the bottom of the page, will give a certain number of seconds, which added to the above found arch, the sum will be that required.

EXAMPLE.

Required the arch answering to the log. sine 9.687098 ? The next less log. sine in the table is 9.687087, to which answers 29° 6' 40'; now the difference between the two log. sines is 11, to which, at the bottom of the table, corresponds 3", which added to the above found arch, gives 29° 6' 43" for that required.

Although the table of logarithmic sines extends only to six decimal places of figures, yet the change of the differences to each tenth second near the beginning of the quadrant is very considerable; and therefore, if the angle is under one third, or one-half of a degree, neither the sine, nor its corresponding angle, can be found accurately by the preceding rules. In that case, therefore, others must be substituted, some of which are as follow.

I.

Find the natural sine of the given arch, and its logarithm will be the logarithmic sine required, and conversely.

II.

Reduce the next less minute and tenth second to seconds; then to the arithmetical complement of its logarithm, add its log. sine, and the log. of the given arch expressed in seconds, the sum, rejecting radius, will be the log. sine required.

HI.

The log. sine of a small arch may be found by the method of Interpolation, or, as it is sometimes called, that of Second Differences, which is exemplified in vol. 1. page 206. See also vol. 11. page 11.

Many other methods might be given, also, by curvilinear projections, &c. However, the three first methods only will be illustrated, in the following example.

EXAMPLE.

By the common method of proportion, the log. sine of 0° 8′ 26′′ will be found to be 7.389705.

The converse of this, that is, the method of finding the arch from its given logarithmic sine, will be obvious.

PROBLEM III.

To find the Log. Tangent, or Co-tangent, of a given Arch, and conversely, by a Table of Log. Sines.

RULE.

From the log, sine of the given arch, its index being increased by 10, subtract the log. co-sine of the same arch, and the remainder will be the log. tangent. Again, from the log. co-sine of any arch, subtract the log. sine, the remainder will the log. co-tangent of that arch; or, the log. tangent subtracted from twice the log. of the radius, will give the log. co-tangent. The degrees, and parts of a degree, answering to a given log. tangent, or co-tangent, may be found by an indirect process from a table of log. sines.

EXAMPLE.

Required the log. tangent and co-tangent of 24° 36′ 40′′?

The equation of second difference may also be found by means of Table XXXVII.

PROBLEM IV.

To find the Log. Secant, or Co- secant, fa given Arch, and conversely, by a Table of Log. Sines.

RULE.

Subtract the log. co-sine of the given arch from twice the log. of the radius, and the remainder will be the log. secant: and the log. sine being subtracted from 20,000000, the remainder will be the log. co-secant of the given arch.

The measure of the arch answering to a given log. secant, or COsecant, may be found by subtracting the given log. from 20,000000, and finding the remainder in the table of log. sines, or co-sines, according as the given log. is that of a co secant, or secant.

EXAMPLE.

Required the log. secant and co-secant of 39° 22' 30"? Given arch 39° 22' 30". log. co-sine 9.888185 log. sine

9.802359

Log, secant

10.111815 log. co-sec. 10.197641

PROBLEM V.

To find the Log. Versed Sine, or Co-versed Sine, of a given Arch, and conversely, by a Table of Log. Sines.

RULE.

To twice the log. sine of half the given arch, or of half its complement, according as the versed sine, or co-versed sine is wanted, add the constant log. 0.301030, and the sum will be the log. versed sine, or co-coversed sine, required.

The arch answering to a given log. versed sine or co-versed sine is obtained by a contrary process.

EXAMPLE.

Required the log. versed sine, and co-versed sine of 146° 36′ 40′′?

Complem. of given arch 56 36 40 Constant log. 0.301030

Half

28 18 20 log, sine

Log. co-versed sine of 146 36 40

9.675937

9.675937

9.652904

TABLE XLVII.

Logarithmic Tangents.

This table is similar to that of logarithmic sines, and the rules for finding the logarithmic sine of an arch, and conversely, are applicable to this table.

Required the arch answering to the log. tangent 9.643258 ?

The logarithmic tangent of an arch under 20', or above 89° 40′, be found as directed in Problem 11, page 28.

may

TABLES

XLVIII. XIX. AND L.

These tables were originally drawn up for the purpose of facilitating the indirect method of finding the latitude, from two altitudes of the Sun and the elapsed time. In the first volume of this work, their use is shown in various other calculations. Hitherto they were blended into one table; here they are divided into separate tables, drawn up according to a new plan, so as to be similar to some of the other tables in this work; and extended to every tenth second of time. It was intended to have added the proportional part answering to each second, but the size of the page would not admit of this addition. However, the proportional part answering to any given seconds, may be found by multiplying the difference between two adjacent logs. by the excess of the given seconds above the next less tenth second, and pointing off the right hand figure; and the arch answering to a given log is found by the converse operation*.

* These tables have been copied from the second edition of this work, printed in 1801, into an Epitome of Navigation, printed in 1805 without mentioning from whom they took it; and strange to tell, the compiler of that book says, in his explanation of the Tables, "they were formerly comprised in one table, but it has been thought more convenient to separate them, which we have accordingly done."

Now 28 x 3" 84, then the right hand figure 4, being pointed off, and the remainder 8, added to 5.01459, gives 5.01467, the log. rising answering to 6h. 24' 53".

II.

Required the time answering to log. middle time 4.92799?

Given log. middle time

Next preceding log. being that of 1h. 40′ 10′′

Difference

4.92799

4.92766

33

Now, a cypher being annexed to 35 makes 330, which divided by 68, the difference between the logs. of 1h. 40' 10" and 1h. 40′ 20′′, the quotient is 5" nearly, which added to 1h. 40′ 10′′ gives 1h. 40′ 15", the time required.

REMARK.

Table XLVIII. or that of half elapsed time, is the logarithmic cosecant of the corresponding time expressed in degrees. Table XLIX. is the sum of the log. sine of the given arch, and 0,30103, the index being diminished by 5; and Table L. is the log. versed sine of the corresponding arch, reduced to degrees.

star.

TABLE LI.

Proportional Logarithms.

This table was calculated by Dr. Maskelyne, in order to render the operation more easy for finding the apparent time at Greenwich, answering to a given distance between the Moon and the Sun,or a fixed It is extended to three hours, on account of the distances being set down in the Ephemeris at such intervals. As degrees and hours are similarly divided, each hour may, therefore, be esteemed a degree. This table is constructed by subtracting the logarithm of any given quantity, less than three hours, or three degrees, reduced to seconds, from that of 10800", the number of seconds in three hours, or three degrees.

The purposes to which this table may be applied are many. It is particularly useful in calculations where sexagesimals are concerned, and where the arch may be substituted for its sine or tangent.

The multiplication of a sexagesimal by a common number is per formed by subtracting the proportional log, of one minute, from the sum of the proportional logs, of the factors.