With a 4 m staff, why is the instrument shifted after 3.825 and 3.455?

The reading nears the staff length, so the next point could not be read from the same set-up

Every fourth reading must be a change point

What does the first reading of -0.855 m on A mean?

The staff was held inverted and A is above the line of sight

A is 0.855 m below the instrument

Which three equal quantities verify the rise and fall level book?

SumBS - SumFS, SumRise - SumFall, and last RL - first RL

SumBS - SumIS, SumFall, and first RL

SumFS + SumBS, SumRise + SumFall, and last RL

RL of A = 380.500, RL of B = 369.460, 13 points at 30 m. What is the gradient?

Falling 11.040 m in 360 m, about 1 in 32.6

Falling 11.040 m in 390 m, 1 in 35.3

Rising 11.040 m in 360 m, 1 in 32.6

Falling 11.040 m in 30 m, 1 in 2.7

Asked 3/31
2081 Baisakh · 6 marks

Level book, RL of A 380.500, 3.945 on B

Short answer

With rise and fall method, RL of B is 369.460 m. The fall from A (380.500 m) to B is 11.040 m over 360 m (12 intervals of 30 m), giving a falling gradient of 1 in 32.6.

The following consecutive readings were taken with a level and a 4 meter levelling staff on a continuously sloping ground at a common interval of 30 meters.

-0.855 (on A), 1.545, 2.353, 3.115, 3.825, 0.455, 1.380, 2.055, 2.855, 3.455, 0.585, 1.015, 1.850, 2.755, 3.945 (on B).

The R.L. of A was 380.500. Make a level book and apply usual checks. Determine the gradient of the line AB.

Arranging the readings. The staff is 4 m long, so the instrument must be shifted when the reading nears 4 m: after 3.825 and after 3.455. The first reading ( $-0.855$ m) is negative: the staff was held inverted (for example under a beam), so the point A is above the line of sight. The set-ups are:

Set-up 1: BS $-0.855$ on A; IS 1.545, 2.353, 3.115; FS 3.825 on CP1.
Set-up 2: BS 0.455 on CP1; IS 1.380, 2.055, 2.855; FS 3.455 on CP2.
Set-up 3: BS 0.585 on CP2; IS 1.015, 1.850, 2.755; FS 3.945 on B.

RL of A $= 380.500$ m. The 15 readings are on 13 points (the FS and BS at a change point are on one point), 30 m apart.

Level field book (rise and fall method)

Station	BS	IS	FS	Fall	RL (m)
A	-0.855				380.500
2		1.545		2.400	378.100
3		2.353		0.808	377.292
4		3.115		0.762	376.530
5 (CP1)	0.455		3.825	0.710	375.820
6		1.380		0.925	374.895
7		2.055		0.675	374.220
8		2.855		0.800	373.420
9 (CP2)	0.585		3.455	0.600	372.820
10		1.015		0.430	372.390
11		1.850		0.835	371.555
12		2.755		0.905	370.650
B			3.945	1.190	369.460

Arithmetic check

$\Sigma BS - \Sigma FS = 0.185 - (11.225) = -11.040$ ; $\Sigma \text{Rise} - \Sigma \text{Fall} = 0.000 - 11.040 = -11.040$ ; last RL $-$ first RL $= 369.460 - 380.500 = -11.040$

All three are equal, so the reduction is correct.

Gradient of the line AB

There are 13 points, so 12 intervals of 30 m: $AB = 12 \times 30 = 360$ m.

\text{Fall from A to B} = 380.500 - 369.460 = 11.040\ \text{m}

\text{Gradient} = \frac{11.040}{360} = \frac{1}{32.6}\ (\text{falling from A to B})

Answer: RL of B = 369.460 m; the line AB falls at a gradient of 1 in 32.6 (about 1 in 33).

✓ Answer checked by Claude Opus 5.5 (high effort)

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