EXAMPLES. 1. A line bears N. 201°E. dist. 117Ch. required the corresponding latitude and departure. Ch. Ch. Ch. Dist. 100 corresp. Lat. 93.67 and Dep. 35.02 17 Whole Dist. 117 Lat. 109.59N. Dep. 40.97E. 2. What is the difference of latitude and departure of a line bearing N. 781°W. dist. 243 perches? Per. Per. Per. Dist. 100 corresp. Lat. 20.36 and Dep. 97.90 Whole Dist. 243 Lat. 49.48N. Dep. 237.90W. When the distance is expressed by chains or perches and decimals of a chain or perch. Find as above the latitude and departure corresponding to the given bearing and to the whole chains or perches. Then considering the decimals as a whole number, find the latitude and departure corresponding to it, removing the decimal point in each, two figures to the left hand if there be two decimals, or one figure to the left if there be but one decimal; these added to the former will give the difference of latitude and departure required. EXAMPLES. 1. If a line bear S. 4130W. dist. 57.36.Ch. what will be the corresponding difference of latitude and depar Ch. Ch. Ch. Dist. 57.00 corresp. Lat. 42.53 and Dep. 37.96 .36 Whole Dist. 57.36 Lat. 42.80S. Dep. 38.20W. 2. Required the latitude and departure corresponding to a line which bears N. 72°W. dist. 124.37 perches. Dist. 100.00 cor. Lat. 30.90 and Dep. 95.11 Whole Dist. 124.37 Lat. 58.43N. Dep. 118.29W. Note. If the number of whole chains or perches be less than 10, and there be but one decimal figure, the latitude and departure may be taken out at one view, by considering the mixed number as a whole one, and taking out the latitude and departure corresponding to it and the given bearing, and removing the decimal point in ́each, one figure to the left hand. Thus if a line bear N.2310W. dist. 9.3Ch. its difference of latitude will be 8.53Ch. N. and its departure 3.71Ch. W. PROBLEM. The bearings and distances of a survey being given, to find the area without the necessity of first protracting it. RULE 1. 1. Rule a table as in the annexed examples: In the first vertical column on the left hand, place the numbers that designate the stations, in the second the bearings, 2. Find by the traverse table the latitudes and departures answering to the several courses and distances, which place in the four succeeding columns, under N. or S., E. or W., according as they are north or south, east or west. Add up the northings and southings, and if the sums are not equal, find their difference, which will be the error of the survey in difference of latitude, which call by the same name as the least sum. Proceed in the same manner with the eastings and westings, and find the error of the departures. 3. Divide each of these errors by the sum of the dis tances in the third column, extending the division to four decimal places. 4. Multiply the first distance, omitting the decimals, if any, by each of these quotients, and the products will be the corrections in difference of latitude and departure, depending on the first course and distance. Take the first two decimals in each of these corrections, increasing the second by a unit if the third exceed 5, and place them in the 8th and 9th columns according to their name, and in the same horizontal column with the first course. and distance. Proceed in the same manner to obtain the other corrections. If the sums of these errors are not equal to the errors in difference of latitude and departure respectively, which in consequence of the decimals neglected will sometimes be the case, alter some of them by a unit in the second decimal to make them so. 5. Apply these corrections to their corresponding dif ferences of latitude and departure, by adding when of the same name and subtracting when of different names, and the corrected differences of latitude and departure will be obtained, which place in the four succeeding co-' lumns. P 6. Choose such a place in the columns of corrected eastings or westings as will admit of a continual addition of the one and substraction of the other; or, which amounts to the same thing, begin at the most easterly or most westerly point of the survey, and proceed to find the meridian distances for the several lines, in the order in which they were surveyed. Thus, beginning at the place chosen, the first departure will be the first meridian distance, which place in the column of meridian distances opposite the said departure; to this add the same departure, setting the sum under the former meridian distance, and in the same horizontal column with it. If the next departure be of the same name with that just used, add it to the meridian distance last formed, and again to that sum; but if it be of a different name subtract it twice, and set the sums or remainders in the column of meridian distances opposite the departure. Proceed in a similar manner with each departure, and if the additions and subtractions be rightly performed, the last meridian distance will come out nothing. 7. Multiply each of the upper numbers in the column of meridian distances by the corresponding latitude, and place the products in the columns of north or south area according as the latitude is north or south. Half the difference of the sums of the numbers contained in these columns will be the area of the survey*. * DEMONSTRATION.-Let ABCDEFGH, Fig. 78, represent the boundary of a survey, and let NS be a meridian passing through the most westerly station. From the points A, B, C, D, E, F, and H, let fall on the meridian NS, the perpendiculars Aa, Bb, Cc, Dd, Ee, Fe, Hh; and from the same points, parallel to NS, draw the lines, Ag, Bn, Cq, Dq, Ep, Fr, Hm; then the bearings, distances, latitudes, departures, meridian distances, and areas will be as in the table page 117. Note 1. In a true survey the sum of the northings and southings will be equal, and also those of the eastings and westings; but in practice, on account of little errors that are unavoidable in measuring the lengths of lines and taking their bearings, these sums will rarely be found exactly equal. Ag and Bg the corresponding latitude and departure, the latitude being north and the departure east. The third course being south, the distance CD is the difference of latitude, and there is no departure. The fifth course being west, the distance EF is the departure, and there is no difference of latitude. The sum of the northings is Ag+Fr+Gh-ab+eG+Gh-ab+eh bh+ha+eheb+ah, and the sum of the southings is Bn+CD+ Dq+Hm=bc+cd+de+ah=eb+ah; the sum of the northings is therefore equal to that of the southings. Also the sum of the eastings Bg+Cn+Hb+Am=Bg+Cn+AaBb+Cn=Cc, and the sum of the westings is Eq+EF+Gr=Fq+Gr=eq=Cc; consequently the sum of the eastings is equal to that of the westings. is Now beginning at G the most westerly point of the survey, the east departure Hh will be the first meridian distance; to this add the same departure Hh or ma, and the sum Hh+ma will be the next meridian distance; the next departure Am being also east, add it twice, and the sums Hh+Aa and Aa+bg will be the two next meridian distances. Proceeding thus agreeably to the rule to add each of the eastings twice and subtract each of the westings twice, the meridian distances will be found as in the table. But Prob. 9. the product of Aa+ Bb, the upper meridian distance in the first horizontal column, by Ag, the corresponding latitude, gives twice the area of AaBb, which by the rule is to be placed in the coluinn of north areas, because the latitude is north; also the product of Bb+Cc, by Bn, gives twice the area of BbcC, to be placed in the column of south areas, because the latitude is south; and so of the others. Now the sum of the south areas is 2.BbcC+2.CcdD+2.DdeE+ 2.AahH-2.Bbe EDCB+2.AahH=2.ABCDEFGH+2.FeG+2.Gh H+2.AHhbB+2.AahH⇒2.ABCDEFGH+2.FeG+2.GhH+2.Aa bB: From this sum subtracting the sum of the north areas, which is 2. AabB+2.Feg+2.GhH, the remainder is 2.ABCDEFGH; that is, |