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as a plane, of which the pole is the centre, and the star's polar distance the radius. When the star is above the pole, its altitude exceeds that of the pole, or the latitude of the place, by its polar distance; and when it is below the pole its altitude is less than the latitude by the same quantity. When six hours distant from the meridian, its altitude is nearly the same as the altitude of the pole; and in all other situations its polar distance may be considered as the hypothenuse of a right angled plane triangle, of which the perpendicular is the difference between its altitude and that of the pole, and the acute angle included by those sides is the star's least distance from the meridian. If from the sum of the sun's right ascension and the given time, the star's right ascension be subtracted, the remainder will be its meridian distance; and if this be less than 6, or more than 18 hours, the star will be above, but if between 6 and 18 hours, below the pole.

Reduce the meridian distance into degrees, and if less than 90°, consider it as a course; if between 90° and 270°, consider its supplement as a course; and if it exceed 270°, consider the difference between it and 360° as a course; with which and the star's polar distance in minutes, enter Table 2; and add the corresponding diff lat to the star's altitude when it is below, but subtract it when it is above the pole, and the sum or remainder will be the required latitude, which is always north.

Note. An error of a few minutes in the apparent time, or in the right ascension, will, in any case, produce but a trifling error in the result; but the nearer the star is to the meridian, the less effect will any mistake in the time have on the latitude; but the right ascenşions may always be taken to the nearest minute.

EXAMPLE. If the altitude of the pole star be 46° 18', at 9h 26m P. M., October 7, 1836, in longitude 16° W, height of the eye 13 feet, required the latitude ?

The Greenwich time of the observation is 10h 30m, and the sun's right ascension reduced to that time is about 12h 54m, the star's lh lm, and its polar distance 94'.

Hence 9h 26m + 12h 54m ih lm = 21h 19m = the star's meridian distance, and it is therefore above the pole. Now 21h 19m = 320° nearly, and 360°

320o =

40°, with which, as a course, and 94 as a distance, in Table 2, we have nearly 51' in the latitude column, to be subtracted from the star's true altitude. Observed altitude

46° 18'
Dip and refraction
True altitude

46 14
Correction, Table 2

11 Latitude

3N

4

45

EXAMPLES FOR EXERCISE,

In the following examples the latitude is required to the nearest minute ?

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To compute the latitude from two altitudes of the sun observed on a given

day, and the time elapsed between the observations. Find the time at Greenwich corresponding to the middle time between the observations, by Problem 4, page 210; reduce the sun's declination to that instant, and thence find his polar distance, by Problem 9, page 224. Compute the true altitudes by Problem 7, page 222, and if the ship has sailed in the interval between the observations, the first altitude must be reduced to what it would have been if taken at the same place with the second. The correction for this purpose may be obtained thus : Take the angle included between the ship’s course and the sun's bearing at the first observation as à course, in Table 1, or Table 2, or the supplement of that angle, if it exceeds a right angle; and to the distance sailed in the interval, as a distance, the required correction will be found in the latitude column, being the minutes which the ship has sailed towards or from the sun. This correction must be added to the first altitude if the angle is less than a right angle, (in which case the ship will have sailed towards the sun,) and subtracted if it is greater, and the altitude will then be obtained which the sun would have had at the first observation, if it had been taken where the second was observed.

Having got the polar distance and the true altitudes reduced to the same place, take half the interval between the two observations, and reducing it to degrees, call it the half elapsed time; and proceed to compute the latitude at the place of the second observation by the following rule.

1. Add the sine of the half elapsed time to the sine of the polar distance and the sum, rejecting 10 from the index, will be the sine of arc first.

2. Add the secant of arc first to the cosine of the polar distance and the sum, rejecting 10 from the index, will be the cosine of arc second, 3. Add together the cosecant of arc first, the cosine of half the sum of the true altitudes, and the sine of half their difference, and the sum, rejecting the tens from the index, will be the sine of arc third.

4. Add together the secant of arc first, the sine of half the sum of the true altitudes, the cosine of half their difference, and the secant of arc third, and the sum, rejecting the tens from the index, will be the cosine of arc fourth.

5. The difference of arc second and arc fourth is arc fifth, when the zenith and the elevated pole are on the same side of the great circle passing through the places of the sun at the times of observation, otherwise, their sum is arc fifth.

6. Add the cosine of arc third to the cosine of arc fifth, and the sum, rejecting 10 from the index, will be the sine of the latitude.

