees 3 minutes 27 seconds ...
actual angle 39 degress (Bauval)
Southern shaft of the upper chamber J-B
tangent 99+99 : 121+77 = 1 : 1
ideal angle 45 degrees
actual angle 45 degrees (Gantenbrink, Bauval)
Southern shaft of the lower chamber J-F
sine 7 : 11 (based on the pseudo-triple 490-594-770)
ideal angle 39 degrees 31 minutes 16 seconds ...
actual angle 39 degrees 30 minutes (Gantenbrink, Bauval)
Point F should lay behind the Gantenbrink door and may hold
a statue of Khufu reborn as Osiris, facing the culminations
of Sirius and Orion.
John Legon
>
>Ahmes: How can you know about these shafts??
>
>Me: People broke an entrance into Khufu's pyramid a long time ago.
>Nowadays we try to understand, preserve and restaure this magnificient
>building. As well as we can. A pity we have no plans and papyri at
>hand. Would be a great help.
>
>Ahmes: Well, if you preserve and restaure the pyramids I will help
>you. Here in my house I keep a precious papyrus from Khufu's time,
>showing his horizon. I will tell you some numbers, then you draw
>a plan for John. Fine?
>
>Me: Very fine, thank you so much. (...)
John: My dear professor Ahmes, your plan for the north shaft from the
upper chamber is exactly the same as mine. You have good information...
First, we construct a square with a diagonal of 280 cubits, equal to the
height of Khufu's pyramid, and place this square centrally beneath the
apex. Since the sides of this square will measure 198 cubits, the lower
side will be (280 - 198) or 82 cubits above the base. This gives the
floor-level of the upper chamber, which is from 1691.4 to 1693.7 inches
+/- 0.6 inches above the base according to the surveyor Petrie (82.03 to
82.14 cubits; I use Petrie's finding for the royal cubit of 20.62 inches
or 0.52375 ms).
The intersections between the north and south sides of this square and
the north and south sides of the pyramid then define the exit points of
the shafts from the upper chamber. They are both 99 cubits horizontally
from the centre, and therefore both 99 x 14/11 equals 126 cubits below
the apex (as you have shown in Franz's diagram, professor). This level
will be (280 - 126) equals 154 equals 14 x 11 cubits above the base, and
thus coincides with the 105th course of the pyramid (Petrie: 3174.7 to
3176.0 inches above the base, mean 154.0 cubits). As mentioned before,
robotics expert Gantenbrink found intersections 80.63 ms +/- 4 cms above
the base (~ 3174.4 inches).
Next, we place another square with the side 198 cubits on the north side
of the first square, to give the point (H) in your diagram, (99 + 198)
or 297 cubits horizontally from the apex. Connecting this point to the
exit point for the northern shaft as already defined, gives the slope of
the shaft of 126 rise on 198 base, or 7 rise on 11 base (the line H-D).
What is so interesting - and as you have also discovered - is that the
northern shaft from the lower chamber is aligned towards the same point
(H) in your diagram. Sadly, Gantenbrink's measurement of the angle of
this shaft was very uncertain, because of fractures in the shaft due to
'Erdbebenverschiebungen', and because Gantenbrink's little robot could
only go a short distance up the shaft. You can find more about this,
I'm sure, in the Mitteilungen [MDAIK 50 (1994), p. 285-294]. (Please
ignore the footnotes on p. 290).
Because the lower (Queen's) chamber is located in the centre of Khufu's
pyramid, I think the shafts from the lower chamber were symmetrical, and
that the angle of the northern shaft was equal to that of the southern
shaft. Yes, I thought this angle might be 11 slope on 7 rise, giving a
very simple numerical variation on the profile of the upper shaft, but
my (published) ideal value was 14 rise on 17 base.
I put forward this value because of the simplicity of the construction.
We just bisect the distance of (99 + 198) cubits from the apex (C) to
the point (H), and mark off the result of 148.5 cubits below the apex to
give the level of the shaft outlets, (280 - 148.5) equals 131.5 cubits
above the base. This level coincides with the 90th course of Khufu's
pyramid, which is one of those peculiar thick courses (level of 90th
course, Petrie, 2711.1 inches or 131.48 cubits). Taking a line to the
point (H) gives a profile of 1 rise on (2 - 11/14) base, or 14 rise on
17 base, with an angle of 39* 28' 21". (14-17-22 pseudo-triple).
Well, that would be interesting, but... The angles of the shafts were
surely determined by simple geometry and not by stellar alignments:)
>(Continuation follows)
>
Many thanks once again, Franz.
John Legon
cir...@access.ch
cir...@access.ch wrote: ...
While transscribing my interview with Ahmes two days ago I made
a mistake. Here is the correct version:
Me: ... while your creation myth comes to life again in the physics
of our days -- only that we speak of a 'vacuum sea' instead of the
Primeval Water Nun, of 'virtual particles' instead of mud particles,
and of a 'singularity' instead of the Primeval Mound.
Ahmes: You should work more carefully. Always mind what you write.
Me: I copy my papyri with the help of an ushebti and still make a
lot of mistakes. You copy a scroll ten cubits long without any help
and make only few mistakes. I find this amazing. You really know
what you are saying.
Ahmes: Why, of course. Now please let me reply to Steve and John.
Thank you both for your many hints and questions.
Here are my answers.
Sorry, Steve, I still don't really understand your rafter problem.
I wish you would render it in a simple way and look out for a simple,
correct and pleasing solution. As I tell my pupils: looking out for
such a solution is always a good exercise for the mind.
I sure like your kind words about our measures and scales. You are
right: we solve many problems in a geometric way. First we use grids
and plans and models. When you work with a grid you can count the
measurements and even define some points of a round figure. Please
imagine a grid measuring 10 x 10 royal cubits and inscribe a circle
of the radius 5 royal cubits.
Me: It may look like this:
. . . . . o . . . . .
. . o . . . . . o . .
. o . . . . . . . o .
. . . . . . . . . . .
o . . . . . . . . . o
. . . . . . . . . . .
. o . . . . . . . o .
. . o . . . . . o . .
. . . . . o . . . . .
Ahmes: The circle passes 12 points of the grid. These points define
as many arcs. 4 arcs measure 40 fingers each while the other 8 arcs
measure 90 fingers each, giving a circumference of 880 fingers.
Divide this number by the 280 fingers of the diameter and you will
obtain 3 '7 as a very good first approximation of the secret number
living in the circle.
Our royal cubit is divided in 7 palms. This number allows a simple
definition of the circle: if the diameter maesures 1 royal cubit
or 7 palms the circumference measures 22 palms or 3 cubits 1 palm.
My above grid measures 10 x 10 royal cubits or 70 x 70 palms. How
long are the diagonals of the grid? Let me draw up one of my number
columns:
1 1 2
2 3 4
5 7 10
12 17 24
29 41 58
70 - - 99 140 and so on
The side of my grid measures 10 cubits or 70 palms. Hence the
diagonal measures 99 palms or 14 cubits 1 palm.
You see, John? We like to work with simple numbers, and we choose
our measures in such a way that we obtain whole numbers, even when
we carry out a more demanding calculation.
Have a nice day, Ahmes
cir...@access.ch
cir...@access.ch wrote: ...
Me: Good morning, professor Ahmes.
Ahmes: Good morning. May I reply to John?
Me: Of course.
Ahmes: Good morning, John.
You propose the numbers 14-17-22 instead of 490-594-770. Well,
your numbers are a fine and useful approximation. They contain
the magic numbers 7 and 11
14 = 2 x 7 22 = 2 x 11
and are based on the value 1 '4 '6 for the square root of 2 (long
side 22, short side 14; 22x22 = 484, 14x14 = 196; 484-196 = 288
= 12x12x2; middle side 12 x square root of 2 = 12 x 1 '4 '6 = 17).
Why do I speak of 'magic' numbers? I will show you.
My old papyrus says: the base of the pyramid measures 11 x 40 royal
cubits while the height measures 7 x 40 royal cubits. Furthermore it
says: the numbers 7 and 11 define a pair of complementary measures:
royal cubit = 7 palms = 28 fingers = 308 Horus marks
magic rod = '7 x 11 royal cubits = 196 Horus marks
Why the name 'magic rod'?
Please have a look at the following formulas and calculations:
diameter of a circle 1 mr circumference 2 rc
radius of a circle 1 mr area 2 x mr x rc
diameter of a sphere 1 mr surface 2 x mr x rc
volume of 3 spheres mr x mr x rc
side of a square 10 mr or 9 rc
diagonal 9 rc or 20 mr
a given length 5 rc or 10 rc or 15 rc or 20 rc
golden minor 3 mr or 6 mr or 9 mr or 12 mr
cult pyramid base 40 royal cubits height 40 magic rods
tomb pyramid base 440 royal cubits height 440 magic rods
Now use the height 440 magic rods as the diameter of a circle.
How long is the circumference of the semicircle? 440 royal cubits.
What is the area of the circle? 2 x radius in magic rods x as many
royal cubits = 2 x 220 magic rods x 220 royal cubits = 2 x 140 royal
cubits x 220 royal cubits = 61,600 square cubits.
And what is the section area of the pyramid? '2 x height x base
= '2 x 280 royal cubits x 440 royal cubits = 61,600 square cubits.
Square the circle. The side of the square will measure 390 magic rods
while the diagonal will measure 351 royal cubits.
Now transform the pyramid's base into a circle of the same area.
The radius will measure 390 magic rods.
As you say, the side of the pyramid on the height 154 royal cubits
measures 198 royal cubits. Hence the remaining height measures 198
magic rods. How long is the diagonal on this level? 198 royal cubits
divided by 9 royal cubits multiplied by 20 magic rods = 440 magic
rods or 280 royal cubits.
In the lower chamber is found a niche whose vertical axis divides
the length of the chamber - 10 royal cubits - in the golden ratio.
The golden minor measures 6 magic rods. Remember my numbers?
Me: We all are very interested in the southern shaft of the lower
chamber. Vizier Stadelmann says it measures 22 x 22 centimeters
(one of our measures). Now my calculating ushabti says this may be
2/3 or "3 of a 'magic rod' while the diagonal may be 3/5 or '2 '10
of a royal cubit. Can you affirm this? What does your papyrus say?
Professor? Hello? Professor Ahmes?
John Legon
>
>Ahmes: Good morning, John.
>
calculations. I read through your numbers and ideas and find much in
agreement with my own way of thinking. You raise some questions...
>
>In the lower chamber is found a niche whose vertical axis divides
>the length of the chamber - 10 royal cubits - in the golden ratio.
>The golden minor measures 6 magic rods. Remember my numbers?
>
golden ratio everywhere in Khufu's pyramid, but I'm not convinced!
Anyway, it's not a very accurate result. You should check your figures.
>Me: We all are very interested in the southern shaft of the lower
>chamber. Vizier Stadelmann says it measures 22 x 22 centimeters
>(one of our measures). Now my calculating ushabti says this may be
>2/3 or "3 of a 'magic rod' while the diagonal may be 3/5 or '2 '10
>of a royal cubit. Can you affirm this? What does your papyrus say?
John: I can tell you what Petrie says, in his 'The Pyramids and Temples
of Gizeh' (1st.ed p.70): "The N. channel is 8.6 (inches) high, and about
8 wide in the chamber wall.... The S. channel is 8.8 (inches) high..."
For the north shaft in the upper chamber, he says (p.84): "The remains
of the original channel show it to have varied from 8.9 to 9.2 (mean
9.0) in width, and to have been 8.72 and 8.74 in height." For the south
shaft he says: "Its width at the top is 8.35 and 8.65, and its height
8.7 to 8.9 (inches)."
I think these shafts were supposed to be just 3 palms square, with a
diagonal of (near enough) 17 fingers. But your own figures also fit.
>
>Professor? Hello? Professor Ahmes?
>
Au revoir, professor...
John Legon
cir...@access.ch
John Legon <jo...@nospam.legon.demon.co.uk> wrote:
> John: Hmm. I really wonder about this. Some people want to find the
> golden ratio everywhere in Khufu's pyramid, but I'm not convinced!
Good morning, John. I will ask the professor about this another day,
just now I can't reach him.
Here is again the scheme of the Great Pyramid, with a comparison
of the ideal angles of the shafts and the actual angles measured
by Rudolf Gantenbrink.
77 121 99 99 121 77
o------o-----------o--------C--------o-----------o------H
I I
I I
I I 126
I I
o o B o D o o
I I 34.333
o o F o G o o
I I
I I
I I 119.333
I I
o------A o o o E------o
I I 44
o------o-----------o--------o--------J-----------o------o
Pyramid A-B-C-D-E-A
Northern shaft of the King's Chamber D-H
tangent 7 : 11
ideal angle 32 degrees 28 minutes 16 seconds ...
Gantenbrink 32 degrees 36 minutes 8 seconds
Northern shaft of the so-called Queen's Chamber G-H
tangent 241 : 297
ideal angle 39 degrees 3 minutes 27 seconds ...
Gantenbrink 39 degrees 7 minutes 28 seconds minus max. 2 degrees
Southern shaft of the King's Chamber J-B
tangent 1 : 1
ideal angle 45 degrees
Gantenbrink 45 degrees
Southern shaft of the so-called Queen's Chamber J-F
sine 7 : 11 (pseudo-triple 490-594-770)
ideal angle 39 degrees 31 minutes 16 seconds ...
Gantenbrink 39 degrees 36 minutes 28 seconds +- max. 12 minutes
Gantenbrink's numbers are found in: Die sogenannten Luftkanaele
der Cheopspyramide, Modellkorridore fuer den Aufstieg des Koenigs
zum Himmel; von Rainer Stadelmann, mit einem Beitrag von Rudolf
Gantenbrink; Mitteilungen des Deutschen Archaeologischen Institutes
Abteilung Kairo (MDAIK), Band 50 1994; pages 285-294.
Thank you, John, for telling me about this article.
Mark Lehner wrote a fine book on the pyramids. I like especially
the drawings. However, he is rather sloppy concerning the numbers.
There is no way around the books and articles by Rainer Stadelmann.
You can almost always rely on the mesurements given in centimeters
(not on the ones in cubits, though).
John, do you understand German? If yes, you may order this book:
Rudolf Krauss, Astronomische Konzepte und Jenseitsvorstellungen
in den Pyramidentexten, Harrassowitz Verlag Wiesbaden 1997, Band
59, herausgegeben von Ursula Roessler-Koehler, ISBN 3-447-03979-5.
Regards Franz Gnaedinger Zurich cir...@access.ch
John Legon
Thank you for giving me details about the book by Rudolf Krauss in your
posting this morning, and also for your earlier mention of the book by
Theophile Obenga. I will try to order copies next time I'm in London.
Yes, I can read German though with some difficulty. I have Stadelmann's
'Die aegyptischen Pyramiden', his 'Die grossen Pyramiden von Giza',
and also copies of some of his articles in MDAIK. Many of Stadelmann's
measurements for the Giza Pyramids come, I believe, from Maragioglio and
Rinaldi, 'L'Architettura delle Pyramidi Menfite', who in turn took them
from Petrie. I also have Mark Lehner's 'The Complete Pyramids'. It is
a nice picture-book, but Lehner is not very good with dimensions.
You can read my detailed discussions about the relevant astronomical
conceptions in the Pyramid Texts, in my article in 'Discussions in
Egyptology' vol. 33 (1995), p 45-56. This article also contains the
geometry of the air-shafts in Khufu's pyramid, with the latest data from
Gantenbrink.
Best wishes,
John Legon
cir...@access.ch
cir...@access.ch wrote:
> 77 121 99 99 121 77
> o------o-----------o--------C--------o-----------o------H
> I I
> I I
> I I 126
> I I
> o o B o D o o
> I I 34.333
> o o F o G o o
> I I
> I I
> I I 119.333
> I I
> o------A o o o E------o
> I I 44
> o------o-----------o--------o--------J-----------o------o
>
> Pyramid A-B-C-D-E-A
> Southern shaft of the so-called Queen's Chamber J-F
>
> sine 7 : 11 (pseudo-triple 490-594-770)
>
> ideal angle 39 degrees 31 minutes 16 seconds ...
> Gantenbrink 39 degrees 36 minutes 28 seconds +- max. 12 minutes
According to Gantenbrink, the oblique part of the southern shaft
of the so-called Queen's Chamber measures 57.55 meters what is only
46 millimeteres short of 110 royal cubits:
57.55 meters = 5755 centimeters
110 x 52.36 cm = 5759.6 centimeters
If we assume an ideal length of 110 royal cubits we obtain
the following numbers:
shaft 110 royal cubits or 770 palms
run 594 palms
rise 70 royal cubits or 490 palms
The length of the shaft measures one fourth of the pyramid's base
while the rise measures one fourth of the pyramid's height.
Another meaningless coincidence?
For building the shaft one might have used a wooden triangle of
these measurements:
vertical cathete 245 Re marks or 229.075 centimeters
horizontal cathete 297 Re marks or 277.695 centimeters
hypothenuse 385 Re marks or 359.975 centimeters
Rudolf Gantenbrink carried out 24 measurements and found a maximal
deviation of 12 minutes or 1/5 degree. I guess that such a precision
can be attained by means of the above triangle and a plummet.
John Legon
[snip diagram]
>> Southern shaft of the so-called Queen's Chamber J-F
>>
>> sine 7 : 11 (pseudo-triple 490-594-770)
>>
>> ideal angle 39 degrees 31 minutes 16 seconds ...
>> Gantenbrink 39 degrees 36 minutes 28 seconds +- max. 12 minutes
>
Or pseudo-triple 14-17-22, based upon simple geometry. This gives an
ideal angle of 39 degrees 28 minutes 21 seconds, which is well within
the range of variation in Gantenbrink's measurement.
>According to Gantenbrink, the oblique part of the southern shaft
>of the so-called Queen's Chamber measures 57.55 meters what is only
>46 millimeteres short of 110 royal cubits:
>
> 57.55 meters = 5755 centimeters
> 110 x 52.36 cm = 5759.6 centimeters
>
>If we assume an ideal length of 110 royal cubits we obtain
>the following numbers:
>
> shaft 110 royal cubits or 770 palms
> run 594 palms
> rise 70 royal cubits or 490 palms
>
>The length of the shaft measures one fourth of the pyramid's base
>while the rise measures one fourth of the pyramid's height.
>
>Another meaningless coincidence?
>
Quite possibly yes, in this instance, I'm afraid. When the results of
Gantenbrink's investigation were first reported, it was widely stated
that the length of the shaft was 65 metres, not 57.55 metres. When I
spoke to Gantenbrink about this, he told me that there were mistakes in
his paper in MDAIK 50, and that his corrections to the text were never
inserted...
It so happens that the entire lengths of both the north and south shafts
from the Queen's Chamber had been probed previously using rods, by the
late Joseph Lepre. According to Lepre: "The north channel is 240 feet
long, while the south channel is somewhat longer, being 250 feet in
length." Lepre also says: "Much ado has been made about these air ducts
and the fact that they were sealed off inside the Queen's Chamber and
cut to within 20 feet of debouching at approximately the ninetieth
masonry course."
See: J.P. Lepre, _The Egyptian Pyramids_ (Jefferson and London, 1990),
page 111, and the diagram on page 64. Although Lepre's numbers are not
accurate, they seem to exclude a shaft-length of only 57.55 metres.
Incidentally, there are scuff marks or scratches on the right side of
the shaft near the stone 'door', which could have been made by Lepre's
rod. It also seems likely that the left-hand copper attachment on the
door was knocked off by the same rod. See photo in MDAIK 50 (1994)
Tafel 56-b, and plate 15 in 'Message of the Sphinx/Keeper of Genesis' by
Hancock and Bauval.
My contribution to a mailing-list concerning this subject was forwarded
to Gantenbrink by Doug Weller. It would be nice if Doug could elicit a
reply...
Regards,
John Legon. jo...@legon.demon.co.uk
cir...@access.ch
John Legon <jo...@nospam.legon.demon.co.uk> wrote:
> Quite possibly yes, in this instance, I'm afraid. When the results of
> Gantenbrink's investigation were first reported, it was widely stated
> that the length of the shaft was 65 metres, not 57.55 metres. When I
> spoke to Gantenbrink about this, he told me that there were mistakes in
> his paper in MDAIK 50, and that his corrections to the text were never
> inserted...
Thank you, John. It's funny that Rainer Stadelmann gives the same 57.55
meters in the 1997 edition of DIE AEGYPTISCHEN PYRAMIDEN, Vom Ziegelbau
zum Weltwunder, Verlag Philipp von Zabern. It would be nice if someone
could give us the correct measurement.
