- 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 (of 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.
MultiplicationGiven 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.
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 it 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.
As we know, to extract one firstly square root , must group the digits of two in two from the comma, as much towards the right as the left, being the number of the following form:
- ... xx xx xx xx , xx xx xx ...
Taking the pair (that could be only a digit) from the left (xx), number ' ' ' a ' is obtained ' ' finds out so that their square is equal or smaller than the pair. This will be the first number of the solution. Reducing of the pair the square of the whole number thus found, we obtain the rest:
- ra = xx - a2 (If the first pair were 07, the number to would be 2, and the rest 7-4=3)
Later, and of iterative form, the following pair is added to the rest, being a number of the form yxx (and, the previous rest, xx the added pair) that we will call Ra. The following number of the solution will have so to be that the square of the partial solution ab (being ab a number of two digits, not a product) is minor who xxxx (both first pairs of being):
- (ab)2 = (a·10 + b)2 = (a·10)2 + 2·a·10· + b2 < xxxx
- 2·a·10·b + b2 < xxxx - (a·10)2 = R
- (2·a·10 + b)·b < Ra (I)
Operating equally once known the numbers ab, will have to determine the third number of the solution (c) and following (d, and...) that, as easily it is possible to be demonstrated operating to the previous case analogous, will have to fulfill:
- (2·(ab)·10 + c)·c < Rb (II)
- (2·(abc)·10 + d)·d < Rb (III)
- (2·(abcd)·10 + e)·e < Rb (IV)
The indicated products can be obtained easily with the abacus of Napier, but for it an auxiliary rod is necessary so that in each horizontal strip it gathers the squares of the corresponding numbers.
Well-known first number a, we placed in the abacus (or) the rods corresponding to duplo of a. Hecho this, will be enough to add the rod of the squares to find the number so that the equation is fulfilled (I), that will be corresponding to strip b. the This number will have to be sustrar of Ra to find Rb.
Found b, we retired the auxiliary rod of the squares and placed in the board the rod corresponding to 2·b; two cases can occur, if b is minor who 5, the double will have only one number with which it will bstará to place the rod; in opposite case (equal or greater than 5) duplo will be greater of 10, reason why it will be necessary to increase the last rod placed in a unit.
Let us see it with an example. We wished to obtain the square root of number 46 78 53 99. We took the first pair (46) and we immediately determined the inferior square, that turns out to be 36 (49 that is the following one is greater than 46), so that the first number of the solution is 6, and the rest: 46 - 6·6 = 46 - 36 = 10.
We placed the rods of 6·2 = 12 in the board, and next the auxiliary rod of the squares. We compose the rest and the following pair obtaining the number 1078 that will not have to be surpassed by the square of (6b). We read in the abacus (1) value 1024, finding that b = 8 and new rest 1078 - 1024 = 54, descending the following pair, we obtain a value of 5453.
We placed the coresponding rods to the double of 8; being 16 (> 10), we will retire the last rod, the one of the 2, replacing it by the one of the 3 (that is to say, we added a unit to him) and added the rod of the 6. The abacus is as it is in (2a). As it can be observed, the placed numbers are the corresponding ones to the double of the solution found until the moment (68·2 = 136); that is to say, 2abc.. of the previous equations.
Done this, we return to place the auxiliary rod, and operating as in the previous case, we obtain (2b) the third number: 3, being rest 1364. We descend the following pair obtaining a value 136499, we placed rod 6 (3·2) and found following digit 9 and rest 13478. While the rest is different from zero can be continued obtaining significant numbers. For example, to obtain the first decimal, we would lower pair 00 obtaining number 1347800 and would place the rods of 9·2 = 18, being left in the board the following ones: 1-3-6-7(6+1)-8-aid. Making the verification, the first decimal is obtained = 9.
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
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 poliedro s.
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).