CE371 Surveying

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CE 371 Surveying

PROFILE LEVELING &

Trigonometric LEVELING

Dr. Ragab Khalil

Department of Landscape Architecture Faculty of Environmental Design King AbdulAziz University Room LIE15

Lec 10 + Lec 11

Dr. Ragab Khalil KAU – FED – CE371 - Surveying

Overview

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• Profile Leveling

• Two-Peg Test

• Trigonometric Leveling

• Elevation of Inaccessible Points

• Grid Leveling

• Radial Line Leveling

• Borrow-Pit Leveling

Profile

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• Profile leveling yields elevations at definite points along a reference line.

• A profile is a curve resulting from a vertical plane crossing the ground along the survey line.

• Used in designing linear facilities:

Highways, Railways, Transmission lines, Canals, Sewers, Water mains, etc…

• Elevations along the line are taken every 10, 50, or 100 m depending on the purpose of survey and the terrain

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Stationing

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• In route surveying, a system calledstationingis used to specify the relative horizontal position of any point along the reference line.

• Stationing is the establishment offull&plusstations

• Full station: is a point on a survey route whose distance from the starting point is a multiple of 100 m.

• Plus station: is a point established at critical locations between full stations.

Full station

Full

station Full

station

Example

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• Write the distance 325 as:

100 m stationing

50 m stationing

20 m stationing

• Solution

• Distance 325 m as 100 m stationing: 3+25

Intermediate site

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Profile Leveling

BS

FS IS

1.25

BM #1 140.506

0+00 0+30 0+40 0+60 0+90 1+20 1+30 1+50 1+80 2+10

BM #2 138.512 TP #1

1.15 2.37 3.12 3.43 3.27 2.75 2.53 2.38

1.15 2.55 2.45 1.73 1.62

2.25

TP #2

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Profile Leveling Procedures

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1. Mark all full and plus stations along the line. If necessary, choose suitable locations for turning points.

2. Measure distances of all plus stations from the starting point (0+00).

3. Set up the level instrument off the survey route such that BS and FS distances are balanced.

4. For any setup, take a BS at the turning point (or the bench mark) before, then take an IS at every intermediate point until reaching the next turning point (or bench mark) at which a FS is then taken.

5. Compute elevations of all stations and turning points.

6. If a grade is required along the line, compute grade elevations at all stations, then compute the required cut and fill at each station.

Rate of Grade

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• Rate of grade (gradient or percent grade) is the rise or fall in meter per 100 meter.

• A grade of 2.5% means a 2.5 m difference in elevation per 100 m horizontally.

• A grade giving equal volumes of cut and fill is preferred.

43.5 m

150 m

= 43.5 + 3.75

= 47.25 m

=2.5 x 150

=3.75 m

Example

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A profile leveling along the center line of a proposed street starting at elevation 140.000 m with percentage grade = -1.5%. The street starts at station 0+00 up to station 2+10 using 100-m stationing. Compute elevations of all full and plus stations along with the amount of cut and fill.

BM#1= 140.506 BM#2=138.510 Solution

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Point BS IS FS HI Elev. Adj. E

Ground .L

Formati on F.L

GL-FL Cut - Fill

BM #1 1.25 141.756 140.506 140.506

0 + 00 1.15 140.606 140.604 140.000 0.604 C

0 + 30 2.37 139.386 139.384 139.550 -0.166 F

0 + 40 3.12 138.636 138.634 139.400 -0.766 F

0 + 60 3.43 138.326 138.324 139.100 -0.776 F

TP #1 2.75 3.27 141.236 138.486 138.484

0 + 90 2.53 138.706 138.702 138.650 0.052 C

1 + 20 2.38 138.856 138.852 138.200 0.652 C

TP #2 1.15 2.25 140.136 138.986 138.982

1 + 50 2.55 137.586 137.580 137.750 -0.170 F

1 + 80 2.45 137.686 137.680 137.300 0.380 C

2 + 10 1.73 138.406 138.400 136.850 1.550 C

BM #2 1.62 138.516 138.510

S 2.90 4.89

D -1.99 -1.99

E_c= 138.516–138.510 =.006 m E_A=6.1 3= 10.6 mm > E_C ok.

