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Then DY B+ BY A=DY A= 78° 21', Then say;

As the Co-sine of DYA = 78 21 C. A. 0.6947934 To the Co-fine of DYB = 54 52 9.7600311 So is the Tangent of B Y = 29 54 9.7596871 To the Tangent of Y A = 58 36 10.2145116

Whence the Point A in the Ecliptic which riseth with the Star is 8 28° 36'; and the Sun being in this Place about the oth of May, shews that is the Time required.

PROBLEM XXXIII.

P

Æ

Given the Latitude of the Place, and the Oblique Descension of a Star or Planet, to find the Time of its Cosmical Setting

PraEtice. Admit the Oblique Descension of the Star Fomabant be 2929 32', I demand the Time of his Cosmical Setting at Chichester, Lati- E tude 50° 56 ?

The Scheme being drawn, there is formed the Oblique Triangle B AY, in which there is given the Angle at T = 23° 29'; the Angle at B = 399 04'; and the Side B Y = 670 28', the Distance of

the

the Oblique Descension from the Point Aries r; to find the Side AY; which by letting fall the Perpendicular Y D, is performed thus;

As Radius

10.0000000 To the Co-line of B Y = 67° 28' 9.5834491 So is the Tangent of DBY = 39°04' 9.9094022 To the Co-tangent of DY B= 72° 43' 9.4928513

Then say;

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As the Co-sine of DYA = 49 14 C. A. 0.185100I Is to the Co-line of DYB = 72 43 9.4728985 So is the Tangent of BY = 67 28 10.3820615 To the Tangent of AY = 47 38 10.0400601

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S
That is in the Ecliptic, short of Aries i

17 38 to which add fix Signs

6 and substract the Sum

7 17 38 from twelve Signs

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I 2

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8 I 2

22

2 there remains the Ascendant

4 that is 12° 22' of the Sign Leo, in which Point the Sun is, about the 25th Day of July ; and that is the Time required.

PROBLEM XXXIV.

Given the Latitude of the Place, and the Oblique Ascension and Descension of a Star or Planet, to find the Time of its Rising Achronically ; and also of its Achronical Setting.

Practice,

PraEtice.

The Solution of this problem is deduced froin the two foregoing ones ; for since a Star rifts achronically when the Sun sets the Moment it rileth, therefore find the Point of the Ecliptic which riseth with the Oblique Ascension of the Star, and its opposite Point will be that which is fought.

Thus the Point of the Ecliptic which rifeth with the Oblique Ascension of Fomabant is 8 29° 36', whose opposite Point is m 2836', to which Place the Sun arrives about the Tenth Day of November, and that is the Time when this Star riseth achronically.

The Achronical Setting of a Star, is when it setteth with the Sun ; therefore that Point of the Ecliptic which descends with the Oblique Descension of a Star is the Point sought.

Thus the Point of the Ecliptic which descends with the Oblique Descension of Fomahant is en 12° 22', by the last Problem.

Now the Sun comes to this point about the 21st of January ; and then will this Star fet achronically, as required.

PROBLEM XXXV.

Given the Latitude of the Place, the Depression of the Star below the Horizon, and the Time of its Cor. mical Rising, to find the Time of its Helical Rising,

Practice.

The Helical Rising of the fixt Stars and Planets, is only their gecring out of the Sun Beams so far as to VOL. II.

M m

appear

appear visible ; but as the Magnitude and Splendor of the Stars is greater and less, so there is required a less or greater Distance from the Sun, in order to their Visibility. With regard to the Horizon, the Stars and Planets become visible according to their Magnitudes, and require a particular Depression of the Sun below the Horizon, as is hereunder set down.

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Saturn
Jupiter
Mars
Venus
Mercury
The Moon, in the Day-time
Stars of the first Magnitude
Stars of the second Magnitude
Stars of the third Magnitude
Stars of the fourth Magnitude
Stars of the fifth Magnitude
Stas of the sixth Magnitude
Nebulous Stars

5 IO

5 I2

13 00
14

Oo
15 OO
16

00 17 00 18

00

Now let it be required to find the Heliacal Rising of the Star Fomabant of the first Magnitude, at Chichester, Lat. 509 56', where he riseth Cosmically when the Sun enters 289 36'?

Let

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Let the Scheme be constructed as in

P Prob. 32 ; draw the Parallel of 12° Depression ( for Foma. E bant is a Star of the first Magnitude) and thro’the Point of Intersection with the Ecliptic at D, describe the Azimuth ZDN; fo shall S there be constituted the Right-angled Triangle ABD, in which calculate the Angle at A by Prob. 28, which you will find to be 21° 36', the Altitude of the Nonagesima Degree ; besides which there is given the Side B D = 12o, the Depression, or Arch of Vision ; by which to find the Side A D, the Distance of the Sun in D from the Cosmical Place at A, to do which say ;

As the Sine of A = 21° 36' 9.5659948 Is to the Sine of B D = 120 00

9.3178789 So is Radius To the Sine of AD = 34° 23' 9.7518841

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That is in the Ecliptic
To which add his Cosmical Place in

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[ 23 36

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Which Place the Sun occupies June the Thirteenth, and that is the Day fought ; for then will this Star be visible at his Rising in the Morning. VOL. II.

M m 2

PRO

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