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Example 2. Required the correction of the polar star's altitude in January 1854, the right ascension of the meridian being 5 hours and 13 minutes?
Correction of the polar star's altitude in Jan.1854, as required, 0:41:36". which differs but 8 seconds from the true result by spherical trigonometry, as will be shown hereafter ; and which evidently demonstrates that the column of annual variation may be safely employed in reducing the correction of altitude to any future period, for a long series of years, since the error in the space of thirty years only amounts to 8 seconds of a degree, which becomes insensible in determining the latitude at sea.
The corrections of altitude contained in the present Table were computed in conformity with the following principles :
Since to an observer placed at the equator, the poles of the world will appear to be posited in the horizon, the polar star will, to such observer, apparently revolve round the north celestial pole in its diurnal motion round its orbit. In this apparent revolution round the celestial pole, the star's meridional or greatest altitude above the horizon will be always equal to its distance from that pole; which will ever take place, when the right ascension of the meridian is equal to the right ascension of the star. In six hours after this, the star will be seen in the horizon, west of the pole; in six hours more it will be depressed beneath the horizon (on the meridian below the pole), the angle of depression being equal to its polar distance; in six hours after, it will be seen in the horizon east of the pole; and in șix hours more, it will be seen again on the meridian above the pole: allowance being made, in each case, for its daily acceleration.
Now, since the north celestial pole represents a fixed point in the heavens, and that the star 'apparently moves round it in an uniform manner, making determinable angles with the meridian; it is, therefore, easy to compute what altitude the star will have, as seen from the equator, in every part of its orbit; for, in this computation, we have a spherical triangle to work in, whose three sides are expressed by the complement of the latitude, the complement of the polar star's altitude, and the complement
of its declination; in which there are given two sides and the included angle to find the third side; viz., the star's co-declination or polar distance and the complement of the latitude, with the comprehended angle, equal to the star's distance from the meridian, to find the star's co-altitude; the difference between which and 90 degrees will be the correction of altitude, or the difference of altitude between the polar star and the north celestial pole, as seen from the equator.
In January 1854, the mean right ascension of the north polar star will be 1!5523?, and its polar distance 1:28:57 ; now, admitting the right ascension of the meridian to be 5:13", the correction of the polar star's altitude, as seen from the equator, is required ?
Cor. of polar star's alt. in Jan. 1854 = 0:41:28" Now, by comparing this result with that shown in Example 2 (page 18), it will be seen that the correction of altitude, deduced directly from the Table, may be reduced to any period subsequent to 1824, without its being affected by any error of sufficient magnitude to endanger the interest of the mariner in any respect whatever.
Note.For further information on this subject, the reader is referred to the author's Treatise on the Sidereal and Planetary Parts of Nautical Astronomy, page 144 to 156.
Correction of the Latitude deduced from the preceding Table. Although the latitude deduced from Table X. will be always sufficiently correct for most nautical purposes, yet, since observation has shown that it will be something less than the truth in places distant from the equator, the present Table has been computed; which contains the number of minutes and seconds that the latitude, so deduced, will be less than what would result from actual observation at every tenth or fifth degree from the equator, to within five degrees of the north pole of the world.
The elements of this Table are, the approximate latitude, deduced from Table X., at top, and the right ascension of the meridian in the left or right-hand column; in the angle of meeting will be found the corresponding correction, which is always to be applied by addition to the approximate latitude. Hence, if the approximate latitude be 50 degrees, and the right ascension of the meridian 6:40", the corresponding correction will be 1:38. additive.
Remark.-Since the corrections of altitude in Table X. have been computed on the assumption that the motions of the polar star were witnessed from the equator, they ought, therefore, to show what altitude that star will have at any given time, in north latitude, when applied to such latitude with a contrary sign to that expressed in the Table ; this, however, is not the case; because when the altitude of the polar star is computed by spherical trigonometry, or otherwise, it will always prove to be something less than that immediately deduced from Table X. : it is this difference, then, that becomes the correction of latitude in Table XI., and which is very easily determined, as may be seen in the following
Example. Let the right ascension of the meridian in January 1824 be 6.40, and the latitude 60 degrees north ; required the true altitude of the polar star, and thence the correction of latitude ?
Latitude or elevation of the pole
60:0: 0 north.
Altitude of polar star, per Table X. = . : 60:7:41"
Now, to compute the true altitude of the polar star, on spherical principles, at the given time and place, we may either proceed as in last example, or, more readily, as follows :
Difference 28.22.12 Nat. cos. 879897
013106 Log.=4. 117457
Star's true altitude 60:5:16 Nat. sine 866791
0:2:25"; which, therefore, is the correction
Note.- The correction of latitude, thus found, differs 4 seconds from that given in Table XI.: this difference is owing to the star's apparent polar distance having been taken, inadvertently, from the Nautical Almanac of 1824, instead of its mean polar distance; but since this can only lead to a trifling difference, and not to any error, it was not, therefore, deemed necessary to alter or recompute the Table.
This Table may be used for the purpose of finding the approximate time of transit of a fixed star, when a Nautical Almanac is not at hand; it
may also be employed in finding the right ascension of the meridian, or midheaven, when the latitude is to be determined by an altitude of the north polar star : for, if to the sun's right ascension, as given in this Table, the apparent time be added, the sum (rejecting 24 hours if necessary) will be the right ascension of the meridian, sufficiently near the truth for determining the latitude,
Equations to equal Altitudes.-FIRST PART. The arguments of this Table are, the interval between the observations at top or bottom, and the latitude in either of the side columns ; in the angle of meeting stands the corresponding equation, expressed in seconds and thirds : hence the equation to inteival 6 hours 40 minutes and latitude 50 degrees, is 15 seconds and 33 thirds.
The equations in this Table were computed by the following rule, viz. :
To the log. co-tangent of the latitude, add the log. sine of half the interval in degrees; the proportional log. of the whole interval in time (esteemed as minutes and seconds), and the constant log. 8. 8239;* the sum of these four logs., rejecting 29 from the index, will be the proportional log. of the corresponding equation in minutes and seconds, which are to be considered as seconds and thirds.
Let the latitude be 50 degrees, and the interval between the observed equal altitudes of the sun 4 hours ; required the corresponding equation?
• 50:0:0". Log. co-tang. 9. 9238
14?18" Propor. log. 1.0999 The equations in the abovementioned Table were computed by Mrs. T. Kerigan.
Equations to equal Altitudes.-PART SECOND. In this Table, the interval between the observations is marked at top or bottom, and the sun's declination in the left or right-hand margin; under or over the former, and opposite to the latter, stands the corresponding equation, expressed in seconds and thirds : thus, the equation answering to 6 hours 40 minutes, and declination 18:30!, is 2 seconds and 48 thirds.
The equations contained in this Table were computed as follows, viz. :
To the log. co-tangent of the declination, add the log. tang. of half the interval in degrees; the proportional log. of the whole interval in time (esteemed as minutes and seconds), and the constant log. 8. 8239;t the
* † The arithmetical complement of 12 hours considered as minutes.