THE HISTORY OF SURVEYING

DEFINITION OF SURVEYING

Surveying orland surveyingis the technique and science of accurately determining the terrestrial or three-dimensional position of points and the distances and angles between them. These points are usually on the surface of the Earth, and they are often used to establish land maps and boundaries for ownership or governmental purposes. To accomplish their objective, surveyors use elements of geometry, engineering, trigonometry, mathematics, physics, and law.

An alternative definition, per the American Congress on Surveying and Mapping (ACSM), is the science and art of making all essential measurements to determine the relative position of points and/or physical and cultural details above, on, or beneath the surface of the Earth, and to depict them in a usable form, or to establish the position of points and/or details.

Furthermore, as alluded to above, a particular type of surveying known as "land surveying" (also per ACSM) is the detailed study or inspection, as by gathering information through observations, measurements in the field, questionnaires, or research of legal instruments, and data analysis in the support of planning, designing, and establishing of property boundaries. It involves the re-establishment of cadastral surveys and land boundaries based on documents of record and historical evidence, as well as certifying surveys (as required by statute or local ordinance) of subdivision plats/maps, registered land surveys, judicial surveys, and space delineation. Land surveying can include associated services such as mapping and related data accumulation, construction layout surveys, precision measurements of length, angle, elevation, area, and volume, as well as horizontal and vertical control surveys, and the analysis and utilization of land survey data.

THE EARLY DAYS OF SURVEYING1400 B.C.The Egyptians first used it to accurately divide land into plots for the purpose of taxation.

Babylon, Egypt, and the city states of Greece all had standards for commercial measuring devices.

500 B.C. Athens had its own central depository of official weights and measures the Tholos

120 B.C.Greeks developed the science of geometry and were using it for precise land division.

Greeks developed the first piece of surveying equipment (Diopter).

Greeks standardized procedures for conducting surveys.

Some of the earliest surviving measuring devices include gold scales recovered in present-day Greece from the tombs of Mycenaean kings.

THE EARLY DAYS OF SURVEYING

1800 A.D. Beginning of the industrial revolution.The importance of "exact boundaries" and the demand for public improvements (i.e. railroads, canals, roads) brought surveying into a prominent position.More accurate instruments were developed.Science of Geodetic and Plane surveying were developed.

Early History of Surveying

It is impossible to determine when surveying was first used by man. Remove not the ancient landmark, which thy fathers have set - Proverbs 22:28 The word geometry is derived from the Greek meaning earth measurements In Egypt, surveyors were called rope stretchers because they used ropes to measure

Roman surveyors got their name gromatici from the groma

DEFINITION OF ROUTE SURVEYINGA survey of the earths surface along a particular route in the compilation and updating of topographical, geological, soil, and other maps and the correlation of selected contours and objects with geodetic reference points or landmarks during linear surveys, and also in the study of the dynamics of natural and socioeconomic phenomena in a narrow strip of terrain. In a route survey, representations of the actual course of the survey and of the plane horizzontal features (including the terrain, if neccessary) on both sides of it within the limits of direct visibility are plotted on a map board using methods of instrument surveying (plane-table, tachymetric, and aerial photographic surveying) or exploratory surveying.

DIFFERENT TYPES OF HIGHWAY CURVESHORIZONTAL AND VERTICAL CURVES

HORIZONTAL CURVESDefinition:Horizontal Curvesare one of the two important transition elements in geometric design for highways (along withVertical Curves). A horizontal curve provides a transition between two tangent strips of roadway, allowing a vehicle to negotiate a turn at a gradual rate rather than a sharp cut. The design of the curve is dependent on the intended design speed for the roadway, as well as other factors including drainage and friction. These curves are semicircles as to provide the driver with a constant turning rate with radii determined by the laws of physics surrounding centripetal force.

Types of Horizontal Curves:A curve may be simple, compound, reverse, or spiral (figure 3-l). Compound and reverse curves are treated as a combination of two or more simple curves, whereas the spiral curve is based on a varying radius.

Simple The simple curve is an arc of a circle. It is the most commonly used. The radius of the circle determines the sharpness or flatness of the curve. The larger the radius, the flatter the curve.

