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from the North at the first observation. Also, in the triangle DNZ we have on=700.21' the sun's polar distance at the second observation, pz=390.35' the zenith distance, and the angle DNZ=30° or 2 hours, the hour from noon when the second altitude was taken, to find the angle nzd=1329.21'.12" the azimuth from the north at the second observation. The difference of the azimuth is therefore=240.34'.12. The azimuths might have been found independent of the time, for in each of the above triangles, the co-declination, the co-altitude, and the co-latitude are given.

EXAMPLE 11. (Plate III. Fig. 11.) When the sun's declination was 22.40' North, his altitude at 10b.54' in the forenoon was 53o.29', and at 15.17' in the afternoon it was 52°.48'; required the latitude of the place of observation, supposing it to be North.

12b. O' Time of the first alt. 10. 54

The time from noon 1. 6
The time from noon 1. 17 when the second alt. was taken.

Elapsed time 2. 23=350.45' the angle and.

Half elapsed time 11.11'.30"=170.52.30"the angle GND. Supposing the sun's declination constant during the elapsed time, and is an isosceles triangle, and the method of solution is precisely the same as in the former example.

i. For ND=AN=670.20', and the angle GND=170.52'.30', whence the angle gon is found = 82°.54.56", and ad = 32°.54'.22".

II. In the triangle Adz are given az the first co-alt. =36'31', Dz the second co-alt.=370.12', and the side AD=32°.54.22", to find the angle adz=65°.44'.40'; hence the difference between GDN and adz gives Noz=170.10'.16".

The Lyon being greater than the Ladz, this example belongs to Figure 11.

III. In the triangle Dzn there are given nd the polar distance=670.20', oz the second co-altitude=370.12, and the angle ndz=17°.10'.16", to find the co-lat. N2 = 32°.51'.36"; hence the latitude is 570.8'.24".

The azimuth at the time of each observation may be found as in the preceding example.

(P) Kf the times as given in Example II. and such like, were exactly true, one altitude only would at all times determine the Jatitude. But we must not depend upon these times; they only serve to point out the elapsed time between the observations. If the watch go true during the elapsed time, or measure that time truly, it is of no consequence to the solution of the problem, whether it be too fast or too slow.

The methods of solving this important problem directly by trigonometry, as is done in the foregoing examples, have been considered by mathematicians as too troublesome for general practice. They have, therefore, contrived various other rules dependent upon an assumed latitude, in such a manner that when the latitude assumed is true, the result of the operation will exactly agree with it. But when the latitude obtained by calculation differs materially from the assumed latitude, the solution must be repeated. The

rule now chiefly used, and which is printed in almost every Treatise on Navigation, the Requisite Tables, &c., was 'first given by Mr. Douwes, examiner of the marine cadets at Amsterdam; this rule, and tables adapted to it, were first published in England, by Mr. R. Harrison, in the year 1759, but without any account of the construction of the tables. In the year 1760 the learned Dr. Pemberton gave a very circumstantial account of the construction of the tables, and the reason of the process; showed the nature and limits of the approximation, and estimated the errors which may arise from the several circumstances that occur in it. Vide Phil. Trans. 1760.

The following rule depending only on the elapsed time, the two altitudes and declination, is deduced from Examples I. and II. It is in all cases accurate, and in general less troublesome in the process than the common method by assumed latitudes. *

(Q) GENERAL RULE. I. Radius, is to the sine of the polar distance; as the sine of half the elapsed time, is to the sine of half the first arc, the double of which gives arc the first.

II. The co-tangent of half the elapsed time, is to radius; as the cosine of the polar distance, is to the co-tangent of arc the second. This arc is acute or obtuse like the polar distance.

III. Add the two zenith distances and arc the first into one sum. From half this sum subtract the greatest zenith distance.

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* Dr. Brinkley, Professor of Astronomy in the University of Dublin, has published a set of rules for solving this problem, which are annexed to the Nautical Almanac for the year 1822.

Add together the co-secant of arc the first, the co-secant of the least zenith distance, (rejecting the indices,) the sine of the half sum, and the sine of the remainder; half this sum is the logarithmical cosine of half the third arc, double of which gives arc the third. The difference* between arc the third and are the second gives are the fourth.