Thank you also for correcting my scheme: the number 34.333 was wrong,
it should read 34.666 or 34 2/3 or 34 "3 royal cubits.
Ahmes: Work more carefully, I told you so before.
Me: Professor Ahmes? Are you back? Welcome. We are seriously trying
to understand the meaning of the shafts in Khufu's Horizon.
Ahmes: Any idea?
Me: For example John Legon sees them as airshafts.
Ahmes: Well, John is right.
Me: Robert Bauval says the shafts are pointing to the culminations
of several stars (Thuban, alpha Draconis, then pole star; Kochab,
beta Ursae Minoris; Al Nitak, left belt star in Orion; and Sirius).
Ahmes: Quite so.
Me: Rainer Stadelmann believes that the shafts are meant for Khufu's
deification.
Ahmes: Of course.
Me: How can these contrary views join?
Ahmes: The pyramid is full of sleeping powers:
The base means Geb, god of the earth, father of Osiris, Isis, Seth,
Nephtys and Horus.
The four faces mean Osiris, Isis, Seth and Nephtys.
The pyramidion means Horus.
The imaginary circle whose vertical diameter is given by the height
of the pyramid means Re, god of the sun.
The imaginary hemisphere in the frame of the golden pyramid means Nut,
goddess of the sky, mother of Osiris, Isis, Seth, Nephtys and Horus.
The shafts mean the arms of Shu, god of the air, who fills the cosmos
with the breath of life. Moreover, Shu is the father of Geb and Nut.
The pit in the lowest chamber, deep down in the rock, means Tefnut,
goddess of the moisture, wife of Shu, mother of Geb and Nut.
Air means life of the mind: spiritual life for the king. The pit
in the lowest chamber means water: physical life for the king.
Shu held Nut high above Geb, so that her body becomes the starry
heavens. As he gently touches her, his hands mean the stars.
Me: Five fingers - five rays?
Ahmes: The pit in the lowest chamber also means the Primeval Water
Nun. The pyramid also means the Primeval Mound or Hill. The imaginary
circle also means the sun rising from the Primeval Hill. The imaginary
hemisphere also means the sky once enclosed in the Primeval Hill.
Me: Such a simple form and so many meanings?
Ahmes: And of course the pyramid is a tomb. The mummy of the king
is laid into the sarcophagus in the upper chamber. His soul flies
to the northern stars. Here his life is judged. If his heart weighs
less than the feather of Maat, his soul is released and returns into
the pyramid. Now the sleeping powers wake up. Nut receives the soul
of the king. She carries him out and gives birth to him as Re-Osiris.
Me: A combination of Re and Osiris?
Ahmes: Being Osiris, the soul of the king flies to Orion. He loves
his wife in Sirius, and he looks down on earth with his eye the moon.
Being Re, the soul of the king grows up as the sunchild. Later on
he leaves his pyramid in form of the solar disk.
From now on, the king lives in the sky and travels along the h-channel,
wherein the planets, the moon and the sun are moving.
Me: In the stripe of the ecliptic, as Rolf Krauss says on his papyrus:
Astronomische Konzepte und Jenseitsvorstellungen in den Pyramiden-
texten (Harassowitz Verlag Wiesbaden 1997).
Ahmes: I see, you have all ideas. Now please combine them. Our works
are looking quite simple but I can tell you they are most complex.
Me: Simple yet complex?
Ahmes: The very formula for understanding our symbols.
SaddamHusseinsUncle
Watch him guys.
khufu himself may speak when the time is right.....
cir...@access.ch
sar...@tcp.co.uk (SaddamHusseinsUncle) wrote:
> cir...@access.ch is getting deep inspiration.
> Watch him guys.
> khufu himself may speak when the time is right.....
Ahmes: Who is Sargon?
Me: No idea. He may please get more explicit. Now, professor, I like
to ask you about the 'golden pyramid' you mentioned yesterday.
Ahmes: Oh yes. Try out a simple number game of mine:
4 + 4 = 8
4 + 8 = 12
8 + 12 = 20
12 + 20 = 32
20 + 32 = 52
32 + 52 = 84
52 + 84 = 136
84 + 136 = 220
136 + 220 = 356
Please notice the last numbers 220 and 356. Half the base of Khufu's
pyramid measures 220 royal cubits while the slope measures 356 royal
cubits.
220 is the golden major of 356.
Now let us imagine a hemisphere in the frame of a hollow pyramid.
It stays on the base of the pyramid and touches the four faces.
How long is the radius of the hemisphere?
9 + 16 = 25
16 + 25 = 41
25 + 41 = 66
41 + 66 = 107
66 + 107 = 173
107 + 173 = 280
The height of the pyramid measures 280 royal cubits, the radius of
the inscribed hemisphere measures 173 royal cubits, and the hemisphere
touches the four faces in a height of 107 royal cubits.
173 is the golden major of 280, 107 is the golden major of 173 and
the golden minor of 280.
A golden pyramid is defined like this:
golden minor of the slope = half the base
golden major of the height = radius of the inscribed hemisphere
golden minor of the height = height wherein the hemisphere
touches the four faces
area of one face = square of the height (no mistake)
3 x volume of the pyramid = 5 x volume of the hemisphere
(only a tiny mistake)
The pyramid symbolizes the Primeval Hill or Mound, the imaginary
hemisphere symbolizes the sky once enclosed in the Primeval Hill,
a semicircle of the hemisphere symbolizes the sky goddess Nut,
and the vertex of the semicircle or the zenith of the hemisphere
symbolizes her womb wherein she bears the sunchild.
Me: Thank you, professor.
May I remind you that most of our people believe that you have
a poor idea of mathematics? However, there are a few exceptions.
For example professor Jean-Philippe Lauer - whom no one in his
right mind can call a 'pyramidiot' - believes that the pyramid
builders knew the golden section very well.
Ahmes: You may try out another game. Draw a rectangle a. Add a square
to a longer side of this rectangle, thus you obtain rectangle b. Add
a square to a longer side of rectangle b, thus you obtain rectangle c.
Go on like this ...
Me: Let me try out. I begin with a double square:
a a c c c e e e e e e e e
b b c c c e e e e e e e e
b b c c c e e e e e e e e
d d d d d e e e e e e e e
d d d d d e e e e e e e e
d d d d d e e e e e e e e
d d d d d e e e e e e e e
d d d d d e e e e e e e e
Funny, it seems to approach the golden rectangle.
Ahmes: You see? These golden forms are very easily obtained.
Me: Again by simple methods based on addition.
John Legon
A reply to my query about the length of the southern air-shaft...
>
>Thank you, John. It's funny that Rainer Stadelmann gives the same 57.55
>meters in the 1997 edition of DIE AEGYPTISCHEN PYRAMIDEN, Vom Ziegelbau
>zum Weltwunder, Verlag Philipp von Zabern. It would be nice if someone
>could give us the correct measurement.
>
I have only the 1991 edition, but perhaps Stadelmann took his data from
Gantenbrink's contribution to MDAIK 50? Anyway, I hope that Gantenbrink
will soon respond to our request for more information.
>
>Me: Professor Ahmes? Are you back? Welcome. We are seriously trying
>to understand the meaning of the shafts in Khufu's Horizon.
>
>Ahmes: Any idea?
>
>Me: For example John Legon sees them as airshafts.
>
>Ahmes: Well, John is right.
>
John: Thank you, professor.
>Me: Robert Bauval says the shafts are pointing to the culminations
>of several stars (Thuban, alpha Draconis, then pole star; Kochab,
>beta Ursae Minoris; Al Nitak, left belt star in Orion; and Sirius).
>
>Ahmes: Quite so.
>
John: I can understand Bauval's point of view, but..
>Me: Rainer Stadelmann believes that the shafts are meant for Khufu's
>deification.
>
>Ahmes: Of course.
>
John: Model corridors for the ascent of the king to heaven? I don't
understand that. The souls of the deceased did not need to be guided
towards any destination, they were free to fly anywhere at will.
>Me: How can these contrary views join?
>
John: Dear professor, I find your ideas about the Great Pyramid poetic
and fascinating. For me, the Horizon of Khufu symbolised the point of
contact between the circle of the sun's disk (Re) and the Earth (Geb).
That is why it is reasonable to make an equation between the square base
of the pyramid, representing the Earth, and the height of the pyramid,
representing the diameter of the solar disk. You say that the imaginary
hemisphere in the frame of the pyramid represents Nut, the goddess of
the sky, and indeed the analogy with the hemispherical 'vault of heaven'
seems excellent to me.
Because some people do not understand why an Egyptian pyramid should be
ventilated, it might be interesting to quote some extracts from the
Egyptian 'Book of Going Forth by Day', which help to explain how I came
to my conclusion about the air-shafts.
In Spell 182, entitled 'Book for the permanence of Osiris, giving breath
to the Inert One in the presence of Thoth...', Thoth says: 'I have given
the sweet breath of the north wind to Osiris Wennefer as when he went
forth from the womb of her who bore him; I cause Re to go to rest as
Osiris, Osiris having gone to rest at the going to rest of Re... I am
Thoth who foretells the morrow and forsees the future... I give breath
to him who is in the secret places...'
In spell 183: '...I have given breath to Wennefer, even the fair breeze
of the north wind, as when he came forth from his mother's womb. I have
caused (it) to enter into the secret cavern in order to revive the heart
of the Inert One, Wennefer the son of Nut...'
The mouths of the air-shafts in the Queen's Chamber were closed off with
'plates of stone' in the wall-surface. Do we see Horus now enter the
chamber with an adze in his hand and cut open those plates of stone in
an 'opening of the mouth' ceremony? He lets air into the body of his
father and cries 'Let Osiris Live!'
>Ahmes: The pyramid is full of sleeping powers:
>
John: And do those powers still sleep, professor?
[....]
>Ahmes: I see, you have all ideas. Now please combine them. Our works
>are looking quite simple but I can tell you they are most complex.
>
>Me: Simple yet complex?
>
>Ahmes: The very formula for understanding our symbols.
>
John: In terms of geometry, I think the pyramid-builders found truth in
simplicity. However, the theological interpretations were multi-layered.
--
John Legon
John Legon
>Ahmes: Who is Sargon?
>
>Me: No idea. He may please get more explicit. Now, professor, I like
>to ask you about the 'golden pyramid' you mentioned yesterday.
>
>Ahmes: Oh yes. Try out a simple number game of mine:
>
> 4 + 4 = 8
> 4 + 8 = 12
> 8 + 12 = 20
> 12 + 20 = 32
> 20 + 32 = 52
> 32 + 52 = 84
> 52 + 84 = 136
> 84 + 136 = 220
> 136 + 220 = 356
>
>Please notice the last numbers 220 and 356. Half the base of Khufu's
>pyramid measures 220 royal cubits while the slope measures 356 royal
>cubits.
>
>220 is the golden major of 356.
>
I have looked at these numbers before, but I am still not convinced that
the Great Pyramid was intended to embody the golden proportion. The pi-
proportion is a good and convincing result, because the circle was the
symbol of Re, and the equation between the circle of the sun's disk and
the base of the Great Pyramid was appropriate to the Horizon of Khufu
and the conceptions of the solar cult.
Suppose the architect had just wanted to build a pyramid with a height
of 280 cubits and base of 440 cubits, as a pure expression of the number
of Re? He could never have avoided the coincidence of numbers which
modern mathematicians now interpret as indicating the golden proportion.
[....]
>The pyramid symbolizes the Primeval Hill or Mound, the imaginary
>hemisphere symbolizes the sky once enclosed in the Primeval Hill,
>a semicircle of the hemisphere symbolizes the sky goddess Nut,
>and the vertex of the semicircle or the zenith of the hemisphere
>symbolizes her womb wherein she bears the sunchild.
>
>Me: Thank you, professor.
>
>May I remind you that most of our people believe that you have
>a poor idea of mathematics? However, there are a few exceptions.
>For example professor Jean-Philippe Lauer - whom no one in his
>right mind can call a 'pyramidiot' - believes that the pyramid
>builders knew the golden section very well.
You surprise me. Lauer thought that the proportions of the Great Pyramid
were copied from those of the Meydum Pyramid, which came about when the
steps of the internal structure were filled in. He says:
"We thus have evident proof that this slope (of 14:11) resulted from
the actual proportion of the profile of the great step-pyramids of the
IIIrd Dynasty; and consequently nothing entitles us to suppose that the
architect of Cheops, any more than that of the Pyramid of Meydum, was
able to be conscious of the existence of the relations (phi) or (pi)
concealed in these proportions of the pyramid." J-Ph. Lauer, Le mystere
des pyramides (Paris, 1974), page 308 (my translation).
Regards,
John Legon
cir...@access.ch
John Legon <jo...@nospam.legon.demon.co.uk> wrote:
> I have only the 1991 edition, but perhaps Stadelmann took his data from
> Gantenbrink's contribution to MDAIK 50? Anyway, I hope that Gantenbrink
> will soon respond to our request for more information.
I fear it's more easy to get the most intimate details of Pharaoh
Bill Clinton's private life than a single measurement of the GP ;)
> In Spell 182, entitled 'Book for the permanence of Osiris, giving breath
> to the Inert One in the presence of Thoth...', Thoth says: 'I have given
> the sweet breath of the north wind to Osiris Wennefer as when he went
> forth from the womb of her who bore him; I cause Re to go to rest as
> Osiris, Osiris having gone to rest at the going to rest of Re... I am
> Thoth who foretells the morrow and forsees the future... I give breath
> to him who is in the secret places...'
Ahmes: Some of our people here say that Isis and Nephtys may be seen
as the arms of Nut. In a similar way Thoth is seen as an arm of Shu.
Being the god of writing, he is a spiritual man. His breath means:
life of the mind, spiritual life. Being the god of calculating and
measuring, he is the guardian of the unit measure which you wrongly
call a 'royal cubit' while it really represents an arm.
Me: Measuring 52.36 cm (a value affirmed to me by Rainer Stadelmann)
while the Florentine 'braccio' of the Renaissance measures 54.36 cm,
'braccio' meaning 'arm'.
Ahmes: Now for the pyramid shafts again. The northern shaft of the
upper chamber points to the pole star where Thoth weighs the heart
of the dead king against the feather of Maat. A feather belongs to
a bird - an animal flying in the air - while Maat is one of several
emanations of Tefnut, wife of Shu, god of the air.
> >Ahmes: The pyramid is full of sleeping powers
> John: And do those powers still sleep, professor?
Ahmes: A very good question. And a difficult one. I can only say:
we don't know. Anyway, who are we to judge?
The life of the king was judged by the heavenly ones. They decided
if his soul shall die, or live on, or even be deified.
Some people say: Look how the pyramids are decaying. The gods and
goddesses can't live in these buildings. Therefore the kings failed.
Other ones disagree and say: Oh yes, the kings were deified. It must
have happened secretly, whereupon they left the pyramids. The remains
we see today are like the empty shell of an egg when the bird slipped.
Again others say: As long as a pyramid keeps its name, or a single
measurement, even a small part of its shape, the gods and goddesses
may still come down and perform their deifying work. So let us restore
the pyramids as well as we can.
Dear John, will you help us restore our pyramids? in your days?
with the help of your amazing, industrious and diligent ushabtis?
------------------------------------------------------------------------
A postscript for Steve Whittet: I find your idea of evolving measures
beautiful and will reply later; please have a little patience.
Regards Franz Gnaedinger Zurich cir...@access.ch
-----== Posted via Deja News, The Leader in Internet Discussion ==-----
John Legon
>
>Ahmes: Some of our people here say that Isis and Nephtys may be seen
>as the arms of Nut. In a similar way Thoth is seen as an arm of Shu.
>Being the god of writing, he is a spiritual man. His breath means:
>life of the mind, spiritual life. Being the god of calculating and
>measuring, he is the guardian of the unit measure which you wrongly
>call a 'royal cubit' while it really represents an arm.
>
>Me: Measuring 52.36 cm (a value affirmed to me by Rainer Stadelmann)
>while the Florentine 'braccio' of the Renaissance measures 54.36 cm,
>'braccio' meaning 'arm'.
John: My goodness, what long arms people used to have! Mine measure
only about 49 cm from the elbow to the tip of the middle finger, even
though my height is above average...
As I mentioned before, my own favourite value for the Egyptian 'cubit'
comes from the work of Flinders Petrie, who obtained a result of 20.620
inches or 52.375 cm from the dimensions of the Giza pyramids. We all
have our own favourite values depending on our own favourite theories,
but Petrie's finding was quoted by I.E.S. Edwards in his 'The Pyramids
of Egypt', and also by Clarke and Engelbach in their work on 'Ancient
Egyptian Masonry'.
Often the difference between your value and mine makes no difference at
all, but let me explain why I prefer my value. In the precise survey of
the Great Pyramid by J.H. Cole (1925), the circumference of the base was
found to measure 921.455 metres. Assuming a theoretical circumference
of 2 x pi x 280 cubits, the length of the 'cubit' would be 52.3765 cm.
In my view, however, the builders did not set out a perfectly square
base, but kept one side with the length of just 440 cubits. This was
the longest, southern side, with a length of 230.454 metres (according
to Cole), giving another value for the cubit of 52.3759 cm.
In the survey by Petrie, the circumference of the base is 36275.2 inches
or 921.389 metres. For the pi-proportion, the length of the cubit would
be 20.6192 inches or 52.3727 cm. I take the mean value of 20.620 inches
or 52.375 cm, which was given by Petrie in the first instance, and find
that this works very well for the Giza site plan also.
Of course, our friend Lauer says that the value of 3.1416 for pi was
still manifestly unknown at the time of the Fourth Dynasty, so my own
findings are entirely speculative...
>
>> >Ahmes: The pyramid is full of sleeping powers
>> John: And do those powers still sleep, professor?
>
>Ahmes: A very good question. And a difficult one. I can only say:
>we don't know. Anyway, who are we to judge?
[...]
>Again others say: As long as a pyramid keeps its name, or a single
>measurement, even a small part of its shape, the gods and goddesses
>may still come down and perform their deifying work. So let us restore
>the pyramids as well as we can.
>
>Dear John, will you help us restore our pyramids? in your days?
>with the help of your amazing, industrious and diligent ushabtis?
John: Yes, professor. Already I have discovered many things and have
written about them in obscure 'journals' which sit on dusty shelves in
libraries where no-one will ever find them. But now, with this strange
'world-wide-web' device, perhaps it is time to put up a few pages with
diagrams so that we can have a better discussion about them.
cir...@access.ch
John Legon <jo...@nospam.legon.demon.co.uk> wrote:
> You surprise me. Lauer thought that the proportions of the Great Pyramid
> were copied from those of the Meydum Pyramid, which came about when the
> steps of the internal structure were filled in. He says:
> "We thus have evident proof that this slope (of 14:11) resulted from
> the actual proportion of the profile of the great step-pyramids of the
> IIIrd Dynasty; and consequently nothing entitles us to suppose that the
> architect of Cheops, any more than that of the Pyramid of Meydum, was
> able to be conscious of the existence of the relations (phi) or (pi)
> concealed in these proportions of the pyramid." J-Ph. Lauer, Le mystere
> des pyramides (Paris, 1974), page 308 (my translation).
Thank you for the reply, John. I read somewhere that Jean-Philippe Lauer
found evidence for the use of the golden section but I can't remember
where so I better let go my argument. However, Lauer found the triangle
15-20-25 royal cubits in the dimensions of the King's Chamber, and he
dared publish it in his book on the pyramids.
My astrologer (softly touching her crystal sphere, calling several
times): Professor Ahmes?
Ahmes (finally responding): You call me again, here in the House
of Life? Well, I guess you like to hear more about Khufu's Horizon?
Here I go.
The simple shape of the pyramid stays for the Primordial One.
The number One gave way to the number Two: number of polarity.
While the base measures 440 royal cubits, the height measures
280 royal cubits or 440 magic rods (a measure I explained before).
Lower the pyramid by less than five fingers and you will obtain the
pyramid of the Primeval Sky Goddess. Half the base equals the golden
major of the slope, while the radius of the inscribed hemisphere,
symbol of the heavenly vault, equals the golden major of the height.
Heighten the pyramid by less than four fingers and you will obtain
the pyramid of the Primeval Sun God. Use the height as the diameter
of a large circle, symbol of the solar disk: its area equals the
section area of the pyramid.
