Theodolite and Angles Measurement

Transcription

1 Building & Construction Technology Engineering Department Theodolite and Angles Measurement Lecture 1 Theodolite and Angles Measurement Lecture No. 1 Main Objectives Lecturer Date of Lecture General advices to make best benefit of this lecture Total number of pages 10 Download link for the PDF version Introducing theodolite equipment Learning the measurements of horizontal and vertical angles Understanding the different types of horizontal and vertical angles Ehsan A. Hasan Msc in transportation engineering Assistant lecturer at Technical College/Al-Musaib Listen carefully to lecturer Take necessary written notes Ask questions to simplify and remove ambiguity Page 1

2 Theodolite and Angles Measurement Lecture Units of Angles Measurement 1- Degree unit: A circle is divided to 360 equal degrees 1 degree = 60 minutes, 1 minute = 60 seconds For example, 2 degrees 5 minutes 7 seconds is written 2 5'7" 2- Radian unit: A circle is equal to 2π 1 degree = π radians 180 Example 1: 1- Convert each of the following angles from degrees to radians 45, 180 Sol: 45 * π 180 = π 4 π 2, π 3 Sol: 180* π 180 = π 2- Convert each of the following angles from radians to degrees π π = 90, π π = Vertical Angles Vertical angles types Angle of elevation: the angle that is measured upward from a horizontal reference line and is considered positive or plus (+) angle Angle of depression: the angle that is measured downward from the horizontal reference line and is considered negative or minus (-) angle Page 2

3 Theodolite and Angles Measurement Lecture 1 Zenith angle: an angle measured with respect to the upward vertical direction (zenith direction) Figure (1) elevation and depression vertical and zenith angles Zenith Angle = 90 Vertical Angle Example 2: Sol: 1- Calculate the zenith angle for each of the following vertical angles 8 45, Calculate the vertical angle for the zenith angle of and state weather the angle is elevation of depression 1- Zenith angle = = (note : 90 = ) Zenith angle = = Vertical angle = = (angle of depression ) 3.0 Horizontal Angles Interior angle: the horizontal angle measured on the inside of a closed polygon Page 3

4 Theodolite and Angles Measurement Lecture 1 Σ I.A = 180 (n-2) I.A: all interior angles in a closed polygon n: number of sides Exterior angle: the horizontal angle measured outside of the closed polygon Σ E.A = 180 (n+2) E.A: all exterior angles of the closed polygon n: number of sides At any point: Σ exterior + interior = 360 Angle to the right: an angle measured in a clockwise direction from the back station to the forward station Angle to left: an angle measured counterclockwise from the back station to the forward station Backsight: the pointing of instrument toward the back station Foresight: the pointing of instrument toward the forward station Figure (2) exterior and interior horizontal angles in a polygon Page 4

5 Theodolite and Angles Measurement Lecture 1 Deflection angle: the horizontal angle between the extension of a back or proceeding line and the succeeding or next line Deflection angle to the right (R): when measured clockwise Deflection angle to the left (L): when measured counterclockwise Figure (3) deflection angles to the right or left Azimuth: the clockwise horizontal angle between the line and the (north) or the (south) direction Bearing: the angle from the nearest north (N) or south (S) to the line, with the added designation of east (E) and west (W) Example 3: Convert each of the following angles from azimuth (n) to azimuth (s) and bearing a) 120 b) 200 c)290 d)30 Sol: Page 5

6 Theodolite and Angles Measurement Lecture 1 Example 4: Determine the azimuth direction for the sides 2-3 and 3-1 for the given three-sided traverse Page 6

7 Theodolite and Angles Measurement Lecture 1 Sol: 1- Determine 132 ΣI.A=180(n-2) ΣI.A=180 1 = = X X= = Determine Azimuth for side (2-3) Azim side (1-2)= x= 120 (given) y= x= 120 (alternate interior angles) z= 180, e= 85 (given) z + y + e= 385 d= = 25 = Azim side (2-3) 3- Determine Azim side (3-1) x= y= 25 (alternate interior angles) e= 180 e+ y+ z= 240 Azim side (3-1)= 240 Page 7

