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Surveying-II, CE-207 (T), CURVES, Lecture No 1, , civil engineering, ENGR.AWAIS KHAN, , 1
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Curves, • Curves are usually employed in lines of, communication in order that the change in, direction at the intersection of the straight lines shall, be gradual., • The lines connected by the curves are tangent to it, and are called Tangents or Straights., • The curves are generally circular arcs but parabolic, arcs are often used in some countries for this, purpose., • Most types of transportation routes, such as, highways, railroads, and pipelines, are connected, by curves in both horizontal and vertical planes., 2
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Curves, • The purpose of the curves is to deflect a vehicle, travelling along one of the straights safely and, comfortably through a deflection angle θ to enable, it to continue its journey along the other straight., , θ, , 3
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Curves, , Horizontal Alignment, Plan View, , Vertical Alignment, Profile View
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Curves, Classification of Curves, Curves, , Horizontal Curves, , Vertical Curves, , Simple Curves, Compound Curves, , Circular Curves, , Transition Curves, Reverse Curves, , Spiral Curves, (non-circular curves), 9
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Curves, Classification of Curves, 1) Simple Curves: Consist of single Arc Connecting two straights., 2) Compound curves: Consist of 2 arcs of different radii, bending, in the same direction and lying on the same sides of their, common tangents, their centers being on the same side of the, curve., 3) Reverse curves: Consist of 2 arcs of equal or unequal radii,, bending in opposite direction with common tangent at their, junction (meeting Point), their center lying on the opposite sides, of the curve., , 10
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Curves, , Simple Curve, , Compound Curve, , Reverse Curve, , 11
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Curves, Nomenclature of Simple Curves, 1) Tangents or Straights: The straight, lines AB and BC which are, connected by the curves are called, the tangents or straights to curves., 2)Point of Intersection: (PI.) The Point, B at which the 2 tangents AB and, BC intersect or Vertex (V) ., , 3)Back Tangent: The tangent line AB, is called 1st tangent or Rear, tangents or Back tangent., 4) Forward Tangent: The tangents, line BC is called 2nd tangent or, Forward tangent., 13
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Curves, Nomenclature of Simple Curves, 5) Tangents Points: The points T1 and, T2 at which the curves touches the, straights., 5.a) Point of Curve (P.C): The, beginning of the curve T1 is called, the point of curve or tangent curve, (T.C)., , 5.b) Point of tangency (C.T): The end, of curve T2 is called point of, tangency or curve tangent (C.T)., 6) Angle of Intersection: (I) The angle, ABC between the tangent lines AB, and BC. Denoted by I., 14
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Curves, Nomenclature of Simple Curves, 7) Angle of Deflection (∅): Then, angle B`BC by which the forward, (head tangent deflect from the, Rear tangent., 8) Tangent Length: (BT1 and BT2), The distance from point of, intersection B to the tangent points, T1 and T2. These depend upon the, radii of curves., 9) Long Cord: The line T1T2 joining, the two tangents point T1 and T2 is, called long chord. Denoted by l., , 15
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Curves, Nomenclature of Simple Curves, 10) Length of Curve: the arc T1FT2 is, called length of curve. Denoted by, L., 11) Apex or Summit of Curve: The mid, point F of the arc T1FT2 is called, Apex of curve and lies on the, bisection of angle of intersection. It is, the junction of lines radii., 12) External Distance (BF): The, distance BF from the point of, intersection to the apex of the curve, is called Apex distance or External, distance., , 16
