22/11/2019

Leveling is the operation of measuring vertical distance, either

directly or indirectly, to determine the difference in elevation. • Differential Leveling is the operation of determining
differences in elevation of points some distance apart or
Methods of establishing bench marks. Usually, differential
The difference in elevation may be measured by the following leveling is accomplished by direct leveling. Precise
methods: leveling is the precise form of differential leveling.

Direct or Spirit Leveling, by measuring vertical distances • Profile Leveling is the operation – usually by direct
directly. Direct leveling is the most precise method of leveling – of determining elevations of points at short
determining elevations and is the one commonly used. measured intervals along a definitely located line, such
as the center line for a highway or a sewer.
Indirect or Trigonometric Leveling, by measuring vertical
angles and horizontal or slope distances.

From the given data of differential leveling shown: HI = Elevation + Back Sight Reading
HI = 98.500 + 1.256
STA BS HI FS ELEV
HI = 99.756
BM1 1.256 98.500
TP1 1.116 1.886
TP 1 = HI – Foresight Reading
TP2 1.228 1.527 TP 1 = 99.756 – 1.886
TP3 1.189 2.246 TP 1 = 97.870
TP4 1.070 2.017
1.256 1.886
TP5 1.831 2.656
TP6 1.489 2.723
BM2 2.548

a. What is the value of HI in the 2nd setup.

b. Find the elevation of TP5. TP 1
c. Determine the elevation of BM2. BM 1
Elev:98.500 m
 22/11/2019

From the given data of differential leveling shown:

STA BS HI FS ELEV
STA BS HI FS ELEV BM1 1.256 98.500
TP1 1.116 1.886
BM1 1.256 99.756 98.500
HI TP1 = 97.870 + 1.116 TP2 1.228 1.527
TP1 1.116 98.986 1.886 97.870 TP3 1.189 2.246
HI TP1 = 98.986
TP2 1.228 98.687 1.527 97.459 TP4 1.070 2.017
TP3 1.189 97.630 2.246 96.441 TP5 1.831 2.656
96.683 95.613 TP6 1.489 2.723
TP4 1.070 2.017
BM2 2.548
TP5 1.831 95.858 2.656 94.027
TP6 1.489 94.624 2.723 93.135 a. What is the value of HI in the 2nd setup.
b. Find the elevation of TP5.
BM2 2.548 92.076
c. Determine the elevation of BM2.

From the given data of differential leveling shown:

ΔElev. = ΣBS - ΣFS STA BS HI FS ELEV
Elev. of TP 5 = Elev. of BM 1 + ΔElev. BM1 1.256 98.500
TP1 1.116 1.886
STA BS HI FS ELEV Find the elevation of TP 5.
TP2 1.228 1.527
BM1 1.256 98.500 TP3 1.189 2.246
Δ Elev. =
TP1 1.116 98.986 1.886 (1.256 + 1.116 + 1.228 + 1.189 + 1.070) TP4 1.070 2.017
TP2 1.228 1.527 – (1.886 + 1.527 + 2.246 +2.017 +2.656) TP5 1.831 2.656
TP3 1.189 2.246 TP6 1.489 2.723
Δ Elev. = - 4.473
BM2 2.548
TP4 1.070 2.017
Elev. of TP 5 = 98.500 +(- 4.473) = 94.027 a. What is the value of HI in the 2nd setup.
TP5 1.831 2.656 94.027
TP6 1.489 2.723 b. Find the elevation of TP5.
c. Determine the elevation of BM2.
BM2 2.548
 22/11/2019

