Napier's bones 1

Napier's bonesNapier's bones is an abacus created by John Napier for calculation of products and quotients of numbers that wasbased on Arab mathematics and lattice multiplication used by Matrakci Nasuh in the Umdet-ul Hisab[1] andFibonacci writing in the Liber Abaci. Also called Rabdology (from Greek ῥάβδoς [r(h)abdos], "rod" and -λογία[logia], "study"). Napier published his version of rods in a work printed in Edinburgh, Scotland, at the end of 1617entitled Rabdologiæ. Using the multiplication tables embedded in the rods, multiplication can be reduced to additionoperations and division to subtractions. More advanced use of the rods can even extract square roots. Note thatNapier's bones are not the same as logarithms, with which Napier's name is also associated.

The abacus consists of a board with a rim; the user places Napier's rods in the rim to conduct multiplication ordivision. The board's left edge is divided into 9 squares, holding the numbers 1 to 9. The Napier's rods consist ofstrips of wood, metal or heavy cardboard. Napier's bones are three dimensional, square in cross section, with fourdifferent rods engraved on each one. A set of such bones might be enclosed in a convenient carrying case.A rod's surface comprises 9 squares, and each square, except for the top one, comprises two halves divided by adiagonal line. The first square of each rod holds a single digit, and the other squares hold this number's double, triple,quadruple, quintuple, and so on until the last square contains nine times the number in the top square. The digits ofeach product are written one to each side of the diagonal; numbers less than 10 occupy the lower triangle, with a zeroin the top half.A set consists of 10 rods corresponding to digits 0 to 9. The rod 0, although it may look unnecessary, is obviouslystill needed for multipliers or multiplicands having 0 in them.

Napier's bones 2

MultiplicationGiven the described set of rods, suppose that we wish to calculate the product of 46785399 and 7. Place inside theboard the rods corresponding to 46785399, as shown in the diagram, and read the result in the horizontal strip in row7, as marked on the side of the board. To obtain the product, simply note, for each place from right to left, thenumbers found by adding the digits within the diagonal sections of the strip (using carry-over where the sum is 10 orgreater).

From right to left, we obtain the units place (3), the tens (6+3=9), the hundreds (6+1=7), etc. Note that in the hundredthousands place, where 5+9=14, we note '4' and carry '1' to the next addition (similarly with 4+8=12 in the tenmillions place).In cases where a digit of the multiplicand is 0, we leave a space between the rods corresponding to where a 0 rodwould be. Let us suppose that we want to multiply the previous number by 96431; operating analogously to theprevious case, we will calculate partial products of the number by multiplying 46785399 by 9, 6, 4, 3 and 1. Then weplace these products in the appropriate positions, and add them using the simple pencil-and-paper method.

This method can also be used for multiplying decimals. For a decimal value multiplied by an integer (whole number)value ensure that the decimal number is written along the top of the grid. From this position the decimal point simplydrops down the vertical line and 'falls' into the answer.When multiplying two decimal numbers together, the decimal points travel horizontally and vertically until they'meet' at a diagonal line, the point then travels out of the grid in the same method and again 'falls' into the answer.The form of multiplication was also used in the 1202 Liber Abaci and 800 AD Islamic mathematics and knownunder the name of lattice multiplication. "Crest of the Peacock", by G.G, Joseph, suggests that Napier learned thedetails of this method from "Treviso Arithmetic", written in 1499....

Napier's bones 3

DivisionDivision can be performed in a similar fashion. Let's divide 46785399 by 96431, the two numbers we used in theearlier example. Put the bars for the divisor (96431) on the board, as shown in the graphic below. Using the abacus,find all the products of the divisor from 1 to 9 by reading the displayed numbers. Note that the dividend has eightdigits, whereas the partial products (save for the first one) all have six. So you must temporarily ignore the final twodigits of 46785399, namely the '99', leaving the number 467853. Next, look for the greatest partial product that isless than the truncated dividend. In this case, it's 385724. You must mark down two things, as seen in the diagram:since 385724 is in the '4' row of the abacus, mark down a '4' as the left-most digit of the quotient; also write thepartial product, left-aligned, under the original dividend, and subtract the two terms. You get the difference as8212999. Repeat the same steps as above: truncate the number to six digits, chose the partial product immediatelyless than the truncated number, write the row number as the next digit of the quotient, and subtract the partialproduct from the difference found in the first repetition. Following the diagram should clarify this. Repeat this cycleuntil the result of subtraction is less than the divisor. The number left is the remainder.

