EXAMPLE IX.

Suppose that on January 10, 1824, an

EXAMPLE XI.

Suppose that by a back-observation withi 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

EXAMPLE X.

the correction for semi-diameter, parallax;
and dip being (as usual) about 12'; re-
quired the true latitude, supposing the de
clination to be 20° S. and that the sun bore
north at the time of observation?

Observed angle...........106° 127
Dip and semi-diam.....add

Subtract....

Zenith distance‡.

Sun's declination

Suppose that by a back-observation with a sextant the lower limb of the sun's image was brought to the back horizon, and the angle shown by the index was 110° 10′, the sun being then on the meridian and bearing south, the declination being 20° 5' N. the Latitude sun's semi-diameter 16' and the observer 20

feet above the horizon; required the latitude?

Observed angle

Semi-diameter........add

12

110° 10'

16

110 26

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=22. 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 2=3 minutes, and the correction of altitude '18 seconds only.

*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 num Ders in Table VIII.

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 more 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 meridia, 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 750 (the longitude of Philadelphia) is the correction 11', to be added to 5h. 3m.; therefore the time of passing the meridian of Philadelphia is April 19d. 5h. 14m. sea account; or 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 sexagesimally, and putting the product one denomination lower; and by dividing by 4, time may be turned into degrees, &c. Thus 80°×4=320'==5h. 20m. and 15° 16' 461′ 4′′=1h. 1m. 4s.; in like manner th. 20m. or 20m. divided by 4, gives 20°, Sh. 16m. or 196m. divided by 4, gives 49°, which agree with the Table. If the ship be furnished with a chronometer regulated to Greenwich or mean time, this part of the operation will be 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 Nautical 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 into 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 greater 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 preceding 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 28° 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 calculating 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 found in that table from 50 42') and the sum will be the correct centrál altitude,

EXAMPLE I.

Suppose that on the 27th of June, 1820, sea account, in long. 800 W. from Greenwich, the meridian altitude of the moon's upper limb was observed to he 40° 0' bearing south; required the true latitude?

June 27th, sea account, is June 26th, by Nautical Almanac, on which day the moon passed the meridian of Greenwich at 12h. 45m. and the next day at 13h. 46m. the daily difference being 61m. In Table XXVIII. under 60 (which is the nearest number to 61 in the table) and opposite to the long. 800, stand 13m. which added to 12h. 45m. gives the time of passing the meridian, June 26d. 12h. 58m.

Suppose that on the 27th September, 1820, sea account, in long. 90 E. the meridian altitude of the moon's lower limb was observed 50° 0', bearing south, required the true latitude ?

September 27, sea account, is September 26, by the Nautical Almanac, on which day the moon passed the meridian of Greenwich at 16h. 17m. and the preceding day at 15h. 19m. differing 58m. in Table XXVIII. under 58' and opposite the long. 90° are 14m. which subtracted from 16h. 17m. leaves 16h. 3m. the time of passing the meridian of the place of observation,

Suppose that on the 29th November, 1820, sea account, in the longitude of 1500 W. the meridian altitude of the moon's upper limb was observed 60° 26', bearing north, required the true latitude?

Nov. 29, sea account, is Nov. 28, by the Nautical Almanac, on which day the moon passed the meridian of Greenwich at 19h. 20m. and the next day at 19h. 59m. differing 39m. In Table XXVIII. under 39′ or 40′ and opposite the longitude of 1500 stand 17m. which added to 19h. 20m. gives 19h. 37m. the time of passing the meridian of the place of observation.

FROM page 4th of the month of the Nautical Almanac, take out the time of the planet's passing the meridian on the day nearest to that on which the observation was made; this will be nearly the time of passing the meridian of the place of observation.*

Turn the ship's longitude into time, and add it to the time of passing the meridian, when in west longitude, but subtract in east, the sum or difference will be the time at Greenwich nearly. Take out the planet's declination, from the Nautical Almanac, for the times immediately preceding and following the day of observation, and note the difference of the declinations when they are of the same name, but their sum when of different names, and find the interval between these times marked in the Nautical Almanac; take also the difference between the time first marked in the Nautical Almanac and the time of observation at Greenwich (remarking that this time is one day less than the sea account;) then as the former interval of time is to the latter, so is the sum, or difference of declinations, to the correction of the declination taken first from the Nautical Almanac, additive if that declination be increasing, but subtractive if decreasing; the sum or difference will be the declination of the planet at the time of observation. But you must observe that if the correction of declination be greater than the declination first marked in the Nautical Almanac, their difference will be the sought declination, which will be of a different name from the first declination.

From the observed altitude of the planet (taken by a fore observation) subtract the refraction and dip, the latter being in general about four minutes, and the remainder subtracted from 90° will give the correct zenith distance nearly; with which, and the declination, the latitude may be found as by an observation of the sun.

* If you wish to find the time of passing the meridian more accurately, you must take a proportional part of the difference of the times of coming to the meridian given in the Nautical Almanac, in the same manner as in finding the declination of the planet.

This time is also given by a chronometer, as in note page 124, or in the explanation mèfixed to the tables.