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Curves, Nomenclature of Simple Curves, 13) Central Angle: The angle T1OT2, subtended at the center of the curve, by the arc T1FT2 is called central, angle and is equal to the deflection, angle., 14) Mid ordinate (EF): It is a ordinate, from the mid point of the long chord, to the mid point of the curve i.e, distance EF. Also called Versed sine of, the curve., •, , •, , If the curve deflect to the right of the, direction of the progress of survey it is, called Right-hand curve and id to the, left , it is called Left-hand curve., The ∆ BT1T2 is an isosceles triangle and, therefore the angle, ∟ BT1T2 = ∟ BT2T1 =, , ∅, 2, , 17
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Curves, Elements of Simple Curves, ∅, 2, , l = 2 T1E = 2 R sin( ), , r, , d) Length of Curve (L):, , S=rx𝜃, , 𝜃, , L = length of arc T1FT2, L = R ∅ (rad) =, , 𝜋R∅, 180, , Or, , ∅, R, , L /2 𝜋 R = ∅ / 360, L = 2 𝜋 R ∅ / 360 =, , L, , R, 2𝜋R, , 𝜋R∅, 180, , 360o, , e) Apex distance or External distance:, BF = BO – OF, ∅, 2, , In ∆OT1B, cos( ) = OT1 / BO, , 19
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Curves, Elements of Simple Curves, , ∅, , ∅, , BO = OT1 / cos(2)= R / cos(2), BO, , ∅, , = R sec(2), , BF = R sec(∅/2) – R, ∅, , BF = R ( sec(2) – 1), BF = R (, , 1, , cos(∅2), , – 1), , f) Mid ordinate or Versed sine of curve:, EF= OF – OE, In ∆T1OE,, , cos(∅/2) = OE / OT1, OE = OT1 cos(∅/2) = R cos(∅/2), EF = R – R cos(∅/2), ∅, , EF = R (1 – cos(2)), , 20
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Curves, Designation Of Curves, • In U.K a curve is defined by Radius which it expressed in, terms of feet or chains(Gunter chain) e.g 12 chain curve, 24, chain curve., , • When expressed in feet the radius is taken as multiple of 100, e.g 200, 300, 400.. ., • In USA, Canada, India and Pakistan a curve is designated by, a degree e.g 2 degree curve , 6 degree curve., , Degree Of Curves, Degree of curve is defined in 2 ways, , 1) Arc Definition, 2) Chord Definition, 21
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Curves, Degree Of Curves, 100 feet, , 1) Arc Definition:, “ The degree of a curve is the, central angle subtended by 100 feet, of arc”., , R, , D = Degree of Curve, Then, , R=, , =, , 100, 2𝜋R, , 5729.58, 𝐷, , D, , O, , Let R = Radius of Curve, , 𝐷, 360, , R, , 100 feet, D, , R, , R, 2𝜋R, , (feet), , 360o, , It is used in highways., 22
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Curves, Degree Of Curves, 2) Chord Definition:, “ The degree of curve is the central, angle subtended by 100 feet of chord”., From ∆OPM, 𝑀𝑃, 50, 𝐷, =, sin( ) =, 𝑂𝑀, 𝑅, 2, R=, , 50, 𝐷, sin( 2 ), , 100 feet, M, , N, 50 ‘, R, , (feet), , It is used in Railway., , P 50 ‘, R, , D/2, D, , Example: D = 1o, , 1) Arc Def:, , R = 5729.58 / D, , O, , = 5729.58 feet, 2) Chord Def:, , R = 50 / sin (D/2), = 5729.65 feet, , 23
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Simple Curves, Method of Curve Ranging, • There are a number of different methods by which a, centerline can be set out, all of which can be summarized in, two categories:, • Traditional methods: which involve working along the, centerline itself using the straights, intersection points and, tangent points for reference., • The equipment used for these methods include, tapes and, theodolites or total stations., • Coordinate methods: which use control networks as, reference. These networks take the form of control points, located on site some distance away from the centerline., • For this method, theodolites, totals stations or GPS receivers, can be used., 24
