Unit 4 ( CURVES )

ENGINEERING SURVEY 2

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UNIT 4

CURVES

OBJECTIVES

General Objective

: To know and understand the basic concepts of curves.

Specific Objectives : At the end of the unit you should be able to :

Explain the basic concept of curves.



To identify the terminologies of curves.



To differentiate between circular curves, transition curves and vertical curves.



Explain the methods of setting out circular curves.



Calculate setting out of circular curves, transition curves and vertical curves. Apply the method of setting out of a circular curves and

 transition curve.


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INPUT

4.1 INTRODUCTION In the geometric design of motorways, railways and pipelines, the design and setting out of curves is an important aspect of an engineer’s work. The initial design is usually based on a series of straight sections whose positions are defined largely by the topography of the area. The intersections of pairs of straights are then connected by horizontal curves. In the vertical design, intersecting gradients are connected by curves in the vertical plane. Curves can be listed under three main headings as follows: 1. Circular curve of constant radius 2. Transition curves of varying curves (spirals) 3. Vertical curves.

4.2

CIRCULAR CURVES

Horizontal, circular or simple curves are curves of constant radius required to connect two straights set out on the ground. Such curves are required for roads, railways, kerb lines, pipe lines and may be set out in several ways, depending on their length and radius. Figure 4.1 illustrates how two tangents are joined by a circular curve and shows some related circular curve terminology. The point at which the alignment changes from straight to circular is known as the BC (beginning of curve).The BC is located at a distance T (sub tangent) from PI (Point of tangent intersection). The length of a circular curve (L) is dependent on the central angle (∆) and the value of R (radius). The tangent deflection angle (∆) is equal to the curve’s central angle (Figure 4.2). The point at which the alignment changes from circular back to tangent is known as the EC (end of curve). Since the curve is symmetrical about the PI, the EC is also located at distance T from the PI. From a study of geometry, we recall that the radius of a circle is perpendicular to the tangent at the point of tangency. Therefore, the radius is perpendicular to the back tangent at the BC and to the forward tangent at the EC. The


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terms BC and EC are also referred to by some agencies as PC (point of curve) and PT (point of tangency) and by others as TC(tangent to curve) and CT(curve to tangent).

Figure 4.1 Circular Curve Terminologies (Source: Surveying With Construction Application, B.F. Kavanagh)

4.2.1

Circular Curve Geometry

Most curve problems are calculated from field measurements (∆ and the chainage of PI) and from design parameters(R). Given R (which is dependent on the design speed) and ∆, all other curve components can be computed. An analysis of Figure 4.2 will show that the curve deflection angle (PI, BC, EC) is ∆/2 and that the central angle at O is equal to ∆, the tangent deflection. The line (O-PI), joining the centre of the curve to the PI, effectively bisects all related lines and angles. a) Tangent: In Triangle BC, O, PI, T ∆ = tan R 2 ∆ T = R tan 2


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b) Chord : In triangle BC, O, B 1 C 2 = sin ∆ R 2 C = 2 R sin

∆ 2

c) Mid- ordinate: OB ∆ = cos R 2 ∆ OB = R cos 2

but

OB = R- M ∆ 2 ∆  M = R1 − cos  2 

R - M = R cos

d) External: In triangle BC, O, PI O to PI = R + E R ∆ = cos R +E 2  1   E = R − 1  cos ∆  2   ∆   = R sec −1 2  


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Figure 4.2 Geometry Of The Circle. (Source: Surveying With Construction Application, B.F. Kavanagh)

e) Arc:(Figure 4.3) L ∆ = 2πR 360 ∆ L = 2πR 360

where ∆ is expressed in degrees and decimals of a degree.

Figure 4.3 Relationship Between The Degree Of Curve (D) And The Circle. (Source: Surveying With Construction Application, B.F. Kavanagh)


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The sharpness of the curves is determined by the choice of the radius R; large radius curves are relatively flat, whereas small radius curves are relatively sharp. D is defined to be that central angle subtended by 100 ft of arc. (in railway designs, D is defined to be that central angle subtended by 100 ft of chord.) From Figure 4.3, D and R: D 100 = 360 2πR 5729 .58 D= R

Arc: L ∆ = 100 D ∆ L = 100 D

f) Deflection angle

Figure 4.4 Deflection angle. (Source: Land Surveying, Ramsay J.P. Wilson) In ∆ T1AO, curve T1A = R x 2δ Curve T1A = Chord T1A

1

∴ δ 1(rad) = Curve T1A / 2R = Chord T1A ∴ δ 1(minutes) = (Curve T1A x 180 x 60) / 2π R = (1718.9 x chord T1A) / R


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EXAMPLE 4a

Refer to Figure 4.5, Given ∆ = 16 ° 38’ R = 1000 ft and PI at 6 + 26.57, calculate the station of the BC and EC. Calculate also lengths C, M and E.

