Details, Explanation and Meaning About Napier's bones

Napier's bones Guide, Meaning , Facts, Information and Description

This article started off as a machine translation of an article from the Spanish Wikipedia. It needs lots of revision and editing before it is usable here. This is a work in progress.

Napier's bones are an abacus invented by John Napier for calculation of products and quotients of numbers. Also called Rabdology (from Greek ραβδoς [rabdos], rod and λóγoς [logos], word). Napier published his invention of the rods in a work printed in Edinburgh at the end of 1617 entitled Rabdologiæ. Using the multiplication tables embedded in the rods, multiplication can be reduced to addition operations and division to subtractions. More advanced use of the rods can even be used to extract square roots.
The abacus consists of a board with a rim in which the Napier's rods will be placed to conduct the operations of multiplication or division. The board has its left edge divided into in 9 squares in which numbers 1 to 9 are written. The Napier's rods are strips of wood, metal or heavy cardboard. Napier's bones are three dimensional, square in cross section, with four different rods engraved on each one. A set of such bones might be enclosed in a convenient carrying case. The surface of the rod is divided into 9 squares, and each square, except for the top one, is divided into two halves by a diagonal line. In the first square of each rod a single-digit number is written, and the other squares are filled with double, triple, quadruple and so on until the last square contains nine times the number written in the top square. The digits of each product are written one to each side of the diagonal and in those cases in which they are less than 10, they are written in the lower square, writing a zero in the top square. A set consists of 9 rods corresponding to digits 1 to 9. In the figure the rod 0 has been represented; although for obvious reasons it is not necessary for calculations.

Table of contents
1 Multiplication
2 Division
3 Extracting Square Roots
4 Modifications
5 Card abacus
6 External links

Multiplication

Given the described set of rods, suppose that we wish to calculate the product of 46785399 and 7. Place inside the board the rods corresponding to 46785399, as shown in the diagram, and read the result in the horizontal strip in row 7, as marked on the side of the board. To find the product, simply add the numbers within the diagonal sections of the strip.

From right to left, we obtain the ones place (3), the tens (6+3=9), the hundreds (6+1=7), etc. If some digit of the multiplicand contained a zero, one would leave a space between the rods where the 0 rod would be. Let us suppose that we want to multiply the previous number by 96431; operating analogous to the previous case, we will calculate partial products of the number by multiplying 46785399 by 9, 6, 4, 3 and 1. Then we place these products in the appropriate positions, and then add them using the simple pencil-and-paper method.

Division

Note: This section and those below it need revision.

Division can be performed in a similar fashion. Let's divide 46785399 by 96431, the two numbers we used in the earlier 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 and 1 to 9 by reading the displayed numbers. Note that the dividend has eight digits, whereas the partial products (save for the first one) all have six. So you must temporarily ignore the final two digits of 46785399, namely the '99', leaving the number 467853. Next, look for the greatest partial product that is less 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 the partial product, left-aligned, under the original dividend, and subtract the two terms. You get the difference as 8212999. Repeat the same steps as above: truncate the number to six digits, chose the partial product immediatly less than the truncated number, write the row number as the next digit of the quotient, and subtract the partial product from the difference found in the first repetition. Following the diagram should clarify this. Repeat this cycle until 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 fractional form of the answer .

If you prefer, we can also find as many decimal points as we need by continuing the cycle as in standard long division. Mark a decimal point after the last digit of the quotient and append a zero to the remainder so we now have 163640. 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 product less than 163640 is 96431, from row 1. Subtracting 96431 from 163640, we're left with 67209. Appending a zero, we have 672090 to consider for the next cycle (with the partial result 485.1) The second digit after the decimal point is 6, as the biggest partial product less than 672090 is 578586 from row 6. The partial result is now 485.16, and so on.

Extracting Square Roots

Extracting the square root uses an additional bone which looks a bit different from the others as it has three columns on it. The first column has the first nine squares 1, 4, 9, ... 64, 81, the second column has the even numbers 2 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 0/1 0/2 0/3 0/4 0/5 0/6 0/7 0/8 0/9 0/1     2   1
2 0/2 0/4 0/6 0/8 1/0 1/2 1/4 1/6 1/8 0/4     4   2
3 0/3 0/6 0/9 1/2 1/5 1/8 2/1 2/4 2/7 0/9     6   3
4 0/4 0/8 1/2 1/6 2/0 2/4 2/8 3/2 3/6 1/6     8   4
5 0/5 1/0 1/5 2/0 2/5 3/0 3/5 4/0 4/5 2/5   10   5
6 0/6 1/2 1/8 2/4 3/0 3/6 4/2 4/8 5/4 3/6   12   6
7 0/7 1/4 2/1 2/8 3/5 4/2 4/9 5/6 6/3 4/9   14   7
8 0/8 1/6 2/4 3/2 4/0 4/8 5/6 6/4 7/2 6/4   16   8
9 0/9 1/8 2/7 3/6 4/5 5/4 6/3 7/2 8/1 8/1   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 99

