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Example 3.

Required the arch corresponding to the given log. tang. 10. 155436 ?

Solution.--The next less log. tangent in the Table is 10. 155423, corresponding to which is 55:2:25"; the difference between this log. tangent and that given, is 13; answering to which, in the column of proportional parts abreast of the tabular log., is 3!; which, being added to the abovefound arch, gives 55:2:28" for that required.

Example 4. Required the arch corresponding to the given log. co-tang. 9. 792048 ?

Solution. The next greater log. co-tangent in the Table is 9.792057, corresponding to which is 58:13?15”; the difference between this log. cotangent and that given, is 9; answering to which, in the column of proportional parts abreast of the tabular log., is 2”; which, being added to the above-found arch, gives 58:13! 17" for that required.

Remark.

The arch corresponding to a given log. tangent may be found by means of a Table of log.sines, in the following manner ; viz.,

Find the natural number corresponding to twice the given log. tangent, rejecting the index, to which add the radius, and find the common log. of the sum ; now, half this log. will be the log. secant, less radius, of the required arch; and which, being subtracted from the given log. tangent, will leave the log. sine corresponding to that arch.

Example. Let the given log. tangent be 10. 084153; required the-arch corresponding thereto by a table of log. sines? Given log.tang. .084153 x 2 = .168306, Nat. num. = 1.473349

to which add the radius = 1.000000

the the

Sum =

2.473349, common log. of which is 0.393286; the half of this is 0. 196643,

secant, less radius of the required arch. Given log. tangent =

10.084153

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Corresponding log. sine =

9.887510, answering to which is 50:31: ; and which, therefore, is the required arch corresponding to the given log. tangent.

The arch corresponding to a given log. tangent may also be found in the following manner, which, it is presumed, will prove both interesting and instructive to the student in this department of science.

Find the natural tangent, that is, the natural number corresponding to the given log. tangent, to the square of which add the square of the radius; extract the square root of the sum, and it will be the natural secant corresponding to the required arch; then, say, as the natural secant, thus found, is to the natural tangent, so is the radius to the natural sine : now, the degrees, &c. answering to this in the Table of Natural Sines, will be the arch required, or that corresponding to the given log, tangent.

Example.

Let the given log. tangent be 10:084153; it is required to find the arch corresponding thereto by a Table of Natural Sines ?

Solution.-Given log. tangent = .084153; the natural number corresponding to this is 1. 213816; which, therefore, is the natural tangent answering to the given log. tangent.

B

E

In the annexed diagram, let B C represent the natural tangent = 1.213816, and A B the radius F = 1.000000. Now, since the base and perpendicular of the right-angled triangle ABC are known, the hypothenuse or secant AC may be determined by Euclid, Book I., Prop. 47. Hence v BC = 1.2138162 + A B2 = 1.0000002 = AC= 1.572689, the natural secant corresponding to the given log. tangent. Having thus found the natural secant A C, the natural sine D E may be found agreeably to the principles of similar triangles, as demonstrated in Euclid, Book VI., Prop. 4; for, as the natural secant A C is to the natural tangent B C, so is the radius AD = AB to the natural sine DE: hence,

As AC 1.572689 : BC 1.213816 :: AD 1.000000 : DE= 771810, the corresponding natural sine; now, the arch answering to this, in the Table of Natural Sines, is 50:31! ; which, therefore, is the arch corresponding to the given log. tangent, as required.

Note.-The Table of log. tangents may be very readily deduced from Tables XXXV. and XXXVI., as thus :-To the log. secant of any given arch, add its log. sine; and the sum, abating 10 in the index, will be the log. tangent of that arch; the difference between which and twice the radius, will be its co-tangent.

Example.

Required the log. tangent, and co-tangent, of 25:27:35".?

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The Table of log. tangents may also be computed in the following manner; viz.,

From the log. sine of the given arch, the index being increased by 10, subtract its log. co-sine, and the remainder will be the log. tangent of that arch; the difference between which and twice the radius, will be its log. co-tangent.

Example.