SOLUTION: ∆ 2 =1000 tan 8°19 ' =146 .18 ft

T = R tan

PI at -T BC = +L EC =

6 + 26.57 1 46.18 4 + 80.39 290.31 7 + 70.70

L = 2πR

∆ 360

= 2π ×1000 × = 290 .31 ft

16 .6333 360

Figure 4.5 (Source: Surveying With Construction Application, B.F. Kavanagh)


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∆ 2 = 2 ×1000 ×sin 8°19 '

C = 2 R sin

= 289 .29 ft ∆  M = R1 −cos  2  =1000 (1 −cos 8°19 ' ) =10 .52 ft ∆   E = Rsec −1 2   =1000 (sec 8°19 '−1) =10 .63 ft

4.2.2

Compound Circular Curves


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A compound circular curves are curves formed when of two (usually) or more circular arcs between two main tangents turn in the same direction and join at common tangent points. Figure 6.4 shows a compound curve consisting of two circular arcs joined at a point of compound curve (PCC). The lower chainage curve is number 1, whereas the higher chainage curve is number 2. The parameters are R1, R2, ∆1, ∆2 (∆1 + ∆2 = ∆), T1 and T2. If four of these six or seven parameters are known, the others can be solved. Under normal circumstances, ∆1, ∆2, or ∆, are measured in the field, and R1 and R2 are given by design considerations, with minimum values governed by design speed. Although compound curves can be manipulated to provide practically any vehicle path desired by the designer, they are not employed where simple or spiral curves can be used to achieve the same desired effect. Practically, compound curves are reserved for those applications where design constraints (topographic or cost of land) preclude the use of simple or spiral curves, and they are now usually found chiefly in the design of interchange loops and ramps. Smooth driving characteristics required that the larger radius be more than 1-1/3 times larger than the smaller radius (this ratio increases to 11/2 when dealing with interchange curves). Solutions to compound curve problems vary, as several possibilities exist as to which of the data are known in any one given problem. All problems can be solved by use of the sine law or cosine law or by the omitted measurement traverse technique. If the omitted measurement traverse technique is used, the problem becomes a five-sided traverse (Figure 4.6) with sides R1, T1, R2 and (R1- R2) and with angles 90°, 180° - ∆° + 90°, 180°+ ∆2° and ∆1°. An assumed azimuth that will simplify the computations can be chosen.

Figure 4.6 Compound Circular Curves


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(Source: Surveying With Construction Application, B.F. Kavanagh) 4.3 Reverse Curves Reverse curves are seldom used in highway or railway alignment. The instantaneous change in direction occurring at the point reverse curve (PRC) would cause discomfort and safety problems for all but the slowest of speeds. Additionally, since the change in curvature is instantaneous, there is no room to provide super elevation transition from cross-slope right to cross-slope left. However, reverse curves can be used to advantage where the instantaneous change in direction poses no threat to safety or comfort. The reverse curve is particularly pleasing to the eye and is used with great success on park roads, form paths, waterway channels, and the like. The curve can be encountered in both situations illustrated in Figure 4.7 a. and b. the parallel tangent application is particularly common (R1 is often equal to R2). As with compound curves, reverse curves have six independent parameters ( R1, ∆1, T1, R2, ∆2, T2); the solution technique depends on which parameters are unknown, and the techniques noted for compound curves will also provide the solution to reverse curve problems.

Figure 4.7 Reverse Curves (a-Non parallel curve, b- Parallel tangents) (Source: Surveying With Construction Application, B.F. Kavanagh) 4.4 Transition Curves The centrifugal force acting on a vehicle as it moves along a curve increases as the radius of the curve decreases. A vehicle moving from the straight with no centrifugal force acting upon it, into a curve would suddenly receive the maximum amount of


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centrifugal force for that radius of curve. To prevent this sudden lateral shock on passengers in the vehicle, a transition curve is inserted between the straight circular curve (Figure 4.8). The transition curve is a curve of constantly changing radius. The radius (R) of transition curves varies from infinity at its tangent with the straight to a minimum at its tangent point with the circular curve. The centrifugal force thus builds up gradually to its maximum amount.

S = shift

Figure 4.8 The Transition Curves. (Source: Land Surveying, Ramsay J.P. Wilson) The purpose of a transition curve then is to achieve a gradual change of direction from the straight (radius ∞) to the curve (radius R) and permit the gradual application of super-elevation to counteract centrifugal force. The central fugal force tending to thrust a vehicle sideways on a curve is resisted by the friction between the wheels and the surface. If the outer edge of the surface is raised or super elevated, the resultant forces tend to reduce the frictional force necessary to hold the vehicle on the surface. At a particular slope α the frictional force necessary can be eliminated by the formula below: 2

α

v ta n = g R


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where v is the velocity and g is the acceleration due to gravity. As vehicle speeds vary, the fractional resistance is always necessary and a vehicle may stop on the curve. The super elevation must not be too great.

4.4.1

Spiral Curve and Composite Curve

A spiral is a curve with a uniformly changing radius. Spirals are used in highway and railroad alignment changes from tangent to circular curves, and vice versa. The length of the spiral curve is also used for transition from normally crowned pavement to fully superelevated pavement. Figure 4.9 shows how the spiral curve is inserted between tangent and circular curve alignment. It can be seen that at the beginning of the spiral (T.S. = tangent to spiral) the radius of the spiral is the radius of the tangent line (infinitely large) and that the radius of the spiral curve decreases at a uniform rate until, at the point where the circular curve begins (S.C = spiral to curve) the radius of the spiral equals the radius of the circular curve. The spiral curve, used in horizontal alignment, has a uniform rate of change of radius (curvature). This property permits the driver to leave a tangent section of highway at a relatively high rate of speed without experiencing problems with safety or comfort. A composite curve is a curve that forms by combination of two transition curves or through combination of two transition curves and a circular curve.

Figure 4.9 Spiral Curves (Source: Surveying With Construction Application, B.F. Kavanagh)


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4.5 Vertical Curves Vertical curves are used in highway and street vertical alignments to provide a gradual change between two adjacent grade lines. Some highway and municipal agencies introduce vertical curves at every change in grade-line slope, whereas other agencies introduce vertical curves into alignment only when the net change in slope direction exceeds a specific value (for example 1.5% or 2%). In Figure 4.10, g1 is the slope of the lower chainage grade line, g2 is the slope of the higher chainage grade line, BVC is the beginning of the vertical curve, EVC is the end of the vertical line, and PVI is the point of intersection of the two adjacent grade lines. The length of vertical curve (L) is the projection of the curve onto a horizontal surface and, as such, corresponds to plan distances. The algebraic change in slope direction is A, where A = g2 – g1.

