260. METHOD III.-The First Meridian passes through the Most Northern Station of the Polygon, as through Station E (see fig. 176). column; S. lat. X W. mer. dist. is put in north area column; N. lat. X W. mer. dist. is put in south area column. The proof of the above rule will appear from the following (see fig. 176). Draw the meridian E W through the point or station E; let p F, g H, r D, s K, R s, C w, and D x, be the departures respectively. b b b/ This includes figure I r v K+ AV KS, S K being the east meridian distance of K; then S K (KA) mer. dist. of the middle of the line A K, which is or east, if S K is more than AK; but if SK is more than ✈ A K, then the meridian distance will be + or east, and if the mer. dist. S K is A K, then the mer. dist. of line K A 0. equal to h We now suppose that S K is less than K A; therefore mer. distance to middle of KA = SK Ag west or negative, and (S K AG) ‚g K = figure g K s y —▲ AgK= figure g K vy + K vs ▲ AgK; but the meridian distance being negative, the product must be nega tive; that is, the above product ▲ Ag K g K vy K v S, which is equal to the A y v, because we have to deduct g K vy + K v s, which ▲ have been including the figure K Irs; consequently north by west is to be added or put in south area column. Let this area be equal to b, and entered in the south area column. The mer. dist. of A is the same as that of B, and is found by adding A g to the last mer. dist. to the middle of A K. That mer. dist. X by A B, gives an area to be added figure g A B bb', which is put in south area column. Also the mer. dist. in middle of B C is west, which multiplied by B C, will give the area BCwb = b, which put in south area column. In like manner we find the area C D x w = b', which put in south area column; and the area of D E x is west of the meridian b, and is to be put in south area column. Hence it appears that those areas derived from east meridian distances are put under their respective heads, S. and N.; but those having west meridian distances, are put in their contrary columns. 261. Calculating the Offsets and Inlets. (See fig. 17e.) Sum of addition, 5100 2820 20562 18228 Sum of subtraction, 18228 Difference, 2334, to be added to the area of the polygon. The letters a, b, etc., show between what points on the line the areas are calculated. When the area, and not the double area, of the polygon is given, then we take half the double area of the difference of the offset and inlet columns, and add of subtract to or from the area of the polygon, as may be the case. In making out the bases, we subtract 150 from 190; put the difference, 40, in base column, and opposite which, in offset column, put 14; then 40 X 14 will give double the area of the Abetween 150 and 190. Again, take 190 from 297; the difference, 107, is put in base column, opposite to which, in offset column, is put 78 - 1464; then 107 X 78 double the area of the trapezium between 190 and 297. This method of keeping field notes facilitates the computation of offsets and plotting detail. We begin at the bottom of the page or line, and enter the field notes as we proceed toward the top or end of the line. The chain line may be a space between two parallel lines, or a single line, as in fig. 17e. If the field book is narrow, only one line ought to be on the width of every page, and that up the middle (see sec. 211). On the first day of May, 1838, I commenced the survey of part of Flaskagh, in the parish of Dunmore, and county of Galway, Ireland, sur veyed for John Connolly, Esq. MICH'L MCDERMOTT, C. L. S. } The angles have been taken by a theodolite, the bearing of one line determined, from which the following bearings have been deduced (see fig. 17e). Land kept on the right. We begin at the most northerly station, as by this means we will always add the south latitudes and subtract the north latitudes. Explanation. On line 1, at distance 210, took an offset to the left, to where a boundary fence or ditch, etc., jutted. The dotted line along said fence shows that the face next the dots is the boundary. At 297, offset of 64 links to Mr. JAMES ROGER'S schoolhouse. At 340, offset of 70 links to south corner of do. The width 30, set down on the end of do. At 400, offset to the left of 14 links to a jutting fence. From 150 to 400, the boundary is on the inside or right, as shown by the characters made by dots and small circles joined. See characters in plates. From this point, 400, the boundary continues to the end of the line, to be on the left side of fence. At 804, met creek 30 links wide, 5 deep, clear water, running in a southern direction. At 820, met further bank of do. At 830, dug a triangular sod out of the ground, making the vertex the point of reference. Here I left a stick 6 inches long, split on top, into which split a folded paper having line 1-830 in pencil marks. This will enable us to know where to begin or close a line for taking the detail. At 960, offset to the right 20 links. At 1000, met station F, where I dug 3 triangular sods, whose vertexes meet in the point of reference. This we call leveling mark. The distance, 1000 links, is written lengthwise along the line near the station mark. The station mark is made in the form of a triangle, with a heavy dot in the centre. Distances