 Napier's bones

Computing devices in Rabdology Napier's bones Promptuary Location arithmetic v · abacus created by John Napier for calculation of products and quotients of numbers that was based on Arab mathematics and lattice multiplication used by Matrakci Nasuh in the Umdetul Hisab^{[1]} and Fibonacci 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 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 extract square roots. Note that Napier'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 or division. The board's left edge is divided into 9 squares, holding the numbers 1 to 9. The Napier's rods consist of 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.
A rod's surface comprises 9 squares, and each square, except for the top one, comprises two halves divided by a diagonal 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 of each product are written one to each side of the diagonal; numbers less than 10 occupy the lower triangle, with a zero in the top half.
A set consists of 10 rods corresponding to digits 0 to 9. The rod 0, although it may look unnecessary, is obviously still needed for multipliers or multiplicands having 0 in them.
Contents
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 obtain the product, simply note, for each place from right to left, the numbers found by adding the digits within the diagonal sections of the strip (using carryover where the sum is 10 or greater).
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 hundred thousands place, where 5+9=14, we note '4' and carry '1' to the next addition (similarly with 4+8=12 in the ten millions place).
In cases where a digit of the multiplicand is 0, we leave a space between the rods corresponding to where a 0 rod would be. Let us suppose that we want to multiply the previous number by 96431; operating analogously 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 add them using the simple pencilandpaper 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 simply drops 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 known under the name of lattice multiplication. "Crest of the Peacock", by G.G, Joseph, suggests that Napier learned the details of this method from "Treviso Arithmetic", written in 1499....
Division
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 from 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 leftmost digit of the quotient; also write the partial product, leftaligned, 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 immediately 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 numbers in each row, ignoring the second and third columns from the square root bone and record these. (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 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 eighth 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 46785399.
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.
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 append 25 to the root and compare that to the remainder; if it is less than or equal to the remainder, then the next digit will be at least five and round up is needed. In the example above, we see that 6839925 is less than 11669900, so we need to round 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, 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.
Diagonal modification
During the 19th century, Napier's bones underwent a transformation to make them easier to read. The rods began to be made with an angle of about 65° so that the triangles that had to be added were aligned vertically. In this case, in each 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 rods touched, making the two components of each digit of the result much easier to read. Thus, in the picture it is immediately clear that:
 987654321 × 5 = 4938271605
Genaille–Lucas rulers
Main article: Genaille–Lucas rulersIn 1891, Henri Genaille invented a variant of Napier's bones which became known as Genaille–Lucas rulers. By representing the carry graphically, the user can read off the results of simple multiplication problems directly, with no intermediate mental calculations.
The following example is calculating 52749 × 4 = 210996.
Card abacus
In addition to the previouslydescribed "bones" abacus, Napier also constructed a card abacus. Both devices are reunited in a piece held by the Spanish National Archaeological Museum, in Madrid.
The apparatus is a box of wood with inlays of bone. In the top section it contains the "bones" abacus, and in the bottom section is the 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 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 positioning of these cards, multiplications can be made up to the limit of a number 100 digits in length, by another number 200 digits in length.
In addition, the doors of the box contain the first powers of the digits, the coefficients of the terms of the first powers of 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 is probable that it originally belonged to the Spanish Academy of Mathematics (which was created by Philip II) or was a gift from the Prince of Wales. The only thing that is sure is that it was conserved in the Palace, of where it was passed 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 in Kensington, where it received much attention, up to the point at which several societies consulted the Spanish representation 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 wellknown in "England, country of origin of its inventor" (though of course he originated in Scotland).
See also
 Genaille–Lucas rulers
 Slide rule
References
 ^ 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 Education 14 (1): 19–31.
 ^ Diccionario Enciclopédico HispanoAmericano, Mountainer y Simón Editores, Barcelona, 1887, Tomo I, pp. 19–20.
External links
 Java implementation of Napier bones in various number systems at cuttheknot
 Napier and other bones and many calculators
Categories: Mathematical tools
 Mechanical calculators
 Scottish inventions
 Multiplication
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Look at other dictionaries:
Napier's bones — Na pi*er s bones , Napier s rods Na pi*er s rods A set of rods, made of bone or other material, each divided into nine spaces, and containing the numbers of a column of the multiplication table; a contrivance of Baron Napier, the inventor of… … The Collaborative International Dictionary of English
Napier's bones — Math. a form of multiplication table originally marked on sticks of bone or ivory that could be rearranged to carry out the operations of multiplication or division. Also called Napier s rods. [1650 60; after their developer, J. NAPIER] * * * … Universalium
Napier's bones — noun a set of graduated rods formerly used to do multiplication and division by a method invented by John Napier • Syn: ↑Napier s rods • Usage Domain: ↑plural, ↑plural form • Hypernyms: ↑calculator, ↑calculating machine … Useful english dictionary
Napier's bones — noun Date: circa 1658 a set of graduated rods (as of wood or bone) invented by John Napier and used for multiplication and division based on the principles of logarithms … New Collegiate Dictionary
Napier's bones — noun A set of numbered rods used for multiplication and division (see the Wikipedia article for a fuller explanation) … Wiktionary
Napier's bones — plural noun Mathematics slips of ivory or other material divided into sections marked with digits, formerly used to facilitate multiplication and division … English new terms dictionary
Napier's rods — Napier s bones Na pi*er s bones , Napier s rods Na pi*er s rods A set of rods, made of bone or other material, each divided into nine spaces, and containing the numbers of a column of the multiplication table; a contrivance of Baron Napier, the… … The Collaborative International Dictionary of English
Napier's rods — noun a set of graduated rods formerly used to do multiplication and division by a method invented by John Napier • Syn: ↑Napier s bones • Usage Domain: ↑plural, ↑plural form • Hypernyms: ↑calculator, ↑calculating machine … Useful english dictionary
Napier, John — born 1550, Merchiston Castle, near Edinburgh, Scot. died April 4, 1617, Merchiston Castle Scottish mathematician and champion of Protestantism. He divided his life between attacks on the church of Rome and the pursuit of numerical calculations.… … Universalium
Napier's rods — noun Another name for Napier’s bones … Wiktionary
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Computing devices in Rabdology Napier's bones Promptuary Location arithmetic