Example 4b: g1 = +1.5% and g2= -3.2% A = g2 – g1 = -3.2-1.5 = -4.7

The geometric curve used in vertical alignment designs is the vertical axis parabola. The parabola has the desirable characteristics of (1) a constant rate of change of slope, which contributes to smooth alignment transition, (2) ease of computation of vertical offsets, which permits easily computed curve elevations


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Figure 4.10 Vertical Curves (Profile View Shown) (Source: Surveying With Construction Application, B.F. Kavanagh) The origin of the axes is placed at the BVC (Figure 4.11), the general equation becomes y = ax2 + bx, and because the slope at the origin is g1, the expression for slope of the curve at point becomes dy = slope dx

= 2ax + g1 The general equation can finally be written as y = ax2 + g1x

Figure 4.11 Types of Vertical Curve (Source: Surveying With Construction Application, B.F. Kavanagh)


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Activity 4a

4.1 Fill in the blanks with related circular curve terminology. B C

I J

G F E

A

D

Figure 1 4.2 Solve the puzzle by using the clues as shown below.

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1

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Horizontal: 1) The __________ curve is a curve of constantly changing radius. 2) __________ curves are used in highway and street vertical alignment to provide a gradual change between two adjacent grade lines. 3) __________ curves are curves of constant radius required to connect two straights set out on the ground. 4) Circular curve is also known as ____________ curves. Vertical: 5) The ________curves can be encountered in both situations which are a non parallel curve and parallel tangents. 6) A _______ curve of two (usually) or more circular arcs between two main tangents turning in the same direction and joining at common tangent points. 7) A ________ is a curve with a uniformly changing radius.


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Feedback 4a

4.1

Figure 1 A – Back tangent


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B – Point of intersection C – Deflection angle D – Radius E – Mid ordinate F – Long chord G – Sub tangent H – End of Curve I – External J – Length of curve

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INPUT 2 4.6

SETTING OUT CURVES

This is the process of establishing the centre-line of the curve on the ground by means of pegs at 10m to 30m intervals. In order to do this, the tangent and intersection points must be first fixed in the ground in their correct positions. The straights OI1, I1I2, I2I3,etc., will have been designed on the plan in the first instance(Figure 4.12). Using railway curves, appropriate curves will now be designed to connect the straights. The tangent points of these curves will then be fixed making sure that the tangent lengths are equal, i.e. T1 I1 = T2I1 and T3 I2 = T4I2. The coordinates of the origin, point O, and all the intersection points will only now be carefully scaled from the plan. Using these coordinates, the bearings of the straights are computed and using the tangent lengths on these bearings, the coordinates of the tangent points are also computed. The difference of the bearings of the straights provides the deflection angles(Δ) of the curves which, combined with the tangent length, enables computation of the curve radius, through chainage and all-setting-out data. Now the tangent and


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intersection points are set out from existing control survey stations and the curves ranged between them using the methods detailed below.

Figure 4.12 Curve Setting Out (Source: Engineering Surveying, W.Schofield)

4.6.1

Setting Out By Offsets With Tangent Angle Method

The following methods of setting out curves is the most popular and it is called Rankine’s deflection or tangential angle method, the latter term being more definitive. In figure 4.13, the curve is established by a series of chords T1X, XY, etc. Thus, peg 1 at X is fixed by sighting to I with the theodolite reading zero, turning off the angle δ 1 and measuring out the chord length T1 X along this line. Setting the instrument to read the second deflection angle gives the direction T1 Y, and peg 2 is fixed by measuring the chord length XY from until it intersects at Y. The procedure is now continued, the angles being set out from T1 I and the chords measured from the previous station. It is thus necessary to be able to calculate the setting out angles δ as follows: Assume OA bisects the chord T1 X at right-angles, then Angle AT1 O =90°- δ ∴ angle IT1A= δ

1

, but angle IT1 =90°

1

By radians arc length T1X= R2δ

1

∴ δ 1 rad = (arc T1X /2R) ≈ (Chord T1X / 2R)


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∴ δ 1min = (chord T1X x 180º x 60) /2Rπ = 1718.9(Chord / R) or δ º = (Dº x Chord ) / 200 where degree of curve is used.

Figure 4.13 Tangent Angle Method (Source: Engineering Surveying, W.Schofield)

Example 4c: The centre-line of two straights is projected forward to meet at I, the deflection angle being 30°. If the straights are to be connected by a circular curve of radius 200 m, tabulate all the setting-out data, assuming 20-m chords on a through chainage basis, the chainage of I being 2259.59 m. Solution: Tangent length = R tan Δ/2 = 200 tan 15° = 53.59 m ∴ Chainage of T1 = 2255.59 - 53.59 = 2206 m ∴ 1st sub-chord = 14 m Length of circular arc = RΔ = 200(30°) rad = 104.72 m From which the number of chords may now be deduced


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1st sub-chord = 14 m 2nd, 3rd, 4th, 5th chords = 20 m each Final sub-chord = 10.72 m Total = 104.72 m {Check} ∴ Chainage of T2; = 2206 m + 104.72 m = 2310.72 m Deflection angles: For 1st sub-chord = 1718.9 (14/200) = 120.3 min = 2° 00' 19" Standard chord = 1718.9 (20/200) = 171.9 min = 2° 51' 53" Final sub-chord = 1718.9 (10.72 /200) = 92.1 min = 1° 32' 08" Check: The sum of the deflection angles = Δ/2 = 14° 59' 59" ≈ 15° Chord number

