EXAMPLE XI. Required the true time of new Moon at LONDON, in April, 1764, New Style, and also whether there were an eclipse of the Sun or not at that time; and likewise the elements necessary for its protraction, if there were at that time an eclipse. By the Precepts. Mean time of Sun's mean Moon's mean Sun's mean distance from the node. D. H. M. S. 8. D. M. S. S. D. M. S. S. D. M. 3. March, 1764 2 8 55 368 220 0 10 13 35 2111 4-54 48 29 12 44 30 29 6 19 0 25 49 010 40 14 4 10 40 Arg. 1st eqa. Mean n. Moon 31 21 39 399 1 26 19 11 9 24 21 0 5 35 2 First equation 1 34 57 Second equa. 32 1 50 19 9 1 26 19 11 10 59 18 Third equation 31 22 25 30 9 20 27 111 10 59 18 0 5 35 2 4 37 Arg. 3d eq. Arg. 4th eq. 31 22 30 7 Sun from node. 05 35 2 True n. Moon 31 22 30 25 3 48 Equa. of days 31 22 26 35 The true time is April 1st, 10 hours, 26 minutes, 35 seconds in the morning, tabular time. The mean distance of the Sun at that time being only 5 degrees, 35 minutes and 2 seconds past the ascending node, the Sun was at that time eclipsed. Now proceed to find the elements necessary for its protraction. The true time being found as above. To find the Moon's horizontal parallax, or semidiameter of the Earth's disk as seen from the Moon. Enter Table 15th with the signs and degrees of the Moon's anomaly, (making proportion because the anomaly is in the table calculated only to every 6th degree,) and from it take out the Moon's horizontal parallax, which for the above time is 54 minutes and 53 seconds, answering to the anomaly of 11s. 9d. 24m. 21 seconds. To find the Sun's distance from the nearest solstice, namely, the beginning of Cancer, which is 3 signs, or 90 degrees from the beginning of Aries. It appears from Example 1st for calculating the Sun's true place, the calculation being made for the same time, that the Sun's longitude from the beginning of Aries, was then Os. 12d. 10m. 12 seconds; that is, the Sun's place was then in Aries, 12 degrees, 10m. 12 seconds, therefore from Subtract the Sun's longitude or place, S. D. M. S. 3000 12 10 12 Remains Sun's distance from the solstice, 2 17 49 48 which is equal to 77 degrees, 49 minutes, 48 seconds, each sign containing 30 degrees. To find the Sun's declination, enter Table 5th, with the signs and degrees of the Sun's true place, namely, 12 degrees, 10m. 12s. and making proportions for the 10m. 12 seconds, take out the Sun's declination, answering to his true place, and it will be found to be 4 degrees 49 minutes north. To find the Moon's latitude. This depends on her true distance from her ascending node, which is the same as the Sun's true distance from it at the time of new Moon, and is thereby found in Table 14th. But we have already found, [see Example,] by calculating the Sun's true place, that, at the true time of new Moon in April, 1764, the Sun's equated distance from the node was Os. 7d. 42m. 14s.; therefore enter Table 14th with the above equated distance, (making proportions for the minutes and seconds,) her true latitude will be found to be 40 minutes and 18 seconds north ascending. To find the Moon's horary motion from the Sun. With the Moon's anomaly, namely, 11s. 9d. 24m. 21s., enter Table 15th, and take out the Moon's horary motion, which, by making proportions in that table, will be found 30 minutes 22 seconds. Then with the Sun's anomaly, namely, 9s. 1d. 26m. 19s. (in the present case,) take out this horary motion, 2 minutes and 28 seconds from the same table; subtract the latter from the former, and the remainder will be the Moon's horary motion from the Sun; namely, 27 minutes and 54 seconds. To find the angle of the Moon's visible path with the ecliptic. This, in the projection of eclipses, may be always rated at 5 degrees and 35 minutes, without any sensible error. To find the semidiameters of the Sun and Moon. These are found in the same table, (15,) and by the same Argument, as their horary motions. In the present case, the Sun's anomaly gives his semidiameter 16 minutes and 6 seconds, and the Moon's anomaly gives her diameter 14 minutes and 27 seconds. To find the semidiameter of the penumbra. Add the Sun's semidiameter to the Moon's, and their sum will be the semidiameter of the penumbra; equal to 31 minutes and 3 seconds. Collect these elements together, that they may the more readily be found when they are wanted in the construction of this eclipse. Thus : D. H. M. S. 1st. The true time of new Moon in April, 1 10 30 25 7th. Angle of Moon's visible path with ecliptic, 5 35 0 8th. Sun's semidiameter, 0 16 6 14 57 31 3 9th. Moon's semidiameter, 10th. Semidiameter of the penumbra, To project an eclipse of the Sun geometrically.