PART II. CHAIN-SURVEYING; By the First and Second Methods: OR DIAGONAL AND PERPENDICULAR SURVEYING. (88) The chain alone is abundantly sufficient, without the aid of any other instrument, for making an accurate survey of any surface, whatever its shape or size, particularly in a district tolera bly level and clear. Moreover, since a chain, or some substitute for it, formed of a rope, of leather driving reins, &c., can be obtained by any one in the most secluded place, this method of Surveying deserves more attention than has usually been given to it in this country. It will, therefore, be fully developed in the following chapters. CHAPTER I. SURVEYING BY DIAGONALS: OR By the First Method. (89) Surveying by Diagonals is an application of the First Method of determining the position of a point, given in Art. (5,) to which the student should again refer. Each corner of the field or farm which is to be surveyed is "determined" by measuring its distances from two other points. The field is then "platted" by repeating this process on paper, for each corner, in a contrary order, and the "content" is obtained by some of the methods explained in Chapter IV of Part I. The lines which are measured in order to determine the cor ners of the field are usually sides and diagonals of the irregular polygon which is to be surveyed. They therefore divide it up into triangles; whence this mode of surveying is sometimes called "Chain Triangulation." A few examples will make the principle and practice perfectly clear. Each will be seen to require the three operations of measur ing, platting, and calculating. Fig. 49. B (90) A three-sided field; as Fig. 49. Field-work. Measure the three sides, AB, BC, and CA. Measure also, as a proof line, the distance from one of the corners, as C, to some point in the opposite side, as D, at which a mark should have been left, when measuring from A to B, at a known distance from A. A stick or twig, with a slit in its top, to receive a piece of paper with the distance from A marked on it, is the most convenient mark. Platting. Choose a suitable scale as directed in Art. (44). Then, by Arts. (42) and (49), draw a line equal in length, on the chosen scale, to one of the sides; AB for example. Take in the compasses the length of another side as AC, to the same scale, and with one leg in A as a centre, describe an arc of a circle. Take the length of the third side BC, and with B as a centre, describe another arc, intersecting the first are in a point which will be the third corner C. Draw the lines AC and BC; and ABC will be the plat, or miniature copy-as explained in Art. (35)~ of the field surveyed. Instead of describing two arcs to get the point C, two pairs of compasses may be conveniently used. Open them to the lengths, respectively, of the last two sides. Put one foot of each at the ends of the first side, and bring their other feet together, and their point of meeting will mark the desired third point of the triangle. To "prove" the accuracy of the work, fix the point D, by setting off from A the proper distance, and measure the length of the line DC, by Art. (43). If its length on the plat corresponds to its measurement on the ground, the work is correct.* Calculation. The content of the field may now be found as directed in Art. (65), either from the three sides, or more easily though not so accurately, by measuring on the plat, by Art. (13), the length of the perpendicular CE, let fall from any angle to the opposite side, and taking half the product of these two lines. Example 1. Figure 49, is the plat, on a scale of two chains to one inch, of a field, of which the side AB is 200 links, BC is 100 links, and AC is 150 links. Its content by the rule of Art. (65), is 0.726 of a square chain, or OA. OR. 12P. If the perpendicular CE be accurately measured, it will be found to be 721 links. Half the product of this perpendicular by the base will be found to give the same content. Ex. 2. The three sides of a triangular field are respectively 89.39, 54.08, and 45.98. Required its content. Ans. 100A. OR. 10P. (91) A four-sided field; as Fig. 50. Field-work. Measure the four sides. Measure also a diagonal, as AC, thus dividing the four-sided field into two triangles. Mea- A Fig. 50. sure also the other diagona, or BD, for a "Proof line." Platting. Draw a line, as AC, equal in length to the diagonal, to any scale, by Arts. (42) and (19). On each side of it, construct a triangle with the sides of the field, as directed in the preceding article. Το prove the accuracy of the work, measure on the plat the length of the "proof line," BD, by Art. (43), and if it agrees with the length of the same line measured on the ground, the field work and platting are both proved to be correct. It is a universal principle in all surveying operations, that the work must be tested by some means independent of the original process, and that the same re sult must be arrived at by two different methods. The necessary length of this proof line can also easily be calculated by the principles of Trigonometry. Calculation. Find the content of each triangle separately, as in the preceding case, and add them together; or, more briefly, multiply either diagonal (the longer one is preferable) by the sum of the two perpendiculars, and divide the product by two. Otherwise reduce the four-sided figure to one triangle as in Art. (78); or, use any of the methods of the preceding chapter. Example 3. In the field drawn in Fig. 50, on a scale of 3 chains te the inch, AB 588 links, BC=210, CD=430, DA = 274, the diagonal AC 626, and the proof diagonal BD = 500. The Sotal content will be 1A. OR. 17P. = = Ex. 4. The sides of a four-sided field are AB = 12.41, BC = 5.86, CD 8.25, DA=4.24; the diagonal BD = 11.55, = = and the proof line AC 11.04. Required the content. = Ans. 4A. 2R. 38P. Ex. 5. The sides of a four-sided field are as follows: AB= 8.95, BC= 5.33, CD 10.10, DA 6.54; the diagonal from A to C is 11.52; the proof diagonal from B to D is 10.92. Required the content. Ans. Ex. 6. In a four-sided field, AB = 7.68, BC = 4.09, CD = 10.64, DA=7.24, AC=10.32, BD=10.74. Required the content. Ans. C 3.00 F Field-Work. Measure all the sides of the fieid. Measure also diagonals enough to divide the field into triangles; of which there will always be two less than the number of sides. Choose such diagonals as will divide the field into triangles as nearly equilateral as possible. Measure also one or more diagonals for "Proof lines." It is well for the surveyor himself to place stakes in advance at all the corners of the field, as he can then select the best mode of division. Platting. Begin with any diagonal and plat one triangle, as in Art. (90). Plat a second triangle adjoining the first one, as in Art. (91). Plat another adjacent triangle, and so proceed, till all have been laid down in their proper places. Measure the proof lines as in the last article. Calculation. Proceed to calculate the content of the figure, precisely as directed for the four-sided field, measuring the perpendiculars and calculating the content of each triangle in turn; or taking in pairs those on opposite sides of the same diagonal; or using some of the other methods which have been explained. Example 7. The six-sided field, shown in Fig. 51, has the lengths of its lines, in chains and links, written upon them, and is divided into four triangles, by three diagonals. The diagonal BE is a "proof-line." The Figure is drawn to a scale of 4 chains to the inch. The content of the field is 5A. 3R. 22P. Ex. 8. In a five-sided field, the length of the sides are as follows: AB 2.69, BC=1.22, CD = 2.32, DE=3.55, EA 3.23. The diagonals are AD = 4.81, BD=3.33. Required its content. = Ans. = (93) A field may be divided up into triangles, not only by mea suring diagonals as in the last figure, but by any of the methods shown in the four figures of Art. (71). The one which we have been employing, corresponds to the last of those figures. Still another mode may be used when the angles cannot be seen from one another, or from any one point within. Take three or more convenient points within the field, and measure from them tc the corners, and thus form different sets of triangles. |