Chord length (m)

Chainage (m)

Deflection angle o,„

Settingout angle o,„

Remarks

1 14 2220.00 2 00 19 2 00 19 peg 1 2 20 2240.00 2 51 53 4 52 12 peg 2 3 20 2260.00 2 51 53 7 44 05 peg 3 4 20 2280.00 2 51 53 10 35 58 peg 4 5 20 2300.00 2 51 53 13 27 51 peg 5 6 10.72 2310.72 1 32 08 14 59 59 peg 6 The error of 1" is, in this case, due to the rounding-off of the angles to the nearest second and is negligible. 4.6.2 Setting Out By Offset From The Tangent The position of the curve (in Figure 4.14) is located by right-angled offsets Y set out from distances X, measured along each tangent, thereby fixing half the curve from each side. The offsets may be calculated as follows for a given distance X. Consider offset Y3, for example. In ΔABO, AO2 = OB2 - AB2 ∴ (R-Y3)2= R2 – X32 and Y3=R-(R2-X32)½ thus for any offset Yi, at distance Xi, along the tangent Yi = R - (R2 - Xi2) ½


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Figure 4.14 Setting Out By Offset From Tangent (Source: Engineering Surveying, W.Schofield)

4.6.3

Setting Out By Offset With Sub-Chords

In Figure 4.15 assume T1A is a sub-chord of length x, from equation Offset CA = (½ chord x chord) / Radius = (chord2)/2R, the offset CA = O1 == X2 / 2R. As the normal chord AB differs in length from T1A, the angle subtended at the centre will be 2θ not 2δ . Thus, the offset DB will not in this case equal 2CA. Construct a tangent through point A, then from the figure it is obvious that angle EAB = θ, and if chord AB = y, then offset EB = y2 / 2R. Angle DAE = δ , therefore offset DE will be directly proportional to the chord length, thus: DE = (O1 / x) y = (x 2 y)2Rx = (xy)/ 2R Thus the total offset DB = DE + EB = ( y / 2R) (x +y) = (Chord / 2R) (sub-chord + chord) Thus having fixed B, the remaining offsets to T2; are calculated as y2/R and set out in the usual way. If the final chord is a sub-chord of length x1, however, then the offset will be


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(x1/ 2R)((x1 + y) A more practical approach to this problem is actually to establish the tangent through A in the field. This is done by swinging an arc of radius equal to CA, i.e. x2 / 2R from Ti. A line tangential to the arc and passing through peg A will then be the required tangent from which offset EB, i.e. y2/lR, may be set off.

Figure 4.15 Setting Out By Offset With Sub-chords (Source: Engineering Surveying, W.Schofield) 4.6.4

Setting Out By Offset With Long-Chords

In this case (Figure 4.16) the right-angled offsets Y are set off from the long chord C, at distances X to each side of the centre offset Y0. An examination of Figure 6.16 shows the central offset Y0 equivalent to the distance T1A on Figure 6.14, thus: Yo = R –[R2 - (C/2)2]1/2 Similarly, DB is equivalent to DB on Figure 6.14, thus: DB=R-(R2 – X12) 1/2 and offset Y1, = Y0 - DB :. Y1 = Y0- [R - (R2 - X12)'/2] and for any offset Yi, at distance Xi, each side of the mid-point of T1 T2,:


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Y1, = Y0 - [R - (R2 - X2)] 1/2

Figure 4.16 Setting Out By Offset With Long-chords (Source: Engineering Surveying, W.Schofield) 4.7 Setting Out A Circular Curve. Circular curves may be set out in a variety of ways, depending on the accuracy required, its radius of curvature and obstructions on site. Methods of setting out are as follows: • Using one theodolite and a tape by the tangent angle method. This method can be used on all curves, but is necessary for long curves of radius unless they are set out by coordinates. • Using two theodolites. This method can be used on smaller curves where the whole length is visible from both tangent points and where two instruments are available. • Using tapes only by the method of offsets from the tangent. This method is used for minor curves only. • Using tapes only by the method of offsets from the long chord. This method is used for short radius curves.

Normally, a circular curve is set up by using a theodolite and a tape (tangent angle method). Before the curve can be set out the tangent points must be located on the ground. For any particular pair of straights there is only one point on each straight for a curve of given radius or degree to leave the first tangentially in order to join the other tangentially. These tangent points cannot be scaled off a plan with sufficient accuracy and they must be located by field observations. The tangent points are represented in Figure 4.17 by the points A and B. The method of locating these tangent points is summarised in


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(a) to (h) as follows: a) Set up the theodolite near A and extend straight towards P. b) Set up on the straight BF and produce it to meet the line at P. c) Mark the intersection of tangents at P. d) Measure angle EPF and obtain angle θ. (e) Calculate tangent length PA by using the formula T = R tan ½θ. f) Place pegs at A and B on the lines. (From P measure the lengths PA and PB = T and line in the points A and B on the straights with the theodolite still set up at P. Mark the pegs at A and B distinctively representing tangent points. This can be done by painting the peg or by placing three pegs, the centre peg representing the tangent point.) g) Set up at A and measure angle PAB, which should equal ½θ. h) Complete the chaining of the first straight to A. NOTE : (i) The chaining of the first straight is completed by measuring the distance from the last chain peg and noting the actual chainage of the tangent point A. ii) In many cases the intersection points P may have been previously marked on the ground. In such cases the field work consists of pegging the straights and measuring θ without having to locate the intersection by the method described above. After the tangent points have been pegged as described above, the points on the curve must be located. The interval between chainage pegs on the curve should be measured along the actual arc. As chords are used in locating the pegs, the difference in length should be calculated strictly as they are slightly shorter than the arc distances. This would be done in precise work, e.g. underground railways. In most practical cases where R exceeds twenty times the chord length this difference is negligible. The tangent point A will seldom fall exactly at a peg interval. Since the chainage must be continuous, the chord AG to the first point on the curve may be shorter than the regular chord length which is c, usually equal to the peg interval or half the peg interval if additional pegs are needed to mark the curve clearly on the ground. There will generally also be a sub-chord at the end of the curve. Let these sub-chord lengths be denoted by c' and c". The method of locating the points on the curve is summarised in (a) to (k} as follows: a) Obtain the first sub-chord c' = c — EA. Assuming E is the position of the last chainage peg on the straight, then EA + c’ =c and as EA has been measured and c is known, the length of the sub-chord can be