Elements of Simple Curve:1. Point of Intersection (PI)- The point of intersection marks the point where the back and forward tangents intersect. The surveyor indicates it one of the stations on the preliminary traverse.2. Intersecting Angle (I)- The intersecting angle is the deflection angle at the PI. The surveyor either computes its value from the preliminary traverse station angles or measures it in the field.3. Radius (R)- The radius is the radius of the circle of which the curve is an arc.4. Point of Curvature (PC)- The point of curvature is the point where the circular curve begins. The back tangent is tangent to the curve at this point.5. Point of Tangency (PT)- The point of tangency is the end of the curve. The forward tangent is tangent to the curve at this point.6. Length of Curve (L)- The length of curve is the distance from the PC to the PT measured along the curve.7. Long Chord (LC)- The long chord is the chord from the PC to the PT. 8. Tangent Distance (T)- The tangent distance is the distance along the tangents from the PI to the PC or PT. These distances are equal on a simple curve.9. External Distance (E)- The external distance is the distance from the PI to the midpoint of the curve. The external distance bisects the interior angle at the PI.10. Central Angle ()- The central angle is the angle formed by two radii drawn from the center of the circle (0) to the PC and PT. The central angle is equal in value to the I angle.11. Middle Ordinate (M)- The middle ordinate is the distance from the midpoint of the curve to the midpoint of the long chord. The extension of the middle ordinate bisects the central angle.

CompoundA compound curve is two or more simple curves which have different centers, bend in the same direction, lie on the same side of their common tangent, and connect to form a continuous arc. The point where the two curves connect (namely, the point at which the PT of the first curve equals the PC of the second curve) is referred to as the point of compound curvature (PCC). Since their tangent lengths vary, compound curves fit the topography much better than simple curves. These curves easily adapt to mountainous terrain or areas cut by large, winding rivers. However, since compound curves are more hazardous than simple curves, they should never be used where a simple curve will do. ReverseA reverse curve is composed of two or more simple curves turning in opposite directions. Their points of intersection lie on opposite ends of a common tangent, and the PT of the first curve is coincident with the PC of the second. This point is called the point of reverse curvature (PRC). Reverse curves are useful when laying out such things as pipelines, flumes, and levees. The surveyor may also use them on low-speed roads and railroads. They cannot be used on high-speed roads or railroads since they cannot be properly superelevated at the PRC. They are sometimes used on canals, but only with extreme caution, since they make the canal difficult to navigate and contribute to erosion

SpiralIn engineering construction, the surveyor often inserts a transition curve, also known as a spiral curve, between a circular curve and the tangent to that curve. The spiral is acurve of varying radius used to gradually increase the curvature of a road or railroad. Spiral curves are used primarily to reduce skidding and steering difficulties by gradual transition between straight-line and turning motion, and/or to provide a method for adequately superelevating curves. The spiral curve is designed to provide for a gradual superelevation of the outer pavement edge of the road to counteract the centrifugal force of vehicles as they pass. The best spiral curve is one in which the superelevation increases uniformly with the length of the spiral from the TS or the point where the spiral curve leaves the tangent.

The curvature of a spiral must increase uniformly from its beginning to its end. At the beginning, where it leaves the tangent, its curvature is zero; at the end, where it joins the circular curve, it has the same degree of curvature as the circular curve it intercepts.

VERTICAL CURVESDefinition:A vertical curve provides a transition between two sloped roadways, allowing a vehicle to negotiate the elevation rate change at a gradual rate rather than a sharp cut. The design of the curve is dependent on the intended design speed for the roadway, as well as other factors including drainage, slope, acceptable rate of change, and friction. These curves are parabolic and are assigned stationing based on a horizontal axis.

Function and Type:When two grade lines intersect, there is a vertical change of direction. To insure safe and comfortable travel, the surveyor rounds off the intersection by inserting a vertical parabolic curve. The parabolic curve provides a gradual direction change from one grade to the next. A vertical curve connecting a descending grade with an ascending grade, or with one descending less sharply, is called a sag or invert curve. An ascending grade followed by a descending grade, or one ascending less sharply, is joined by a summit or overt curve.

