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 the departure is east, and the sign, when it is west. 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 cach 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 22 Let us again resume the example which we have been considering, B and write the difference of latitude and the double meridian distances W of the courses, in the following table. e m Stations. 1 2* Demonstration of the Rule. N 3 4 +2ba +2gp +2nh +2ed Area. + 2cAB +cB +Bs -yD - Df It is now evident, that cB multiplied by 2ba=cA, will give double the area of the triangle cAB. But cB 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. A 2 Bs C Area. E 2ms CD 2 cm DA 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: and as the same may be shown for any figure whatever, we may regard the rule as demonstrated for all cases. We will now make the calculations in numbers. Having balanced the work, we can place it in the following table. Observing in the field notes that station 2 is the most westerly point of the land, we assume the meridian which passes through this point, as the one from which the meridian distances are calculated. We mark the principal sta tion with a star. Opposite station 2, we enter, in the column of double meridian distances, headed D. M. D., the departure of the course from 2 to 3, which is the double meridian distance of that course, and plus. To this we add the departure of the course, and also the departure of the next course their sum is the double meridian distance of the course from 3 to 4. To the last sum add the departure opposite station 3, and the minus departure opposite station 4: their algebraic sum is the double meridian distance from 4 to 1. To the last sum add the last departure, which is minus, also the next departure which is likewise minus: this will give the double meridian distance of the course from 1 to 2, which is also equal to its departure. Then forming the products, adding them together, taking their difference, and dividing it by 2, according to the rule, we obtain the content of the ground. N 144. It only remains to make a plot of the ground. For this purpose, draw any line, B as NS, to represent the meridian passing through the principal station, on which take any point, as B, to represent that station. A S FIRST METHOD OF PLOTTING. Having fixed upon the scale on which the plot is to be made, lay off from B on the meridian, a distance Bs equal to the difference of latitude of the first course, and at s crect a perpendicular to the meridian, and make it equal to the departure of the first course: then draw BC, which will be the first course. Through C draw a meridian, and make Cf equal to the difference of latitude of the second course, and through f draw a perpendicular ƒD, and make it equal to the departure of the second course: draw CD, and it will be the second course. Lay down, in the same manner, the courses DA and AB, and the entire plot will be completed. SECOND METHOD OF PLOTTING. The work may be plotted in another manner, thus. At the principal station B, lay off an angle equal to the bearing from B to C, which will give the direction of BC. Then, from the scale of equal parts, make BC equal to the first course: this will give the station C. Through C draw a meridian, and lay off an angle equal to the bearing from C to D, and then lay off the course CD. Do the same for the bearing at D and the course DA; also, for the bearing at A and the course AB, and a complete plot of the ground will thus be obtained. If the work is all right, the last line AB will exactly close the figure. This plot is 2. It is required to determine the content and plot of a piece of land, of which the following are the field notes, viz. I Stations. 1 2 Dif. 1 N 46° W 20 ch 13.77 2* N 51° E 13.80 8.54 IN B2 W 3 Error in Northing. 4 5 6 5 S 3310 W 18.80 6 N 7410 W 30.95 8.27 Sum of courses.. Lat. ... 15.72 Bearings. Distances. N 46° W | 20 ch. N 51° E 13.80 E 21.25 S 56° E 27.60 S 3310 W 18.80 N 74° W | 30.95 A Dop. 132.40130.58131.16:54.97154.65 30.58 54.65 W Dep. D.M.D. AREA. 14.51+13.88-14.56 14.56 202.0928 +8.61 +10.81 10.81 93.0741 +21.20 42.82 BALANCED. Lat. -15.2922.82 86.84 10.31-15.63-10.36 99.30 29.838.43-29.91 59.03 497.6229 0.58 0.32 Error in Westing Ans. 1044 IR 16P Plot of the above example. D AREA. 1327.7836 1552.0590 792.789812879.8426 792.7898 2)2087.0528 1043.5264 E REMARK. When a bearing is due east or west, the error |