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Take the Sun's semidiameter, 16' 6", in your compasses, from the scale CA, and setting one foot in the path of London at m, namely at 47 minutes past X, with the other foot describe the circle UY, which shall represent the Sun's disc as seen from London at the greatest obscuration.-Then take the Moon's semidiameter, 14' 57", in your compasses, from the same scale; and setting one foot in the path of the penumbra's centre at m, 47 minutes after X; with the other foot describe the circle TY for the Moon's disc, as seen from London, at the time when the eclipse is at the greatest; and the portion of the Sun's disc which is hid or cut off by the Moon's, will shew the quantity of the eclipse at that time; which quantity may be measured on a line equal to the Sun's diameter, and divided into 12 equal parts for digits.

Lastly, take the semidiameter of the penumbra 31' 3", from the scale CA in your compasses; and setting one foot in the line of the penumbra's central path, on the left hand from the axis of the ecliptic, direct the other foot toward the path of London; and carry that extent backward and forward till both the points of the compasses fall into the same instant in both the paths; and that instant will denote the time when the eclipse begins at London.-Then, do the like on the right hand of the axis of the ecliptic; and where the points of the compasses fall into the same instant in both the paths, that instant will be the time when the eclipse ends at London.

These trials give 20 minutes after IX in the morning for the beginning of the eclipse at London, at the points N and 0; 47 minutes after X, at the points m and n, for the time of greatest obscuration; and 18 minutes after XII, at R and S, for the time when the eclipse ends; according to mean or equal time.

From these times we must subtract the equation of natural days, viz. 3 minutes 48 seconds, in leapyear April 1, and we shall have the apparent times;

namely, IX hours 16 minutes 12 seconds for the beginning of the eclipse, X hours 43 minutes 42 seconds for the time of greatest obscuration, and XII hours 14 minutes 12 seconds for the time when the eclipse ends.-But the best way is to apply this equation to the true equal time of new Moon, before the projection be begun; as is done in example I. For the motion or position of places on the Earth's disc answers to apparent or solar time.

In this construction, it is supposed that the angle under which the Moon's disc is seen, during the whole time of the eclipse, continues invariably the same and that the Moon's motion is uniform and rectilinear during that time.-But these suppositions do not exactly agree with the truth; and therefore, supposing the elements given by the tables to be accurate, yet the times and phases of the eclipse, de. · duced from its construction, will not answer exactly to what passes in the heavens; but may be at least two or three minutes wrong, though done with the greatest care. Moreover, the paths of all places of considerable latitudes are nearer the centre of the Earth's disc, as seen from the Sun, than those constructions make them; because the disc is projected as if the Earth were a perfect sphere, although it is known to be a spheroid. Consequently, the Moon's shadow will go farther northward in all places of northern latitude, and farther southward in all places of southern latitude, than it is shewn to do in these projections. According to Mayer's tables, this eclipse will be about a quarter of an hour sooner than either these tables, or Mr. Flamstead's, or Dr, Halley's, make it: and Mayer's tables do not make it annular at London.

The projection of Lunar Eclipses.

When the Moon is within 12 degrees of either of her nodes, at the time when she is full, she will be eclipsed, otherwise not.

We find by example II. page 351, that at the time of mean full Moon in May, 1762, the Sun's distance from the ascending node was only 4° 49' 35", and the Moon being then opposite to the Sun, must have been just as near her descending node, and was therefore eclipsed.

The elements for constructing an eclipse of the Moon are eight in number, as follows:

1. The true time of full Moon: and at that time. 2. The Moon's horizontal parallax. 3. The Sun's semidiameter. 4. The Moon's. 4. The Moon's. 5. The semidiameter of the Earth's shadow at the Moon. 6. The Moon's latitude. 7. The angle of the Moon's visible path with the ecliptic. 8. The Moon's true horary motion from the Sun.-Therefore,

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1. To find the true time of full Moon. Work as already taught in the precepts. Thus we have the true time of full Moon in May, 1762, (see example II. page 351,) on the 8th day, at 50 minutes 50 seconds past III o'clock in the morning.

2. To find the Moon's horizontal parallax. Enter Table XVII. with the Moon's mean anomaly (at the above full) 9° 2° 42′ 42′′, and thereby take out her horizontal parallax; which by making the requisite proportion, will be found to be 57′ 20′′.

3, 4. To find the semidiameters of the Sun and Moon. Enter Table XVII, with their respective anomalies, the Sun's being 10' 7° 27′ 45′′ (by the above example), and the Moon's 9' 2° 42′ 42′′; and thereby take out their respective semidiameters: the Sun's 15' 56", and the Moon's 15' 39".

5. To find the semidiameter of the Earth's shadow at the Moon. Add the Sun's horizontal paral

lax, which is always 10", to the Moon's, which in the present case is 57' 20", the sum will be 57' 30", from which subtract the Sun's semidiameter 15' 56", and there will remain 41' 34" for the semidiameter of that part of the Earth's shadow which the Moon then passes through.

6. To find the Moon's latitude. Find the Sun's true distance from the ascending node (as already taught in page 361) at the true time of full Moon; and this distance, increased by six signs, will be the Moon's true distance from the same node; and consequently the argument for finding her true latitude, as shewn in page 363.

Thus, in example II. the Sun's mean distance from the ascending node was 0° 4° 49′ 35′′, at the time of mean full Moon: but it appears by the example, that the true time thereof was 6 hours 33 minutes 38 seconds sooner than the mean time, and therefore we must subtract the Sun's motion from the node (found in Table XII, page 342) during this interval, from the above mean distance 0° 4° 49′ 35", in order to have his mean distance from it at the true time of full Moon.-Then to this apply the equation of his mean distance from the node found in Table XV. by his mean anomaly 10° 7° 27' 45"; and lastly, add six signs: so shall the Moon's true distance from the ascending node be found as follows:

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which is the Moon's true distance from her ascending node at the true time of her being full; and consequently the argument for finding her true latitude at that time.-Therefore, with this argument, enter Table XVI. making proportion between the latitudes belonging to the 6th and 7th degree of the argument at the left hand (the signs being at the top) for the 10' 32", and it will give 32′ 21′′ for the Moon's true latitude, which appears by the table to be south-descending.

7. To find the angle of the Moon's visible path with the ecliptic. This may be stated at 5° 35', without any error of consequence in the projection of the eclipse.

8. To find the Moon's true horary motion from the Sun. With their respective anomalies take out their horary motions from Table XVII. in page 346; 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′ 52". Now collect these elements together for use.

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2. Moon's horizontal parallax 3. Sun's semidiameter

4. Moon's semidiameter

5. Semidiameter of the Earth's shadow

at the Moon

D. H. M. S.

8 3 50 50

0 57 20

0 15 56 O 15 39

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0 41 34

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