Example 1. From the tables we find that log 35745 = 4.5532153 and log 35746=4.5532275. The difference of these logarithms is '0000122. To find, then, log 35745 73, we have by above rule log 35745.73-log 35745 log 35746-log 35745 73 ..log 35745 73-log 35745=. of '0000122=0000089, 100 ..log 35745 73=log 35745+0000089=4.5532242. Example 2. From the same data to find conversely the number whose logarithm is 4.5532242, i.e. the value of 104-5532242 This process of interpolation is aided by the introduction of Subsidiary Tables. Thus, at the side of the main table, a column headed Diff. will be found. The difference of the logarithms, corresponding to a difference of 1 in the numbers, at any particular part of the table, is here tabulated. And below this difference is tabulated the differences corresponding to 1, 2, 3, 4, 5, 6, 7, 8, 9 tenths respectively. Dividing these latter differences by 10, we have those corresponding to 1, 2, 3, 4, 5, 6, 7, 8, 9 hundredths respectively. Thus for the above example, we have tabulated Diff. for 1= 122; for 7 tenths = 85, for 3 hundredths = 3.7 .. Diff. for 73 = 89 nearly. 272. Examples in interpolation may be written in the following form : Given log 35745 and log 35746 find log 35745.73. From log 35746=4.5532275 Take log 35745 = 4.5532153 .. when diff. in nos. = 1, then .. when diff. in nos. 122 = diff. in logs, 73, then 73 of 122 = diff. in logs. [Now ·73 of 122 = '7 of 122 + '03 of 122 = 85 + 4 = 89,] .. when diff. in nos. = 73, then 89 diff. in logs. Add log 35745 = 4.5532153 .. log 35745 73 = 4.5532242 = Conversely, given log 35745 and log 35746, find the number whose logarithm is 4.5532242. .. when diff. in logs = 89, then '73 = diff. in nos. .. 4·5532242 = log 35745·73. 273. In the same way we may interpolate in the tables of the ratios of angles, and of the logarithms of the ratios. 274. It is important to observe that, as the secondary ratios cosine, cotangent, cosecant vary incongruently with the angles, their differences and the differences of their logarithms will have the opposite sign to the corresponding differences of the angles. Example. To find cos 30° 13′ 2′′, cos 30° 13' 2" - cos 30° 13' cos 30° 13'-cos 30° 13′ 2′′ cos 30° 13' - cos 30° 14' This result is obtained by subtracting the difference, corresponding to 2", from cos 30° 13'. Or we might add the difference, corresponding to 58", to cos 30° 14′. 275. One half of the ratios of angles between 0 and 90°, viz. all the sines and cosines and all the tangents up to 45°, are less than 1. The logarithms of these ratios are, therefore, negative. In order to avoid tabulating so many negative quantities, 10 is added to the logarithms of all ratios in the tables. In this way all the tabulated logarithms are reduced to positive numbers. [The number 4, instead of 10, would answer the same purpose: for the smallest tabulated logarithm, viz. log sin l', is greater than - 4.] The symbol L is used to denote a logarithm increased by 10. $ 5. ADAPTATION OF FORMULE TO LOGARITHMIC CALCULATION. 276. Numbers which require for notation several significant figures may be called long numbers. The practical use of logarithms is to enable us to avoid having to perform any operations, involving long numbers, except addition and subtraction. 277. For logarithmic purposes an expression requires to be put into factor-form. We here use the word factors to denote the parts of an expression which, being either multiplied or divided, make up the expression. Thus : If If PA× B; log P = log A+ log B. 1 Strictly, of course, not B, is the factor in the second case. B' 278. A formula is best adapted to logarithmic calculation, when the factors of the required value involve only the addition and subtraction of long numbers. The following articles will illustrate this statement. 279. To solve a right-angled triangle, whose sides a and b (containing the right-angle) are given. I. To find c, we have c2 = a2 + b2. We might find a2 and b2 by multiplication, which would be a troublesome process if a and b were long numbers. We may use the logarithmic table to avoid this process; but not conveniently. Thus : then Suppose Now a = 3456 4 ft. and b = 4543.5 ft. = log 1.1947 07725 and log 2.064331478, .. log c = 3.7565423 and c = 5708.8. This work requires 6 references to the table; showing that the formula c2 = a2+b2 is not well adapted for logarithmic purposes. 9 .. ·0001813 = diff. for of 60" i.e. 42′′ .. A 37° 15′ 42′′ and B = 90° - A = 52° 44′ 18′′. = III. Having found A, we may find c as follows: 280. α .. log clog a = log cosec 37° 15′ 42′′. L cosec 37° 15′ = 10.2180336 L cosec 37° 16' 10.2178676 .. diff. for 1': = = 7 ⚫0001660 .. diff. for 42′′ = 1 10 1000000= ·0001162, .. log c=3·5386240 + ·2179174 = 3·7565414, .. c=5708.8 as before. To find the side a of a right-angled triangle, when the hypothenuse c and third side b are given. In this case the required value is expressed by factors which involve addition and subtraction only. Hence it is better adapted for logarithmic calculation. Thus 2 log a = log (c + b) + log (c-b). c = 5708.8 and b = 4543·5, .. c+b=10252.3 and c-b=1165.3, .. log (c+b) = 4·0108213 and log (c—b) = 3·0664377, .. 2 log a=7·0772590, .. log a = 3.5386295, .. a=3456.4. 281. To solve a triangle when two angles and a side are given. [See Arts. 163, 164.] Here A = 180° – B-C'; which gives A by mere subtraction. = a sin B sin A .. log b = log a + log sin B - log sin A. This is well adapted to logarithmic calculation. Similarly c may be found. |