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Venus. When these planets are examined with a good telescope, after their inferior conjunctions with the Sun (see T. T. for 1815, p. 22), they appear like the new Moon, shining with a fine luminous crescent. As their elongation from the Sun increases, this crescent augments in breadth; and when they reach their point of greatest western elongation, nearly half of their illuminated disc becomes visible. During their return to the Sun, the enlightened part of their disc is gradually turned towards the Earth, till their luminous side is opposite the Earth at the time of their superior conjunction; when they would appear like the full Moon, if they were not at that time rendered, invisible by the intensity of the solar light. As they proceed from this point to their greatest eastern elongation, the enlightened part of their disc gradually diminishes till they become apparently about half illuminated, when they attain that elongation. In their return to the Sun, the enlightened part now continues to diminish, as the dark side of the planet is turned towards the Earth, which is completely the case, at the moment of their inferior conjunctions.

From these remarks, it is evident that the phases of the inferior planets are subject to constant variation like those of the Moon; and the following method will enable the reader to calculate these phases, or to find the greatest breadth of the enlightened part, at any time; the whole diameter of the disc being supposed to be divided into twelve equal parts or digits.

In order to accomplish this, let S, E, and V, in the following Figure, be the respective positions of the Sun, the Earth, and Venus, at any given time; then that side of Venus which is turned towards the Sun will be completely illuminated by his rays; but only a part of that enlightened face can

evidently be seen from the Earth. Draw EvVv from the Earth through the centre of Venus, and join VS and ES; and draw rs perpendicular to EC and po perpendicular to rs; then pq will be the circle of illumination. Now as pV, a part of that circle, will be seen obliquely, it will, according to the laws of vision, and the immense distance of the planet compared with its magnitude, be projected into an elliptic arc on the plane of projection rs, and the point p will, to a spectator at E, appear at o; so that ro will be the versed sine of the arc rp, and consequently so the versed sine of the arc sp.

Fig. 1.

E

Now the angle pVs is equal the angle v'VS; for pVv is equal v'Vq, and vVs equal to qVS, being both right angles; and therefore, by addition, the sum of the two angles pVv, vVs, is equal to the sum of the two vVq, qVS; but pVv + vVs = pVs, and v'Vq + qVS = v'VS; and consequently os, which is the versed sine of the former pVs, is also the versed sine of the latter v'VS. As this proof is not limited by any particular position of Venus, it follows as a necessary consequence,

that the breadth of the illuminated part of the planet's disc, as seen from the Earth, is always the versed sine of the exterior angle v'VS of the triangle SEV formed by the lines supposed to join the centres of these three bodies. But the exterior angle 'VS is equal to the sum of the two interior angles VSE and VES: and hence we have this easy rule for finding the breadth of the enlightened part of the disc of either Venus or Mercury, as seen from the Earth.

Add together the angles formed at the Sun and the Earth by the three lines supposed to join these two bodies to each other, and to the planet, and multiply the natural versed sine of this sum by 6, and the result will be the breadth of the illuminated part of the planet's apparent disc, in digits, or 12th parts of its diameter.

It may be necessary to observe, for the information of such of our readers who have not made much progress in the study of astronomy, that the angle at the Sun is equal to the difference of the heliocentric longitudes of the Earth and the planet; and that at the Earth is the difference of the geocentric longitudes of the Sun and the planet. These are to be found in the Nautical Almanac for the time required. The longitude of the Earth is found by adding 6 signs to the longitude of the Sun, as given in the Nautical Almanac, when that longitude is less than 6 signs, and subtracting them when it is greater than that quantity thus, if I denote the longitude of the Sun, and l' that of the Earth, we shall have l' = 1+63., as l is less or greater than 6 signs. If the longitude of the Sun, therefore, was 2. 12° 42′, that of the Earth will be 8. 12° 42'; and if that of the Sun was 9. 3° 7', the longitude of the Earth would be 35. 3° 7'. It may also be observed, that when the arc of which the versed sine is required is greater than 90°, which is the

case when that angle exceeds 3 signs, this versed sine is found by subtracting the versed sine of its supplement from 2. To find the versed sine of 4'. 15° 20', we have 6.—(4o. 15° 20′) = 1. 14° 40', or 44° 40', the versed sine of which is 28879; and therefore, 2-28879171121 = the versed sine of 4. 15° 20′ or 135° 20′, as required.

Let it be required to ascertain the illuminated part of Venus on the 1st of January 1819. Then the heliocentric longitude of Venus, taken from the Nautical Almanac for that day, is 3'. 13° 50′, and the longitude of the Sun 9o. 10° 19′, consequently that of the Earth is 3. 10° 19; the geocentric longitude of Venus at the same time, also, is 9. 0° 56'; therefore,

Heliocentric longitude of Venus
Longitude of the Earth

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3S. 13° 50' subtract 3 10 19

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Hence the natural versed sine of 12° 54′, multiplied by 6,·02524 × 6 = 0.15144, the breadth of the enlightened part as stated above, under the head of Occurrences.

Again, for the 1st of July 1819, we have
The heliocentric longitude of Venus
The longitude of the Earth

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1s. 3° 28'

9 8 45

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Now 180°-141° 12′ 38° 48', or 141° 12′, the versed sine of which is 220662; and this taken from 2 leaves 1-779338, for the versed sine of 141° 12'; and which multiplied by 6, gives 10-67603 digits for the enlightened part of the disc of Venus at the time required.

The appearance of Venus, however, will be rendered more palpable by the following delineations. Let ABDE, Figs. 2 and 3, be the projected hemisphere, or disc of Venus, and EF the breadth of the enlightened part as found by the preceding method, as that is less or greater than the radius of the circle ABDE.

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Then having described the circle, and drawn the diameters AD and BE at right angles to each other, set off EF, and describe semi-ellipses through the three points A, E, and D, having AD for their transverse, and CE for their semi-conjugate diameters, and they will be correct representations of Venus at these two stages of her illumination.

Now, if it were required to find the ratio between the dark and enlightened part of the planet's apparent disc, since, from the nature of the ellipse and circle, there is always a constant ratio between CF and CE or CB; and as the semi-ellipse and the semicircle may be conceived to be made up of an indefinite number of these lines, or rather small rectangles, the sum will have the same ratio as each pair; and consequently the ellipse and circle are also in the same proportion to each

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