These shapes are combined in the pyramid built and approximated
as follows:
A) half base 220, slope 356, according to the golden sequence
4, 8, 12, 20, 32, 52, 84, 136, 220, 356 ...
height 280, radius of the inscribed hemisphere 173, according
to the golden sequence 9, 16, 25, 41, 66, 107, 173, 280 ...
B) base 440 royal cubits, height 280 cubits or 440 magic rods,
number of Re 3 '7
Now the Great Gallery is a combination of two galleries:
A) divide the slope 356 royal cubits by 4 and you will obtain
89 royal cubits for the ceiling; the run mesures 80 royal
cubits while the rise measures 39 royal cubits, according
to the triple 39-80-89
B) divide the height 440 magic rods by 3 '7 and you will obtain
140 magic rods for the ceiling; the run measures again 80 royal
cubits while the rise measures 39 '5 royal cubits, according
to the pseudo-triple 539-1100-1225
Me: Please let me check these numbers. 89 royal cubits measure 4660
cm while 140 magic rods measure 4664.8 cm. The length of the ceiling
measures 4671 cm according to Rainer Stadelmann or 4663 cm according
to Karlheinz Schuessler. Now for the angle. The run measures 80 royal
cubits and the rise 39 or 39 '5 royal cubits, giving a mean of 39 '10
royal cubits or an ideal angle of 26 degrees 2 minutes 49 seconds ...
while the actual angle measures 26 degrees 2 minutes 30 seconds
according to Rainer Stadelmann.
Ahmes: John, I tell you our secrets for you seem to be a fine mind,
having sympathy for our way of reasoning. I sure like your arguments
for the shafts being airshafts.
Ludwig Borchardt: Zahlenmystik! Numerologie! Pyramidenspinner!
Me: Me?
Ludwig Borchardt: we0gfübfmbq$üigk
My astrologer: Sorry, my crystal sphere has a problem translating
German. Hieroglyphs are fine, but this German, you know.
John Legon
>
>My astrologer (softly touching her crystal sphere, calling several
>times): Professor Ahmes?
>
>Ahmes (finally responding): You call me again, here in the House
>of Life? Well, I guess you like to hear more about Khufu's Horizon?
>
>Now the Great Gallery is a combination of two galleries:
>
> A) divide the slope 356 royal cubits by 4 and you will obtain
> 89 royal cubits for the ceiling; the run mesures 80 royal
> cubits while the rise measures 39 royal cubits, according
> to the triple 39-80-89
>
John: That's a nice triple, and the 'run' is indeed 80 royal cubits...
> B) divide the height 440 magic rods by 3 '7 and you will obtain
> 140 magic rods for the ceiling; the run measures again 80 royal
> cubits while the rise measures 39 '5 royal cubits, according
> to the pseudo-triple 539-1100-1225
>
John: Isn't that becoming rather complicated?
>Me: Please let me check these numbers. 89 royal cubits measure 4660
>cm while 140 magic rods measure 4664.8 cm. The length of the ceiling
>measures 4671 cm according to Rainer Stadelmann or 4663 cm according
>to Karlheinz Schuessler. Now for the angle. The run measures 80 royal
>cubits and the rise 39 or 39 '5 royal cubits, giving a mean of 39 '10
>royal cubits or an ideal angle of 26 degrees 2 minutes 49 seconds ...
>while the actual angle measures 26 degrees 2 minutes 30 seconds
>according to Rainer Stadelmann.
>
John: These results appear very good and convincing, professor, and yet
I still don't believe in your golden numbers! First of all, we need to
determine where these measurements came from. The dimension of 4663 cm
was originally, I'm sure, just one of those approximate measures given
by I.E.S. Edwards in his book, The Pyramids of Egypt, where he says that
the length of the gallery is 153 feet. When this book was translated
into German as 'Die aegyptischen Pyramiden', the dimension of 153 feet
became 46.63 ms, which looks very accurate but isn't.
Stadelmann's measurement is much better, being (almost certainly) taken
from a plan by Maragioglio and Rinaldi, who in turn took the dimension
from Flinders Petrie. In case you were wondering, the original figure
was 1838.6 inches. This is 89.2 cubits (for your cubit of 52.36 cm), so
the dimension is not exactly 89 cubits (or 140 magic rods). Of course,
Petrie did not crawl upside-down on the ceiling of the gallery to get
this dimension, but had to calculate it from several other measurements.
Which makes me wonder why you have to take one of the most inaccessible
dimensions inside the entire pyramid to support the idea that the number
of 89 cubits is significant?
But there is another problem relating to the angle of slope. I am sorry
to say that Stadelmann was wrong when he said that the slope is the same
as that of the ascending corridor (Die aegyptischen Pyramiden, p. 115),
because the measurements of Petrie (and others) show that it is not.
The angle of slope was increased by about 15 minutes of arc, giving a
different relationship between 'rise' and 'run'.
>Ahmes: John, I tell you our secrets for you seem to be a fine mind,
>having sympathy for our way of reasoning. I sure like your arguments
>for the shafts being airshafts.
>
>Ludwig Borchardt: Zahlenmystik! Numerologie! Pyramidenspinner!
>
>Me: Me?
>
>Ludwig Borchardt: we0gfübfmbq$üigk
>
John: Translating Borchardt is sometimes not worth the effort...
Lucy: Death to Borchardt!
John: Oh be quiet, Lucy, go back to sleep.
cir...@access.ch
John Legon <jo...@legon.demon.co.uk> wrote:
> My goodness, what long arms people used to have! Mine measure
> only about 49 cm from the elbow to the tip of the middle finger, even
> though my height is above average...
John, I still have problems understanding the Egyptian measures.
Being 172 cm tall, the distance from my elbow to the tip of my middle
finger measures 45 cm, what is fairly well a short cubit measuring
about 44.88 cm. Now the royal cubit measures about 52.36 cm. When I
multiply 172 by 52.36 / 44.88 or 7/6 I obtain 200 cm. Does this mean
a king was 2 meters tall? Or may it be an imaginary measure alluding
to the mythical giants of Zep Tepi, the legendary First Time?
Here follows another attempt for a comprehension of the royal cubit.
Possibly a silly one. Or maybe not so silly.
1 finger = 1.87 cm
average breadth of my fingers = 2 cm
1 palm = 4 fingers = 7.48 cm
breadth of my palm = length of the back of my hand = 8 cm
1 fist = 5 fingers = 9.35 cm
breadth of my fist = 10 cm
small span = 12 fingers = 22.44 cm
my under arm, from elbow to wrist = 25 cm (arm bended)
great span = 14 fingers = 26.18 cm
my upper arm, from shoulder to elbow = 29 cm (arm bended)
The ratio 26.18 / 29 is about 9/10.
When I multiply my 172 cm by 9/10 I obtain 154.8 cm.
Can you imagine so small people?
I may remind you of Khufu's rather small sarcophagus
Does anyone know the measurements of the Egyptian mummies?
The length of my stretched arm, from shoulder to fist (knuckles),
measures 59 cm. When I multiply 59 cm by 9/10 I obtain 53.1 cm,
about the length of a royal cubit.
Please notice: I find different measurements when I bend or stretch
my arm:
under arm (arm bended) 25 cm
upper arm (arm bended) 29 cm
back of my hand 8 cm
sum 62 cm
arm (stretched, from the shoulder to the knuckles of my fist)
59 cm
difference 3 cm
When I wear a shirt and bend my arm the sleeve glides back.
This effect is either due to a special geometry of the bending arm,
or else I make myself a case for Alan ;-)
The length of my upper arm and my under arm (stretched) measures
51 cm. When I multiply 51 cm by 9/10 I obtain 45.9 cm. The short
cubit measures 6 palms = 24 fingers = 44.88 cm.
Provisional measures, ideal scheme:
royal cubit = stretched ARM including FIST = 7 palms = 28 fingers
short cubit = stretched arm without fist = 6 palms = 24 fingers
great span = upper arm (arm bended) = 14 fingers
short span = under arm (arm bended) = 3 palms = 12 fingers
The ROYAL cubit being longer than the normal one would now mean:
the KING, owner of the royal cubit, uses his FIST = has the POWER.
He swings the MACE - a wooden 'arm' with a 'fist' made of stone,
sign of his power (as seen on the Narmer Palette).
The arm as limb, as weapon, as measure? May it be that the origin
of the royal cubit was a ritual mace, the king's sceptre, sign of
his power?
In this case, the first palm menas the king's fist, or macehead,
or power. The deities are: RE; TEFNUT (goddess of moisture) in
her appeareance as MAAT (goddess of truth, order and justice);
SHU (god of the air, while air means life of the mind, spiritual
life); GEB (god of the earth) and/or AMUN (god of the universe),
both symbolizesd by a goose. As if the king were saying: I got
my power from Amun-Re, I reign in the Maat, over the moist land,
over the earth, inspired by Shu.
Other ideas are welcome.
Regards Franz Gnaedinger Zurich cir...@access.ch
-----== Posted via Deja News, The Leader in Internet Discussion ==-----
cir...@access.ch
John Legon <jo...@legon.demon.co.uk> wrote:
> Stadelmann's measurement is much better, being (almost certainly) taken
> from a plan by Maragioglio and Rinaldi, who in turn took the dimension
> from Flinders Petrie. In case you were wondering, the original figure
> was 1838.6 inches. This is 89.2 cubits (for your cubit of 52.36 cm), so
> the dimension is not exactly 89 cubits (or 140 magic rods). Of course,
> Petrie did not crawl upside-down on the ceiling of the gallery to get
> this dimension, but had to calculate it from several other measurements.
> Which makes me wonder why you have to take one of the most inaccessible
> dimensions inside the entire pyramid to support the idea that the number
> of 89 cubits is significant?
Measuring the length of the ceiling ain't such a problem: measure
the length of the wall and subtract the sum of the 'Einkragungen'.
Or use a plummet, mark the ends of the ceiling on the floor and
measure the distance of the marks.
Getting reliable measurements from a professor of Egyptology is
almost impossible. I am begging since years for a single measurement
of the gallery - oblique height of the blocks which build the layers
of the 'Kraggewoelbe' - and got some kind letters but never the
measurement I asked for. Maybe you find it in your books? According
to my thesis it should measure 44 fingers or about 82.28 centimeters
(on the average) while the vertical height would measure 49 fingers
or 91.63 cm.
Building the gallery was a demanding task, for one had to consider
5 dimensions: width, horizontal and oblique height, oblique and
vertical height.
Combining the royal cubit (52.36 cm) with the hypothetical measure
of the 'magic rod' (7/11 royal cubits = 196 Horus marks = 33.32 cm)
makes it easy:
horizontal length oblique height 4 royal cubits 4 palms
oblique length vertical height 7 magic rods 1 magic rod
The ceiling consists of 40 blocks having a length of 140/40 = 3.5
royal cubits each while the run of every block measures 2 royal cubits.
Gallery B is based on the triple 39-80-89. Gallery A is based on
the pseudo-triple 539-1100-1225 or 2156-4400-4900. A funny triangle:
oblique height 44x44 rise 44x49 run 44x100 slope 49x100
tangent 49/100 sine 11/25 cosine 44/49
cosine = 4 royal cubits / 7 magic rods = 4 palms / 1 magic rod
Jean-Philippe Lauer says that the pyramid builders tried to find
measurements made of whole numbers. I agree with him. Bu I like
to add that they had clever ways for obtaining whole numbers,
even when defining diagonals, angles and round forms.
Let me ask professor Ahmes on this topic.
My astrologer (touching her crystal sphere again): Professor Ahmes?
Hello? Professor? Yes?
Bobby: Some are mathematicians
Some are carpenters' wifes
Don't know how long they've studied
Don't know what they do with their lifes
Ahmes: Who are you?
Bobby: You may call me Bobby
Ahmes: Well, Bobby, you got a funny voice, but I sure like your
melody. By the way, my dear daughter Aisha is a gifted singer ...
Bobby: Flowers on the hillside blooming crazy
.......................................
Blue river running slow and lazy
.....................................
Aisha (listening for a while, then chiming in)
Bobby (lowering his voice)
Aisha (improvising on Bobby's theme, with a clear and pretty voice,
coming from a far away time)
John Legon
>John, I still have problems understanding the Egyptian measures.
>Being 172 cm tall, the distance from my elbow to the tip of my middle
>finger measures 45 cm, what is fairly well a short cubit measuring
>about 44.88 cm. Now the royal cubit measures about 52.36 cm. When I
>multiply 172 by 52.36 / 44.88 or 7/6 I obtain 200 cm. Does this mean
>a king was 2 meters tall? Or may it be an imaginary measure alluding
>to the mythical giants of Zep Tepi, the legendary First Time?
>
Thank you, Franz, for your interesting ideas about the origins of the
royal cubit. I too have wondered how this measure came into being, so
here are a few more ideas...
I think we should start with the measurements of the Egyptian people,
which have indeed been studied in great detail. In an analysis of the
dimensions of some 60 mummies, Gay Robins found that the mean height of
the adult Egyptian male was 166 cm [see G. Robins, 'The length of the
forearm in canon and metrology', Goettinger Miszellen 59 (1982), 61-75,
68]. Now the ratio between your height and the length of your arm from
elbow to tip of middle finger is 172/45 or 3.82. For me it is 185/49 or
3.78. We apply the mean value of 3.8 to the mean height of the ancient
Egyptian male, and find a length for the Egyptian forearm of 166/3.8 or
43.7 cm - assuming the same human proportions.
This dimension is confirmed by anatomical measurements, but is rather
shorter than the small cubit of 24 fingers or about 45 cm. Well, some
cubit rods have only 23 fingers in the small cubit, giving a length of
43 cm. This suggests that the small cubit was not really defined as a
standard, but was a variable measure which depended on the forearm of
the individual. It apparently varied between about 43 and 45 cm for the
adult Egyptian male.
How then, do we account for the royal cubit of about 52.4 cm? Clearly,
it was not an anatomical measurement, unless we assume an individual of
quite exceptional height even for modern times. It could represent the
idealised forearm-length of a mythical 'god-king' prototype, but I am
more interested in the underlying metrology.
Firstly, we must note that archaeologists Junker and Verner both found
evidence for the use of a unit equal to 1/6 of the royal cubit in some
monuments of the Old Kingdom. This unit is identical to the so-called
great palm of 8.73 cm, which is found on cubit rods of the late period.
It is in fact close to the natural palm-width, when measured across the
open palm as opposed to the fingers. We now find that there are just
five of these palms in our anatomical small cubit, and six such palms in
the royal cubits:
small cubit = 5 x 8.73 = 43.65 cm
royal cubit = 6 x 8.73 = 52.38 cm
Thus it seems to me that the royal cubit was obtained by adding an extra
palm-width to the prototypal small cubit. The reasons for introducing
this longer cubit may have been purely practical. Firstly, a cubit of
five palms is not convenient for use with simple fractions. A cubit of
six palms, however, is easily divided into halves and thirds - the two
most important fractions. The separation of the cubit from the natural
arm-length may also have been intended to establish a standard, for as
long as the Egyptians could use their own arms to measure dimensions,
consistent results could never be obtained.
In any event, the institution of the royal cubit with six large palms
evidently happened by the time of the First Dynasty, because there is
evidence for this type of cubit in the earliest archaic mastabas. At a
later date, it would seem, the royal cubit was refined with a division
into seven smaller palms, each containing four fingers. At first sight
this number of palms seems hopelessly impractical, since it does not
divide naturally into halves, thirds, or quarters. As you have already
said, Franz, this seven-part division was very good for measurements of
the circle and square, and this may indeed be why it was chosen.
Yet other factors to consider are the Egyptian foot-measurement, which
was identified a long time ago as being equal to half the royal cubit,
and the question of the Egyptian canon of art - the guide-line system of
the Old Kingdom and the later grid. Contrary to Erik Iversen and Gay
Robins, I have shown that the canon was based on the six-part division
of the royal cubit, as follows:
In the guide-line system, the height of the standing male from the floor
to the so-called hair-line was divided by horizontal lines into three
equal parts. In the later grid, this hair-line level is equivalent to
18 squares, while the height to the top of the head equals 19 squares.
For the mean anatomical height of 166 cm, the hair-line level will be:
166 cm x 18/19 = 157.26 cm = 3 x 52.4 cm = 3 royal cubits (near enough).
Thus each of the three equal parts up to the hair-line in the guide-line
system was equal to one royal cubit, and each of the grid squares in the
later system was equal to 1/6 royal cubit.
[J. Legon, 'The Cubit and the Egyptian Canon of Art', in _Discussions in
Egyptology_ 35 (1996), 61-76.]
John Legon
Steve Whittet
The idea that a building should have an architectual modular
of proportinate unit fractions was probably more important to
the Egyptians Greeks and Romans than it is to us. One reason
might be that they worked with unit fractions on a daily basis
and knew how to use them.
I find the Egyptian unit fractions translated into a system of
sequential proportionate measures very useful in architecture
and would note that they are commensurate with both the foot
and meter. It would be pretty hard to find a modern artifact
which doesn't have components proportionate to one or more
of the divisions you find on an Egyptian ruler.
The exact length of the Royal cubit probably varied more
amongst the ancient Egyptians than its value does with
modern Egyptologists. Once you realise the measures
tend to be unit fractions of one another its hard not
to note the proportional arrangements.
>
>> Stadelmann's measurement is much better, being (almost certainly) taken
>> from a plan by Maragioglio and Rinaldi, who in turn took the dimension
>> from Flinders Petrie.
I think elsewhere you mention Coles survey, I wonder if this
is available through Amazon. I suspect that it would be both
accurate and subject to some caveats.
The use of his measures to check ancient Egyptian units
presumes the Great Pyramid was intended to be 440 royal cubits
in length at its base.
This begs the question, where do we cut the pane of its base?
Cole measured to the base sockets. This is comparable to establishing
the proportions of a modern house by measuring to its footing as
opposed to its foundation or wall.
The more likely point for a proportion to be established is at the
point the finished pavement meets the finished casing. An Egyptologists
reconstructions are only as good as the points he chooses to measure from.
The same applies to the inside measures. Some time back I received
Cad drawings with the field measures from the exploration of the
shafts.
My perception would be that the basic unit used was the
Hayt of 10 royal cubits, 280 fingers, 210 inches, 70 palms,
56 hands 40 spans, 35 cubels, 18 Roman pes, 17.5 English feet,
17 Greek feet, 5.25 m, ...The system is basically septenary.
>> In case you were wondering, the original figure
>> was 1838.6 inches. This is 89.2 cubits (for your cubit of 52.36 cm), so
>> the dimension is not exactly 89 cubits (or 140 magic rods).
We might do best if we begin with known ancient measures
in septenary multiples and consider the implications of
what we find there before proceeding to more sophisticated
analysis.
>> Of course,
>> Petrie did not crawl upside-down on the ceiling of the gallery to get
>> this dimension, but had to calculate it from several other measurements.
>> Which makes me wonder why you have to take one of the most inaccessible
>> dimensions inside the entire pyramid to support the idea that the number
>> of 89 cubits is significant?
>
>Measuring the length of the ceiling ain't such a problem: measure
>the length of the wall and subtract the sum of the 'Einkragungen'.
>Or use a plummet, mark the ends of the ceiling on the floor and
>measure the distance of the marks.
>
>Getting reliable measurements from a professor of Egyptology is
>almost impossible. I am begging since years for a single measurement
>of the gallery - oblique height of the blocks which build the layers
>of the 'Kraggewoelbe' - and got some kind letters but never the
>measurement I asked for. Maybe you find it in your books? According
>to my thesis it should measure 44 fingers or about 82.28 centimeters
>(on the average) while the vertical height would measure 49 fingers
>or 91.63 cm.
44 fingers is 11 palms and 49 fingers is 7 spans both of which
are perfectly good known ancient units. The 7:11 proportion
is retained but transformed by a shift in the units used
which is allowed by using an architectural modulor.
>
>Building the gallery was a demanding task, for one had to consider
>5 dimensions: width, horizontal and oblique height, oblique and
>vertical height.
>
>Combining the royal cubit (52.36 cm) with the hypothetical measure
>of the 'magic rod' (7/11 royal cubits = 196 Horus marks = 33.32 cm)
>makes it easy:
>
> horizontal length oblique height 4 royal cubits 4 palms
> oblique length vertical height 7 magic rods 1 magic rod
>
>The ceiling consists of 40 blocks having a length of 140/40 = 3.5
>royal cubits each while the run of every block measures 2 royal cubits.
