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A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: Hav-Doniol
From: Hewitt Schlereth
Date: 2015 Jun 17, 07:28 -0700
From: Hewitt Schlereth
Date: 2015 Jun 17, 07:28 -0700
Hanno, I think it's a case of "so easy when you know how." To abandon what my beloved Mrs White toiled to make clear to me in grade school ... Well, it would feel like a betrayal.
I remember looking up Napier's Bones on Wikipedia. They are a grid with a column of the numbers 1 thru 9 on the left; a row 1 thru 9 across the top. Each box in the grid has the number you get by multiplying across from left to right down the columns.
The the number in each box is printed into its 10s and 1s piece. 10s piece in the NW corner, 1s in the SE. They are separated by a diagonal line running NE-SW.
Then you make a copy of this grid (table) with the row across the top consisting of the digits of the number you want to multiply.
Here's the example given on the Internet site
"To multiply two numbers, arrange the bones as described above. The above illustration shows this process for 
. The computation proceeds from right to left, starting with the rightmost bone in the row determined by themultiplier. In this case, the last digit in the 7s row of the 6-bone is 2, so write down 2. Now add the two adjacent numbers in the same row to the left (i.e., the ones in the parallelogram) to obtain 
, which is the next digit, so we now have 72. The next sum is 
, so write down the 2 two obtain 272 and carry the 1. Proceeding to the next digit, it is 
(because of the carry), so write down the 4 to obtain 
and carry the 1. The leftmost digit is then 
(from the carry), giving the final answer 
."
. The computation proceeds from right to left, starting with the rightmost bone in the row determined by themultiplier. In this case, the last digit in the 7s row of the 6-bone is 2, so write down 2. Now add the two adjacent numbers in the same row to the left (i.e., the ones in the parallelogram) to obtain , which is the next digit, so we now have 72. The next sum is , so write down the 2 two obtain 272 and carry the 1. Proceeding to the next digit, it is (because of the carry), so write down the 4 to obtain and carry the 1. The leftmost digit is then (from the carry), giving the final answer ."If I understand this correctly, for Doniol you put one number across the top. Then you go down the left column beginning with the rightmost digit of the number you are multiplying by, and multiply each digit in the top row, stacking and shifting the results to the left, just as my dear old teacher showed me.
I just don't see where the savings in time or accuracy come from. I have to 1) draw up a matrix (table, grid), 2) multiply row and column digit by digit, 3) stack the results, just as in my old method, 4) finally, add the columns of the stacks, just as in my old method.
For me, the blunders come in the last step. So, for me Napier doesn't offer any advantage. Beyond step 1, everything is the same. So, for me the table is, A) extra work, B) one more place to blunder.
Besides, I couldn't dump Mrs White for "Bones" Napier. It would break her heart. :-)
Hewitt
And, well, you would not need "cheating" as you professed, Vedic can be done w/o a calculator easily. If I, the slowest multiplicator on the list, can you surely can multiply with Vedic easily, too. Ever tried it?Hewitt,sorry, I do not understand - a table for / of what, please?HOn Mon, Jun 15, 2015 at 12:26 PM, Hewitt Schlereth <NoReply_Schlereth@fer3.com> wrote:Hanno, don't these methods require you to make a table before you do the multiplying - ie, a preliminary chance to make blunders even ante getting to to the problem?I'm with Gary and Greg on this one. Though I'm very slow at long multiplication, I do get there. (Then of course there's my backup - baggies of batteries and two-bit calculators.)
Hewitt
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