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Example 2.— June 25th p.m. at ship. Position by D.R. lat. 15° 42' N., long. 63° 1' 30" E. Time by chronometer 5h. 47m. 545., supposed to be correct. Required the altitude of Mars.

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33 36

Hav. 7.72948
74° 18'00" Sin. 9.98349
40 40

Sin. 9.96506
Arc I. Hav. 7:67803

Nat. hav. 0.00477
38° 22 40"

Nat. hav. O'10 803
Zen. Dist. 39° 15' 0" Nat. hav. O`11280

90
True Alt. 50 45

00

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Example 3.-May 27th, 3h. 5m. 56s. p.m. A.T.S. Lat. 33° 54' S., long. 108° 44' E. Find the true altitude of the sun.

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H.

M.

S.

P
3
5 56 Hav.

9*19234
I'
56° 6'
0" Sin.

9'91909
Þ III

18 36 Sin. 9996924
Arc I.

Hav. 9:08067 Nat. hav. 0'12041
l' - P 55° 12' 36"

Nat. hav. 0.21471
Zan. Dist. 70° 44' 45" Nat. hav. 0'33512

90

00 True Alt. 19 15 15

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N.B.-The formula used in solving the above will be found the most expeditious for finding the third side when two sicies and the included angle are given.

Method II.-To the log. haversine of the supplement of the hour angle add the log. sine of the co-latitude and the log. sine of the polar distance ; the sum (rejecting 20 from the index) is the log. haversine of an angle, which call x.

Then add together the co-latitude and polar distance and place x beneath this sum ; find their sum and difference, and half sum, and half difference. Add together the log. sine of the half sum and the log. sine of the half difference (rejecting to from the index); the result is the log. haversine of the true zenith distance, which subtract from goo for the true altitude.

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where h is the hour angle, l' the co-latitude, p the polar distance, x an auxiliary angle, and z the zenith distance.

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LATITUDE BY DOUBLE ALTITUDES

The latitude may be found with sufficient accuracy by two altitudes of the same object taken at any time of the day, the interval or elapsed time between the observations being measured by the chronometer or a good watch.

There are several methods of solving this problem, but we shall only give two, IVORY'S METHOD,” and what is known as the “ DIRECT METHOD."

For Ivory's METHOD proceed as follows-

1. Subtract the time of the first observation from the time of the second

observation; the result will be the mean interval.

2.

Turn the mean interval into an apparent interval (see CONVERSION OF

TIMES). 3. Divide the elapsed time, or apparent interval, by 2 for the half elapsed

time (H.E.T.). 4. To the time by chronometer (corrected) at first observation add half the

mean interval; the result will be the middle mean time at Greenwich.

5. Correct the sun's declination (Nautical Almanac, p. II.) for middle me n

time at Greenwich.

6. Correct the observed altitudes and get the true altitudes.

7. Reduce the first true altitude to what it would have been if it had been

taken where the second altitude was taken. To do this the ship's course and distance between the observations is required (see CORREC

TION OF ALTITUDES). 8. Find the half sum and also half difference of true altitudes.

The above applies to the sun ; if a star be observed the mean interval between the observations would have to be converted into a sidereal interval (see CONVERSION OF TIMEs).

When a planet is observed the mean interval will have to be converted into a planetary interval as follows : Obtain the sidereal interval as before ; then multiply the planet's variation in right ascension in one hour of longitude (Nautical Almanac) by the mean interval ; the product will be the correction to apply to the sidereal interval in order to find the planetary interval; to be added when the planet's right ascension is decreasing and to be subtracted when the planet's right ascension is increasing.

Example.Greenwich date July 4d. 13h., mean interval 3h. 6m. ; find the planetary interval for Jupiter.

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When the moon is observed the mean interval must be converted into a lunar interval as follows: Obtain the sidereal interval; then multiply the moon's variation in right ascension in im. by the mean interval and subtract the product from the sidereal interval, and the result will be the lunar interval.

To COMPUTE THE VARIOUS ARCS

For Arc 1.--Add together log. co-secant of half elapsed time and log. secant of declination; the sum will be log. co-secant of arc I., which take out.

