Rhind Mathematical Papyrus (1 of 8) / 1979-2003 by Franz Gnaedinger, Zurich, fg@seshat.ch / www.seshat.ch

 

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PERI AIGYPTON HAI MATHAEMATIKAI PROTON TECHNAI SYNESTAESAN Aristotle, metaphysics, book 1, chapter 1

 

CORRECT METHOD OF RECKONING, FOR GRASPING THE MEANING OF THINGS AND KNOWING EVERYTHING THAT IS, OBSCURITIES ... AND ALL SECRETS Ahmes, opening lines of the Rhind Mathematical Papyrus (translated by Gay Robins & Charles Shute)

 

 

 

How long are the diagonals of a square?

 

By drawing a grid and measuring the diagonals of various squares one may find the following numbers:

 

side 2 diagonal a little less than 3

side 3 diagonal a little more than 4

side 5 diagonal about 7

 

side 5 diagonal slightly more than 7

side 7 diagonal slightly less than 10

side 12 diagonal practically 17

 

side 12 diagonal practically 17

side 17 diagonal practically 24

side 29 diagonal practically 41 and so on

 

These numbers generate a simple pattern (add a pair of numbers and you obtain the number below, double the first number of a line and you obtain the last number)

 

1 1 2

2 3 4

5 7 10

12 17 24

29 41 58

70 99 140

169 239 338

408 577 816

985 1393 ....

.... .... ....

 

If a square measures 70 by 70 royal cubits, the diagonals measure practically 99 royal cubits, and if a square measures 99 by 99 royal cubits, the diagonals measure practically 140 royal cubits. - How long is the diagonal of a square if the side measures 118 royal cubits or 826 palms? You may proceed as follows:

 

side 118 = 99 + 12 + 7

140 + 17 + 10 = 167 diagonal

 

If the side of a square measures 118 royal cubits, the diagonal measures about 167 royal cubits (mistake 6.43 centimeters).

 

side 826 = 577 + 169 + 70 + 7 + 3

816 + 239 + 99 + 10 + 4 = 1168

 

If the side of a square measures 118 royal cubits = 826 palms, the diagonal measures about 1168 palms or 166 cubits 6 palms (mistake 1 centimeter).

 

Divide 1393 by 985 and you obtain 1;24,51,10,3,2... in the sexagesimal number system of Babylon. Leave out the small numbers 3,2... and keep the value 1;24,51,10. This excellent value for the square root of 2 is found on the Babylonian clay tablet YBC 7289 dating from around 1650 BC.

 

 

 

Cube and equilateral triangle

 

Let me draw up an analogous number pattern:

 

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

(Archimedes) 153 265 459

418 724 1254

209 362 627

571 989 1713

1560 2702 ....

(Archimedes) 780 1351 ....

 

If a cube measures 41 by 41 by 41 fingers, the diagonals of the faces measure about 58 fingers (first number pattern) while the cubic diagonals measure about 71 fingers.

 

If the side of an equilateral triangle measures 194 fingers, its height measures practically 168 fingers or 24 palms or 6 royal cubits; the radius of the inscribed circle measures practically 56 fingers or 2 royal cubits, and the radius of the circumscribed circle 112 fingers or 4 royal cubits.

 

The side of an equilateral hexagon measures 10 royal cubits or 70 palms or 280 fingers. Calculate the diameter of the inscribed circle. Proceed as follows:

 

side 280 = 194 + 56 + 19 + 15

336 + 90 + 33 + 28 = 485 diam. inscribed circle

 

side 280 = 112 + 112 + 56

194 + 194 + 97 = 485 diameter inscribed circle

 

side 280 = 194 + 71 + 15

336 + 123 + 26 = 485 diameter inscribed circle

 

If the side of a regular hexagon measures 10 royal cubits or 70 palms or 280 fingers, the diameter of the inscribed circle measures practically 485 fingers or 17 cubits 2 palms 1 finger (mistake half a millimeter).

 

The edge of a cube measures 10 royal cubits. Calculate the cubic diagonal. It measures again 485 fingers or 17 cubits 2 palms 1 finger (with a tiny error of only half a millimeter).

