GEODESY.] 



MATHEMATICS. SPHERICAL TRIGONOMETRY. 



663 



(17)- The Measurement of a Base Line. 



A space of open ground which is nearly level must be 



chosen, the line to be measured being indicated by 



stations, and stages erected, if necessary, to secure the 



horizontality of the base ; the measure may be made by 



Fig. 9. 



6 



\! 



rods of glass, or steel, proper corrections being applied 

 for temperature. A more convenient contrivance for se : 

 curing accuracy in the measures has been devised of late 

 for the Irish Survey : it is of the following kind : A B, 

 C D, are rods of platina and iron riveted together at the 

 middle point P, and are exactly the same length for a 

 given temperature : A]>, ttq, are marks affixed to the 

 ends of each. Now the metals have different expansions 

 for the same temperature ; suppose then that A B, for a 

 change in temperature, becomes 06, while C D becomes 

 cd, then Ap will assume the position ap, and let Aa = fa 



Cc = 6, then if Cp be taken so long, that -_-^ = _? then 



Cp co 



p will not be changed by the change of temperature ; 

 since, within very large limits, the expansion of a metal 

 u proportional to the increase of temperature, and .*. 



~ is constant ; the same arrangement being made at the 



< ', 



end, B D, the distance pq, will not be affected by change 

 of temperature. 



(18). To Correct for want of Straightntu in the Base Line. 

 The nature of the ground may be such as to render it 

 impossible to measure a perfectly straight base line of 



Fig. 10. 



sufficient length. 



This was the case 



in some of the 



French surveys, 



where the actual 



measurement was 



of two straight ln\es inclined to each other at an angle of 



very nearly 180. 



Let A U be the line the length of which is required, 

 the measurements are A C 6. C B = a. and B C N 

 in circular measure => n* suppose. Then it is plain 

 that A B (a + 6). Let us suppose A B = a + 6 x ; 

 our object is to find r. 

 A&=a*+V> 2aicos. (v 6) - a' + 6 s + 2 ab cos. 



Now (Plane Trig., Art. 50, cor.) cos. = 1 -g if we 



omit 0" . . . 



.-. A W = a 1 + 2 ai + V ab0* = 



= a-f b 



alO* 



Now - 



,= -000004848 



180- X GO X CO" 

 z -.",'' t X 00000000001175 



a + 6 



The correction to be applied to the sum of o and 6, to 

 obtain the true distance, A 13. 

 (19). To measure the Angles of the Triangles of a Survey. 



Any two of the three angular points of one of these 

 triangles are rarely in a horizontal plane passing through 

 the third ; the angle required is, of course, such a hori- 



zontal angle. Now the angles are measured either by a 

 theodolite, or by a repeating circle in the case of the 

 former instrument, the vertical elevation of each object 

 is observed, and the horizontal angle between them 

 that is to say, if A B C (Fig. 11) are the stations, A N M 

 the horizontal plane through A, U N and C M perpen- 

 diculars from B and C on A M N, then by the theodolite 

 we observe the angles BAN, CAM, and MAN; the 

 last is the horizontal angle required. The theodolite is 

 the instrument that has been used in the English surveys. 

 But if a repeating circle be employed, the angle BAG 

 is the one observed ; and it is necessary to deduce from 

 this the horizontal angle M A N. The repeating circle 

 is the instrument used in the French surveys. 



(20). To determine the Correction for reducing an Angle to 



the Horizon. 



Let ABCMN be the same as in last 'article; with 

 Fig. 11. 



centre A and any radiua describe a sphere, which meets 

 the lines A B, A C, A N, A M in />, g, n, m, respectively. 

 pn and mq, if joined by great circles, clearly meet in Z 

 vertically over A, since the circles must be perpendicular 

 to the horizontal plane. Then mn, or the angle mZn 

 (Spherical Geometry, Prop. V., Cor. 4) is the angle re- 

 quired, and for its determination we have given pq = A, 

 pn = h, qm = h'. Suppose mZn = A + SA, then our 

 object is to determine BA. 

 Now in triangle pZq we have, by formula 3, 

 cos. pq cos. pZ cos. qZ 



OOS. pZq ? 3 



sin. ;'/ cos. qZ 

 Now pZ = 90- A, and qZ = 90 -h' 



cos. A sin. h sin. h 

 . . cos. (A + SA) = . , 



cos. h cos. h 



Now we will suppose IA so small that we can omit 

 &A 1 ... and h and h' so small that we can omit every 

 power and product higher than 7i 2 , hh', and h' 2 . 



.'. (Cor. Plane Trig., Art. 50) sin. SA = SA, cos. A 



= 1, sin. h = h, sin. h' = h', cos. h = 1 , and cos. 



Now cos. (A + A) = cos. A cos. SA sin. A sin. & 



cos. A hh' 

 Hence cos. A SA sin. A = = = cos. 



.'. A sin. A = hh' ] ? +h '\os. A 



Now let h + h' =*p, and h h' = q. .'. 2h = p + q, 

 and 2h' = p q 



.. 4hh' = p* g 2 , and 2 (A 2 -f h' 1 ) = p 2 +q 3 

 .'. 4 Asin. A =p 2 2 (p 2 -j- <j 2 ) cos. A = 

 p 1 (1 cos. A) g 2 (1 + cos. A) 



/. 4 SA 2 sin. ^cos. = 



sin. 



A~ 



cos. - 



A 



/"I 



A 



sin. - 



