Fly levelling, BM 1280 m; falling gradient

Reading of the notes. The printed "(B)" is taken as the mark of the first BS, which is on the BM of RL 1280.00 m. The four BS readings are 0.63, 2.05, $-2.424$ and 2.56 m, and the three FS readings are 2.444, 1.35 and $-2.42$ m (negative readings are taken with the staff held inverted). The set-ups are BS 0.63, FS 2.444; BS 2.05, FS 1.35; BS $-2.424$, FS $-2.42$; and the last BS 2.56 m on the last change point CP3, from which the pegs are set out.

a) Level field book (rise and fall method)

Arithmetic check

$\Sigma BS - \Sigma FS$ (set-ups 1 to 3) $= (0.63+2.05-2.424) - (2.444+1.35-2.42) = 0.256 - 1.374 = -1.118$ m.

$\Sigma \text{Rise} - \Sigma \text{Fall} = 0.700 - 1.818 = -1.118$ m, and RL of CP3 $-$ RL of BM $= 1278.882 - 1280.000 = -1.118$ m. All agree.

b) Pegs on a falling gradient of 1 in 200

HI of the last set-up $= 1278.882 + 2.56 = 1281.442$ m. Fall per 30 m $= 30/200 = 0.15$ m. The pegs are at 30, 60, 90 and 120 m from CP3.

RL of a peg $=$ RL of CP3 $- 0.15 \times n$, and staff reading $=$ HI $-$ RL (the readings increase by 0.15 m per peg).

Check: the staff reading on the first peg $= 2.56 + 0.15 = 2.71$ m, and each following peg adds 0.15 m, so the readings are 2.71, 2.86, 3.01, 3.16 m.

Answer: RL of CP3 = 1278.882 m (HI = 1281.442 m). RLs of the 4 pegs = 1278.732, 1278.582, 1278.432, 1278.282 m; staff readings on the pegs = 2.71, 2.86, 3.01, 3.16 m.