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Simple Curves, Method of Curve Ranging, The methods for setting out curves may be divided into 2, classes according to the instrument employed ., 1) Linear or Chain & Tape Method, 2) Angular or Instrumental Method, Peg Interval:, Usual Practice--- Fix pegs at equal interval on the curve, , 20 m to 30 m ( 100 feet or one chain), 66 feet ( Gunter’s Chain), Strictly speaking this interval must be measured as the Arc, intercept b/w them, however it is necessarily measure along, the chord. The curve consist of a series of chords rather than, arcs., Along the arc it is practically not possible that is why measured, along the chord., , 25
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Simple Curves, Method of Curve Ranging, Peg Interval:, For difference in arc and chord to be negligible, , Length of chord >, , 𝑅, of curve, 20, , R = Radius of curve, Length of unit chord = 30 m for flate curve ( 100 ft), (peg interval), , 20 m for sharp curve (50 ft), 10 m for very sharp curves (25 ft or less), , 26
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Simple Curves, Method of Curve Ranging, Location of Tangent points:, To locate T1 and T2, , 1) Fixed direction of tangents, produce, them so as to meet at point B., 2) Set up theodolite at point B and, measure T1BT2 (I)., , Then deflection angle ∅ = 180o –I, 3) Calculate tangents lengths by, ∅, 2, , BT1 = BT2 = R tan( ), , 4) Locate T1 and T2 points by measuring, the tangent lengths backward and, forward along tangent lines AB and BC., 27
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Simple Curves, Method of Curve Ranging, Procedure:, • After locating the positions of the tangent points T1, and T2 ,their chainages may be determined., • The chainage of T1 is obtained by subtracting the, tangent length from the known chainage of the, intersection point B. And the chainage of T2 is found, by adding the length of curve to the chainage of T1., • Then the pegs are fixed at equal intervals on the, curve., • The interval between pegs is usually 30m or one chain, length., • The distances along the centre line of the curve are, continuously measured from the point of beginning of, the line up to the end .i.e the pegs along the centre, line of the work should be at equal interval from the, beginning of the line up to the end., 28
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Simple Curves, Method of Curve Ranging, Procedure:, • There should be no break in the regularity of their spacing, in passing from a tangent to a curve or from a curve to, the tangent., • For this reason ,the first peg on the curve is fixed at such a, distance from the first tangent point (T1) that its chainage, becomes the whole number of chains i.e the whole, number of peg interval., • The length of the first sub chord is thus less than the peg, interval and it is called a sub-chord., • Similarly there will be a sub-chord at the end of the curve., • Thus a curve usually consists of two sub-chords and a no., of full chords., , 29
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Simple Curves, Method of Curve Ranging, Important relationships for Circular, Curves for Setting Out, • The ∆ BT1T2 is an isosceles triangle and, therefore the angle, ∟ BT1T2 = ∟ BT2T1 =, , ∅, 2, , The following definition can be given:, • The tangential angle 𝛼 at T1 to any point, X on the curve T1T2 is equal to half the, angle subtended at the centre of curvature, O by the chord from T1 to that point., • The tangential angle to any point on the, curve is equal to the sum of the tangential, angles from each chord up to that point., • I.e. T1OT2 = 2(α + β + 𝛾) and it follows, that BT1T2= (α + β + 𝛾)., , 30
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Simple Curves, Problem 01: Two tangents intersect at chainage of 6 +26.57. it is, proposed to insert a circular curve of radius 1000ft. The deflection, angle being 16o38’. Calculate, a) chainage of tangents points, b) Lengths of long chord , Mid ordinate and External distance., Solution:, ∅, , Tangent length = BT1 = BT2 = R tan(2), BT1 = BT2 = 1000 x tan(16o38`/2), = 146.18 ft, 𝜋R ∅, Length of curve = L = 180o, 𝜋 x 1000 x 16o38`, L=, = 290.31ft, 180o, Chainage of point of intersection, minus tangent length, chainage of T1, plus L, Chainage of T2, , = 6 + 26.56, = -1 + 46.18, = 4 + 80.39, = + 2 + 90.31, = 7 + 70.70, , 31