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obtained. b) Calculate δ for chord length c. This can be calculated from sinδ = (c/2R) or δ = 1718.9 ( c/R) minutes. c) Calculate δ ’ for the first sub-chord. This can be calculated in the same way as for δ , but for flat curves it can be obtained with sufficient accuracy from δ ’ = (c’ /c) δ d) Calculate the final sub-chord and its δ ". Calculate from θ and the radius the length of the curve (L= Rθ). Then the chainage of A + L =: chainage of B. The amount, by which the chainage of B exceeds and exact number of peg intervals, plus the initial sub-chord, is the length of the end subchord c". (e) Draw up a table of deflection angles to the various points. i) This will take the following form: 1st deflection angle to G = (c’/c) δ = δ ” 2nd deflection angle to H = δ ’ + δ 3rd deflection angle to K = δ ’+δ + δ . ii) The final deflection angle to tangent point B must equal ½θ, allowance being made for the sub-chords, i.e. ½θ = δ ' + δ + δ + . . . δ + δ ". f) From A set out δ ’for the line AG. The instrument is set up at A and P is sighted at a reading of 0° 00' 00" and the horizontal circle is clamped with the lower clamp. The first deflection angle δ ’ is set on the vernier or optical micrometer using the upper clamp and tangent screw only, so that the line of sight is along AG. g) Place G a distance of c' m from A on line AG. The zero of the tape is held at A and the distances marked with a peg, which is then moved on to the line AG as defined by the theodolite sighting.

h) From A set out δ ’+ δ for the line AH. This is the second deflection angle PAH obtained from the table in (e).


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i) Set out GH =c. The zero of the tape is now held at Q and the chord length or peg interval along the tape is marked with a peg which is moved on to the line AH as given by the theodolite. j) Repeat the same process to set out the remaining pegs. i) Continue until the last peg on the curve has been placed and measure the remaining distance to B which should equal the calculated length c" of the final subchord. Also set out the final deflection angle, which should pass through tangent point B, indicating no disturbance of the instrument. ii) As a final check on the accuracy, locate point B by the deflection angle and sub-chord c". If this position does not coincide with the tangent point B, the distance between the two is the actual error of tangency. If this is large, indicating an error, the whole process must be repeated. Where calculations are inaccurate by a few millimetres in the final placing of the pegs, it is usual to adjust the last few pegs to secure tangency. In accurate tunnel work the degree of precision must, of course, be greater. k) The first chainage peg on BF will be c — c" from -B. Having calculated the distance of the first chainage peg F on the second straight, chaining may be proceeded with after moving the instrument to B or some other convenient point on this straight.

Figure 4.17 Setting out a circular curve (Source: Land Surveying, Ramsay J.P. Wilson)


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Example 4d: Two straights AP and PB intersect with an angle of deflection of 12º 20' as illustrated in Figure 6.17. They are to be connected by a circular curve of radius 600 m. The chainage of the intersection point is 12 + 73.16. Calculate the setting-out data required to peg the curve at a continuous chainage with pegs at 25m intervals. Solution : a) Calculate the tangent lengths from T == R tan ½θ T = 600 tan 6° 10' = 600 x 0.108046 = 64.83 m. b) Calculate, the arc length from L = Rθ L = 600 x 12° 20' x (2π / 360) = 600 x 0.21526 = 129 .16 m. c) Calculate the chainages: Chainage of P = 12 + 73.16 Less T = 64.83 Chainage of A = 12 + 08.33 Add L = 129.16 Chainage of B = 13 + 37.49 m. d) Calculate the sub-chords. The last peg on the straight is at chainage 12 + 00, therefore the next peg must be at chainage 12 + 25. There is still 8-33 m on the straight to the tangent point, so there will be 25.00 — 8.33 = 16.67 m along the curve to the first peg on the curve. Thus 16.67 m is the length of the first sub-chord. As the curve length is 12 9.16 m and the first sub-chord is 16.67 m, there is 129.16 – 16.67 = 112.49 of arc left. Four 25m standard chords make up the next 100m, leaving a final sub-chord of 12.49 m. The measurement of the arc distance by these chords is sufficiently accurate for most practical purposes although theoretically the measured distance is shorter than the arc distance. e) Calculate the deflection angles from δ = 1718.9( c/R) δ = 1718.9 x (25/600) = 71.62 = 1° 11. 62’ = 1° 11' 37.2”.