 

 

 

Double square

 

Yet another number pattern:

 

1 1 5

2 6 10

1 3 5

4 8 20

2 4 10

1 2 5

3 7 15

10 22 50

5 11 25

16 36 80

8 18 40

4 9 20

13 29 65

42 94 210

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

 

If a double square measures 4 by 8 royal cubits or 34 by 68 royal cubits or 72 by 144 royal cubits, the diagonals measure about 9 or 76 or 161 royal cubits, with an increasing accuracy. - Let a rectangle measure 10 by 20 royal cubits or 280 by 560 fingers. Calculate the diagonal. Proceed as follows:

 

short side 280 = 144 + 89 + 34 + 13

322 + 199 + 76 + 29 = 626 diagonal

 

short side 280 = 199 + 76 + 5

445 + 170 + 11 = 626 diagonal

 

If a double square measures 10 by 20 royal cubits or 70 by 140 palms or 280 by 560 fingers, the diagonal measures practically 626 fingers or 22 cubits 2 palms 2 fingers (error 2 millimeters).

 

 

 

Doubling the volume of a cube

 

1 1 1 2

2 2 3 4

4 5 7 8

9 12 15 18

3 4 5 6

 

3 4 5 6

7 9 11 14

16 20 25 32

36 45 57 72

12 15 19 36

 

12 15 19 24

27 34 43 54

61 77 97 122

138 174 219 276

46 58 73 92

 

46 58 73 92

104 131 165 208

235 296 373 470

531 669 843 1062

177 223 281 354

 

177 223 281 354

400 504 635 800 and so on

 

A cube measures 400 by 400 by 400 units. Double the volume and the cube will measure 504 by 504 by 504 units. The numbers 504 and 400 provide an excellent approximate value for the cube root of 2: 504/400 = 63/50.

 

 

 

Calculating the circle

 

Imagine a grid measuring 10 by 10 royal cubits:

 

 

. . . . . d . . . . .

. . e . . . . . c . .

. f . . . . . . . b .

. . . . . . . . . . .

. . . . . . . . . . .

g . . . . + . . . . a

. . . . . . . . . . .

. . . . . . . . . . .

. h . . . . . . . l .

. . i . . . . . k . .

. . . . . j . . . . .

 

 

The side of the square measures 10 royal cubits or 70 palms or 280 fingers. The diagonal measures practically 99 palms. The points a b c d e f g h i j k l mark a circle whose radius measures 5 royal cubits or 35 palms or 140 fingers. The eight short arcs measure about 40 fingers each, the four longer arcs measure practically 90 fingers each, giving a circumference of about 880 fingers or 220 palms and yielding a very fine approximate value for pi, namely 22/7 or 3 1/7.

 

Imagine a grid which measures 10 by 10, 50 by 50, 250 by 250, 1250 by 1250 ... ever smaller units. A circle inscribed in it will pass the 4 ends of the axes, furthermore 8, 16, 24, 32 ... inner points of the grid. Their distances from the axes and from the center of the grid are defined by the following triples, beginning with the 'Sacred Triangle' 3-4-5:

 

3-4-5 or 15-20-25 or 75-100-125 or 375-500-625 ...

7-24-25 or 35-120-125 or 175-600-625 ...

44-117-125 or 220-585-625 ...

336-527-625 ...

 

If you know a triple a-b-c and wish to know the next, you may calculate according to the following terms:

 

4a plus/minus 3b 4b plus/minus 3a 5c

 

Use the positive results ending on 1, 2, 3, 4, 6, 8 or 9 (neither on zero nor on five).

 

Combine the 12, 20, 28, 36 ... points by straight lines and you will obtain a sequence of irregular polygons. Their side lengths are whole number multiples of the square roots of 2 or 5 or 2x5. The square roots of 2 and 5 can be approximated by means of two of the number patterns above.

