will come out,845098040, &c. (as before), which value will be true to 11 places of figures, by taking the first term of the series, only.

Again, let the common logarithm of the next prime number, which is 11, be required. Here a may be taken = 10, b = 11, and c = 12; but fewer terms of the series will suffice, if other three numbers, composed of 11 and the inferior primes be taken, whereof the common difference is a unit. Thus, because 98 = 2 × 7 × 7, 99 = 3 × 3 × 11 (9 × 11), and 100 = 2 × 2 × 5 × 5 (or 10 x 10), let there be taken a 98, b99, and c = 100; and then, by the first term of the series only, the log. of 99 will be found true to 14 places; whence that of 11 (log. 99— log. 9) is also known.

But, notwithstanding all these artifices and compendiums, a method, similar to that in page 21, for finding the logarithms of large numbers, one from another, by addition and subtraction only, still seems wanting in the calculation of tables; I shall, therefore, here subjoin such a method.

1. Let A, B, and C denote any three numbers in arithmetical progression, not less than 10000 each, whereof the common difference is 100.

2. From twice the logarithm of B, subtract the sum of the logarithms of A and C, and let the remainder be divided by 10000.

3. Multiply the quotient by 49,5, and to the product add part of the difference of the logarithms of A' and

1

B; then the sum will be the excess of the logarithm of A +1 above that of A.

4. From this excess let the quotient, found by Rule 2., be continually subtracted, that is, first from the excess itself, then from the remainder, then from the next remainder, &c.

5. To the logarithm of A add the said excess, and to the sum add the first of the remainders; to the last sum add the next remainder, &c. then the several sums, thus arising, will exhibit the logarithms of A + 1, A + 2, A + 3, &c. respectively.

Thus, let it be proposed to find the logarithms of all the whole numbers between 17900 and 18100; those of the two extremes 17900 and 18100, and that of the mean (18000) being given.

rithm of

have

C to 4,257678575

2 log. Blog. A-log. C

10000

=,00000000134 (sce

Rule 2.), which multiplied by 49,5, and the product added

to

log. Blog. A

100

gives ,00002426107 for the excess of

the logarithm of A + 1 above that of A (by Rule 3.) From whence the work, being continued according to Rule 4 and 5, will stand as follows:

Note. The logarithms found according to this method, in numbers between 10000 and 20000, are true to 8 or 9 places of figures: those of numbers between 20000 and 50000 err only in the 9th or 10th place; and those of above 50000 are true to 10 places, at least.

Having explained the manner of constructing a table of logarithms, and that by various methods, I now come to show the use of such a table in the business of trigonometry.

First, in the right-angled plane triangle ABC (fig. 2.), let there be given the hypothenuse AC 17910 feet, and the angle A 35° 20′; to find the perpendicular BC and the base AB.

=

radius

; therefore,

Here, because radius: sine 35°20':: 17910: BC (by Thesine 35° 20′ x 17910 or. 1. p. 7.), we have BC= because the addition and subtraction of logarithms answers to the multiplication and division of the natural numbers (see p. 43. 44.), we have log. BC = log. sine 35° 20′ + log• 17910- log. radius.

But, by the tables of artificial or logarithmic sines*, the log. sine of 35° 20′ will appear to be 9,7621775; to which add 4,2530956, the log. of 17910, and from the sum (14,0152731) take 10, the log. of radius, and there results 4,0152731 = the log. of BC; which, in the tables, answers to 10358, the length of BC required.

Again, for AB, it will be, as radius: sine of C (54° 40′) :: AC (17910) : AB (by Theorem 1.) Whence, by adding the logarithms of the second and third terms together, and subtracting that of the first, as above, we have AB14611. See the operation.

* A table of artificial sines is nothing more than a table of the logarithms of the numbers expressing the natural sines, to the rad us 10000000000; whose logarithm is 10.

H

Moreover, in the oblique plane triangle ABC (fig. 3.), let there be given AB 75, AC 60, and the included angle A = 48°; to find the other two angles. Then (by Theorem 5.) it will be,

Which 14° added to 66°, the half sum of the angles C and B, gives the greater C = 80°; and, subtracted therefrom, leaves the lesser B = 52°,

Lastly, in the right-angled spherical triangle ABC (fig. 17.), let there be given the hypothenuse AC = 60°, and the angle A 23° 29'; to find the base and perpendicular. Then (by Theor. 1. p. 28.) the operation will be as follows.