CHAP. III. cor The rection is The moon's longitude, as just computed, will be The difference will be Therefore, at the time for which these longitudes were computed, the moon will be past her full by 14" of are: to correct the time, then, we must find how much time will be required for the moon to gain 14"; which, by the problem of the couriers, is The unit for t is one hour, and the denominator of the fracsubtractive tion is the difference of the hourly motions of the sun and because the moon, as determined by the tables; the result is 29 seconds moon is past of time to be subtracted. conjunction, otherwise it The Greenwich time will be, 1851, July 12d. 19h. 15m. Os. Subtract would be ad 29 12 19 14 31 But the time given by the lunation table was 19 h. 14 m., differing only 31 seconds from the true time; the approximate and true time, however, do not commonly coincide as near as this: if they did, none but the most rigid astronomer would use the lunar tables for the time of conjunction or opposition. To be very exact we must correct the moon's latitude for what it will vary in 31 seconds; that is, in this case, increase it 4.5. The moon's latitude, at the time of full moon, is, therefore, 37' 17".4. We have now all the elements necessary for computing the eclipse, or, at least, we have all the materials for finding them, and, for convenience, we collect the elements together: 1. True time of full moon, July, 3. Angle of the moon's visible path with the ecliptic,* d. h. m. S. 12 19 14 31 39' 39" 5 38 26 *This is the angle of the base of a right-angled triangle, whose base 15 4 54 43 6. Moon's semidiameter, 7. Semidiameter of and earth's shadow, Whenever the moon's latitude, at the time of full moon, is less than this last element, the moon must be more or less eclipsed; and it is by computing and comparing these two elements, viz., 4 and 7, that all doubtful cases are decided. TO CONSTRUCT A LUNAR ECLIPSE. CHAP. III. When the very little la From any convenient scale of equal parts, take the 7th element in your dividers (54 43) = 543, and from C, as a center moon has with that distance, describe the semicircle BDHE (Fig. 55). titude deTake CA= the 2d element, and describe the semidiameter scribe a full of the earth's shadow. From C the center of the shadow, draw Cn at right angles to B E the ecliptic, above BE when south latithe latitude is north, as in the present example, but below, tude, deif south. Fig. 55. circle. When large scribe only the lower semicircle. Take the moon's latitude from the scale of equal parts, and set it off from C to n. Through n draw Dn H, the moon's path, so that the line shall incline to B E, the ecliptic, by an angle equal to the 3d element. Conceive the moon's is the hourly motion of the moon from the sun (28′ 31′′), and the perpendicular, the moon's hourly motion in latitude (2' 49"). See page 266, figure 54. CHAP. III center to run along the line from D to H, and from C draw C'm perpendicular to DH. When the moon is ascending in her orbit, DH must incline the other way, and Cm must lie on the other side of Cn. The eclipse commences when the moon arrives at D. It is the time of full moon when it arrives at n; the greatest obscuration occurs when it arrives at m, and the eclipse ends at H. The duration is the time employed in passing from D to H; and to find the duration apply DH to the scale, and thus The 5th ele- find its measure. Divide this measure by the 5th element, ment is the and we shall have the hours and decimal parts of an hour in motion the duration. Also apply Dn to the scale and find its measure. Divide this measure by the 5th element, for the time of describing Dn, also divide the measure nH for the time of describing nH. moon's angu. lar from the sun. The time of describing Dn, subtracted from the time of full moon, will give the time of the beginning of the eclipse; and the time of describing n H, added to the time of full moon, will give the time when the eclipse ends. With lunar eclipses the time of greatest obscuration is the instant of the middle of the eclipse, provided the moon's motion from the sun, for this short period of time, is taken as uniform, as it may be without sensible error. In reference to this example Dn=36′ and nH=44′. These distances, divided by 28' 31", give 1 h.14 m. 16 s. for the time of describing Dn, and 1h. 32 m. 40 s. for n H: whole time, or duration, 2 h. 27 m. 20 s. That is, in 1851, July 12 d. 18 h. 0 m. 15 s. mean astronomical time, the eclipse begins; but this time corresponds with July 13, at 6 h. 0 m. in the morning; and at this time, the sun will be above the horizon of Greenwich, and, of course, the full moon, which is always opposite to the sun, will be below the horizon, and the eclipse will be invisible to all Europe. CHAP. III. Visible in In the United States, however, the eclipse will be visible; the U.S. for, at these points of absolute time, the sun will not have risen nor the moon have gone down; but, to be more definite, we demand the times of the beginning, middle, and end of the eclipse, as seen from Albany, N. Y. To answer this demand, all we have to do, is to subtract from the Greenwich time the difference of meridians between the two places, which, in this case, is 4h. 55 m.; and the result is, In the same manner we would compute the time for any other place. The quan eclipse how For the quantity of the eclipse we take the portion of the moon's diameter, which is immersed in the shadow, tity of the at the time of greatest obscuration, and compare it with found. the whole diameter of the moon; and in the present example, we perceive, that more than half of the diameter is eclipsed-about 7 digits when the whole is called 12, or 0.6 when the diameter is 1. All these results, however, except the time of full moon, are approximate, because we cannot, nor do we pretend to construct to accuracy; but any mathematician can obtain accurate results by means of the triangles D CH and Cnm, and the relative motion of the moon from the sun. of the dura In the right-angled triangle C'nm, right-angled at m, Cn The exact is the latitude of the moon = 37′ 17′′.4= 2237′′.4, and the computation angle n C'm = 5° 38′ 26′′; with these data we find mn=tion_of_the 220", and C'm = 2212′′ In the right-angled triangle C Dm, or its equal CmH, we have Or, Or, - Cm2; mH2 = (CH+Cm) (CH— Cm). CH is the 7th element =3283", and Cm 2212".6. = Therefore, m H= √ (5495) (1071)=2426′′ This eclipse. CHAP. III. divided by 1711", the 5th element, gives the time of half The trigo the duration of the eclipse 1 h. 25 m.; therefore the whole duration is 2 h. 50 m., which is 2 m. 40 s.more than the time we obtained by the rough construction. The distance nm, as just determined, is 220", and the time of describing this space, at the rate of 1711" per hour, requires 7 m. 52 s., which taken from and added to the semiduration, gives 1 h.17m. 8 s. from the beginning of the eclipse to full moon, and 1h. 32 m. 52 s. from the full moon to the end of the eclipse. For the magnitude of the eclipse, we add the moon's seminometrical diameter in seconds (904′′) to Cm (2212"), and from the computation of the magni- sum subtract the semidiameter of the shadow in seconds tude of the (2379), and the remainder is the portion of the moon's di eclipse. The con struction a ameter not eclipsed. Subtract this quantity from the moon's diameter, and we shall have the part eclipsed. Divide this by the whole diameter, and the quotient is the magnitude of the eclipse, the moon's diameter being unity. Following these directions, we find the magnitude of this eclipse must be 0.587. In all these computations we were guided by the construcsufficient tion; which will always prove a sufficient index, and all that guide to car- should be required. ry out the We determine, in may trigonometri any case, whether the eclipse will or cal computa. will not be total, by the following operation: tions. Subtract the 's semidiameter from the semidiameter of the shadow, and if the moon's latitude, at the time of full moon, is less than the remainder, the eclipse will be total, otherwise not. To find the duration of total darkness.- Diminish the semidiameter of the shadow by the semidiameter of the moon, and from the center of the shadow describe a circle, with a radius equal to the remainder; a portion of the moon's path must come within this circle; that portion, measured or divided by the hourly motion, will give the time of total darkness. When the moon's latitude is north, as in the present example, the southern limb of the moon is eclipsed—and conversely. |