So in this example, we get a quotient of 485 with a remainder of 16364. We can just stop here and use the fractionalform of the answer .

If you prefer, we can also find as many decimal points as we need by continuing the cycle as in standard longdivision. Mark a decimal point after the last digit of the quotient and append a zero to the remainder so we now have163640. Continue the cycle, but each time appending a zero to the result after the subtraction.Let's work through a couple of digits. The first digit after the decimal point is 1, because the biggest partial productless than 163640 is 96431, from row 1. Subtracting 96431 from 163640, we're left with 67209. Appending a zero, wehave 672090 to consider for the next cycle (with the partial result 485.1) The second digit after the decimal point is6, as the biggest partial product less than 672090 is 578586 from row 6. The partial result is now 485.16, and so on.

Napier's bones 4

Extracting square rootsExtracting the square root uses an additional bone which looks a bit different from the others as it has threecolumns on it. The first column has the first nine squares 1, 4, 9, ... 64, 81, the second column has the even numbers2 through 18, and the last column just has the numbers 1 through 9.

Napier's rods with the square root bone

1 2 3 4 5 6 7 8 9 √

1 0/10/2

0/30/4

0/50/6

0/70/8

0/90/1     2   1

2 0/20/4

0/60/8

1/01/2

1/41/6

1/80/4     4   2

3 0/30/6

0/91/2

1/51/8

2/12/4

2/70/9     6   3

4 0/40/8

1/21/6

2/02/4

2/83/2

3/61/6     8   4

5 0/51/0

1/52/0

2/53/0

3/54/0

4/52/5   10   5

6 0/61/2

1/82/4

3/03/6

4/24/8

5/43/6   12   6

7 0/71/4

2/12/8

3/54/2

4/95/6

6/34/9   14   7

8 0/81/6

2/43/2

4/04/8

5/66/4

7/26/4   16   8

9 0/91/8

2/73/6

4/55/4

6/37/2

8/18/1   18   9

Let's find the square root of 46785399 with the bones.First, group its digits in twos starting from the right so it looks like this:

46 78 53 99Note: A number like 85399 would be grouped as 8 53 99

Start with the leftmost group 46. Pick the largest square on the square root bone less than 46, which is 36 from thesixth row.Because we picked the sixth row, the first digit of the solution is 6.Now read the second column from the sixth row on the square root bone, 12, and set 12 on the board.Then subtract the value in the first column of the sixth row, 36, from 46.Append to this the next group of digits in the number 78, to get the remainder 1078.At the end of this step, the board and intermediate calculations should look like this:

Napier's bones 5

1 2 √

1 0/10/2

0/1     2   1

2 0/20/4

0/4     4   2

3 0/30/6

0/9     6   3

4 0/40/8

1/6     8   4

5 0/51/0

2/5   10   5

6 0/61/2

3/6   12   6

7 0/71/4

4/9   14   7

8 0/81/6

6/4   16   8

9 0/91/8

8/1   18   9

_____________

√46 78 53 99 = 6 -36–10 78

Now, "read" the numbers in each row, ignoring the second and third columns from the square root bone and recordthese. (For example, read the sixth row as : 0/6 1/2 3/6 → 756)Find the largest number less than the current remainder, 1078. You should find that 1024 from the eighth row is thelargest value less than 1078.

