Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

EXAMPLE III.

[ocr errors]

хс XT

Хт

Suppose the distance from an unknown star at T was measured to be 25o.42.10", from a star at x, whose latitude is 39o.33.30" South, longitude 35.11o.13'.44"; and 51°6'.56" from another known star at c, whose latitude is 229.51'.57" North, and 28.189.57.57" longitude ; required the latitude and longitude of the unknown star.

First. In the triangle xmc, you will have given

xm=the complement of x's latitude,

cm=the distance of c from the same pole of the ecliptic, and the angle xmc=the difference between their longitudes=vo.

Hence xc will be found=65o.47'.42", and the angle mxc= 153o.17.21".

Secondly. In the triangle xtc, are given

To find the angle cxt=490.14 and tc

Now mxc~CXT=1049.3.21"=mXT.
Thirdly. In the triangle mxt, are given

To find tm=74o.1'.46", the distance of the star
T from the pole of the ecliptic, and the angle

xmt=110.41.58", the difference between the lonLmXT

gitude of x and T. (Y) By the assistance of the foregoing problem, and a knowledge of any one fixed star's true situation, the places of all the rest may be determined. Indeed, it is by this problem principally that astronomers have rectified the places of all the fixed stars; and thence, by a similar mode of proceeding, found the true places of the planets. Leadbetter tells us in his Astronomy, page 230, that he made use of this method for constructing his astronomical tables. The preceding examples contain every variation that can happen with respect to the stars' latitudes and longitudes, but the same things may be determined from their right ascensions and declinations, as in the following example:

EXAMPLE IV. (Plate III. Fig. 9.) Suppose the distance of a star or planet at y, was observed to be 65o.47.42", from Sirius the Dog-star at x, whose right ascension is 99.0.21", and declination 16°26'35" South; and to be 510.6'.56" distant from Procyon at t in the Little Dog, whose right ascension is 112.6'.47", and declination 59.45'.3" North; it is required to find the right ascension and declination of this unknown star or planet

XT

а

First. In the triangle xnt, we have The angle xnt=13°.6'.26''=the difference between the right ascensions of x and t, measured by the arc RQ.

XN=1060.26.35"=x's distance from the pole n.

TN= 84°.14.57"=T's distance from the same pole. Hence two sides and the included angle are given, to find the third side xt=25°.42'.10", and the angle Nxt=31o.22'.

Secondly. In the triangle yxt, we have given Yx=65°.47'.42" To find the angle ext=490.14'. YT=51. 6. 56

Then YXT

NXT=17. 52. XT= 25. 42.10

=YXN. The Zyxt being greater than the 2 Nxt shews the right ascension of the unknown star at y to be less than the right ascension of that at x. Thirdly. In the triangle xyn, are given

To find yn the complement of the dexy= 65°.47-42"

clination by, and the angle ynx the xn=106. 26.35

difference between the right ascenLYXN= 17. 52.

sions of y and x, BR.

YN=44°.14', consequently by=450.46' N. You will find ynx=239.38.55", hence the right ascension

of y=AB=750.21'.26". (Z) In the same manner we may find the distance between any two places on the earth; suppose x and t to be two places whose latitudes and longitudes are known, then in the triangle XNT, xn is the distance of x from the pole, TN the distance of T from the same pole, and the angle xnt, measured by the arc RQ, the difference of longitude between the two places, hence the distance xt is easily found. In a similar manner the distance between T and y, or y and x may be obtained.

EXAMPLE V.

What is the extent of Africa, from Cape Verd, latitude 14°. 46' N., longitude 170.47'W. to Cape Guardafui, latitude 11°. 47' N., longitude 519.35' E.?

Answer. 670.19' the arc of a great circle contained between the places, or 4678 English miles, of 69, to a degree, or 4039 geographical miles, of 60 to a degree.

PROBLEM XIII. (Plate III. Fig. 1.) (A) Given the latitude of the place, the sun's declination and

[ocr errors]

1

[ocr errors]

EXAMPLE, I. In latitude 510.32 North the sun's altitude was observed to be 50°, and his declination 230.28' North; required his azimuth from the North.

In the triangle szn, we have given Sn=660.32'. the co-declination, or sun's polar distance; for

DS is the sun's declination North. SZ=40. 0. the co-altitude of the sun; for cs is his altitude

above the horizon Ho. Zn=38. 28. the co-latitude of the place, or what on the

elevation of the pole wants of 90°. To find the angle szn, measured by the arc of the horizon co, the sun's distance from the North.

