Babylonian Mathematics / 1979-2001 by Franz Gnaedinger, Zurich, fg(a)seshat.ch, fgn(a)bluemail.ch / www.seshat.ch

 

 

 

Babylonian clay tablet YBC 7289

 

On the Babylonian clay tablet YBC 7289 is drawn a square, with the side measurement indicated as 30, while the numbers 42;25,35 and 1;24,51,10 are written on its diagonal. If the side of a square measures 30 units, the diagonal measures practically 42;25,35 units, while the corresponding value for the square root of 2 equals 1;24,51,10 in the sexagesimal number system of Babylon. This excellent value can be found by means of the following number pattern:

 

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 ....

 

Divide 1393 by 985 and you obtain 1;24,51,10,3,2... Let the small numbers go and keep 1;24,51,10.- Imagine a circle inscribed in the square 30 by 30. Calculate the circumference. Using a pocket calculator you will find an amazingly good approximation:

 

diameter 30 circumference 1,34;14,52

 

This number is based on an excellent value for pi: 84823 / 27000 = 3;8,29,44

 

This value is even better than 355/113. Could the Babylonians possibly have found such a fine value? Yes, they could easily have done so by using the number sequences of their Egyptian colleagues:

 

4 (plus 3) 7 10 13 16 19 22 25

1 (plus 1) 2 3 4 5 6 7 8

 

3 (plus 22) 25 47 69 ... 311 333 355 377

1 (plus 7) 8 15 22 ... 99 106 113 120

 

333 (plus 355) 688 1043 ... 2463 ... 84823

106 (plus 113) 219 332 ... 784 = 28x28 ... 27000

 

Let a cube measure 30 by 30 by 30 units. The volume equals

 

30 x 30 x 30 = 27000 cubic units

 

while the squared cubic diagonal equals

 

30x30 plus 30x30 plus 30x30 = 2700 square units

 

Imagine the cube holding a sphere and calculate the volume of the sphere. It would simply measure 84823/6 cubic units. Imagine a sphere holding the cube and calculate the surface of the sphere. It would simply measure 84823/10 square units. Both results can easily be given in the sexagesimal form.

 

 

 

Plimpton 322

 

Papyrus was a precious and highly expensive material, and there is little writing space on a Babylonian clay tablet. These limitations may have led the Egyptian and Babylonian mathematicians of the second millenium BC to convey their knowledge by way of telling examples. I dare also say that the 15 triples mentioned in the famous Babylonian clay tablet Plimpton 322 are a 'book' of mathematics in a very condensed form. Let me have a look at the 15 triples and invent some tasks to go with the numbers. First a few Sumerian measures:

 

1 gish (about 360 m) = 6 es (rope, 60m) = 60 gar (6m) = 120 gi (cane, 3 m) = 720 kush (cubit, 50 cm) = 1440 shubad (span, 25 cm) = 21600 shusi (finger, 1.7 cm)

 

 

First triple (120) - 119 - 169 (kush)

 

The sides of a rectangular triangle measure 119-120-169 kush. How long is the diameter of the inscribed circle? Simply

 

119 + 120 - 169 = 70 kush

 

The circumference measures practically 220 kush.

 

4 (plus 3) 7 10 13 16 19 22

1 (plus 1) 2 3 4 5 6 7

 

3 (plus 22) 25 ... 377 ... 531

1 (plus 7) 8 ... 120 ... 169

 

Let the side of a square measure 119 = 7 x 17 kush. The diagonal measures practically.7 x 24 = 168 kush. The circumference of the inscribed circle measures practically 22 x 17 = 374 kush.

 

Let the side of a square measure 1 es = 120 kush. The diagonal measures practically 170 kush. Calculate the circumference of the inscribed circle using the Babylonian pi-values 3 and 25/8, the better value 377/120, and the excellent value 1;8,29,44 found via YBC 7289. You will obtain 3 es, 3 es 15 kush, 3 es 17 kush, and 3 es 16 kush 29;44 shusi respectively.

 

Let the side of a square measure 169 kush. The circumference of the inscribed circle measures practically 531 kush, the diagonal of the square practically 119 + 120 = 239 kush:

 

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 ....

 

The lines of this basic number pattern contain the triples and quadruples ( )-1-1, 1-2-2-3, 3-4-5, 8-9-12-17, 19-20-21, 49-50-70-99, 119-120-169, 288-289-408-577 ...

