the altitude given by the instrument, and the dip and parallax subtracted therefrom, and by subtracting 90° from the remainder, the true zenith distance will be obtained.

To find the Latitude by the meridian altitude of any object. Having obtained the true meridian zenith distance by either of these methods, you must then find the declination of the object at the time of observation. This may be found for the sun by the Nautical Almanac or by means of Tables ÏV. and V. in the manner before explained. The declination of a fixed star may be easily found by inspection in Table VIII. The declination of the moon or a planet may be found in the Nautical Almanac in a manner which will be hereafter explained. Having the meridian zenith distance and declination, the latitude is to be found by the following rules. CASE I.

When the object rises and sets.

RULE. If the object bear south, when upon the meridian, call the zenith distance north; but if the bearing be north you must call the zenith distance south. Place the zenith distance under the declination, and if they are of the same name add them together, but if they are of different names, take their difference; this sum or difference will be the latitude which will be of the same name as the greatest number.

CASE II.

When the object does not set, but comes to the meridian above the horizon twice in 24 hours.

Many stars are always above the horizon of certain places of the earth, and in high latitudes the sun is sometimes above the horizon for several days, in which case the meridian altitude may be observed twice in 24 hours; that is, once at the greatest height above the pole, and again at the lowest height upon the meridian below the pole. In the former case, the latitude is to be found by the preceding rule, but in the latter, by the following

RULE. Add the complement of the declination to the meridian altitude; the sum will be the latitude, of the same name as the declination.

NOTE. When the sun or star is on the equator, or has no declination, the zenith distance will be equal to the latitude of the place, which will be of the same name as the zenith distance.

When the sun or star is in the zenith, the declination will be equal to the latitude, and it will be of the same name as the declination.

To find the latitude by the meridian altitude of the sun or star.

EXAMPLE I.

EXAMPLE II.

Suppose that at the end of the sea day, Suppose that at the end of the sea day, June 21, 1824, in the longitude of 60 W. April 14, 1824, in the longitude of 140° È. the meridian altitude of the sun's lower from Greenwich, the altitude of the sun's lib bearing south, was found by a fore-lower limb by a fore-observation was 60 observation to be 40 6-required the la-25' when on the meridian and bearing south, titude, supposing the correction of the ob- the correction for dip, semi-diameter and served altitude for parallax, dip, and semi-parallax being 12 miles-required the lati diameter to be 12 miles? Observed altitude...........

Par. dip, and semi-diam. add

tude?

40° 6'
12

Observed altitude..
Correction.....

add

.60° 25' 12

In this rule the sun is supposed to be the fixed point, and the zenith is referred to it. Thus, if the 1 bears south from an observer (or from his zenith) the zenith bears north from the sun, and it is thus Aster bearing which is used in the rule.

⚫ The refraction being smell is here neglected.

S

EXAMPLE VI.

EXAMPLE III. Suppose that at the end of the sea day, By a back-observation with a quadrant of May 15, 1824, in the meridian of Green-reflection, the meridian altitude of the sun's wich, the meridian altitude of the sun's lower lower limb was 25° 12' when the declination limb bearing north was found by a fore-ob-was 21° 14' S. and the eye of the observer 40 servation to be 30° 6', the correction for feet above the horizon, the sun bearing S.parallax, dip and semi-diam. being 12 miles-required the lat. of the place of observation? required the latitude?

Observed altitude....

Par. dip, and senti-diam.... add

Observed altitude
Semi-diameter...

250 12/

...300 6'
12

.sub.

16

24 56

25 2

Refraction

subtract

0

Subtract from.

.90 00

0 N.

Declination

21 14 S.

Latitude

43 46 N.

EXAMPLE VII.

Suppose that on January 1, 1824, an observer 17 feet above the water finds by a fore-observation that the altitude of Sirius

Suppose that at the end of the sea day, is 53° 33' when passing the meridian to the Nov. 17, 1824, in the longitude of 80° E. southward; required the latitude of the from Greenwich, by a fore-observation the place of observation? meridian altitude of the sun's lower limb Observed altitude was 50 6', bearing south; the eye of the Dip of horizon observer being 17 feet above the surface of

observation the altitude of the sun's lower limb on the meridian below the pole 8° 14': required the latitude?

The sun being below the pole at 12 hours
By a fore-observation, the meridian alti-before the end of the sea day June 13, the

tude of the sun's lower limb was found to
be 40° 20′ bearing south of the observer, the
declination being 9° 56′ N. and the eye 26
feet above the horizon-required the lati-
tude of the place?
Observed altitude
Semi-diameter.

