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Remark.-The numbers in the general Traverse Table 'were computed agreeably to the following rule; viz.,

As radius is to the distance, so is the co-sine of the course to the difference of latitude; and so is the sine of the course to the departure.

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Example. Given the course 35 degrees, and the distance 147 miles ; to compute the difference of latitude and the departure.

1

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This Table contains the meridional parts answering to each degrée and minute of latitude from the equator to the poles ; the arguments of which are, the degrees at the top, and the minutes in the left or right hand marginal columns; under the former, and opposite to the latter, in any given latitude, will be found the meridional parts corresponding thereto, and conversely. Thus, if the latitude be 50:48:, the corresponding meridional parts will be 3549.8 miles.

Remark.-The Table of meridional parts may be computed by the fol

lowing rule ; viz.,

Find the logarithmic co-tangent less radius of half the complement of any latitude, and let it be esteemed as an integral number ; now, from the

common logarithm of this, subtract the constant log. 2. 101510 *, and the remainder will be the log. of the meridional parts answering to that latitude,

Example 1. Required the meridional parts corresponding to latitude 50:48?? Given lat. = 50:48! ; complement = 39:12' +2= 19:36', the half complement; hence, Half comp.=19:36. log. co-tangent less radius = .448448, the log. of which is

5.651712 Constant log.

2. 101510 Meridional parts corresponding to given lat. 3549.78=log.=3.550202

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Example 2.

Required the meridional parts corresponding to latitude 39:30! ? Given lat. = 89:30! ; comp. = 0:30! +2 = 0.15., the half complement; hence, Half comp. =

= 0:15. log. co-tangent less radius = 2.360180, the log. of which is

6.372945 Constant log.

2, 101510

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Meridional parts corresponding to given lat. 18682.49=log.=4.271435

TABLE XLIV.

The Mean Right Ascensions and Declinations of the principal fixed

Stars.

This Table contains the mean right ascensions and declinations of the principal fixed stars adapted to the beginning of the year 1824.—The stars are arranged in the Table according to the order of right ascension in which they respectively come to the meridian ; the annual variation, in right ascension and declination, is given in seconds and decimal parts of a second; that of the former being expressed in time, and that of the latter motion.

The stars marked t, have been taken from the Nautical Almanac for the year 1824.-The stars that have asterisks prefixed to them are those from which the moon's distance is computed in the Nautical Almanac for the purpose of finding the longitude at sea.

The measure of the arc of 1 minute (page 54,) is .0002908882; which being multiplied by 10000000000, (the radius of the Tables) produces 290.8882000000 ; and, this being multiplied by the modulus of the common logarithms, viz., 43429448190, gives 126.331140109823580 ;. the common log. of which is 2.101510, as above.

The places of the stars, as given in this Table, may be reduced to any future period by multiplying the annual variation by the number of years and parts of a year elapsed between the beginning of 1824, and such future period : the product of right ascension is to be added to the right ascensions of all the stars, except ß and 8, in Ursa Minor, from whose right ascensions it is to be subtracted : but the product of declination is to be applied, according to the sign prefixed to the annual variation in the Table, to the declinations of all the stars without any exception ;---thus, To find the right ascension and the declination of a Arietis, Jan. Ist, 1834.

22:37:33 N.

R. A. of a Arietis, per Tab. 1:57:16, and its dec.
+3".35

Ann, var.+ 17". 40.

Annual var.
Number of years

after 1824 =

Num. of yrs.

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Product . +33".5

+0.33".5

Prod. + 174".03 + 2.54!!

Rt. asc. of a Arietis, as req. 1:57"49o.5, and its declination 22:40:274 N.

Should the places of the stars be required for any period antecedent to 1824, it is evident that the products of right ascension and declination must be applied in a contrary manner.

The eighth column of this Table contains the true spherical distance and the approximate bearing between the stars therein contained and those preceding, or abreast of them on the same horizontal line; and the ninth, or last column of the page, the annual variation of that distance expressed in seconds and decimal parts of a second.-By means of the last column, the tabular distance may be reduced very readily to any future period, by multiplying the years and parts of a year between any such period and the epoch of the Table, by the annual variation of distance; the product being applied by addition or subtraction to the tabular distance, according as the sign may be affirmative or negative, the sum or difference will be the distance reduced to that period,

Example.
Required the distance between a Arietis and Aldebaran, Jan. Ist, 1844 ?