 

 

Second triple (3456) - 3367 - 4825 (kush)

 

Let a rectangle measure 3367 by 3456 kush or 6734 by 6912 shubad. Square the circumscribed circle. The diagonal of the rectangle measures 4825 kush, the radius of the circle 4825 shubad, and the side of the square of the same area practically 8552 shubad or 4276 kush, according to the following number sequences for pi and the square root of pi:

 

4 (plus 3) 7 10 13 16 19 22 25 28 .. 49 = 7x7

1 (plus 1) 2 3 4 5 6 7 8 9 .. 16 = 4x4

 

9 (plus 19) 28 47 66 ... 256 = 16x16

3 (plus 6) 9 15 21 ... 81 = 9x9

 

3 (plus 22) 25 47 69 ... 1521 = 39x39

1 (plus 7) 8 15 22 ... 484 = 22x22

 

1 (plus 16) 17 33 49 ... 177 ... 225 241 257

1 (plus 9) 10 19 28 ... 100 ... 127 136 145

 

7 (plus 39) 46 85 124 ... 475 514 553 592

4 (plus 22) 26 48 70 ... 268 290 312 334

 

257 (plus 553) 810 1363 1916 2469 .... 8552

145 (plus 312) 457 769 1081 1393 .... 4825

 

Two neighboring values in a sequence above generate a new sequence that contains a still better and easily convertible value:

 

553 (plus 592) 1145 1737 .... 7065 7657

312 (plus 334) 646 980 .... 3986 4320

 

7657 / 4320 = 382850 / 216000 = 1;46,20,50

 

If you wish to square a circle you may multiply the radius by 1;46,20,50. Thus you obtain the side of a square of practically the same area.

 

 

Third triple (4800) - 4601 - 6649 (kush)

 

Let a rectangle measure 4601 by 4800 kush. The diagonal of the square of the same area measures practically 6646 kush (square root of 2 x 4601 x 4800). The circumference of the circumscribed circle measures practically 20879 kush (29 gish minus 1 kush).

 

3 (plus 22) ... 289 ... 355 289 (plus 355) ... 20879

1 (plus 7) ... 92 ... 113 92 (plus 113) ... 6646

 

 

Fourth triple (13500) - 12709 - 18541 (kush)

 

Turn a triangle of these numbers in a square of the same area. The side measures practically 9262 kush. Now turn the rectangle 13500 by 12709 kush in a circle of the same area. Multiply 9262 kush by 1;24,51,10 and divide the result by 1;46,20,50. Thus you will obtain the radius. It measures practically 7390 kush.

 

 

Fifth triple (72) - 65 - 97 (kush)

 

The diagonal of the rectangle 72 by 65 kush measures 97 kush or 97 x 30 = 2910 Sumerian fingers. The circumference of the circumscribed circle measures practically 9142 fingers.

 

3 (plus 22) ... 311 ... 355 311 (plus 355) ... 4571 (9142)

1 (plus 7) ... 99 ... 113 99 (plus 113) ... 1455 (2910)

 

 

Sixth triple or triangle (360) - 319 - 481 (kush)

 

Imagine a rectangular triangle with these measurements. The diameter of the inscribed circle measures 319 + 360 - 481 = 198 kush, and the circumference practically 622 kush (pi value 311/99).

 

 

Seventh triangle (2700) - 2291 - 3541 (kush)

 

The diameter of the inscribed circle measures 1450 kush. Square the circle. The side of a square of the same area measures practically 1285 kush.

 

 

Eighth triangle (960) - 799 - 1249 (kush)

 

The diameter of the inscribed circle is 510 kush. The side of the squared circle measures practically 452 kush.

 

 

Ninth triangle (600) - 481 - 769 (kush)

 

diameter inscribed circle 312 kush, circumference 980 kush

 

3 (plus 22) 25 47 69 ... 245 (980)

1 (plus 7) 8 15 22 ... 78 (312)

 

The area of the circle is practically 76,440 square kush, the diagonal of the squared circle practically 391 kush.

 

 

Tenth triangle (6480) - 4961 - 8161 (kush)

 

Use the periphery 4 x 99 x 99 shubad as diameter of a circle. The circumference measures practically 123,163 shubad:

 

333 (plus 355) ... 2463 .... 84823 .... 123163

106 (plus 113) ... 28x28 .... 27000 .... 4x99x99

 

 

Eleventh triple (60) - 45 - 75 (kush)

 

Let a rectangle measure 60 by 45 kush. The circumscribed circle also holds the rectangle 72 by 21 kush. Basic triples 3-4-5 and 7-24-25. Let a grid measure 150 by 150 kush. The inscribed circle passes 20 points of the grid. Draw a circle around a building using the grid 150 by 150 and the triples 45-60-75 and 21-72-75 kush.