* The parallax being small is bere neglected, and the sun's semi-diameter is supposed to be 16'. †The north polar distances of these bright stars are given for every 10 days in the Nautical Almanac when great accuracy is required, the declinations deduced from these may be used instead of the numbers in Table VIII.

The parallax being small is neglected.

EXAMPLE IX.

EXAMPLE XI. .

Suppose that on January 10, 1824, an Suppose that by a back-observation with observer 18 feet above the water, finds the a sextant the lower limb of the sun's image altitude of the north star, when on the me- was brought to the back horizon, and the ridian below the pole, to be 36° 23′ by a angle shown by the index was 106° 12', the fore-observation; required the latitude of altitude of the observer being 22 feet and the place of observation?

Observed altitude...
Subtract dip. 4'. ref. 1'.

True altitude

Comp. declin. Tab. VIII.*

Latitude.....

36° 23'
5

36 18

1 38 N.

37 56 N.

the correction for semi-diameter, parallax,
and dip being (as usual) about 12'; re-
quired the true latitude, supposing the de-
clination to be 200 S. and that the sun bore
north at the time of observation?

Observed angle.......... .....1060 12/
Dip and semi-diam.....add

12

It was observed in the directions for finding the meridian altitude of an object, that an error would arise if the ship were in motion, or the sun's declination should vary. The amount of this correction may be estimated in the following manner.

Find the number of miles and tenths of a mile northing or southing made by the ship in one hour, and also the variation of the sun's declination in an hour expressed also in miles and tenths. Add these together, if they both conspire to elevate or depress the sun, otherwise take their difference, which call the arch A. Find in Table XXXII. the arch B, expressed in seconds, corresponding to the latitude and declination; then the arch A, divided by twice the arch B, will express the time in minutes from noon when the greatest (or least) altitude was observed. Moreover, the square of the arch A, divided by five times the arch B, will be the number of seconds to be applied to the observed altitude to obtain the true altitude which would have been observed if the ship had been at rest.

Thus if the ship sailed towards the sun south 12 miles per hour, the declination increasing northerly 1' per hour, we should have A=11+1=12. If the latitude was 420 N. declination 20 S. we should have by Tab. XXXII. B=2". In this case the time from noon is 4=3 minutes, and the correction of altitude 14-18 seconds only.

*The north polar distances of these bright stars are given for every 10 days in the Nautical Almanac ; then great accuracy is required, the declinations deduced from these may be used instead of the num

Lers in Table VIIL

The refraction and parallax being only a few seconds are neglected.

The refraction being small is neglected.

TO FIND THE LATITUDE

BY THE

MERIDIAN ALTITUDE OF THE MOON.

THE latitude may be found at sea by the moon's meridian altitude morc accurately than by any other method, except by the meridian altitude of the sun; but to do this it is necessary to find the time of her passing the meridian, and her declination at that time. To facilitate these calculations we have given the Tables XXVIII. XXIX. and XXX. The uses of which will evidently appear from the following rules and examples.

To find the true time of the moon's passing the meridian.

In the sixth page of the Nautical Almanac, find the time of the moon's coming to the meridian of Greenwich for one day earlier than the sea account; and also the time of her coming to the meridian of Greenwich the next day, when you are in west longitude, but the preceding day when in east longitude; take the difference between these times, with which you must enter the top column of Table XXVIII. and against the ship's longitude in the side column will be a number of minutes to be applied to the time taken from the Nautical Almanac, for the day immediately preceding the sea account, by adding when in west longitude, but subtracting when in east longitude; the sum or difference will be the true time of passing the meridian of the given place.

EXAMPLE.

Required the time of the moon's passing the meridian of Philadelphia, April 19, 1820, sea account?

The day preceding the sea account is April 18, on which day the moon passed the meridian of Greenwich at 5h. Sm.; and being in west longitude, I find also the time of her passing the meridian the next day, 5h. 55m.; the difference of these two numbers is 52m.; with this I enter Table XXVIII. and at the top find 52′; under this and opposite 75° (the longitude of Philadelphia) is the correction 11', to be added to 5h. Sm.; therefore the time of passing the meridian of Philadelphia is April 19d. 5h. 14m. sea account; ór April 18d. 5h. 14m. P. M. civil account.

To find the moon's declination when on the meridian.

Find the time of the moon's coming to the meridian as above; turn the ship's longitude into time, (by Table XXI.f) and add it thereto if in west longitude, but subtract it in east, the sum or difference will be the time at Greenwich.-Take out the moon's declination from page 6th of the Nautical Almanac, for the nearest noon and midnight; and note the difference of the declinations if of the same name, but their sum if of different names; enter Table XXX. with this sum or difference at the top, and the time at Greenwich in the side column, under the former and opposite the latter will be the correction to be applied to the declination which stands first in the

Taking the time one day earlier than the sea account reduces it to astronomical time used in the Nautical Almanac.