Computations:In order to achieve a smooth change of direction when laying out vertical curves, the grade must be brought up through a series of elevations. The surveyor normally determines elevation for vertical curves for the beginning (point of vertical curvature or PVC), the end (point of vertical tangency or PVT), and all full stations. At times, the surveyor may desire additional points, but this will depend on construction requirements.1. Length of Curve- The elevations are vertical offsets to the tangent (straightline design grade) elevations. Grades G1 and G2 are given as percentages of rise for 100 feet of horizontal distance. The surveyor identifies grades as plus or minus, depending on whether they are ascending or descending in the direction of the survey. The length of the vertical curve (L) is the horizontal distance (in 100-foot stations) from PVC to PVT. Usually, the curve extends L stations on each side of the point of vertical intersection (PVI) and is most conveniently divided into full station increments. A sag curve is illustrated in figure 3-20. The surveyor can derive the curve data as follows (with BV and CV being the grade lines to be connected). Determine values of G1 and G2 , the original grades. To arrive at the minimum curve length (L) in stations, divide the algebraic difference of G1 and G2 (AG) by the rate of change (r), which is normally included in the design criteria. When the rate of change (r) is not given, use the following formulas to compute L:

If L does not come out to a whole number of stations from this formula, it is usually extended to the nearest whole number. Note that this reduces the rate of change. Thus, L = 4.8 stations would be extended to 5 stations, and the value of r computed from r = . These formulas are for road design only. The surveyor must use different formulas for railroad and airfield design.

2. Station Interval- Once the length of curve is determined, the surveyor selects an appropriate station interval (SI). The first factor to be considered is the terrain. The rougher the terrain, the smaller the station interval. The second consideration is to select an interval which will place a station at the center of the curve with the same number of stations on both sides of the curve. For example, a 300-foot curve could not be staked at 100-foot intervals but could be staked at 10-, 25-, 30-, 50-, or 75-foot intervals. The surveyor often uses the same intervals as those recommended for horizontal curves, that is 10, 25, 50, and 100 feet. Since the PVI is the only fixed station, the next step is to compute the station value of the PVC, PVT, and all stations on the curve. PVC = PVI - L/2PVT = PVI + L/2Other stations are determined by starting at the PVI, adding the SI, and continuing until the PVT is reached.

3. Tangent Elevations- Compute tangent elevations PVC, PVT, and all stations along the curve. Since the PVI is the fixed point on the tangents, the surveyor computes the station elevations as follows:

Elev PVC = Elev PVI + (-1 x L/2 x G1)Elev PVT = Elev PVI + (L/2 x G2)

The surveyor may find the elevation of the stations along the back tangent as follows:Elev of sta = Elev of PVC + (distance from the PVC x G1).

The elevation of the stations along the forward tangent is found as follows:

Elev of sta = Elev of PVI + (distance from the PVI x G2)

4. Vertical Maximum- The parabola bisects a line joining the PVI and the midpoint of the chord drawn between the PVC and PVT. In figure 3-19, line VE = DE and is referred to as the vertical maximum (Vm). The value of Vm is computed as follows: (L= length in 100-foot stations. In a 600-foot curve, L = 6.)

In practice, the surveyor should compute the value of Vm using both formulas, since working both provides a check on the Vm, the elevation of the PVC, and the elevation of the PVT.

5. Vertical Offset- The value of the vertical offset is the distance between the tangent line and the road grade. This value varies as the square of the distance from the PVC or PVT and is computed using the formula:

Vertical Offset = (Distance)2 x Vm

A parabolic curve presents a mirror image. This means that the second half of the curve is identical to the first half, and the offsets are the same for both sides of the curve.

6. Station Elevation- The surveyor computes the elevation of the road grade at each of the stations along the curve. The elevation of the curve at any station is equal to the tangent elevation at that station plus or minus the vertical offset for that station, The sign of the offset depends upon the sign of Vm (plus for a sag curve and minus for a summit curve).

7. First and Second Differences- As a final step, the surveyor determines the values of the first and second differences. The first differences are the differences in elevation between successive stations along the curve, namely, the elevation of the second station minus the elevation of the first station, the elevation of the third station minus the elevation of the second, and so on. The second differences are the differences between the differences in elevation (the first differences), and they are computed in the same sequence as the first differences.The surveyor must take great care to observe and record the algebraic sign of both the first and second differences. The second differences provide a check on the rate of change per station along the curve and a check on the computations. The second differences should all be equal. However, they may vary by one or two in the last decimal place due to rounding off in the computations. When this happens, they should form a pattern. If they vary too much and/or do not form a pattern, the surveyor has made an error in the computation.

SAMPLE PROBLEMS1. Two tangents intersect at Station 4 + 016.770. The deflection angle to the right is 40 00' 00". It is decided to design the highway for a maximum speed of 90 km/hr, and using AASHTO* recommendation for superelevation and friction a minimum radius of 270 meters and a maximum degree of curve, Da is to be 22 . Calculate T, La, R and the stationing of the P.C. and P.T. using 20 meter arc length. * American Association of State Highway and Transportation Officials Green Book

Solution:

2. Calculate the radius of the curve that will pass through Point P, using 1atitudes and departures. Determine the stationing of the P.C. and P,T, and o f Point P on the curve.