 

The sides of the irregular polygons are slightly shorter than the respective arcs. We may hope to counterbalance this by choosing values for the square roots of 2 and 5 that are slightly bigger than the actual numbers. Calculate the first polygon by means of the ratios 10/7 and 9/4 and you will find the ratio 22/7 for pi. Calculate the second polygon by means of the ratios 17/12 and again 9/4 and you will find 157/50 for pi. The average is about 311/99. These values allow to draw up a number sequence providing many more fine values. Write 3 above 1 and add continuously 22 above 7:

 

3 (plus 22) 25 47 ... 157 ... 311 333 355 377

1 (plus 7) 8 15 ... 50 ... 99 106 113 120

 

key figure 1 / key figure 2 / polygon 1 / polygon 2 / polygon 3 / polygon 4 / polygon 5 // polygon a / polygon b / polygon c / polygon d

 

By the way: the above triples can also be found by means of a number column. Begin with 1 and 1, use a factor of minus 4, and consider every second line:

:

1 1 -4

2 -3 -8

-1 -11 4

-12 -7 48

-19 41 76

22 117 -88 and so on

 

 

x = -3 y = -8/2 = -4 r = 5

x = -7 y = 48/2 = 24 r = 25

x = 117 y = -88/2 = -44 r = 125 and so on

 

 

 

Cumbersome? No, a clever tool

 

Many historians of mathematics believe that unit fractions are cumbersome. Are they? I tried to work with them and much to my surprise found an easy way to handle those funny numbers: round all results, and the mistakes will even out in the long run, allowing you to work with whole numbers only, and yielding fine results all the same.

 

In my first number column are found the numbers 70 and 99, yielding the value 99/70 for the square root of 2. Now let me transform this ratio into a pair of unit fraction series:

 

99/70 = 1 + 1/5 + 1/7 + 1/14 or simply 1 '5 '7 '14

99/70 = 1 + 1/3 + 1/15 + 1/70 or simply 1 '3 '15 '70

 

Let the side of a square measure 360 royal cubits or 2520 palms. If you wish to calculate the diagonal you may multiply the side by one of the above series and round all the numbers:

 

360 royal cubits x 1 '5 '7 '14

 

360 x 1 = 360

360 x '5 = 72

360 x '7 = 51 (rounded)

360 x '14 = 26 (rounded)

sum 509

 

side 360 royal cubits

diagonal 509 royal cubits (mistake about 6 centimeters)

 

2520 palms x 1 '5 '7 '14

 

2520 x 1 = 2520

2520 x '5 = 504

2520 x '7 = 360

sum 3564

 

side 2520 palms

diagonal 3564 palms (mistake about 14 millimeters)

 

Khafre's pyramid at Giza was originally 274 royal cubits tall, while the base measured 411 royal cubits and the slope 324 '2 royal cubits, according to the Sacred Triangle 3-4-5. How long was the diagonal of the base?

 

411 x 1 = 411

411 x '5 = 82 (rounded)

411 x '7 = 59 (rounded)

411 x '14 = 29 (rounded)

sum 581

 

side 411 royal cubits

diagonal 581 royal cubits (mistake about 13 centimeters)

 

411 royal cubits equal 2877 palms:

 

2877 x 1 = 2877

2877 x '5 = 575 (rounded)

2877 x '7 = 411

2877 x '14 = 205 '2 ??

 

Here we have a rounding problem. 2877 divided by 14 equals 205 plus '2. Should we round up to 206? or down to 205? No, we shall solve the problem by using the alternative series 1 '3 '15 '70:

 

2877 x 1 = 2877

2877 x '3 = 959

2877 x '15 = 192 (rounded)

2877 x '70 = 41 (rounded)

sum 4069

 

side 2877 palms

diagonal 4069 palms (mistake about 23 millimeters)

 

411 royal cubits equal 2877 palms or 11508 fingers:

 

11508 x 1 = 11508

11508 x '3 = 3836

11508 x '15 = 767 (rounded)

11508 x '70 = 164 (rounded)

sum 16275

 

side 11508 fingers or 411 royal cubits or

diagonal 16275 fingers or 581 cubits 7 fingers

 

(mistake 4 millimeters)

 

The base length of Khafre's pyramid measured 411 royal cubits, while the diagonal of the base measured 581 cubits 7 fingers (with a tiny mistake of only four millimeters).

 

 

 

 

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