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Drawing and Use of a Profile

Plotted profiles are used to:

1. Determine depth of cut/fill on proposed highways, railroads and airports.

2. Study grade-crossing problems.

3. Determine the most economical grade, location, depth of sewers, pipelines, tunnels, and irrigation ditches.

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Drawing and Use of a Profile

0+00 0+30 0+40 0+60 0+90 1+20 1+50 1+80 2+10

136 137 138 139 140 141

Distance 1:1000 Elevations 1:100

Ground Level Formation Level Cut

Fill Cut

Cut

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Collimation error

• Occurs when the line of sight (as defined by the cross-hairs) is not horizontal

• Leads to an incorrect staff reading

horizontal line

error

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Two-Peg Test

30 30 رتم

رتم

b1 d1

DH

e e

30 رتم 30

30 رتم رتم

e 3e

b2

d2 DH = (b1 –e) –(d1 - e) = b1-d1

DH = (b2 –e) –(d2 - 3e) = b1-d1

b2 –d2 + 2e = b1-d1 𝑒 =𝑏1 − 𝑑1 − (𝑏2 − 𝑑2)

2

Adj d²=d²−3e

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example

• A two-peg test is done with the following results:

b1=1.543 m, d1=1.586 m,b2=1.529 m, d2=1.588 m, X=50.000 m. Compute the error in mm per m. Is the error accepted. Compute the adjusted d2 value.

• Solution

• Error e = [(1.543-1.586)-(1.529-1.588)] /2 = 0.008 m = 8 mm per 50 m

• Error per 30 m=30(8/50)=4.8 mm > 2 mm per 30 m (Adjustment is needed)

• Adjusted d2 rod reading = 1.588-3(0.008) = 1.564 m.

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Trigonometric Leveling

Trigonometry can be used to compute difference in elevation between two points by measuring horizontal distanceHbetween the two points and the vertical anglea (or zenith anglez).

ق

H hi

a

V r

A C B

D

E z

S a= 90-z

V= H. tan (a)

ElevB= ElevA+ hi + V - rB V= H. cot (z)

(For distances up to 300 m)

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Trigonometric Leveling Example

A theodolite is set up at point A whose elevation is 100.00 m. A level rod is put at point B whose horizontal distance from A is 120.00 m. If rod reading is 2.29 m, zenith angle is 65^o, height of instrument is 1.50 m, findElevB. Solution:

V = 120.00 cot(65^o) = 55.96 m

ElevB= 100.00 + 1.50 + 55.96 - 2.29 = 155.17 m

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Curvature And Refraction in Trigonometric Leveling

• For long horizontal distances, the effect of curvature and refraction should be taken into account.

ElevB= ElevA+ hi + V –rB+.0675 H²

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Trigonometric Leveling _Long distances Example

A theodolite is set up at point A whose elevation is 100.00 m. A level rod is put at point B whose horizontal distance from A is 500.00 m. If rod reading is 2.29 m, zenith angle is 65^o, height of instrument is 1.50 m, findElev_B. Solution:

V = 500.00 cot(65^o) = 233.15 m Combined curvature and refraction effect CR = 0.0675(0.5)²= 0.017 m

ElevB=100.00 + 1.50 + 233.15 - 2.29 + 0.017 = 332.38 m

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Trigonometric Leveling With Unknown Distance

• If the horizontal distance is unknown, we can useStadia method or we have to take two level rod readings and two zenith angles.

𝐻 =_tan(𝑎^𝑟¹^−𝑟²

1)−tan(𝑎2)

ق

H hi

a₁

r₁ V1

A C B

D

E z₁

S r₂

V2

a2

z2

ElevB= ElevA+ hi + V1–r1

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Trigonometric Leveling With Unknown Distance --- stadia

A theodolite is set up at point A whose elevation is 50.00 m. A level rod is put at point B. The three hair readings at B were 2.162/1.780/1.398 m and taken at a zenith angle of 82^o. Compute horizontal distance AB and elevation of B. Height of instrument is 1.50 m.

• Solution

• Vertical angle (a) = 90-82 = 8^o

• I=U-L= 2.162-1.398= 0.764

• H= 100*I*{cos (a)}²= 100x 0.764 x (cos (8))²= 74.92 m

V= 74.92 x tan (8) = 10.53 m ElevB= ElevA+ hi + V –r

ElevB= 50 + 1.50 + 10.53 –1.78 = 60.25 m

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Trigonometric Leveling With Unknown Distance--- two readings

A theodolite is set up at point A whose elevation is 50.00 m. A level rod is put at point B. A rod reading of 1.78 m is taken at a zenith angle of 82^o. A second rod reading of 0.45 m is taken at zenith angle equals 83^o. Compute horizontal distance AB and elevation of B. Height of instrument is 1.50 m.

• Solution

a1= 90-82 =8º a2= 90-83 =7º 𝐻 = ^1.78−0.45

tan(8)−tan(7)=74.90 m V1= 74.90 x tan (8) = 10.53 m

ElevB= 50 + 1.50 + 10.53 –1.78 = 60.25 m

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Elevation of Inaccessible Points

• Case 1. Base of the object accessible

• Case 2. Base of the object inaccessible, Instrument stations in the vertical plane as the elevated object.