IV. Add the sine of the polar distance, the sine of the least zenith distance, and double the sine of half the fourth arc into one sum, and reject 30 from the index. Consider the remainder as a common logarithm, and find the natural number answering to it, to as many places of figures as the tables extend to, if the index be 9; if 8, to one less; if 7, to two less; and if 6, to three less (for this index can never exceed 9, nor fall short of 6). Subtract double this natural number from the natural cosine of the difference between the polar distance and the least zenith distance, the remainder will be the natural sine of the true latitude.

EXAMPLE III.

In a supposed latitude of 50°.41' N. at 10h.17'.30per watch, the true altitude of the sun's centre was 170.13', and at 116.17.30" it was 19o.41', the sun's declination on that day was 20° South; required the true latitude independent of the supposed one?

Here 116.17.30"-105.17.30"=1b the elapsed time, hence the elapsed time=30m. = 70.30'. ., I.

II. Rad, sine of 90° 10.00000 Cot į elap. time 70.30 10.88057 : sine pol. dist. 110° 9.97299 : rad, sine of 900.

10.00000 : : sine {elap. time 70.30 9•11570 :: cos. pol. dist. 110o.

9.53405 : sine į arc 1st 70.2.43" 9•08869

: cot arc 2d 92°.34'.41" 8.65348 Arc Ist=140.5.26"

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1. When the supposed latitude, or latitude by account, is less than, or nearly equal to, the declination, and both the same way, the magnetic azimuths of the sun ought to be attended to at the time of taking the altitudes. For when the magnetic azimuth at taking the least altitude is greater than the magnetic azimuth at taking the greatest altitude, counting from the elevated pole ;. the supplement of arc the second must be added to the supplement of arc the third to obtain arc the fourth.

2. Should the double natural number used near the conclusion of the problem, at any time exceed the natural cosine of the difference between the least zenith distance and polar distance, the latitude will be of a different name with the supposed latitude.

3. Hence it appears that in places near the equator where the azimuths increase and decrease slowly, the above method of solution requires a little more attention. However, should any doubt remain whether the latitude found by the general rule be true or not: find the fourth arc as directed in the first note, and use it instead of that found by the rule; one of the two results will most certainly be the true latitude, and some circumstances will generally occur that will seldom fail to point out which is the proper one.

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IV. Sine polar distance 110°

9.97299 Sine least zen. dist. 70°.19

9.97385 fourth arc

20.33.35" sine x 2 = 17.29986 Nat. number of this com. log.

176=7.24668

2

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Note. This example, by the approximating rules, requires a repetition of the process, vide page 18. Requisite Tables, Example 2.

EXAMPLE IV. In a supposed latitude of 510.0' North, when the sun's declination was 22o.23' South, some time past noon the sun's corrected altitude was found by observation to be 14°.46' and 16.22' afterwards, his altitude was 8°.27'; required the true latitude independent of the supposed one?

Answer. The true latitude is 500.33'.44" N.

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EXAMPLE V.

In a supposed latitude of 470.19' North, when the sun's declination was 12o.16' North, some time in the afternoon the true altitude of the sun's centre was observed to be 51°.59', and 21.50' afterward it was 499.9'; required the true latitude independent of the supposed one?

Answer. The true latitude is 49o.19.30" North.

EXAMPLE VI.

In a supposed latitude of 30°.0 North, when the sun's de clination was 10°.24' North, at 85.54' in the forenoon the sun's true altitude was 41°.30'; and at 1h. 24' in the afternoon his altitude was 61°.47'; required the true latitude independent of the supposed one?

Answer. The latitude is 31o.33.26" North.

EXAMPLE VII.

In a supposed latitude of 40°.o' North, when the sun's declination was 20:46' South, at 9b.20' in the forenoon the sun's altitude corrected was 330.11', and at 11.20' in the afternoon his altitude was 42°.44'; required the true latitude independent of the supposed one?

Answer. The latitude is 40°.50.10" North.

EXAMPLE VIII.

In a supposed latitude of 50°.40' North, when the sun's declination was 20° South, the true altitude of the sun's centre was 19o.41', and one hour afterward his altitude was 170.13'; required the true latitude independent of the supposed one?

Answer. The true latitude is 50° North.

EXAMPLE IX.

In a supposed latitude of 60° North, when the sun was on the equinoctial, or had no declination, the true altitude of his centre was 289.53', and two hours afterwards his corrected altitude was 200.42'; required the true latitude independent of the supposed one?

Answer. The true latitude is 59o.59.32".

In this example the polar distance is equal to 90°; and in all such examples, the elapsed time will be arc the first, and

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