surface; PB, SA the horizontal lines at right angles to CA and CB; also suppose A and B to be. the true places of the objects observed, and a and b their apparent places. Then the bps will be the refraction at P, and the Lasp that at s*. In the quadrilateral figure csop, the angles at p and s are right angles, therefore the SOP+4c-two right angles, but the three angles of the triangle sop=two right angles, hence OSP +4SOP+4OPS SOP+C, consequently OSP +40PS = LC, which is measured by the intermediate arc PS. Now (ASP+BPS) − (≤BPb + ≤ Asa) = (LOSP + ≤ops) — (BPb+Asa)=<c−(BPb+2 Asa) = La Sp+bps the sum of both the refractions. Hence the following (Y) RULE. Subtract the sum of the two depressions from the contained arc, and half the remainder is the mean refraction. (Z) If one of the objects (A) instead of being depressed, be elevated, suppose to the point e, the 4 of elevation being e SA; then the sum of the angles msp, mps will be greater than Zops+ ZOPS+ OSP (the c, or contained arc Ps) by the of elevation e SA. Hence 2c+Le SA=Lm SP+Lm Ps, from each of these equals take the BPb, then c+LesA-LBPb = Lm SP+4bPs the sum of the two refractions; that is, subtract the depression from the sum of the contained arc and elevation, and half the remainder is the mean refraction. Perhaps it may be necessary to remark, that previous to the observations the error of the instrument must be accounted for (Q. 79.) EXAMPLE XLVIII. (A) The refraction between Dover castle and Calais church was thus determined †. Let c be the centre of the earth, Ps the surface; D the station on Dover castle; A the top of the great balustrade of Calais steeple; EDF the horizontal line; also let PG=SD; then the FDGC, or half the arc Ps (Q. 80.). The distance from Dover to Calais is 137455 feet, hence 364950 feet : 60′ :: 137455 feet : 22′. 35" the c, hence FDG 11. 17′′ The height of D above low-water spring tides=469 The height of A (communicated from France)=140 feet. feet. AG=328/ * In observing these angles two instruments are used, one at p and another at s; and the reciprocal observations are made at the same instant of time by means of signals, or by watches previously regulated for that purpose. The observer at p takes the depression of s, at the same moment which the observer at s takes the depression of P. † Trigonometrical Survey, vol. i. page 173. The Σ The triangle DGA may be considered as isosceles, and DG or DA=137455 feet, the distance between Dover and Calais. Hence AG rad. :: DA : secant ZDAG= 89°. 55'. 52"-2, the double of which, deducted from 180°, leaves 8'. 15" for the ZGDA, to which add the FDG=11'. 17", and the whole angle FDA=19. 32" supposing there was no refraction; but the. LFDA was determined from observation to be 17'. 59′′, hence the refraction was (19. 32"-17. 59") 1. 33", being about of the contained arc. (B) Mr. Huddart is of opinion, that a true correction for the effect of terrestrial refraction cannot be obtained by taking any part of the contained arc *; for, different points, though nearly at the same distance from the observer, will have various refractions. OF THE REDUCTION OF ANGLES TO THE CENTRE OF THE STATION. (C) In surveys of kingdoms and counties, where signals on the steeples of churches, vanes of spires, &c. are used for points of observation, the instrument cannot be placed exactly at the centre of the signal, and consequently the angle observed will be different from that which would have been found at the centre. The correction is generally very small, and is only necessary where great accuracy is required. The observer may be considered in three different positions with respect to the centre, viz. he is either in a line with the centre and one of the objects, or a line drawn from the centre through his situation would, if produced, pass between the objects; or a line drawn from the centre to the place of the observer, when produced, would pass without the objects. (D) FIRST, let the observer be at D, in a line between the objects B and C, viz. on one side of the triangle ABC; B being the proper centre of the station and the ABC that required. It is plain that the CDA, being the exterior of the triangle ADB, is too large by the interior 2 D (E) SECONDLY. Let the observer be at o, within the tri Philosophical Transactions for 1797, page 29th et seq. angle angle ABC, and let в be the centre of the station, and the ABC that required. Now