tions of churches, and other remarkable objects on the right and left of the railway, by intersection; it being unnecessary, at the same time, to measure their distances, especially if those distances from the railway be considerable, their positions being only required to shew more clearly the locality of the proposed railway to those interested therein. Also, at the same time, the angles of fences, or other parts of the actual survey that may be in the same direction of the churches, &c., may be conreniently taken, which will thus save the trouble of measuring tie lines. As the railway again leaves the direction of the main line B C at C, a third main line will be required (this line is not shewn in the figure); which may be either tied to B C, or its angle of inclination thereto be taken with the theodolite, according to convenience. The filling up of the part of the survey to the right and left of the part of B C, towards C may be readily effected by the help of the prismatic compass, or box sextant, on account of the fences being straight, and the fields large; the angles which these straight fences make with B C, being aken, and the lengths of the fences ineasured to the meetings of other straight fences. CHAPTER VI. SECTION I. THE VARIOUS METHODS OF LAYING OUT GIVEN QUANTITY OF LAND IN ANY PROPOSED REGULAR OR IRREGULAR FORM; OF SEPARATING ANY REQUIRED QUANTITY OF LAND FROM AN ENCLOSURE OR COMMON; AND OF DIVIDING LAND OF VARIABLE VALUE AMONG SEVERAL CLAIMANTS, IN PROPORTION TO THEIR RESPECTIVE CLAIMS. PRELIMINARY PROBLEMS. PROBLEM I. To lay out a given quantity of land in the form of a square. RULE.-Reduce the given quantity to square links, by the table of square measure page 15; and the square root of the result is a side of the required square, in links. EXAMPLES. 1. Lay out 2a. 1r. 5. p. in the form of a square. 2a. 200,000 square links. = 25,000 (links nearly, = 8 B 228,437(478 side of the square 16 A B DC. 87)684 609 948)7537 7584 2. Required the side of a square, that shall contain 3a. 3r. 28p. Ans. 626. links. Note. A given quantity of land is sometimes required to be laid out in the form of a square for pleasure grounds, &c. PROBLEM II. To lay out any proposed quantity of land in the form of a rectangle, the length of which is given. RULE.—Divide the square links in the quantity of land by the given length, and the quotient will be the breadth of the required rectangle.--If the dimensions are required in feet, the given area must be reduced to square feet. EXAMPLES. 1. Lay out 4a. Or. 12p. in a rectangular form, the length of which shall be 812 links. 4a. Or. 12p. = 407,500 square links, and 812) 407500(501.8 links = AC, the 4060 required breadth of А the rectangle ABDC. 1500 812 B 6880 384 2. The breadth of a rectangular piece of ground is 140 feet; required its length, when its area is 2 acres. acres. 622 feet, the length required. 3. The length of a rectangular pleasure ground is proposed to be 1000 feet, required its breadth so that it contain 12} Ans. 5442 feet. NOTE. Besides the extensive use of this Problem in the division of commons, &c., it is also used to determine the length of frontage of a given quantity of building ground, when its depth or breadth is given; or, to determine the depth, when the length of frontage is given. PROBLEM III. To lay out a given quantity of land in the form of a rect. angle, the length of which shall have a given proportion to its breadth: that is, the length shall be to the breadth as 3 to 1 or as 4 to 3, or as 5 to 2, &c. RULE.*-Divide the given area by the product of the two terms of the proportion, and the square root of the quotient, multiplied separately by the terms of the proportion, will give the required length and breadth. EXAMPLES. 1. Lay out 3 acres of land in the form of a rectangle, the length of which shall be to its breadth as 3 to 2. 3 x 2 = 6) 300000 square links in 3 acres 50000 (2236 42) 100 84 links. 223.6 x 3 = 670·8 length. 223.6 x 2 = 447.2 breadth. 443) 1600 1329 4466)27100 26796 304 • Let me and n x be the length and breadth of a rectangle, which aro in the ratio of m:n, and a its area; then ma X na = m n 2 = an orc= 2. Lay out 64 acres in a rectangular form, the ratio of whose length and breadth shall be as 5:2. 5 X 2 = 10)625000 5 x links. 250 x 5 = 1250 = length. N62500 = 250. 250 X 2= 500 = breadth. 3. It is required to lay out a rectangular fish pond, the length of which shall be four times its breadth, and its area 10 acres. Ans. Length 2000 and breadth 500 links. Note. Proportions of this kind are frequently required in laying out orna. mental grounds, ponds, &c., in rectangular forms. PROBLEM IV. To lay out a triangle, of a given area and a given base, one of the sides of which shall have a given position. RULE 1.—Divide twice the area by the base for the perpendicular, which erect at any point in the base, and at the extremity of the perpendicular and at right angles to it, range a line till it meet the side given in position, which point of meeting is the vertex of the triangle required. EXAMPLES. 1. It is required to lay out a triangle ABC to contain 2a. 2r. 16p. on a base A B of 1200 links, the position of A H, of which the side A C is a part, being given. By Rule 1. CH 2a. 2r. 16p. = 260000 links 2 1200) 520000 A B D 433} links=CD A On any point d of the given base AB= 1200 links, erect the perpendicular dc = 433} links; range cC perpendicular to do ; till it meet A H in C; then, B C being joined, A B C is the triangle required. For let CD be drawn perpendicular to AB, then, by the nature of the construction, CD=cd. Q. E. D. RULE 2.-Measure HB perpendicular to AH, the line given in position, to the end B of the given base, and divide twice the given area by H B, and the quotient is A C, which, being measured off, will give the point C. 2. Solve the last example by Rule 2, the perpendicular BH being 860 links. 260000 x 2 = 604.65 links : A C. 860 3. Required the perpendicular of a triangle, containing 26 2r. 28p., its given base being 3112 links. Ans. 1714, links, nearly. Note. This Problem is extensively used in the division of lands, commons, &c., as shall be hereafter shewn. PROBLEM V. To lay out a trapezium of given area, the positions and lengths of two of its adjoining sides being given, and also the position of its third side. RulE.-Join the ends of the two adjacent sides, thus forming a triangle, the area of which must be found, which area must be subtracted from the given area, and the remaining area must be laid out in the form of a triangle, on the line joining the two given sides, as base, by Prob. IV. EXAMPLE. P Let the given area of the trapezium ABDC be 6 acres, A B, AC the sides that are given in length and position, and ČH the position of the third side. Join BC; measure the perpendicular CP, which is found = 840 links, H and CB = 880 links, A B being 1200 links; whence the area of the triangle ABC =600 x 840=504000 A square links ; and 600000 504000 = 96000 = area of the triangle B C D; therefore 96000 x 2 = 880=218 links = perpendicular DQ; which, it will be seen, ; is applied as in Prob. IV.; thus constituting the required trapezium ABDC. CD may also be found as by Rule 2, Prob. IV. NOTE. This Problem is also much used in the division of lands, commons, &c. PROBLEM VI. To lay out a rhomboid of given area, the lengths of its sides being given. RULE.—Divide the area by the longer side, subtract the square of the quotient from the square of the shorter side, and the square root of the remainder is the distance of the perpendicular from the end of the larger side, which perpendicular, being made equal to the quotient first found, gives the breadth of the rhomboid. TT |