from which lines started or on which lines closed, are marked with a crow's foot or broad arrow, made by 3 short lines meeting in a point. Along the line write the number of the line and its bearing. Line 2 may be drawn in the field book as in this figure, or it may be continued in the same line with line 1, observing to make an angle mark on that side of the line to which line 2 turns. This may be seen in lines 4 and 5, where the angle mark is on the right, showing that line 5 turns to the right of line 4. Line 2, total distance to station G 1700 links. The distance from the station to the fence, on the continuation of line 2, is 10 links, which is set correctly on the line. Key offset. See where line 2 starts from end of line 1. At the end of line 1, offset to corner of fence 10. At 10 links on line 2, offset to corner 2. This is termed the key offset, and is always required at each station for the computation of offsets and inlets. Running from one line to another. We mention the distance of the points of beginning and closing as follows: Line 5. Line 1. This shows that the line started from 830, on line 1, and closed on 600, line 5. It also shows, from the manner in which distances 804, 820 and 830 are written, that the line turns to the right of line 1. When we use a distance, as 830, etc., we make 2 broad arrows opposite the distance. This will enable us to mark them off on the plotting lines for future reference. We take detail on this line-it will serve as a check when the scale is 2, 3, or 4 chains to 1 inch scale. We number it and enter it on the diagram, which must always be on the first page of the survey. The diagram will show the number of the line; the distances on which it begins and ends; the reference distances. This will enable the surveyor to lay down his plotting or chain lines, and test the accuracy of the survey. Having completed the plotting plan, we then fill in the detail, and take a copy or tracing of it to the field, and then compare it with the locality of the detail. This comparison is made by seeing where a line from a corner of a building, and through another corner of a fence or building, intersects a fence; then from the intersection we measure to the nearest permanent object. We draw the line in pencil on the tracing, and compare the distance found by scale with the measured distance. Some surveyors can pace distances near enough to detect an error. On the British Ordnance Survey, the sketchers or examiners seldom used a chain, unless in filling in omitted detail. On Supplying Lost Lines or Bearings. 263. It would be unsafe to depend on this method, unless where the line or lines would be so obstructed as to prevent the bearings and distances to be taken. The surveyor seeing these difficulties, will take all the available bearings and measure the distances with the greatest accuracy, leaving no possible doubt of their being correctly taken. Then, and not till then, can he proceed to supply the omissions. Case 1. In fig. 176, we will suppose that all the lines and bearings have been correctly taken, but the distance I K has been obliterated, and that its bearing is given to find the distance I K. Let the bearing of I K be S. 60 W. From sec. 259, method 2, we have calculated the departure of K from the line A B 17,4257 departure of I from do. 35,6123 consequently the departure of line I K is KL 18,1866 We have the angle K I L = 60°, therefore the < I K L = 30°, and its departure The product of the last two numbers will give (by sec. 167) IL By E. I, 47, from having I K and K L we find 10,50 IK or IL-9,1933, divided by the lat. or cos. of 60° or,86603 ,5000 9,0933 10,50 = I K Case 2. The bearing and distance of the line I K is lost. From the above sec., method 2, we have Lat. K A ་་ Lat. A B Lat. B C Lat. C D Lat. D E = 7,7140 N. 54,9556 N. 44.4560 S. Lat. I L = Therefore, by E. 1, 47, K L2 + LI2 = KI2; consequently K I is found. But I K • cos. < KIL: Therefore I L. cosine <KI L, which take from table of lat. and dep., I L ΙΚ and it gives << KIL KL 9,0933 10,50 ,8662 the bearing of the line K I 60°. Consequently the bearing is S. 60° W., cos.<IK L; ... the < I K L = 30°, and N. 60° E. from station K. Case 3. Let there be two lines wanted whose bearings are known to be S. 60° W. and N. 80° W. Here the station K may be obstructed by being in a pond, in a building, or that buildings are erected on part of the lines I K and A K (see fig. 176). We find fron case 2 that A is south of F Now we have A a and a I, ... we find the line A I. And A a divided by a I gives the tangent of < A I a =,2085. And the AI a = 11° 47'. 51,8830 44,4560 7,4270 35,6123 I a divided by the cosine 10° A I — 35,6123÷,9789 — 36,38. = By sec. 194, we have sine <AKI: AI:: sine < AIK: A K. sine < A KI: AI:: sine < KAI: KI. Case 4. Let all the sides be given, and all the bearings, except the bearings of I K and A K, to find these bearings. By the above methods we can find the departure a i of the point Ì, east of the meridian A B. We also have the difference of lat. of the points A and I Now having the sides A 1, A K and K I, by sec. 205, we can find the angles K A I and K I A. And the < A I a and < A I K are given; their sum < A I K is given; ... the bearing of the line I K is given. |