Find the true bearing of the lines with the following observed magnetic bearings and declinations. 

Find the true bearing of the lines with the following observed magnetic bearings and declinations.  (a) Bearing of line  = 189°12. Declination = 2° W (b) Bearing of line  = 72° 44′ Declination = 1° 12 ‘ E solutions : (a) True bearing of line= magnetic bearing of line – Declination (West) = 189° 12′ … Read more

Compute the back bearings for the following reduced bearings.(a)  S 79° 04′ E(b)N 88°12’W (C) N 28° 12′ E (d) S 4° 12′ W 

Compute the back bearings for the following reduced bearings.(a)  S 79° 04′ E(b)N 88°12’W (C) N 28° 12′ E (d) S 4° 12′ W  SOLUTIONS : (a) The fore bearing = S 79’04 ‘E. back bearing of= N 79° 04’N.  ( It is a opposite direction remaining value is same ) (b) The fore bearing … Read more

Find the back bearings of the following lines from the observed Fore Bearings:(a)  79° 24′(b)  354° 24′(c)  178 ° 4′(d)  204 ° 12′

Find the back bearings of the following lines from the observed Fore Bearings:(a)  79° 24′(b)  354° 24′(c)  178 ° 4′(d)  204 ° 12′ solutions : (a) 79°24′ Fore bearing = 79° 24′ Back bearing = FB +180° ( as the FB is less than 180° ) = 79°24′ +180° = 259°24′ (b) 354°24′ Fore bearing= … Read more

Convert the Following whole circle bearing to quadrantal bearing : (a) 38°30′ (b) 142°15′ (C) 248°45′ (d) 325°30′

Convert the Following whole circle bearing to quadrantal bearing : (a) 38°30′ (b) 142°15′ (C) 248°45′ (d) 325°30′ SOLUTIONS : (a) WCB = 38°30′ Rule: If WCB Is between 0° and 90 ° QB or RB = WCB = N 38°30’E (b) WCB= 142° 15′ Rule: If WCB Is between 90° and 180° Q B … Read more

Convert the following reduced bearings to whole circle bearings: (a) N 46° 30′ E (b) N60°30’W (c) S 80° 00′ E, (d) S 20°00′ W

Convert the following reduced bearings to whole circle bearings:(a) N 46° 30′ E(b) N60°30’W(c) S 80° 00′ E,(d) S 20°00′ W SOLUTIONS: ( a ) N 46° 30′ E Rule: For 1 Quadrant WCB Of = RB= 46° 30′   ( It is First Quadrant ) (b) S 80°  00′ E Rule : It is a … Read more

plot the following cross staff Survey of a field ABCDEFG and calculate its area in hectares

plot the following cross staff Survey of a field ABCDEFG and calculate its area in hectares Soultion: show the field ABCDEFG . The calculations of area are listed in the Following table S. NO figure chainage Base offset mean Offset Area Remark  1 ΔAQB o to 200 200 0 to 130 65 13,000  2. Trapezium … Read more

Plot the following cross-staff survey of a field and calculate its area in hectares. All distances noted in the field book are in metres 

Plot the following cross-staff survey of a field and calculate its area in hectares. All distances noted in the field book are in metres    solution : shows the plotting of field notes of the given area not to scale. The area Can be Found out in the Following manner. 1. Area of triangle AHB … Read more

Following figure shown a page in a field book of cross staff survey. Calculate the area enlarged between the stations:

Following figure shown a page in a field book of cross staff survey. Calculate the area enlarged between the stations:   1. Area of triangle AHD = 1 / 2 x 30 x 22.50 = 337.50 m² 2. Area of Trapezium HDFJ = ( 88.0-30.0) = ( 22.50+33.80/ 2 ) = 1632 . 70m² 3. … Read more

Q.A survey line ABC crossing a river at right angles cuts its banks at B and C. To determine width of river BC, a line BE, 30 m long was set out roughly parallel to the river. A point D was found in CE produced and middle point of DB determined, EF was then produced to G, making FG equal to EF, and DG produced to cut the survey line in H. GH and HB were found to be 20 and 40 metres long respectively. Find the distance from B to C.

Q.A survey line ABC crossing a river at right angles cuts its banks at B and C. To determine width of river BC, a line BE, 30 m long was set out roughly parallel to the river. A point D was found in CE produced and middle point of DB determined, EF was then produced … Read more