Page:Cyclopaedia, Chambers - Volume 1.djvu/640

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ECL

Ricciolus, Kepler, &c. attribute to the Light of the Earth's Atmofphere -transmitted thither. See Atmosphere.

Lastly, She grows fenfibly paler and dimmer, before ever me enters wkhin the Earth's Shadow, which is attri- buted to the Earth's ^Penumtra. See Penumtra.

Afironomy of Lunar EcLirsES, or the Method of cal- culating their limes, 'Places, Magnitudes, and other ^Phtsnomena. Preliminary i. To find the Length of the Earth's Jbado-wy Cove. Find the Sun's Diflance from the Earth for the given Time ; See Sun and Distance. Then, as the Sun's Diameter is known in Semi-diameters of the Earth, the Length of the Cone will be found from the Rules given under Shadow.

Suppofe, e. gr. the Sun's greateft Diflance from the Earth, 34996' Semi-diameters of the Earth j and the Sun's Semi-diameter to be to that of the Earth as 153 to r. Then will the Length of the ihadowy Cone be found

230+.

Hence, as the Moon's kail Diflance from the Earth, is fcarce 64 Semi-diameters j the Moon, when in Oppofition to the Sun, in or near the Nodes, will fall into the Earth's Shadow, tho' the Sun and Moon be in their Apogees. And much more, if they be in or near their Perigees, by Reafon the Shadow is then longer $ and the Moon nearer the Bafe of the Cone.

2. To find the apparent Semi-diameter of the Earth's Shadow, in the 'Place of the Moon's 'Pajfage, for any given Time. Find the Sun and Moon's Diitance from the Earth •, and thence, their Horizontal Parallaxes : Add the Parallaxes together, and from the Sum, fubftracT: the appa- rent Semi-diameter of the Sun. The Remainder is the apparent Semi-diamet«r of the Shadow.

Thus, fuppofe the Moon's Parallax 56' 4S", the Sun's 6": The Sum is 56' 24" 5 from which the Sun's apparent Semi-diameter 16' 5" fubftrac~red, leaves 40' 19" tor the Semi-diameter of the Shadow.

Note, M. de la Hire omits the Sun's Parallax, as of no Coniideration 5 but increafes the apparent Semi-diameter of the Shadow by a whole Minute, tor the Shadow of the Atmofphere 5 which would give the Semi-diameter of the Shadow, in our Inftance, 41' 13".

3. The Moon's Latitude, AL, at the Time of her Oppofition, together with the Angle at the Node B, being given ; to find the Arch between the Centres A I, and the Arch IL. (Fig- IS-} Since in the fpherical Triangle AIL, Rectangular at A 5 the Side A L is given, as alfo the Angle A L I, as being the Complement of LAI, or B to a right Angle ; the Arch between the Centres A I, is found by fpherical Trigonometry : And fince the Angle LAI, is equal to B, each of them, with I A B, making a right Angle : And the Moon's Latitude A L is given $ the Arch L I will likewife be found by fpherical Trigonometry. See Spherical Triangle.

To determine the Sounds of an Eclipse of the Moon.

Since there is no Eclipfe pofiible, but when the Aggregate of the Semi-diameters of the Shadow, and the Moon is greater than the Moon's Latitude, (for without this, the Moon will not come in the Shadow) add the apparent Semi-diameters of the Moon in 'Perigceo, and of the Shadow, fuppofing the Sun in Apogceo $ by which you will have the Side M 0. Then, in the fpherical Triangle M N O, {Fig* 3tf.) having given the Angle at the Node, whofe Quantity is the Moon's greateil Latitude in the Conjunctions 5 the right Angle E 5 and the Leg M O j find the Moon's Diflance from the Node N O : Which is the utmoA Bound, beyond which the Eclipfe cannot reach. After the fame Manner, adding the apparent Semi-diameters of the Moon in Apogteo, and of the Shadow of the Sun in 'Perigceo, for the Sake of having the LH, in the Triangle NLH; the Diflance of the Moon, from the afcending Node H N, will be found by fpherical Trigonometry ; which is the Bound within which the Moon will nece&arily be Ecliffed.

Thus, e. gr. the Semi-diameter of the Shadow, when the Sun is in Apog£o, and the Moon in Terigceo, according to Kepler, is 49' 40". And the apparent Semi-diameter of the Moon in 'Perigtfo 1.5' 22". Confequently M O 66', or l d 6' -j and therefore there will be no Eclipfe at all, if the Moon's Latitude be greater than i° 6'. Now, as the fame Angle N is fuppofed by Kepler to be 5 18'.

