and the distance 79 four-pole chains. Under 35°. 15' or 35 degrees, and opposite 79, we find 64.52 for the latitude, and 45. 59 the departure, which signify that the end of that station differs in latitude from the beginning 64.52 chains, and in departure 44. 59 chains. Note, We are to understand the same things if the distance is given in perches or any other measures, the method of proceeding being exactly the same in every case. Again, let the bearing be 543 degrees, and distance as before; then over said degrees we find the same numbers, only with this difference, that the lat. before found, will now be the dep. and the dep. the lat. because 543 is the complement of 35 degrees to 90, viz. lat 45. 59. dep. 64. 52. 2. Suppose the same course, but chains 90 links, or as many perches. the distance 7 Here we find the same numbers, but the decimal point must be removed one figure to the left. Thus, under 354, and in a line with 79 or 7.9' are Lat. 6. 45 the 5 in the dep. being increased by 1, because the 9 is rejected; but over 543 we get Lat. 4. 56 THEO. II. When the first Meridian passes through the Map. If the east meridian distances in the middle of each line be multiplied into the particular southing, and the west meridian distances into the particular northing, the sum of these products will be the area of the map. Let the figures abkm be a map, the lines ab, bk, to the southward, and km, ma, to the northward, NS the first meridian line passing through the first station a The meridian 2zdxao Distances east Stuxox(by) =Area 2 am Saw These four areas am+aw+xp+gl will be the area of the whole figure cmswiprle, which is equal to the area of the map abkm. Complete the figure. The parallelograms am and ow, are made of the east meridian distances dz and tu, multiplied into the southings ao and ox. The parallelograms xp and gl are composed of the west meridian distances ef and bh. multiplied into the northings xg and ga (my) but these four parallelograms are equal to the area of the map; for if from them be taken the four triangles marked Z, and in the place of those be subsisted the four triangles marked O, which are equal to the former; then it is plain the area of the map will be equal to the four parallelograms. Q. E. D. THEŎ. III. If the meridian distance when east, be multiplied into the southings, and the meridian distance when west be multiplied into the northings, the sum of these less by the meridian distance when west, multipli ed into the southings, is the area of the survey. Let a b c be the map. The figure being completed, the rectangle af, is made of the meridian distance eq, when east, multiplied into the southing an; the rectangle yk, is made of the meredian distance xw, multiplied into the northing cz or ya. These two rectangles, or parallelograms, af+yk, make the area of the figure dfnyikd, from which taking the rectangle oy, made of the meridian distance tu when west, into the southing oh or bm, the remainder is the area of the figure dfohiked, which is equal to the area of the map. Let bou=Y, urih=L, ric=0, wrc=Z, akw=K, and efb=B, ade=A. I say that Y+Z+B=K+L+A. Y=L+O, add Z to both, then Y+Z=L+Ö+Z ; but Z+0=K, put K instead of Z+O, then Y+Z=L +K, add to both sides the equal triangles B and A, then Y+Z+B=L+K+A. If therefore B+Y+Z be taken from abc, and in lieu thereof we put L+K+A, we shall have the figure dfohikd=abc, but that figure is made up of the meredian distance when east, multiplied into the southing, and the meridian distance, when west, multiplied into the northing less by the meridian distance, when west, multiplied into the southing. Q. E. D. COROLLARY. Since the meridian distance (when west) multiplied into the southing, is to be subtracted, by the same reasoning the meridian distance when east, multiplied into the northing, must be also subtracted. |