1 2 √ (value)

1 0/10/2

0/1     2   1 121

2 0/20/4

0/4     4   2 244

3 0/30/6

0/9     6   3 369

4 0/40/8

1/6     8   4 496

5 0/51/0

2/5   10   5 625

6 0/61/2

3/6   12   6 756

7 0/71/4

4/9   14   7 889

8 0/81/6

6/4   16   8 1024

9 0/91/8

8/1   18   9 1161

_____________

√46 78 53 99 = 68 -36–10 78 -10 24

-----

54

As before, append 8 to get the next digit of the square root and subtract the value of the eighth row 1024 from thecurrent remainder 1078 to get 54. Read the second column of the eighth row on the square root bone, 16, and set thenumber on the board as follows.The current number on the board is 12. Add to it the first digit of 16, and append the second digit of 16 to the result.So you should set the board to

12 + 1 = 13 → append 6 → 136Note: If the second column of the square root bone has only one digit, just append it to the current number onboard.

The board and intermediate calculations now look like this.

Napier's bones 6

1 3 6 √

1 0/10/3

0/60/1     2   1

2 0/20/6

1/20/4     4   2

3 0/30/9

1/80/9     6   3

4 0/41/2

2/41/6     8   4

5 0/51/5

3/02/5   10   5

6 0/61/8

3/63/6   12   6

7 0/72/1

4/24/9   14   7

8 0/82/4

4/86/4   16   8

9 0/92/7

5/48/1   18   9

_____________

√46 78 53 99 = 68 -36–10 78 -10 24

-----

54 53

Once again, find the row with the largest value less than the current partial remainder 5453. This time, it is the thirdrow with 4089.

1 3 6 √

1 0/10/3

0/60/1     2   1 1361

2 0/20/6

1/20/4     4   2 2724

3 0/30/9

1/80/9     6   3 4089

4 0/41/2

2/41/6     8   4 5456

5 0/51/5

3/02/5   10   5 6825

6 0/61/8

3/63/6   12   6 8196

7 0/72/1

4/24/9   14   7 9569

8 0/82/4

4/86/4   16   8 10944

9 0/92/7

5/48/1   18   9 12321

_____________

√46 78 53 99 = 683 -36–10 78 -10 24

-----

54 53

-40 89

-----

13 64

The next digit of the square root is 3. Repeat the same steps as before and subtract 4089 from the current remainder5453 to get 1364 as the next remainder. When you rearrange the board, notice that the second column of the squareroot bone is 6, a single digit. So just append 6 to the current number on the board 136

136 → append 6 → 1366to set 1366 on the board.

Napier's bones 7

1 3 6 6 √

1 0/10/3

0/60/6

0/1     2   1

2 0/20/6

1/21/2

0/4     4   2

3 0/30/9

1/81/8

0/9     6   3

4 0/41/2

2/42/4

1/6     8   4

5 0/51/5

3/03/0

2/5   10   5

6 0/61/8

3/63/6

3/6   12   6

7 0/72/1

4/24/2

4/9   14   7

8 0/82/4

4/84/8

6/4   16   8

9 0/92/7

5/45/4

8/1   18   9

_____________

√46 78 53 99 = 683 -36–10 78 -10 24

-----

54 53

-40 89

-----

13 64 99

Repeat these operations once more. Now the largest value on the board smaller than the current remainder 136499 is123021 from the ninth row.In practice, you often don't need to find the value of every row to get the answer. You may be able to guess whichrow has the answer by looking at the number on the first few bones on the board and comparing it with the first fewdigits of the remainder. But in these diagrams, we show the values of all rows to make it easier to understand.As usual, append a 9 to the result and subtract 123021 from the current remainder.

