Explanation of the Logarithm Technique
ABACUS: MYSTERY OF THE BEAD
The Bead Unbaffled - An Abacus Manual

  Calculating Logarithms on a Soroban - Contributed by Deji Adegbite

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Iíve been working on how I can extract the logarithms of numbers on a Soroban. I came up with the following technique. Please reply me and tell me what you think. The technique is still in its infant stage. Iíd like you to help me fine-tune it.

Actually Iím still looking for ways to take it further. Now I will be introducing you to it. I will use logarithm of 6 (base 10) as example and I will explain. Please Iíd like comments, suggestion, and further ideas. You can introduce it to as many soroban fans as possible.

That is the logarithm of 6 (base 10). You can check with a calculator Ė mine gives log 6 = 0.778.

The technique is surprisingly accurate especially if you consider that you are using beads. It still needs to be refined and perfected. Iíll explain the whole thing over again. Hereís the outline.
 

1.        0

2.        1         1

3.        1         1

4.        2         3

5.        2         3

6.        2         3

7.        3         6

8.        3         6

9.        4        12

10.      4        12

11.      5        24

12.      5        24

13.      6        49

14.      6        49

15.      6        49

16.      7        99

17.      7        99

18.      7        99

19.      8       199

20.      8       199

21.      8       199

22.      9       398

23.      9       398

24.      9       398

25.     10      797

26.     10      797

27.     10      797

28.     11    1594

10

sqr root(10) = 3.162

3.162,

3.162 x 1.778 = 5.622

1.778

sqr root(1.778) = 1.333  

1.333 x 5.622 = 7.494

sqr root(1.333) = 1.154

1.154 x 5.622 = 6.487

sqr root(1.154) = 1.074

1.074 x 5.622 = 6.038

sqr root(1.074) = 1.037

1.037 x 5.622 = 5.830

1.037

sqr root(1.037) = 1.018

1.018 x 5.830 = 5.934

1.018

sqr root(1.018) = 1.009

1.009 x 5.934 = 5.987

1.009

sqr root(1.009) = 1.004

1.004 x 5.987 = 6.010

1.004

sqr root(1.004) = 1.002

1.002 x 5.987 = 5.998

1.002

sqr root(1.002) = 1.001

1.001 x 5.998 = 6.003

 

 

sqr root(3.162) = 1.778

 

5.622

5.622

 

5.622

 

5.622

 

5.622

 

5.830

5.830

 

5.934

5.934

 

5.987

5.987

 

5.987

5.987

 

5.998

5.998

 

The explanation of the steps are as follows.

  1. Set up 10 on the right hand side of the soroban.
     
  2. Extract the square root of 10 which is 3.162. On the left hand side of the soroban, set up 1 at 2 rods to the end of the soroban. Then about 6 rods away, set up another 1.
     
  3. Set up another 3.162 and find its square root which is 1.778
     
  4. Multiply the 3.162 by the 1.778 to get 5.622. Move to the left hand side, because the product we just got is less than 6 (the number whose logarithm weíre looking for), take the 1 on the left hand side of the soroban double it and add 1. This gives us 3 (2 x 1 = 2; 2 +1 = 3). Add 1 to the 1 that is nearer to the left edge of the soroban to give 2.
     
  5. Clear off the 3.162.
     
  6. Get the square root of 1.778 to get 1.333
     
  7. Multiply the 1.333 by the 5.622. This gives us 7.494 Remember to still leave 1.333 and the 5.622 intact. The product is greater than 6 (the number whose log we want to find), double the 3 on left hand side to get 6 (donít add 1 after doubling because 7.494 is greater than 6). Then, add 1 to the 2 to get 3
     
  8. Once again, since 7.494 is greater 6, clear it off and get the square root of 1.333 which is 1.154. Remember that we need the square root only to 3 decimal places and also remember to clear off the remainder.
     
  9. Multiply the 1.154 by 5.622. This gives 6.487. Always remember to leave the 2 numbers that you are multiplying intact. 6.487 is greater than 6 so just double the 6 to get 12 (donít add 1 after doubling) and add 1 to the 3 to get 4.
     
  10. Get the square root of the 1.154 to get 1.074
     
  11. Multiply the 1.074 by the 5.622 to get 6.038 which is greater than 6 (the number whose log we are trying to find), we will double the 12 (weíll not be adding 1) to get 24. Add 1 to the 4 to get 5
     
  12. Clear the 6.038 and get the square root of 1.074. At this point here is how we will calculate the square root. Because there is a 0 between the 1 and the 74, just half the 74 part to get the square root. With this technique, we will be saying that the square root of 1.074 is 1.037. Thus, you donít have to set up anything again. Mentally half the 74 part.
     