ΔElev. = ΣBS - ΣFS

Elev. of TP 5 = Elev. of BM 1 + ΔElev. a) What is the value of HI in the 2nd setup?
ΔElev. = ΣBS - ΣFS Elev. BM1 + BS (BM1) + BS (TP1) – FS (TP1)
STA BS HI FS ELEV
Elev. of BM 2 = Elev. of TP 5 + ΔElev. 98.500 + 1.256 + 1.116 – 1.886 = 98.986
BM1 1.256 99.756 98.500
STA BS HI FS ELEV Find the elevation of BM 2.
TP1 1.116 98.986 1.886 97.870 b) Find the elevation of TP 5.
BM1 1.256 98.500 98.687
Δ Elev. = TP2 1.228 1.527 97.459 Δ Elev. = (1.256 + 1.116 + 1.228 + 1.189 + 1.070)
TP1 1.116 98.986 1.886 (1.831 + 1.489) – (2.723 + 2.548) = - 1.951 TP3 1.189 97.630 2.246 96.441 – (1.886 + 1.527 + 2.246 + 2.017 + 2.656)
= - 4.473
TP2 1.228 1.527 96.683
Elev. of BM 2 = 94.027+(- 1.951) = 92.076 TP4 1.070 2.017 95.613
TP3 1.189 2.246 TP5 1.831 95.858 2.656 94.027
Elev of TP 5 = 98.500 + (- 4.473) = 94.027

TP4 1.070 2.017 TP6 1.489 94.624 2.723 93.135 c) Find the elevation of BM 2.
TP5 1.831 2.656 94.027 BM2 2.548 92.076 Δ Elev. = 1.831 + 1.489 – (2.723 + 2.548) = - 1.951
TP6 1.489 2.723 ΔElev. = ΣBS - ΣFS Elev. of BM 2 = 94.027 + (- 1.951) = 92.076
BM2 2.548 92.076 Elev. of BM 2 = Elev. of TP 5 + ΔElev.

Problem 2:
The following data shows the difference in elevation between A 3
and B. • A line of levels 6 km long is run from BM1 to BM2
Trial Diff.in Elev.(m) No. of Measurements
with a computed elevation of 165.8m. The average
1 520.14 1
2 520.20 3 BS and FS distances are 120m and 180m
3 520.18 6 respectively. It was found out however that the line
of sight of the instrument is inclined upward by
a. What is the probable weight of trial 2? 0.002m in a distance of 10m.
b. What is the most probable difference in elevation?
c. Compute the elevation of B if the elevation of A is 900m with B
higher than A. a. Compute the error in every setup.
b. Determine the total error.
c. Find the correct elevation of BM2.
 22/11/2019

Problem 4:
Always ADD the total error (etotal) to the To adjust a dumpy level by peg method of adjustment, two
NOTE: (for error/set-up) value of measured/erroneous elevation points A and B were set up. With the instrument at Point
*Line of sight because the sign (+/-) of the error per set-up
Inclined Upward : e = -e1 + e2
A, the rod reading at A was 1.623m and the foresight at B
will dictate whether it will be added or
Inclined Downward : e = e1 - e2 subtracted. was 2.875m. The level is transferred at B and the
backsight at B was 1.622m while the rod reading at A
was 0.362m.
e1 e2
FS
BS a. Find the true difference in elevation between A and B.
b. What is the error in the line of sight?
120 m 180 m c. Determine the correct reading at A that will give the
TP1 TP2 level line of sight with the instrument still at point B.

1.623 2.875
0.362 1.622

A A

B B
 22/11/2019

Problem 4:
To adjust a dumpy level by peg method of adjustment, two
1.623 points A and B were set up. With the instrument at Point
2.875 A, the rod reading at A was 1.623m and the foresight at B
1.252 was 2.875m. The level is transferred at B and the
A
B backsight at B was 1.622m while the rod reading at A
was 0.362m.

a. Find the true difference in elevation between A and B.

0.362 b. What is the error in the line of sight?
1.622
1.260 c. Determine the correct reading at A that will give the
A
level line of sight with the instrument still at point B.
B

b. What is the error in the line of sight? Problem 4:

e
To adjust a dumpy level by peg method of adjustment, two
1.623 points A and B were set up. With the instrument at Point
2.879 2.875
2.875 A, the rod reading at A was 1.623m and the foresight at B
1.252
1.256 was 2.875m. The level is transferred at B and the
A
B backsight at B was 1.622m while the rod reading at A
was 0.362m.

a. Find the true difference in elevation between A and B.

