A complete treatise on practical land-surveying, in seven parts; . 300 X 300 = 90000, twice the area of the trapezoid DFGE; 100-1- 300 X 300 = 400 X 300 = 120000, twice the area of the trapezoid FHKG; and 400 X 300 = 120000, twice the area of the triangle HLK; then 15000 -f 140000 + 90000 -f 120000 -f- 120000 = 485000, twice the area of the whole offset; and 485000 ~ 2 = 242500 square links = 2a. 1r. 28p. the area required. p 4 216 LAXD-SURVEYING. (Part V. Computation of the Area by Casting. From the angle M, let fall the perpendicular M N, which you •11 £ j , «« i. , i 323 X 1500 484500 will
A complete treatise on practical land-surveying, in seven parts; . 300 X 300 = 90000, twice the area of the trapezoid DFGE; 100-1- 300 X 300 = 400 X 300 = 120000, twice the area of the trapezoid FHKG; and 400 X 300 = 120000, twice the area of the triangle HLK; then 15000 -f 140000 + 90000 -f 120000 -f- 120000 = 485000, twice the area of the whole offset; and 485000 ~ 2 = 242500 square links = 2a. 1r. 28p. the area required. p 4 216 LAXD-SURVEYING. (Part V. Computation of the Area by Casting. From the angle M, let fall the perpendicular M N, which you •11 £ j , «« i. , i 323 X 1500 484500 will find to measure 323 links; then — — 2 2 242250 square links = 2a. 1r. the area required; which differs only four-tenths of a perch from the area found by offsets. PROBLEM V. Lay down a curve-line Offset from the following Dimensions ; re-duce it to a right-angled Triangle by the Parallel Rider ; andfind its Area both by equidistant Ordinates and Casting. AN 0 1200 M 190 1000 K 260 800 G 270 600 E 250 400 C 180 200 0 000 From A> g° LHFDB Having laid down the figure, erect the perpendicular A P,for a temporary line. Part V.) LAND-SURVEYING. c2l7 Lay the ruler from A to E; move it parallel to C ; and markthe temporary line at 1. Lay the ruler from 1 to G; move it parallel to E ; and markthe temporary line at 2. Lay the ruler from 2 to K; move it parallel to G; and markthe temporary line at 3. Lay the ruler from 3 to M; move it parallel to K; andmark the temporary line at 4. Lay the ruler from 4 to N; move it parallel to M; and markthe temporary line at 5. Draw a line from N to 5 (P) ; and NAP, will be the right-angled triangle required. Computation of the Area by equidistant Ordinates. See Prob. 9,Part III Here the sum of the first and last ordinates is nothing; (J80+ 270 -f 190) X 4 = 640 X 4 = 2560, four times the sum ofthe even ordinates; and (250 -f-260) X 2 = 510 x 2 = 1020, *^ A « A ~1J JT « A 2560 + 1020 + 200 twice the sum ot the odd ordinates;
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