If the perpendiculars of two triangles fall on one and the same base, it is unnecessary to put down the base twice, but insert the second perpendicular opposite to the number of the triangle in the table, and join it with the other perpendicular by a brace, as Nos. 1 and 2, 4 and 5, 6 and 7, 9 and 10, &c. Proceed after this manner till you have measured all the triangles, and then, by prob. 6, find the contents in perches of each respective triangle, which severally place in the table opposite to the number of the triangle, in the fourth column, under the word contents. But where two perpendiculars are joined together in the table by a brace, having both one and the same base, find the contents of each (being a trapezium) in perches, by prob. 11, which place opposite the middle of those perpendiculars, in the fourth column, under the word contents. Having thus obtained the contents of each respective triangle and trapezium which the map contains, add them all together, and their sum will be the contents of the map in perches, which being divided by 160 gives the contents in acres. Thus, for This being divided by 160 will give 25A. 3R. 22P., the contents of the map. Let your map be laid down by the largest scale your paper will admit, for then the bases and perpendiculars can be measured with greater accuracy than when laid down by a smaller scale, and if possible measure from scales divided diagonally. If the bases and perpendiculars were measured by four-pole chains, the contents of every triangle and trapezium may be had as before in problems 6 and 11, and consequently the whole contents of the map. If any part of your map has short or crooked bounds, as those represented in plate 7, fig. 5, then by the straight edge of a transparent horn draw a fine pencilled line, as AB, to balance the parts taken and left out, as also another BC: these parts, when small, may be balanced very nearly by the eye, or they may be more accurately balanced by method the third. Join the points A and C by a line, so will the contents of the triangle ABC be equal to that contained between the line AC and the crooked boundary from A to B, and to C: by this method the number of triangles will be greatly lessened, and the contents become more certain; for the fewer operations you have the less subject will you be to err, and if an error be committed the sooner it may be discovered. The lines of the map should be drawn small and neat, as well as the bases, the compasses neatly pointed, and the scale accurately divided ; without all which you may err greatly. The multiplications should be run over twice at least, as also the addition of the column of contents. From what has been said it will be easy to survey a field by reducing it into triangles and measuring the bases and perpendiculars by the chain. To ascertain the contents only it is not material to know at what part of the base the perpendicular was taken; since it has been shown (in cor. to theo. 13 geom.) that triangles on the same base and between the same parallels are equal : but if you would draw a map from the bases and perpendiculars, it is evident that you must know at what part of the base the perpendicular was taken, in order to set it off in its due position; and hence the map is easily constructed. PROBLEM XVI. Pl. 8. fig. 5. To determine the area of a piece of ground, having the map given, by reducing it to one triangle equal thereto, and thence finding its contents. Let ABCDEFGH be a map of ground which you would reduce to one triangle equal thereto. Produce any line of the map, as AH, both ways; lay the edge of a parallel ruler from A to C, having B above it ; hold the other side of the ruler, or that next you, fast; open till the same edge touches B, and by it, with a protracting pin, mark the point b on the produced line ; lay the edge of the ruler from b to D, having C above it, hold the other side fast, open till the same edge touches C, and by it mark the point c on the produced line. A line drawn froin c to D will take in as much as it leaves out of the map. Again, lay the edge of the ruler from H to F, having G above it; keep the other side fast, open till the same edge touches G, and by it mark the point g on the produced line; lay the edge of the ruler from g to E, having F above it, keep the other side fast, open till the same edge touches F, and by it mark the point f on the produced line. Lay the edge of the ruler from f to D, having E above it, keep the other side fast, open till the same edge touches E, and by it mark the point e on the produced line. A line drawn from D to e will take in as much as it leaves out. Thus have you the triangle cDe, equal to the irregular polygon ABCDEFGH.* If, when the ruler's edge is applied to the points A and C, the point B falls under the ruler, hold that side next the said points fast, and draw back the other to any convenient distance; then hold this last side fast, and draw back the former edge to B, and by it mark b on the produced line; and thus a parallel may be drawn to any point under the ruler as well as if it were above it. It is best to keep the point of your protracting pin in the last point in the extended line till you lay the edge of the ruler from it to the next station, or you may mistake one point for another. This may also be performed with a scale or ruler which has a thin-sloped edge, called a fiducial edge, and a fine-pointed pair of compasses. Thus, Lay that edge on the points A and C; take the distance from the point B to the edge of the scale, so that it may only touch it, in the same manner as you take the perpendicular of a tri * The demonstration of this is evident from prob. 19, Geom., page 63 of this book. angle; carry that distance down by the edge of the scale parallel to it to b, and there describe an arc on the point b, and if it just touches the ruler's edge the point b is in the true place of the extended line. Lay then the fiducial edge of the scale from b to D, and take a distance from C that will just touch the edge of the scale; carry that distance along the edge till the point which was in C cuts the produced line in c; keep that foot in c and describe an arc, and if it just touches the ruler's edge the point c is in the true place of the extended line. Draw a line from c to D and it will take in and leave out equally: in like manner the other side of the figure may be balanced by the line eD. Let the point of your compasses be kept to the last point of the extended line till you lay your scale from it to the next station, to prevent mistakes from the number of points. That the triangle cDe is equal to the right-lined figure ABCDEFGH will be evident from problems 18, 19, geom. ; for thereby, if a line were drawn from b to C, it will give and take equally, and then the figure bCDEFGH will be equal to the map. Thus the figure is lessened by one side, and the next balance line will lessen it by two, and so on, and will give and take equally. In the same manner an equality will arise on the other side. The area of the triangle is easily obtained, as before, and thus you have the area of the map. It is best to extend one of the shortest lines of the polygon; because if a very long line be produced, the triangle will have one angle very obtuse, and consequently the other two very acute; in which case it will not be easy to determine exactly the length of the longest side, or the points where the balancing • lines cut the extended one. This method will be found very useful and ready in small enclosures, as well as very exact; it may be also used in large ones, but great care must be taken of the points on the extended line, which will be crowded, as well as of not missing a station. PROBLEM XVII. A map with its area being given, and its scale omitted to be either drawn or mentioned, to find the scale. Cast up the map by any scale whatsoever, and it will be EXAMPLE. A map whose area is 126A. 3R. 16P. being given, and the scale omitted to be either drawn or mentioned, to find the scale. Suppose this map was cast up by a scale of 20 perches to an inch, and the contents thereby produced be 31A. 2R. 34P. As the area found, 31A. 2R. 34P.=5074P. to 20 X 20=400, So is the given area of the map 126A. 3R. 16P.=20296P. To the square of the scale by which it was laid down. 5074 : 400 :: 20296 : 1600, the square of the required scale. Root. Answer. The map was laid down by a scale of 40 perches to an inch. PROBLEM XVIII. How to find the true contents of a survey, though it be taken by a chain that is too long or too short. Let the map be constructed, and its area found, as if the chain were of the true length. And it will be, As the square of the true chain map, map EXAMPLE. If a survey be taken with a chain which is 3 inches too long, or with one whose length is 42 feet 3 inches, and the map thereof be found to contain 9:20A. 2R. 20P.; required the true contents. As the square of 42ft. Oin. the square of 504 inches = 254016 Is to the contents of the map, 920A. 1R. 20P.=147260P., So is the square of 42ft. 3in. = the square of 507 inches 257049 To the true contents. |