8 Theodolite and Angles Measurement Lecture Theodolite Definition: a precision instrument used for angular measurement with accuracy varies from (0.1 seconds to 1 minute) Types : 1- repeating theodolite 2- direction theodolite Main components: Reading systems : Figure (4) typical optical mechanical theodolite components 1- Watts Microptic Figure (5) typical reading system in a theodolite instrument (Watts Microptic) Page 8

9 Theodolite and Angles Measurement Lecture 1 2- Wild T16 3- Wild T2 Figure (6) typical reading system in a theodolite instrument (Wild T16) Figure (7) typical reading system in a theodolite instrument (Wild T2) Measuring horizontal angles by theodolite 1- Set up : level the instrument over the angle vertex 2- Backsight: aim the instrument at the point that marks the left-hand side of the angle and lock the motion 3- Zeroing: Set the reading at zero 4- Foresight: free the motion and aim at the point that marks the right-hand side of the angle 5- Reading: Read the angle Page 9

10 Theodolite and Angles Measurement Lecture 1 Measuring vertical angles by theodolite 1- Set up: level the instrument at the angle vertex 2- Zenith sight: Aim the instrument vertically toward the zenith and lock the motion 3- Zeroing: Set the reading at zero 4- Lower sight: Aim the instrument toward the lower sight and take the reading 5- Calculation: Calculate the elevation or depression angle from the zenith angle Errors in reading angles 1- Instrumental errors 2- Personal errors 3- Natural errors Page 10

11 2 nd class Building & Construction Technology Engineering Department Traverse Surveying Lecture 2 Traverse Surveying Lecture No. 2 Main Objectives Introducing the concept and objective of traverse surveying in horizontal control Learning the basic traverse closure computations Learning the inversing of traverse Learning traverse area computations Lecturer Date of Lecture General advices to make best benefit of this lecture Total number of pages 17 Download link for the PDF version Ehsan A. Hasan Msc in transportation engineering Assistant lecturer at Technical College/Al-Musaib Listen carefully to lecturer Take necessary written notes Ask questions to simplify and remove ambiguity Page 1

12 2 nd class Traverse surveying Lecture Traverse Surveying Definition: the type of surveying includes a connected sequence of lines whose lengths and directions are measured Courses: interconnected series of lines Consist of: Stations: series of points on the ground Objectives: 1- Measure the distances between stations and angles between courses 2- Local horizontal control over areas 3- Measuring various points precisely without accumulating error 2.0 Traverse Types Open: does not form a closed figure nor end at a point of known position Traverse Loop: starts and ends at the same point forming a closed figure Closed Connecting: looks like open traverse but it starts and ends at a point of known position 3.0 Loop Traverse Closure Computations Error of closure: Due to some errors in field measurements, the given coordinates of the starting point in a loop traverse will not precisely be the same as the computed coordinates from field measurement Page 2

13 2 nd class Traverse surveying Lecture 2 A: the given coordinates A': the field measured coordinate Figure (1): Error of Closure in a Loop Traverse 3.1 Computations of Angular Error 1- Calculate the sum of interior angles based on the number of sides (σ1) σ1= Σ I.A= (n-2) (180) n= number of sides for the polygon 2- Calculate the sum of interior angles based on field measurement (σ2) n= number of angles 3- Calculate the sum of angular error (Σē) 4- Determine error per angle (ē) σ2= Σ ( n) Σē= σ2 - σ1 ē = Σē no.of angles 5- Correct each angle Corrected angle = field angle - ē Page 3

14 2 nd class Traverse surveying Lecture 2 Example no.1: Compute the angular error and adjust the angles of the given traverse (figure 2) Figure (2): Traverse for Example No.1 with Unadjusted Field Data Solution: σ1= ΣI.A= (n-2) (180) = 540 σ2= Σ ( A + B + C + D + E) =540 02' 30" Σē = ' 30" ' 00" = 00 02' 30" ē = 2 30 " = 30" per angle 5 Correction of angles: A corrected = A field measured - ē A corrected = 64 53' 30" - 30" = 64 53' 00" B corrected = ' 15"- 30" = ' 45" Same implies for C, D and E Page 4