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i) The initial sub-chord is 16.67 m, so its deflection angle will be in the proportion of: (16-67 /25 ) x (1º11.62’) = 47.76’ ∴ δ = 47' 45.6”. ii) The final sub-chord is 12.49 m, so its deflection angle will be in the proportion of: (12.49 / 25) x 1° ll.62’ = 35.78’. δ = 35’ 46.8” g) Tabulate the deflection angles. The deflection angles are tabulated as follows: Instrument at A = 12 + 08.33 To peg at P

Chord Length

12 + 25

16.67

12 + 50

25

12 + 75

25

13

25

13 + 25

25

B = 13 + 37.49

12.49

Bearing °’” 00 00 00 + 47 45.6 00 47.45.6 + 1 11 37.2 1 59 22.8 + 1 11 37.2 3 11 00.0 + 1 11 37.2 4 22 37.2 + 1 11 37.2 5 34 14.4 + 35 46.8 6 10 01.2 = ½θ (Check)

Table 4.1 Calculation Of Setting Out Circular Curve (Source: Land Surveying, Ramsay J.P. Wilson) NOTE:

There is always likely to be minor rounding off of errors such as the 1.2”, which is negligible. To keep these errors to a minimum the calculation is always carried out to 0.1”, but the observed bearings are rounded off to 1" for more accurate work and frequently to 10" or oven 20", depending on the theodolite being used for setting out.


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4.8 Obstructions To Setting Out Obstructions on site may prevent normal setting out in a variety of ways. Most problems of this kind can easily be overcome if setting out is by means of coordinates, but two common problems which often arise are the following: a) Where the intersection point is inaccessible. b) Where there are obstructions to sighting the deflection angle to every point on the curve from the initial tangent point. 4.8.1 The Inaccessible Intersection Point. It may not be possible to measure θ at the intersection point if it is inaccessible e.g. on mountain roads. By setting out a line such as XY in figure 4.18, and by measuring its length and the angles α and β the triangle XPY can be solved for the lengths PX and PY and θ can be deduced. The tangent points can then be located from X and Y and the curve set out in the usual way.

Figure 4.18 : An inaccessible intersection-point (Source: Land Surveying, Ramsay J.P. Wilson)

4.8.2

Obstructions To Sighting The Deflection Angles.

Where obstructions prevent the sighting to every peg on the curve, the following procedure must be adopted as illustrated in Figure 4.19.


(a)

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Pegs 2, 3 and 4 have been placed turning off deflection angle δ each time. Peg 5 cannot be placed from peg 1 owing to an obstruction. i) Triangle 1X4 is isosceles, therefore angle X14 = angle 14 X = 3δ . ii) The angle between the chord 1-4 produced and the tangent X4 produced is also 3δ and the angle δ is required to be turned off this tangent to locate peg 5. iii) An angle of 180° + 4δ is required to be turned off line 4-1 in order to locate the direction 4-5.

b) Set up the theodolite at peg 4, sight peg 1 at a zero setting and turn of at an angle equal to 180° + δ (5 - 1) = 180° + 4δ ; i.e. δ must be multiplied by the number of standard chord lengths between the two points being sighted to. NOTE: i) The longest possible backsight should always be used to orient the theodolite.

ii) If a sub-chord exists between the instrument and the point sighted to, the angle to be turned off will be 180° + (δ x Number of standard chords between the pegs sighted) + δ ', the deflection angle of the sub-chord. iii) The rule for obtaining the angle applies between any two pegs on any one circular curve.

Figure 4.19 : Obstructed Deflection Angles (Source: Land Surveying, Ramsay J.P. Wilson)

4.9 COMPUTING AND SETTING OUT A TRANSITION CURVE 4.9.1 Introduction It is normal to set out a transition curve using deflection angles from the tangent point or by deflection distances for short transitions, in the same way as for circular


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curves. The deflection angles for transitions are not equal as are those for circular curves. The chord length used is often half that used on the circular curve. In practice the settingout data are usually extracted from tables which relate to various design speeds. The only calculations needed are for the tangent lengths using the observed deflection angle θ . Once the tangent points have been established, the transitions are set out from both tangent points to T1 and T2, the limits of the circular curve. Then from T1 or T2 the direction of the tangent to the circular curve is obtained by turning off 2/3φ ' (Figure 4.20) from the chord to the transition and the circular curve deflection angles are set out as before.

Figure 4.20 : Transition Curve Detail (Source: Land Surveying, Ramsay J.P. Wilson)

4.9.2

Setting-out Calculations.

If the tabulated data are not available, the length of the transition must first be obtained from the formula below: The rate of change of radial acceleration

=

v3 RL

This forms part of highway design and is dependent on traffic speed, available space and the radius to be adopted. The setting out surveyor will be provided with the transition length and the radius or degree of curve. With this information and the observed deflection angle θ , the following calculations are needed before setting out the pegs: (a) Shift. This is calculated from:


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L2 S= 24R The radius R = 5729.58 /D, if only the degree of curve is given. (b) Tangent lengths. This is calculated from: P − To = ( R + S ) tan

1

2

θ+

L 2

This distance will be taped back from the intersection point and the two tangents T0, and T3 pegged. (c) Deflection angles. The deflection angles for 10-, 15- or 20-m chords are calculated from: d = 572 .958

l2 min utes RL

where l is the continuous chainage along the transition. For small angles of deflection the summation of the chord lengths may be taken to equal the lengths l. L

The final deflection angle to locate T1 is 572.958 R minutes, and φ ' is three times this value. (d) Length of circular curve. This is obtained from Rθ ' in the usual way, where: θ ' = (θ — 2φ ') and L R

2φ '' = 57.2958

degrees.

If 2φ ' is greater than θ , it is not possible for the transitions to be contained within the straights. Longer transitions leading to a curve of smaller radius will have to be used. NOTE:

Example 4e.


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Calculate the setting-out data for a 75m transition curve to connect an 8° circular curve joining two straights with an angle of deflection of 20° 00', using 15m chords.

