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The radii of curvature R, R', are respectively the radii of the circles of least and greatest curvature at the given place, the radius of any oblique section being always intermediate. The difference between these radii is 2 ae (1-sin2 1) = 2 ae cos2 1, and it is greatest when l = 0.
EXERCISE. Find the lengths of the radii of curvature R, R', and r, at latitude 55° 40'. Ans. R 20,925,000, R' 20,970,000, and r = 20,947,500.
598. PROBLEM XIX.-Given the latitude and longitude of a place, and the bearing and distance of another place from the former, to find the difference of latitude and longitude of the two places, and also the bearing of the former in reference to the latter.
Let A and A' be the two places, P the pole, and PAM, PA'M' two meridians; also
the latitudes of A and A', PAPA', their difference of longi
A MAA', the azimuth of A' taken
A' = M'A'A, the azimuth of A taken
D = the distance AA' in feet,
the radius of curvature of the arc perpendicular to the meridian at A ;
D2 sin2 A tan l
D cos A
2 R2 sin 1"
which gives the difference of latitude in seconds were the earth spherical. But a correction must be introduced on account of the earth's ellipticity, and the formula then be
Dc COS A D2 sin2 A tan /
1 — 1' = ( +
D sin A
And the value of A' in degrees is found by the equation sin(l + "')
A' 180° + A—P
The last two formulas are the same for a sphere or spheroid, as no correction is required, on account of ellipticity, for the difference of longitude or the azimuth.
EXAMPLE. In the trigonometrical survey in France, it was found that the latitude of Panthéon was 48° 50′ 48′′-59, the azimuth of Dammartin observed at Panthéon 133° 44' 23"-03, and the distance between these two places 33494-32 metres; required their difference of latitude and longitude, and also the azimuth of Panthéon in reference to Dammartin.*
The distance being given in metres, it must first be reduced to feet.
1 metre: 33494.32 m.
L. D 109891
And it was found in the example to article 597 that R' = 20,962,000.
Also, p=1+ e + e cos 2 l = 1 +380 + 3ỗo × ·13388 = 1.003779.
To find the difference of latitude.
D cos A
D2 sin2 Atan /
R' sin 1"
L. R' (c)
3.2808992 feet: D in feet.
2 R'2 sin 1"
L. 750-714 = 2·875474 1.699
-) (1 + e + e cos 2 l)
749" 01512′ 29′′-015
*This example is taken from Francœur's "Geodesie," p. 211, in which, however, it is computed by a different process.
The arithmetical complements of the logarithms of the quantities in the denominators of the terms of the formula, are indicated above by (c), and are of course additive.
Since cos A is negative, the first part of the calculation gives a negative result, and
-750"-714+1′′·699 —— - 749"-015 - 12′ 29′′-015, which is only 0"-96 greater than that obtained by a different formula in Francœur's "Geodesie," p. 212, and with a value of e30365 instead of 300, which is adopted in article 584, as nearly a mean of the values that are most worthy of confidence.
Latitude of Panthéon
l' = 49
Latitude of Dammartin
= 1 = = 48° 50′ 48′′-59
To find the difference of longitude.
D sin A
R' cos 'sin l′′’
where '49° 3′ 17′′-6.
L. cos l'
L. sin 1"
= 7.321432 9.816465
To find the azimuth of Panthéon in reference to
A' 180° + A—P
P=1192"-13, 1 (l+1') = 48° 57' 3"-1,
1 (1-1)=-6' 14"-51.
L. cos (l—l') (c) =
0° 14' 59"-05
= 313 29
This value of A' is measured from the south in the same direction as A; but if it is to be measured in an opposite direction, and also from the south, then will
A' 360°-313° 29′ 23′′-98 — 46° 30′ 36′′·02. The values of P and A' differ only by 0"-02 from the results obtained by Francœur by means of different formulas.
When the value of R' is computed for a given latitude, it may be used without sensible error for all other places whose latitudes differ by less than 3 or 4 degrees from the former.
EXERCISE. The distance of Black Down from Dunnose is 314,307.5 feet; the latitude of Dunnose, 50° 37′ 7′′-3; the azimuth of Black Down, observed at Dunnose, contained between the meridian to the south and the direction of Black Down, is 95° 5' 7"-5, the latter place lying westward, and a little north of Dunnose; required the difference of latitude and longitude of these two places, and the azimuth of Dunnose in reference to Black Down.
Ans. l'′ = 50° 41′ 14′′·07, P : 1° 21′ 2′′·3, and A′ — 85° 57′ 37′′-25, reckoned from the south towards the east.*
THE RELATIVE AND ABSOLUTE HEIGHTS OF THE
599. The distances of the stations of a trigonometrical survey are generally so great that refraction sensibly affects the angles of elevation or depression-that is, the vertical angles of position of one station observed at another. (See art. 232, Part I.) The vertical angle of position of one station in reference to another being corrected for refraction, and the distance between them being known, the difference of altitude of the two stations can then be found.
*These are the results given in the " Encyclopædia Britannica," article Trigonometrical Surveying, and differ from those given in the survey of Britain, from the less correct values of the elements of the earth's figure assumed in the survey.
600. PROBLEM XX.-Given the observed angles of elevation or depression of two stations in reference to each other, and their distance, to find the correct angles and also the refraction, considering it equal for both angles.
Let C and W be the two stations, and O the earth's centre, or rather the centre of curvature (585), OC, OC', two radii of curvature through C and W, and CH, WH, two perpendiculars to them, that is, two horizontal lines through C and W. W appears from refraction to be elevated to w; and similarly C appears at c when seen from W; also, let C' be taken, so that OC' — OC, and let OW' OW; also, let angle
HCwe, the observed angle of depression of W,
COW o, the angle at centre of curvature,
C'CW =, the real angular depression of W below C, and CWW'v', the real angular elevation of C above W; then is 2 ro — (e + e'), and v = 10 − (e + r) ; when r is not required, then v = (e-e'). When e or e is an angle of elevation, its sign must be changed; also, if z and h are the distances between C and W, and the height C'W of C above W, and if v is expressed in seconds, then is
h = vz sin 1", or Lh6·6855749 + Lv + Lz.
EXAMPLE. In the British trigonometrical survey, it was found that two stations of a large triangle, connecting the borders of England and Scotland, were 235018-6 feet, or 44-511 miles distant; from Cross-Fell, one of these stations, Wisp Hill, the other station, was seen depressed 30′ 48′′, and from Wisp Hill, Cross-Fell was depressed 2′ 31′′; required the refraction and the height of Cross-Fell above Wisp Hill.
Here e 30′ 48′′, e' = 2′ 31′′, z=235018.6. And if at this latitude (about 55°) an arc of one minute