3. Solve for length of the curve intersection distance AB, based on the following diagram.

4. Calculate BC, area ABC (shaded) and angle d:

5. BC and EC stations, length of chord (C), middle ordinate (M), and external distance (E), if:

R=1000ftPI Sta. at 6 + 26.57

Solution:

6. Refer to the figure below. T1 = 380 ft., R1= 500 ft., 1= 2112, = 3950 and the PI is at station 136 + 42.85. Using the chord definition for degree of curve, compute the T2, R2, and 2 and the stationing of the C and the E.

7. Two tangents intersect at n = 70, P. I. = Stn 32 + 51 -82. Set up the field notes for the following compound curve (symmetrical).

1st curve A = 20, D = 8R (20 rn chords)2nd curve A = 300, D = 12R (10 m chords)3rd curve A = 200, D = 8R (20 m chords)

Solution:

8. Calculate the sub-tangent distance, T, for the symnetrically-compounded curve shown below.

Solution:

9. Determine the radius of the central curve of a symmetrical compound curve which passes through point "A, and determine the stationing of point "A".

10. A compound curve, with large radius curve firstI= 60 ; IL= 34 ; Is= 26 ; RL= 500 ft; Rs= 350 ft and PCC= 565 +35Find: Notes to layout the compound curve (include deflection s and chords). The station of the PI.Steps:1. Determine ts, tL, Das, DaL2. Determine Ls and LL.3. From PCC, determine TC and CT.4. Develop field notes from step 3 info.5. Solve equation D to E to obtain TL ( Ts, id Rs first)6. V or PI for the compound curve = TL + TC (Ts + TC , if Rs first)

11. Refer to the figure below. 1 and 2 are two parallel lines, s= 250m. We would like to connect the two lines by reversed curves consisted of two simple circular curves, R1= 200 m and R2= 300 m. Compute the main elements of the reversed curves, T1 and T2, 1,2, and , t1 and t2.

Solution:s= BF + EGBF= R1(1-cos1)EG= R2(1-cos2)s= (R1 + R2) (R1 +R2) cos1cos1 = [ (R1 + R2) s]/ (R1 + R2)1= 2=600000t1= R1tan(1/2) = 115.47mt2= R2tan(2/2) = 173.21m

12. Two parallel railway tracks, centre lines being 60 m apart, are to be connected by a reverse curve, each section having the same radius. If the maximum distance between the tangent points is 220 m calculate the maximum allowable radius of the reverse curve that can be used.

Solution:

13. The first branch of a reverse curve has a radius of 200 m. If the distance between the tangent points is 110 m, what is the radius of the second branch so that the curve can connect two parallel straights, 18 m apart ? Also calculate the length of the two branches of the curve.

Solution:

14. It is proposed to introduce a reverse curve between two straights AB and CD intersecting at a point I with CBI = 30 and BCI = 120. The reverse curve consists of two circular arcs AX and XD, X lying on the common tangent BC. If BC = 791.71, the radius RAX = 750 m, and chainage of B is 1250 m, calculate (i) the radius RXD, (ii) the lengths of the reverse curve, and (iii) The chainage of D.

Solution:

15. The perpendicular distance between two parallel tangents of a reversed curve is 35m. The azimuth of the common tangent is 300. If the radius of the first curve is 150m, determine the radius of the second curve.

16. Deflection angle between tangents= 30, station TS= 7+08, radius of circular curve R= 382m, length of the transition curve= 120m. prepare a setting out table to set out the transition and circular curve at points every 20m.

Solution:

17. Two straights AB and BC intersect at chainage 1530.685 m, the total deflection angle being 3308. It is proposed to insert a circular curve of 1000 m radius and the transition curves for a rate of change of radial acceleration of 0.3 m/s3 , and a velocity of 108 km/h. Determine setting out data using theodolite and tape for the transition curve at 20 m intervals and the circular curve at 50 m intervals.Solution:

18. Two straights having a total deflection angle of 6545 are connected with a circular curve of radius 1550 m. It is required to introduce a curve of length 120 m at the beginning and end of the circular curve without altering the total length of the route. The transition curve to be inserted is a cubic spiral, and the chainage of the point of intersection is 5302.10 m. Calculate (i) the distance between the new and the previous tangent points, (ii) the setting out data for transition curve taking peg intervals 20 m, and (iii) the data for locating the midpoint of the new circular curve from the point of intersection.

