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PROBLEM XXX. To find the meridian line, when the time, the sun's declination, and the latitude of the place, are known.

Adjust the instrument. Set the semicircle of altitude to the co-latitude, and the nonius of the declination semicircle to the declination, and set the nonius of the horary circle to the apparent time. Turn the horizontal circle till the sights are directed to the sun. The semicircle of altitude is then in the plane of the meridian.

This problem gives the position of the meridian more accurately than Problem xxviii. It is much more ready, where the time can be had, than the method of equal altitudes, and it is near enough to the truth of small instruments. The nearer the observation is made to the time of noon, the better; because the sun then changes its azimuth the quickest.

PROBLEM XXXI. To find the declination of the sun, or any celestial object, when the latitude of the place and position of the meridian are known.

Rectify the instrument for the co-latitude, as in the foregoing problem, and place the semicircle of altitude in the meridian. Then direct the sights to the object, partly by moving the declination semicircle on the axis of the equatorial circle, and partly by moving the nonius of the semi

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circle last-mentioned. This nonius will then shew the declination.

PROBLEM XXXII. To find the right ascension of any celestial object, when the time, the latitude, and the position of the meridian, are known.

Rectify the instrument for the co-latitude, and place the semicircle of altitude in the meridian. Move the declination and equatorial circles till the cross hairs nearly concide with the object, then place the sight a little to the westward thereof, and observe by your clock the time when the object passes the vertical wire.

By the clock you have obtained the sun's time, by the nonius of the equatorial you have the hour of the object.

Take the difference between the sun's time and the star's time; and if the time of the star be less than that of the sun, add that difference to the sun's right ascension at the time of observation; from the Nautical Almanack, the sum (rejecting 24 hours, if it exceeds that number) is the right ascension of the object.

But if the star's time exceeds that of the sun, the difference must be taken from the sun's right ascension; the remainder (adding 24 hours to the sun's right ascension, if necessary) is the right ascension of the star.

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From which rejecting twenty four hours, you obtain the star's right ascension

August 31, 1778.

Observed a star, the time by the
equatorial was 10h A. M. i. e.
astronomically
Sun's time by the clock 5h 55m 4s
which astronomically is:
Difference between the two to
As the star's time exceeds that of
the sun, this difference is to be
subtracted from the sun's right

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22 00 00

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cension of the star

PROBLEM XXXIII.

To direct the line of sight to any star or planet.

Adjust the instrument; that is, place it so that the circles of altitude and declination may be in the plane of the meridian: set the semicircle of altitude to the co-latitude, and the circle of declination of the given body.

Take the difference between the right ascension of the sun and given body, and if the right ascension of the body be greater than that of the sun, subtract the difference; if not, add to the time of observation. The remainder in one case, or the sum in the other will be the hour and minute to which the nonius on the horary circle is to be set; which being done, the sight will point to the star or planet sought.

If the time be too small to admit of having the difference taken from it, borrow 24 hours, and reckon the remainder from XII at noon.

PROBLEM XXXIV. To find the longitude.*

Adjust the instrument to the co-latitude and me ridian, and take the time of the transit of the moon's limb, and also of a proper star, as near as possible to the moon's parallel, and the longitude may be deter. mined by the following rule.

Rule. Find from the Nautical Ephemeris, the increase of the moon's right ascension in 12 hours; observe, as before directed, the interval of time between the passage of a given fixed star, and of the moon's enlightened limb, over the meridian at each place, and the difference of these intervals will shew the increase of the moon's right ascension in the time of the star's passage from one

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* Vinse's Practical Astronomy, p. 90 and 169.

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meridian to the other. But as the increase of the moon's R. A. in 12 hours is to 180°, so is the above difference of intervals, to the difference of longitude.

Example, on Nov. 30, 1792. 13h 12' 57.62" meridian transit of the moon's 2d limb 29.08 meridian transit of

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at Greenwich,
by the clock,


31.46 difference of R. A. at Greenwich.
8.05 meridian transit of the
moon's 2d limb
30.13 meridian transit of a m

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at York, by the clock.

The clocks being

22.08 diff. of R. A. at York supposed to be go-
31.46 ditto at Greenwich.

ing nearly sidereal
time, no correction
is necessary.

9.38 increase of moon's right ascension between Greenwich and York by observation, which reduced into degrees gives 141". Now the increase of the moon's right ascension in twelve hours was 23340"; hence as 23340" to 180°, so is 141" to 1° 5' 14", the difference of longitude between Greenwich and York.

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Those stars that are nearest in right ascension and declination to the moon, are best suited for this

purpose.

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