1

tude, his corrected altitude 38°.19', and his azimuth S. 72°.13' E.; required the hour of the day and the latitude of the place?

Answer. The hour angle is 52°.30' 3h.30', the hour from noon, hence the time is 8h.30^ A. M. and the latitude 51°.32' North.

When the altitude of the sun and his declination are equal, the triangle is isosceles, and the hour angle is equal to the azimuth. But when the altitude of the sun is less than the declination, the case is ambiguous.

2. Given the sun's declination=19°.39' North, his altitude corrected=38°.19′ at 3h.30′ in the afternoon; required the latitude of the place, supposed to be north, and the sun's azimuth?

Answer. This problem is ambiguous, the azimuth may be N. 72°.13'.30" W. and the latitude 9°.16' N. or the azimuth may be N. 107°.46'.30′′ W. and the latitude 51°.30′.54′′ N. But when the sun's altitude is less than the declination, the problem will be limited.

3. Given the latitude of the place=51°.32′ N. the sun's declination 19°.39′ North, the sun's azimuth S. 72°.13′ W.; required the sun's altitude and the hour of the day?

Answer. The sun's altitude is 38°.18'.30", the hour angle 52°.30.20", and the time from noon 3h.30'. If the latitude of the place be less than the sun's declination, when both have the same name, the case will be ambiguous; see the following example.

4. Given the latitude of the place=13°.30' North, the sun's declination=23°.28′ N. the sun's azimuth N. 67°.30' East; required the sun's altitude and the hour of the day?

Answer. The angle opposite to the complement of latitude may be either acute 78°.20', or obtuse=101°.40'. In the former case the sun's altitude is 7°. 7′ nearly, and the time from noon 5.52'.12′′; and in the latter case the sun's altitude is 57°. 4.34", and the time from noon 2h.12.45". Hence it follows that in this latitude on the longest day, the sun appears upon the same point of the compass twice in the forenoon and twice in the afternoon.

5. Given the latitude 13°.30' North, the sun's corrected altitude 60°.30′ at two o'clock in the afternoon; required the sun's azimuth and its declination?

Answer. The angle opposite to the complement of latitude may be either acute 80°.52′, or obtuse 99°.8'; in the former case the declination is 10°.21.42"; and the azimuth 87°.12', and in the latter case the declination is 20°-37'.34", and azimuth 71°.51'.30".

6. In latitude 51°.32′ North at 3.30′ in the afternoon, the

sun's azimuth was S. 72°.13' West; required the sun's altitude, his declination, and the angle formed by the intersection of the vertical circle, and the azimuth circle passing through the sun?

Answer. The sun's altitude - 38°.18'.53", the declination= 19°.39.19", and the angle zenith SN formed by the vertical and azimuth circle=38°.58.28".

7. The sun's corrected altitude at 8b.30' in the forenoon was 38°.19', the angle of position, viz. formed by the vertical and azimuth circles, was 38°.58'; required the sun's azimuth, declination, and the latitude of the place?

Answer. The latitude is 51°,32'.34": but it cannot be determined by the data whether it be north or south, the sun's declination 19°.39′.19′′, and the azimuth=72°.13' from the east or west.

=

PROBLEM XVI. (Plate III. Fig. 10 and 11.)

(N) Given two altitudes of the sun and the time between the observations, to find the latitude of the place.

EXAMPLE I. (Plate III. Fig. 10.)

Given the sun's declination 19°.39' North, his altitude in the forenoon 38°. 19', and at the end of one hour and a half the same morning the altitude was 50°.25'; required the latitude of the place, supposing it North?

On a supposition that the sun's declination is the same at both observations, (which will not materially affect the result of the operation, when the observations are taken at no great distance from each other, and the time be not near the equinoxes;) the triangle AND is isosceles, a being the place of the sun at the first observation, and D its place at the second, and ANND. Draw NG perpendicular to AD, then will AG=GD and GND AND half the elapsed time. (K. 143.)

1. In the right-angled triangle NGD,

There is given ND=70°. 21′ the sun's distance from the pole at each observation, and the angle GND=11°.15' time, to find GD and AD.

*

Rad x sine GDsine GND X sine ND.

(log sine GND+log sine ND)-10=log sine GD.

or, rad

elapsed

sine ND-70°.21':: sine GND=11°15′; sine GD=10o,

35'. 13", the double of which is 21°.10′.26′′-
=AD.

2. To find the L GDN.

*

Rad x cos ND Cot GND X cot GDN

(10+log cos ND)-log cot GND=log cot GDN. or, cot GND=11°.15': rad::cos ND=70°.21′: cat GDN 86°.10'.24".

3. To find the L ADZ.

In the oblique-angled triangle ADZ, there is

First co-alt Az=51°.41'

Given Sec. co-alt Dz-39°.35'

Hence ADZ 570.41'.33"

And ADZ115°.23'. 6′′

The ADZ being greater than the GDN, this example belongs to Figure 10th.

And ZADZ GDN 115°.23'.6"-86°.10'.24" 29°.12.42" =LNDZ.

4. To find Nz the co-latitude.

In the oblique triangle DzN, there is
The co-decl ND 70°.21'

Given Second co-alt Dz=39°.35'

And the Ndz=29°12′.42"

To find Nzthe co-lat

= 38°.28'.11".

This may be done by the rules Case III. (N.245.) Or by Formula I. Case VIII. page 203. viz.

Hence the latitude is 51°.31'.49" North.

(0) The principal use to which this problem is applied, is in questions of a similar nature with the preceding; but several other things may be determined from the same data; such as the hour from noon when each altitude was taken, the azimuth at each observation, the difference of azimuths, &c. It will be sufficient in this place to point out the method of solution of these less useful cases.

The hour from noon when the second altitude was taken, is measured by the angle DNZ, for O is the place of the sun when on the meridian. Now in the triangle DNZ, we have ND= 70°.21', DZ 39°.35', ZN = 38°.28'.10", and the angle NDZ29°.12′.42′′, to find the angle DNZ the hour from noon = 30° or 2 hours; consequently the second altitude was taken at 10 o'clock, at the first at 8h.30′ in the forenoon. Again, in the triangle ANZ we have AN=70°.21', the sun's polar distance at the first observation, Az=51°.41′ his zenith distance, and the angle ANZ 3b.30′ or 52°.30′ the hour from noon when the first altitude was taken, to find the angle AZN 107°.47' the azimuth

from the North at the first observation. Also, in the triangle DNZ we have DN 70°.21' the sun's polar distance at the second observation, DZ=39°.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 132°.21′.12′′ the azimuth from the north at the second observation. The difference of the azimuth is therefore 24°.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 53°.29', and at 1.17′ 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=35°.45′ the angle AND.

Half elapsed time 1h.11′.30"-17°.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 NDAN=67°.20', and the angle GND=170.52'.30", whence the angle GDN 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. 37°.12', and the side AD=32°54′22′′, to find the angle ADZ=65°.44′.40"; hence the difference between GDN and ADZ gives NDZ=17°.10′.16′′.

The Z GDN being greater than the ADZ, this example belongs to Figure 11.

III. In the triangle DZN there are given ND the polar distance=67°.20', pz the second co-altitude=37°.12', and the angle NDZ 17.10.16", to find the co-lat. Nz 32°.51′.36′′;

hence the latitude is 57°.8'.24".

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

(P) If the times as given in Example II. and such like, were exactly true, one altitude only would at all times determine the latitude. 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.

* 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.