15 2 nd class Traverse surveying Lecture Computations of Error of Closure and Relative Accuracy 1- Calculate latitude and departure for traverse courses L: length of course β: bearing angle of course Latitude = y = L cos (β) Departure = x = L sin (β) Figure (3): Definition of Latitude and Departure of a Line 2- Adjust signs for latitude and departure Page 5

16 2 nd class Traverse surveying Lecture 2 Figure (4): Sign Convention for Latitudes and Departures 3- Calculate error of closure (Ec) and relative accuracy Ec = (Σ y) 2 + (Σ x) 2 Relative accuracy = 1: ( p Ec ) Ec: error of closure Σ y: sum of latitudes Σ x: sum of departures P: total traverse length Example no.2: Determine the error of closure (Ec) and relative accuracy for the given traverse (figure 5) Page 6

17 2 nd class Traverse surveying Lecture '45" 64 53'00" '45" '5" 96 38'15" Figure (5): Traverse Data for Example No.2 Solution: For course AB: Latitude = L COS (β) = COS (79 49'00") = Since bearing for AB is in quadrant (SW), latitude sign is (-) Latitude AB= Departure= L SIN (β) = SIN (79 49'00") = Since bearing for AB is in quadrant (SW), departure sign is (-) Departure AB= The same implies for the other courses (BC, CD, DE) Page 7

18 2 nd class Traverse surveying Lecture 2 Ec= (Σ Y) 2 + (Σ X) 2 Ec= ( 0.68) 2 + (1.05) 2 = 1.25 ft P= total traverse length = = ft Relative accuracy = 1: ( p Ec ) Relative accuracy = 1: ( ) = 1:2990 Table 1: Computation of Latitudes and Departures for Example No Correcting Course Latitudes and Departures (Compass Rule) Latitude correction = Σ y L Departure correction = Σ x L p p Σ Y, Σ X: sum of errors in latitude and departure P: total traverse length L: length of course Page 8

19 2 nd class Traverse surveying Lecture 2 Example no.3: Correct the latitudes and departures for the traverse given the data in table no.1 Sol: Σ Y= Σ X= P = ft For course AB: Latitude correction = Σ y p = ( 0.68) = L Departure correction = Σ x = = p L Adjusted latitude= ( ) = Adjusted departure= (-0.19) = Same implies for other courses (BC, CD, DE, EA) Table 2: Correction Data for Traverse Latitudes and Departures in Example No.3 Page 9

20 2 nd class Traverse surveying Lecture Computation of Stations Coordinates in Traverse 1- Assume a value for the station closest to the west of (x) or (E), and a value for the station closest to the south of (y) or (N) 2- Calculate the coordinates (y,x) or (N, E) for the other stations based on: N2, E2: the y and x coordinates of station 2 N1, E1: the y and x coordinates of station 1 Lat1-2: the latitude of course 1-2 Dep1-2: the departure of course 1-2 N2= N1 + Lat1-2 E2= E1 + Dep1-2 N A = N B + Lat B-A N A = = E A = E B + Dep B-A E A = = Figure 6: Typical Computation of Coordinates Example no. 4: Calculate the coordinates for the stations of the given traverse in figure 7 Page 10

21 2 nd class Traverse surveying Lecture 2 Course latitude departure AB BC CD DE EA Figure 7: Traverse Plot with Latitudes and Departures of the Courses for Example No.4 Sol: Assume station C (closest to the west) has an East coordinate (E) of (100.00) Compute East coordinates for the rest of stations EC= (assumed) ED= EC+DepC-D ED= = EE= ED+DepD-E EE= = EA= EE+ DepE-A EA= = EB= EA+ DepB-A EB= = Page 11

22 2 nd class Traverse surveying Lecture 2 Assume station E (closest to the south) has a North coordinate (N) of (100.00) Compute North coordinates for the rest of stations NE= (assumed) NA= NE + LatE-A NA= = NB= NA + LatA-B NB= = NC= NB + LatB-C NC= = ND= NC+ LatC-D ND= = Figure 8: Traverse Plot with Stations Coordinates for Example No.4 Page 12