Longitude may be turned into time without the help of Table XXI. by multiplying by 4 sexagesi mally, and putting the product one denomination lower; and by dividing by 4, time may be turned into degrees, &c. Thus 30X4=320-5b. 20m. and 15o 16 × 461′ 4′′-1h. 1m. 4s. ; in like manner th. 20m. or 80m. divided by 4, gives 20, Sh. 16m. or 196m. divided by 4, gives 49°, which agree with the Table. If the ship be farnished with a chronometer regulated to Greenwich or mean time, this part of the operation will he saved, for by applying the equation of time, Table IV. A., with a contrary sign to that in the Table, the apparent time at Greenwich will be obtained, as in the explanation prefixed to the Tables If the time at Greenwich be exactly noon or midnight, the true declination will be given by the Nau tical Almanac, without the trouble of referring to Table XXX.

Nautical Almanac; additive, if that declination be increasing; subtractive, if decreasing; the sum or difference will be the true declination at the time of passing the meridian.

NOTES.

1. By the above rule, the day of the month on which the moon passes the meridian must be taken one less than the sea account: and when you add the longitude (turned inte time) to the time of passing the meridian, and the hours of the sum exceed 24, you must subtract 24h. and add one to the day of the month; if the longitude be subtractive and greater than the time of passing the meridian, you must, previous to the subtraction, add 24 hours to the time of passing the meridian, and subtract one from the day of the month; the sum or difference will be the time at Greenwich. If this time be less than 12 hours, you must take out the declination for the preceding noon and the following midnight; but if the time exceed 12 hours, you must take out the declination for the preceding midnight and the following noon.

2. When one of the declinations taken from the Nautical Almanac is north and the other south, the difference between the correction of Table XXX. and that declination which stands first in the Nautical Almanac, will be the true declination, which will be of the same name as that first declination, when the correction of Table XXX. is less than the first declination, but if greator of a contrary name.

3. In the same manner we may find the declination for any time in the day, by making use of the given time instead of the time of the moon's passing the meridian. 4. In the above rules the second differences of the moon's motion are neglected. In cases where very great accuracy is required, the calculation may be made as in Problem I. of the Appendix.

EXAMPLE.

Required the moon's declination at the time of her passing the meridian of Philadelphia, April 19, 1820, sea account?

The time of passing the meridian of Philadelphia was found in the prereding Example to be April 19d. 5h. 14m. sea account, or April 18th. 5h. 14m. by Nautical Almanac account; this being added to the longitude of Philadelphia, in time 5h. 1m. nearly, the sum is the time at Greenwich, April 18th. 10h. 15m. The declination April 18th. at noon, was 28° 26' N. and on April 18th. at midnight 27° 48′ N. the difference being 38', this being found at the top of Table XXX. and the time 10h. 15m. in the side column, the number corresponding is 33', which subtracted from the first declination 23° 26' leaves the declination required 27° 53′ N.

At the time of the moon's passing the meridian you must observe the altitude of her upper or lower limb, and correct this altitude for semi-diameter, dip, parallax, and refraction, and you will obtain the central altitude, with which and the declination you may find the latitude by the rules before given. Or you may correct the observed altitude by the following approximate method which shortens the calculation, and is sufficiently accurate, especially when the dip is about 4 or 5', which is nearly the value in common observations at sea.

To find the latitude by the moon's meridian altitude, obtained by a fore-ob

servation.

To the observed altitude of the moon's lower limb add 12 minutes, but if her upper limb was observed, subtract 20 minutes; with this altitude enter Table XXIX. and take out the minutes corresponding and add thereto, the sum will be the central altitude of the moon;* with this altitude and the moon's declination found as above, the latitude may be found as by a meridian altitude of the sun.

In ealculating accurately the moon's central altitude, you must proceed in the following manner: Find the time of the moon's passing the meridian reduced to Greenwich time as above, take out the moon's horizontal parallax and semi-diameter for this time, from the seventh page of the month of the Nautical Almanac, increase the semi-diameter by the correction in Table XV. add this augmented semidiameter to, or subtract it from the observed altitude according as the lower or upper limb was observed by a fore-observation) subtract the dip of the horizon taken from Table XIII. and add the correction for parallax and refraction (which may be easily found by Table XIX. by subtracting the correction ford in that table from 50 42) and the sum will be the correct central altitude.