--Make a scale of any convenient length, A. C. and divide it into as many equal parts as the earth's semi-disk contains minutes of a degree; which at the time of the eclipse in April, 1764, was 54 minutes and 53 seconds ; then with the whole length of the scale as a radius, describe the semicircle A M B upon the centre C, which semicircle will represent the northern half of the earth's enlightened disk, as seen from the Sun. Upon the centre C, raise the straight line CH perpendicular to the diameter A CB, then will ACB be a part of the ecliptic, and CH its axis. Being provided with a good sector, open it to the radius C A in the line of chords, and taking from thence the chord of 23 degrees and 28 minutes in your compasses, set it off both ways from H to g, and g to h, in the periphery of the semi-disk, and draw the straight line g V h, in which the north pole of the disk will be always found. When the Sun is in Aries, Taurus, Gemini, Cancer, Leo and Virgo, the north pole of the earth is enlightened by the Sun; but while the Sun is in the other six signs, the south pole is enlightened. When the Sun is in Capricorn, Aquarius, Pisces, Aries, Taurus, and Gemini, the northern half of the earth's axis, C XII P lies to the right hand of the axis of the ecliptic, as seen from the Sun; and to the left hand, whilst the Sun is in the other 6 signs. Open* the sector, till the radius [or distance of the two 90s] of the signs be equal to the length of V h, and take the sign of the Sun's distance from the solstice, 77 degrees, 49 minutes, and 48 seconds, in your compasses from the line of sines, and set off that distance from V to P in the line of g V h, because the earth's axis lies to the right hand of the axis of the ecliptic in this case, [the Sun being in Aries,] and draw the straight line C XII P for the earth's axis, of which P is the north pole. If the earth's axis had lain to the left hand from the axis of the ecliptic, the distance VP would have been set off from V towards g. * To persons acquainted with Trigonometry, the angle contained between the earth's axis and that of the ecliptic, may be found more accurately by calculation. RULE. As radius is to the sine of the Sun's distance from the solstice, so is the tangent of the distance of the poles (23 degrees and 28 minutes) to the tangent of the angle contained by the axis. Then set off the chord of the angle from H to h, and join C.H, which will cut F Gin P. the place of the north pole. To draw the parallel of latitude of any given place, as suppose for London in this case, or the path of that place on the earth's enlightened disk, as seen from the Sun, from sunrise to sunset, take the following method. Subtract the latitude of London in this case, 51 degrees and 30 minutes, from 90 degrees, and the remainder, 38 degrees and 30 minutes, will be the co-latitude, which take in your compasses from the line of chords, making C A or CB the radius, and set it from h to the place where the earth's axis meets the periphery of the disk to VI and VI, and draw the occult or dotted line VI K VI, then from the points where this line meets the earth's disk, set off the chord of the Sun's declination, 4 degrees and 49 minutes, to E and F, and to E and G, and connect these points by the two occult lines, F XII G and ELE. Bisect L, K, XII, in K, and through the point K draw the black line VIK VI, then making CB the radius of a line of sines on the sector, take the co-latitude of London, (38+ degrees,) from the sines in your compasses, and set it both ways from K to VI and VI. These hours will be just in the edge of the disk at the equinoxes, but at no other time in the whole year. With the extent K VI taken into your compasses, set one foot in K in the black line below the occult one as a centre, and with the other foot describe the semicircle VI, 7, 8, 9, 10, &c. and divide it into 12 equal parts; then from these points of division draw the occult 7, p, 8, 0, 9, n, parallel to the earth's axis, C, XII, P. With the small extent K XII as a radius, describe the quadrantal arc XII f, and divide it into six equal parts, as XII, a, a, b, bc, cd, de, ef, and through the division points a, b, c, d, e, draw the occult lines VII, e, V, VIII, d, IV, IXC, III, X, b, II, and XI, a, I, all parallel to VI, K, VI, and meeting the former occult lines 7, p, 8, 0, &c. in the points VII, VIII, IX, X, XI, V, IV, III, II, and I, which points will mark the several situations of London on the earth's disk at these hours respectively, as seen from the Sun, and the elliptic curve VI, VII, VIII, &c. being drawn |