23 2 nd class Traverse surveying Lecture Inverse Computations for Traverse Inversing: the process of computing the new directions and lengths of traverse courses resulting from the correction of the traverse. Dep β= tan-1 Lat L = Lat Cosβ = Dep Sinβ = Lat2 + Dep 2 β: the new (corrected) bearing angle Dep: corrected departure Lat: corrected latitude L: new course length Dep= ET ES Lat= NT - NS ES and ET: the eastings (x coordinates) of stations S and T NS and NT: the northings (y coordinates) of stations S and T Figure 9: Geometry of the process of inversing Page 13

24 2 nd class Traverse surveying Lecture 2 Example No.5: Calculate the adjusted bearing and length of the course AB in the traverse which has the given data: Sol: Corrected latitude= Corrected departure= β= tan = Since latitude and departure is both negative, then the course direction in the South West quadrant The bearing is S 79 49'47"W L= = ft Example No.6: Compute the adjusted bearing and length of the course BC of the traverse using the given data: Sol: Dep = EC - EB Dep= = Lat= NC NB Lat= = β= tan = L= ( ) 2 + (174.00) 2 = ft Station B: N E Station C: N E Since the departure is negative and the latitude is positive, the course BC is in the North West quadrant β= N 73 35'57" W Note: converting the angle unit from degrees decimals to degree minutes seconds is illustrated in the annex A Page 14

25 2 nd class Traverse surveying Lecture Traverse Area Computations 1- Arrange the coordinates by the following sequence n: total number of stations 2- Calculate summations of diagonal coordinates (Σa, Σb) Σa= Σ(E1 N2)+(E2 N3)+(E3 N4).. Σb= Σ(N1 E2)+(N2 E3)+(N3 E4) 3- Calculate the difference between the two summations (Σc) 4- Calculate the area Example No.7: Determine the area in (ft 2 ) of the given traverse Σc= Σa Σb Area = Σc 2 Figure 10: traverse with stations coordinates for example 7 Page 15

26 2 nd class Traverse surveying Lecture 2 Sol: Σa= Σ( )(734.98)+(691.17)(908.98)+(100.00)(239.96)+(208.89)(100.00)+( )(857.00)= Σb= Σ(857.00)(691.17)+(734.98)(100)+(908.98)(208.89)+(239.96)( )+(100.00)( )= Σc= Σa Σb = ft 2 Area = Σc = = ft Page 16

27 2 nd class Traverse surveying Lecture 2 ANNEX A Conversion of angular units between decimals and degrees minutes seconds 1- Converting Degrees, Minutes, & Seconds to Degrees & Decimals Example: Convert the angle 37 42' 17" to the degree decimal unit Sol: 1- Divide the seconds fraction (17") on 60 17/60 = Add the value to the minutes fraction (42') Divide the minutes fraction ( ) on /60 = Add the value to the degrees 37 42' 17" = Converting Degrees & Decimals to Degrees, Minutes, & Seconds Example: Convert the angle to degrees minutes and seconds Sol: 1- Subdivide the angle from the whole degrees: = Multiply the decimal value by 60: = (14 is the whole minutes) 3- Subdivide the whole minutes from the number : = Multiply the decimal value by 60: = 4.41 (4 is the whole seconds) = 5 14' 4" Page 17

28 Building & Construction Technology Engineering Department Tacheometry Lecture 3 Tacheometry Lecture No. 3 Main Objectives Lecturer Date of Lecture General advices to make best benefit of this lecture Total number of pages 9 Download link for the PDF version Introducing tacheometry methods and equipment Measuring horizontal and vertical distances using theodolite and rod Learning Tacheometry using reduction tacheometer Learning Tacheometry using theodolite and subtense bar Ehsan A. Hasan Msc in transportation engineering Assistant lecturer at Technical College/Al-Musaib Listen carefully to lecturer Take necessary written notes Ask questions to simplify and remove ambiguity Page 1

