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solstice, which let be=x, in the following manner. Let the time from the first to the second observation be =a, and the time from the second to the third observation be=na, then we have AV=a+x, and VE=na-x. Let FH=c, FG=d, and let the parameter of the para

bola be p. Hence x2=VFxp, and VF=2. In like

Р n'a2-2nax+x3

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Р

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-d: wherefore p= с

P and by reducing the equation we have x=x.

d

a cn2-d x=2xcn+d'

which gives this analogy. As cn+d: cn2—d : :

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But if the third observation be nearer to the solstice a cn2+d

than the second, x=2xcn-d

To illustrate this by an example. We find that in the year 1500, Bernard Walker, in the city of Nuremberg, observed the chords of the sun's zenith distances as follows:

June 2. 45467

8.44975

9. 44934

12. 44883

16. 44990

Now if we take the observations of the 2, 9, 16, of June, at 7 days apart, where a=7, and n=1, we have c=533, d=56—c-d-477, and c+d=589; and 589: 477:: 3.5: x=2, 20h, 2′, which added to the middle

observation, will give the time of the solstice June

In like manner, if we take the

8, 12, 16a, the solstice will be

8, 9, 16,

9, 12, 16,

114, 201, 2′, O. S. observations of the 114, 20, 23', O. S.

11, 20, 20,

11, 20, 38,

And taking the mean of these four determinations, we have the time of the solstice June 11, 20, 26', O. S. or June 21a, 201, 26′, N. S.

Gassendus also, at Marseilles, in the year 1636, measured, by a stile 55 feet high, the proportion of the gnomon to its solstitial shade, on the days before and after the summer solstice, and the proportions of the measures were as follows.

June 19. 31766

20. 31753

21. 31751

By comparing the measures

22. 31759

19, 20, 21, of June the time of the solstice is,June 20a, 16h, 22,

20, 21, 22,

19, 20, 22,

made on the

19, 21, 22,

20, 16h, 48,

20d, 17, 15,

20d, 175, 25',

By the mean of all, the time of the solstice was June 20a, 16h, 56′, N. S. which is 1a, 3h, 30′ sooner than by the observation of 1500, which was 136 years before it; of this 14, 1, 3'6.8 was owing to the precession of the equinox, which makes the tropical year less than the civil year by 11', 18".

TO FIND THE LENGTH OF THE TROPICAL, PERIODICAL, AND ANOMALISTICAL YEARS.

Let two observations be selected among those of greatest authenticity on record, of the time when the sun had like positions with regard to his longitude, which determine the length of the tropical year; or

with regard to some noted star, and from thence determine the length of the sidereal year; or with regard to the line of the apsides, and from thence find the length of the anomalistical year. These observations should be as distant as possible, that the whole time elapsed being divided by the number of revolutions, we may obtain the more accurate result. According to Mayer's tables, these years are as follows:

Tropical
Sidereal

365 5h 48′ 42′′

365 6 9 7

Anomalistical 365 6 15 29

The tropical year being 20' 25" shorter than the sidereal, shows that the sun has returned to the same point of the ecliptic, with respect to the beginning of any of the signs, before he has arrived at the same point with respect to the stars. And consequently any cardinal point of the ecliptic, such as the equinoxes or solstices, has a retrograde motion towards the west, at the rate of about 50′.3 per annum. For, 365d, 6h, 9', 7": 360°:: 20', 25":50".3=the precession of the equinoxes in a year.

A sidereal revolution being performed in 6′ 22′′ less time than the anomalistical, shows that the sun's apogee, from whence this revolution is computed, advances about 15′′.7 of a degree towards the east per annum. For, 365d 6h 15′ 29′′: 360°:: 6′ 22′′: 15".7- the annual motion of the sun's apogee towards the east.

The length of the year, deduced from a comparison of distant observations made at or near the time of the sun's passing his apogee, is less than that deduced from the like observations when he is about passing his perigee, therefore a mean between them should be chosen; or the length of the year should be determined

by observations made when the sun is at his mean distance.

TO FIND THE SUN'S LONGITUDE.

Having a clock regulated to mean time, observe the time when a star, whose right ascension is known, passes the meridian, and also when the sun passes the meridian: the difference of these times gives the right ascension of the sun. From the right ascension of the sun and the known obliquity of the ecliptic, find, by spherical trigonometry, the longitude; from which his place in the ecliptic will then be known.

TO FIND THE GREATEST EQUATION OF THE CENTER. At the times when the sun is near his mean distances, find his longitude, and the difference between these two longitudes will give his real motion for that interval. From his periodical time find also his mean motion for that interval; then half the difference between his true and mean motions will be the greatest equation of the center.

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and also, in the following year 1770, March,

The difference of time is

Od 23h 49′ 12′′ M.T. Long. 6° 9° 32′ 0′′.6

29d Oh 4′ 50′′ M.T.Long. Os 8° 50′ 27′′.6

178d Oh 15' 38" Diff. Long. 5$ 290 18' 27" The tropical year is 365d 5h 48′ 42′′-365.2421527. The observed interval 178d Oh 15' 38", which is 178.01085648. Then say as 365.2421527 : 178. 01085648:: 360°: 175°.455948, mean motion,=5° 25° 27 21", which answers to 5s 29° 18' 27". Half their difference, viz. 1° 55' 33" is the greatest equation of the center, which was required.

TO FIND THE ECCENTRICITY, APHELION AND PERIHELION DISTANCES OF THE EARTH'S ORBIT.

Say, as the diameter of a circle in degrees is to the same in equal parts, so is the greatest equation of the center in degrees, to the eccentricity in equal parts; then the sum and difference of the mean distance and eccentricity will respectively give the aphelion and perihelion distances of the earth. As 3.1415926: 1 :: 360° : 114.591599 parts equal to the diameter. Then as 114.591599: 1000 &c. the mean distance :: 1° 55' 33"1.9268166: 0.0168066-the eccentricity of the earth's orbit. Hence 1.016806 the aphelion distance, and 0.983194 the perihelion distance.

=

TO FIND THE SUN'S MEAN ANOMALY,

FOR ANY GIVEN TIME.

Let the epocha of the sun's passage through his apogee be accurately determined; then say, as a solar year of 365 5h 48′ 42′′: 360°: the time since the sun's passing through his apogee: the degrees of his mean anomaly. Or the sun's mean motion for the given time may be taken from the tables of the sun's mean motion, and this will be his mean anomaly. The tables of his mean motion give his distance from the first point of Aries, and this motion lessened by the motion of the apogee gives his mean anomaly, or mean longitude; which corrected by the equation of the center will give his true longitude.

TO FIND THE SUN'S RIGHT ASCENSION.

The latitude of the place being known, find the sun's meridian altitude, from whence his declination is obtained; with which and the known obliquity of the ecliptic his right ascension is found by trigonometry, thus. As tangent obliquity: radius: tangent declination sine right ascension.

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