Note: 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 the sixth 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:

  1 2
1 0/1 0/2 0/1     2   1
2 0/2 0/4 0/4     4   2
3 0/3 0/6 0/9     6   3
4 0/4 0/8 1/6     8   4
5 0/5 1/0 2/5   10   5
6 0/6 1/2 3/6   12   6
7 0/7 1/4 4/9   14   7
8 0/8 1/6 6/4   16   8
9 0/9 1/8 8/1   18   9
         _____________
        √46 78 53 99    =    6
         36
         --
         10 78

Now "read" the number in each row (ignore the second and third columns from the square root bone.) For example, read the sixth row as

0/6 1/2 3/6 → 756

Now find the largest number less than the current remainder, 1078. You should find that 1024 from the eighth row is the largest value less than 1078.

  1 2 (value)
1 0/1 0/2 0/1     2   1 121
2 0/2 0/4 0/4     4   2 244
3 0/3 0/6 0/9     6   3 369
4 0/4 0/8 1/6     8   4 496
5 0/5 1/0 2/5   10   5 625
6 0/6 1/2 3/6   12   6 756
7 0/7 1/4 4/9   14   7 889
8 0/8 1/6 6/4   16   8 1024
9 0/9 1/8 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 eigth row 1024 from the current remainder 1078 to get 54. Read the second column of the eighth row on the square root bone, 16, and set the number 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 → 136

Note: If the second column of the square root bone has only one digit, just append it to the current number on board.

The board and intermediate calculations now look like this.

  1 3 6
1 0/1 0/3 0/6 0/1     2   1
2 0/2 0/6 1/2 0/4     4   2
3 0/3 0/9 1/8 0/9     6   3
4 0/4 1/2 2/4 1/6     8   4
5 0/5 1/5 3/0 2/5   10   5
6 0/6 1/8 3/6 3/6   12   6
7 0/7 2/1 4/2 4/9   14   7
8 0/8 2/4 4/8 6/4   16   8
9 0/9 2/7 5/4 8/1   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 third row with 4089.

  1 3 6  
1 0/1 0/3 0/6 0/1     2   1 1361
2 0/2 0/6 1/2 0/4     4   2 2724
3 0/3 0/9 1/8 0/9     6   3 4089
4 0/4 1/2 2/4 1/6     8   4 5456
5 0/5 1/5 3/0 2/5   10   5 6825
6 0/6 1/8 3/6 3/6   12   6 8196
7 0/7 2/1 4/2 4/9   14   7 9569
8 0/8 2/4 4/8 6/4   16   8 10944
9 0/9 2/7 5/4 8/1   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 remainder 5453 to get 1364 as the next remainder. When you rearrange the board, notice that the second column of the square root bone is 6, a single digit. So just append 6 to the current number on the board 136

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

  1 3 6 6
1 0/1 0/3 0/6 0/6 0/1     2   1
2 0/2 0/6 1/2 1/2 0/4     4   2
3 0/3 0/9 1/8 1/8 0/9     6   3
4 0/4 1/2 2/4 2/4 1/6     8   4
5 0/5 1/5 3/0 3/0 2/5   10   5
6 0/6 1/8 3/6 3/6 3/6   12   6
7 0/7 2/1 4/2 4/2 4/9   14   7
8 0/8 2/4 4/8 4/8 6/4   16   8
9 0/9 2/7 5/4 5/4 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 is 123021 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 which row has the answer by looking at the number on the first few bones on the board and comparing it with the first few digits 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 0/1 0/3 0/6 0/6 0/1     2   1 13661
2 0/2 0/6 1/2 1/2 0/4     4   2 27324
3 0/3 0/9 1/8 1/8 0/9     6   3 40989
4 0/4 1/2 2/4 2/4 1/6     8   4 54656
5 0/5 1/5 3/0 3/0 2/5   10   5 68325
6 0/6 1/8 3/6 3/6 3/6   12   6 81996
7 0/7 2/1 4/2 4/2 4/9   14   7 95669
8 0/8 2/4 4/8 4/8 6/4   16   8 109344
9 0/9 2/7 5/4 5/4 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 integer portion 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) must be the largest perfect square smaller than 46785899. Why? The square root of 46785399 is going to be something like 6839.xxxx... This means 68392 is smaller than 46785399, but 68402 is bigger than 46785399 -- the same thing as saying that 68392 is the largest perfect square smaller than 4678599.

