Erbil Polytechnic University

1Survey I

Lecture 4

Out lines

❖ Leveling.

❖ Instruments.

❖ Types of levels.

❖ Basic Principle of Leveling

❖ Types of leveling.

2

31- Level is an instrument which is used for observing staff

reading on leveling staff kept over different points.

2- Staff. Is simply a large ruler, available in lengths of 3, 4 or 5

meters and usually made of aluminum with telescopic sections.

Reading the Staff

Color Alternates every meter each graduation is 100mm, each

“E” is 50mm

meter height & 1/10m is located in lower 50mm each Part of the E is 10mm

millimeters are interpolated staff is read to the millimeter.

Leveling Instruments.

4Staff Reading

Stadia Reduction

The middle line is the line of

Collimation, the short lines are

called stadia lines usually stadia

multiplier is 100

Collimation is an imaginary

line that passes through the level

instrument at the cross hairs.

5Basic Principle of Leveling

6Basic Principle of Leveling

B/S 2.510

F/S 1.202

Diff Between 1.308

B/S 2.301

F/S 1.517

Diff Between 0.784

7Basic Principle of Leveling

Basic Rules for Leveling

➢ Always start and finish a levelling run on a Benchmark (BM )

and close the loop.

➢ Keep fore sight and back sight distances as equal as possible.

➢ Keep lines of sight short (normally < 50m).

➢ Never read below 0.5m on a staff (refraction).

➢ Use stable, well defined change points.

➢ Beware of shadowing effects and crossing waters.

8

9Leveling Types

1- Height of Instrument (H.I) method

H.I = Elev. B.M + B.S

10Example

The following staff readings

0.915, 2.672, 3.145, 2.112, 1.215, 1.725, 2.315, 3.315, 3.623, 1.292, 1.550, 2.355

The level was shifted after second and 6th and 9th reading , the elevation of B.M

= 323.300

Calculate the elevations of the points.

11Solution .

1- H.I method

H.I = B.M Elev. + B.S

Elev. of any point = H.I – Staff Reading ( I.S or F.S)

No. Point B.S I.S F.S H.I Elev. (m) Remarks

1 A 0.915 324.215 323.300 B.M

2 B 3.145 2.672 324.688 321.543

3 C 2.112 322.576

4 D 1.215 323.473

5 E 2.315 1.725 325.278 322.963

6 F 3.315 321.963

7 G 1.292 3.623 322.947 321.655

8 H 1.550 321.397

9 I 2.355 320.592

12Solution.

Check Check the calculations:

ΣBS – ΣFS = Last Elev. - First Elev.

(7.667 - 10.375) = (320.592 - 323.300) = - 2.708

13Rise and Fall method

Difference in height (ΔH)= Last reading – Next reading =

To find the elevation of other points

Next Elev. = Last Elev. + R or Next Elev. = Last Elev. – F

To check the calculations:

ΣBS – ΣFS = Σ Rises – Σ Falls = Last Elev. - First Elev.

( + ) R

( - ) F

14Leveling.

Example

The following staff reading

0.915, 2.672, 3.145, 2.112, 1.215, 1.725, 2.315, 3.315, 3.623, 1.292, 1.550,

2.355

The level was shifted after second and 6th and 9th reading , the elevation of

B.M = 323.300

Calculate the elevations of the points. Using :

1- Rise and Fall method.

15Solution.2- Rise and Fall method

# point B.S I.S F.S Rise Fall Elev. (m) Remarks

1 A 0.915 323.300 B.M

2 B 3.145 2.672 1.757 321.543

3 C 2.112 1.033 322.576 T.P

4 D 1.215 0.897 323.473

5 E 2.315 1.725 0.510 322.963 T.P

6 F 3.315 1.000 321.963

7 G 1.292 3.623 0.308 321.655 T.P

8 H 1.550 0.258 321.397

9 I 2.355 0.805 320.592

∑ = 7.667 = 10.375 = 1.930 = 4.638

One Loop(-)

(-)

(-)

(-)

(-)

(+)

(+)(=)

(=)

(=)

(=)

(=)

(=)

(=)

16Solution.

2- Rise and Fall method Check the calculations:

ΣBS – ΣFS = Σ Rises – Σ Falls = Last Elev. - First Elev.

7.667 - 10.375 = 1.930 - 4.638 = 320.592 - 323.300 = - 2.708

17Example:

The following staff readings were taken on an ascending land, 1-

1- Find the elevations of another points.

2- Check the calculations.

The staff reading on the B.M with elev. 445.325 m was 1.5651.565, 2.315, 1.891, 1.565, 1.016, 0.730, 3.150, 2.635,1.925,1.252, 0.835, 2.831,

2.085, 1.731, 1.205, and 0.895

# Points B.S I.S F.S H.I Elev. Remarks

1 A 1.565 446.890 445.325 B.M

2 B 2.315 444.575 First #

3 C 1.891 444.999

4 D 1.565 445.325

5 E 1.016 445.874

6 F 3.150 0.730 449.310 446.160 T.P1

7 G 2.635 446.675

8 H 1.925 447.385

9 I 1.252 448.058

10 J 2.831 0.835 451.306 448.475 T.P2

11 K 2.085 449.221

12 L 1.731 449.575

13 M 1.205 450.101

14 N 0.895 450.411 Last #

18Solution.1- H.I method

19Solution.2- Rise and Fall method

# Points B.S I.S B.S Rise Fall Elev. Remarks

1 A 1.565 445.325 B.M

2 B 2.315 0.750 444.575 First #

3 C 1.891 0.424 444.999

4 D 1.565 0.326 445.325

5 E 1.016 0.549 445.874

6 F 3.150 0.730 0.286 446.160 T.P1

7 G 2.635 0.515 446.675

8 H 1.925 0.710 447.385

9 I 1.252 0.673 448.058

10 J 2.831 0.835 0.417 448.475 T.P2

11 K 2.085 0.746 449.221

12 L 1.731 0.354 449.575

13 M 1.205 0.526 450.101

14 N 0.895 0.310 450.411 Last #

∑ 7.546 2.460 5.836 0.750

References 20

Ghilani, C. D. and P. R. WOLF (2014). Elementary Surveying: An Introduction to Geomatics . New Jersey, PEARSON.

Uren, J. and B. Price (2010). Surveying for Engineers. UK, PALGRAVE MACMILLAN.

Barry F. Kavanagh – 7th – ed. SURVEYING with ConstructionApplications. PEARSON.

21End of Lecture 4