>
>Gallery B is based on the triple 39-80-89. Gallery A is based on
>the pseudo-triple 539-1100-1225 or 2156-4400-4900. A funny triangle:
>
> oblique height 44x44 rise 44x49 run 44x100 slope 49x100
>
> tangent 49/100 sine 11/25 cosine 44/49
>
> cosine = 4 royal cubits / 7 magic rods = 4 palms / 1 magic rod
The triples can be obtained in whole numbers by placing
different unit bases on each leg of the square and its hypotenuse
for example put a foot on one leg of a framing square and a
remen on the other the hypotenuse is 19.2"
Ok, so what does that mean? 19.2" is where that funny little diamond
you find on Stanley (but not Lufkin) tape measures lands. Framers
use it to space rafters so as to span 5 rafters with a sheet of plywood
instead of 6 saving some materials. I don't have a name for it
but with a length of 26 fingers its cube holds 25 wine gallons
of 282 cubic inches.
Once you have your unit triple you then can run three separate
modulars based on each unit.
>Jean-Philippe Lauer says that the pyramid builders tried to find
>measurements made of whole numbers. I agree with him. But I like
>to add that they had clever ways for obtaining whole numbers,
>even when defining diagonals, angles and round forms.
Yes, I think that is evident.
...
steve
Doug Weller
09:10:13 +0100, jo...@nospam.legon.demon.co.uk said...
>
> I have only the 1991 edition, but perhaps Stadelmann took his data from
> Gantenbrink's contribution to MDAIK 50? Anyway, I hope that Gantenbrink
> will soon respond to our request for more information.
>
>
This is a speech he gave at the Ordo Et Mensura Conference in Munich last
year.
Ascertaining and evaluating relevant structural points using the Cheops
Pyramid as an example
by Rudolf Gantenbrink
In 1992 and 1993 I had the opportunity of undertaking three archaeological
campaigns on the Cheops Pyramid in collaboration with the German
Archaeological Institute in Cairo. An important objective of the work
then undertaken was to re-survey certain important basic points within the
corridor chamber system. I had noticed that several measurements that I
required to produce a digital computer model were incomplete or wrong and
a re-survey was therefore required. I had for some time been considering
the extent to which the measurements obtained from a structure permit
conclusions as to the construction stages applied. In doing so, I assumed
that present day designs by engineers or architects also contained
traceable construction stages which quite substantially led to conclusions
as to the basic thinking behind the design. In addition, the analysis of
whole-number values is of decisive importance.
An example will make this clear. A present-day architect is hardly likely
to plan a house from the start which is 20.63 m long and 10.95 m wide but
if we follow the rules of logic will initially have a house of 11 by 20
meters in mind. Even if a whole-number criterion of this kind could not
be put into practice on structural grounds, the first mental step is
nonetheless always in whole figures. When I find whole-number values in a
structure, they must essentially be regarded as primary construction steps
to which the non-whole-number, secondary construction steps are
subordinate.
Let us take the simplified layout of a modern house as an example (Fig.
1a). We can already derive some information from it. The architect has
worked not with inches but with meters and has conceived it from the
outside inwards. No precisely defined spatial requirement or, possibly,
the given length of a piece of furniture predominates, but the layout of
the house itself. It therefore forms the architect's first design step.
The points relevant to the structure lie at the four corners of the
building. The dimensions of the internal space, on the other hand, are
obtained only through structural components, namely the requisite wall
thickness. The internal space is consequently a secondary design product
which is already subject to structural constraints. If we follow the
simple and proven principle, the whole-number values can also be set out
as grids. Whole-number planning is evident in a great many Egyptian
structures (e.g. in the Oracle Temple in Siwa). However, 100%
correspondence is not always evident. There are a number of possible
causes for this:
1. inaccuracies in the means of measuring methods then used.
2. imprecise execution of the works
3. changes in plan during construction.
4. regional departures in the unit of measurement
5. geological shifting (e.g. the Oracle Temple in Siwa).
These factors necessitate a certain tolerance in calculating,
corresponding to the general structural performance of the culture
concerned.
However, these should be kept well beneath the limit where all dimensions
become freely interpretable quantities. In the case of the Cheops
Pyramid, I allowed a maximum departure from the grid of 5cm. This
extremely small value was possible only through the precise execution of
the work on this pyramid. In addition, the grid had to be calibrated to
the unit of measurement actually applied. This means that the mean of the
longest sections to be assumed as whole-number (in the case of the Cheops
Pyramid, the pyramid base) is divided by the value that the builder aimed
for. Since the master builder of the Cheops Pyramid will hardly have laid
out the base as 439 x 439 or 441 x 441 cubits, we must assume that the
basic dimension he had in mind was 440 x 440 cubits, which brings us to a
unit generally recognised for Cheops of 1 cubit = 0.5236 meters. In the
case of the Cheops Pyramid, the process comes up against further limits,
imposed by the size of the structure. A grid of 440 x 280 cubits can no
longer be precisely set out on paper, even in format A0, since simply the
shrinking of the paper through humidity can make a whole cubit´s
difference. This task can therefore only be executed on the computer, i.e.
by precise mathematical visualisation. I have produced a three-
dimensional model for this on the CAD system AUTO-CAD; it contains the
measurement results of Petrie, Maragiolio and Rinaldi, the Edgar Brothers
and the data I ascertained for myself in 1993 and 1994. This model is
therefore the best and most exact basis at present for ascertaining points
within the Cheops Pyramid relevant to its design. The illustrations given
below are based on this computer model. However, before we can now start
with an evaluation, i.e. the search for whole-number values and ratios, we
must know which points were of special importance within the structure,
namely at which points were heights or distances actually measured during
construction?
This is a search process for structural indices which I should like to
clarify here by means of selected examples. Fig. 2 shows the northern
wall of the Queen's Chamber. Access to this chamber and the entrance to
the northern air shaft have only one reference in common, namely the same
height of the ceiling. The distance between the floor in the chamber and
the floor of the shaft, on the other hand, is not a whole-number value.
We find the same in the King's Chamber. This initial reference to the
fact that measurements were taken at the ceiling or practically before the
corridor was covered over, on the smooth upper surface of the side walls
free of building debris, is confirmed at further places in the structure.
The ceiling height similarly constitutes a common quantity at the lower
end of the Great Gallery (Fig. 3).
At the floor, on the other hand, neither the 90° angle A towards the
landing nor the vertical B of the jointing forms a defined common
intersection of the three structures. Since the heights of the two
galleries are different, there is certainly no common intersection at the
angle bisection, even if one were to assume that the ancient Egyptians
were in a position to determine the bisector. We obtained the same
picture at the upper end of the Great Gallery (Fig. 4). A clearly
defined, common reference point for all three structures lies in the
height of the corridor ceilings. On the floor, however, there is no
defined conclusion, either for the Great Gallery nor for the other
galleries. Here, too, the corridors, or their theoretical extension, do
not have the same heights, so that here no conclusion is possible at the
angle bisects, either. A further indication that the ceiling height was
very probably a primary construction quantity is evident from the Cheops
Pyramid's rock chambers. The ceilings are the only completed element of
these chambers, a finding which occurs also in other rock chambers of the
Old Kingdom. The walls and floor, on the other hand, are still in
unfinished state and cannot therefore possibly have served as a reference
for measurements.
These indices point to the fact that the primary construction points must
generally be looked for at the ceilings and not on the floors, which
hitherto have simply been regarded as a reference height for no good
reason. The indices that we have compiled so far are highly meaningful,
but they still do not amount to proof. The thesis is proved only when the
analysis of the points ascertained in this way reveal meaningful design
steps and a thought-out system. If all the structural points ascertained
in the past are entered into a computer model of the Cheops Pyramid
reduced to essentials, they all end up as whole numbers (shown in red, see
Fig. 5). Even the lower intersection of the upper air shafts A, even
though only theoretically existing, forms a whole number. At the exit B
of the ascending corridor from the pyramid, the intersection is not a
whole number. We have to take one step back to understand the reason for
this.
The example of the simplified, modern house layout (Fig. 1b) shows that
the upper left reference point is not technically executed although it was
needed both for the paper design, because after all one logically starts
here by drawing a rectangle, and for measurements during construction.
Could the missing point for the Cheops Pyramid therefore also lie outside
the structure? Crossing of the paths with the missing point is shown in
Fig. 6. Both corridors have exactly the same angle, so that what applies
to the ascending one must also apply to the descending one. The length of
the X coordinate is a prime number in the ascending corridor. Because
this is indivisible, the only possibility of nonetheless obtaining a
division into easily measured sub-quantities is to move down into the next
smaller measurement unit.
With the Egyptian cubit, this is the so-called "palm" of 4 fingers. This
division, which looks like a staircase in the illustration, clearly shows
that the diagonal XY possesses only two whole-number points, namely the
commencing point and the terminal point. We can also see that even moving
down into the next smaller measurement unit does not create an
intersection point with the pyramid's flank, which in any event could
never coincide with a whole-number cubit value.
A further movement back into the finger measurement unit therefore makes
little sense and, as I have checked, similarly fails to produce a
definable crossing point. We must therefore admit here that the Egyptians
of the Old Kingdom simply did not master the intersection of three
different angles and they were clearly compelled to place the imaginary
structural point outside. At the base of the pyramid, the result is
therefore a shift from 7 to 11 cubits, a quantity that we shall encounter
again in due course. Now that we have found the last, missing point, we
can start on the evaluation. Fig. 7 shows that the same quantities occur
time and again in the distances between the whole-number construction
points. Distances of the same length are in each case shown in the same
colour. The quantities are repeated up to three times and, with the
exception of two values, are all divisible by 11. This first reference to
an existing system in planning the pyramid is impressively confirmed when
analysing the heights of the construction points (Fig. 8). The heights of
all construction points are essentially either one 40th of the basic width
of the pyramid or one 40th of their height or, in two cases, even one 40th
of both (these designate the construction points of the upper air shafts).
This is all the more surprising when we consider that the base of 440
cubits when divided by 40 produces 11 and the height of 280 cubits when
divided by 40 produces 7. Two exceptions apart, as we have already seen,
all distances of the same length are divisible by 7 or 11 or both. This
clear sub-division into 40ths leads us to suspect that the architect of
the Cheops Pyramid worked in a scale of 1:40 when placing his plan on
papyrus (?). Especially significant, however, is the fact that the
construction points ascertained are clearly really relevant to the design,
i.e. the steps that we have taken to discover them were the right ones.
After having first created the basis for understanding the structure of
the Cheops Pyramid, I should finally like to show, by means of an example,
how our knowledge of the points relevant to the design can be put to use.
In our measuring campaigns in 1992, I attached particular value to
measuring the exit points of the upper airshafts. Although the surveys by
Maragiolio and Rinaldi showed only poorly measured quantities, even with
confusion between south and north, I could already clearly see that these
exit points lay at the same height (Fig. 9). Our measurements confirm
this suspicion, the departure from the whole-number cubit grid remaining
within the tolerance limits set.
Since the theoretical, lower intersection point of the shafts is also
staggered against the axis of the pyramid by a whole number, it is clear
that the Bauval theory that these shafts were aligned on certain stars
does not add up. The shafts are clearly designed by whole numbers and
according to clear geometrical rules. On their being aligned on certain
stars, the same exit height would, moreover, be all too much of a
coincidence. It is interesting that the shifting of the lower
construction point of the shafts from the pyramid axis amounts to exactly
22 cubits, i.e. 2 x 11 cubits. This shift resulted in quite substantial
problems during execution of the works, because the exit points clearly
had to lie at equal height. For this, not only had two angles to be
determined but so had the ratio of the two angles to each other and to the
axis of the pyramid, in order for them to be precisely executed
structurally.
Fig. 10 shows the mathematical geometrical principle with which this
problem was easily solved. A grid of 11 x 11 cubits was placed above the
pyramid. The grid therefore corresponds to a scale of 1:40 referred to the
pyramid base. This grid is irrelevant to the height of the pyramid. In
actual fact, the Cheops grid, as I ascertained during my ongoing work, is
not square but rectangular, in a ratio of 7 to 11 cubits, i.e. one 40th of
the height to one 40th of the base. We are using the square grid here
only to clarify the design process more effectively. The right northern
shaft is clearly designed in a ratio of 11:7 grid points and the left
southern one in a square ratio of 7:7 grid points. By reversing the ratio
of 11:7 to 7:11, I obtained the counter-angle in the diagonal (shown by a
broken line), which lies at 90° to the northern shaft. The angle, the
counter-angle and the square counter-ratio can therefore be geometrically
determined. When we remember that the ancient Egyptians had to rely on
only imperfect drawing aids when planning this gigantic structure, i.e.
when drawing they could not determine with sufficient precision in a grid
whether the line drawn and the grid actually intersected or only seemed to
do so through the small scale, we can see that draughtsmanship alone was
insufficient to enable such precise building work. An additional means
was therefore needed to prove the assumed intersecting points
mathematically. We do know from the Rhind Papyrus that the ancient
Egyptians calculated with fractions. I believe that this simple,
mathematical knowledge can also be imputed to the Old Kingdom. Ratios of
the kind arising with the shafts can of course also be expressed in
fractions. In doing so, the X quantity forms the denominator and the Y
quantity the numerator. We therefore say that 11/7 forms the counter-
angle of 7/11 (cf. Fig. 10).
A simple arithmetical operation (Fig. 11) which today curiously can only
be found in alchemical literature and over which every mathematician only
shakes his head, suddenly becomes a simple method of bisecting the angle
when geometrically demonstrated. We are no longer familiar with this
process today, because we have, in the meantime, as we know, discovered
far better methods for bisecting the angle. Subtracting the numerator
from the ratio and counter-ratio with simultaneous addition of their
denominators in principle always produces the bisector of the angle when
geometrically demonstrated. The ancient Egyptians seem to have known this
process, since the bisector of the angle or, to put it more precisely, the
functional ratio of 2:9 grid points indisputably defines the shift of the
shafts from the axis of the pyramid.
As we can see from Fig. 12, the lower part of the pyramid right down to
the base logically continues to sub-divide in the ratio 7:11. I believe
that the finger print of the architect of the Cheops Pyramid now becomes
clearly visible. Already at this stage, he allows us to draw conclusions
with far-reaching consequences. The master builder of the Cheops Pyramid
was familiar neither with Pythagoras' theorem nor that of Tange, or they
would otherwise have made use of this knowledge instead of messing about
with primitive mathematical systems.
On the other hand, it is also clear that they did not embark on a reckless
building spree but that the structure was already carefully planned before
work commenced, with the consistent application of expertise that was
still relatively simple for the period. We also discovered that the upper
air shafts mark the commencement of interior design because their
construction, contrary to all other spatial elements, remains whole-
numbered right down to the smallest possible scale. As I have intimated
above, simply creating the right basic prerequisites for analysis is
itself a thoroughly rewarding process. The quantities relevant to the
structure must first be separated from those which are of only secondary
importance to the building work, since only then can the contrasting of
unrelated quantities be prevented, like comparing chalk with cheese. I
hope that with this paper I have made a small contribution towards lifting
the structure of the Cheops Pyramid out of the numerically esoteric taboo
area of science and restoring it to what it really is, namely one of the
most informative structures of antiquity.
--
Doug Weller Moderator, sci.archaeology.moderated
Submissions to:sci-archaeol...@medieval.org
Requests To: arch-mo...@ucl.ac.uk
Co-owner UK-Schools mailing list: email me for details
Steve Whittet
says...
>
>In article <6ufidp$bvo$1...@nnrp1.dejanews.com>, cir...@access.ch wrote:
>
>>John, I still have problems understanding the Egyptian measures.
>>Being 172 cm tall, the distance from my elbow to the tip of my middle
>>finger measures 45 cm, what is fairly well a short cubit measuring
>>about 44.88 cm. Now the royal cubit measures about 52.36 cm. When I
>>multiply 172 by 52.36 / 44.88 or 7/6 I obtain 200 cm. Does this mean
>>a king was 2 meters tall? Or may it be an imaginary measure alluding
>>to the mythical giants of Zep Tepi, the legendary First Time?
Begin with the fact that the Egyptians used unit fractions
and did a lot of their mathematics by doubling or halving.
The assumption that the units in their fully developed form began
with parts of the body and worked up may be incorrect. They
may be divisions of the body as a whole.
One of the things which archaeologists have found which
suggest a systematic approach to unit measures are the
grids which define the cannon of proportions for figures
in inscriptions. The whole unit is the height of the individual
and the units which divide it are palms of four fingers,
hands of five fingers, fists of six fingers and spans
of seven fingers.
The grids can be used to date inscriptions depending on
what cannon of proportions is used. Looking at the later
inscriptions an individuals height is 14 spans of 98 fingers,
between 1.833 and 1.8375 m; similar to what the Romans
called an orguia (1.84 m.) (a little over 6 feet)
The Romans divided this decimally into 100 digitus.
The Egyptians divided it into fingers, nails, palms,
hands, fists, spans, cubels, feet, remen, cubits
and royal cubits.
>Thank you, Franz, for your interesting ideas about the origins of the
>royal cubit. I too have wondered how this measure came into being, so
>here are a few more ideas...
>
>I think we should start with the measurements of the Egyptian people,
>which have indeed been studied in great detail. In an analysis of the
>dimensions of some 60 mummies, Gay Robins found that the mean height of
>the adult Egyptian male was 166 cm [see G. Robins, 'The length of the
>forearm in canon and metrology', Goettinger Miszellen 59 (1982), 61-75,68].
The only problem with that is that the King was the model for the
canon of proportions and drawn in an ideal form a bit larger than life.
I think you should begin with the measured grids they actually used.
I am looking at S Gideon "The Eternal Present, the beginnings of
Architecture" p 487 where there is illustrated such a grid "from
Iverson, after Lepisus". I think there are now several hundred
such grids which have been found in Egypt.
One early grid is 19 hands or 95 fingers high. and this would
give a height of about 5'10.
>Now the ratio between your height and the length of your arm from
>elbow to tip of middle finger is 172/45 or 3.82.
In the older form the arm is shown as seven hands measured
from the inside of the armpit to the ends of the fingers
just under 26". The arm has horizontal lines indicating
a wrist and shoulder band. The inside measure of the
wrist and shoulder nands is 3.5 hands, one cubit. The
outside measure is 7 palms, one royal cubit. The two
feet are slightly different lengths. Both are 3 hands
one plus the nail of the toe or 16 fingers overall
and the other the entire toe back to the joint.
>For me it is 185/49 or 3.78. We apply the mean value
>of 3.8 to the mean height of the ancient Egyptian male,
>and find a length for the Egyptian forearm of 166/3.8 or
>43.7 cm - assuming the same human proportions.
Note that the Egyptians did not align their grid with
the extremities of the arm or foot but laid the body
on the grid in such a way that the proportions of the
body could be measured from the grid. I think this suggests
that the units of the grid did not come from units of the
body such as the arm or foot
We might further note that our units of measure best agree
with the units used in the later rather than the earlier grids.
The feet are drawn properly so that the foot which is farther
away is drawn slightly smaller than the foot which is closer.
The Roman foot was 16 digits as the right Egyptian foot is drawn
and the Greek foot was 2/3 finger longer which agrees with the way
the left foot is drawn.
>
>This dimension is confirmed by anatomical measurements, but is rather
>shorter than the small cubit of 24 fingers or about 45 cm. Well, some
>cubit rods have only 23 fingers in the small cubit, giving a length of
>43 cm. This suggests that the small cubit was not really defined as a
>standard, but was a variable measure which depended on the forearm of
>the individual. It apparently varied between about 43 and 45 cm for the
>adult Egyptian male.
The small cubit clearly was defined as a standard, One hundred
small or ordinary cubits are the side of the standard egyptian
field known as the khet. One hundred Royal cubits are the side
of the setat. The khet is 1/2 acre and the setat 2/3 acre.
>How then, do we account for the royal cubit of about 52.4 cm?
The royal cubit adds the 1/6 part of the cubit to get seven palms
>Clearly,
>it was not an anatomical measurement, unless we assume an individual of
>quite exceptional height even for modern times.
The royal cubit of 20.62" is measured by placing your elbow
on a flat surface and extending your fingers as far up a tape
measure extending vertically from that surface as you can reach.
>It could represent the
>idealised forearm-length of a mythical 'god-king' prototype, but I am
>more interested in the underlying metrology.
It's probable that the body measures were measured from
the units of the grid. Later a systematic arrangement
of the units in a pleasing manner such as that mentioned
by Vitruvius was adopted as both a canon of proportion
and a standard of measure.
