Thus, it will be found by computation that the external offset space EGFD is internal offset space CIHD is Difference Areas of principal triangles = 50704 sq. links 31870 = 18834 470978 489812 Sum or area of given field = 4 ac. 3 ro. 23-6 pls. The crooked boundary, DHI, may be reduced to a straight line DL, meeting CI produced in L by article 130 in Practical Geometry; and then a triangle CLD is formed equal to the irregular space CIHD, the area of which is = LS CD. The length of LS can be found by means of the scale used in constructing the figure. The offset space EGFD, with the curvilineal boundary, can also be reduced to a triangle EKD of equal area, which can be calculated like that of triangle CLD. The straight lines EK, KD, can be determined with sufficient accuracy by the eye, so as to cut off as much space from the inside of the curved boundary EGFD, as is added on the outside. A ruler made of transparent horn is used for this purpose, or a fine wire stretched on a whalebone bow. = 23. The practice of constructing a plan, or, as it is also called, a plot, of any surface, the dimensions of which are taken, and reducing the crooked and curved boundaries in the manner stated above, is very common with the best surveyors, on account of its expedition and sufficient accuracy. It is also usual to measure on the plan the altitudes of the principal triangles, and to calculate their areas by the simple rule in article 258 of Mensuration of Surfaces. = Thus, by drawing the perpendicular BV on AC, and measuring it, the area of triangle ABC is AC BV; and in a similar manner the areas of the other principal triangles are found. TO SURVEY WITH THE CHAIN, CROSS, AND THEODOLITE. 24. Although it frequently happens that the most expeditious mode of surveying is by the chain and cross, yet in the case of large surveys the theodolite is very advantage ously combined with them for measuring angles. When some of the angles of a triangle are known, its area can be found without knowing all its sides; and the tedious process of measuring them all by the chain is thus dispensed with, unless the measuring of offsets or some other cause requires all the sides to be measured; but circumstances will determine what is the best mode of proceeding. It is often useful to measure more lines and angles than are necessary for determining the area, for the purpose of serving as a check to ensure accuracy in the results. 25. PROBLEM II.-To survey a field by taking a single station within it, and measuring the distances to its different corners, and the angles at the station contained by these distances. The field is thus divided into triangles, in each of which two sides and the contained angle are known; and their areas may therefore be found by article 259 in Mensuration of Surfaces, and their sum will be the area of the field. EXERCISES. 1. From a station O within a pen- E tagonal field, the distances to the different corners A, B, C, D, E, were measured and found to be respectively 1469, 1196, 1299, 1203, and 1410; and the angles AOB, BOC, &c., contained by them were in order 71° 30', 55° 45', 49° 15', and 81° 30'; required the area of the field. B A Ans. 39 ac. 0 ro. 30-2 pls. 2. From a station near the middle of a field of six sides, ABCDEF, the distances and angles, measured as in the preceding exercise, were as below: AO 4315 links Angle AOB 60° 30′ = BOC= 47° 40' ... ... ... = = 26. PROBLEM III.-To survey a polygonal field by measuring all its sides but one, and all its angles except the two at the extremities of that side. From the data it will always be possible, by applying trigonometrical calculation, to find two sides and the contained angle of each of the component triangles, the areas of which can be calculated as in last problem. Let ABCDE be the polygonal field; and let the sides AB, BC, CD, DE, be given, and also the angles B, C, and D. Join CE and CA; then, in triangle ABC, find AC and angle C (First Part, 211); and in triangle CDE, find CE and angle C; then angle ACE BCD-(ACB + DCE). There are therefore now known two sides and a contained angle in each triangle; and hence their areas can be found, the sum of which is that of the given field. = Find the area of the subjoined field ABCDE, from these measurements: EXERCISE. Side AB=388 Angle B = 110° 30′ C=117° 45' E BC 311 = CD=425 ... DE=548 ... ... D B 27. PROBLEM IV.-To survey a field from two stations in it, by measuring the distance between them, and all the angles at each station contained by this distance, and lines drawn from the stations to the corners of the field. From the data all the lines drawn from one of the stations to the corners of the field can be calculated by case 1 in Oblique Trigonometry; and then the areas of the triangles contained by these lines, and the sides of the figure, can be calculated as in the last problem. Let ABCD be the given figure, and OQ the stations; measure all the angles at O and Q; D then in triangle DOQ, the angles are known, and the side OQ; hence find OD; similarly in triangle OQC find OC; then find OB in triangle OBQ; and, lastly, OA in triangle OAQ. Then the areas of the four triangles AOB, BOC, COD, DOA, can be found as in the last problem. EXERCISES. 1. Find the area of the field ABCD, from these measurements: and hence and OQ ... ... and hence m 2. Find the area of the field ABCDEF, from the subjoined measurements :— F Angle w 36° 10' e = 115° 16' r = 94° 30' Ans. 61 ac. 1 ro. 6-4 pls. B A Angles at O. m = 21° 20' n = 49° 10' = 57° 12' u = 29° 40' v = 64° 25' w=79° 16' and the distance OQ = 500 links. Angles at Q. r = 10°40' s = 18° 40' = 42° 0' x= z= 67° 5' Ans. 12 ac. 2 ro. 39.83 pls. 28. It is evident that, by the preceding method, a field may be surveyed from two stations situated without the field, its area computed, and a plan of it made. SURVEYING WITH THE PLANE TABLE. 29. By means of the plane table a plan of a field or estate is expeditiously made during the survey, from which the contents may be computed by the method described in article 23. 30. This instrument consists of a plain and smooth rectangular board fitted in a moveable frame of wood, which fixes the paper on the table, PT, in the adjoining figure. The centre of the table below is fixed to a tripod, having at the top a ball-and- P socket joint, so that the table may be fixed in any required position. BAT The table is fixed in a horizontal position by means of two spirit levels lying in different directions, or by placing a ball on the table, and observing the position of it in which the ball remains at rest. R The edges of one side of the frame are divided into equal parts, for the purpose of drawing on the paper lines parallel or perpendicular to the edges of the frame; and the edges of the other side are divided into degrees corresponding to a central point on the board for the purpose of measuring angles. A magnetic compass box, C, is fixed to one side of the table for determining the bearings of stations and other |