« ForrigeFortsett »
Find the number of degrees at the head or foot of the table; and the minutes in the left or right-hand column, and in the common angle of meeting, formed by the title placed at the head or foot of the column, and the minutes, you will find the fine, tangent or secant required.
But if the given arch contains any parts of a minute, intermediate to those found in the tables; take the difference between the fines, &c. of the given degrees and minutes, and of the minute next greater.
Then as i minute is to that difference, so is the given intermediate part of a minute in decimals to a fourth pro. portional, Which being added to the fine, c. first found will give the fine, &c. required.
Example. The Log. fine of 1°. 48'. 281.12"), is required. The Log. fine of 1° 49'. is
8.5010798 The Log. fine of 1°. 48'. is
8.4970784 Their difference is
40014 Then as 1 : 40014: : 0.47 (281.12H1.): 18807 nearly. Therefore to the Log. fine of 1°. 48'.
8.4970784 Add the difference found
40014 The fum is the Log. fine of 1°. 48'. 281. 12111 8.5010798 2. To find the arch answering to any given fine, tangent,
or secant. Take the next less found in the tables, and subtract it from that given, obferving the degrees and minutes answering to it, and divide this remainder by the difference between it and the next greater, adding as many cyphers as are necessary to the dividend, and the quotient will be the decimal part of a minute to be added to the degrees and minutes before found.
Example 1. Suppose it were required to find the arch answering to the Log, fine
9.8393859 The next less is the Lag. fine of 43° 41'. - 9.8392719 The difference is some
1140 Also the difference between the Log. fine of 43° 41', and 43° 42'. is 1322.
Then 1322) 1140.000(0!. 862=514.43", which being added to 439. 41', gives 43° 41'. 511. 43', for the arch required.
Example 2. Let it be required to find the arch correr ponding to the Log. tangent
9.6766687 The tang. of 250.24', the next less is 9.6765426 The difference is
Also the difference between the Log, tangent of 25° 24'. and the Log, tangent of 25". 25. is 3260.
Then, 3260 1261.000000': 3873 231. 13". Therefore 250:24. 23. 13"), is the arch answering to the given tangent.