>
>Firstly, we must note that archaeologists Junker and Verner both found
>evidence for the use of a unit equal to 1/6 of the royal cubit in some
>monuments of the Old Kingdom.
This is the scale of the remen or board measure times the difference
between the Roman pes (11.44" - 11.66") and English foot (12")
1/6 Royal cubit = 3.44"
The Egyptian scale of units includes several different base units
board measure 17 mm
finger = 18.75 mm
inch = 25.4 mm
17 x 28/25.4 = 18.75
"3 (3/4")
1 '3 (1 1/2' thickness of a 2x4)
2
2 "3
3 '3 (3 1/2" width of a 2x4)(1/6 Royal cubit)
4
4 "3
5 '3
6
6 "3
7 '3
8
8"3
9 '3
10
10 "3
11 '3
12
12 "3
13 '3
14
14"3
15 '3
16
> This unit is identical to the so-called great palm of 8.73 cm,
>which is found on cubit rods of the late period.
Yes, The Egyptians had several variants on the cubit and royal cubit
over the years just as the Greeks and Romans and English did later,
most of these seem to be tied to 100 feet being equal to 1 second
of the earths equatorial circumference.
>It is in fact close to the natural palm-width, when measured across the
>open palm as opposed to the fingers. We now find that there are just
>five of these palms in our anatomical small cubit, and six such palms in
>the royal cubits:
>
> small cubit = 5 x 8.73 = 43.65 cm
> royal cubit = 6 x 8.73 = 52.38 cm
>
>Thus it seems to me that the royal cubit was obtained by adding an extra
>palm-width to the prototypal small cubit.
yes
The reasons for introducing
>this longer cubit may have been purely practical. Firstly, a cubit of
>five palms is not convenient for use with simple fractions. A cubit of
>six palms, however, is easily divided into halves and thirds - the two
>most important fractions. The separation of the cubit from the natural
>arm-length may also have been intended to establish a standard, for as
>long as the Egyptians could use their own arms to measure dimensions,
>consistent results could never be obtained.
The Egyptians divided everything into unit fractions, they took
its, '2, '3, "3, "'4, '4, '5,'6,'7 etc; and all these became
unit measures.
>
>In any event, the institution of the royal cubit with six large palms
>evidently happened by the time of the First Dynasty, because there is
>evidence for this type of cubit in the earliest archaic mastabas. At a
>later date, it would seem, the royal cubit was refined with a division
>into seven smaller palms, each containing four fingers. At first sight
>this number of palms seems hopelessly impractical, since it does not
>divide naturally into halves, thirds, or quarters. As you have already
>said, Franz, this seven-part division was very good for measurements of
>the circle and square, and this may indeed be why it was chosen.
Yes, like the unit triples, unit proportions for pi and phi and
also for doubling cubes, trisecting angles seem a good practical
solution to the problem of irrationals.
>Yet other factors to consider are the Egyptian foot-measurement, which
>was identified a long time ago as being equal to half the royal cubit,
>and the question of the Egyptian canon of art - the guide-line system of
>the Old Kingdom and the later grid. Contrary to Erik Iversen and Gay
>Robins, I have shown that the canon was based on the six-part division
>of the royal cubit, as follows:
>
>In the guide-line system, the height of the standing male from the floor
>to the so-called hair-line was divided by horizontal lines into three
>equal parts. In the later grid, this hair-line level is equivalent to
>18 squares, while the height to the top of the head equals 19 squares.
yes, thats true as far as it goes.
>For the mean anatomical height of 166 cm, the hair-line level will be:
>166 cm x 18/19 = 157.26 cm = 3 x 52.4 cm = 3 royal cubits (near enough).
No, the hairline level is about 66", three royal cubits is at most 63"
what you have is three royal cubits plus one palm.
>
>Thus each of the three equal parts up to the hair-line in the guide-line
>system was equal to one royal cubit, and each of the grid squares in the
>later system was equal to 1/6 royal cubit.
Not quite, the system is a bit more subtle and sophisticated than that.
Since you have the same diagram I do, see what happens if you take the
cubit and royal cubit as marked by the wrist bracelet and shoulder band
use proportional dividers if you have them.
>
>[J. Legon, 'The Cubit and the Egyptian Canon of Art', in _Discussions in
>Egyptology_ 35 (1996), 61-76.]
>
>
>John Legon
regards,
steve
John Legon
>In article <akxfrHA3...@legon.demon.co.uk>,
> John Legon <jo...@legon.demon.co.uk> wrote:
>
>> Stadelmann's measurement is much better, being (almost certainly) taken
>> from a plan by Maragioglio and Rinaldi, who in turn took the dimension
>> was 1838.6 inches. This is 89.2 cubits (for your cubit of 52.36 cm), so
>> Petrie did not crawl upside-down on the ceiling of the gallery to get
>> this dimension, but had to calculate it from several other measurements.
>> Which makes me wonder why you have to take one of the most inaccessible
>> dimensions inside the entire pyramid to support the idea that the number
>> of 89 cubits is significant?
>
>Measuring the length of the ceiling ain't such a problem: measure
>the length of the wall and subtract the sum of the 'Einkragungen'.
>Or use a plummet, mark the ends of the ceiling on the floor and
>measure the distance of the marks.
>
from the top 'corbel' of the vault at either end. But he had to climb
rather precariously up a ladder to do it.
Petrie found that the horizontal length of the grand gallery is equal to
the height of the upper end (great step) above the base of the pyramid.
He only worked in inches, but we have now seen that the level of the
step (and floor of King's Chamber) is 82 cubits above the base. Since
the laps of the corbelled vault overhang, when added together, by one
cubit at either end and on both sides, the 'run' or horizontal length of
the ceiling should be (82 - 1 - 1) equals 80 cubits.
For the slope of the gallery, we have:
Angle of slope, mean axis (Petrie) = 26 degrees 16 minutes 40 seconds
Angle of floor and ramps (Smyth) = 26 degrees 17 minutes 37 seconds
For the horizontal length of the ceiling of 80 cubits, the mean sloping
length should therefore be 89.22 cubits, which is really pretty close to
Petrie's measurement of 1838.6 inches. Nothing to do with those strange
golden numbers of yours, in my view...
>Getting reliable measurements from a professor of Egyptology is
>almost impossible. I am begging since years for a single measurement
>of the gallery - oblique height of the blocks which build the layers
>of the 'Kraggewoelbe' - and got some kind letters but never the
>measurement I asked for. Maybe you find it in your books? According
>to my thesis it should measure 44 fingers or about 82.28 centimeters
>(on the average) while the vertical height would measure 49 fingers
>or 91.63 cm.
The only figures I can find for the oblique heights of the laps of the
vault (i.e. measured at right-angles to the slope), come from the work
of Piazzi Smyth (Life and Work at the Great Pyramid, Edinburgh, 1867).
I think Smyth had great trouble taking these measurements, because he
put some of them in brackets. However, for the mean oblique heights of
the seven laps on the east and west sides of the vault at the north end
we have, in inches: 32.2, 32.2, 31.8, 32.2, 32.0, 31.0, 31.0
And at the south end: 32.3, 30.9, 31.7, (31.6), (32.0), (31.0), (32.0)
So the larger laps measure nearly 44 fingers or 32.4 inches in oblique
height, but not quite.
>
>Building the gallery was a demanding task, for one had to consider
>5 dimensions: width, horizontal and oblique height, oblique and
>vertical height.
>
>Combining the royal cubit (52.36 cm) with the hypothetical measure
>of the 'magic rod' (7/11 royal cubits = 196 Horus marks = 33.32 cm)
>makes it easy:
>
> horizontal length oblique height 4 royal cubits 4 palms
> oblique length vertical height 7 magic rods 1 magic rod
>
>The ceiling consists of 40 blocks having a length of 140/40 = 3.5
>royal cubits each while the run of every block measures 2 royal cubits.
>
That's funny, Smyth says there are only 36 blocks, and others before him
said only 30 or 31. Perhaps they just got tired of counting? Also, I
think you mean 3.5 magic rods, not royal cubits. I've forgotten what
the correct number of blocks is supposed to be...
>Gallery B is based on the triple 39-80-89. Gallery A is based on
>the pseudo-triple 539-1100-1225 or 2156-4400-4900. A funny triangle:
>
> oblique height 44x44 rise 44x49 run 44x100 slope 49x100
>
> tangent 49/100 sine 11/25 cosine 44/49
>
> cosine = 4 royal cubits / 7 magic rods = 4 palms / 1 magic rod
>
Yes, some very good numbers, except that the actual angle of slope of
the gallery is about 26 degrees 17 minutes, and not 26 degrees 6 minutes
or less. I have considered these factors myself, and think they played
a part in the geometry, but they were not the 'final' solution for the
slope of the gallery...
>Jean-Philippe Lauer says that the pyramid builders tried to find
>measurements made of whole numbers. I agree with him. Bu I like
>to add that they had clever ways for obtaining whole numbers,
>even when defining diagonals, angles and round forms.
That is exactly what I think.
>
>Let me ask professor Ahmes on this topic.
>
>My astrologer (touching her crystal sphere again): Professor Ahmes?
>Hello? Professor? Yes?
>
>Bobby: Some are mathematicians
> Some are carpenters' wifes
> Don't know how long they've studied
> Don't know what they do with their lifes
>
>Ahmes: Who are you?
>
>Bobby: You may call me Bobby
>
John: Who on Earth is Bobby?
John Legon
<whi...@shore.net> wrote:
>I think elsewhere you mention Coles survey, I wonder if this
>is available through Amazon. I suspect that it would be both
>accurate and subject to some caveats.
Cole's survey paper was published in 1925 and is now very difficult to
find, even in specialist libraries...
>
>The use of his measures to check ancient Egyptian units
>presumes the Great Pyramid was intended to be 440 royal cubits
>in length at its base.
>
I think that conclusion has been fairly widely accepted.
>This begs the question, where do we cut the pane of its base?
>
>Cole measured to the base sockets. This is comparable to establishing
>the proportions of a modern house by measuring to its footing as
>opposed to its foundation or wall.
>
No, Cole based his survey on the casing-stones on the pavement, and the
traces of the casing-edge on the pavement where the casing-stones have
been removed. He included the corner-sockets in his survey only as a
means of checking that the reconstructed corners of the casing, on the
finished pavement, lay on the diagonals of the corner-sockets beneath
them, just as Petrie had done before him.
>The more likely point for a proportion to be established is at the
>point the finished pavement meets the finished casing. An Egyptologists
>reconstructions are only as good as the points he chooses to measure from.
>
How true that is!
Regards
John Legon
John Legon
I think most of your arguments are based on the assumption that Erik
Iversen's metrological system for the Egyptian canon of art is correct.
Dr Gay Robins, however, has extensively demolished Iversen's hypotheses
in various articles and in her recent book, 'Proportion and Style in
Ancient Egypt' (London, 1994), 40-56. She is now probably the leading
authority of the subject, and I think her arguments against Iversen are
pretty solid. In my view, Iversen constructed an artificial canonical
system that has little or no scientific credibility.
If you prefer to take Iversen as your authority, then that is your
privilege.
Best wishes,
John Legon
cir...@access.ch
dwe...@ramtops.demon.co.uk (Doug Weller)
kindly forwarded a lecture held by Rudolf Gantenbrink.
I agree with many points, however not with the following one:
> with far-reaching consequences. The master builder of the Cheops Pyramid
> was familiar neither with Pythagoras' theorem nor that of Tange, or they
> would otherwise have made use of this knowledge instead of messing about
> with primitive mathematical systems.
Thank you, Steve and John, for the many replies and measurements.
I study them and shall reply - Oh my astrologer. She tells me Ahmes
is waiting ... ;)
Me: Good morning, professor Ahmes. I have a question. Some days ago
you told us that the royal cubit measure means an arm. Now John Legon
and Steve Whittet propose other explanations. Their arguments sound
fairly reasonable to me. So I don't know who is right.
Ahmes: Well, probably everyone. You certainly know that Egypt was
united in a long war which finally was won by the followers of Horus.
Yet one can win a war and fail all the same. The early kings succeeded
only because there were some bright people around who said: Why do we
go on fighting? It would be so much better if we could join our forces
and religions and symbols and customs and even our measures ...
The Nile Valley was inhabited by various tribes who were using many
measures. Now the wise people we remember under the name of Thoth
Aisha: ... and of Seshat
Ahmes: ... Yes, my dear daughter, also under the name of Seshat,
goddess of writing, calculating, measuring, astronomy and of the
calendar.
Aisha: One of your many papyri calls her 'the first one writing'
- theferfore she must have been the teacher of her husband, and
as wise as him
Ahmes: You hear my daughter? :)
Well, the wise people we remember as Thoth and S e s h a t
Aisha: :)
Ahmes: ... combined the various measures in a clever system which
satisfied everyone.
The royal cubit stays for the arm of the king and also for his mace:
6 palms for the arm, 1 palm for the fist; 6 palms for the wooden rod,
1 palm for the macehead, sign of his power.
The first deities are Re and Maat. This means that a king got his
power from Re while he has to follow Maat, goddess of truth, order
and justice.
Well, this is the general explanation of the royal cubit measure
while the many subdivisions - some of them inscribed on the cubit
rods - represent the various earlier measures.
Me: I see. The many tribes were united in a single kingdom and the
many measures in a single system. Sounds rather convincing to me.
Now for something else. Many of our professors believe that the
pyramid builders had no idea of the formula of Pythagoras, I mean
the one of Imhotep ...
Ahmes: You speak of the triples 3-4-5, 5-12-13, 8-15-17, 7-24-25
... and the quadruples 1-2-2-3, 2-3-6-7, 8-9-12-17 ...? Of course
we know them, and so did Imhotep and his successors. Let me give
you some examples:
The sarcophagus of Djoser is a most ingenious combination of ten
triples while the squares of his Holy Town are based on the pseudo-
triple 17-17-24.
Sneferu's northern Appearance is based on the triple 20-21-29: height
200 royal cubits, half base 210 royal cubits, slope 290 royal cubits.
Khufu's Horizon is defined by the pseudo-triple 70-55-89, the chamber
containing his sarcophagus by the Sacred Triangle 15-20-25 royal
cubits. The gallery combines the triple 39-80-89 and the pseudo-triple
44x49 - 44x100 - 49x100. The pseudo-triples 70-70-99 and 99-99-140
are fine numbers for a square. The Sothis arm of Shu follows the
pseudo-triple 490-594-770 while the descending gangway is based on
the pseudo-triple 72-144-161: the vertical distance floor-ceiling
measures 72 fingers, the horizontal one 144 fingers, and the slope
161 fingers.
Finally, Khafre's pyramid is based on the Sacred Triangle 3-4-5.
Do your people really believe that one could build a large pyramid
without knowing these triples? Without making use of such pseudo-
triples? And without using the quadruple 1-2-2-3 which defines half
a cube and its diagonal?
Me: Sorry, professor, they do.
Ahmes: No one knowing better?
Me: A few ones. For example professor Wolf Meyer-Christian. However,
his fine work is almost completely neglected.
Ahmes: May I encourage him?
cir...@access.ch
John Legon <jo...@legon.demon.co.uk> wrote: a reply to Steve
Thank you, John, for the measurements of the gallery (in a previous
post). I appreciate your help.
Here follows another interview with Ahmes.
Me: Good morning, professor Ahmes. You keep this precious papyrus
from Khufu's time. Can you tell us a little more, please?
Ahmes: Well, I may tell you something on the number 2.
This number stays for polarity and for the powers of creation.
Khufu's Horizon is ruled by a pair of measures:
divine measure: magic rod 196 Horus marks
human measure: royal cubit 308 Horus marks
base 440 royal cubits height 440 magic rods
The seemingly so simple pyramid combines a pair of shapes:
Golden pyramid: the inscribed hemisphere symbolizes
the Primeval Sky Goddess who lives on in Hathor and Nut
Pyramid based on the number of Re: the circle whose vertical
diameter is given by the height of the pyramid symbolizes
the Primeval Sun God who lives on in Re
The gallery combines 2 ideal galleries which are based on the triple
39-80-89 and on the pseudo-triple 44 x 49 -- 44 x 100 -- 49 x 100.
The 2 gangways have almost the same angle while they are defined
in different ways: one ascends in the angle of the gallery while
the other one follows the diagonal of the double square
Me: ... which has a tangent of 1/2. Hence the angle should measure
26 degrees 33 minutes 54 seconds. What does Rainer Stadelmann say?
The angle of the descending corridor measures 26 degrees 34 minutes
23 seconds. Well, the mistake is less than half a minute.
Ahmes: Now please consider my number column for the double square:
1 1 5 1 1 5
2 6 10 2 6 10
1 3 5 1 3 5
4 8 20 4 8 20
2 4 10 2 4 10
1 2 5 1 2 5
3 7 15 3 7 15
10 22 50 10 22 50
5 11 25 5 11 25
16 36 80 16 36 80
8 18 40 8 18 40
4 9 20 4 9 20
13 29 65
42 94 210 and so on
21 47 105
68 152 340
34 76 170
17 38 85
55 123 275
178 398 890
89 199 445
288 644 1440
144 322 720
72 161 360 and so on
The last lines define the descending gangway:
vertical distance floor-ceiling 72 fingers = 360 Geb marks
horizontal distance floor-ceiling 144 fingers = 720 Geb marks
oblique measurement 161 fingers = 805 Geb marks
oblique height of the gangway 322 Geb marks
Me: Let me check the last number. 322 Geb marks = 64.4 fingers =
2.3 royal cubits = 2.3 x 52.36 cm = 120.428 cm. Rainer Stadelmann?
The height of the descending corridor measures 120 centimeters.
Ahmes: My number column contains a pair of golden sequences:
1 3 4 7 11 18 29 47 76 123 199 322 and so on
1 1 2 3 5 8 13 21 34 55 89 144 and so on
Me: Our teachers made a fuss about the golden section. We remember
and even fear it as something mysterious and very very difficult.
Ahmes: Really? Tell me a pair of numbers.
Me: 5 and 9?
Ahmes: Go like this and you will obtain another golden sequence:
5 + 9 = 14
9 + 14 = 23
14 + 23 = 37
23 + 37 = 60
37 + 60 = 97
60 + 97 = 157 and so on
Me: I guess our teachers could win more children for geometry
and mathematics if they would begin with simple number games.
Ahmes: Oh yes. Our youngest pupils play all kinds of games. Sooner
or later they come up with questions. We answer them and teach the
children by and by, always on their level of understanding, making
sure they can follow us. It's much better to know something simple
very well than having a halfbaked idea of something complicated.
How can the children learn about geometry when they can't follow
our reasoning and even begin to fear our wonderful discipline?
- Why are our scribes and artisans so good? Becasue the love
their professions.
-----------== Posted via Deja News, The Discussion Network ==----------
http://www.dejanews.com/ Search, Read, Discuss, or Start Your Own
Steve Whittet
In article <6v1v2n$c2t$1...@nnrp1.dejanews.com>, cir...@access.ch says...
>
>In article <lvHgJWAq...@legon.demon.co.uk>,
> John Legon <jo...@legon.demon.co.uk> wrote: a reply to Steve
>
>Thank you, John, for the measurements of the gallery (in a previous
>post). I appreciate your help.
>
>Here follows another interview with Ahmes.
>
>
>Me: Good morning, professor Ahmes. You keep this precious papyrus
>from Khufu's time. Can you tell us a little more, please?
>
>Ahmes: Well, I may tell you something on the number 2.
>
>This number stays for polarity and for the powers of creation.
>
>Khufu's Horizon is ruled by a pair of measures:
>
> divine measure: magic rod 196 Horus marks
> human measure: royal cubit 308 Horus marks
>
> base 440 royal cubits height 440 magic rods
base 756 feet height 481 feet
308/196=1.5714285 = 11/7, the rise to run is 14/7
756/481=1.5717255
comparing how the same number of units in two different
units of measure are arranged on a larger scale of measures
is interesting; for example
base circuit 1760 royal cubits 1 mile 1760 yards
The great pyramid could be laid out a number of different ways
but it seems reasonable to begin with a fairly large unit measure,
something like the [hayt]. The pyramids base is 44 [hayt] and its
height is 28 [hayt]. (A hayt is a cord or rod of 10 Royal cubits
which is itself '10 the side of the [setat] or Royal Egyptian acre.