0.362 b. What is the error in the line of sight?
1.622
1.260
1.256 c. Determine the correct reading at A that will give the
A
level line of sight with the instrument still at point B.
B
 22/11/2019

c. Determine the correct reading at A that will give the level line
of sight with the instrument still at point B. Problem 5:
A dumpy level was tested whether the line of sight is truly horizontal
e
when the bubble is at the center by using an alternate peg method as
1.623
2.879 2.875 follows. The level was set up midway between points A and B which is
1.256 100 m apart. The rod reading on A and B are recorded as 3m and 2m
A respectively. The instrument was then transferred to a point C which is in
B line with points A and B but not in between A and B with A nearer than B.
Rod readings were then taken at A and B to be 2.75 and 1.50 m
respectively. Distance from C to A is 20 m.

a. What is the difference in elev. between A and B?

e b. What is the error at point B with the instrument still at point C?
0.362
0.362 TRA c. What should be the reading at B so that the line of sight will be truly
1.618 1.622
1.256 horizontal?
A
B

A dumpy level was tested whether the line of sight is truly horizontal when the bubble is at the center
by using an alternate peg method as follows. The level was set up midway between points A and B
Problem 5:
which is 100 m apart. The rod reading on A and B are recorded as 3m and 2m respectively. The
instrument was then transferred to a point C which is in line with points A and B but not in between A dumpy level was tested whether the line of sight is truly horizontal
A and B with A nearer than B. Rod readings were then taken at A and B to be 2.75 and 1.50 m when the bubble is at the center by using an alternate peg method as
respectively. Distance from C to A is 20 m. follows. The level was set up midway between points A and B which is
100 m apart. The rod reading on A and B are recorded as 3m and 2m
respectively. The instrument was then transferred to a point C which is in
line with points A and B but not in between A and B with A nearer than B.
Rod readings were then taken at A and B to be 2.75 and 1.50 m
e e
respectively. Distance from C to A is 20 m.
2m
3m a. What is the difference in elev. between A and B?
y b. What is the error at point B with the instrument still at point C?
c. What should be the reading at B so that the line of sight will be truly
B horizontal?
A
50 m 50 m
 22/11/2019

A dumpy level was tested whether the line of sight is truly horizontal when the bubble is at the center
by using an alternate peg method as follows. The level was set up midway between points A and B
Problem 5:
which is 100 m apart. The rod reading on A and B are recorded as 3m and 2m respectively. The
instrument was then transferred to a point C which is in line with points A and B but not in between A dumpy level was tested whether the line of sight is truly horizontal
A and B with A nearer than B. Rod readings were then taken at A and B to be 2.75 and 1.50 m when the bubble is at the center by using an alternate peg method as
respectively. Distance from C to A is 20 m. follows. The level was set up midway between points A and B which is
100 m apart. The rod reading on A and B are recorded as 3m and 2m
respectively. The instrument was then transferred to a point C which is in
line with points A and B but not in between A and B with A nearer than B.
Rod readings were then taken at A and B to be 2.75 and 1.50 m
e1 e2 respectively. Distance from C to A is 20 m.
2.75 m 1.50 m
a. What is the difference in elev. between A and B?
1m b. What is the error at point B with the instrument still at point C?
C B c. What should be the reading at B so that the line of sight will be truly
A horizontal?
20 m 50 m 50 m

A dumpy level was tested whether the line of sight is truly horizontal when the bubble is at the center
by using an alternate peg method as follows. The level was set up midway between points A and B
which is 100 m apart. The rod reading on A and B are recorded as 3m and 2m respectively. The
+ 6
instrument was then transferred to a point C which is in line with points A and B but not in between STA BS HI FS IFS ELEV
A and B with A nearer than B. Rod readings were then taken at A and B to be 2.75 and 1.50 m
respectively. Distance from C to A is 20 m.
BM1 0.93 329.63 - 328.70
327.43
- -
1 2.20 From the given profile leveling notes:
2 3.00 326.63
TP1 2.87 329.09 3.41 326.22
3 1.70 327.39 a. What is the diff. in elev. bet. 2 and 5?
4 1.20 327.89 326.63 - 326.29 = 0.34
e1 e2 5 2.80 326.29 b. Find the height of instrument @ TP2.
1.50 m TP2 3.12 330.93 1.28 327.81
2.75 m HI @ TP2 = 330.93
6 1.60 329.33
1m 7 2.00 328.93 c. What is the elevation of BM2?
C B 8 3.60 327.33 Elev. BM 2 = 326.04
TP3 0.85 329.43 2.35 328.58
A
TP4 1.35 327.28 3.50 325.93
20 m 50 m 50 m
9 0.90 326.38
BM2 1.24 326.04