There are four BS and three FS readings. What does this show?

Four set-ups; the last BS is on CP3 and pegs are set out from the last set-up

Three set-ups, with BM in the last

Four set-ups, with the last BS taken on a peg

On a rise of 1 in 80 with pegs 30 m apart, how is the staff reading at each peg found?

Rise per peg = 30/80 = 0.375 m; staff reading = HI - peg RL, decreasing by 0.375 m per peg

Rise per peg = 80/30 = 2.67 m; reading = HI + RL

Rise per peg = 0.375 m; readings increase by 0.375 per peg

Peg 5 must be at 3012.476 m. What is the first peg RL and the staff reading at peg 5?

Peg 1 = 3010.976 m; staff reading at peg 5 = 1.002 m

Peg 1 = 3010.609 m; reading = 1.369 m

Peg 1 = 3011.343 m; reading = 1.002 m

Peg 1 = 3010.976 m; reading = 2.502 m

2079 Bhadra · 6 marks

Fly levelling for road pegs, BM 3010 m

Short answer

After fly levelling from BM 3010.000 m, RL of CP3 is 3010.609 m and HI at last set-up is 3013.478 m. Correct RLs of pegs 1–5 set out on a rising 1 in 80 gradient: 3010.976, 3011.351, 3011.726, 3012.101, 3012.476 m; staff readings: 2.502, 2.127, 1.752, 1.377, 1.002 m.

During the construction of a road, 5 pegs are to be set out at the centre lines of the road. For this purpose, fly level is run from a benchmark of RL 3010 m and the following readings were obtained: Backsight: (-)1.234, 2.594, 1.327, 2.869; Foresight: 0.456, 1.123, 0.499. From the last position of the instrument, first 5 pegs at 30 m intervals are to be set out on a uniform rising gradient of 1 in 80. Enter the readings on a level field book and work out the staff readings on the top of the pegs. If the last peg is to be established to have a RL of 3012.476 m, find the correct RLs of each station.

Reading the problem. The fly levelling starts at a BM of RL 3010 m. There are four BS and three FS readings, so the instrument was set up four times; the fourth set-up (BS 2.869 m on the last change point CP3) is the last position, from which the five pegs are set out. The first BS ( $-1.234$ m) is negative: the staff was held inverted.

Level field book (rise and fall method)

Station	BS	FS	Rise	Fall	RL (m)	HI (m)
BM	-1.234				3010.000	3008.766
CP1	2.594	0.456		1.690	3008.310	3010.904
CP2	1.327	1.123	1.471		3009.781	3011.108
CP3	2.869	0.499	0.828		3010.609	3013.478

Check: $\Sigma BS - \Sigma FS$ for the three complete set-ups $= (-1.234+2.594+1.327) - (0.456+1.123+0.499) = 2.687 - 2.078 = 0.609$ m $=$ RL of CP3 $-$ RL of BM $= 3010.609 - 3010.000 = 0.609$ m.

HI of the last set-up $= 3010.609 + 2.869 = 3013.478$ m.

Setting out the pegs on a rising gradient of 1 in 80

Rise per 30 m peg interval $= 30/80 = 0.375$ m. Staff reading at a peg $=$ HI $-$ RL of the peg.

Taking peg 1 at the RL of the last change point (so the last peg is at 3012.109 m):

Peg	Distance (m)	RL (m)	Staff reading (m)
1	0	3010.609	2.869
2	30	3010.984	2.494
3	60	3011.359	2.119
4	90	3011.734	1.744
5	120	3012.109	1.369

Correct RLs when the last peg must be at 3012.476 m

The last peg is to be established at RL 3012.476 m, which is 0.367 m higher than 3012.109 m. Keeping the same gradient (1 in 80), the RLs of all pegs are therefore raised by 0.367 m: the first peg is $3012.476 - 4 \times 0.375 = 3010.976$ m.

Peg	Distance (m)	Correct RL (m)	Staff reading (m)
1	0	3010.976	2.502
2	30	3011.351	2.127
3	60	3011.726	1.752
4	90	3012.101	1.377
5	120	3012.476	1.002

Answer: RLs of the stations: BM 3010.000, CP1 3008.310, CP2 3009.781, CP3 3010.609 m; HI at the last set-up = 3013.478 m. Correct RLs of pegs 1 to 5 = 3010.976, 3011.351, 3011.726, 3012.101, 3012.476 m; staff readings on the pegs = 2.502, 2.127, 1.752, 1.377, 1.002 m.

Assumption: peg 1 is the first peg of the gradient line and the pegs are 30 m apart from peg 1 to peg 5 (120 m).

✓ Answer checked by Claude Opus 5.5 (high effort)

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All Surveying I chapter 5 questions →

Fly levelling for road pegs, BM 3010 m

Level field book (rise and fall method)

Setting out the pegs on a rising gradient of 1 in 80

Correct RLs when the last peg must be at 3012.476 m

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