29 Tacheometry Lecture Tacheometry Definition Tacheometry means the quick measurement of distances 2.0 Tacheometry Objectives 1- Measuring vertical distances 2- Measuring horizontal distances 3- Measuring the difference between the elevations of two points 4- Reducing time and extensive work required for measurements 3.0 Accuracy of Measurement The accuracy of measurement is between 1/500 to 1/ Equipment of Tacheometry Measurement 1- Level with rod 2- Theodolite with rod 3- Reduction tacheometer with rod 4- Theodolite with subtense bar 5.0 Tacheometry using theodolite with rod 5.1 Stadia Reading Figure 1: Theodolite Telescope with Stadia Hairs Page 2

30 Tacheometry Lecture 3 Stadia Interval (S) = Upper Stadia Lower Stadia 5.2 Measuring of Horizontal and Vertical Distances Using Theodolite and Rod Figure 2: measuring of horizontal and vertical distances using theodolite and rod H= K S cos 2 α V= 1 2 K S sin2α H: Horizontal distance V: Vertical distance K: stadia interval factor = 100 S: Stadia interval (the difference between upper stadia reading and lower stadia reading Page 3

31 Tacheometry Lecture 3 α: Zenith angle 5.3 Measuring The Difference Between Two Points Elevations Using Theodolite and Rod Figure 3: measuring the difference between two points elevations If α is positive: If α is negative: Elev. (B) Elev. (A) = H.I + V R.R Elev. (B) Elev. (A) = H.I V R.R Elev. (B): elevation of point B Elev. (A): elevation of point A H.I: Height of instrument V: Vertical distance R.R: rod reading of the middle stadia hair Page 4

32 Tacheometry Lecture 3 Example No.1: If the height of instrument above point (A) is 1.55 m and the zenith angle is (-7 15') and the readings of the three hairs are (1.00, 1.32 and 1.64) estimate the elevation of point (B) and the horizontal distance between the two points knowing that point (A) elevation is m. Solution: R.R = 1.32 m S= = 0.64 m V= 1 2 K S sin2α = sin (14 30') = 8.01 m Since α is negative: Elev (B) Elev (A) = H.I V R.R Elev (B) = =34.87 m H= k s cos 2 α = cos 2 (7 15') =62.98 m 6.0 Tacheometry using reduction tacheometer Reduction tacheometer: an instrument with reduction diagrams that facilitate the measurement of horizontal and vertical distances Objectives of reduction tacheometer: The measurement of vertical and horizontal distances without the need for solving equations or measuring zenith angles Reduction curves types: 1- Zero curve: the lower curve Page 5

33 Tacheometry Lecture 3 2- Distance curve: the upper curve 3- Height curve: the middle curve 6.1 Measurement of Horizontal and Vertical Distances Using Reduction Tacheometer H: horizontal distance V: vertical distance K: stadia interval factor = 100 usually S1 = (Distance curve reading zero curve reading) S2= (Height curve reading zero curve reading) H= K S1 V= K S2 Example No.2: Determine the horizontal and vertical distances between point (A) and point (B), using the following tachometer readings: Zero curve reading = 1.00 Height curve reading = (k=20) Distance curve reading = (k=100) Sol: S1= = H= k s1 = = 56.5 m S2= = V= k s2 = = 4.5 m Page 6

34 Tacheometry Lecture Tacheometry Using Theodolite and Subtense Bar Objectives: 1- Measuring the horizontal distance between theodolite and subtense bar 2- High accuracy of measurements 3- Determination of the difference in elevation between two points Figure 4: subtense bar 7.1 Measuring of Horizontal Distances Using Theodolite and Subtense Bar Figure 5: measuring of horizontal distance using theodolite and subtense bar Page 7

35 Tacheometry Lecture 3 D= cot ( θ 2 ) D: the horizontal distance from theodolite to the center of subtense bar θ: the angle measured by theodolite between the subtense bar edges ( 1T2) 7.2 Measuring the Difference Between Two Points Elevations Using Theodolite and Subtense Bar Figure 6: measuring the difference in elevations between two points using theodolite and subtense bar If α is positive: If α is negative: Elev. B Elev. A= H.I1 + D.tanα H.I2 Elev. B Elev. A= H.I1 D tanα H.I2 Page 8

