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A. D. M. of the Sun Of the Perih.
S

S
The Radix 1781 109 21 20 06:3

8 46 50 Current Years 191 11 29 24 Ito

O 15 So M, Mot, for 1800 C9 20 47 2013 9 02 40 November

45 Biffextile $ 25th 10 25 15 49

3 903 25 Place of Perihel. The M. Motion 716 00 09 Equation Subst.

1 33 50 17 16 oo og Sun's M. Mor.

3 8 03 25 M. of Peri, subst. Sun's true Place 7 14 36 19

4 6 26 44 Mean Anomaly.

That is, the Sun will at that Moment occupy the 15th Degree of mil

, or Scorpio, the 27th Minute of that Degree, and the twentieth Second of that Minute.

The Time I here ( and shall all along ) mean, is the Mean, Astronomical or absolute Time ; such as is measured by the equable Motion of a PendulumClock, Watch, &c. and not the Apparent or Vulgar Time, which is ineasured by the unequal Motion of the Sun, on a Dial, &c. The Latter sometimes differing considerably from the Former, and requires a farther two-fold Equation to have the Sun's Place adapted to it.

You must take notice also in the Use of these Tables, not to confound the common Day, which begins at Midnight, with the Astronomical Day, that begins at the Noon following. For the 7th common Day at 8 a-Clock in the Morning, is but the 20th Hour of the Sixth Astronomical Day ; and the 5th Hour of an Astronomical Day is the 17th of a Vulgar or Common Day, or 5 at Night. So much for finding the Sun's Place in the Ecliptic for any given Moment of Time.

PRO

S

or

PROBLEM II.
To find the Latitude of any Place.

Practice.
This (amongst
many other ways
is moft easily
and very exactly A

E done, by taking

B
the Altitude or
Elevation of the
North Pole a-

H
bove the Ho-
rizon of the
Place, by means
of a very good
Quadrant,

D Sectant. For this Altitude or Elevation of the North Pole is equal to the Latitude of the Place, which is equal to its distance from the Equator, as is evident from the adjacent Scheme.

For therein Æ Q is the Equator; HO the Horizon of the Place whose Uertex is V; B D is the Axis of the Earth, pointing directly to the North Pole at P; draw AV parallel to HO, and supposing a Spectator at S observing the Height or Elevation of the North Pole, which he finds by his Quadrant to be the Angle PSA = 50° 56'; I say this is equal to the Angle VCÆ, or the Arch , the Distance of the Place V from the Equator.

For since À V is parallel to H0, the Right Line PD shall intersect each in equal Angles, and so the Angle P S A = CH; But since V C A + B CV= BCH + B CV= a Right Angle ; therefore BCV being taken from each side of the Equation, VCÆ

BCH

= BCH (= PSA) = 50' 56', the Latitude of this City of Chichester, and thus is the Latitude of any other Place to be found.

;

PROBLEM III.

To find the greatest Declination of the Sun, or the Obliquity of the Ecliptic, or the Angle which the Sun's (or Earth's ) Orbit maketh with the Equator.

Practice. Observe

very correctly the Meridian Altitude of the Sun on the Days of the Summer and Winter Solstice, then subduct the Latter from the Former ; and half the Remainder or Difference, will give the Obliquity, or Angle of the greatest Declination from the Equinoctial; for Example.

Suppose ( in the foregoing Scheme) that V represents the Tower of London, where the Elevation of the North Pole B CH=VCÆ 519 32' exactly. In the Summer Solstice, the Sun enters the Tropic of Cancer, and his Meridian Altitude is then there O E = 614 57'. But in the Winter Solstice the Sun be ing in the Beginning of Capricorn, his Meridian Alti, tude is then OL= 14° 59'.

Whence from O E= 61 57

Subduct OL = 14 59

Difference EL= 46 58 The Half therefore EC Æ = 23 29 = Obliquity of the Ecliptic, or the Sun's greatest Declination from the Equinoctial.

VOL. II.

Gg

PRO

PROBLEM IV.

The Sun's Place, and greatest Declination given, to find his present Declination.

P

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Practice. I demand the Sun's Declination on the 12th Day of May, Anno Dom. 1735, his Place being then in I 1° 46', and the greatest Declination 239 297?

In the Right-angled Spherical Triangle A B C, rightangled at A, there is given the Angle A B C = 23° 29', and the Side or Hy- H pothenuse BC= 619 461 the Sun's Distance from the Equinoctial, to find the Side AC, the present Declination, by

N
Case 2, Var. 2, in
the Synopsis.

Analogy.
As Radius

10.0000000 Is to the Sine of BC= 61° 46' 9.9449899 So is the Sine of B= 23° 29' 9.6006997 To the Sine of the Declin. AC = 20° 34' 9.5456896 as was fought.

S

PRO

PROBLEM V.

The Sun's Place, and greatest Declination given, to find the Right Afcension.

Pratice, Suppole May the First, 1735, the Sun's Place be I 1° 46', what is the Right Ascension ?

In the Triangle above, there is given B C = 610 46', and the Angle B = 239 29', to find the Side A B, or Right Ascension.

Analogy. As the Co-tangent of BC = 61° 46' 9.7299295 Is to Radius So is the Co-fine of B= 23° 29' 9.9624527 To the Tangent of R. Af. AB = 59° 39' 10.2325232

Or thus, by the Analogy in the Synopsis, Case 2,

10.0000000

Var. i.

CSB:R::ct BC:ct AB.

Varied thus
R:tBC::cs B :t A B.

PPOBLEM VI.

Given the Sun's greatest and present Declinaţion, to find the Sun's Place, and Right Ascension, i

Practice. Suppose the present Declination of the Sun be 20° 34', what is his Place, and Right Ascension?

In the foregoing Spherical Triangle, there is given the Angle B = 23° 29', and the Side A C= 20° 34'; to find the Sides B C, and AB ; the Analogies for which are these, as s B:s AC:: R VOL. II.

:SBC

Gg 2

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