Tabular dist. between the two given stars = . 35:32:7"
Annual var. of distance. 0".02
Number of years after 1824 =

20

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Product

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0.40

True spherical distance between the two given stars, as required. .

35:326".60.

Logo

Remark. The true spherical distance between any two stars, whose right ascensions and declinations are known, may be computed by the following rule; viz.,

To twice the log. sine of half the difference of right ascension, in degrees add the log. sines of the polar distances of the objects; from half the sum of these three logs. subtract the log. sine of half the difference of the polar distances, and the remainder will be the log. tangent of an arch; the log. sine of which being subtracted from the half sum of the three logs., will leave the log. sine of half the true distance between the two given stars.

Example. Let it be required to compute the true spherical distance between a Arietis and Aldebaran, January 1, 1844. R. A. of a Arietis red. to 1844 = 1.5323!, and its dec. = 22:43:21" N. R. A. of Aldebaran red, to 1844 = 4.26.58.6, and its dec.=16. 11. 28 N. Difference of right ascension - 2.28.35'.6=

37:8:54"-2=18:34:27" 4:}18:34:27"

5. Twice the }19.0063060

{67. 16.39 {lio } 9.9649129 N. polar dist. of Aldebaran = {73. 48.32

9.9824236

Hall difference of R. A, in de

grees N. polar dist. of a Arietis =

Sum . . 38.9536425

Diff. of Polar dists. 6:31:53" Half=19.4768212 ... 19.4768212.5

Half diff. of ditto 3:15"561". Log S. 8.75561771

Arch 79:14:27".5826 log. tang. 10.7212035 Log. S. 9.9922976.3

Half the

req.

dist..

.

17:46:3”.4424. Log. S. 9.4845236.2

True spher. dist. between

the two given stars

35:32'6".$848 on Jan. I, 1844.

Now, by comparing this computed distance with that directly deduced from the Table, as in the preceding example, it will be seen that the difference amounts to very little more than the fifth part of a second in twenty. years; which evidently demonstrates that the tabular distances may be reduced to any subsequent period, for a considerable series of years, with all the accuracy that may be necessary for the common purposes of navigation.

Note.-The tabular distances will be found particularly useful in determining the latitude, at sea, by the altitudes of two stars, as will be shown hereafter,

TABLE XLV.

Acceleration of the Fired Stars; or to reduce Sidereal to Mean Solar

Time.

Observation has shown that the interval between any two consecutive transits of a fixed star over the same meridian is only 23.56"4!.09, whilst that of the sun is 24 hours :—the former is called a sidereal day, and the latter a solar day; the difference between those intervals is 3"55'.91, and which difference is called the acceleration of the fixed stars.

This acceleration is occasioned by the earth's annual motion round its orbit: and since that motion is from west to east at the mean rate of 59.8".3 of a degree each day; if, therefore, the sun and a fixed star be observed on any day to pass the meridian of a given place at the same instant, it will be found the next day when the star returns to the same meridian, that the sun will be nearly a degree short of it ; that is, the star will have gained 3:56.55 sidereal time, on the sun, or 3"55'.91 in mean solar time; and which amounts to one sidereal day in the course of a year :-for 3:55'.91 x 365d 52 48*48:= 23.56" 4::-hence in 365 days as measured by the transits of the sun over the same meridian, there are 366 days as measured by those of a fixed star.

Now, because of the earth’s equable or uniforın motion on its axis, any
given meridian will revolve from any particular star to the same star again
in
every

diurnal revolution of the earth, without the least perceptible difference of time shewn by a watch, or clock, that goes well :—and this presents us with an easy and infallible method of ascertaining the error and the rate of a watch or clock :-to do which we have only to observe the instant of the disappearance of any bright star, during several successive nights, behind some fixed object, as a chimney or corner of a house at a

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