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Simple Curves, Problem 01:, Solution:, ∅, 2, , Length of chord = l = 2 R sin( ), , l = 2 x 1000 x sin(36o38`/2) = 289.29 ft, ∅, 2, , Mid ordinate = EF = R (1 – cos( ), EF= 1000 x ( 1 – cos(36o38`/2)) = 10.52 ft, Ex. distance = BF = R ( sec, , ∅, 2, , - 1), , ∅, 2, , BF = 1000 x ((1/cos( ) – 1) = 10.63 ft, , 32
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Simple Curves, Problem 02: Two tangents intersect at chainage of 14 +87.33,, with a deflection angle of 11o21`35``. Degree of curve is 6o., Calculate chainage of beginning and end of the curve., Solution:, D = 6o, R = 5729.58 / D ft = 954.93 ft, ∅, , Tangent length = BT1 = BT2 = R tan(2), BT1 = BT2 = 954.93 x tan(11o21`35``/2), BT1 = BT2 = 94.98 ft, 𝜋R ∅, Length of curve = L = 180o, 𝜋 x 954.93 x 11o21`35``, L=, = 189.33 ft, 180o, Chainage of intersection point B, = 14 + 87.33, minus tangent length BT1, = - 0 + 94.96, Chainage of T1, = 13 + 92.35, plus L, = + 1 + 89.33, Chainage T2, = 15 + 81.68, , 33
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Simple Curves, Method of Curve Ranging, 1) Linear or Chain & Tape Method, • These methods use the chain surveying tools only., • These methods are used for the short curves which doesn’t, require high degree of accuracy., • These methods are used for the clear situations on the road, intersections., a) By offset or ordinate from Long chord, b) By successive bisections of Arcs, c) By offset from the Tangents, d) By offset from the Chords produced, 34
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Simple Curves, Method of Curve Ranging, 1) Linear or Chain & Tape Method, a) By offset or Ordinate from long chord, , ED = Oo = offset at mid point of T1T2, PQ = Ox = offset at distance x from E , so that EP = x, OT1 = OT2 = OD = R = Radius of the curve, , Exact formula for offset at any point on the, chord line may be derived as:, By Pathagoras theorem, ∆OT1E, OT12 = T1E2 + OE2, l, OT1 = R , T1E =2, OE = OD – DE = R – Oo, R2 = (l/2)2 + (R – Oo)2, , 35
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Simple Curves, Method of Curve Ranging, 1) Linear or Chain & Tape Method, a) By offset or Ordinate from long chord, , When the radius of the arc is larger as compare to the length of, the chord, the offset may be calculated approximately by, formula or, , Ox =, , 𝑥 (𝐿 −𝑥), -------- 2 (Approximate formula), 2𝑅, , In eqn 1 the distance x is measured from the mid point of the, long chord where as eqn 2 it is measured from the 1st tangent, point T1., • This method is used for setting out short curves e.g curves for, street kerbs., ., 37
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Simple Curves, Method of Curve Ranging, 1) Linear or Chain & Tape Method, a) By offset or Ordinate from long chord, , Working Method:, To set out the curve, • Divided the long chord into even, number of equal parts., • Set out offsets as calculated from the, equation at each of the points of, division. Thus obtaining the required, points on the curve., , • Since the curve s symmetrical along ED,, the offset for the right half of the curve, will be same as those for the left half., 38
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Simple Curves, Problem 03: calculate the ordinate at 7.5 m interval for a circular, curve given that l = 60 m and R = 180 m, by offset or ordinate, from long chord., Solution:, Ordinate at middle of the long chord = verse sine = Oo, l, 60, Oo = R – √ (R2 – ( 2 )2) = 180 – √ (1802 – ( 2 )2, Oo = 2.52 m, Various coo2.52 rdinates may be calculated by formula, l, Ox = √(R2 –x2) – (√(R2 – ( )2)), 2, , x = distance measured from mid point of long chord., 60, O7.5 =, –, – ( )2)) = 2.34 m, 2, 60, O15 = √(1802 –152) – (√(1802 – ( )2)) = 1.89 m, 2, 60, O22.5 = √(1802 –22.52) – (√(1802 – ( )2)) = 1.14 m, 2, 60, O30 = √(1802 –302) – (√(1802 – ( )2)) = 0 m, 2, , √(1802, , –7.52), , (√(1802, , X (m), , Ox (m), , 0, , 2.52, , 7.5, , 2.34, , 15, , 1.89, , 22.5, , 1.14, , 30, , 0, , 39
