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Example.

Required the logarithmic rising answering to 1:43"27: ?

1:43"25!, answering to which is Odd seconds 2, pro. part answering to which is .

. 5.00040

28

Given time = 1:43:27:, corresponding logarithmic rising

5.00068

The converse of this, that is, finding the time corresponding to a given log. will appear obvious; thus,

Let the given logarithmic rising be 5.69088, to find the corresponding time.

The next less tabular log. is 5.69071, answering to which is 3.57"30!; the difference between this log. and that given is 17, answering to which in the column of proportional parts, abreast of the tabular log., is 3 seconds; now, this being added to the time found, as above, gives 3:57"33'; which, therefore, is the time corresponding to the given logarithmic rising

Note. The numbers in this Table were computed by the following rule, viz :

To twice the logarithmic sine of half the meridian distance, in degrees, add the constant log. 6.301030, and the sum, rejecting 20 from the index, will be the logarithmic rising.

Example.

Required the logarithmic-rising answering to 4:10:45: ? Given meridian distance = 4" 10"45, in degrees = 62:41:15":

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Logarithmic rising answering to the given meridian distance = 5.733323

The numbers in the present Table may be also computed by means of the natural versed sines contained in Table XXVII., as thus ;

Reduce the meridian distance to degrees, and find the natural versed sine corresponding thereto; the common log, of which will be the logarithmic rising

Example. Required the logarithmic rising answering to 4" 2230!, or 65:37:30? Meridian distance in degrees = 65:37:30", natural versed sine = 587293, log. 5.768855; which, therefore, is the logarithmic rising answering to the given meridian distance.

In this method of computing the logarithmic rising, the natural versed sine is to be conceived as being multiplied by 1000000, the radius of the Table, and thus reduced to a whole number.

TABLE XXXIII.

To reduce Points of the Compass to Degrees, and conversely. This Table is divided into six columns; the two first and two last of which contain the names of the several points and quarter points of the compass ; the third column contains the corresponding number of points and quarter points reckoned from the meridian; and the fourth column the degrees and parts of a degree answering thereto.---The manner of using this Table is obvious; and so is the method by which it was computed :for since the whole compass card is divided into 32 points, and the whole circle into 360 degrees ; it is evident that any given number of points will be to their corresponding degrees in the ratio of 32 to 360; and vice versa, that any given number of degrees will be to their corresponding points as 360 is to 32 :Hence, to find the degrees corresponding to one point.As 82? : 360: :: 1?:11:15!; so that one point contains 11 degrees and 15 minutes ;-two points, 22 degrees 30 minutes, &c. &c.

Table XXXIV.

Logarithmic Sines, Tangents, and Secants, to every Point and Quarter

Point of the Compase. In this Table the points and quarter points are contained in the left and right hand marginal columns, and the log. sines, tangents, and secants, corresponding thereto, in the intermediate columns.

If the course be given in points, it will be found more convenient to take the log. sine, tangent, or secant of it from this Table, than to reduce those points to degrees, and then find the corresponding log. sine, &c. &c. in either of the following Tables - The manner of using this Table must appear obvious at first sight.

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TABLE XXXV.

Logarithmic Secants.

In the first 10 degrees of this Table, the logarithmic secants are given to every tenth second, with proportional parts, answering to the intermediate seconds, in the right hand marginal column.—Thence to 88 degrees, the log. secants are given to every fifth second, with proportional parts, adapted to the intermediate seconds, in the right hand column of each page :--and because the numbers increase rapidly between 80 and 88 degrees, producing very considerable differences between any two adjacent logs.; therefore betwixt those limits, there are two pages allotted to a degree; every page being divided into two parts of 15 minutes each, so that no portion whatever of the proportional parts might be lost, and that the whole might have room to be fully inserted.- In the two last degrees, viz. from 88 to 90, the log. secants are given to every second. - The Table is so arranged as to be extended to every second in the semicircle, or from 0 to 180 degrees; as thus : the arcs corresponding to the log. secants are given in regular succession at top from 0 to 90 degrees, and then continued at bottom, reckoning towards the left hand, from 90 to 180 degrees the arcs corresponding to the co-secants are placed at the bottom of the Table, in numerical order, from the right hand towards the left (like the secants in the second quadrant), from 0 to 90 degrees, and then continued at top, agreeably to the order of the secants in the first quadrant, from 90 degrees to the end of the semicircle.--This mode of arrangement, besides doing away with the necessity of finding the supplement of an arch when it exceeds 90 degrees, possesses the peculiar advantage of enabling the reader to take out the log. secant, or co-secant of any arch whatever, and conversely, at sight, as will appear evident by the following problems.

Note.-The log. co-secant of a given degree, or secant of a degree above 90, will be found in the same page with the next less degree in the first column under 0". at top, it being the first number in that column; and the log. co-secant of a given degree and minute, or secant of a degree and minute above 90, will be found on the same line with the next less minute in the column marked 60% at bottom of the Table,

PROBLEM 1. .

To find the Logarithmic Secant, and Co-secant of any given Arch,

expressed in Degrees, Minutes, and Seconds.

Rule.

If the given arch be comprised within the limits of the two last degrees of the first quadrant, that is, between 88 and 90 degrees, the Table will directly exhibit its corresponding log. secant or co-secant;-but when it falls without those limits, then find the log. secant, or co-secant, in the angle of meeting made by the given degree and next less fifth or tenth second at top, and the minutes in the left hand column; to which, add the proportional part corresponding to the odd seconds from the right hand column abreast of the angle of meeting, if a secánt be wanted, or a co-secant above 90 degrees ; but subtract that part when a co-secant is required, or & secant above 90 degrees ; and the sum, or difference, will be the log. secant or co-secant answering to the given arch.

Example 1.

Required the logarithmic secant, and co-secant, corresponding to 23:14:23! ?

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Given arch = 23:14:23. Corres, log: co-secant = 10.403866

Erample 2.

Required the logo secant, and co-secant, corresponding to 113:23:47"}

To find the Log. Secant:

113:23:45", ans. to which is 10.401121
Odd seconds
2 propor. part to which is

10.
Given arch = 113:23:47". Corres. log. secant =

10.401111

To find the Log. Co-secant :

113:23:45", ans. to which is 10.037260
Odd seconds
2 propor. part to which is +

2 Given arch = 113:23:47". Corres. log. co-secant=10.037262 Note.-In that part of the Table which lies between 10 and 80 degrees, the size of the page would not admit of the indices being prefixed to any other logs. than those contained in the first column of each page; nor, indeed, is it necessary that they should be, since they are uniformly the same as those contained in the said first column; viz., 10 for each log. secant or Co-secant.

PROBLEM II.

To find the Arch corresponding to a given Logarithmic Secant or

Co-secant:

RULE.

If the given log. secant, or co-secant, exceeds the secant of 88 degrees, viz., 11.457181, its corresponding arch will be found at frst sight in the Table; but if it be under that number, find the arch answering to the next less secant, or next greater co-secant; the difference between which and that given, being found in the column of proportional parts, abreast of the tabular log., will give a certain number of seconds, which, being added to the above-found arch, will give that required.

Example 1. Required the arch corresponding to the given log. secant 10. 235421 ?

Solution. The next less secant, in the Table, is 10: 235412, corresponding to which is 54:26:25"; the difference between this log. secant, and that given, is 9; answering to which, in the column of proportional parts abreast of the tabular log., is 3" ; which, being added to the abovefound arch, gives 54:26:48" for that required.

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