Required the log. tangent, and co-tangent, of 32:39:40". ?

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To reduce the Time of the Moon's Passage over the Meridian of

Greenwich, to the Time of her Passage over any other Meridian.

The daily retardation of the moon's passage over the meridian, given at the top of the Table, signifies the difference between two successive transits of that object over the same meridian, diminished by 24 hours; as thus: the moon's passage over the meridian of Greenwich, July 22d, 1824, is 21.7", and that on the following day 22.979; the interval of time between these two transits is 25:2", in which interval it is evident that the moon is 1:27 later in coming to the meridian; and which, therefore, is the daily retardation of her passage over the meridian.

This Table contains the proportional part corresponding to that retardation and any given interval of time or longitude; in computing which, it is easy to perceive that the proportion was,

As 24 hours, augmented by the daily retardation of the moon's transit over the meridian, are to the said daily retardation of transit, so is any given interval of time, or longitude, to the corresponding proportional part of such retardation. The operation was performed by proportional logs., as in the following

Example.

Let the daily retardation of the moon's transit over the meridian be 1.2; required the proportional part corresponding thereto, and 9:40% of time, or 145 degrees of longitude ?

As 24 hours + 1?2? (daily retard.)= 252 2" Ar. comp. pro. log. 9. 1432 Is to daily retardation of transit = 1. 2 Propor. log. 0.4629 So is given interval of time = . 9.40 Propor. log. • 1. 2700

Το corresponding proportional part = 23°57'= Pro. log.= 0. 8761; and in this manner were all the numbers in the Table obtained.

The corrections or proportional parts contained in this Table are expressed in minutes and seconds, and are extended to every twentieth minute of time, or fifth degree of longitude : these are to be taken out and applied to the time of the moon's transit, as given in the Nautical Almanac, in the following manner :

Find, in page VI. of the month in the Nautical Almanac, the difference between the moon's transit on the given day (reckoned astronomically) and that on the day following, if the longitude be west ; but on the day preceding, if it be east. With this difference enter the Table at the top, and the given time in the left-hand, or the longitude in the right-hand column ; in the angle of meeting will be found a correction, which, being applied by addition to the time of transit on the given day, if the longitude be west, but by subtraction, if east, the sum, or difference, will be the reduced time of transit.

Example 1.

Required the apparent time of the moon's passage over a meridian SO degrees west of Greenwich, July 22, 1824 ?

21! 70:

Mn's pas. over mer. of Greenw.on giv. day is 2127"

Ditto on the day following = 22.9

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Apparent time of the moon's transit over the given meridian = 21:20:13!

Example 2. Required the apparent time of the moon's passage over a meridian 120 degrees east of Greenwich, August 20th, 1824?

20.54" 0:

Mn's pas. over mer, of Greenw.on giv, day is 20:54

Ditto on the day preceding=19.54

Retardation of the moon's transit =

1? 0; ans. to which

and 120 degs.is - 19.12

Apparent time of the moon's transit over the given meridian = 20:34:48:

Table XXXIX.

Correction to be applied to the Time of the Moon's Transit in finding

the Time of High Water.

Since the moon is the principal agent in raising the tides, it might be expected that the time of high water would take place at the moment of her passage over the meridian; but observation has shown that this is not the case, and that the tide does not cease flowing for some time after : for, since the attractive influence of the moon is only diminished, and not entirely destroyed, in passing the meridian of any place, the ascending impulse previously communicated to the waters at that place must, therefore, continue to act for some time after the moon's meridional passage.

The ascending impulse, thus imparted to the waters, ought to cause the time of the highest tide to be about 80 minutes after the moon's passage over the meridian ; but owing to the disturbing force of the sun, the actual time of high water differs, at times, very considerably from that period.

The effect of the moon in raising the tides exceeds that of the sun in the ratio of about 24 to l; but this effect is far from being uniform : for, since the moon's distance from the earth bears a very sensible proportion to the diameter of this planet, and since she is constantly changing that distance, (being sometimes nearer, and at other times more remote in every

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