Example. Let BD 30 chains; ED 12 chains; and the desired area 24.8 acres. Then BF35.22 chains. The scale of the figure is 30 chains to 1 inch=1:23760. (506) When the field is given by Bearings, find the angle B from the Bearings; then is Example. Let BA bear S. 75° E., and BC N. 60° E., and let five acres be required to be Fig. 348. B F parted off from the field by a perpendicular to BA. Here the angle B = 45°, and BF10.00 chains. The scale of the figure is 20 chains to 1 inch = 1:15840. (507) To part off a quadrilateral. Produce the converging sides to meet at B. Calculate the Fig. 369. parted off, and it will give that of the triangle FGB, and the method of the preceding case can then be applied. (508) To part off any figure. If the field be very irregularly shaped, find by trial any line which will part off a little less than the required area. This trial line will represent HK in the preceding figure, and the problem is reduced to parting off, according to the required condition, a quadrilateral, comprised between the trial line, two sides of the field, and the required line, and containing the difference between the required content and that parted off by the trial-line. C. BY A LINE RUNNING IN ANY GIVEN DIRECTION. (509) To part off a triangle. By construction, on the ground or the plat, proceed nearly as in Art. (505), setting out a line in the required direction, calculating the triangle thus formed, and obtaining BF by the same formula as in that Article. (510) If the field be given by Bearings, find from them the angles CBA and GFB; then is × BFG sin. (B + F) BF sin. B. sin. F 2). Example. Let BA bear S. 30° E.; BC, N. 80° E.; and a fence be required to run from some point in BA, a due North course, and to part off one acre. Required the distance from B to the point F, whence it must start. Ans. The angle B 70°, and F 30°. Then BF 6.47. Fig. 350. E G B The scale of Fig. 350 is 6 chains to 1 inch 1:4752. Fig. 351. F (511) To part off a quadrilateral. Let it be required to part. off, by a line running in a given direction, a quadrilateral from a field in which are given the side AB, and the directions of the two B other sides running from A and from B. On the ground or plat produce the two converging sides to meet at some point E. Calculate the content F A of the triangle ABE. Measure the side AE. From ABE subtract the area to be cut off, and the remainder will be the content of the triangle CDE. From A set out a line AF parallel to the given direction. Find the content of ABF. Take it from ABE, and FAE' thus obtain AFE. Then this formula, EDAE AEODE, will fix the point D, since AD AE— ED. (512) When the field and the dividing line are given by Bearings, produce the sides as in the last article. Find all the angles from the Bearings. Calculate the content of the triangle ABE, by the formula for one side and its including angles. Take the Example. Let DA bear S. 2010 W.; AB, N. 51° W., 8.19; BC, N. 731° E.; and let it be required to part off two acres by a fence, DC, running N. 45° W. Ans. ABE=32.50 sq. chains; whence CDE=12.50 sq. chs. Also, AE=8.37; and finally AD 8.375.49 2.88 chains. The scale of Fig. 351 is 5 chains to 1 inch 1:3960. If the sum of the angles at A and B was more than two right angles, the point E would lie on the other side of AB. The necessary modifications are apparent. (513) To part off any figure. Proceed in a similar manner to that described in Art. (508), by getting a suitable trial-line, producing the sides it intersects, and then applying the method just given. D. BY A LINE STARTING FROM A GIVEN POINT IN A SIDE. (514) To part off a triangle. Let it be required to cut off from a corner of a field a triangu Fig. 352. by half the base, and the quotient will be the height of the triangle. Set off this distance from any point of AB, perpendicular to it, as from A to C; from C set out a parallel to AB, and its intersection with the second side, as at D, will be the vertex of the required triangle. Otherwise, divide the required content by half of the perpendi cular distance from A to BD, and the quotient will be BD. This original formula is very convenient for logarithmic computation. (515) If the field be given by the Bearings of two sides and the length of one of them, deduce the angle B (Fig. 352) from the 2 × ABD AB. sin. B' Bearings, as in Art. (243). Then is BD If it is more convenient to fix the point D, by the Second Method, Art. (6), that of rectangular co-ordinates, we shall have BE—BD. cos. B; and ED BD. sin. B. The Bearing of AD is obtained from the angle BAD; which is known, since ED ED EA AB — BE tang. BAD. Example. Eighty acres are to be set off from a corner of a field, the course AB being N. 80° W., sixty chains; and the Bearing of BD being N. 700 E. Ans. BD=53.33; BE 46.19; ED = 26.67 ; and the Bearing of AD, N.1748 W. The scale of Fig. 352 is 40 chains to 1 inch 1:31680. If the field were right angled at B, of course BD 2 ABD (516) To part off a quadrilateral. Imagine the two converging sides of the field produced to meet, as in Art. (511). Calculate the content of the triangle thus formed, and the question will then be reduced to the one explained in the last two articles. (517) To part off any figure. Proceed as directed in Art.(513). Otherwise, proceed as follows. The field being given on the ground or on a plat, find on which side of it the required line will end, by drawing or running "guess lines" from the given point to various angles, and roughly measur ing the content thus parted off. # If, as in the figure, A being the given point, the guess line AD parts off less than the required content, and AE parts off more, then the desired division line AZ will B end in the side DE. Subtract the D Fig. 353. area parted off by AD from the ----------- A required content, and the difference will be the content of the tri angle ADZ. Divide this by half the perpendicular let fall from the given point A to the side DE, and the quotient will be the base, or distance from D to Z. Or, find the content of ADE and make this proportion; ADE: ADZ:: DE: DZ. (518) The field being given by Bearings and distances, find as before, by approximate trials on the plat, or otherwise, which side the desired line of division will terminate in, as DE in the last figure. Draw AD. Find the Latitude and Departure of this line, and thence its length and Bearing, as in Art. (440). Then calculate the area of the space this line parts off, ABCD in the figure, by the usual method, explained in Part III, Chapter VI. Subtract this area from that required to be cut off, and the remainder will be the area of the triangle ADZ. Then, as in Art. (515), 2 ADZ ᎠᏃ AD. sin. ADZ This problem may be executed without any other Table than that of Latitudes and Departures, thus. Find the Latitude and Departure of DA, as before, the area of the space ABCD, and thence the content of ADZ. Then find the Latitude and Departure of EA, and the content of ADE. Lastly, make this proportion: ADE : ADZ :: DE : DZ.* Example. In the field ABCDE, &c., part of which is shown in Fig. 353, (on a scale of 4 chains to 1 inch =1:3168), one acre is to be parted off on the west side, by a line starting from the angle A. Required the distance from D to Z, the other end of this dividing line.† The only courses needed are these. AB, N. 53° W., 1.55; BC, N. 20° E.,2.00; CD, N. 5310 E., 1.32; DE, S. 57° E., 5.79. A rough measurement will at once shew that ABCD is less than an acre, and that ABCDE is more; hence the desired line will fall * The problem may also be performed by making the side on which the division line is to fall, a Meridian, and changing the Bearings as in Art. (244). The difference of the new Departures will be the Departure of the Division line. Its position can then be easily determined, by calculations resembling those in Part VII, Chapter IV, Arts. (443), &c. + If the whole field has been surveyed and balanced, the balanced Latitudes and Departures should be used. We will here suppose the survey to have proved perfectly correct. |