1 3 6 6 √

1 0/10/3

0/60/6

0/1     2   1 13661

2 0/20/6

1/21/2

0/4     4   2 27324

3 0/30/9

1/81/8

0/9     6   3 40989

4 0/41/2

2/42/4

1/6     8   4 54656

5 0/51/5

3/03/0

2/5   10   5 68325

6 0/61/8

3/63/6

3/6   12   6 81996

7 0/72/1

4/24/2

4/9   14   7 95669

8 0/82/4

4/84/8

6/4   16   8 109344

9 0/92/7

5/45/4

8/1   18   9 123021

_____________

√46 78 53 99 = 6839 -36–10 78 -10 24

-----

54 53

-40 89

-----

13 64 99

-12 30 21

--------

1 34 78

You've now "used up" all the digits of our number, and you still have a remainder. This means you've got the integerportion of the square root but there's some fractional bit still left.Notice that if we've really got the integer part of the square root, the current result squared (6839² = 46771921) mustbe the largest perfect square smaller than 46785899. Why? The square root of 46785399 is going to be somethinglike 6839.xxxx... This means 6839² is smaller than 46785399, but 6840² is bigger than 46785399—the same thing assaying that 6839² is the largest perfect square smaller than 46785399.

Napier's bones 8

This idea is used later on to understand how the technique works, but for now let's continue to generate more digitsof the square root.Similar to finding the fractional portion of the answer in long division, append two zeros to the remainder to get thenew remainder 1347800. The second column of the ninth row of the square root bone is 18 and the current numberon the board is 1366. So compute

1366 + 1 → 1367 → append 8 → 13678to set 13678 on the board.The board and intermediate computations now look like this.

1 3 6 7 8 √

1 0/10/3

0/60/7

0/80/1     2   1

2 0/20/6

1/21/4

1/60/4     4   2

3 0/30/9

1/82/1

2/40/9     6   3

4 0/41/2

2/42/8

3/21/6     8   4

5 0/51/5

3/03/5

4/02/5   10   5

6 0/61/8

3/64/2

4/83/6   12   6

7 0/72/1

4/24/9

5/64/9   14   7

8 0/82/4

4/85/6

6/46/4   16   8

9 0/92/7

5/46/3

7/28/1   18   9

_____________

√46 78 53 99 = 6839. -36–10 78 -10 24

-----

54 53

-40 89

-----

13 64 99

-12 30 21

--------

1 34 78 00

The ninth row with 1231101 is the largest value smaller than the remainder, so the first digit of the fractional part ofthe square root is 9.

1 3 6 7 8 √

1 0/10/3

0/60/7

0/80/1     2   1 136781

2 0/20/6

1/21/4

1/60/4     4   2 273564

3 0/30/9

1/82/1

2/40/9     6   3 410349

4 0/41/2

2/42/8

3/21/6     8   4 547136

5 0/51/5

3/03/5

4/02/5   10   5 683925

6 0/61/8

3/64/2

4/83/6   12   6 820716

7 0/72/1

4/24/9

5/64/9   14   7 957509

8 0/82/4

4/85/6

6/46/4   16   8 1094304

9 0/92/7

5/46/3

7/28/1   18   9 1231101

_____________

√46 78 53 99 = 6839.9 -36–10 78 -10 24

-----

54 53

-40 89

-----

13 64 99

-12 30 21

--------

1 34 78 00

-1 23 11 01

----------

11 66 99

Subtract the value of the ninth row from the remainder and append a couple more zeros to get the new remainder11669900. The second column on the ninth row is 18 with 13678 on the board, so compute

Napier's bones 9

13678 + 1 → 13679 → append 8 → 136798and set 136798 on the board.