The base covers 13 English acres
The sloped side or apothem is a stadium (600 Greek feet)
Herodotus says
"the base is a square eight of 100 [feet] each way"
(The square of the base (13.12 acres) is equal
in area to 8 square units which measure 100 units)
this side of 100 units is 267' making the unit about 32"
or roughly in the same proportion to the royal cubit as
the measure of the base to the measure of the height
(16 English perche = 264'
(1 English perche or rod = 16.5')
(1 [hayt] (knotted cord or rod = 17.18 feet
17.18/16.5 x 12 = 11.52 the same proportiuon as feet to pes.
"...and the height the same"
(The height (with the pyramidion of 4 royal cubits)
is 8 times the square of a unit whose side is 100 feet.)
it needs to be remembered that the pyramid has a pyramidion
4 royal cubits in height which gives the height with and
without the pyramidion as 284 cubits.
284^2=80656 making the square of the height 8 [setat]
1.)(284:280 the only pair of amicable numbers known to antiquity.)
2.)(The source for the height of the pyramidion is Agatharchydes of Cygnus)
The surface of the pyramid is 26 acres, twice its base
Most of the proportions of the interior chambers and their location
in the overall shape are also in unit [hayt]
>The seemingly so simple pyramid combines a pair of shapes:
>
> Golden pyramid: the inscribed hemisphere symbolizes
> the Primeval Sky Goddess who lives on in Hathor and Nut
>
> Pyramid based on the number of Re: the circle whose vertical
> diameter is given by the height of the pyramid symbolizes
> the Primeval Sun God who lives on in Re
>
>The gallery combines 2 ideal galleries which are based on the triple
>39-80-89 and on the pseudo-triple 44 x 49 -- 44 x 100 -- 49 x 100.
Essentially the proportions are in the ratio of 7:11
>
>The 2 gangways have almost the same angle while they are defined
>in different ways: one ascends in the angle of the gallery while
>the other one follows the diagonal of the double square
>
>Me: ... which has a tangent of 1/2. Hence the angle should measure
>26 degrees 33 minutes 54 seconds. What does Rainer Stadelmann say?
>The angle of the descending corridor measures 26 degrees 34 minutes
>23 seconds. Well, the mistake is less than half a minute.
If the rise to run is as simple as 1:2 why carry it out to
minutes and seconds?
Wilkinsen lists a range of values for cubit rules
found in Egypt
Wooden royal cubit found at Memphis by Jomard 20.47291"
That of the museum of Turin 20.57869" (English inches)
another 20.61806"
another 20.65843"
Wilkinsons measure of the Nilometer 20.625"
Mr Perrings calculations at the Pyramid 20.625"
Jomards average of measures of wooden rules 523.506mm(20.61")
Wilkensons account of the measure of the Temple of Karnak (20.65")
Mr Harris cubit from Thebes 20.65"
that of the Nilometer 20.74840"
The Elephantine Nilometer according to Jomard 20.7484"
A cubit on a stone at E'sooan found by Wilkinson 21"
Using Stecchinis value for the Royal cubit, which is based
on ancient units of measure, weight and volume we get 525mm
120 cm = 64 fingers = 16 palms = 4 feet
taking the finger as 18.75mm and the royal cubit as 525 mm (20.67")
>
>Ahmes: My number column contains a pair of golden sequences:
>
> 1 3 4 7 11 18 29 47 76 123 199 322 and so on
> 1 1 2 3 5 8 13 21 34 55 89 144 and so on
>
>Me: Our teachers made a fuss about the golden section. We remember
>and even fear it as something mysterious and very very difficult.
>
>Ahmes: Really? Tell me a pair of numbers.
>
>Me: 5 and 9?
>
>Ahmes: Go like this and you will obtain another golden sequence:
>
> 5 + 9 = 14
> 9 + 14 = 23
> 14 + 23 = 37
> 23 + 37 = 60
> 37 + 60 = 97
> 60 + 97 = 157 and so on
The golden ratio occurs often in nature, have you any thoughts
as to why it is so easily generated?
>
>Me: I guess our teachers could win more children for geometry
>and mathematics if they would begin with simple number games.
>
>Ahmes: Oh yes. Our youngest pupils play all kinds of games. Sooner
>or later they come up with questions. We answer them and teach the
>children by and by, always on their level of understanding, making
>sure they can follow us. It's much better to know something simple
>very well than having a halfbaked idea of something complicated.
>How can the children learn about geometry when they can't follow
>our reasoning and even begin to fear our wonderful discipline?
>- Why are our scribes and artisans so good? Becasue the love
>their professions.
The idea of living the life in maat seems both subtle and
sophisticated. To do what is right and proper according
to a standard you believe in means whatever you do it
will be well balanced and harmonious and make you happy.
>
steve
cir...@access.ch
whi...@shore.net (Steve Whittet) wrote:
> Wooden royal cubit found at Memphis by Jomard 20.47291"
> That of the museum of Turin 20.57869" (English inches)
> another 20.61806"
> another 20.65843"
> Wilkinsons measure of the Nilometer 20.625"
> Mr Perrings calculations at the Pyramid 20.625"
>
> Jomards average of measures of wooden rules 523.506mm(20.61")
> Wilkensons account of the measure of the Temple of Karnak (20.65")
> Mr Harris cubit from Thebes 20.65"
> that of the Nilometer 20.74840"
> The Elephantine Nilometer according to Jomard 20.7484"
> A cubit on a stone at E'sooan found by Wilkinson 21"
> Using Stecchinis value for the Royal cubit, which is based
> on ancient units of measure, weight and volume we get 525mm
Hi Steve,
thank you for the many informations. Rainer Stadelmann and Rudolf
Gantenbrink agree on a royal cubit measure of 52.36 centimeters
used for building the Great Pyramid while the royal cubit measure
of the New Kingdom was slightly longer, about 52.5 cm. By the way,
there is a funny relation of the GP cubit and the meter: please
imagine a circle whose circumference measures 6 x 52.36 cm.
- How long is the diameter of this circle??
My astrologer: I got a connection to Ahmes.
Me: Thank you. Good morning, professor.
Ahmes: Good morning.
Me: May I ask you something?
Ahmes: Why, yes, of course.
Me: I wonder if only Khufu's Horizon is so very fascinating.
Or does your old papyrus also mention other pyramids?
Ahmes: Yes, for example Sneferu's northern Appearance.
The base of this pyramid measures 420 royal cubits and the height
200 royal cubits. The diagonal of the base measures 594 cubits,
according to the pseudo-triple 70-70-99. The slope measures 290
royal cubits, according to the triple 20-21-29. The edge measures
358 royal cubits, according to the pseudo-quadruple 100-105-105-179.
Now please imagine a sphere in the frame of this pyramid. How long
is the radius? 84 royal cubits, according to the triple 80-84-116.
The imaginary sphere means the sun enclosed in the Primeval Hill.
Divide the base of Sneferu's pyramid by 4 and you will obtain 105
royal cubits. Divide the base of Khufu's pyramid by 5 and you will
obtain 88 royal cubits. A rectangle measuring 88 x 105 cubits has
a diagonal of 137 royal cubits, according to the triple 88-105-137.
Multiply 137 royal cubits by 2 and 3. You will obtain 274 and 411
royal cubits: height and base of the pyramid of Khafre who was a
son of Khufu and a grandson of Sneferu.
I mention this kind of a pyramid in my own papyrus where it has
a base of 12 royal cubits and a height of 8 royal cubits.
Me: We know it as problem no. 59 of the Rhind Mathematical Papyrus.
Ahmes: In my depot I keep a wooden model of this pyramid. Base and
height measure 12 and 8 fingers while the shape is given by a Sacred
Triangle:
half base 3 x 2 fingers
height 4 x 2 fingers
slope 5 x 2 fingers
The area of the base measures 12 x 12 = 144 fingers
The area of one face measures '2 x 12 x 10 = 60 square fingers.
The area of all the four faces measures 4 x 60 = 240 square fingers.
And the whole volume? It measures '3 x 12 x 12 x 8 = 384 square fingers.
Now please imagine a sphere in the frame of this pyramid. Its radius
measures 3 fingers, according to the triple 3-4-5, while the distance
of the center of the sphere from a corner of the base measures 9
fingers, according to the quadruple 1-2-2-3 or 3-6-6-9.
The diameter of the sphere measures 6 fingers. Now let me calculate
the surface and volume by means of the formulas
diameter x diameter x re
'6 x diameter x diameter x diameter x re
Do you remember my sequence for the number of the circle?
3 (plus 22) 25 47 69 91 113 ...
1 (plus 7) 8 15 22 29 36 ...
I use the value '36 x 113 and obtain the following numbers:
6 fingers x 6 fingers x '36 x 113 equal 113 square fingers
'6 x 6f x 6f x 6f x '36 x 113 equal 113 cube fingers
Ain't this a fine surprise?
The surface and volume of the pyramid result in the same number,
and so do the surface and volume of the inscribed sphere.
Now replace every finger of my model by a unit of 959 fingers
or 34 '4 royal cubits and you will obtain Khafre's pyramid.
Furthermore, there are some lines regarding Menkaure's pyramid.
But now I have to go and give a lesson. Fare-well!
John Legon
>Ahmes: Good morning.
>
>Me: May I ask you something?
>
>Ahmes: Why, yes, of course.
>
>Me: I wonder if only Khufu's Horizon is so very fascinating.
>Or does your old papyrus also mention other pyramids?
>
>Ahmes: Yes, for example Sneferu's northern Appearance.
>
>The base of this pyramid measures 420 royal cubits and the height
>200 royal cubits. The diagonal of the base measures 594 cubits,
>according to the pseudo-triple 70-70-99. The slope measures 290
>royal cubits, according to the triple 20-21-29. The edge measures
>358 royal cubits, according to the pseudo-quadruple 100-105-105-179.
>Now please imagine a sphere in the frame of this pyramid. How long
>is the radius? 84 royal cubits, according to the triple 80-84-116.
>The imaginary sphere means the sun enclosed in the Primeval Hill.
>
John: But professor, surely you know that Rainer Stadelmann says that
the angle of Sneferu's Northern Appearance is just 45 degrees (see Die
aegyptischen Pyramiden, p. 100). As you said yourself, there is no way
around the dimensions given by Stadelmann. Tell me where the 20-21-29
triple angle comes from, and I will tell you why you are wrong :)
>Divide the base of Sneferu's pyramid by 4 and you will obtain 105
>royal cubits. Divide the base of Khufu's pyramid by 5 and you will
>obtain 88 royal cubits. A rectangle measuring 88 x 105 cubits has
>a diagonal of 137 royal cubits, according to the triple 88-105-137.
>Multiply 137 royal cubits by 2 and 3. You will obtain 274 and 411
>royal cubits: height and base of the pyramid of Khafre who was a
>son of Khufu and a grandson of Sneferu.
>
John: That is a very interesting result, and I am glad that you realise
that the sides of Khafre's pyramid measure 411 cubits. This is exactly
the measurement that I have put foward for some time. It gives the true
length of the royal cubit for the Giza pyramids, since Petrie found that
the sides of Khafre's pyramid measure 8474.9 +/- 1.5 inches. That's
215.26 ms in our silly metric units. So, the length of the royal cubit
will be 215.26/411 equals 0.523747 metres.
Now, we apply the same length of cubit to the base of the Great Pyramid,
and find that it perfectly expresses the 'secret number living in the
circle' for the height of just 280 cubits. Remember?
[...]
>
>Furthermore, there are some lines regarding Menkaure's pyramid.
John: I look forward to reading them.
>But now I have to go and give a lesson. Fare-well!
Auf wiedersehen!
John
Martin Stower
> John: That is a very interesting result, and I am glad that you realise
> that the sides of Khafre's pyramid measure 411 cubits. This is exactly
> the measurement that I have put foward for some time. It gives the true
> length of the royal cubit for the Giza pyramids, since Petrie found that
> the sides of Khafre's pyramid measure 8474.9 +/- 1.5 inches. That's
> 215.26 ms in our silly metric units. So, the length of the royal cubit
> will be 215.26/411 equals 0.523747 metres.
Yes! Height = 2 x 137 cubits, base-side length = 3 x 137 cubits. It's a
curiously unround modulus. Had the nearest nice round figure of 140 cubits been
used, we'd get this result: height = 280 cubits (the same as Khufu), base-side
length = 420 cubits (the same as Dahshur North). What occurs to me is that this
might be one of several signs of a concern for economy in Khafre's pyramid.
Shaving three cubits off the modulus saved a volume equal to more than half of
that of Menkaure's pyramid. I wonder if they actually told Khafre . . .
Martin Stower
cir...@access.ch
John Legon <jo...@legon.demon.co.uk> wrote:
> John: But professor, surely you know that Rainer Stadelmann says that
> the angle of Sneferu's Northern Appearance is just 45 degrees (see Die
> aegyptischen Pyramiden, p. 100). As you said yourself, there is no way
> around the dimensions given by Stadelmann. Tell me where the 20-21-29
> triple angle comes from, and I will tell you why you are wrong :)
Ahmes: Good morning, John. You are quite right, the angle of the casing
blocks at the base are a little greater, for optical reasons. You know,
a very long line like for example the edge of a pyramid must be slightly
rounded for to appear straight.
Me: Yes, this was also considered by the builders of the Acropolis.
By the way, Rainer Stadelmann himself says that the height of the Red
Pyramid measures 200 'Ellen'.
John: (...)
> So, the length of the royal cubit
> will be 215.26/411 equals 0.523747 metres.
>
> Now, we apply the same length of cubit to the base of the Great Pyramid,
> and find that it perfectly expresses the 'secret number living in the
> circle' for the height of just 280 cubits. Remember?
Ahmes: A fine number game. However, you might consider that Khafre
chose a slightly longer unit. Anyway, go on with your number games.
Playing with our numbers and measurements is by far the best and even
the only way to get an idea of our geometry.
Now for Menkaure's pyramid. You are looking forward to hear about it,
you told me? Well, here I go.
There are a few damaged lines on my old papyrus regarding this pyramid.
I read them as follows:
shorter base 196 royal cubits or 308 magic rods
longer base 200 royal cubits
diagonal 280 royal cubits or 440 magic rods
height 126 royal cubits or 198 magic rods
edge 126 royal cubits or 296 magic rods
These numbers are based on the pseudo-triples 49-50-70 and 70-70-99,
furthermore on the pseudo-quadruple 539-550-693-1036. Fine numbers.
However, it seems that Menkaure's pyramid wasn't built as precisely
as Khufu's Horizon.
The height of Khufu's pyramid and the ideal diagonal of the base
of Menkaure's measure 280 royal cubits.
The periphery of the base of Khufu's pyramid measures 4 x 440 = 1760
royal cubits while the one of Menkaure's pyramid measures 200 + 196
+ 200 + 196 = 792 royal cubits.
Now please transform the periphery of the base into a circle.
How long will the radius be?
Divide 1760 royal cubits by 2 and again by 3 '7. Thus you will obtain
280 royal cubits --- the height of Khufu's pyramid.
Divide 792 royal cubits by 2 and again by 3 '7. Thus you will obtain
126 royal cubits --- the ideal height of Menkaure's pyramid.
Now please draw an imaginary circle around the top of the pyramid,
using the height as radius. The circumference equals the periphery
of the base. - What may this imaginary circle mean?
Me: The solar disk? the sun rising from the Primeval Hill?
or the sun king and his deification?
Ludwig Borchardt: Pyramidenspinner!
Me: You again?
Ludwig Borchardt: Numerologe! Pyramidenspinner!
Ahmes: What's eating this man?
Me: He hates pyramid geometry.
Because of him, many a fine paper can't be published. And if someone
succeeds all the same, like for example professor Wolf Meyer-Christian
("Der 'Pythagoras' in Aegypten am Beginn des Alten Reiches?" published
in: Mitteilungen des Deutschen Archaeologischen Institutes Abteilung
Kairo MDAIK, Band 43, 1987) hardly anyone dars quote him.
Ludwig Borchardt: Zahlenmystiker! Numerologe! Pyramidenspinner!
Me: Please, professor Borchardt, explain me how the builder of the
Great Pyramid at Giza was able to measure out a square of some 230
x 230 meters without having an idea of geometry?
Ludwig Borchardt:
Me: I am listening
John and Ahmes: We too
Ludwig Borchardt:
Me, John and Ahmes (waiting for an answer, in vain)
Steve Whittet
>
>In article <VjPsKAAd...@legon.demon.co.uk>,
> John Legon <jo...@legon.demon.co.uk> wrote:
>
>> John: But professor, surely you know that Rainer Stadelmann says that
>> the angle of Sneferu's Northern Appearance is just 45 degrees (see Die
>> aegyptischen Pyramiden, p. 100). As you said yourself, there is no way
>> around the dimensions given by Stadelmann. Tell me where the 20-21-29
>> triple angle comes from, and I will tell you why you are wrong :)
>
>Ahmes: Good morning, John. You are quite right, the angle of the casing
>blocks at the base are a little greater, for optical reasons. You know,
>a very long line like for example the edge of a pyramid must be slightly
>rounded for to appear straight.
>
>Me: Yes, this was also considered by the builders of the Acropolis.
>By the way, Rainer Stadelmann himself says that the height of the Red
>Pyramid measures 200 'Ellen'.
>
>John: (...)
>
>> So, the length of the royal cubit
>> will be 215.26/411 equals 0.523747 metres.
The Royal cubit used for measuring buildings and other royal
properties varied between about 523 and 525 mm. Giving it to
6 points of decimal based on a single example should probably
be avoided.
>>
>> Now, we apply the same length of cubit to the base of the Great Pyramid,
>> and find that it perfectly expresses the 'secret number living in the
>> circle' for the height of just 280 cubits. Remember?
The Pi ratio of 3 '7 does appear to be used in some buildings.
It could be argued that this is an inherent property of the
septenary system of measurements and the Egyptians use of unit
fractions. How would you explain to the sceptics in the crowd
the difference between the value of Pi used in the buildings and
the value of Pi used in the mathematical papyri?
>
>Ahmes: A fine number game. However, you might consider that Khafre
>chose a slightly longer unit. Anyway, go on with your number games.
>Playing with our numbers and measurements is by far the best and even
>the only way to get an idea of our geometry.
>
>Now for Menkaure's pyramid. You are looking forward to hear about it,
>you told me? Well, here I go.
>
>There are a few damaged lines on my old papyrus regarding this pyramid.
>I read them as follows:
>
> shorter base 196 royal cubits or 308 magic rods
> longer base 200 royal cubits
> diagonal 280 royal cubits or 440 magic rods
> height 126 royal cubits or 198 magic rods
> edge 126 royal cubits or 296 magic rods
IES Edwards gives the following dimensions for Egyptian pyramids
Zoser 411 ft EW by 358 ft NS
Sekhemkhet 395 ft sq
Khaba 276 ft sq
Sneferu 473 ft sq (Meidum)
Sneferu 620 ft sq (Bent)
Sneferu 719 ft sq (Northern)
Cheops 756 ft sq
Djedefre 320 ft sq
Chephren 708 ft sq
Mycernius 356 ft sq
Userkaf 247 ft sq
Sahure 257 ft sq
Neferirkare 360 ft sq
Neferefre 197 ft sq
Niuserre 274 ft sq
Isesi 265 ft sq
Unas 220 ft sq
Teti 210 ft sq
Pepi 1 250 ft sq
Merenre 263 ft sq
Pepi II 258 ft sq
Ibi 102 ft sq
Nebhepetre 70 ft sq
Ammenemes 1 296 ft sq
Sesostris 1 352 ft sq
Ammenemes II 263 ft sq
Sesostris II 347 ft sq
Sesostris II 350 ft sq
Ammenemes III 342 ft sq
Khendjer 170 ft sq
Do any of these measures help us better understand the value
of the royal cubit. Theoretically they should all be even multiples
of royal cubits rather than feet.
to try a few
feet rc value of rc
395 230 20.6087
276 161 20.57143
473 275 20.64
620 361 20.60942
719 418 20.64115
756 440 20.61818
320 186 20.64516
708 412 20.62136
356 207 20.63768
247 144 20.58333
...etc
>These numbers are based on the pseudo-triples 49-50-70 and 70-70-99,
>furthermore on the pseudo-quadruple 539-550-693-1036. Fine numbers.
>However, it seems that Menkaure's pyramid wasn't built as precisely
>as Khufu's Horizon.
One issue that is sometimes raised is whether ther deviation
from an expected norm is intentional. How does the above list
of cubit values suggest this question should be answered?
>
>The height of Khufu's pyramid and the ideal diagonal of the base
>of Menkaure's measure 280 royal cubits.