ZAOC+20AC+2OCA=2ABC+20AB+20A€ +2OCA+ZOCB, each of the sums being equal to two right angles; therefore the AOC=ABC+20AB+2оCB, that is, the ZAOC is greater than the ABC by the sum of the angles oab and OCB. Therefore ABC LAOC-(ZOAB+ZOCB). * (F) THIRDLY. Let the observer be at E, without the triangle ABC, and let a be the centre of the station, and the ¿CAB that required. Now A CEB + LECA + LACO + LOCB + LEBO + LOBC = CABACO +2оCB+ZABE+ZEBO+ZOBC, each of the sums being equal to two right angles; therefore CAB+ZABE=CEB+ ZECA, and consequently EXAMPLE XLIX. Let A and B represent the vanes on two steeples; E the situation of the theodolite upon the steeple A, and o its situation upon the steeple B. Then, suppose It is required to find the angles CAB and ABC, the distance AB being 5000 feet. SOLUTION. The CEB CAB nearly, and ZAOC=ZABC nearly, with these angles and AB, find AC and BC (as in Example I. Chap. III.) 4581.8 and 5811.6. Then, AC sin. CEA :: AE: sin. LECA 5′, 50′′ AB sin. BEA :: AE sin. ZABE 7.29′′. Hence CAB 74°. 30′. 21" (F. 84.) BC sin. COB :: BO: sin. ZOCB=4. 10". AB sin. ZAOB :: BO: sin. ZOAB=0'. 56′′. Hence ABC 49°. 21'. 54" (E. 83.). With the corrected angles CAB, ABC, and the distance AB, the sides AC and BC may be determined. 7 OF THE REDUCTION OF ANGLES ANOTHER. FROM ONE PLANE TO (G) Angles which are inclined to the horizon, may be reduced to the corresponding horizontal angles, in cases where If the proper centre of the station were at o, and the observer at B, it is plain that the angles OAB and CB must be added to the ABC to obtain the LAOC. very great accuracy is required. Let the lines Ps, PB, Es be three chords of terrestrial arcs, that is, let the points P, B, and s, be all equally distant from the centre of the earth, and let the point D be elevated so as to be farther from the centre of the earth than any of the points P, B, s, by the quantity SD, it is required to reduce the triangle BDP to the triangle ESP. 772 (H) The line sp may be supposed to be perpendicular to each of the chords SB and SP without sensible error, though strictly speaking the angles DSB and DSP are each equal to 90° the arc which the chords SB and SP subtend (Q. 80. and Note). Likewise the chords SB and SP may be used instead of their corresponding arcs (G. 77.). By inspection of the figureit is plain that BD is greater than BS, and PD greater than PS; but the base PB is common to the two triangles BDP and BSP, therefore the BSP is greater than the BDP, in the same manner as the ZAOC is greater than the ABC in a preceding case. (E. 83.) B G P n (I) In the right-angled triangles DSB and DSP. SD2=BD2- Bs2==DP2- Ps2 (Euclid I. and 47.). Hence BD2DP2BS-PS2, to each of these equals add the square of PB, then PB2+BD2-DP2=PB2+BS- PS2, and if each of these equal quantities be divided by 2PB, we get PB2BD2 DP2 2PB -- PB2BS2- PS2 2PB (K) Now if a perpendicular be drawn from D upon the base PB, PB2 + BD2 — DP2 GB= 2PB (Euclid II. and 13.). But rad. : BD :: cos. DBG=cos. DBP GB, hence GB= and if a perpendicular be drawn from the point s upon the base PB, the segment intercepted between ps and the perpendicular Exactly in the same manner the SPB may be found. BSP. (M) But the BDP taken on the elevated situation D, may be reduced to its corresponding BSP, by using only the observed BDP, and the angles of depression mDB, and DP. The 4mDB=4DBS and the NDP DPS; hence the three angles necessary for this reduction are BDP, DBS, and DPS. (N) That is, The square of the side subtending any acute L of a triangle, is less than the squares of the sides containing that acute ; by double the rectangle of those sides, multiplied by the cosine of the acute divided by radius. Therefore, in the triangle BDP G -). By making the values of PB2 equal to each other, and reducing the equation we get (BD2-Es2) + (PD2 - Ps2) = (2Bn × PDX COS. BSP -)-(2BS X PS X -). But BD2-BSPD2 — ps2= COS. BDP rad. rad. SD (Euclid I. and 47.), therefore SD2 x rad. (BDX PDX COS. BDP) - (Bs x Ps x cos. BSP), And hence cos. BSP= rad. BS X PS BD :: sine DBS: SD rad. PD :: sine DPS: SD. Hence, rad2: BDX PD :: sine DBS x sine DPS : SD3. And, SD2 x rad. BDX PDX sine DBS X sine DPS X 1 rad.' Again, rad. : BD :: COS. DBS: BS rad. PD :: cos. DPS: Ps. Hence, rad.2: BDX PD :: COS. DBS X COS. DPS: BS X PS, And BS XPS BDX PD X Cos. DBS X COS. DPS X 1 rad.2 Therefore by substitution the cosine of the BSP = COS |