Solution: (a) To calculate the radius and shift: Radius , R =

5729 .28 D

5729 .58 8 = 716 .20 m =

ShiftS =

L2 24 R '

75 2 24 × 716 .20 = 0.327 m =

(b) To calculate the tangent lengths: L 2  = (716 .20 + 0.33 ) tan 10 + 37 .5

Tangent , T = ( R + s ) tan

1

2

θ+

= 163 .84 m

(c) To calculate and tabulate the deflection angles and deflection distances: δ = 572 .958

l2 l3 Minutes andx = m RL 6 RL

572 .958 l2 716 .20 × 75 = 0.01067 l 2 =


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δ L L2 15 225 2'24" 30 900 9'36" 45 2025 60 3600 38'24" 75 5625 60'00" Table 4.2 Deflection Angle And Deflection Distances. (Source: Land Surveying, Ramsay J.P. Wilson)

Chord 15 15 15 15 15

x 0.010 0.084 0.283 1.309

: Either the deflection angles or the deflection distances are calculated and used for setting out. In many cases, where the distances are short they form the more convenient setting out method. NOTE

(d) To check the final deflection angle and calculate φ ': L minutes R 75 = 572.958 716 .20

Final 8 = 572.958

= 60' = 1° 00", which checks the final deflection angle calculated and tabulated in Table 6.2. ∴ φ '' =

L radians 2R

= 3° 00' ∴ 2φ ' = 6° 00" but 2φ '' = 57.2958 L

R

degrees

= 6° 00", which checks the above value.

4.9.3

(e) To calculate the length of the circular curve: θ consumed by both transitions, = 2φ ' = 6°. ∴ L curve = 716.20 x (20° - 6°) x (2π /360) = 716.20 X 0.24435 = 175.00 m. Setting out curves containing a transition. To illustrate this procedure the following describes the process of setting out the


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curve, the details of which are found in the question above: (a) Setting out the transitions. The first transition is set out from To. The other tangent point T3 is set out along the second straight the same distance from P as T0, equal to 163.84 m. The final transition is then set out between T3 and T2 as before. (b) Setting out the circular curve. Assuming 15m chords are to be used, there will be eleven 15m chords and one 10m subchord to make up the arc of 175 m. The standard deflection angle will be: 15 of 4  = 36 ' 100

set up at T1 for the sub-chord:

10 of 4  = 24 ' 100

Set up at T1 (Figure 4.20), sight T0 and swing through 180° +2/3φ '+ δ to sight to the first peg on the curve, completing the setting out to T2 in the usual way for circular curves. This initial angle is 180° + 2° + 36' = 182° 36' 00". 4.10 PROCEDURE FOR COMPUTING A VERTICAL CURVE 1. Compute the algebraic difference in grades: A = g2-g1. 2. Compute the chainage of the BVC and EVC. If the chainage of the PVI is known, ½ L is simply subtracted and added to the PVI chainage. 3. Compute the distance from the BVC to the high or low point (if applicable): x=

− g1L A

and determine the station of the high/low point. 4. Compute the tangent grade-line elevation of the BVC and the EVC. 5. Compute the tangent grade-line elevation for each required station. 6. Compute the midpoint of chord elevation: Elevation of BVC + elevation of EVC 2

7. Compute the tangent offset (d) at the PVI (i.e., distance VM in Figure 4.21): d =

difference in elevation of PVI and midpoint of chord 2


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8. Compute the tangent offset for each individual station (see line ax2 in Figure 4.21): 2

 x   or 4d x 2 tan gentoffset = d  L  L2  2

( )

where x is the distance from the BVC or EVC (whichever is closer) to the required station. 9. Compute the elevation on the curve at each required station by combining the tangent offsets with the appropriate tangent grade-line elevations—add for sag curves and subtract for crest curves.

Figure 4.21 Geometric Properties Of The Parabola (Source: Surveying With Construction Application, B.F. Kavanagh)

EXAMPLE 4f The techniques used in vertical curve computations are illustrated in this example.


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Given that L= 300 ft, g1 = -3.2%, g2 = + 1.8%, PVI at 30 + 30, and elevation =465.92, determine the location of the low point and elevations on the curve at even stations as well as at the low point. Solution : 1. A = 1.8 - (-3.2) = 5.0 2.

PVI - ½L = BVC BVC at (30 +- 30) - 150 = 28 + 80.00 PVI + ½L = EVC EVC (30 + 30) + 150 = 31 + 80.00 EVC - BVC = L (31 + 80) - (28 + 80) = 300 Check

3.

Elevation of PVI = 465.92 150 ft at 3.29r = 4.80 (see Figure 6.20) Elevation BVC == 470.72 Elevation PVI = 465.92 150 ft at 1.8% = 2.70 Elevation EVC = 468.62

4. Location of low point


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− g1 L A 3.2 × 300 = 5 = 192 .00 ft ( fromtheBVC ) x=

5. Tangent grade-line computations are entered in Table 4.3. Example: Elevation =470.72 =470.08

at 29 +00 =470.72 - 0.64

- (0.032 ×20)

Figure 4.22: Sketch For Example 4 (Source: Surveying With Construction Application, B.F. Kavanagh) Station BVC 28 + 80 29 + 00 30 + 00 PVI 30 + 00 LOW30 + 72 31 + 00 EVC 31 + 80 30 + 62 30 + 72 30 + 82

Tangent Elevation + Tangent Offset 470.72 (0/150)2 x 1.8.75 =0 470.08 (20/150)2 x 1.875 = 0.03 466.88 (120/150)2 x 1.875 = 1.20 465.92 (150/150)2 x 1.875 = 1.875 466.68 (108/150)2 x 1.875 = 0.97 467.18 (80/150)2 x 1.875 = 0.53 468.62 (0/150)2 x 1.875 = 0 466.50 (118 / 150) 2 x 1.875 = 1.16 466.68 (108 / 150) 2 x 1.875 = 0.97 466.86 (98 / 150) 2 x 1.875 = 0.80

= Curve Elevation 470.72 470.11 468.08 467.80 467.65 467.71 468.62 467.66 467.65 467.66

*Where x is distance from BVC or EVC, whichever is closer.