This idea is used later on to understand how the technique works, but for now let's continue to generate more digits of the square root.

Similar to finding the fractional portion of the answer in long division, append two zeros to the remainder to get the new remainder 1347800. The second column of the ninth row of the square root bone is 18 and the current number on the board is 1366. So compute

1366 + 1 → 1367 → append 8 → 13678
to set 13678 on the board.

The board and intermediate computations now look like this.

  1 3 6 7 8
1 0/1 0/3 0/6 0/7 0/8 0/1     2   1
2 0/2 0/6 1/2 1/4 1/6 0/4     4   2
3 0/3 0/9 1/8 2/1 2/4 0/9     6   3
4 0/4 1/2 2/4 2/8 3/2 1/6     8   4
5 0/5 1/5 3/0 3/5 4/0 2/5   10   5
6 0/6 1/8 3/6 4/2 4/8 3/6   12   6
7 0/7 2/1 4/2 4/9 5/6 4/9   14   7
8 0/8 2/4 4/8 5/6 6/4 6/4   16   8
9 0/9 2/7 5/4 6/3 7/2 8/1   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 of the square root is 9.

  1 3 6 7 8  
1 0/1 0/3 0/6 0/7 0/8 0/1     2   1 136781
2 0/2 0/6 1/2 1/4 1/6 0/4     4   2 273564
3 0/3 0/9 1/8 2/1 2/4 0/9     6   3 410349
4 0/4 1/2 2/4 2/8 3/2 1/6     8   4 547136
5 0/5 1/5 3/0 3/5 4/0 2/5   10   5 683925
6 0/6 1/8 3/6 4/2 4/8 3/6   12   6 820716
7 0/7 2/1 4/2 4/9 5/6 4/9   14   7 957509
8 0/8 2/4 4/8 5/6 6/4 6/4   16   8 1094304
9 0/9 2/7 5/4 6/3 7/2 8/1   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 remainder 11669900. The second column on the ninth row is 18 with 13678 on the board, so compute

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

  1 3 6 7 9 8
1 0/1 0/3 0/6 0/7 0/9 0/8 0/1     2   1
2 0/2 0/6 1/2 1/4 1/8 1/6 0/4     4   2
3 0/3 0/9 1/8 2/1 2/7 2/4 0/9     6   3
4 0/4 1/2 2/4 2/8 3/6 3/2 1/6     8   4
5 0/5 1/5 3/0 3/5 4/5 4/0 2/5   10   5
6 0/6 1/8 3/6 4/2 5/4 4/8 3/6   12   6
7 0/7 2/1 4/2 4/9 6/3 5/6 4/9   14   7
8 0/8 2/4 4/8 5/6 7/2 6/4 6/4   16   8
9 0/9 2/7 5/4 6/3 8/1 7/2 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.

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, say 54782.917. Everything is the same, except you start out by grouping the digits to the left and right of the decimal point in groups of two.

That is, group 54782.917 as

5 47 82 . 91 7

and proceed to extract the square root from these groups of digits.

Modifications

During the 19th century, Napier's bones underwent a transformation to facilitate the reading. The rods began to make with an inclination of the order of 65°, so that the triangles that had to be added were aligned vertically. In this case, in each square of the rod slogan the unit to the right and the ten (or the zero) to the left.

The rods were made of way so that the vertical and horizontal engraving was more visible than the meetings between the rods, facilitating themselves much the reading when being the pair of components of each digit of the result in a rectangle. Thus, in the figure it is appraised immediately that:

987654321 x 5 = 4938271605

Card abacus

In addition to the previously-described "bones" abacus, Napier also constructed a card abacus. These abaci are reunited in an apparatus held by the Spanish National Archaeological Museum.

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

In the doors of the box are in addition the first powers to the numbers digits, the coefficients of the terms of the first powers of binomial and regular the numeric data of poliedros.

It is not known who was the author of this riquísima jewel, nor if it is of Spanish responsibility or it came from the foreigner, although it is probable that originally Academy of Mathematics belonged to the Spanish Academy of Mathematics created by Felipe II or that brought like gift Prince of Wales . The only thing that can make sure is that it was conserved in Palace, of where passed to National library and later to the National Archaeological Museum, where still it is conserved. In 1876, the Spanish government sent the apparatus to the exhibition of scientific instruments celebrated in Kensington , where it called the attention extraordinarily, until the point of which several societies consulted to the Spanish representation about the origin and use of the apparatus, which motivated that D. Felipe Picatoste wrote a monograph that later was sent to all the nations, surprising the fact that the abacus was only well-known in England, country of origin of its inventor.

Source: Hispano-American Encyclopedic Dictionary , Montaner i Simon (1887).

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