of horizontal parallax. The proportional part for the excess of the given above the next less tabular altitude, is contained in the right-hand column of each page; and that answering to the seconds of parallax is given in the intermediate part of the Table. This correction is to be taken out of the Table in the following manner; viz. Enter the Table with the moon's apparent altitude in the left-hand column, or the altitude next less if there be any odd minutes; opposite to which, and under the minutes of the moon's horizontal parallax, will be found the approximate correction. Enter the compartment of the "Proportional parts to seconds of parallax," abreast of the approximate correction, with the tenths of seconds of the moon's horizontal parallax in the vertical column, and the units at the top; in the angle of meeting will be found the proportional part for seconds, which add to the approximate correction. Then, Enter the last or right-hand column of the page, abreast of the approximate correction or nearly so, and find the proportional part corresponding to the odd minutes of altitude. Now, this being added to or subtracted from the approximate correction, according to its sign, will leave the true correction of the moon's apparent altitude.. And since the apparent altitude of a celestial object is depressed by parallax and raised by refraction, and the lunar parallax being always greater than the refraction to the same altitude, it hence follows that the correction, thus deduced, is always to be applied by addition to the moon's apparent altitude. Example 1. Let the moon's apparent altitude be 8:38, and her horizontal parallax 57:46; required the corresponding correction? Correction to alt. 8:30 and horiz. parallax 57:0% is 50:14" Do. to 8 min. of alt. (8 x 0.5=4".0) = + 0. 4 Correction of the moon's apparent altitude, as required 51 4" Example 2. Let the moon's apparent altitude be 33:16, and her horizontal parallax 59:34"; required the corresponding correction? Correction to alt. 33:10 and horiz. parallax 59'0" is 47:56% + 28 3 Correction of the moon's apparent altitude, as required 48:21" TABLE XIX. To reduce the True Allitudes of the Sun, Moon, Stars, and Planets, to their Apparent Altitudes. This Table is particularly useful in that method of finding the longitude by lunar observations, where the distance only is given, and where, of course, the altitudes of the objects must be obtained by computation. : The Table consists of two pages, each page being divided into two parts the left-hand part contains four columns; the first of which comprehends the true altitude of the sun or star; the second the reduction of the sun's true altitude; the third the reduction of a star's true altitude; and the fourth the common difference of those reductions to 1 minute of altitude for sun or star. The other part of the Table is appropriated to the moon; in which the true altitude of that object is given in the column marked "Moon's true altitude," and her horizontal parallax at top or bottom; the two last or right hand columns of each page contain the difference to 1 minute of altitude, and 1 second of parallax respectively; by means of which the reduction may be easily taken out to minutes of altitude and seconds of horizontal parallax. The first part of the Table is to be entered with the sun's or star's true altitude (or the altitude next less when there are any odd minutes, as there generally will be,) in the left-hand column; abreast of which, in the proper column, will be found the approximate reduction; from which let the product of the difference to 1 minute by the excess of the odd minutes above the tabular altitude, be subtracted, and the remainder will be the true ⚫ reduction of altitude for sun or står. Example 1. Let the true altitude of the sun's centre be 8:15; required the reduction to apparent altitude? Correction corresponding to altitude 8 degrees 6:15 Cor. for min. of alt.; viz. diff. tol min. of alt.=0". 70x 15'10".5=-10 Given the true altitude of a star 19:45; the reduction to apparent altitude is required? Cor. for min. of alt. ; viz. diff. to 1 min. of alt.=0′′. 15 × 45:= 6′′.75=-7 Required reduction = 2:37 The reduction of the moon's true altitude is to be taken from the second part of the Table, by entering that part with the true altitude in the proper column (or the altitude next less when there are any odd minutes) and the horizontal parallax at top or bottom; in the angle of meeting will be found a correction; to which apply the product of the difference to 1 minute by the excess of the odd minutes above the tabular altitude by subtraction, and the product of the difference to 1 second by the odd seconds of parallax by addition and the true reduction will be obtained, as may be seen in the following Example. Let the true altitude of the moon's centre be 29:13, and her horizontal parallax 58:37"; required the corresponding reduction to apparent altitude? Correc. corres. to alt. 29 degs., and horiz. parallax 58; = Required reduction 49:22" 5".3=-5 33′′.3=+33 49:50% Remark. The reduction of the sun's true altitude is obtained by increasing that altitude by the difference between the refraction and parallax corresponding thereto : then, the difference between the refraction and parallax answering to that augmented altitude, will be the reduction of the true altitude. Example. Let the true altitude of the sun's centre be 5 degrees; required the reduction to apparent altitude? The correction for reducing a star's true altitude to its apparent, is obtained in the same manner, omitting what relates to parallax. Thus, if the true altitude of a star be 8 degrees, and the corresponding refraction 6:29, their şum, viz., 8:6429" will be the augmented altitude; the refraction answering to this is 6:24", which, therefore, is the reduction of the true to the apparent altitude of the star. The correction for reducing the true altitude of the moon to the apparent, is found by diminishing the true altitude by the difference between the parallax and refraction answering thereto; then the difference between the parallax and refraction corresponding to the altitude so diminished, will be the reduction of the true to the apparent altitude. As thus :— Let the true altitude of the moon's centre be 10 degrees, and her horizontal parallax 57 minutes; required the reduction to apparent altitude? Since the solution of the Problem for finding the longitude at sea, by celestial observation, is very considerably abridged by the introduction of an auxiliary angle into the operation, the true central distance being hence readily determined to the nearest second of a degree by the simple addition of five natural versed sines; this Table has, therefore, been computed; and to render it as convenient as possible, it is extended to every tenth minute of the moon's apparent altitude, and to each minute of her horizontal parallax; with proportional parts adapted to the intermediate minutes of altitude, and to the seconds of horizontal parallax. This Table was calculated in the following manner :— To the moon's apparent altitude apply the correction from Table XVIII., and the sum will be her true altitude; from the log. cosine of which (the index being augmented by 10) subtract the log. cosine of her apparent altitude, and the remainder will be a log., which, being diminished by the constant log. .300910,* will give the logarithmic cosine of the auxiliary angle. Example. Let the moon's apparent altitude be 4 degrees, and her horizontal parallax 55 minutes; required the corresponding auxiliary angle? Moon's apparent altitude. 4: 0 0 Log. cosine 9.998941 Auxiliary angle, as required 60: 1:21"= Log, cosine The correction of the auxiliary angle for the sun's or star's apparent altitude, given at the bottom of each page of the Table, was computed by the following rule-viz. From the log. cosine of the sun's or star's true altitude subtract the log. cosine of the apparent altitude, and find the difference between the remainder and the constant log. .000120.† Now this difference, being subtracted from the log. cosine of 60 degrees, will leave the log. cosine of an arch; the difference between which and 60 degrees will be the correction of the auxiliary angle depending on the apparent altitude of the sun or star. Example. Let the sun's or star's apparent altitude be 3 degrees; required the correction of the auxiliary angle? This is the log. secant, less radius, of 60 degrees diminished by .000120, the difference between the log. cosines of a star's true and apparent altitude betwixt 30 and 90 degrees. This is the difference between the log. cosines of a star's true and apparent altitude, between 30 and 90 degrees. |