15th with the Moon's mean anomaly, (at the time of the above full Moon,) namely, 9 signs, 2 degrees, 42 min utes, 42 seconds, and with it take out her horizontal parallax, which, by making the requisite proportions will be found to be 57 minutes and 23 seconds. To find the semi-diameters of the Sun and Moon, enter Table 15th, with their respective anomalies, the Sun's being 10 signs, 7 degrees, 27 minutes, 45 seconds, and the Moon's 9 signs, 2 degrees, 42 minutes, 42 seconds, (in this case,) and with these take out their respective semidiameters, the Sun's 15 minutes and 56 seconds, and the Moon's 15 minutes and 38 seconds. To find the semidiameter of the earth's shadow at the Moon, add the Sun's horizontal parallax, which is always 9 seconds, to the Moon's, which in the present case is 57 minutes and 23 seconds, the sum will be 57 minutes and 32 seconds; from which subtract the Sun's semidiameter, 15 minutes and 56 seconds, and there will remain 41 minutes and 36 seconds for the semidiameter of that part of the earth's shadow, which the Moon then passes through. To find the Moon's latitude. Find the Sun's true distance from the Moon's ascending node, (as already taught,) in the first Example for finding the Sun's true place, at the true time of full Moon, and this distance increased by 6 signs, will be the Moon's true distance from the same node, and consequently the argument for finding her true latitude. The Sun's mean distance from the ascending node was at the true time of full Moon, O signs, 4 degrees, 49 ininutes, 35 seconds; but it appears by the Example that the true time thereof, was 6 hours, 33 minutes and 38 seconds sooner than the mean time, and therefore we must subtract the Sun's motion from the node during this interval, from the above mean distance 0 signs, 4 degrees, 49 minutes and 35 seconds, in order to have his mean distance from the node at the time of true full Moon. Then, to this apply the equation of his mean distance from the node, found in Table 13th, by his mean anomaly, 10 signs, 7 degrees, 27 minutes, 45 seconds; and lastly, add6 signs, and the Moon's true distance from the ascending node, will be found as follows: Sun from node at mean time of full Moon, 6 hours, His motion from the node in < 33 minutes,. 38 seconds, Subtract the sum, Remains his mean dist. at true full Moon, Sun's true distance from the node, Moon's true distance from the node, S. D. M. S. 04 49 35 15 35 1 26 2 17 3 04 32 32 138 0 06 10 32 6000 6 6 10 32 And it is the argument used to find her true latitude at that time. Therefore, with this argument, enter Table 14th, making proportions between the latitudes belonging to the 6th and 7th degree of the argument for the 10 minutes and 32 seconds, and it will give 32 minutes and 21 seconds for the Moon's true latitude, which appears by the table to be south descending. To find the angle of the Moon's visible path with the ecliptic. This may be always stated at 5 degrees and 35 minutes without any error of consequence, in the projection of either solar or lunar eclipses. To find the Moon's true horary motion from the Sun. With their respective anomalies, take out their horary motions from Table 15th; and the Sun's horary motion, subtracted from the Moon's, leaves remaining the Moon's true horary motion from the Sun, in the present case, 30 minutes and 52 seconds. The above elements are collected for use. D. H. M. S. 1st. The time of full Moon in May, 1762, 83 50 50 2d. Moon's horizontal parallax, D. ..... 057 23 32 21 6th. Moon's true latitude south descending, 7th. Angle of Moon's visible path with ecliptic, 5 35 0 0 30 52 8th. Moon's true horary motion from Sun, These elements being found for the construction of the Moon's eclipse in May, 1762, proceed as follows :Make a scale of any convenient length, as W X and divide it into 60 equal parts, each part standing for a minute of a degree. Draw the right line A CB for part of the ecliptic, and C A perpendicular thereto for the southern part of its axis, (the Moon having south latitude.) Add the semidiameters of the Moon and earth's shadow together, which in this case, make 57 minutes and 14 seconds; and take this from the scale in your compasses, and setting one foot in the point Cas a centre, with the other describe the semicircle S D B, in one point of which the Moon's centre will be at the beginning of the eclipse, and the other at the end. Take the semidiameter of the earth's shadow, (41 minutes and 36 seconds,) in