Solution:R = 1550 m = 6545 = 65.75/2= 325230

II O = 6545 + 5707 30 = 1225230

19. Part of a proposed rural road consists of two straights, which intersect at an angle of 1724618. These are to be joined using a wholly transitional horizontal curve having equal tangent lengths. The ransitional curves are to be cubic parabolas. The design speed for the road is to be 85 kph and the rate of change of radial acceleration 0.3 m/s3. Calculate the minimum radius of curvature of the curve and comment on its suitability. Tabulate the data required to set out the transition curves by offsets taken at exact 20m intervals along their tangent lengths.

20. On a proposed road having a design speed of 100 kph and a carriageway width of 7.30 m, a composite curve consisting of two transition curves and a central circular arc of radius 750 m is to join two intersecting straights having a deflection angle of 093428. The rate of change of radial acceleration for the road is to be 0.3 m s-3. The superelevation should be introduced at a rate of no more than 1%. -Calculate the amount of superelevation that must be built into the central circular arc. -Check that the transition curves are long enough for the superelevation to be introduced. Calculate the amount of superelevation that should be constructed along the entry transition curve at 20 m intervals from the entry tangent point.

Solution:The amount of superelevation that must be built into the central circular arcmaximum allowable SE = SE= SE= 0.344mthen to express this as a %s%= = =4.71%The radius of 750m is greater than the desirable min value of 720m for a 100kph SE of 4.71% is less than the value of 5%. Hence the 0.344m SE should be built into the central circular arc.Checking that the transition curves are long enough The length of each transition curve required for comfort and safety is obtained equationLT= == 95.26mThe superelevation value of 0.344 m must be introduced and removed over a of 95.26 m, which represents a gradient of0.344/95.26 = 0.36%Since this is less that the maximum allowable rate of introduction of 1%, the transitions are long enough.The amount of SE that should be constructed along the entry transition curve at 20m intervals from the entry tangent point.rl= K = RLTK= 95.26(750) = 71 445mAt 20m along the curve from entry tangent pointr= K/20 = 71 445/20 = 3572.25 mSE at 20m along the curve = (7.30)(1002)/(282.8)(3572.25) = 0.07ms% at 20m along the curve = 1002/(2.828)(3572.25) = 0.99%Because this is less than the min allowable value of 2.5% for drainage, a value of 2.5% must be used therefore SE built at 20m along the curve = 2.5% of B = 0.025 * 7.30 = 0.18m

21. g1= +3.0%, g2= -2.4%, V station is 46+70 and V elevation is 853.48, L= 600 ft, compute the curve for stakeout at full stations.Solution:r= (-2.4-3.0)/6 = -0.90% stationBVC station= (46+70)- (6+00/2) = 49+70EVC station = (43+70) + (6+00) = 49 + 70Elevation of BVC = 853.48 (3.0)(3)= 844.48For each point, compute X and substitute in the equation below to compute Y:Y= 844.48 +3.00(X) + (-0.90/2)X2For example: at station 44 + 00: X= 0.3Then X= 1.3, 2.3, 3.3, 4.3, 5.3, end at station 49 +70: X=6 or L.

22. At one section of a highway an equal tangent vertical curve must be designed to connect grades of +1.0% and -2.0%. Determine the length of curve required assuming that the SSD= 220.6m.

Solution::

23. Given the information show below, compute and tabulate the curve for stakeout at full 100 stations.

Solution:L = STAEVC STABVCL = 4970 4370 = 600 or 6 full stationsr = (g2 g1) / Lr = (-2.4 3) / 6r = -0.90 r/2 = -0.45 % per stationSTABVC = STAVertex L / 2 = 4670 600/2 = STABVC= STA 43 + 70STAEVC = STAVertex + L / 2 = 4670 + 600/2 = STAEVC= STA 49 + 70ElevBVC = Elevvertex g1 (L/2) = 853.48 3.00 (3) = 844.48ElevEVC = Elevvertex g2 (L/2) = 853.48 2.40 (3) = 846.28 r/2 = -0.45 % per stationElevx = ElevBVC + g1x + (r/2)x2 Elev 44 + 00 = 844.48 + 3.00(0.30) 0.45(0.30)2 = 845.34Elev 45 + 00 = 844.48 + 3.00(1.30) 0.45(1.30)2 = 847.62Elev 46 + 00 = 844.48 + 3.00(2.30) 0.45(2.30)2 = 849.00etc.Elev 49 + 00 = 844.48 + 3.00(5.30) 0.45(5.30)2 = 847.74Elev 49 + 70 = 844.48 + 3.00(6.00) 0.45(6.00)2 = 846.28 (CHECKS)Stationx (stations)g1xr/2 x2Curve Elevation