36 Tacheometry Lecture 3 Elev. B: elevation of point B Elev. A: elevation of point A H.I1: height of theodolite H.I2: height of subtense bar D: horizontal distance between A and B α: zenith angle Example No.3: A theodolite was installed above point (A) which has an elevation of (62.45 m) and its height was measured and found to be (1.65 m), and the subtense bar was installed above point (B) with height of (1.5 m), the angle between the subtense bar edges was ( ), and the zenith angle was ( ), calculate the horizontal distance between (A) and (B) and the elevation of (B) Sol: D= cot ( ) = m 2 Elev. B= Elev. A + H.I1- D.tanα H.I2 = ( tan ( ))-1.5 = m Page 9

37 Building & Construction Technology Engineering Department EDM & Total Station Lecture 4 EDM & Total Station Lecture No. 4 Main Objectives Lecturer Date of Lecture General advices to make best benefit of this lecture Total number of pages 7 Download link for the PDF version Introducing the principles of EDM Understanding EDM objectives and applications Learning total station principles and applications Learning the field procedure for using total station instrument Ehsan A. Hasan Msc in transportation engineering Assistant lecturer at Technical College/Al-Musaib Listen carefully to lecturer Take necessary written notes Ask questions to simplify and remove ambiguity Page 1

38 EDM & Total Station Lecture Electronic Distance Measurement (EDM) An EDM uses electromagnetic (EM) energy to determine the length of a line Objectives: 1- Increase the speed of measuring distances 2- Increase the accuracy of measuring distances 3- More economical work 4- The ability to determine distances over lakes, rivers, swamps etc. 2.0 Principles of EDM instruments EDM instruments generate and project an electromagnetic beam of light waves from one end of the measured line to the other, the beam is reflected back to the transmitting instrument Figure 1: Principle of Measuring Distances by EDM 3.0 Distance Determination by EDM EDM instruments determine the distance by using signal structure and phase shift Page 2

39 EDM & Total Station Lecture 4 r Figure 2: Signal Propagation from EDM P= r 360 λ P: partial wavelength r: returning angle of the signal to the EDM λ : Wavelength D= (Nλ+P) 2 D: total distance from EDM to reflector N: number of signal full phases from EDM to reflector and back to EDM Example no.1: Determine total length between EDM and reflector assuming that the wavelength is 20 ft, total number of phases for signal is 10, and signal returning angle is 82 30'38.9" Sol: P= 82 30'38.9" 20 = ft '00" Page 3

40 EDM & Total Station Lecture 4 D= = ft The distance measured by EDM is a slope distance not the horizontal distance unless the EDM and the reflector are at the same elevation Figure 3: The Slope Distance Measured by EDM 4.0 Total Station Instrument (TSI) Total Station Instrument: is an instrument with combined digital electronic theodolite, EDM and Microprocessor Digital theodolite Total Station Instrument EDM Microprocessor Page 4

41 EDM & Total Station Lecture 4 Functions of (TSI) Measuring Calculating Horizontal Angles Vertical Angles Slope Distances Horizontal Distances Vertical Distances Coordinates Azimuths of Lines Figure 4: Total Station Parts Page 5

42 EDM & Total Station Lecture Distance Reduction Using Total Station Total station instrument measures the slope distance and zenith angle in order to calculate the vertical and horizontal distances Figure 5: Distance Reduction Using Total Station H= S sin (z) V= S cos (z) H: horizontal distance from total station instrument to the reflector V: vertical distance from total station instrument to the reflector S: slope distance Z: zenith angle Page 6

43 EDM & Total Station Lecture Total Station Field Operating Procedure 1- Set up and level TSI over a point 2- Enter the following data (the instrument coordinates and elevation, the height of the instrument and reflector) 3- Aim the telescope on the reflector and make sure of the proper alignment of both the TSI and reflector 4- After alignment, the return signal level is automatically optimized and displayed 5- Make the required measurement by pressing the appropriate buttons and follow the manufacturer instructions Page 7

