101 For an example, we will resume the same example that has already been considered. As 37.25 Error in Northing, . Sum of courses, 37.25 12.76 13.44 12.69 12.63 12.76 12.63 : 0.18 7.43 13.11 13.11 12.67 12.67 error in lat. of 1st course. error in lat. of 2d course. 7.60 As 37.25 0.68: 10.40 0.19 error in lat. of 4th course. As 37.25 0.06: 10 : 0.02* error in dep. of 1st course. 139. REMARK I. In finding the error in latitude or departure, for a particular course, the last figure is sometimes doubtful; in which case it is best to mark it, as in the third proportion for error în latitude, and the first for error in departure; and then, if the figures taken do not balance the work,. let each be increased or diminished by 1. 140. REMARK II. It has already been observed (Art. 136), that if the measurements on the field are correctly made, the sums of the northings and southings will be equal to each other, as also those of the eastings and westings. It is the opinion of some surveyors, that when the error in latitude or departure exceeds one link for every five chains of the courses, the field notes ought not to be relied on. This, perhaps, is a higher degree of accuracy than can be attained. The error, however, should always be made considerably less than one link to a chain. Of the double meridian distances of the courses. 141. After the work has been balanced, the next thing to be done is to calculate the double meridian distance of each course. For this purpose, a meridian line is assumed, lying either wholly without the land, or passing through any point within it. It is, however, most convenient to take that meridian which passes through the most. easterly or westerly station of the survey; and these two stations are readily determined by inspecting the field notes. Having chosen the meridian, let the station through which it passes, be called the principal station, and the course which begins at this point, the first course. Care, however, must be taken, not to confound this with the course which begins at station 1, and which is the first course that is entered in the field notes. It has already been remarked (Art. 132), that all departures in the direction east, are considered as plus, and all departures in the direction west, as minus: then, through whatever station of the survey the assumed meridian be taken, we shall have for the calculation of the double meridian distances, the following RULE. I. The double meridian distance of the first course is equal to its departure. II. The double meridian distance of the next course is equal to the double meridian distance of the first course, plus its departure, plus the departure of the second course. III. The double meridian distance of the third course is equal to the double meridian distance of the second, plus its departure, plus the departure of the third course. IV. And, the double meridian distance of any course is equal to the double meridian distance of the preceding course, plus its departure, plus the departure of the course itself. REMARK. It should be recollected that plus is here used in its algebraic sense, and that when double the meridian distance of a course and the departure which is to be added to it, are of different names, that is, one east and the other west, they will have contrary algebraic signs; hence, their algebraic sum will be expressed by their difference, with the Demonstration of the Rule. N S Let the figure ABCD, which we have already surveyed with the compass, be resumed. By inspecting the field notes, it will be seen that B, or station 2, is the most westerly B station. Through this point let the assumed meridian NS be supposed b e A to pass. Then, B will be the principal station, and BC the first course. By what has been already said, every departure towards the east is to be considered as plus, and every departure towards the west, as minus. S Now, since p, k, d and a, are the middle points of the courses BC, CD, DA and AB, we have, by similar triangles, 2 qp=2 sx=sC=the first departure. 2 Cr 2 hk Cy=the second departure. 2 fg=2 gA=Af=the third departure. 2 At=2 ab=Ac=the fourth departure. We also have, 2 qp=sC=doub. mer. dis. of BC. 2 qp+2xC+2 Cr=2 kn=doub. mer. dis. of CD. 2 kn+2 kh-2 gf=2 de=doub. mer. dis. of D.A. 2 de 2 gA-2 At=2 ab=doub. mer. dis. of AB. The departure of the courses BC, CD, are east, and therefore positive; while the departures of the courses DA, AB, are west, and consequently negative. Since the course of reasoning just pursued is applicable to all figures, we may regard the rule as demonstrated for every case which can occur. REMARK. The double meridian distance of the last course should be equal to the departure of that course. A verification of the work is, therefore, obtained by comparing this double meridian distance with the departure of the course. 142. To apply the above rule to the particular example already considered, rule a new table, as below, in which are entered the balanced northings and southings, and the bal In this table there is but a single column for the difference of latitude, and a single column for the departures. The sign shows when the difference of latitude is north, and the sign, when it is south. The sign also shows when We see, from inspecting the notes, that 2 is the most westerly, and 4 the most easterly station. Either of them may, therefore, be taken for the principal station. Let us assume 2 for the principal station, and distinguish it by a star, thus *. Having done so, we enter the departure 8.21 in the column of double meridian distances, which gives the double meridian distance of the first course. The double meridian distances of the other courses are calculated according to the rule; and as the last, opposite to station 1, is equal to the departure of the course, the work is known to be right. Of the Area. 143. Having calculated the double meridian distance of each course, the next and last operation for finding the content of the ground, is explained in the following RULE. I. Multiply the double meridian distance of each course by its northing or southing, observing that like signs in the multiplicand and multiplier give plus in the product, and that unlike signs give minus in the product. II. Place all the products which have a plus sign in one column, and all the products which have a minus sign in another. III. Add up each of the columns separately and take their It is now evident, that cB multiplied by 2ba=cA, will give double the area of the triangle cAB. But cВ and ba are both plus; hence, the product will be plus, and must be put in the column of plus areas. Double the area of the triangle BSC, is equal to Bs multiplied by 2qp, which product is also plus. The area of the trapezoid ms CD is equal to yD=ms multiplied by nh (Geom. Bk. IV, Prop. VII); hence, double the area is equal to yD into 2nh. But since yD is minus, and 2nh plus, it follows that the product will be negative; hence, it must be placed in the column of negative areas. Double the area of the trapezoid cADm, is equal to Df=mc multiplied by 2de: but, since Df is negative and 2de positive, the product will be negative. It is now evident that the difference between the two columns is equal to twice the content of the figure ABCD: |