330

ASTRONOMY.

the Sun , 15' 53",54, we skill have for a remainder the seinidiameteruf the shadow 44' 2",36. On account ol the refraction of our at-mosphere it is necessary to add as many seconds as the semidiame-ter of theEarth’sshadow contains minutes of a degree in order to re-duce its real to its apparent seinidiameter. In the example be-fore us from this addition it becomes 44' 40". The seinidiameterof the shadow, found hy the preceding rule may vary from 37'46“ to 46' 19"; it is greatest of all w hen the Moon is in perigee, andthe Earth in aphelion. To determine the inclination of thelunar orbit to the ecliptic, and also her relative horary motion inher orbit, we must make use of the following proportions: Thedifference of the two horary motions in longitude is to the horary-motion of the Moon in latitude, as Radius is to the tangent of theinclination: and, The cosine of the inclination is to Radius , asthe difference of the horary motions of Sun and Moon is to therelative horary motion of the Moon in her orbit.

Example. In the lunar eclipse of September 2, 1830, we find,that tin- hull Moon , or true opposition, will take place at lOh. 36m.19,9s. tiiat the horary motion of the Moon will be 36' 10",97 inlongitude, and 3' 21",51 in latitude, and the Sun ’s horary motion2' 25",43 : the difference of the horarv motions in longitude, 33'45",54, is to the horary motion in latitude, 3' 21",51, as Radius isto the tangent of the inclination, 5° 40' 53". Also, the co-:.ine ofthe inclination, 84° 19' 7", is to Radius , as the diiierence of the•horary motions in longitude, 33’ 45",54, is to the relative horarvmotion of the Moon in her orbit, 33' 55",53. The time of the FuilMoon, or true opposition, being known, as also the latitude of theMoon for the same time, the inclination of her orbit, (which de-pends on the horary motion of the Moon in longitude and lati-tude,) and the horary motion of the Sun , we must next seek the timeo: the middle of the eclipse. Let O (Fig. 2) be the point of theecliptic opposed to the Sun, or the centre of the Earth ’s shadowat the distance of the Moon ; OG the seinidiameter of theshadow, ELS the relative orbit of the Moon , L the place of theMoon at the instant of the opposition, OL the latitude of theMoon , or her distance from the ecliptic KG, OM the perpen-dicular let fall upon the relative orbit EMS. At the commence-ment of the eclipse, the Moon being in E, her disk touches theconiines of the shadow at P; thus E is the place of the Moon atthe beginning of the eclipse. In like manner S is the place ofthe Moon at the end of the eclipse. The triangles MOE, MOS,are equal, because they have the side OM common, the equalsides OE and OS, (which are composed of the sum of the semi-diameters) being also both right-angled at M. In like mannerthe side EM is equal to the side MS ; hence the point M is theplace where the Moon is at the middle of the eclipse, and not thepoint L, which is the Moon ’s place at the time of the ecliptic op-position. In the triangle LOM, formed by the circle of lati-tude OL, and by the perpendicular OM, the angle LOM isequal to the inclination of the lunar orbit to the ecliptic. To bemore explicit, the perpendicular to the orbit and the perpendicularto the ecliptic differ from each other only by the quantity of differ-■ence that subsists between the orbit and the ecliptic; hence, ifwe continue the orbit EMS till it meets the ecliptic, as at V', (Fig.3,) we shall be convinced that the angle of these two lines will betlie complement of the angle MOK, (Fig. 2,) which complementrs the angle MOL, or the inclination of the orbit. With thisangle and the side LO we may find LM by making use of the fol-lowing proportion: Radius is to the sine of the inclination, as theecliptic latitude OL is to the interval LM in parts of a degree.To reduce it to time use the following proportion : The Moon ’srelative horary motion is to one hour, as the space LM is to thetime that there is between the opposition and middle of the eclipse.We must snbtrart this interval of time from the moment of the op-position, if the latitude of the Moon be increasing, and add it iftiie latitude be decreasing, in order to obtain the time of themiddle of the eclipse.

Example. The latitude of the Moon in opposition will be5' 12”,7 N. D. by the Tables. Radius is to the sine of the incli-nation, 5° 40' 53", as the latitude, 2' 12',7, is to the interval ML13", 13 in parts of a degree. To reduce it to time we mustfar, As the relative horary motion of the Moon in her orbit,33' 55",53, is to one hour,so is 13 ", 13 to 23",22 of time, which addedto tiie time of tire ecliptic opposition, because the latitude is de-crea-ing, (the Moon not yet having arrived at her node,) the sumwill be lOn. 30.n. 43,12s. the middle of the eclipse; that is to

say, the 2d of September, 1830, N. S. at 36m. 43 Js. past ten atnight. The same quantities which have been made use of to lindthe difference LM between the opposition and the middle of theeclipse will serve to determine die nearest approach OM of tiielunar orbit to the centre of tiie slu (low ; for in the triangle LOM,right-angled at M, we know LO, which is the latitude at the limeot the opposition, and the angle I.OM equal to the inclination oftiie relative orbit; whence the side OM is found by the followingproportion : as Radius is to the latitude LO, so is the sine of theangle L, or the co-sine of the relative inclination, to tiie nearestapproach of the centres OM.