For Arc II.-Add together log. co-secant of arc I., log. cosine of half sum of true altitudes, and log. sine of half difference of true altitudes; the sum will be log. sine of arc II , which take out.

For Arc 111.-Add together log. secant of arc I., log. sine of half sum of true altitudes, log. cosine of half difference of true altitudes, and log. secant of arc II. ; the sum will be log. cosine of arc III., which take out.

For Arc IV.-Add together log. sine of declination, and log. secant of arc I.; the sum will be log. cosine of an arc, which take out; it will be arc IV. if latitude and declination are of the same name ; but if latitude and declination are of contrary names, subtract it from 180° for arc IV.

For Arc V.—Generally the difference of arcs III. and IV. will be arc V.; but if the latitude of the ship and the declination of the object are of the same name and latitude less than declination, the sum of arcs III. and IV. will be arc V.

For the Latitude.--Add together the log. secant of arc II. and the log. secant of arc V.; the sum will be the log. co-secant of the latitude, which take out. It will be of the same name as the latitude by dead reckoning.

If it is uncertain whether the latitude or the declination is the greater, arc V. will have two values; consequently each value of arc V. will give a different latitude. Having found the latitude for both values of arc V., take that as the correct one which is nearer to the latitude by dead reckoning.

When the altitudes are equal, compute arcs I. and IV. as above, but arc II. will be o° 0' 0".

For Arc III.-Add together the log. secant of arc I. and the log. sine of the altitude; the sum will be the log. cosine of arc III., which take out.

For the Co-Latitude.—When the polar distance exceeds the co-latitude the difference of arcs III. and IV. will be the co-latitude; when the polar distance is less than the co-latitude the sum of arcs III. and IV. will be the co-latitude, which subtract from goo for the latitude.

When the declination is oo o' o" the half elapsed time will be arc I., and arc IV. will be 90°. For the remaining arcs proceed as before.

The latitude just found is on the assumption that there has been no change in the declination, which is only true in the case of a star. When, however, the sun, planet, or moon is the object observed, a correction for the change of declination in the half elapsed time will have to be applied to the approximate latitude.

The computation of the latitude correction is as follows: Add together the log. of the change of declination in the half elapsed time, the log. secant of the approximate iatitude, the log. co-secant of the half elapsed iime, and the log. sine of arc II. (all to four places of decimals ); the sum will be the log. of the correction in seconds of arc, which take out.

The correction is to be added to the approximate latitude when the second altitude is the greater, and the polar distance decreasing, or when the second altitude is the less and the polar distance increasing; otherwise it is subtractive.

THE DIRECT METHOD Students will find this method much shorter, and there is no ambiguity in finding any of the arcs or angles.

When the sun, planet, or moon is the object observed, find the apparent, planetary, or lunar interval in the manner explained above. For a star find the sidereal interval, also the half sidereal interval.

Correct the declination for the Greenwich time at the first and second observations and find the polar distance at each observation.

Find the true altitudes at the first and second observations, and also the true zenith distances, not forgetting to correct the first altitude for the course and distance run in the interval between the observations. These constitute all the data used in solving this problem by the Direct Method.

When the polar distance at each observation is the same, as in the case of a star, the calculation is considerably curtailed by the introduction of Napier's Rules, as will be seen in the two examples given.

We shall now give an example of Ivory's Method and several examples of the Direct Method, explaining each portion of the latter as we proceed.

Example 1. -- February 28th; latitude by dead reckoning 50° N.; longitude 30° W.; the following observations of the sun were taken for latitude by double altitudes :

O 21

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40.8

Ship time nearly. Time by Chron. (corrected). Obs. Alts. of Sun's L.L.
H. M.

H. M. S.
Ist obs. 9 24 a.m. Ist obs. II 24 43

Ist obs.

40 bearing S. 46° E. 2nd obs.

54 p.m.
2nd obs. 3 54

2nd obs. 28 5 20 west of Meridian, height of eye 24 feet; ship's course and distance in the interval between the observations N. 77° E., 44 miles : find the latitude when the second observation was taken. For construction of problem see Fig. I.

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