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Simple Curves, Method of Curve Ranging, 1) Linear or Chain & Tape Method, b) By Successive Bisection of Arcs, •, , Let T1 and T2 be the tangents, points. Join T1 and T2 and bisect it, at E., , •, , Setout offsets ED(y’), determined, point D on the curve equal to, ∅, , ED = y’ = R ( 1 – cos (𝟐), •, , Join T1D and DT2 and bisect them, at F and G respectively., , •, , Set out offset HF(y’’) and GK(y’’), each eqn be, ∅, , FH = GK = y’’= R ( 1 – cos (𝟒), , Obtain point H and K on a curve., By repeating the same process,, obtain as many pints as required, on the curve., 40
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Simple Curves, Method of Curve Ranging, 1) Linear or Chain & Tape Method, 3) By Offsets from the Tangents, In this method the offsets are setout either radially or perpendicular to, the tangents BA ad BC according to as the center O of the curve is, accessible or inaccessible., a) By Radial Offsets: (O is Accessible), , Working Method:, •, , Measure a distance x from T1 on back, tangent or from T2 on the forward tangent., , •, , Measure a distance Ox along radial line, A1O., , •, , The resulting point E1 lies on the curve., 41
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Simple Curves, Method of Curve Ranging, 1) Linear or Chain & Tape Method, 3) By Offsets from the Tangents, In this method the offsets are setout either radially or perpendicular to, the tangents BA ad BC according to as the center O of the curve is, accessible or inaccessible., a) By Radial Offsets:, , EE1 = Ox , offsets at distance x from T1 along, tangent AB., In ∆OT1E, , OT12 + T1E2 = OE2, , T1, , E, , x, , B, , A, Ox, E1, , R, , OT1 = R, T1E = x, OE = R + Ox, , R, , R2 + x2 = (R + Ox)2, R + Ox = √ (R2 + Ox2), Ox =√(R2 + X2) – R ---------A (Exact Formula), , O, 42
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Simple Curves, Method of Curve Ranging, 1) Linear or Chain & Tape Method, 3) By Offsets from the Tangents, a) By Radial Offsets:, Expanding, , √(R2, , + x), , x2, x4, , Ox = R ( 1 + 2R2 + 8R4 + …..) - R, , T1, , Taking any two terms, , x2, , R, Ox = 2 R2, , B, , A, , eqn (A) Ox = R ( 1 + 2R2 ) – R, x2, , E, , x, , Ox, E1, , R, R, , x2, Ox =, -------B ( Approximate formula), 2R, O, Used for short curve, 43
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Simple Curves, Method of Curve Ranging, 1) Linear or Chain & Tape Method, 3) By Offsets from the Tangents, a) By Radial Offsets:: Example, , 44
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Simple Curves, Method of Curve Ranging, 1) Linear or Chain & Tape Method, 3) By Offsets from the Tangents, In this method the offsets are setout either radially or perpendicular to, the tangents BA ad BC according to as the center O of the curve is, accessible or inaccessible., b) By Offsets Perpendicular to Tangents, , (O is Inaccessible), Working Method:, •, , Measure a distance x from T1 on back, tangent or from T2 on the forward tangent., , •, , Erect a perpendicular of length Ox., , •, , The resulting point E1 lies on the curve., 45
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Simple Curves, Method of Curve Ranging, 1) Linear or Chain & Tape Method, 3) By Offsets from the Tangents, b) By Offsets Perpendicular to Tangents : Example, , 47
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Simple Curves, Method of Curve Ranging, 2) Angular or Instrumental Methods, 1) Rankine’s Method of, Tangential Angles, 2) Two Theodolite Method, 1) Rankine’s Method of, Tangential Angles, • In this method the curve is, set out tangential angle, often called deflection, angles with a theodolite,, chain or tape., 52