  13. Multiply the 5.622 by the 1.037. This gives us 5.830. Notice that this product is less than 6 so this time, after doubling the 24 we well add 1. This gives us 24 x 2 = 48; 48 + 1 = 49. Now add 1 to the 5 to get 6.
     
  14. Now clear off the 5.622.
     
  15. Get the square root of 1.037. Once again, simply half the 37 part to get the answer 1.018 which is the square root of 1.037.
     
  16. Multiply the 5.830 by the 1.018 to get 5.934. Notice that the product is getting closer to 6. Since 5.934 is less than 6, we will double the 49 on the left and add 1. This gives us 99. Now add 1 to the 6 to get 7.
     
  17. Clear off the 5.830.
     
  18. Get the square root of 1.018. Remember to just half the 18 part to get the square root i.e. 1.009.
     
  19. Multiply the 1.009 by the 5.934 (remember to keep both the 1.009 and the 5.934 intact). This gives us 5.987 which is less than 6 (the number whose log we are trying to find).  Since the product 5.987 is less than 6, double the 99 and add 1. This should give us 2 x 99 = 198; 198 + 1 = 199. Now, add 1 to the 7 to get 8.
     
  20. Clear off the 5.934.
     
  21. Get the square root of 1.009 to get 1.004. Remember to just half the 009 part to get 004.
     
  22. Multiply 5.987 by 1.004 to get 6.010.  Now since 6.010 is less than 6, double the 199 on the left to get 398 (donít add 1 after doubling because the product we got 6.010 is greater than 6). Now add 1 to the 8 to get 9.
     
  23. Now clear off the 6.010.
     
  24. Get the square root of 1.004 to get 1.002.
     
  25. Multiply 5.987 by 1.002 to get 5.998. Since 5.998 is less than 6, double the 398 on the left and add 1. This gives us 797. Now add 1 to the 9 to get 10.
     
  26. Clear off the 5.987.
     
  27. Now, get the square root of 1.002 to get 1.001.
     
  28. Multiply 1.001 by 5.998 to get 6.003 which is too large so weíll double the 797 without adding 1. This gives us 1594. Now add 1 to the 10 to get 11.
     
  29. Now clear off all the numbers to the right hand side of the soroban and leave only the ones on the left i.e. 11 and 1594.

Weíre there. To get the log of 6 (base 10), raise 2 to the power of 11 (2^11 = 2048). Now divide 1594 by the result Ė i.e. 1594/2048. Donít get scared. There is an easy way to do this. Just continue with the following steps.

The last few steps are listed below.

i)       11                    1594

ii)       10                    1594/2 = 797

iii)         9                    797/2 = 398.5

iv)         8                    398.5/2 = 199.25

v)          7                     199.5/2 = 99.625

vi)          6                   99.625/2 = 49.8125

vii)         5                  49.8125/2 = 24.90625

viii)         4                 24.90625/2 = 12.453125

ix)           3                  12.453125/2 = 6.2265625

x)            2                  6.2265625/2 = 3.11328125

xi)           1                   3.11328125/2 = 1.556640625

xii)           0                 1.556640625/2 = 0.7783203125

Thatís all about the technique though it still needs to be fine-tuned and perfected. Feel free to contact me if you donít understand it. As you can see, it is quite accurate for a simple machine like the soroban. Remember to introduce it to as many soroban fans out there as you possibly can. I still need more ideas, comments and suggestion Ė donít forget to ask questions too. Some other ideas I need are; how I can reduce the number of rods used, setting up the soroban for the operation and what aspects of the technique can be done mentally. Also, as is obvious, the technique might not be very good for overtly high number bases (I tried get log 135 base 256). It may also not be very good for numbers with more than 1 decimal digit e.g. 1.563, 8.995, 3.95, etc.

Now just two more things Ė

First, My next research is on trigonometric ratios. That is, how to calculate the sine, cosine and tangent of angles on a soroban. I hope to get ideas and assistance from you guys out there. If there is anyone willing to assist me, let me know and Iíll send you some ideas Iíve had so far though Iím still working on just the Sine of angles.

Secondly, I intend to write an article - ďInteresting Abacus Techniques.Ē Iím looking for people who will volunteer to place these articles on their websites. Please, if you are willing to assist me, just let me know. I will let you know how far Iíve gone as I compile the articles.  Iíll be so grateful.

The Soroban Ė Simple, yet intriguing.

 

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© February 2008
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Deji Adegbite   Lagos  Nigeria