S SURVEYING. Now the angles SAO, SOA together are less than both the angles BAO, BOA by the sum of the angles SAB, SOB; therefore ASO is greater than ABO by that sum; hence the angle ABO 70° 39′-670° 33'. And BO, AO calculated with the corrected angles 45° 43' and 70° 33', are 2065 3 and 2720.2 yards. It is not necessary that the angles ONA, OSB should be very accurately taken; but the distances NA, SB must be carefully measured. 248. If A, B, C, be three objects whose distances from each other are AB = 4516, AC 4809, BC 3018 yards; and suppose at the station S we observe the angles CSB = 117° 56', BSA 110° 12'; it is required to find the distances from the station to the three objects. Construction. If the triangle ABC be laid down with the three given distances, and segments of circles described upon any two sides to contain the angles they subtend (172), the intersection of the arcs will evidently be the station, whether it falls within, or without the triangle. But the following method is rather more simple.-About AB describe a circle so that the segment ABS shall contain the angle 110° 12':. make the angle BAR 62° 4' the supplement = A R B of 117° 56' (CSB), join CR; and S, where it intersects the circle, is the station. For if AS, SB, BR are drawn, the angle ASB is = 110° 12' by construction; and RSB being equal to RAB (70) or 62° 4', the angle CSB which is its supplement, will be 117° 56' the other observed angle. Calculation. The three sides 4516, 4809, 3018 give the angle ABC = 76° 28′ (229). Angle ABR (= ASR the supplement of ASC) = 48° 8' BAR ᎪᏒᏴ ...... side AB give BR=4251.3. =62 4 = 69 48, these with the The angle RBC 48° 8' + 76° 28' including sides, give RCB 32° 47', and 124° 36' which, with the two CRB = 22° 37'. Now SAB SRB 22° 37′; therefore all the angles of the triangles Whence (221), the distances S., SB, SC, are found to be 3530, 1851, 1672 yards, respectively. When the station is without the triangle (suppose at R) it is evident the circle must be described so that the outward segment ARB shall contain the whole observed angle ARB; then if the angles ABS, BAS be made respectively equal to the observed angles ARC, BRC, and CR drawn through S, R will be the station. If the whole observed angle ARB should be equal to the supplement of the angle ACB, the circle will pass through the point C; in which case the problem is indeterminate: for the angles standing on the chords BC, AC would be the same in all points of the arc ARB, (70.) 249. The last problem will be found useful in reconnoitring a country with a map or plan; for the angles taken to any three objects which are laid down, will determine the situation. of the observer. A small pocket sextant is the most convenient instrument for measuring the angles. And it appears from the preceding construction that it is not necessary to describe a circle. For example, if the station be within the triangle, then the angles BAR, ABR being made equal to the supplements of the observed angles BSC, ASC, the intersection of AR and BR gives the point R; then if the angle ABS be made equal to the angle ARC, BS will meet RC in S the station. On the contrary, when the place of observation is without the triangle, the angles ABS, BAS, are made equal to the observed angles ARC, BRC, respectively, then CR being drawn through S, and the angles ABR, BAR made equal to ASR, BSR, BR and AR will meet CR in R the station. A R In this latter case however, when the point S falls near the object C, the construction may give the point R considerably wide of the truth. 250. In making a Survey we found two spots N and S conveniently situated for stations; and at S took the angles NSA 52° 58′, NSB 55° 4′, to the spires A and B: But at N an intervening height hid the spire B; we therefore observed the angle between the wind-mill W and station S, and found it 38° 4′, and then took the angle SNA which was 41° 46′. Now AW, AB, BW, being respectively equal to 5232, 4490, 2678 yards, it is required to find the distance SN? Construction. With the three given sides lay down the triangle AWB. Then about AB and AW describe circles so that the segment ASB shall contain an angle of 108° 2′ (52° 58′ + 55° 4′); and the segment ANW an angle of 79° 50′ (38° 4′ +41° 46'). Draw the chord AD to subtend an angle (AND) = 41° 46′, and the chord AG to subtend an angle (ASG) = 52° 58′; join DG; and the intersections S, N, will be the stations. For if SB, SA; NA, NW are drawn, the angles at S and N D B to the three objects will be equal to the observed angles, by the construction, and Art. 70. Calculation. Draw DW, GB. Then all the angles of the triangles ADW, AGB are given; As sin. ADW: 5232 (AW) :: sin, DWA : 35406 AD. AG. The sides of the triangle AWB give the angle WAB = 30° 47′ DAG 133 55 with this angle and the ineluding sides we get ADG = 29° = AWN; therefore in the triangle AWN all the angles and the side AW are given, whence AN=2577; then, as the angles of the triangle ASN are also given, we get SN = 3217 yards. And the method of construction and calculation will vary Jittle from the preceding, howsoever posited the stations may be in respect of the three given objects. Of Surveying with the Compass. 251. In this operation we do not measure the angles subtended by distant objects in the same manner as with a Theodolite, but take their angular distances or bearings from the magnetical meridian. Thus if NS represents the magnetic needle or meridian, W the west, and E the east; and suppose the sights on the Compass are directed to the windmill A: then if the angle ACN is 40°, for example, the wind-mill is A N said to bear NW 40°, or 40° westward from N the magnetical north. Or if the sights are directed to the spire B, and the angle SCB is 64° then the spire bears SE 64°. If CH represents the direction of the true meridian, the angle NCH is called the variation of the magnetical needle; which, at this time, is about 23° or 24° westward at London. 252. Let A and B be two stations bearing SW 61° and NE 61° from each other; Then from G and P draw two lines parallel to NE 51, and SE 70°, and their intersection will determine the sta. tion C. And the intersection of the SW 54° line from A, with that of the SE 31° line from C gives the position of the object R. 253. Since the magnetical meridians are considered as parallels, it is evident that the bearings of any two objects already laid down will give the place of the observer; but every intersection should be as near a right angle as circumstances will admit. The bearings of all conspicuous objects however, ought to be taken at every station, by which means a great num. ber may be fixed from several intersections. They are easily This is a very expeditious method of laying down the relative situations of the prominent points of a small tract of country, The compasses most convenient are about 3 inches in diameter; and may be carried in the pocket. fitted to the top of a stick or staff which must be stuck upright in the ground that the needle may play freely. These compasses are divided into degrees only, and consequently much accuracy cannot be expected in Surveys of this kind: they |