Note 1. When there is any doubt whether the zenith and elevated pole are on the same side of the great circle passing through the places of the sun, the latitude may be computed on both suppositions, and that considered as the true latitude which most nearly agrees with the latitude by account. This additional computation will give very little trouble, for it is only arc fourth and its cosine that will require alteration.

Note 2. By this method the latitude may be determined from two altitudes of the same fixed star, with the interval of time between the observations; but if the interval be in solar, it must then be reduced to sidereal time, which may be done with sufficient exactness by increasing the observed interval one second of time for every ten minutes, or six seconds for

every

hour.' Note 3. It will expedite the calculation, if all the logarithms which are found at the same opening of the book be taken out at the same time; and any little mistake in the observations will produce a less error in the result, if the greater altitude be observed when the sun is not far distant from the meridian.

A very ingenious and simple method of solving this problem by construction has been given by Dr. Kelly, in his useful Practical Introduction to Spherics and Nautical Astronomy.

EXAMPLE. If on the 28th of February, 1868, in latitude by estimation 48 N, longitude 37 W, at 8h 9m 4s A. M. the altitude of o be 27° 31', bearing SE; and after running NE, 8 miles an hour, till 11h 30m 18s A. M. the altitude of be 32° 40', height of the eye 20 feet, required the latitude at the time of taking the second observation ?

As the angle between the sun's bearing and the course is 117 points, the ship has sailed within 45 points of the direction opposite to the sun; therefore if the first altitude had been observed at the

sun.

place at which the second was observed, it would have been less than it was observed to be by what the ship has sailed directly from the

Now the distance run in the interval is nearly 27 miles, with which as a distance, and 44 points as a course, in Table 2, we find in the latitude column 18' for the correction of the first altitude, or the distance which the ship has sailed directly from the sun.

First alt. O.. 27°31' 0" Second alt. Q 32° 40' 0"
Dip
4 24 Dip

4 24
27 26 36

32 35 36 Cor. of alt. 1 41 Cor. of alt.

1 22 27 24 55

32 34 14 Semidiameter + 16 11 Semidiameter + 16 11 27 41 6

32 50 25 Cor. for ship's run

true altitudes.

27 23 6 18 0 True altitude 27 23 6 Sum of alts... 60 13 31 half sum.. 30° 6' 45"

Difference 5 27 19 half diff 2. 43 39

}

}

Diff....

Feb. 27, 20h 9m 4s

times of observation. 23 30 18

3 21 14, half difflh 40m 37s = 25° 9', half elapsed time.
Sum ... 43 39 22, half sum 21 49 41, middle time.
Longitude in time

+ 2 28
Middle time at Greenwich.... 0 17 41 February 28.

22' 40"

....

2's declination February 28, 1824, (per Nautical Almanac, or Table 19)

8° 14' 16'S Reduction to 1868, (Table 20)

7 51

8 6 25 Correction for mid. time at Greenwich, (Table 30).... 17 O's corrected declination

8 6 8 S

90 Polar distance

98 6

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8

To compute the latitude.

9•628378 sin 25° 9' H. E.T.
9.995646 sin 98 6 polar dist. cos 9.148915
9•624024 sin 24 53 arc 1, sec 10.042313

...... cosec 10:375953 98 36 arc 2,

cos 9:191228

10.042313 sec arc 1.
9.700498 sin

30° 7' COS 9.937019 9.999506 cos

2 44 sin 8.678405 10:002103 sec arc 3. 5 28 sin 8.991377 56 17 arc 4. cos 9.744420

:.. COS 9.997897 42 39 arc 5.

COS

9.866586

EXAMPLES FOR EXERCISE.
The true latitude is required in each of the following examples,
both altitudes being taken at the same place ?

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In each of the following examples the true latitude is required at the time of taking the second observation ?

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hm s
9 52 14 A.M.
2 12 20 P.M.
1 30 20 P.M.
3 33 25 P.M.
8 0 0 A.M.
1 0 0 P.M.
9 30 OAM.
2 40 16 P.M.
1 50 14 P.M.
4 0 14 P.M.
8 45 10 A.M.
0 30 16 P.M.
8 30 0 A.M.
1 45 0 P.M.

47 26 Q 48 13 Q 15 17

6 30 2 28 4 2 69 10 0 20 12 Q 8 56 0 31 52 2 12 58 Q 21 30 0 39 1 Q 7 56 o 17 6

SbE! E

20

42 14 N

NE
10 miles per hour

N EN
3.5 miles per hour

NW b W
6 miles per hour

N W
5 miles per hour

NW b W
6 miles per hour

W&N
4 miles per hour

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SW b W

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NNE

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16 March 29, 1823 68 12 S

20 E

ENE

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