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of stars. For instance (Art. 56), an error as great as 5" was discovered at Takal K'hera in Central India by Colonel Everest, arising from the attraction of a distant table-land. Sir Henry James has shown that a deflection of about the same amount occurs at Arthur's Seat, Edinburgh (Phil. Trans. 1857). We have mentioned that the attraction of the Himmalaya Mountains produces a deflection amounting to as much as 28" at the northern extremity of the Great Indian Arc (Art. 61). We have calculated elsewhere (see Art. 62 and Phil. Trans. for 1859) that the deficiency of matter in the vast ocean south of India causes such deflections as 6", 9", 10":5, 19"7 at various stations: and (Art. 63), shown that it is not improbable that extensive but slight variations of density prevail in the interior of the Earth, the causes of which are not visible to us as mountain masses and vast oceans are, sufficient to produce errors in the plumb-line quite as great as and even greater than most of those already enumerated. These seem abundantly to account for the variety in the calculated semi-axes and ellipticities in the last Article, derived as they are from uncorrected observations.

101. Mr Airy has entered very thoroughly into a comparison (see Figure of the Earth, Encyc. Metrop.) of the various arcs measured in different parts of the world. He has used them according to their importance and value, as determined by the circumstances under which they were measured and observed. His result satisfactorily shows that the ellipticity of the mean spheroid is about 3oo. The conditions, therefore, required for supposing the Earth to have received its present average form from having been once in a fluid state, are altogether satisfactorily fulfilled.

The same result has been obtained by another process, first used by the late M. Bessel and adopted by Captain A. Clarke, R.E. in the Volume of the British Ordnance Survey. This method we shall now explain, first introducing one or two propositions which we shall require for its application. Let the form of the meridian line be such, that

p= A + 2B cos 21 + 2C cos 47

is the radius of curvature at a point of which the latitude is l.

PROP. To prove that if the meridian be an ellipse,

6 CA = 5B2.

102. Let p'x'y' be the radius of curvature and co-ordinates to a point in latitude l, in an ellipse, *****=1, d=1

e =

= 72

e=1

:1-5 al'

'

22

y

azt

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=

Comparing this with p= A +2B cos 21+ 2C cos 41,
3
3 3

15 А

C ae; 64

32

128

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.:. 6CA

45
64

ea?

45 64

B= 5B2. 64 9

103. COR. The expansions of ac' and y' are as follows:

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PROP. To find an expression for measuring the departure of the curve of the meridian from an ellipse at any point, when the meridian is not elliptical.

104. Let xy be the co-ordinates to any point of which l is the latitude, and the radius of curvature as above.

dy
doc

dạy

Now cotl=

di = cosec1 dx2

;

dx

dac dl

P

p

1

1

psin l, and did = p cos l;

dy
(A - B) cos? +}(B-C) cos 31 +{C cos 51,
yA
y= (A + B) sinl+- (B+C) sin 31+Csin 51.

Let a and b be the semi-axes of the curve, whether it be an ellipse or not. Hence these values of ac and y give

a

+

2

C 15

a=A-B+(B-C)+c= A - B-1
6=A+B-}(B+C) +{C= A+

-C
+2), neglecting B.C &c.

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+

B

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2 15

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If we put these for a and e' in the expressions for a' and y' in Art. 103, we shall have the co-ordinates of a point (lati

tude 7) in the ellipse constructed upon the same axes as those of the actual curve. After reduction, we get

2
1 B2

B2

1 B
15
9A

A

cos 51, 6 A

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B

5 B?
A

3l

1 B 6 A

sin 5l;

+=(A-B-10+ cos2+(18-15 con 31+7

+
y=(1+B=10+) sinl+(18-12 sin 8+

B-
0–2 = 1- cos 31 +{cos 51),
y-c-
–y'=(0-2 min + sin 81 + { sin st).

1 51

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2

COS 15

1

l +

Let ds and or be the distances between the points (xy) and (ac'y') measured along the arc of the ellipse and the normal;

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PROP. To obtain a formula for correcting the amplitude of an arc, 80 as to make its measured length accord with a given

curve.

105. Let s be the length of the arc and p the radius of curvature as before; then, by integration,

8= Al+B sin 21+ 4Csin 42+ constant. Let 7-10,1+14 be the limits of s, 1 being the latitude of the middle point, and $ the amplitude of the arc;

.. s= A$+2B cos 21 sin $ + C cos 41 sin 20. Suppose now that x,' are the small corrections which must be applied to the observed latitudes, l-16,1+, to make them accord with the measured length s.

Then l - 7$ + x and 1 + $+ ' and $ + ' - must be put instead of 1-10, 1+1$, and $ in the above formula. Hence, neglecting the squares of small quantities, 8=A ($+x;' – x) + 2B cos 21 {sin 0+ (x' — «) cos o}

+ Ccos 41 {sin 20+2 (x,' — 2) cos 20};

.: (oc' - x) (A + 2B cos o cos 27)
=s - A$-2Bcos 21 sin $-C cos 41 sin 20.

Put

A + 2B cos o cos 21= A=u;

Cu sin $ cos 21 - sin 26 cos 41. A

А

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=

and the above formula will become

x'=m+aU+BV +Z+x, where m, a, b, y are functions of the observed latitudes, the measured length, and numerical quantities only.

106. As an example which the student may work out for himself, the following is selected from the Volume of the Ordnance Survey.

Observed

Station.

Measured Arcs in feet.

Latitudes.

Amplitudes.

Damargida 18°3' 15"-292
Kalianpur 24 7 11 262
Kaliana 29 30 48 322

6° 3'55".970

2202904•7

11 27 33 .030

4164042.7

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