Log. Sin. N 89655337

Sin. MO 82832433

Whole Sine 100000000

Perigee, and the Moon's Apogee is 43' 50", and the Moon y s Semi-diameter in her Apogee 15'. Confequently, L H is 58' 5o //P . And, therefore, there will be an Eclipfe^ if the Moon's Latitude don't exceed 58' 50". But here, as before, the Argument of the Latitude is found io° 4c/.

To determine the £>ttantity of an Eclipfe, or the Number of digits Ecliffed.

Add theMoon'sSemi-diametcr IK, ( Fig. $ 5.) to the Semi- diameter of the Shadow AM 5 then will AM-j-IK=;lM + IK— AI + MK. From this Sum, therefore, fub- flraft the Arch between the Centres A I, the Remainder gives the Scruples, or Parts of the Diameter Ecliffed M K. Say, therefore, as the Diameter K H, is to the Scruples, or Parts thereof Eclipfed MK: fo is 12, to the Digits Ecliffed.

Thus, fuppofing K H, 30' 44", and confequently 1 K, 15' 22"; AM, 4/ ifi and A L, 43' 14": The Moon's Semi-diameter will be 15' 22 //, and that of the Shadow 41' i3 // ;, the Sum whereof, is 5<J / 35 // . From which the Arch between the Centres 43' 14^, being fubftraaed, leaves 13 21 Scruples, or 801 Seconds. Then as 1844:801:: 12 : S^Dig- or 5 Dig. 13'.

To find the Scruples of half Duration of an Eclipfe, or the Arch of the Lunar Orbit, which her Centre de- fcribes from the Beginning of the Eclipfe to the middle thereof.

Add the Semi-diameters of the Shadow AP, and the Moon P N together ■ the Sum gives A N. From the Square of A N, fubflracl the Square of A I j the Remain- der is the Square of I N, the Square Root of this ReHdue is the Arch 1 N required.

To find the Scruples of half 'Duration of a Total Qarknefs, in a Total Eclipfe.

Subflracl the Moon's Semi-diameter S V, (Fig.37.') from the Semi-diameter of the Shadow AVj the Remainder is AS: In the Triangle A IS, which is Rectangular at I, therefore, we have the Arch AS given by the lafl Me- thod ; and the Arch between the Centres A I : Whence the Arch IS is found, as in the lail Problem.

To find the 'Beginning, Middle, and End of a Lunar Eclipfe.

Say, as the Moon's Horary Motion from the Sun, (See Horary) is to 31500 Horary Seconds ^ fo are the Seconds of the Arch L I, {Fig. 35.) to the Horary Seconds equivalent thereto. Subflracl thefe Scruples, or Seconds, in me n>ft and third Quadrant of the Anomoly, from the Time of full Moon, and add it to the fame in the fecond and fourth ; the Refult is the Time of the middle of the Eclipfe. Then fay, as the Moon's Horary Motion from the Sun is to 3600 Scruples, or Seconds, fo arc the Seconds of half Duration I N, to the Time of half Duration ; The double of which gives the whole Duration. Laflly, fubftract the Time of half the Duration, from the Time of the mid- dle of the Eclipfe, the Remainder will be the Beginning of the Eclipfe. And add the fame to the fame, the Sum will be its End.

Suppofe, e. gr. LI = 4 ; ^'—.i^ 1 ', IN 1530", Time of Full Moon 9 h 23' 49'', Horary Motion of the Moon from the Sun, 30" 12', or 1812', then will

Log. Hor. » from O 32581581 Log. 359

Log. of half Duration 36386378. The Number correfponding to which, in the Tables, is 4351''', or,

i b ii' 31"

Log. ofSincON 95177096. The Number cor- refponding to which in the Tables is n° ^> 5c f' # jf therefore, the Moon's Diflance from the afcending Node be greater than 12 ■ no Eclipfe can happen. And, in like Manner, the Semi-diameter of the Shadow in the Sun's

Duration of Eclipje 2. 11 25 2

Time of mid. Eclipfe 9 k 15 31

HalfDurat. SubftraCt. i h 12 31

Begin, of Eclipfe 8 h 3 12

Time of mid. Eclipfe 9 h 15 43

HalfDurat. added i h 12 31

End of Eclipfe io h 28 14

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