Call the co-latitude zn the base, and draw the perpendicular sf. Then (W.231.) tang I ZN=z=199.14

9.54269 : tang (SN+sz) = 53o.16)

10.12710 :: tang 1 (SNSZ) = 139.16

9:37250 : tang ft = 420.10

9.95691 Take the difference between ft and half the complement of latitude. Then Radius, sine of 90°.

10.00000 : tang ft~ zt=22°.56'

9.62645 :: tang of the sun's altitude cs= 50°

10.07619 : sine of azimuth from the east or west=30°.17' 9.70264 Hence 90° +30°.17'=120°.17 the sun's azimuth from the North.

(B) A slight review of Figure 1. Plate III. will naturally point out to us the following observations:

If the latitude of the place and the declination of the sun be of the same name, the sun is less than 90° from the elevated pole; therefore the co-declination is the polar distance.

If the sun or star be in the equinoctial, it is 90° from the pole, and forms a quadrantal triangle. If the declination be of a contrary name with the latitude, the distance of the object from the elevated pole will be equal to the declination increased by 90°

(C) The preceding solution is on a supposition that the sun's declination, as taken from the Nautical Almanac, remains the same in all places, and at all times of the day. But this cannot be strictly true, for the declination is subject to a daily variation, and is adapted to the meridian of Greenwich. (B. 270.) An azimuth will not be materially affected by this change, but it will make a considerable difference when we want to find the time of the day.

EXAMPLE II. Suppose the sun's altitude at London, latitude 510.32 North, to be 1o.14' when the declination is 20°.11' South, what is the azimuth from the North?

Answer. 1250.34'.

EXAMPLE III.

Given the latitude of the place 510.32' North, the sun's altitude 25°, his declination 4o.47' South; required the azimuth from the North ?

Answer. 137o.16'.

EXAMPLE IV.

Suppose in latitude 51°32' North, the altitude of the star Arcturus in Boötes was 44°.30', when its declination was 20°. 16' N.; required its azimuth from the North at that time.

Answer. 1170.8'.

EXAMPLE V.

:

In latitude 480.51' North, when the sun's declination is 18o. 30' North, and his altitude 520.35'; what is his azimuth from the North?

Answer. 1349.36'.8". (D) This problem is not only useful for finding the variation of the compass, but it will be found very convenient for determining the true meridian, and the four cardinal points. For, if with a line representing the vertical on which the sun is, an angle be made equal to the supplement of the azimuth from the North, the meridian will be obtained: provided care has been taken to correct the sun's altitude by refraction.

(E) An azimuth may be constructed by a scale of chords, sufficiently accurate for Nautical purposes; Thus, With the chord of 60° describe a circle (Plate IV. Fig. 2.), set off the complement of the latitude from z to P, and the complement of the altitude from z to a, draw zon through the centre c and aca at right angles to it. Set off the polar distance from P to p, draw pp through the centre c, and from the point i where it intersects aoa draw ir parallel to zon, and through c the centre draw cs parallel to aoa.

With c as a centre, and the distance oa, cross the line ir in m; through m draw cn, then sn, measured on a scale of chords will be the azimuth from the south (in North latitude) if the point i fall on the left-hand of zn, and from the north if it fall on the right-hand.

See the Mathematical Repository, vol. i. page 283, second edition, 1799, and vol. ii. page 26 and 27. See also Dr. MacPROBLEM XIV. (Plate III. Fig. 1.) (F) Given the latitude of the place, the sun's declination and altitude, to find the hour of the day.

EXAMPLE I. At London, latitude 51°.32' North, the altitude of the sun's centre was 38°.19', and his declination 190.39' North, required the hour of the day?

In the triangle szn, we have given
sn=70°.21' the sun's distance from the elevated pole.
sz=51. 41 the sun's co-altitude.
ZN=38. 28 the co-latitude.

To find the angle zns, measured by the arc ÆD, the distance between the meridian which the sun is on at the time of obseryation, and the meridian of the place of observation.

Since Æz is equal to the latitude on; it follows, that if the latitude and declination be both north or both south, the difference between them is equal to the meridional zenith distance; if the one be north and the other south, their sum is the meridional zenith distance. Hence the following general method of solution : Nat cos of merid zenith dist (51°.32'-19o.39')

310.53=84913 Nat sine of the sun's altitude

38. 19=62001

Remainder

22912

O

Common logarithm of
Secant of the latitude
Secant of the declination

22912=4.36007 519.32' = 20617 19o.39'= '02606

Num. in col.ris., Requisite Tables, 3h.30' 4:59230

Or, Increase the index by 5+

520.29'.20"-9.59230

O

Log. versed sine
Or add the logarithm of
5000000000= radius.

+9•69897

Divide by 2 | 19.29127

Log sine of half the hour angle 26°.14'.40"=9•64563

2

The hour angle

52.29.20=35.29'.57".20".

« ΠροηγούμενηΣυνέχεια »