Overview. Profile 2/27/2018. CE 371 Surveying PROFILE LEVELING & Trigonometric LEVELING

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1 Lec 10 + Lec 11 CE 371 Surveying PROFILE LEVELING & Trigonometric LEVELING Dr. Ragab Khalil Department of Landscape Architecture Faculty of Environmental Design King AbdulAziz University Room LIE15 Overview Profile Leveling Two-Peg Test Trigonometric Leveling Elevation of Inaccessible Points Grid Leveling Radial Line Leveling Borrow-Pit Leveling 2/34 Profile 3/34 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 1

2 Stationing 4/34 In route surveying, a system called stationing is used to specify the relative horizontal position of any point along the reference line. Stationing is the establishment of full & plus stations 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 5/34 Write the distance 325 as: 100 m stationing 50 m stationing 20 m stationing Solution Distance 325 m as 100 m stationing: 3+25 Distance 325 m as 50 m stationing: 6+25 Distance 325 m as 20 m stationing: Intermediate site 6/34 BS IS FS BM # TP #1 TP #2 BM # Profile Leveling 2

3 Profile Leveling Procedures 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. 7/34 Rate of Grade 8/34 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. = = m 43.5 m 150 m =2.5 x 150 =3.75 m Example 9/34 A profile leveling along the center line of a proposed street starting at elevation 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= BM#2= Solution 3

4 Point BS IS FS HI Elev. Adj. E Formati Ground on F.L.L BM # D GL-FL Cut - Fill C F F F TP # C C TP # F C C BM # S E c = =.006 m E A = = 10.6 mm > E C ok. 10/34 Drawing and Use of a Profile 11/34 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. Drawing and Use of a Profile 12/34 Elevations 1: Cut Fill Ground Level Formation Level Cut Cut Distance 1:1000 4

5 Collimation error 13/34 Occurs when the line of sight (as defined by the cross-hairs) is not horizontal Leads to an incorrect staff reading horizontal line error Two-Peg Test 14/34 DH = (b1 e) (d1 - e) = b1-d1 b1 e d1 e DH 30 متر 30 متر b2 e d2 3e 30 متر 30 متر 30 متر DH = (b2 e) (d2-3e) = b1-d1 e = b2 d2 + 2e = b1-d1 b1 d1 (b2 d2) 2 Adj d2=d2 3e example 15/34 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= m. Compute the error in mm per m. Is the error accepted. Compute the adjusted d2 value. Solution Error e = [( )-( )] /2 = 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 = (0.008) = m. 5

Trigonometric Leveling 16/34 Trigonometry can be used to compute difference in elevation between two points by measuring horizontal distance H between the two points and the vertical angle a (or

6 Trigonometric Leveling 16/34 Trigonometry can be used to compute difference in elevation between two points by measuring horizontal distance H between the two points and the vertical angle a (or zenith angle z). a= 90-z D V= H. tan (a) V= H. cot (z) hi Elev B = Elev A + hi + V - r B C A z a S H B E V ق r (For distances up to 300 m) Trigonometric Leveling Example A theodolite is set up at point A whose elevation is m. A level rod is put at point B whose horizontal distance from A is m. If rod reading is 2.29 m, zenith angle is 65 o, height of instrument is 1.50 m, find Elev B. 17/34 Solution: V = cot(65 o ) = m Elev B = = m Curvature And Refraction in Trigonometric Leveling For long horizontal distances, the effect of curvature and refraction should be taken into account. 18/34 Elev B = Elev A + hi + V r B H 2 6