44 Building & Construction Technology Engineering Department Horizontal and Vertical Curves Lecture 5 Horizontal and Vertical Curves Lecture No. 5 Main Objectives Learning the elements and geometry of horizontal curves Understanding the computations of horizontal curves Learning the elements and geometry of vertical curves Understanding the computations of vertical curves Lecturer Date of Lecture General advices to make best benefit of this lecture Total number of pages 15 Download link for the PDF version Ehsan A. Hasan Msc in transportation engineering Assistant lecturer at Technical College/Al-Musaib Listen carefully to lecturer Take necessary written notes Ask questions to simplify and remove ambiguity Page 1

45 Horizontal and Vertical Curves Lecture Horizontal Curves Horizontal curve: The curve that connects the back tangent to the forward tangent in the plan of a roadway It is a single arc of a circle Figure 1: horizontal curve from a plan view Horizontal Curve elements and computations: Figure 2: the horizontal curve elements and geometry Page 2

46 Horizontal and Vertical Curves Lecture 5 T: tangent distance, the distance from PC to PI R: radius of curve Δ: intersection angle L: length of curve T= R tan ( 2 ) L= π R 180 LC: long chord, the straight line connects PC to PT E: external distance, distance from the curve to PI LC= 2 R sin ( 2 ) E= R ( 1 cos 2 1) M: middle ordinate, distance from midpoint of the curve to midpoint of the long chord (LC) M= R (1-cos 2 ) Location of PC and PT: PC station = PI station - T PT station= PC station + L Degree of curvature: Define the sharpness of a simple curve The central angle that is facing an arc of 100 ft Page 3

47 Horizontal and Vertical Curves Lecture 5 Figure 3: degree of curvature Da: degree of curvature in degrees Da= 5730 R R: radius in feet The higher the degree of curvature the sharper is the curve Figure 4: a sharp curve and a flat curve Page 4

48 Horizontal and Vertical Curves Lecture 5 Example No.1: A simple horizontal curve of radius 300 ft connects two tangents that form an intersection angle of 74 46'36" compute the elements of the curve, including the tangent distance, the length of the arc, the long chord, the external distance and the middle ordinate. Sol: Convert Δ to decimals (refer to appendix a from lecture 2) Δ= Δ 2 = T= 300 tan( )= ft L= π = ft Lc= sin( )= ft E= 300 ( 1 cos ) = ft M= 300 (1-cos )= ft Example No.2: A simple curve is to be laid out so that its middle ordinate is 30 m long. If the tangents intersect at an angle of 50, what is the minimum radius required? Sol: 30= R (1-cos 50 2 ) R= 320 m Example No.3: What is the degree of curvature if the radius= 300 ft? Sol: D= = 19.1 Page 5

49 Horizontal and Vertical Curves Lecture 5 Example No.4: For a simple horizontal curve if the tangent distance is ft and arc length L= ft, the location of PI is established at st find the locations of PC & PT? Note: each 1 station is equal to 100 ft Sol: St PI= St PC= St.PI - T = St PC= St. PT= St.PC + L = ( ) + ( ) = Vertical Curves The curve that connects two tangents in the profile of a roadway It is a segment of parabola Figure 5: vertical alignment, a profile view Crest (summit) [g2 g1= (-) negative] Types of vertical curves Sag [g2 g1= (+) positive] Page 6

50 Horizontal and Vertical Curves Lecture 5 Figure 6: crest and sag vertical curves Page 7

51 Horizontal and Vertical Curves Lecture 5 Vertical curve elements and computations: 1- Location and elevation of (PVC, PVT & PVI) Figure 7: a typical sag curve with elements PVC: point of vertical curve, the point on the back tangent where the vertical curve begins PVC station = PVI station - L 2 PVC elevation = PVI elevation ± g1. L 2 + for sag curve, - for crest curve PVI: point of vertical intersection, the point where the two tangents meet L: length of vertical curve measured on its horizontal projection g1: gradient of the back tangent g2: gradient of the forward tangent PVT: point of vertical tangency, the point where the curve joins the forward tangent Page 8