Example. In tiie eclipse of September 2, 1830, the latitude ofthe Moon LO will be 2' 12",7, and tiie inclination of the relativeorbit MOL 5° 40' 53". Then say, As Radius is to the latitude,2' 12",7, so is co-sine of MOL, 84" 19' 7", to 2' 12",04S, tiienearest approach oi the centres of the Moon and Earth ’s shadow.This quantity v ill be of use by and by in finding the commence-ment, end, and size, of tiie eclipse. When we know the middleof the eclipse, tiie nearest approach of tiie centres of the Moon andEarth ’s shadow, the apparent seinidiameter of tiie shadow, and thesemidiameter of the Moon , there remain only two triangles to be re-solved ; tiie first to obtain the time of the beginning and end of theeclipse, the second the beginning and end of total darkness. LetOM (lig. 2) be the nearest approach of the centres, or tiie per-lendicular let fall from the centre O of the shadow upon the re-ad ve orbit EMS of the Moon ; GPAK. the semi-circumferenceof the shadow, EP tiie radius of the lunar disk, E the centre of theMoon at the moment she enters the shadow at P, S tiie centre oftiie Moon when she quits the shadow, or at the end of tiie eclipse.Tiie distance OE of tiie centres of the Moon and shadow is com-posed of the quantities OP and PE, the first of which is the semi-diameter of the shadow, the other the semidiameter of (he Moon as seen from the centre of the Earth . In like manner tiie distanceOS, at th end of the eclipse, is composed of the quantities ORand RS, that is to say, that it is also the sum of the seniidiame-ters of the shadow and Moon ; therefore OS is equal to OE. Inthe triangle OHM, right-angled at M, knowing tiie perpendicularOM, and the ; um OE of the semidiameters of the Moon and sha-dow, we must next find the third side M E in parts of the ecliptic;afterwards we must reduce it to time by the following proportion :As the relative horary motion of the Moon in her orbit is to onehour, so is tiie side .VIE to the semiduration of the eclipse. Thissemiduration being subtracted from the time of the middle of theeclipse, the difference will be the time of its beginning ; but ifadded, the sum will be the time of its end.

Example. In tiie. lunar eclipse of September 2, 1830, the per-pendicular MO will be 2’ 12",048, the semidiameter OP of theshadow 44' 46", tiiat oi the Moon )C' 19",8; the sum of tiie semi-diameters, therefore, (the refraction of the atmosphere being takeninto the account,) is t° 1' 5",8. Hence in the triangle EMO, byknowing OE and OM, we may find ME hy the following easy

method of M. de La Lande :

Sum of tiie sides OE and OM, 1° 3' 17",85.Log. 3,5795378

Difference of the sides OE and OM, 58' 53",75...Log. 3,5482358

Sum of the two Logarithms. 7,1277736

Half tiie sum, or Logarithm of EM, 61' 3",42.3,5638863

To reduce the quantity 61' 3",42 to time, say, as 33' 55",5 3,the Moon ’s relative horary motion in her orbit, is to one hour, sois E.V1 61' 3",42 to 111. 47m. 59,05s. the semiduration of theeclipse. By subtracting tiiis quantity from the time of the middleof tiie eclipse, lOh. 36m. 43,12s. we shall find the beginning to lie at8h. 48m. 44,07s.; and by adding it tiie time of the end will befound to be at 12b. 24m. 42,17s. But our eclipse beina total,we have yet two other phienomena to find, i. e. the beginning andend of total darkness. Let D (Jig. 2) lie the place ot the Moon at the instant of her total immersion into the Earth ’s shadow, andV 7 her place at the cud of total darkness ; then we have anothertriangle OM D or OMV to solve, whose hypothenu-e OI> orOV is equal to the difference oi the semidinmotci-s of tiie Moon and shadow. This operation is the same with (he triangle EMO.The following example will be sufficient to elucidate it. in tiielunar eclipse of September 2, 1830, the perpendicular MO willlie 2' 12",048, the seinidiameter of the shadow 44' 46", and tiieseinidiameter oi the Moon 16' 19",8 ; the difference of the semi-

diameters