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(K) The hour found by calculation, as above, is the distance of the star from the meridian of the place of observation ; consequently when the star is eastward of the meridian at the time of observation, this hour angle must be subtracted from the time of the star's culminating, and when westward it must be added.

*'s right asc. (1822, Tab. VIII.)+24h. =315.82.59" O's right ascen. Sept. 7th, 1822, (Naut Alm.)=11. 2. O

Time of *'s culminating nearly

=20. 27. 59

o's right ascension, Sept. 7th, 1822, at noon =llh 2' 0" o's right ascension, Sept. 8th, 1817, at noon =11. 5. 36

Increase of the O's right ascension in 24 hours = 0. 3. 36 Then 24h.3.36" : 24h:: 205.27.'59" : 20h.24.55" time

of Procyon's culminating at Greenwich. Then, 15° : 10"::30°.18' : 20", hence 20h.24.55" +20'=209.25'.15" the time of the *'s culminating in longitude 30°.18'. E.

Lastly. 20°.25'.15" — 46.2.1"=165.23'.14" the correct time.

(L) In order to attain the greatest accuracy several altitudes of a star should be observed in succession, and if two or more separate stars be observed, and the error of the watch be deduced from each star, the medium result will be more correct than that derived from the observation of a single star.

If an equal number of stars can be observed on each side of the meridian, and nearly equidistant therefrom, the errors arising from the imperfection of the instrument, &c. will be rendered almost insensible.

EXAMPLE II.

On the 29th of January 1822, in latitude 530.24' North, and longitude 25°.18' West; suppose the following altitudes of Procyon to the west, and Alphacca to the east of the meridian, were observed by two separate persons at the same instant of time; required the error of the watch.

The right ascension of Alphacca being 156.27.9" declination 270.19.12" N., and of Procyon 7h.29'.59'' and 59.40'.29" N. The sun's right ascension at noon on the 29th=20h.46'.11" (Naut. Alm.) and on the 30th=204.50'.17".

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Mean 14. 58.38 Mean 19. 53. 55 Mean 42. 8. 55

Answer. The hour 2 by Alphacca's corrected altitude is 3". 40'.47", and the * culminates at 18h.37'.31", hence the correct time=14h56'.54".

The hour 2 by Procyon's corrected altitude is 4".15'.55", and the * culminates at 104.41'.42", hence the correct time=14". 57.37". The mean time between the two is 14h.56'.55", consequently the watch is 1'.43" too fast.

EXAMPLE III. In latitude 48o.56' North, longitude 66° West, suppose the altitude of Aldebaran, when west of the meridian, to be 22o.20'. 15" on the 15th of April 1822; required the apparent time of observation ?

The right ascension of Aldebaran being 45.25'.43" and declination 16°.8'.36" N. Also, the O's right ascension at noon on the 15th (Naut. Alm.) = 15.32.23", and on the 16th= 19.36'.5'.

Answer. The hour / by the *'s corrected altitude=45.57'. 33", and the * culminates at 25.52'.9', hence the correct apparent time=74,49'.42".

(M) SCHOLIUM. The three preceding problems (XIII, XIV, and XV.) are all included in one triangle (Zenith ns Plate III. Fig. 1.), but they have been considered separately, on account of their importance, and to prevent confusion.

Any three of the five quantities, viz. the sun's or star's declination, the latitude of the place, the altitude of the object, the azimuth, and the hour of the day being given, the rest may be found.

A sixth quantity might be taken into consideration, were it materially useful, viz. the opposite to the complement of the latitude, being formed by the intersection of the vertical circle Zsc, and the azimuth circle Nsd.

PRACTICAL EXAMPLES. 1. Given the sun's declination 190.39' North, in North lati

1

tude, his corrected altitude 38° 19', and his azimuth=S. 720.13 E.; required the hour of the day and the latitude of the place ?

Answer. The hour angle is 52°.30ʻ=9h.30', the hour from noon, hence the time is 8h.30A. M. and the latitude 510.32 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=19o.39' North, his altitude corrected=389.19' at 3h.30' in the afternoon; required the latitude of the place, supposed to be north, and the suu’s azimuth?

Answer. This problem is ambiguous, the azimuth may be N. 720.13.30" W. and the latitude 9o.16' N. or the azimuth may be N. 1079.46'.30" W. and the latitude 510.30'.54" N. But when the sun's altitude is less than the declination, the probl will be limited.

3. Given the latitude of the place=51°32' N. the sun's declination=190.39' North, the sun's azimuth=S. 720.13' W.; required the sun's altitude and the hour of the day?

Answer. The sun's altitude is 38o.18.30", the hour angle 620.30'.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=130.30' North, the sun's declination=230.28'N. the sun's azimuth = N. 670.30' 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=789.20', 'or obtuse=101o.40'. In the former case the sun's altitude is 7o. 7' nearly, and the time from noon 56.52'.12"; and in the latter case the sun's altitude is 57o. 4.34", and the time from noon 25.12.45". 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 139.30' 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=999.8'; in the former case the declination is 100.21'.42"; and the azimuth 870.12', apd in the latter case the declination is 20°.37.34", and azimuth 710.51.30".

6. In latitude 51°.32' North at 3h.30' in the afternoon, the

PROBLEM XVI.

sun's azimuth was S. 720.13'.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 = 38o.18'.53", the declination= 190.39'.19", and the angle zenith sn formed by the vertical and azimuth circle=38°.58.28".

7. The sun's corrected altitude at 86.30' 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 de. clination – 199.39'.19", and the azimuth=720.13' from the east or west.

(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 190.39' North, his altitude in the forenoon 38o. 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 p its place at the second, and AN=ND. 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'= } elapsed time, to find GD and ad.

Rad x sine Gpsine GND X sine np.

•: (log sine GND +log sine ND)-10=log sine go. or, rad : sine ND=700.21'::sine GND=11°15' : sine gd=10°. 35'. 13", the double of which is 21°.10ʻ.26"=AD.

2. To find the LGDN. Rad x cos ND=cot GND X Cot GDN ::: (10 +log cos ND)- log cot GND=log cot GDN. or, cot GND=11o.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=390.35'

Hencebadz=570.41'.33"

And Ladz=1150.23'. 6" And

AD=210.10'26" ) The L Adz being greater than the LGDN, this example belongs to Figure 10th. And LADZ

LGDN -- 1150.23'.6" -86°.10'.24" = 29o.12.42" = _NDZ.

4. To find nz the co-latitude. In the oblique triangle dzn, there is

7 To find szthe co-lat Given Second co-alt dz-399.35 And the _ NDZ=29°.12.42"

= 380.28'.11". This may be done by the rules Case III. (N.245.) Or by Formula I. Case VIII. page 203. viz.

cos / NDZ . tang ND Tang<=

; and cosNz= .cos(DZ~). rad

cos o Cns / NDZ=29° 12.42" 9.94092 ||Cos nd= =700.21 Tang ND = 70°.21 10•44725 Cos (02~0)=28°.10.2" 9 94526

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COS ND

9.52669

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Hence the latitude is 510.31'.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 anz, for O is the place of the sun when on the meridian. Now in the triangle DNZ, we have ND= 70°.21, Dz=399.35', zn = 380.28'.10", and the angle ndz= 29o.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 85.30' in the forenoon. Again, in the triangle and we have an=700.21', 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=1070.47' the azimuth

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