Table 4.3 Parabolic Curve Elevation by Tangent Offset

6. Mid-chord elevation:


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470.72 (BVC) + 468.62 (EVC) = 469 .97 2

8. Tangent offsets are computed by multiplying the distance ratio squared, [x/(L/2)]2, by the maximum tangent offset (d). See Table 4.3. 9. The computed tangent offsets are added (in this example) to the tangent elevation in order to determine the curve elevation.

Activity 4b


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4.3) Match the formulae used to calculate the following

Transition deflection angle

P − To = ( R + S ) tan

Shift

x=

Tangent length

d = 572 .958

Length of the circular curve

2

θ+

s=

L deg rees R l2 min utes RL

L2 24 R

4.4)

State the methods of setting out a circular curve without using a theodolite.

4.5)

How is the problem of an inaccessible intersection point overcome?

4.6)

The tangent length of a simple curve was 202.12m and the deflection angle for a 30m chord 2° 18'. Calculate the radius, the total deflection angle, the length of curve and the final deflection angle.

Let me think first………

Feedback 4b

L 2

l3 6 RL

2φ' = 57 .2958

Transition deflection distance

1


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4.3) The formulae used to calculate the following:

Transition deflection angle

P − To = ( R + S ) tan

Shift

x=

Tangent length

d = 572 .958

Length of the circular curve

4.4)

2

θ+

L 2

l3 6 RL

2φ' = 57 .2958

Transition deflection distance

1

s=

L deg rees R

l2 min utes RL

L2 24 R

a) Setting Out By Offset With Sub-Chords. b) Setting Out By Offset With Long-Chords. c) Setting Out By Offsets From Tangent.

It is easy�

4.5) If the intersection point is inaccessible, this problem can be solved by setting out a line such as XY in figure 1. By measuring its length and the angles α and β the triangle XPY can be solved for the lengths PX and PY and θ can be deduced. The tangent points can then be located from X and Y and the curve set out in the usual way.


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Figure 1. 4.6)

2° 18' = 138' 138'= 1718.9( 30/R ) R = 373.67 202.12 = R tan ∆ /2 202.12 = 373.67 tan ∆ /2 ∆ = 56° 49' 06" Length of curve = R ∆ rad = 373.67 x 0.991667 rad = 370.56m Using 30m chords, the final sub-chord = 10.56m ∴ final deflection angle = (138' x 10.56)/30 = 48.58' = 0° 48' 35"

Well done! Keep it Up!

Self Assessment


1)

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Refer to the figure below, given ∆ = 12° 51’, R = 400m , PI at 0 + 241.782. Calculate the station of the BC and EC.

Figure 2 2)

What is transition curve? Why is it important in engineering survey? 3) A downgrade of 1 in 20 = 5 in 100= -5% An upgrade of 1 in 25 = 4 in 100 = +4% Calculate the algebraic difference of the two gradients.

4) The straight lines ABI and CDI are tangents to a proposed circular curve of radius 1600m. The length AB and CD are each 1200m. The intersection point is inaccessible so that it is not possible directly to measure the deflection angle; but the angles at B and D are measured as ∠ ABD = 123° 48', ∠ BDC = 126° 12' and the length BD is 1485m. Calculate the distances from A and C of the tangent points on their respective straights


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and the deflection angles for setting out 30m chords from one of the tangent points. (Figure 3)

Figure 3

I must work hard….

Feedback to Self Assessment


1)

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T = R tan (∆ /2) = 400 tan 0°25’ 30” = 45.044m L = (2π R ∆ )/360 = 2π x 400 x (12.850/360) = 89.170m PI at 0+241.782 -T 45.044 BC = 0+196.738 +L 89.710 EC = 0+286.448 2) The transition curve is a curve of constantly changing radius. It is inserted between the straight circular curves to prevent this sudden lateral shock on passengers in the vehicle. In engineering survey, the purpose of a transition curves then is to achieve a gradual change of direction from the straight (radius ∞) to the curve (radius R) and permit the gradual application of super-elevation to counteract centrifugal force.

3)

g2 = -5% and g1= +4% A = g2 – g1 = -5%-4% = -9% 4) ∆ 1 = 180° -123° 48' = 56° 12'

∆ 2 = 180° -126° 12' = 53° 48' ∴ ∆ =∆ 1 + ∆ 2 = 110° φ = 180°- ∆ = 70° Tangents length IT1 and IT2 = R tan ∆ /2 = 1600 tan 55° = 2285m By sine rule in triangle BID: BI =

BD sin ∆2 BDsin sinφ∆1

ID = sin53 φ  48 ' 1485 sin =  1485sin sin7056 15 ' = = 1275sin .2m 70  = 1314 m


Thus

AI = AB + BI = 1200 + 1275.2 =2475.2m CI = CD + ID = 1200 + 1314 = 2514m ∴ AT1 = AI -IT1 = 2475.2 -2285 = 190.2m CT2 = CI -IT2 = 2514 -2285 = 229m

Deflection angle for 30m chord = 1718.9 x (30/1600) = 32.23' = 0° 32' 14"

Yes ! This time I get an A+.

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Unit 4 ( CURVES )

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