In the rise and fall method, how is the rise or fall between consecutive points found?

Previous reading minus present reading from the same set-up; positive is a rise

Present minus previous; positive is a rise

On a rising gradient of 1 in 200 with 30 m spacing, what are the first peg RL and staff reading?

RL 1470.14 m, staff reading 2.27 m

RL 1470.14 m, staff reading 2.57 m

RL 1469.84 m, staff reading 2.57 m

RL 1470.14 m, staff reading 1.82 m

SumBS - SumFS for the first four set-ups is -0.01 m. What must it equal?

SumRise - SumFall and RL of CP4 - RL of BM

2076 Chaitra · 6 marks

Fly levelling, BM 1470 m; set 4 pegs

Short answer

From BM 1470.000 m, RL of CP4 is 1469.99 m (HI 1472.41 m). For a rising 1 in 200 gradient, peg RLs at 30 m intervals: 1470.14, 1470.29, 1470.44, 1470.59 m; corresponding staff readings: 2.27, 2.12, 1.97, 1.82 m.

During fly levelling the following note is made. B.S: 0.62, 2.05, 1.42, 2.63 and 2.42 m. F.S.: 2.44, 1.35, 0.53 and 2.41 m. The first B.S was taken on a BM of RL 1470 m. From the last B.S. it is required to set 4 pegs each at a distance of 30 m on a rising gradient of 1 in 200. Enter these notes in the form of a standard level book and calculate the R.L. of the top of each peg by the rise and fall method. Also calculate the staff readings on each peg.

Entering the notes. There are 5 BS and 4 FS readings, so the instrument was set up five times. The first BS (0.62 m) is on the BM (RL 1470 m); the readings in order are BS 0.62, FS 2.44; BS 2.05, FS 1.35; BS 1.42, FS 0.53; BS 2.63, FS 2.41; and the last BS 2.42 m is on the last change point CP4, from which the pegs are set out.

Level field book (rise and fall method)

Station	BS	FS	Rise	Fall	RL (m)	HI (m)
BM	0.620				1470.000	1470.620
CP1	2.050	2.440		1.820	1468.180	1470.230
CP2	1.420	1.350	0.700		1468.880	1470.300
CP3	2.630	0.530	0.890		1469.770	1472.400
CP4	2.420	2.410	0.220		1469.990	1472.410

Arithmetic check

$\Sigma BS - \Sigma FS$ (set-ups 1 to 4) $= (0.62+2.05+1.42+2.63) - (2.44+1.35+0.53+2.41) = 6.72 - 6.73 = -0.01$ m.

$\Sigma \text{Rise} - \Sigma \text{Fall} = 1.81 - 1.82 = -0.01$ m, and RL of CP4 $-$ RL of BM $= 1469.99 - 1470.00 = -0.01$ m. All agree.

Setting out 4 pegs at 30 m on a rising gradient of 1 in 200

HI of the last set-up $= 1469.99 + 2.42 = 1472.41$ m. Rise per 30 m $= 30/200 = 0.15$ m. The pegs are at 30, 60, 90 and 120 m from CP4.

RL of a peg $=$ RL of CP4 $+ 0.15 \times n$ , and staff reading $=$ HI $-$ RL.

Peg	Distance from CP4 (m)	RL (m)	Staff reading (m)
1	30	1470.14	2.27
2	60	1470.29	2.12
3	90	1470.44	1.97
4	120	1470.59	1.82

Answer: RL of CP4 = 1469.99 m (HI = 1472.41 m). RLs of the tops of the 4 pegs = 1470.14, 1470.29, 1470.44, 1470.59 m; staff readings on the pegs = 2.27, 2.12, 1.97, 1.82 m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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Fly levelling, BM 1470 m; set 4 pegs

Level field book (rise and fall method)

Arithmetic check

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