Why are 4.685 and 0.625 both entered at point B?

B is a change point: 4.685 is the FS before shifting and 0.625 the BS after shifting the instrument

0.625 is a check reading by a second observer

Which gives the correct check for this book?

Sum BS - sum FS = Sum rise - sum fall = last RL - first RL = -10.775 m

Asked 2/31
2068 Baisakh · 8 marks

Level book, 5 m staff, B at 1200.800

Short answer

By rise and fall method, reduced levels: A=1205.100 m, B=1200.800 m, C=1196.940 m, D=1194.325 m. The line falls 10.775 m over 240 m, giving a gradient of 1 in 22.3 (4.49% downward).

The following consecutive readings were taken with a level and a 5 meter levelling staff on a continuously sloping ground at a common interval of 20 meters: 0.385 (Point A), 1.030, 1.925, 2.825, 3.730, 4.685 (Point B), 0.625, 2.005, 3.110, 4.485 (Point C), 0.975, 1.382, 1.836, 2.702, 3.59 (Point D). The reduced level of the point B was 1200.800 m. Rule out a page of a level field book and enter the above readings. Calculate the reduced levels of the points by rise and fall method and the gradient of the line joining the first and the last point.

Assumptions: the readings 4.685 (B) and 4.485 (C) are foresights on change points, and the next readings 0.625 and 0.975 are backsights on the same points after the instrument is shifted. Chainage is taken from A at 20 m per staff position, so A to B = 100 m, B to C = 60 m, C to D = 80 m (total 240 m).

B is the 6th reading, so its RL is fixed first (1200.800 m) and the others are found upwards and downwards from it.

Point	Chainage	BS	IS	FS	HI	Fall	RL (m)
A	0	0.385			1205.485		1205.100
2	20		1.030			0.645	1204.455
3	40		1.925			0.895	1203.560
4	60		2.825			0.900	1202.660
5	80		3.730			0.905	1201.755
B	100	0.625		4.685	1201.425	0.955	1200.800
7	120		2.005			1.380	1199.420
8	140		3.110			1.105	1198.315
C	160	0.975		4.485	1197.915	1.375	1196.940
10	180		1.382			0.407	1196.533
11	200		1.836			0.454	1196.079
12	220		2.702			0.866	1195.213
D	240			3.590		0.888	1194.325

HI is shown on the row where a new set-up starts. RL = previous RL + rise - fall.

Arithmetic checks

\begin{aligned} \Sigma BS - \Sigma FS &= 1.985 - 12.760 = -10.775 \text{ m}\\ \Sigma \text{Rise} - \Sigma \text{Fall} &= 0.000 - 10.775 = -10.775 \text{ m}\\ \text{Last RL} - \text{First RL} &= 1194.325 - 1205.100 = -10.775 \text{ m} \end{aligned}

All three are equal, so the arithmetic is correct.

Gradient of the line AD (first to last point)

RL of A = 1205.100 m, RL of D = 1194.325 m. Difference in level = 10.775 m (falling). Horizontal distance = 240 m.

\text{Gradient} = \frac{10.775}{240} = \frac{1}{22.27}

Answer: RLs: A 1205.100, B 1200.800, C 1196.940, D 1194.325 m. The line AD falls at about 1 in 22.3 from A to D (4.49 %).

✓ Answer checked by Claude Opus 5.5 (high effort)

Level book, RL of B 1180.750 m · Surveying I, chapter 5 · asked 2×
Level book, staff readings at B and C (1251.250) · Surveying I, chapter 5 · asked 3×
Level book, staff readings at B and C (1251.000) · Surveying I, chapter 5 · asked 3×
Level book: profile levelling readings · Surveying I, chapter 5
Level book, RL of A 380.500, 3.945 on B · Surveying I, chapter 5 · asked 3×
Level book, 20 m interval, RL 1280 to 1279.924 · Surveying I, chapter 5
Level book, RL of B 1380.500 · Surveying I, chapter 5 · asked 3×

All Surveying I chapter 5 questions →

Level book, 5 m staff, B at 1200.800

Arithmetic checks

Gradient of the line AD (first to last point)

Related questions