>
>The periphery of the base of Khufu's pyramid measures 4 x 440 = 1760
>royal cubits while the one of Menkaure's pyramid measures 200 + 196
>+ 200 + 196 = 792 royal cubits.
>
>Now please transform the periphery of the base into a circle.
>How long will the radius be?
2 pi, the area of the square divided by the area of the circle
will be pi/4
>
>Divide 1760 royal cubits by 2 and again by 3 '7. Thus you will obtain
>280 royal cubits --- the height of Khufu's pyramid.
>
>Divide 792 royal cubits by 2 and again by 3 '7. Thus you will obtain
>126 royal cubits --- the ideal height of Menkaure's pyramid.
>
>Now please draw an imaginary circle around the top of the pyramid,
>using the height as radius. The circumference equals the periphery
>of the base. - What may this imaginary circle mean?
It might mean that the ancient Egyptians were capable of practical
solutions of the three classical problems of Greek Antiquity,
squaring the circle, trisecting the angle and doubling the cube.
All seem to share a common solution in a curve which the Egyptians
may or may not have been capable of generating. Can we get Professor
Ahmes to show us how he might have done this?
Regarding trisecting an angle, an unmarked ruler can make a grid
which will provide a means of reading the angle to whatever
precision is desired.
We know that the interior angles of regular polygons
can be found with the formula (s-2*180-360)
and that the the limit is 180 degrees.
It is possible to construct a rotated array
of regular polygons which divide a circle into
any number of degrees.
Couldn't the Egyptians use a series of rotated regular polygons
as unit fractions of a circle egual to the '3 or "3 of an angle?
This isn't a precise solution but it could be useful as a table.
>
>Me: The solar disk? the sun rising from the Primeval Hill?
>or the sun king and his deification?
>
>Ludwig Borchardt: Pyramidenspinner!
>
>Me: You again?
>
>Ludwig Borchardt: Numerologe! Pyramidenspinner!
>
>Ahmes: What's eating this man?
>
>Me: He hates pyramid geometry.
Could numerology in the pythagorean style have led to
some of the Egyptians mathematical discoveries?
>
>Because of him, many a fine paper can't be published. And if someone
>succeeds all the same, like for example professor Wolf Meyer-Christian
>("Der 'Pythagoras' in Aegypten am Beginn des Alten Reiches?" published
>in: Mitteilungen des Deutschen Archaeologischen Institutes Abteilung
>Kairo MDAIK, Band 43, 1987) hardly anyone dars quote him.
>
>Ludwig Borchardt: Zahlenmystiker! Numerologe! Pyramidenspinner!
>
>Me: Please, professor Borchardt, explain me how the builder of the
>Great Pyramid at Giza was able to measure out a square of some 230
>x 230 meters without having an idea of geometry?
Playing the devils advocate, How much geometry does it take
to lay out a grid if you measure with a knotted cord?
>
>Ludwig Borchardt:
>
>Me: I am listening
>
>John and Ahmes: We too
>
>Ludwig Borchardt:
>Me, John and Ahmes (waiting for an answer, in vain)
>
?
steve
John Legon
<mst...@netcomuk.co.uk> wrote:
>
>Yes! Height = 2 x 137 cubits, base-side length = 3 x 137 cubits. It's a
>curiously unround modulus. Had the nearest nice round figure of 140 cubits
>been
>used, we'd get this result: height = 280 cubits (the same as Khufu), base-side
>length = 420 cubits (the same as Dahshur North). What occurs to me is that
>this
>might be one of several signs of a concern for economy in Khafre's pyramid.
>Shaving three cubits off the modulus saved a volume equal to more than half of
>that of Menkaure's pyramid. I wonder if they actually told Khafre . . .
>
Although it seems evident that the core-masonry of Khafre's pyramid was
not as well-built as Khufu's, using smaller blocks and a poorer quality
of stone, I have never been convinced by the argument that the reduction
in the size of the Old Kingdom pyramids after Khufu was due to economic
reasons. So far as I can see, the resources available to each reigning
king were inherently renewable and derived from the land and its people.
Some people say that the Egyptians started to build smaller pyramids
'when the money ran out', but in fact there was no money, and the main
requirement was the availability of a large labour-force to participate
in the enterprises of the 'state'.
Perhaps by the Sixth Dynasty, we can begin to see some erosion of the
power of pharaoh as the nomarchs in the provinces usurped more and more
of that power for themselves - eventually leading to the collapse of the
Old Kingdom. During the Fourth Dynasty, however, there is no reason to
doubt that the 'economic' factors remained fairly constant.
The experience gained during the reign of Khufu might have led to the
construction of even bigger and better pyramids in the following reigns.
The fact that Khafre's architects decided to build a smaller pyramid,
therefore, suggests the conscious decision to do something different,
not as an economy-measure but to satisfy the requirements of a specific
site plan.
John Legon
cir...@access.ch
whi...@shore.net (Steve Whittet) wrote:
> It might mean that the ancient Egyptians were capable of practical
> solutions of the three classical problems of Greek Antiquity,
> squaring the circle, trisecting the angle and doubling the cube.
> All seem to share a common solution in a curve which the Egyptians
> may or may not have been capable of generating. Can we get Professor
> Ahmes to show us how he might have done this?
Ahmes: Good morning, Steve, and thank you very much for your interest.
You mention a fascinating problem which is haunting my dear colleagues
at Babylon: how can one divide a circle into 360 equal parts? They
succeed in obtaining 120 equal parts, but then? I guess this problem,
trisecting an angle, ain't really solvable by means of ruler and
compass. Therefore I try to get a fine approximation numerically.
You may remember one of my number sequences:
6 (plus 22) 28 50 72 ... 314 ... 424 ... 600
2 (plus 7) 9 16 23 ... 100 ... 135 ... 191
Please have a look at the last number pair: if the radius of a circle
measures 3 x 191 fingers, the circumference measures 6 x 600 fingers
or 3600 fingers or 360 x 10 fingers.
It's funny that the number 191 plays an important role for the
pyramids of Djedefre and Ba-ka
Me: ... at Abu Rawash and Zawyet el-Aryan
Ahmes: ... however you call the places in your time.
Well, have a look at the numbers which I find on my precious papyrus
dating from Khufu's Dynasty.
Height and base of Djedefre's pyramid measure 126 and 203 royal cubits,
according to a golden sequence which contains the sacred number 7:
7 + 21 = 28
21 + 28 = 49
28 + 49 = 77
49 + 77 = 126
77 + 126 = 203
The base measures 203 royal cubits and the diagonal of the base 287
royal cubits, according to the number pattern
7 7 14
14 21 28
38 49 70
84 119 168
203 287 406
The edge measures 191 royal cubits, according to the very fine
pseudo-quadruple 203-203-252-382.
Finally, the radius of the inscribed hemisphere - symbol of the
sky once enclosed in the Primeval Hill - measures 79 royal cubits,
according to the pseudo-triple 79-98-126; a triangle of these numbers
has almost the same angles as half the cross-section of the pyramid.
Ba-Ka's pyramid is defined as follows: base 2 x 191 = 382 royal
cubits, height 191 royal cubits, diagonal of the base 540 royal
cubits, slope 270 royal cubits, radius of the inscribed sphere
- symbol of the sun in the Primeval Hill - 79 royal cubits.
These numbers are found by means of the following patterns:
3 7 6 6 14 12
10 13 20 20 26 40
23 33 46 46 66 92
56 79 112 112 158 224
135 191 270 270 382 540
The edge of Djedefre's pyramid and the height of Ba-ka's pyramid
measure 191 royal cubits, while the radius of the inscribed
hemisphere and sphere measure 79 royal cubits each.
You see: the two pyramids belong together. However, Djedefre's
pyramid was never finished, and the one of his son Ba-Ka was
hardly began.
Aisha, are you calling me? - Well, I got to go.
Have a nice day!
M.C.Harrison
>
> In article <24zS1.306$1w1.2...@news.shore.net>,
> whi...@shore.net (Steve Whittet) wrote:
>
> > solutions of the three classical problems of Greek Antiquity,
> > squaring the circle, trisecting the angle and doubling the cube.
> > All seem to share a common solution in a curve which the Egyptians
> > may or may not have been capable of generating. Can we get Professor
> > Ahmes to show us how he might have done this?
>
> You mention a fascinating problem which is haunting my dear colleagues
> at Babylon: how can one divide a circle into 360 equal parts? They
Well, you could approach the problem this way. Position a large circular
slab so that it is parallel to the plane of orbit of the stars (having
earlier established that the sun moves up and down).
Every day, at exactly the moment the sun sinks beneath a horizontal slab
that you have positioned to the west for this purpose, mark off on the
slab where a specific star is to be found, in line with the centre of
the slab. Repeat this for a few years, turning the slab round, picking
other suitable stars, and you will end up with 365 marks which don't
quite line up, and realise that the year is not actually 360 days long
after all and so you jack in astrology and start doing proper
measurements and stuff.
You might even work out that the "day" is a few minutes less than the
sunrise, if you try calculating the motion of the stars with your slab,
and even figure out how long the year is, proving that the earth orbits
the sun in at least one sense.
Then again, you might not, and apparently this had to wait until 1600 or
so.
Steve Whittet
nos...@spammers.of.the.world.unite.etc says...
>
>cir...@access.ch wrote:
>>
>> In article <24zS1.306$1w1.2...@news.shore.net>,
>> whi...@shore.net (Steve Whittet) wrote:
>>
>> > solutions of the three classical problems of Greek Antiquity,
>> > squaring the circle, trisecting the angle and doubling the cube.
>> > All seem to share a common solution in a curve which the Egyptians
>> > may or may not have been capable of generating. Can we get Professor
>> > Ahmes to show us how he might have done this?
>>
>> You mention a fascinating problem which is haunting my dear colleagues
>> at Babylon: how can one divide a circle into 360 equal parts?
The simplest way is to find two sets of measures which vary as
the circumference and diameter of a circle.
Lets take a really large circle of say 5000 iter
(75,000,000 royal cubits) in circumference
its diameter is 8,000 milliare. A degree
of its circumference is 70 milliare long
(using Pi = 3.15)
A roman pes is 11.7" it has the property that
A circle with a diameter of 15" has a circumference of 4 pes
A circle with a diameter of 30" has a circumference of 8 pes
A circle with a diameter of 60" has a circumference of 16 pes
A circle with a diameter of 360" has a circumference of 96 pes
If you lay out a circle with a diameter of 360"
its circumference is 64 ordinary cubits.(17.67 inches) and
55 royal cubits (20.563 inches)
>
>Well, you could approach the problem this way. Position a large circular
>slab so that it is parallel to the plane of orbit of the stars (having
>earlier established that the sun moves up and down).
>
>Every day, at exactly the moment the sun sinks beneath a horizontal slab
>that you have positioned to the west for this purpose, mark off on the
>slab where a specific star is to be found, in line with the centre of
>the slab. Repeat this for a few years, turning the slab round, picking
>other suitable stars, and you will end up with 365 marks which don't
>quite line up, and realise that the year is not actually 360 days long
>after all and so you jack in astrology and start doing proper
>measurements and stuff.
>You might even work out that the "day" is a few minutes less than the
>sunrise, if you try calculating the motion of the stars with your slab,
>and even figure out how long the year is, proving that the earth orbits
>the sun in at least one sense.
>
>Then again, you might not, and apparently this had to wait until 1600 or
>so.
It might be just an odd coincidence that our units of measure with
their ancient roots happen to work out so that there are 365,240
feet in a degree of the earths equatorial circumference.
Its odd that the varience between Egyptian, Greek, Roman and
English units all give different whole number aproximations
100 Greek feet = 1 second of degree of the earths equatorial
circumference
The Egyptians start out with a foot of 300 mm
617 Egyptian feet = 600 Greek feet which is the slant side
or Apothem of the Great Pyramid.(612 English feet)
The Greeks make 600 Greek feet = 1 stadium
8 stadia = 1 milliare (4800 Greek feet)(308.4 mm)
The Eomans vary the length of the milliare to make
it 625 feet to a stadium (divisible by paces) and
end up with a milliare of 5000 pes.(296 mm)
We vary the milliare in Elizabethan times to end up
with 5280 feet (304.8 mm) 1/8 mile = 660 feet
More later on the original question
steve
cir...@access.ch
nos...@someplace.somewhere.somehow wrote:
> Well, you could approach the problem this way. Position a large circular
> slab so that it is parallel to the plane of orbit of the stars (having
> earlier established that the sun moves up and down).
>
> Every day, at exactly the moment the sun sinks beneath a horizontal slab
> that you have positioned to the west for this purpose, mark off on the
> slab where a specific star is to be found, in line with the centre of
> the slab. Repeat this for a few years, turning the slab round, picking
> other suitable stars, and you will end up with 365 marks which don't
> quite line up, and realise that the year is not actually 360 days long
> after all and so you jack in astrology and start doing proper
> measurements and stuff.
Thank you for the reply, MC.
Did the Egyptians also use a circle of 360 degrees. Or possibly
one of 365 smaller degrees?
Well, I may ask professor Ahmes again.
You know, I have it easy. While other people study and go to
libraries and work a lot I simply go to my astrologer and she
touches her fine crystal sphere and already
Ludwig Borchardt: Pyramidenspinner!
Me: Au weia. Not again, please. - Professor Ahmes, where are you?
Ahmes: Hello?
Me: Good morning, professor. We are discussing astronomy.
Ahmes: Oh yes, a topic we like very much. I told you the numbers
of several pyramids. Now you may wish to hear something about their
meaning. Listen carefully what I tell you know: the pyramids of Khufu,
Khafre and Menkaure stay for the belt star of Sahu-Osiris, the ones
of Djedefre and Ba-Ka for his left knee and his left shoulder.
Me: Really? That's exactly what one Robert Bauval is saying. However,
our professors don't believe him.
Ahmes: Funny professors.
Me: Robert Bauval says that the pyramids of Djedefre and Ba-Ka
- or Nebka - are called
Djedefre is a Sehed Star and Nebka is a star
Karlheinz Schuessler and Mark Lehner call Djedefre's pyramid
Djedefre Belongs to the Firmament
while Rainer Stadelmann calls it
Sternenzelt
Well, we could have guessed a long time ago that the arrangement
of those pyramids may stay for the beautiful constellation of Orion,
even the more so as the Pharaoh hopes to be reborn not only as Re
but as Re-Osiris.
Now Robert Bauval goes further and says thet Sneferu's Shining Ones
mean the left hand of Sahu-Osiris.
Ahmes: His left elbow.
Me: The Dahshur pyramids then stay for the Hyades?
Amazing. The shape of the Bent Pyramid resembles this group of stars.
Ahmes: From which are born young ones.
Me: You mean the meteor shower of the Taurids? radiating from Epsilon
Tauri? between October 20 and November 30, reaching a maximum on
November 3?
Ahmes: When this happens we look up and say: Sahu-Osiris releases
another soul. For we believe that the souls of the ones who live
in Maat are reborn as stars.
Well, the Shining Appearances mean the elbow of Sahu-Osiris while
the pyramid of Djed Sneferu means his hand which is represented
by another group of stars we call the Seven Ones or the Eight Ones.
Me: You mean that the Maidum pyramid stays for the Plejades?
a group of stars we call "Siebengestirn"?
Ludwig Borchardt: I like this pyramid.
Ahmes: The first pyramid was built in seven steps, according to the
name of the Seven Ones. Then it was enlarged to eight steps, according
to the name of the Eight Ones. Whereupon Sneferu's men began building
a real pyramid around the steps.
Ludwig Borchardt: I begin to see. But why were these relatively small
and dim stars chosen for to represent a mighty king?
Me: The Hyades and Plejades?
Ahmes: Our colleagues at Babylon call these stars the Golden Door,
because the sun, the moon and the planets pass between them.
We, however, consider Sahu-Osiris to be the keeper of the h-channel
wherein the sun, the moon and the stars are swimming.
Me: We call this channel 'ecliptic', and really, it passes between
the Hyades and Plejades. Now I understand how the symbolism of sun
and stars go together.
By the way, one Rolf Krauss wrote a very fine papyrus: Astronomische
Konzepte und Vorstellungen in den Pyramidentexten (Harrassowitz Verlag
Wiesbaden 1997). He identifies your h-channel with our ecliptic.
Ludwig Borchardt: Oh my, I begin to like these ideas. I alwayas was
fond of the Maidum pyramid while people only spoke of the great one.
Am I turning into a Pyramidenspinner myself? Du heiliger Bimbam.
Me: ;)
(seriously again): I wonder if there might also be a representation
of the right arm and leg of Sahu-Osiris-Orion?
Ahmes: His right arm and leg are the River Nile.
Me: Oh, I see. One Plutarch told us that you venerate Osiris in your
magnificient river. Seems he was right.
Now for something else. I'd like to ask you about - Professor Ahmes?
can you hear me? hello? professor?
My astrologer: Sorry, I lost him. And this German too.
Me: Well then, have a nice day.
people study and go to libraries and work a lot I simply go to my
Steve Whittet
Hi Franz,
In article <6vn45o$cfq$1...@nnrp1.dejanews.com>, cir...@access.ch says...
.
>
>Did the Egyptians also use a circle of 360 degrees. Or possibly
>one of 365 smaller degrees?
You ask a good question, the Egyptians divided their year both
ways but what about their circles? I think they used 360 degrees.
>
>Well, I may ask professor Ahmes again.
I would be interested to see if he agrees that while
mathematical rigor is not required if a practical solution
is all that is desired, accuracy in measures takes us a long way.
Take some practical problems where the accuracy would have been as
good as the known value for Pi. Lets use those three classical
problems of antiquity, trisecting the angle, squaring the circle
and doubling the cube.
An interesting point about circles is that while you can't
evenly divide the circumference by the diameter if you use
the same units to measure both, you can evenly divide the
circumference by the diameter if you measure them in different
units. This is similar to what you do when you make up a
table of numbers with common differences.
For example if the diameter is a remen of 15"
the circumference is 4 pes (of 11.78")
if the diameter is 3 remen of 45"
the circumference is 12 pes and also 8 cubits
if the diameter is 24 remen of 360" (10 yards)
the circumference measures 96 pes, or 64 cubits
or 55 royal cubits, take your pick.
Lets begin with the premiss that the accuracy required
in the middle bronze age is limited to what can be measured.
The integration of proportionate systems means that squaring
a circle, trisecting an angle and doubling a cube using only
an unmarked ruler and a compass is simpler than it seems.
Ancient rulers even if unmarked were still made up as measured
unit lengths and widths. Their corners were square and their
edges were parallel. Such a ruler can be taken as a unit
and used to make other units proportional to each other.
Lets set up such a pair of proportional units so that
56 pes = 55 inches
Lets say we have a standard ruler 1 foot long by 1 inch wide.
We make a grid with the ruler by drawing a line along its edges
and setting the compass to walk off the units until we have 56.
We set the compass to that length and intersect the 55 mark from
the origin.
We lay off a diagonal between the origin and the 55 mark
We now have units which are in a proportion of 55:56
15 of the long units are a remen
12 of the short units are a pes
We lay off a circle with a diameter of 7.5 remen
and we have a circumference of 30 pes (which
has 360 of the short units and is thus divisible
into one degree increments). For more precision we
make a bigger circle.
Now that we can divide the circle into degrees
we can trisect an angle just as we would with a
protractor.
For practical purposes, to double the cube of a volume
whose side is measured in feet we take the cube of the
side which measured in remen comes out the same.
To get a circle whose area is equal to that of a square
Try a circle with a radius of 6 pes. Its area is the same
as a square with a side of 1 arkana or 10 Greek feet.
(Professer Ahmes may set the rule as squaring the 8/9ths
of the diameter and then institutionalized it in the
relative proportions of two measures)
Once you know these relationships any practical application
of these problems is a snap.
steve
cir...@access.ch
whi...@shore.net (Steve Whittet) wrote:
> You ask a good question, the Egyptians divided their year both
> ways but what about their circles? I think they used 360 degrees.
Hi Steve,
thank you for the reply and the fine remarks on the use of different
measures. However, I don't know anything on the British measures
and can't reply on this topic. So let me ask professor Ahmes again.
Me: Good morning, professor Ahmes. You keep this precious papyrus
from Khufu's time. Can you please tell me a little more on the
Bent Pyramid? Some numbers?
Ahmes: Well, Sneferu's Epiphanies mean the group of stars we call
the elbow of Sahu-Osiris. In the time of king Narmer it was called
the leg and talon of Horus. You call it - how was it again?
