be divided be equal, it will be only required to divide two of the opposite sides of the rectangle into the given number of equal parts, and range the lines for the several fences to the consecutive points of division, and the land will then be divided as required. This appears to be too simple a case to require a diagram. ČASE II.-If the rectangular space is to be divided among several joint purchasers, who have paid unequal sums for the purchase thereof: then use the following RULE. As the sum of all the sums paid, is to the length of the side of the rectangle from which the division lines abut, so is any one person's sum to the breadth on the side of the rectangle due to that person; then divide both this side and the one opposite to it, by laying off the resulting breadths, and range lines to the corresponding points, and the rectangle will be divided by parallel lines, as required. NOTE. This is obviously a question of single fellowship. Divide the rectangle ABCD, the length of which is 1470 links and its breadth 684 links, among three ioint-purchasers P, Q and R, who paid for the purchase thereof respectively EXAMPLE. D с d £120, A a B £150, £220. Whole breadth 1470 which proves the work. NOTE. In this operation the breadth of the rectangle, which is common to all the three divisions, is not required to be used. PROBLEM II. To divide a triangle of equal value throughout, either equally or unequally, among several claimants, who shall all have the use of the same watering-place, situated at one of the angles of the triangular field. A α b C B EXAMPLE. It is required to divide the triangular field ABC among three persons, whose claims therein are as the numbers 2, 3 and 5, so that they may all have the use of a wateringplace situate at the angle C; AB being = 1000, A C = 685, and C B 610 links. which are the portions of the base A B, belonging to the respective claimants; therefore, if lines be drawn from a and b to C, the triangular field will be divided in the required proportion, each claimant having the use of the watering-place at C. NOTE. In solving this Problem it is not necessary to use the lengths of the sides AC, CB, because all the three triangles A Ca, a Cb, bCB have a com. mon perpendicular; and, therefore, their areas are as their bases. PROBLEM III. To divide a triangular field of equal value throughout, either equally or unequally, among sundry claimants, by fences running from any given point in one of its sides. The method of solving this Problem will be best shewn by the following EXAMPLE. Divide the triangular field ABC, the sides of which measure 30 chains = AB, 23 B C, and 19 A C, equally among three persons, by fences running from an occupation road that meets the side A B at H, which is 14 chains from A, that all the three persons may have the use of the road at H. Divide AB into three equal parts in the points D, E; from H, (the point where the road meets A B) draw HC; parallel to which draw DF, EG meeting AC, BC respectively, in F and G; and join HF and HG, in which directions fences being made, will divide the triangle as required. NOTE. This method of solution is founded on the areas of triangles between the same parallels being equal: but it may be solved by actual measurement on the ground, by means of the following preliminary calculation. whence the distances AF and B G may be measured from A and B; and from the points F and G the fences of division may then be drawn to H. NOTE. When the claims of the several persons are unequal, it will be readily seen that A B is then only required to be divided in the proportion of the several claims, as in the preceding Problem, after which the solution will be the same, as that just given. PROBLEM IV. To divide an irregular field with any number of sides among sundry claimants, so that they may all have the use of a pond, situated at a given point within the field. NOTE. The method of solving this Problem will be best seen from the method adopted in the following Example, in conjunction with the preceding Problems. EXAMPLE. Three persons Q, R, and S, bought a five sided field ABCDE, having a pond therein at P; for which they paid respectively £100, £150, d D By multiplying eaah side by half the perpendicular thereon, the sum of the five products will be the area of the field; thus, 864 x 280 241920 area of APB, 827 x 304 = 251408 = 806 x 240 = 193440 = BPC, CPD, sum = 11.39572 = acres, the area of the field. The sums paid for the field by Q, R, and S, are as the numbers 2, 3, and 4, the sum of which is 9; therefore, 9: 11.39572 :: 2 : 2.53238 = : : Q's share, Let DP be assumed the divisional fence between Q and S's shares; then the area of the triangle D PE = 1·85504 acres is less than Q's share, therefore This difference must be taken from the triangle EPA to complete Q's share; this is best done by dividing the said difference by half the perpendicular Pe of the triangle EPA, and the quotient will be the distance E M: thus, Pe 270) 67736 (250-85 links = E M. = The distance E M, being nearly 251 links must now be measured from E on E A, which will give the point M, and, a straight fence being set out from M to P, will cut off Q's share. The remainder of the triangle EPA, viz. 2·67300 •67734 = 1.99566 is less than R's share, therefore, This difference must be taken from the triangle APB to complete R's share, as before: thus, Pa 280) 1.80291 (643.9 links = A N. = This distance being measured from A towards B, will give the point N; and, the fence NP being now set out, will divide the field as required; the triangles N PB, BPC, CPD making up the exact quantity required for S's share, as may be readily shewn by adding their three areas together, which will prove the accuracy of the work: thus, 864 = AB 643.9 AN sum 5.06476 = S's share, which proves the work. NOTE 1. If some or all of the fences of the field ABCDE had been crooked, the operation of division would have been the same, excepting that the quantities of the offsets would have to be taken into the account, thus making a little additional calculation. It will at once be seen that this method of division may be extended to any number of claimants, whatever be the shape of the ground to be divided, the dotted lines from P to the angles of the field not requiring to be measured. NOTE 2. All the surveyors, who have written on the subject of this Problem, use what they call "guess lines" to effect the division, and correct the resulting errors by dividing the double quantities in excess or defect by these " guess lines" for the perpendiculars to determine the correct positions of the divisional fences. The student will at once perceive that this is a circuitous, blundering, and unscientific method of proceeding, and should be avoided in every case, except where the boundary is very crooked, and the divisional fences are not required to be continuous straight lines. NOTE 3. In this Problem the division is effected and proved without the aid of a plan, but it would, perhaps, be better for the satisfaction of the claimants, as well as for the surveyor himself to plan the whole of the work; especially as an error in the work might thus be more readily detected. PROBLEM V. To set out from a field or common of variable value, a quantity of land, that shall have a given value, by a straight fence in a given direction. NOTE. This Problem presents a great variety of cases, the most simple of which shall be first produced. CASE I.-CABD is a portion of a straight-sided field, right angled at A and B, and PQ a right line, also at right angles to A B, dividing the field into portions of different values; it is required to lay off a quantity of land a A Bb of given value, by a straight fence a parallel to A B. N.B. Lines, such as PQ, are technically called quality lines. |