• Case 3. Base of the object inaccessible, Instrument stations not in the same vertical plane as the elevated object.

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Case 1. Base of the object accessible

B

A = Instrument station B = Point to be observed h = Elevation of B from the

instrument axis

D = Horizontal distance between A and the base of object h1 = Height of instrument (H. I.) Bs = Reading of staff kept on B.M.

= Angle of elevation = L BAC h = D tan 

Elev. of B = Elev. of B.M. + Bs + h

= Elev. of B.M. + Bs + D. tan  If distance is large, then add Cc & Cr

Elev. of B = Elev. of B.M. + Bs + D. tan + 0.0675 D²

Elev. of B = Elev. of A+ hi + h

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Case 2. Base of the object inaccessible, Instrument stations in the vertical plane as the elevated object.

There may be two cases.

(a) Instrument axes at the same level (b) Instrument axes at different levels.

1) Height of instrument axis to the object is lower:

2) Height of instrument axis to the object is higher:

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Case 2. Base of the object inaccessible, Instrument stations in the vertical plane as the elevated object.

(a) Instrument axes at the same level

DPAP, h= D tan 1 DPBP, h= (b+D) tan 2

D tan 1= (b+D) tan 2 D tan 1= b tan 2+ D tan 2 D(tan 1- tan 2) = b tan 2

Elev. of P = Elev. of B.M + Bs + h Elev. of P = Elev. of A+ hi + h

h= D tan ₁

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(b)Instrument axes at different levels.

1) Height of instrument axis to the object is lower:

DPAP, h1= D tan 1

DPBP, h2= (b+D) tan 2

hd is difference between two height hd = h₁–h₂

hd = D tan 1- (b+D) tan 2

= D tan 1- b tan 2-D tan 2

hd = D(tan 1- tan 2) - b tan 2

hd + b tan 2= D(tan 1- tan 2)

h1 = D tan 1 Elev. of P = Elev. of A+ hi + h1 +CR

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Ex.

To determine the elevation of building p above point B in the figure, the following measurements were made: b=50.00 m, hi_B=1.24 m, hi_A =0.94 m, angle ₂ = 13.125^o, angle ₁ = 18.626^o. The instrument in the two positions was at the same level. Compute Elevpabove BM B.

Solution

hd = 1.24-0.94=0.30

𝐷 = 0.3+50 tan(13.125)

tan 18.626 −tan(13.125)=115.12 h2= 165.12 tan (13.125) =38.50 Elev. of P = Elev. Of B+ hi + h2 +CR

Elev. of P - Elev. Of B = 1.24 + 38.50+ 0.0675 (165.12/1000)²=39.74 m

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(b)Instrument axes at different levels.

2) Height of instrument axis to the object is higher:

DPAP, h1= D tan 1

DPBP, h2= (b+D) tan 2

h_dis difference between two height h_d= h₂–h₁

h_d= (b+D) tan 2- D tan 1

= b tan 2+ D tan 2- D tan 1

h_d= b tan 2+ D (tan 2- tan 1) h_d- b tan 2= D(tan 2- tan 1) - h_d+ b tan 2= D(tan 1- tan 2)

h1= D tan 1 Elev. of P = Elev. of A+ hi + h1 +CR

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Case 3. Base of the object inaccessible, Instrument stations not in the same vertical plane as the elevated object.

Set up instrument on A Measure 1 to P L BAC =  Set up instrument on B Measure 2 to P L ABC =  L ACB = 180 –( + ) Sin Rule:

BC= b· sin

sin{180˚ - (+ )}

AC= b· sin

sin{180˚ - (+  h1 = AC tan 1

h2 = BC tan 2

b

Elev. of P = Elev. of A+ hi + h1 +CR

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Borrow-Pit Leveling

• It is a method employed on construction jobs to evaluate quantities of earth, gravel, rock, or other material to be excavated or filled. Also to generate contour maps

• Two methods can be used

 Grid Leveling

 Radial Line Leveling

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Grid Leveling

• Grid leveling is a method for locating contour lines and topographic features by stacking an area in squares of 5, 10, 50, 100 m, or more depending on the project extent, ground roughness, and accuracy required

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Radial Line Leveling

• This method for locating contour lines and topographic features is simpler to perform compared to grid leveling, and it requires less time.

• The level instrument is set up in the middle of the field and the rod person moves along radial lines from the instrument. Radial lines are spaced at equal or unequal central angles.

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Summery

• Profile Leveling

• Two-Peg Test

• Trigonometric Leveling

• Elevation of Inaccessible Points

• Grid Leveling

• Radial Line Leveling

• Borrow-Pit Leveling