The document describes three leveling problems involving taking elevation readings with a level and calculating reduced levels (RLs) using the rise and fall method. Problem 1 involves 11 readings with instrument shifts after the 4th and 8th readings. The difference between the first and last point is -7.92m. Problem 2 involves 10 readings with shifts after the 4th and 7th readings. Checks show the difference between first and last RL is -3.15m. Problem 3 provides 9 readings and asks to calculate RLs using both rise and fall and collimation methods, with checks showing a difference of -6.945m.
The document describes three leveling problems involving taking elevation readings with a level and calculating reduced levels (RLs) using the rise and fall method. Problem 1 involves 11 readings with instrument shifts after the 4th and 8th readings. The difference between the first and last point is -7.92m. Problem 2 involves 10 readings with shifts after the 4th and 7th readings. Checks show the difference between first and last RL is -3.15m. Problem 3 provides 9 readings and asks to calculate RLs using both rise and fall and collimation methods, with checks showing a difference of -6.945m.
The document describes three leveling problems involving taking elevation readings with a level and calculating reduced levels (RLs) using the rise and fall method. Problem 1 involves 11 readings with instrument shifts after the 4th and 8th readings. The difference between the first and last point is -7.92m. Problem 2 involves 10 readings with shifts after the 4th and 7th readings. Checks show the difference between first and last RL is -3.15m. Problem 3 provides 9 readings and asks to calculate RLs using both rise and fall and collimation methods, with checks showing a difference of -6.945m.
The document describes three leveling problems involving taking elevation readings with a level and calculating reduced levels (RLs) using the rise and fall method. Problem 1 involves 11 readings with instrument shifts after the 4th and 8th readings. The difference between the first and last point is -7.92m. Problem 2 involves 10 readings with shifts after the 4th and 7th readings. Checks show the difference between first and last RL is -3.15m. Problem 3 provides 9 readings and asks to calculate RLs using both rise and fall and collimation methods, with checks showing a difference of -6.945m.
Problem No.01 • The following consecutive readings were taken with a level 2.58, 4.72, 8.62, 10.98, 3.12, 2.67, 1.78, 0.78, 4.82, 3.57 and 6.68. The instrument was shifted after the fourth and eighth readings .The first reading was taken on the staff held on the benchmark of reduced level 820.75. • Rule out a page of a level field book and enter the above readings. Calculate the levels of all points and show the usual checks .What is the difference between the first and last reading point. Readings Statio B.S I.S F.S Rise Fall R.L Remar n ks 1 2.58 820.75 B.M 2 4.72 2.14 818.61 3 8.62 3.9 814.71 4 3.12 10.98 2.36 812.35 C.P 5 2.67 0.45 812.8 6 1.78 0.89 813.69 7 4.82 0.78 1.00 814.69 C.P 8 3.57 1.25 815.94 • Checks: ∑B.S - ∑ F.S = ∑ Rises - ∑ Falls = Last R.L - First R.L 10.52 - 18.44 = 3.59 - 11.51 = 812.83 - 820.75 -7.92 = -7.92 = - 7.92 Problem No.02 • The following consecutive readings were taken with a level 1.805, 2.555, 3.805, 3.985, 1.960, 1.700, 1.600, 1.270, 2.395, and 2.60. The instrument was shifted after the fourth and seventh readings .The first reading was taken on the benchmark having the reduced level of 250.00m. • Rule out a page of a level field book and calculate the reduced levels of all points by using rise and fall method. Station Readings Rise Fall R.L Remark s B.S I.S F.S 1 1.805 250.00 B.M 2 2.555 0.75 249.25 3 3.805 1.25 248.00 4 1.960 3.985 0.18 247.82 C.P 5 1.700 0.26 248.08 6 1.270 1.600 0.10 248.18 7 2.395 1.125 247.055 C.P 8 2.60 0.205 246.85 Checks: ∑B.S - ∑ F.S = ∑ Rise - ∑ Fall = Last R.L - First (5.035) (8.185) (0.36) (3.51) 5.035 - 8.185 = 0.36 - 3.51 = 246.85 - 25 -3.15 = -3.15 = -3.15 Problem No.03 HW • 0.375, 1.195, 2.665, 3.855, 0.690, 1.785, 2.980, 0.26, and 1.435.If the reduced level of 1st point is 200 m then find the R.Ls of the other points by rise and fall method and also by collimation method. • Solution: • (1) Rise and Fall Method: Station Readings Rise Fall R.L Remark s B.S I.S F.S 1 0.375 200.00 B.M 2 1.195 0.82 199.18 3 2.665 1.47 197.71 4 0.690 3.855 1.19 196.52 C.P 5 1.785 1.095 195.425 6 0.26 2.980 1.195 194.23 C.P 7 1.435 1.175 193.055 Checks: 1.325 8.27 6.445 ∑B.S - ∑ F.S = ∑ Rises - ∑ Fall = Last R.L - First R. 1.325 - 8.27 = 0.00 - 6.945 = 193.055 - 2 -6.945 = -6.945 = -6.945