Me: Hyades.
Ahmes: Now the southern Epiphany was shaped according to the form
of this group of stars. Here is the cross-section:
o I
o o F G H
o o A B C D E
A-B = D-E = 66 royal cubits B-C = C-D = 114 royal cubits
base A-B-C-D-E = 66 + 114 + 114 + 66 = 360 royal cubits
B-F = D-H = 93 royal cubits A-F = E-H = 114 royal cubits
according to the pseudo-triple 22-31-38 or 66-93-114
F-G = G-H = 114 rc G-I = 108 rc F-I = H-I = 157 rc
according to the pseudo-triple 108-114-157
B-C = C-D = 114 rc C-I = 93 + 108 = 201 rc B-I = D-I = 231 rc
according to the pseudo-triple 38-67-77 or 114-201-231
Now please have a look at the base:
A B C D
E F G H
I J K L
M N O P
A-B = C-D = D-H = L-P = P-O = M-N = M-I = E-A = 66 royal cubits
B-C = F-G = H-L = G-K = O-N = K-J = I-E = J-F = 228 royal cubits
A-D = E-H = D-P = C-O = P-M = L-I = M-A = N-B = 360 royal cubits
A-H = D-E = D-O = P-C = P-I = M-L = M-B = A-N = 366 royal cubits
according to the pseudo-triple 11-60-61 or 66-360-366
Now let me show you how well base and temenos go together:
a b c d
e f g h
i j k l
m n o p
a-b = c-d = d-h = l-p = p-o = n-m = m-i = e-a = 105 royal cubits
b-c = f-g = e-i = f-j = 360 royal cubits (base of the pyramid)
a-d = d-p = p-m = m-a = 570 royal cubits = 10 x 57 royal cubits
b-f = c-g = h-g = l-k = o-k = n-j = i-j = e-f = 105 royal cubits
b-c = f-g = h-l = g-k = o-n = k-j = i-e = j-f = 360 royal cubits
b-g = c-f = h-k = l-g = o-j = n-k = i-f = e-j = 375 royal cubits
according to the triple 7-24-25 or 105-360-375
Now for the temenos and the bending lines of the pyramid:
a b c d
e f g h
i j k l
m n o p
a-d = d-p = p-m = m-a = 570 royal cubits = 10 x 57 royal cubits
b-c = c-g = h-g = l-k = o-k = n-j = i-j = e-f = 171 royal cubits
b-c = f-g = h-l = g-k = o-n = k-j = i-e = j-f = 228 royal cubits
b-g = c-f = h-k = l-g = o-j = n-k = i-f = e-j = 285 royal cubits
according to the Sacred Triangle 3-4-5 or
3 x 57 royal cubits - 4 x 57 royal cubits - 5 x 57 royal cubits
Please consider the number 57. It was used for astronomical reasons.
Go 57 fingers far and 1 finger up. The resulting angle will be '360
of the full circle.
By using an instrument like this you can measure small angles from
1 till 14 degrees:
0 7 fingers
0
oooooooooooooooooooooooooooooooooooooooooooooo
0
57 fingers 0 7 fingers
Me: My calculating ushabti finds an angle of 14.002535 degrees
or 14 degrees 9 seconds. Well, the mistake is very small.
Ahmes: And by considering the marks indicating 5 and 5 fingers
the astronomers of Sneferu measured a decan.
Now for the small pyramid in the complex of the Southern Epiphany.
The base measures 100 x 100 royal cubits, the height 49 royal cubits,
the slope 70 royal cubits, the radius of the inscribed hemisphere -
symbol of the sky in the Primeval Hill, symbol of Nut in the pyramid
- measures 35 royal cubits, and the diameter 70 royal cubits again.
The small but very fine pyramid was built by young Hemon, who then
designed Sneferu's Northern Epiphany and Khufu's Horizon.
cir...@access.ch
cir...@access.ch wrote: ...
Osiris - present in the River Nile and in the aligning pyramids
---------------------------------------------------------------
RIVER NILE Symbol of Osiris (according to Plutarch)
Djed-Pillar symbol of Osiris, his backbone, eternity
possibly a combination of the wooden phallus made by Isis
for her husband Osiris and of an early Nilometer
may the Nile rise every year, may Egypt live on forever
MAIDUM PYRAMID ancient name: Djed Sneferu
*Plejades* - left hand of Sahu-Osiris
Rolf Krauss identifies h-channel with ecliptic (1997)
Babylonians call Hyades and Plejades 'Golden Door':
sun, moon and planets pass between them
BENT PYRAMID OF DAHSHUR ancient name: Sneferu's Southern Epiphany
*Hyades, Epsilon Tauri* Robert Bauval (1994)
RED PYRAMID OF DAHSHUR ancient name: Sneferu's Northern Epiphany
*Aldebaran* Robert Bauval (1994)
left elbow of Sahu-Osiris
DJOSER COMPLEX meaning heaven? stone beam sarcophagus in
southern grave decorated with stars, fascinating geometry,
combining 10 triples (Wolf Meyer-Christian 1987)
Step Pyramid: *Heka* or triangle *Betelgeuse-Heka-Bellatrix* ?
Pyramids at Saqqara: meaning head or crown of Sahu-Osiris?
PYRAMID OF NEBKA OR BAKA AT ZAWYET EL-ARYAN
ancient name: Nebka or Baka is a star
*Bellatrix* left shoulder of Sahu-Osiris Robert Bauval (1994)
KHUFU'S PYRAMID first ancient name: Khufu's Horizon
*Al Nitak* belt of Sahu-Osiris Robert Bauval (1994)
KHAFRE'S PYRAMID ancient name: Khafre is great
*Al Nilam* belt of Sahu-Osiris Robert Bauval (1994)
MENKAURE'S PYRAMID ancient name: Menkaure is divine
*Al Mintaka* belt of Sahu-Osiris (Robert Bauval (1994)
PYRAMID OF DJEDEFRE AT ABU RAWASH
ancient name: Djedefre is a sehed star, Dejedefre belongs
to the firmament, or "Sternenzelt" (Rainer Stadelmann)
*Rigel* left knee of Sahu-Osiris Robert Bauval (1994)
LETOPOLIS a symbol of *Sirius* ?
Regards Franz Gnaedinger Zurich cir...@access.ch
M.C.Harrison
>
> > other suitable stars, and you will end up with 365 marks which don't
> > quite line up, and realise that the year is not actually 360 days long
> > after all and so you jack in astrology and start doing proper
> > measurements and stuff.
>
> Thank you for the reply, MC.
>
> Did the Egyptians also use a circle of 360 degrees. Or possibly
> one of 365 smaller degrees?
360, I gather. In fact, the error is continued to this very day, as we
also have 360 degrees in a circle, when we aren't using radians. Blame
the guys in Mesopotamia, or I gather it was their idea like a lot of
astrology and mathematics (they needed to become quite numerate to build
irrigation and stuff).
M.C.Harrison
>
> It might be just an odd coincidence that our units of measure with
> their ancient roots happen to work out so that there are 365,240
365,222 feet, to an excess significant figure. This would be a
coincidence.
cir...@access.ch
nos...@someplace.somewhere.somehow wrote:
(Me:)
> > Did the Egyptians also use a circle of 360 degrees. Or possibly
> > one of 365 smaller degrees?
(MC:)
> 360, I gather. In fact, the error is continued to this very day, as we
> also have 360 degrees in a circle, when we aren't using radians. Blame
> the guys in Mesopotamia, or I gather it was their idea like a lot of
> astrology and mathematics (they needed to become quite numerate to build
> irrigation and stuff).
Yes, and the astronomical pyramid 'Sneferu's Southern Epiphany'
(bent pyramid of Dahshur) leads to the same number which is present
in the base measuring 360 x 360 royal cubits. Here are again the
main numbers of this very fine pyramid whose shape reminds of the
Hyades and must have been intended:
Grid of temenos, pyramid and bending lines
105+66+114+114+66+105 x 105+66+114+114+66+105
containing the triples
6x11 6x60 6x61 7x15-24x15 25x15 3x57 4x57 5x57
lower run 66 lower rise 93 lower slope 114
according to the pseudo-triple 3x22 3x31 3x38
upper run 114 upper rise 108 upper slope 157
according to the pseudo-triple 108-114-157
The number 57 (2x57=114, 10x57=570) can be used for making
s simple astronomic device:
0 7 fingers
0
ooooooooooooooooooooooooooooooooooooooooooooooooo
0
57 fingers 0 7 fingers
By means of such an instrument you can measure small angles from
1 till 14 degrees.
Here is a scheme of the main pyramids inspired by Robert Bauval:
0
0 Nile
0 Osiris Plutarch (45 - 120)
0 Milky Way Robert Bauval (1994)
0
0
0
0 Maidum
0 Djed Sneferu
0 Plejades
0 left hand of Sahu-Osiris
0 h-channel
0 e-----c-----l-----i-----p-----t-----i-----c Rolf Krauss
0 (1997)
0 Dahshur
0 Sneferu's Southern Epiphany
0 Epsilon Tauri Robert Bauval (1994)
0 Sneferu's Northern Epiphany
0 Aldebaran Robert Bauval (1994)
0 left elbow of Sahu-Osiris
0
0
0 Saqqara pyramids
0 Heka, or the triangle
0 Betelgeuse-Heka-Bellatrix
0 head and crown of Sahu-Osiris
0
0
0
0 Zawyet el Aryan
0 Nebka or Baka is a star
0 Bellatrix
0 left shoulder of Sahu-Osiris
0 Robert Bauval (1994)
0
0
0
0 Giza pyramids
0 Al Nitak Al Nilam Al Mintaka
0 belt of Sahu-Osiris
0 Robert Bauval (1994)
0
0
0
0 Abu Rawash
0 Djedefre is a sehed star
0 Rigel
0 left knee of Sahu-Osiris
0
0
0
0 Letopolis
0 Sirius
0 Robert Bauval (1994)
0
0
Sir Karl Popper did not only ask for falsifiable theses
but also for audacious ones - nicht nur pruefbare und
falsifizierbare sondern auch moeglichste kuehne Thesen.
Regards Franz Gnaedinger Zurich cir...@access.ch
-----------== Posted via Deja News, The Discussion Network ==----------
cir...@access.ch
cir...@access.ch wrote: ...
Dear Mr. Bauval,
if you read my messages, may you please send an e-mail to s...@teliamail.dk ?
Your schoolfriends Souzanna D'Souza and Andre Blondiau sent me an e-mail
asking for your address. I couldn't tell them, so I asked if I may ask you
in public whereupon Mr. Blondiau said yes in the name of his wife. They lost
you a long time ago and saw you on TV lately and asked me very kindly if
I may help them finding you.
The following is a hommage for you. I think we owe you a great insight,
and I like your first book very much, however, I must say that I have
problems with your second one. Please allow me these open words, and
good luck for your further work. (Sorry for my poor English.)
> 0 Robert Bauval (1994)
> 0
> 0
> 0 Letopolis
> 0 Sirius
> 0 Robert Bauval (1994)
> 0
> 0
> Regards Franz Gnaedinger Zurich cir...@access.ch
> -----------== Posted via Deja News, The Discussion Network ==----------
Marc Line
saying:
>Dear Mr. Bauval,
>
>if you read my messages, may you please send an e-mail to s...@teliamail.dk ?
>Your schoolfriends Souzanna D'Souza and Andre Blondiau sent me an e-mail
>asking for your address.
Dear Franz
Mr Bauval does not read this newsgroup. However, I shall be happy to
make him aware of this contact.
Regards
Marc Line
cir...@access.ch
Marc Line <ma...@bosagatedotdemondotco.uk> wrote:
> Dear Franz
>
> Mr Bauval does not read this newsgroup. However, I shall be happy to
> make him aware of this contact.
>
> Regards
>
> Marc Line
Thank you, Marc.
My astrologer: I got professor Ahmes on my fine magic crystal sphere.
Me: Good morning, professor Ahmes. I was told again - once more -
that you and your Babylonian colleagues are no real mathematicians.
Only numberfinders.
Ahmes: Of course.
Me: You agree?
Ahmes: Well, we Egyptians pick them funny numbers from a tree in
a temple at Heliopolis while the Babylonians fish the even numbers
out of the Euphrat and the odd ones out of the Tigris ;-)
Me: I mentioned the value 1:24,51,10
Ahmes: A pretty carp, ey?
Me: I might also speak of the triples on 'Plimpton 322':
120 119 (169)
3456 3367 (4825)
4800 4601 (6649)
13500 12709 (18541)
72 65 (97)
360 319 (481)
2700 2291 (3541)
960 799 (1249)
600 481 (769)
6480 4961 (8161)
60 45 (75)
2400 1679 (2929)
240 161 (289)
2700 1771 (3229)
90 56 (106)
Ahmes: Meaningless numbers.
Me: Sure.
However, I wonder why all these triples, which define angles between
45 and 30 degrees in descending line, are basic triples - except for
60 45 (75) and 90 56 (106)? Why not simply 4 3 (5) and 45 28 (53)?
Ahmes: Play with the numbers, and you shall see.
Me: Can you please give me a hint?
Ahmes: Imagine a rectangle measuring 56 x 90 fingers. The diagonals
measure 106 fingers. Draw a circle around it. The radius measures
53 fingers. How long is the circumference? 333 fingers.
Transform the rectangle into a square of the same area. The side
measures 71 fingers while the circumference of the inscribed circle
measures 223 fingers.
Now for the numbers 60, 45 and 75.
Every number belongs to another triple: 27-36-45, 36-48-60, 21-72-75.
The first triples have again the same form based on the triple 3-4-5
while the last one bases on the triple 7-24-25.
The angle 7v24 doubles the angle 36v45 or 48v60 or 60v75. Exactly,
no mistake.
The periphery of the triangle 45-60-75 measures 180 fingers. If the
periphery of an equilateral triangle measures 180 fingers, the side
measures 60 fingers and the height 52 fingers.
Me: I see.
These fine but certainly meaningless number games and geometrical
transformations require the numbers 45-60-75 and 56-90-106 instead
of the basic triples 3-4-5 and 28-45-53.
Funny, the meaningless numbers 120 119 (169) define almost a square.
Ahmes: A triangle and a square.
Imagine a triangle of these numbers. The diameter of the inscribed
circle measures 119 + 120 - 169 = 70 fingers, the circumference 220
fingers.
Draw a rectangle measuring 119 x 120 fingers. The diagonals measure
169 fingers. Add one finger to a side, thus you will obtain a square
measuring 120 x 120 fingers. The circumference of the inscribed circle
measures 377 fingers.
Me: Let me pick the numbers 360, 319 and 481.
Ahmes: Draw a triangle of these numbers. The diameter of the inscribed
circle measures 360 + 319 - 481 = 198 fingers, the circumference 622
fingers.
Me: I wonder if one triangle might have a meaningless angle.
Ahmes: You may consider a triangle of 1679, 2400 and 2929 units.
The ratio 1679 to 2400 is almost 7 to 10 while the angles measures
35, 55 and 90 degrees.
Me: I like the meaningless transformations. Are there more examples?
Ahmes: Well, have a look at the dull numbers 240, 161 and 289.
Imagine a triangle of these numbers and transform it into a square
of the same area. The side measures 139 fingers and the periphery
556 fingers while a circle of the same circumference has a diameter
of 177 fingers.
Me: I assume that the big numbers 12709-13500-18541 are the dullest
ones of all?
Ahmes: Of course.
Please imagine an isoscele triangle. The height measures 3 x 3 x 3
units while the periphery measures 5 x 5 x 5 units. How long are
the sides?
Let 1 equal 500 m. Then the height equals 13500 m, half the base
measures 12709 m and the slope 18541 m. From this we get one side
measuring 50:50,9,36 and two sides measuring 37:4,55,12.
Me: Thank you, professor Ahmes. You convinced me that all you do
is really meaningless.
Ahmes: Completely and absolutely.
Me: Our professors of Princeton, Oxford and Cambridge are right. You
have NO idea of geometry and mathematics. You are NO mathematicians.
You are just lucky numberfinders.
And in some 5000 years an archaeologist will find a Princeton tablet
saying E = MCC. It will be declared a meaningless formula. Shure enough
the primitive race of the Early Concrete Age had NO idea of real physics.
kc...@iastate.edu
In a piece "Fermat's Last Theorem--Was it a Right Question?" I wrote recently
I also dealt with these triples.
http://www.public.iastate.edu/~choi/fermat.htm
Kwan
In article <706rkq$e6e$1...@nnrp1.dejanews.com>,
kc...@iastate.edu
Your piece was delectable!
In a piece "Fermat's Last Theorem--Was it a Right Question?" I wrote recently
I also dealt with these triples. Here is the correct link.
http://www.public.iastate.edu/~kchoi/fermat.htm
Kwan
In article <706rkq$e6e$1...@nnrp1.dejanews.com>,
cir...@access.ch wrote:
> In article <51MF+GAY0eJ2EwM$@bosagate.demon.co.uk>,
> Marc Line <ma...@bosagatedotdemondotco.uk> wrote:
> My astrologer: I got professor Ahmes on my fine magic crystal sphere.
>
> Me: Good morning, professor Ahmes. I was told again - once more -
> that you and your Babylonian colleagues are no real mathematicians.
> Only numberfinders.
>
> Ahmes: Of course.
>
> Me: You agree?
>
> Ahmes: Well, we Egyptians pick them funny numbers from a tree in
> a temple at Heliopolis while the Babylonians fish the even numbers
> out of the Euphrat and the odd ones out of the Tigris ;-)
cir...@access.ch
kc...@iastate.edu wrote:
> circle and professor Ahmes:
>
> Your piece was delectable!
>
> In a piece "Fermat's Last Theorem--Was it a Right Question?" I wrote recently
> I also dealt with these triples. Here is the correct link.
>
> http://www.public.iastate.edu/~kchoi/fermat.htm
>
> Kwan
Thank you, Kwan, for your kind words.
I am sorry for having to disappoint you in two respects: your lemma
is known since a long time while a full quote of Pierre de Fermat's
famous remark shows that he really meant the problem we associate
his name with:
'On the other hand it is impossible to separate a cube into
two cubes, or a biquadrate into two biquadrates, or generally
any power except a square into two powers with the same exponent.
I have discovered a truly marvellous proof of this, which, however,
the margin is not large enough to contain.'
Quoted from: THE HERITAGE OF IMHOTEP by W.S.Anglin and J.Lambek,
Springer New York 1995 (title improved by me)
Andrew Wiles of Princeton finally proved Fermat's conjecture, also
called Fermat's Last Theorem (FLT), by using many ideas of others
(and his own, of course). However, Wiles did not prove that Fermat
had no proof.
When I wish to buy a bread I can simply go over the street - or I
can walk around the world and hope that the bakery will still be open
when I come back. Who knows if Wiles traveled around the world while
Fermat, who found at least a proof for the special case of n = 4,
simply went over the street?
It's very reasonable to assume that Pierre de Fermat had no general
proof. It's even most reasonable to assume that there is NO simple
proof at all. But who knows? The mathematicians are trained to think
in complicated ways and miss many a simple solution. At least when
it comes to the history of pre-Greek geometry and mathematics.
You began with the equation 3x3 + 4x4 = 5x5 and found another one:
3x3x3 + 4x4x4 + 5x5x5 = 6x6x6. A pretty gem!
There are many fascinating special cases. One is found in problem
no. 32 of the Rhind Mathematical Papyrus, written by one Ahmes in
around 1650 BC as a copy of one or several older papyri which were
about 200 years older.
Ahmes divides 2 by 1 + 1/3 + 1/4 or simply 1 '3 '4 and obtains
1 '6 '12 '114 '228. I interprete his numbers as the measurements
of a right parallelepiped:
length 2 units
breadth 1 '3 '4 units
height 1 '6 '12 '114 '228 units
How long are the diagonals of the volume? Simply
1 '3 '4 units plus 1 '6 '12 '114 '288 units
or
1 1 plus '3 '6 plus '4 '12 plus '114 '228 units
or
2 '2 '3 '76 units
Divide 2 by any number A and you will obtain B. If a right parallel-
epiped measures 2 x A x B units, the diagonal measures A + B units.
As you are interested in the mathematics of ancient Babylon you may
play with the numbers of the clay tablets and papyri. We still know
little about their achievements, and playing with their numbers is
by far the best way to find out more.
Good luck!
Regards Franz Gnaedinger Zurich cir...@access.ch
-----------== Posted via Deja News, The Discussion Network ==----------