1 3 6 7 9 8 √

1 0/10/3

0/60/7

0/90/8

0/1     2   1

2 0/20/6

1/21/4

1/81/6

0/4     4   2

3 0/30/9

1/82/1

2/72/4

0/9     6   3

4 0/41/2

2/42/8

3/63/2

1/6     8   4

5 0/51/5

3/03/5

4/54/0

2/5   10   5

6 0/61/8

3/64/2

5/44/8

3/6   12   6

7 0/72/1

4/24/9

6/35/6

4/9   14   7

8 0/82/4

4/85/6

7/26/4

6/4   16   8

9 0/92/7

5/46/3

8/17/2

8/1   18   9

_____________

√46 78 53 99 = 6839.9 -36–10 78 -10 24

-----

54 53

-40 89

-----

13 64 99

-12 30 21

--------

1 34 78 00

-1 23 11 01

----------

11 66 99 00

You can continue these steps to find as many digits as you need and you stop when you have the precision you want,or if you find that the reminder becomes zero which means you have the exact square root.Having found the desired number of digits, you can easily determine whether or not you need to round up; i.e.,increment the last digit. You don't need to find another digit to see if it is equal to or greater than five. Simply append25 to the root and compare that to the remainder; if it is less than or equal to the remainder, then the next digit will beat least five and round up is needed. In the example above, we see that 6839925 is less than 11669900, so we need toround up the root to 6840.0.There's only one more trick left to describe. If you want to find the square root of a number that isn't an integer, say54782.917. Everything is the same, except you start out by grouping the digits to the left and right of the decimalpoint in groups of two.That is, group 54782.917 as

5 47 82 . 91 7and proceed to extract the square root from these groups of digits.

Napier's bones 10

Diagonal modificationDuring the 19th century, Napier's bones underwent a transformation to make them easier to read. The rods began tobe made with an angle of about 65° so that the triangles that had to be added were aligned vertically. In this case, ineach square of the rod the unit is to the right and the ten (or the zero) to the left.

The rods were made such that the vertical and horizontal lines were more visible than the line where the rodstouched, making the two components of each digit of the result much easier to read. Thus, in the picture it isimmediately clear that:

987654321 × 5 = 4938271605

Napier's bones 11

Genaille–Lucas rulersIn 1891, Henri Genaille invented a variant of Napier's bones which became known as Genaille–Lucas rulers. Byrepresenting the carry graphically, the user can read off the results of simple multiplication problems directly, withno intermediate mental calculations.The following example is calculating 52749 × 4 = 210996.

Card abacus

The two Napier's abacuses at the National Archaeological Museum of Spain(Madrid).

In addition to the previously-described"bones" abacus, Napier also constructed a cardabacus. Both devices are reunited in a pieceheld by the Spanish National ArchaeologicalMuseum, in Madrid.

The apparatus is a box of wood with inlays ofbone. In the top section it contains the "bones"abacus, and in the bottom section is the cardabacus. This card abacus consists of 300stored cards in 30 drawers. One hundred ofthese cards are covered with numbers (referredto as the "number cards"). The remaining twohundred cards contain small triangular holes,which, when laid on top of the number cards, allow the user to see only certain numbers. By the capable positioningof these cards, multiplications can be made up to the limit of a number 100 digits in length, by another number 200digits in length.

Napier's bones 12

In addition, the doors of the box contain the first powers of the digits, the coefficients of the terms of the first powersof the binomial and the numeric data of the regular polyhedra.[2]

It is not known who was the author of this piece, nor if it is of Spanish origin or came from a foreigner, although it isprobable that it originally belonged to the Spanish Academy of Mathematics (which was created by Philip II) or wasa gift from the Prince of Wales. The only thing that is sure is that it was conserved in the Palace, of where it waspassed to the National library and later to the National Archaeological Museum, where it is still conserved.In 1876, the Spanish government sent the apparatus to the exhibition of scientific instruments celebrated inKensington, where it received much attention, up to the point at which several societies consulted the Spanishrepresentation about the origin and use of the apparatus. This motivated D. Felipe Picatoste to write a monograph,that was sent to all the nations, in which he expressed surprise about the fact that the abacus was only well known in"England, country of origin of its inventor" (though of course he originated in Scotland).

References[1] Corlu, M. S.; Burlbaw, L. M.; Capraro, R. M.; Corlu, M. A.; Han, S. (2010). "The Ottoman Palace School Enderun and The Man with

Multiple Talents, Matrakçı Nasuh". Journal of the Korea Society of Mathematical Education Series D: Research in Mathematical Education14 (1): 19–31.

[2] Diccionario Enciclopédico Hispano-Americano, Mountainer y Simón Editores, Barcelona, 1887, Tomo I, pp. 19–20.