52 Horizontal and Vertical Curves Lecture 5 PVT station = PVI station + L 2 PVT elevation = PVI elevation ± g2. L 2 + for sag curve, - for crest curve 2- Determine the elevation of any point on vertical curve Figure 8: geometry for determining the elevation of point a on vertical sag curve r: rate of change in grade r= g2 g1 L Page 9

53 Horizontal and Vertical Curves Lecture 5 Ya= elevation of any point on the curve Ypvc= elevation of the PVC Xa= horizontal distance of the point from the PVC Ya= Ypvc + g1.xa + r 2. x a 2 3- Determine the elevation of highest or lowest point on curve In order to: a- Determine the clearance beneath a bridge b- Determine the location of storm-water drainage inlet in a sag curve The highest or lowest point is the vertical curve turning point X': the distance of the turning point from PVC X' = g1.l g1 g2 Figure 9: highest and lowest points on vertical curves Page 10

54 Horizontal and Vertical Curves Lecture 5 Example No.5: The data for a summit vertical curve are as follows: PVI station= PVI elevation= ft G1= % G2= -2.00% L= 550 ft Determine location and elevation of (PVC) and (PVT) Determine the elevation of stations (9+00, 11+00) along the curve Sol: Figure 10: crest curve of example no.5 PVC station= PVI station - L 2 = ( ) (2+75)= PVC elevation= PVI elevation - g1. ( L ) 2 = (0.06) (275)= ft PVT station= PVI station + L 2 = ( ) + (2+75)= PVT elevation= PVI elevation - g2.( L 2 ) = (0.02)(275)= ft r = g2 g1 L Page 11

55 Horizontal and Vertical Curves Lecture 5 r = for st : = x= (9+00) - ( ) = Y= Ypvc + g1.x + r 2. x2 = (26.82) + (0.06)(72.57) Y= ft the elevation at st.9+00 For st : X= (11+00)-( )= ft Y= (0.06)(272.57) +( (272.57) 2 ) 2 =37.77 ft Example no.6: The data for a sag vertical curve are as follows: PVI station= PVI elevation= ft g1= % g2= +3.00% L= 450 ft Figure 11: sag curve of example No.6 Page 12

56 Horizontal and Vertical Curves Lecture 5 Determine stations and elevations of PVC and PVT and the elevations of stations (20+00, 22+00) on curve Sol: PVC station = PVI station - L 2 = (21+25) ( 450 ) = PVC elevation= PVI elevation + g1.( L 2 ) [ + for sag curve] = ( ) = ft 2 PVT station= PVI station + L 2 = (21+25) + ( 450 ) = PVT elevation= PVI elevation + g2.( L 2 ) = ( ) =89.54 ft Elevation of st Y= Ypvc+ g1.x + r 2. x2 r= g2 g1 L = 0.03 ( 0.05) 450 = X= (20+00) (19+00)= 100 ft Y= ( ) (100 2 ) 2 Y= ft elevation of st Elevation of st x= = 300 ft y= ( ) (300 2 ) 2 y= ft elevation of st Page 13

57 Horizontal and Vertical Curves Lecture 5 Example no.7: The data for a vertical sag curve are as follows PVI station = PVI elevation = ft g1= -4.00% g2= 7.00% L= 600 ft Compute the location and elevation of the lowest point on curve Sol: Elevation & station of PVC Figure 12: sag curve for example No.7 PVC station= PVI station - L 2 = ( ) 300 = PVC elevation= PVI elevation + g1.( L 2 ) = ( ) = ft Location of the lowest point: X'= g1.l g1 g2 = ( ) = ft ( ) Page 14

58 Horizontal and Vertical Curves Lecture 5 Station of the lowest point = PVC station = ( ) = Elevation of the lowest point: Y= Ypvc + g1.x+ r 2. x2 r= g2 g1 L = = Y= ( )+( ) 2 Y= ft Page 15