43 + 70 BVC0.00.000.00844.48

44 + 000.3.90-0.04845.34

45 + 001.33.90-0.76847.62

46 + 002.36.90-2.38849.00

47 + 003.39.90-4.90849.48

48 + 004.312.90-8.32849.06

49 + 005.315.90-2.64847.74

49 + 70 EVC6.018.00-6.20846.28

24. A grade g1of -2% intersects g2 of +1.6% at a vertex whose station and elevation are 87+00 and 743.24, respectively. A 400 vertical curve is to be extended back from the vertex, and a 600 vertical curve forward to closely fit ground conditions. Compute and tabulate the curve for stakeout at full stations.

Solution:

Pt. A STA 85 + 00, Elev. = 743.24 + 2 (2) = 747.24Pt. B STA 90 + 00, Elev. = 743.24 + 1.6 (3) = 748.04

The grade between points A and B can now be calculated as:gA-B = 748.04 - 747.24 = +0.16% 5and the rate of curvature for the two equal tangent curves can be computed as:

Therefore: r1/2 = +0.27 and r2/2 = +0.12

The station and elevations of the BVC, CVC and EVC are computed as:

BVC STA 83 + 00, Elev. 743.24 + 2 (4) = 751.24EVC STA 93 + 00, Elev. 743.24 + 1.6 (6) = 752.84CVC STA 87 + 00, Elev. 747.24 + 0.16 (2) = 747.56

Please note that the CVC is the EVC for the first equal tangent curve and the BVC for the second equal tangent curve.

g1x= -2(1) = -2.00g2x= 0.16(1) = 0.16(r1/2)x2 = (0.27)(1)2 = 0.27(r2/2)x2 = (0.12)(1)2 = 0.12Y1=751.24 2.00 + 0.27 = 749.51Y2= 747.56 + 0.16 +0.12 = 747.84

25. Design an equal-tangent vertical curve to meet a railroad crossing which exists at STA 53 + 50 and elevation 1271.20. The back grade of -4% meets the forward grade of +3.8% at PVI STA 52 + 00 with elevation 1261.50.

Solution:

REFERENCES:http://smweng.com/civil-engineering-definitions-and-history/land-surveying-definedhttps://engineering.purdue.edu/~asm215/topics/history.pdfhttp://encyclopedia2.thefreedictionary.com/Route+Surveyhttps://en.wikibooks.org/wiki/Fundamentals_of_Transportation/Horizontal_Curveshttp://www.globalsecurity.org/military/library/policy/army/fm/5-233/ch3.pdfhttps://en.wikibooks.org/wiki/Fundamentals_of_Transportation/Vertical_Curveshttp://engrwww.usask.ca/classes/CE/271/notes/PART%205%20Simple%20Curves_Updated.pdfhttp://www.slideshare.net/haroldtaylor1113/compound-and-reserve-curveshttp://engrwww.usask.ca/classes/CE/271/notes2012/Part-6.pdfwww.kau.edu.sa/GetFile.aspx?id=169149&fn=Surveying%20-%20Problem%20Solving%20with%20Theory%20and%20Objective%20Type%20Questions.pdfhttp://www.slideshare.net/kailadturla/horizontal-curves-pdfhttps://falmatasaba.files.wordpress.com/2013/04/hid_6-the-transition-curves.pdfhttp://www.tcd.ie/civileng/Staff/Brian.Caulfield/3A1/3A1%20Lecture%2013.pdfcourses.washington.edu/cive316/lectures/Vertical%20Curves.ppsxacstaff.cbu.edu/~gmcginni/classes/.../Vertical%20Curves.ppt

Sheet1STATIONxg1x(r/2)x2Curve ElevationBVC83 + 00000751.24'84 + 001-2.000.27749.51'85 + 002-4.001.08748.32'86 + 003-6.002.43747.67'CVC87 + 004-8.004.32747.56'88 + 0010.160.12747.84'89 + 0020.320.48748.36'90 + 0030.481.08749.12'91 + 0040.641.92750.1292 + 0050.803.00751.36'EVC93 + 0060.964.32752.84'