External links• Java implementation of Napier bones in various number systems (http:/ / www. cut-the-knot. org/ blue/ Napier.

shtml) at cut-the-knot• Napier and other bones and many calculators (http:/ / www. peterkernwein. de/ Rechengeraete-Sammlung/

mechcal. htm)

Article Sources and Contributors 13

Article Sources and ContributorsNapier's bones  Source: http://en.wikipedia.org/w/index.php?oldid=531642567  Contributors: 10metreh, A.T.M.Schipperijn, A3RO, Alex.petralia, Andrew Levine, Antandrus, Arundhati bakshi,Az29, BAxelrod, Beland, Bender235, Bfinn, Breno, Bsherr, Can't sleep, clown will eat me, CommonsDelinker, Cwkmail, Cyde, DVdm, Darkfreakinwolf, Dave souza, Davidiad, Decltype,Dfsmth45, Doradus, Drmies, E2eamon, Ernstblumberg, Everyking, Evil saltine, Flewis, Fnlayson, Fomthebomb, Fredrik, Gandalf61, Giftlite, Glane23, Graham87, Gurch, Hairy Dude, Hex,Ironholds, JahJah, Jgottula, JodyB, Jonathan Drain, Jonemerson, Jossi, Jpgordon, Jung dalglish, Kbs, Korinth, Kuru, LAX, Lradrama, Ludicolo98, Lysdexia, Mais oui!, Mareczek25, Martin451,Materialscientist, Mathmannix, Michael Hardy, Mike Krüger, Mikker, Milogardner, Mintleaf, Miranda, Neilc, Oberiko, OlEnglish, Omnipaedista, PV=nRT, Phil Boswell, Picapica, Pmdboi,RDBury, RJFJR, RJaguar3, Rich Farmbrough, Rossa1, Sam Hocevar, Savantas83, Sceptre, Shadowjams, The Anome, The Thing That Should Not Be, Thingg, Thumperward, Tigerattack, TinaCordon, Topory, Trunks ishida, Versus22, VeryVerily, Wadems, Wernher, WikHead, Woohookitty, Worrydream, Yonaa, Yuvsore2, Zaqarbal, 202 ,יהודה שמחה ולדמן anonymous edits

Image Sources, Licenses and ContributorsImage:Bones of Napier (board and rods).png  Source: http://en.wikipedia.org/w/index.php?title=File:Bones_of_Napier_(board_and_rods).png  License: GNU Free Documentation License Contributors: Fabienkhan, Luigi Chiesa, MaksimImage:Napier-example-1.png  Source: http://en.wikipedia.org/w/index.php?title=File:Napier-example-1.png  License: unknown  Contributors: Luigi Chiesa, Maksim, 1 anonymous editsImage:Napier example 2.png  Source: http://en.wikipedia.org/w/index.php?title=File:Napier_example_2.png  License: GNU Free Documentation License  Contributors: 132nd, Luigi ChiesaImage:Napier-example-3.png  Source: http://en.wikipedia.org/w/index.php?title=File:Napier-example-3.png  License: GNU Free Documentation License  Contributors: Luigi Chiesa, MaksimImage:Napier-example-4.png  Source: http://en.wikipedia.org/w/index.php?title=File:Napier-example-4.png  License: GNU Free Documentation License  Contributors: Kbs, Zeimusu, 1anonymous editsImage:Napier Modification.png  Source: http://en.wikipedia.org/w/index.php?title=File:Napier_Modification.png  License: GNU Free Documentation License  Contributors: 132nd, LuigiChiesaImage:Genaille-Lucas rulers example 5.png  Source: http://en.wikipedia.org/w/index.php?title=File:Genaille-Lucas_rulers_example_5.png  License: Public Domain  Contributors: WorrydreamFile:Ábacos neperianos (M.A.N. Madrid) 01.jpg  Source: http://en.wikipedia.org/w/index.php?title=File:Ábacos_neperianos_(M.A.N._Madrid)_01.jpg  License: GNU Free DocumentationLicense  Contributors: User:Zaqarbal

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