7 Trigonometric Leveling _Long distances Example A theodolite is set up at point A whose elevation is m. A level rod is put at point B whose horizontal distance from A is m. If rod reading is 2.29 m, zenith angle is 65 o, height of instrument is 1.50 m, find Elev B. 19/34 Solution: V = cot(65 o ) = m Combined curvature and refraction effect CR = (0.5) 2 = m Elev B = = m Trigonometric Leveling With Unknown Distance 20/34 If the horizontal distance is unknown, we can use Stadia method or we have to take two level rod readings and two zenith angles. H = r 1 r 2 tan(a 1 ) tan(a 2 ) D r 1 ElevB = Elev A + hi + V 1 r 1 Elev B = Elev A + hi + V 2 r 2 hi C S r 2 z V z 2 V 1 1 B ق 2 a 1 a 2 A H E Trigonometric Leveling With Unknown Distance --- stadia 21/34 A theodolite is set up at point A whose elevation is 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) = = 8 o I=U-L= = H= 100*I*{cos (a)} 2 = 100x x (cos (8)) 2 = m V= x tan (8) = m Elev B = Elev A + hi + V r Elev B = = m 7

8 Trigonometric Leveling With Unknown Distance--- two readings 22/34 A theodolite is set up at point A whose elevation is 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 a 1 = =8º a 2 = =7º H = = tan(8) tan(7) m V 1 = x tan (8) = m Elev B = Elev A + hi + V 1 r 1 Elev B = = m Elevation of Inaccessible Points Case 1. Base of the object accessible 23/34 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. Case 1. Base of the object accessible B 24/34 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 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 D 2 Elev. of B = Elev. of A+ hi + h h = D tan 8

9 Case 2. Base of the object inaccessible, Instrument stations in the vertical plane as the elevated object. 25/34 There may be two cases. (a) (b) Instrument axes at the same level 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: Case 2. Base of the object inaccessible, Instrument stations in the vertical plane as the elevated object. (a) Instrument axes at the same level 26/34 D PA P, h= D tan 1 D PB 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 h= D tan 1 Elev. of P = Elev. of B.M + Bs + h Elev. of P = Elev. of A+ hi + h (b)instrument axes at different levels. 1) Height of instrument axis to the object is lower: 27/34 D PA P, h 1 = D tan 1 D PB P, h 2 = (b+d) tan 2 hd is difference between two height hd = h 1 h 2 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 + h 1 +CR 9

10 Ex. 28/34 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 = o, angle 1 = o. The instrument in the two positions was at the same level. Compute Elev p above BM B. Solution hd = = tan(13.125) D = = tan tan(13.125) h 2 = tan (13.125) =38.50 Elev. of P = Elev. Of B+ hi + h 2 +CR Elev. of P - Elev. Of B = (165.12/1000) 2 =39.74 m (b)instrument axes at different levels. 2) Height of instrument axis to the object is higher: 29/34 D PA P, h 1 = D tan 1 D PB P, h 2 = (b+d) tan 2 h d is difference between two height h d = h 2 h 1 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 ) h 1 = D tan 1 Elev. of P = Elev. of A+ hi + h 1 +CR Case 3. Base of the object inaccessible, Instrument stations not in the same vertical plane as the elevated object. 30/34 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: b sin BC= sin{180 - ( + )} b sin AC= sin{180 - ( + b h1 = AC tan 1 Elev. of P = Elev. of A+ hi + h 1 +CR h2 = BC tan 2 10

Borrow-Pit Leveling 31/34 It is a method employed on construction jobs to evaluate quantities of earth, gravel, rock, or other material to be excavated or filled.

an area in squares of 5, 10, 50, 100 m, or more depending on the project extent, ground roughness, and accuracy required Radial Line Leveling 33/34 This method for locating contour lines and

11 Borrow-Pit Leveling 31/34 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 Grid Leveling 32/34 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 Radial Line Leveling 33/34 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. 11

12 Summery Profile Leveling Two-Peg Test Trigonometric Leveling Elevation of Inaccessible Points Grid Leveling Radial Line Leveling Borrow-Pit Leveling 34/34 12

CE 371 Surveying Circular Curves

Lec. 25 1 CE 371 Surveying Circular Curves Dr. Ragab Khalil Department of Landscape Architecture Faculty of Environmental Design King AbdulAziz University Room LIE15 Overview Introduction Definition of