The Double Longitude of ANY course is equal to the Double Longitude of the preceding course, plus the Departure of that course, plus the Departure of the course itself.* The Double Longitude of the last course (as well as of the first) is equal to its Departure. Its "coming out" so, when obtained by the above rule, proves the accuracy of the calculation of all the preceding Double Longitudes. (289) Areas. We will now proceed to find the Area, or Content of a field, by means of the "Double Longitudes" of its sides, which can be readily obtained by the preceding rule, whatever their number. Fig. 194. (290) Beginning with a three-sided field, ABC in the figure, draw a Meridian through A, and draw perpendiculars to it as in the last figure. It is N plain that its content is equal to the difference of the areas of the Trapezoid DBCE, and of the Triangles ABD and ACE. The area of the Triangle ABD is equal to the product of AD by half of DB, or to the product of AD by FG; i. e. equal to the product of the Latitude of the 1st course by its Longitude. F A& H E The area of the Trapezoid DBCE is equal to the product of DE by half the sum of DB and CE, or by HJ; i. e. to the product of the Latitude of the 2d course by its Longitude. S The area of the Triangle ACE is equal to the product of AE by half EC, or by KL; i. e. to the product of the Latitude of the 3d course by its Longitude. Calling the products in which the Latitude was North, North Products, and the products in which the Latitude was South, South Products, we shall find the area of the Trapezoid to be a South Product, and the areas of the Triangles to be North Pro The last course is a "preceding course" to the first course, as will appear on remembering that these two courses join each other on the ground. ducts. The Difference of the North Products and the South Products is therefore the desired area of the three-sided field ABC. Using the Double Longitudes, (in order to avoid fractions), in each of the preceding products, their difference will be the double area of the Triangle ABC. Fig. 1:5. (291) Taking now a four-sided field, ABCD in the figure, and drawing a Meridian and Longitudes as before, it is seen, on inspection, that its area N would be obtained by taking the two Triangles, ABE, ADG, from the figure EBCDGE, or from the sum of the two Trapezoids EBCF and FCDG. The area of the Triangle AEB will be found, as in the last article, to be equal to the product of the Latitude of the 1st course. by its Longitude. The Product will be North. The area of the Trapezoid EBCF will be found to equal the Latitude of the 2d course by its Longitude. The product will be South. B E F A S D C The area of the Trapezoid FCDG will be found to equal the product of the Latitude of the 3d course by its Longitude. The product will be South. The area of the Triangle ADG will be found to equal the product of the Latitude of the 4th course by its Longitude. The product will be North. The difference of the North and South products will therefore be the desired area of the four-sided field ABCD. Using the Double Longitude as before, in each of the preceding products, their difference will be double the area of the field. (292) Whatever the number or directions of the sides of a field, or of any space enclosed by straight lines, its area will always be equal to half of the difference of the North and South Products arising from multiplying together the Latitude and Double Longi tude of each course or side. We have therefore the following GENERAL RULE FOR FINDING AREAS. 1. Prepare ten columns, headed as in the example below, and in the first three write the Stations, Bearings and Distances. 2. Find the Latitudes and Departures of each course, by the Traverse Table, as directed in Art. (281), placing them in the four following columns. 3. Balance them, as in Art. (284), correcting them in red ink. 4. Find the Double Longitudes, as in Art. (288), with reference to a Meridian passing through the extreme East or West Station, and place them in the eighth column. 5. Multiply the Double Longitude of each course by the corrected Latitude of that course, placing the North Products in the ninth column, and the South Products in the tenth column. 6. Add up the last two columns, subtract the smaller sum from the larger, and divide the difference by two. The quotient will be the content desired. (293) To find the most Easterly or Westerly Station of a survey, without a plat, it is best to make a rough hand-sketch of the survey, drawing the lines in an approximation to their true directions, by drawing a North and South, and East and West lines, and considering the Bearings as fractional parts of a right angle, or 90°; a course N. 45° E. for example, being drawn about half way between a North and an East direction; a course N. 28° W. being not quite one-third of the way around from North to West; and so on, drawing them of approximately true proportional lengths. (294) Example 1, given below, refers to the five-sided field, of which a plat is given in Fig. 175, page 151, and the Latitudes and Departures of which were calculated in Art. (282), page 175. Station 1 is the most Westerly Station, and the Meridian will be supposed to pass through it. The Double Longitudes are best found by a continual addition and subtraction, as in the margin, where they are marked D. L. The Double Longitude of the last course comes out equal to its Departure, thus proving the work. The Double Longitudes being thus obtained, are multiplied by the corresponding Latitudes, and the content of the field obtained as directed in the General Rule. This example may serve as a pattern for the most compact manner of arranging the work. DOUBLE DOUBLE AREAS. E.+W.. -LONGITUDES. N. + | 8. (295) The Meridian might equally well have been supposed to pass through the most Easterly station, 4 in the figure. The Double Longitudes could then have been calculated as in the margin. They will of course be all West, or minus. The products being then calculated, the sum of the North products will be found to be 29.9625, and of the South products 8.1106, and their difference to be 21.8519, the same result as before. (296) A number of examples, with and without answers, will now be given as exercises for the student, who should plat them by some of the methods given in the preceding chapter, using each of them at least once. He should then calculate their content by the method just given, and check it, by also calculating the area of the plat by some of the Geometrical or Instrumental methods given in Part I, Chapter IV; for no single calculation is ever reliable. |