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render the absolute value of the horizontal dip uncertain: in such cases, therefore, the altitudes of the objects must be obtained by computation, as in the following problems; the principles of which will be found amply illustrated in "The Young Navigator's Guide to the Sidereal and Planetary Parts of Nautical Astronomy," page 237.

PROBLEM I.

Given the Latitude and Longitude of a Place, and the Apparent Time at that Place; to find the true and the apparent Altitude of the Sun's Centre.

RULE.

Reduce the given apparent time to the meridian of Greenwich, by Problem III., page 297; to which let the sun's declination be reduced, by Problem V., page 298.

If the latitude of the place and the sun's declination are of different names, let their sum be taken; otherwise, their difference: and the meridional zenith distance of that object will be obtained. Then,

To the logarithmic rising answering to the sun's distance from the meridian, (that is, the interval between the given apparent time and noon,) add the logarithmic co-sines of the latitude of the place and of the sun's reduced declination: the sum, rejecting 20 from the index, will be the logarithm of a natural number; which, being added to the natural versed sine of the sun's meridian zenith distance, found as above, will give the natural co-versed sine of its true altitude.

To the sun's true altitude, thus found, let the correction corresponding thereto in Table XIX., be added; and the sum will be the apparent altitude of the sun's centre.

Example 1

Required the true and apparent altitude of the sun's centre, January 10th, 1825, at 3145 apparent time, in latitude 40:30 N., and longitude 59:2:30 W. ?

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Sun's declination at noon, January 10th,= 21:57:50% S.
Correction of ditto for 6:57"55:=

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2.40

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Log. co-sine 9.967412
Log. co-sine=

Sun's mer. z. dist. = 62°25'10"Nat. vers. sine= 537005

9.881046

Nat. number 210440 Log. 5.323128

True alt.of sun's cen. 14:37:43"Nat.co-V.S.= 747445

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Required the true and apparent altitude of the sun's centre, January 20th, 1825, at 19:16 18: apparent time, in latitude 37:20: S., and longitude 49:45: E.?

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Sun's reduced dec. = 19.58.23 S.
Latitude of the place 37. 20. 0 S. .

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Sun's mer. zen, dist. = 17:21:37"Nat. v. sine=045552

Nat. num. 503075 Log.=5.701633

True alt. of sun's cent. 26:49:55 Nat.co-V.S=548627
Reduc. of do.,Tab. XIX.,= +1.44

App. alt. of sun's centre 26:51:39%

* 24 hours 191618 = 443 42', the sun's horary distance from the meridian,

PROBLEM II.

Given the apparent Time at a known Place, to find the true and apparent Altitude of a fixed Star.

RULE.

Reduce the given apparent time to the meridian of Greenwich, by Problem III., page 297; to which let the sun's right ascension, at noon of the given day, be reduced, by Problem V., page 298.

Let the star's right ascension and declination (Table XLIV.) be reduced to the given period, by the method shown in page 115. To the sun's reduced right ascension let the given apparent time be added, and the sum will be the right ascension of the meridian; the difference between which and the star's reduced right ascension will be the horary distance of the latter from the meridian. Now, with the star's horary distance from the meridian, thus found, its reduced declination, and the latitude of the place, compute the true altitude of that object, by the last problem. Then, to the star's true altitude, thus found, let the correction corresponding thereto, in Table XIX., be added; and the sum will be the star's apparent altitude.

Example 1.

Required the true and apparent altitude of a Arietis, January 1st, 1825, at 11:929 apparent time, in latitude 40°29′ N., and longitude 59:45: W.?

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Star's mer.zen.dist.=17:51:10%Nat. vers. sine=048153

True alt. of the star-36:29:56 Nat. co-V. S. 405193

Reduc.of do.Tab.XIX=+1.17

App.alt. of giv. star=36:31:13′′

Example 2.

Required the true and apparent altitude of Procyon, January 1st, 1825, at 9:31:39 apparent time, in latitude 39:20:30 S., and longitude 75:40 E.?

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Sun's right ascension at noon, January 1st, = 18:47 19:

Correction of ditto for 4:28"*59: =

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+ 0.49

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Star's mer. z. dist. = 45% 0:28′′Nat. vers. sine=292990

Nat. number = 250533 Log.= 5.398865

True alt. of giv. star=27: 9:36"Nat. co-V. S.=543523
Reduc.of do.Tab.XIX. +1.50

App. alt. of giv. star=27:11:26"

PROBLEM III.

Given the Latitude and Longitude of a Place, and the apparent Time at that Place; to find the true and apparent Altitude of a Planet.

RULE.

Reduce the given apparent time to the meridian of Greenwich, by Problem III., page 297; to which time let the sun's right ascension be reduced, by Problem V., page 298; and let the planet's right ascension and declination be reduced to the same time, by Problem VII., page 307.

To the sun's reduced right ascension let the given apparent time be added, and the sum will be the right ascension of the meridian; the difference between which and the planet's reduced right ascension will be the horary distance of the latter from the meridian. Now, with the planet's horary distance from the meridian, thus found, its reduced declination, and the latitude of the place, compute the true altitude of that object, by Problem I., page 404. Then, with the planet's true altitude, thus found, by computation, enter Table XIX., and take out the quantity corresponding to the reduction of a star's true altitude; the difference between which and the planet's parallax in altitude, Table VI., will leave a correction, which, being added to the true, will give the apparent altitude of the planet.

Example 1.

Required the true and apparent altitude of the planet Jupiter, January 3d, 1825, at 9:123: apparent time, in latitude 50:30: N., and longitude 48:45 W.?

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