your compasses from the scale, and setting one foot in the centre C, with the other describe the semicircle KLM for the southern half of the earth's shadow, because the Moon's latitude is south in this eclipse. Make C D equal to the radius of a line of chords on the sector, and set off the angle of the Moon's visible path with the ecliptic, (5 degrees and 35 minutes,) from D to E, and draw the right line CFE for the southern half of the axis of the moon's orbit, lying to the right hand from the axis of the ecliptic C A, because the Moon's latitude is south descending in this eclipse. It would have been the same way on the other side of the ecliptic, if her latitude had been north descending, but contrary in both cases, if her latitude had been either north or south ascending. Bisect the angle ACE by the right line Cg, in which the true equal time of opposition of the Sun and Moon falls, as found from the tables. Take the Moon's latitude, 32 minutes and 21 seconds, from the scale in your compasses, and set it from C to G in the line CGg, and through the point g; at right angles to CFE, draw the right line, PHGFN, for the path of the Moon's centre. Then F shall be the point in the earth's shadow, where the Moon's centre is at the middle of the eclipse; G, the point where her centre is at the tabular time of her being full; and H, the point where her centre is at the instant of her ecliptical opposition. Take the moon's horary motion from the sun, (30 minutes and 52 seconds,) in your compasses from the scale W X, and with that extent make marks along the line of the Moon's path PGN; then divide each space from mark to mark, into 60 equal parts, or horary minutes, and set the hours to the proper dots in such manner, that the dots signifying the instant of full Moon, namely, 50 minutes and 50 seconds after 3 in the morning, may be in the point G, where the line of the Moon's path enters the line that bisects the angle D CE. Take the Moon's semidiameter, 15 minutes and 38 seconds, in your compasses from the scale, and with that extent, as a radius upon the points NF and Pas centres, describe the circle Q for the Moon at the beginning of the eclipse, when she touches the earth's shadow at V; the circle R for the Moon at the middle, and the circle S for the Moon at the end of the eclipse, just leaving the earth's shadow at W. The point N denotes the instant when the eclipse begins, namely, at 15 minutes and 10 seconds after two in the morning; the point F, the middle of the eclipse, at 47 minutes and 45 seconds after three; and the point P, the end of the eclipse, at 18 minutes after five: at the greatest obscuration, the Moon was 10 digits eclipsed. The Moon's diameter, (as well as the Sun's,) is sup posed to be divided into 12 equal parts, (called digits, and so many of these parts as are darkened by the earth's shadow, so many digits is the Moon eclipsed. All that the Moon is eclipsed above 12 digits, shows how far the shadow of the earth is over the body of the Moon, on that edge to which she is nearest, at the middle of the eclipse. RULES FOR FINDING THE TIME WHEN A LUNAR ECLIPSE BEGINS AND ENDS, AND ALSO THE DIGITS ECLIPSED ARITHMETICALLY. To find the parts deficient. Rule. To the Moon's horizontal parallax add the Sun's horizontal parallax, and from the sum subtract the Sun's semidiameter, and the remainder is the semidiameter of the earth's shadow. To the semidiameter of the earth's shadow add the Moon's semidiameter, and from the sum subtract the Moon's latitude, and the remainder is the parts deficient. To find the digits eclipsed. Rule. As the Moon's semidiameter is to six digits, so are the parts deficient to the digits eclipsed. To find the scruples of incidence. Rule. Reduce the sum of the Moon and earth's shadow to seconds, and also the Moon's latitude--find their sum and difference-multiply the sum by the difference, and the square root of the product will be the scrupies of incidence in seconds. To find the time of half duration. Rule. As the hourly motion of the Moon from the Sun is to one hour, so are the scruples of incidence to the time of half the duration of the eclipse. To the time of the middle of the eclipse, found by the tables, add the time of half duration, and the sum will be the time when the eclipse ends. Subtract it, and the remainder will show the time it commences. To calculate the time of total darkness, in total eclipses of the Moon. Rule. Subtract the Moon's semidiameter from that |