 

 

Twelfth triple (2400) - 1679 - 2929 (kush)

 

alternative triple 2020-2121-2929, basic triple 20-21-29

alternative triple 580-2871-2929, basic triple 20-99-101

 

 

Thirteenth triple (240) - 161 - 289 (kush)

 

alternative triple 136-255-289, basic triple 8-15-17

 

The Egyptian method for calculating pi works with every starting triple, not only with the Sacred Triangle 3-4-5 but also for example with the triple 8-15-17:

 

starting triple a-b-c = 8-15-17

 

+- 8a +- 15b (choose the positive value not divisible by 17)

 

+- 8b +- 15a (choose the positive value not divisible by 17)

 

17c

 

8-15-17 136-240-289 2312-4335-4915 .....

161-240-289 2737-4080-4915 .....

495-4888-4915 .....

 

Calculate the first polygon using the ratios 10/7 and 25/6 for the square roots of 2 and 17, and the second one using the ratios 17/12, 25/6 and 35/6 for the square roots of 2, 17 and 34. You will obtain 160/51 and 5443/1734 for pi.

 

6 (plus 22) ... 160 9 (plus 22) ... 5443

2 (plus 7) ... 51 6 (plus 7) ... 1734

 

 

Fourteenth triangle (2700) - 1771 - 3229 (kush)

 

The periphery measures 7700 kush. The diameter of a circle of the same circumference measures about 2450 kush (pi value 22/7), or more precisely 2451 kush:

 

245 (plus 355) 600 955 1310 .... 7345 7700

78 (plus 113) 191 304 417 .... 2338 2451

 

 

Fifteenth triple (90) - 56 - 106 (kush)

 

Let a rectangle measure 90 by 56 kush. The diagonal measures 106 kush. The circumference of the circumscribed circle measures practically 333 kush. Turn the rectangle in a square of the same area. The side of the square and the circumference of the inscribed circle measure practically 71 and 223 kush:

 

3 (plus 22) 25 47 69 ... 223 ... 333

1 (plus 7) 8 15 22 ... 71 ... 106

 

 

 

 

 

PS from November 2002

 

The number patterns for the calculation of the square are generators of triples:

 

1 1 2 2 7 4

2 3 4 9 11 18

5 7 10 20 29 40

12 17 24 49 69 98

29 41 58 118 167 236

70 99 140 ...

169 239 338

408 577 ...

 

 

1 1 2 2x2 1x3 1x1+2x2

2 3 4 4 3 5

 

rectangular triangle 4-3-5

periphery 3x4 = 12

area 1x1x2x3 = 6

radius of the inscribed circle 1x1 = 1

diameter 1x2 = 2

tangents of the half angles 1, 1/2, 1/3

 

 

2 3 4 4x5 3x7 2x2+5x5

5 7 10 20 21 29

 

rectangular triangle 20-21-29

periphery 7x10 = 70

area 2x3x5x7 = 210

radius of the inscribed circle 2x3 = 6

diameter 3x4 = 12

tangents of the half angles 1, 2/5, 3/7

 

 

5 7 10 10x12 7x17 5x5+12x12

12 17 24 120 119 169

 

rectangular triangle 120-119-169

periphery 17x24 = 408

area 5x7x12x17 = 7140

radius of the inscribed circle 5x7 = 35

diameter 7x10 = 70

tangents of the half angles 1, 5/12, 7/17

 

and so on

 

 

2 7 4 4x9 7x11 2x2+9x9

9 11 18 36 77 85

 

rectangular triangle 36-77-85

periphery 11x18 = 198

area 2x7x9x11 = 1386

radius of the inscribed circle 2x7 = 14

diameter 7x4 = 28

tangents of the half angles 1, 2/9, 7/11

 

 

9 11 18 18x20 11x29 9x9+20x20

20 29 40 360 319 481

 

rectangular triangle 360-319-481

periphery 29x40 = 1160

area 9x11x20x29 = 57420

radius of the inscribed circle 9x11 = 99

diameter 11x18 = 198

tangents of the half angles 1, 9/20, 11/29

 

and so on

 

 

The same number patterns generate quadruples:

 

1 1 2 1x2=2 1x1=1 1x2=2 1+2=3 quadruple 2-1-2-3

 

2 3 4 3x4=12 3x3=9 2x4=8 9+8=17 quadruple 12-9-8-17

 

5 7 10 7x10=70 7x7=49 5x10=50 49+50 quadruple 70-49-50-99

 

 

2 7 4 7x4=28 7x7=49 2x4=8 49+8=57 quadruple 28-49-8-57

 

9 11 18 11x18 11x11 9x18 (11x11)+(9x18) q 198-121-162-283

 

20 29 40 29x40 29x29 20x29 (29x29)+(20x29) 1160-841-800-1641

 

 

 

 

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