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Simple Curves, Method of Curve Ranging, 2) Angular or Instrumental Methods, 1) Rankine’s Method of Tangential Angles, Working method:, 1. Fix the theodolite device to be at point, T1 and directed at point B., 2. Measure the deflection angles 𝛿 1 and, the chords C1., 3. Connect the ends of the chords to, draw the curve., Deflection Angles:, The angles between the tangent and the, ends of the chords from point T1., 53
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Simple Curves, Method of Curve Ranging, 2) Angular or Instrumental Methods, 1) Rankine’s Method of, Tangential Angles, • AB = Rear tangent to curve, • D,E,F = Successive point on the, curve, • δ1, δ2, δ3….. The tangential angles, which each of successive chord, …T1D,DE,EF…… makes with the, respective tangents at T1, D, E., • ∆1 , ∆2, ∆3…… Total deflection, angles, • C1, C2, C3…….. Length of the, chord. T1D, DE, EF., 54
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Simple Curves, Problem 04: Two tangents intersect at chainage 2140 m . ∅= 18o24`., Calculate all the data necessary for setting out the curve, with R =, 600 m and Peg interval being 20 m by:, 1) By deflection angle ∅, 2) offsets from chords., Solution:, BT1 = BT2 = R tan, , ∅, 2, , = 600 tan, , 18o24`, 2, , BT1 = BT2 = 97.18 m, Length of curve = L =, , 𝜋R∅, 180𝑜, , =, , L = 192.68 m, Chainage of point of intersection, Minus Tangent length, Chainage of T1, Plus L, Chainage of T2, , 𝜋, , 600 18𝑜24`, 180𝑜, , = 2140 m, = - 97.18 m, = 2042.82 m, = + 192.68 m, = 2235.50 m, 57
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Simple Curves, Method of Curve Ranging, 2) Angular or Instrumental Methods, 2) Two Theodolite Method, • This method is used when ground is not favorable for accurate, chaining i.e rough ground , very steep slope or if the curve one water, • It is based on the fact that angle between tangent & chord is equal to, the angle which that chord subtends in the opposite segments., ∆1 is b/w tangent T1B & T1D => BT1D = T1T2D = ∆1, T1E = ∆2 = T1 T2 E, The total tangential angle or deflection angle, ∆1, ∆2 ∆3 … ,As calculate in the 1st method., , 61
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Simple Curves, Obstacles in Setting Out Simple Curve, • The following obstacles occurring in common practice will be, considered., , 1) When the point of intersection of Tangent lines is inaccessible., 2) When the whole curve cannot be set out from the Tangent, point, Vision being obstructed., 3) When the obstacle to chaining occurs., , 62
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Simple Curves, Obstacles in Setting Out Simple Curve, 2) When the whole curve cannot be set out from the Tangent, point, Vision being obstructed, • As a rule the whole curve is to be set out from T1 however, , obstructions intervening the line of sight i.e Building, cluster of tree,, Plantation etc. In such a case the instrument required to be set up, at one or more point along the curve., , 65
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Simple Curves, Obstacles in Setting Out Simple Curve, 3) When the obstacle to chaining occurs, , 66
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Simple Curves, , •, , •, •, •, , Assignment, Obstacles in Setting Out Simple Curve, (Detail procedure), Page 130 Part II, Example 1 (approximate method), Example 2, Example 3, Page 135 Part II, , 67
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Curves, Compound Curves, • A compound curve consist of 2 arcs of different radii, bending in the same direction and lying on the same side, of their common tangent. Then the center being on the, same side of the curve., RS = Smaller radius, RL = Larger radius, TS = smaller tangent length = BT1, TL = larger tangent length = BT2, 𝛼 = deflection angle b/w common, tangent and rear tangent, 𝛽 = angle of deflection b/w common, tangent and forward tangent, N = point of compound curvature, KM = common tangent, , 68
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Curves, Compound Curves, , Solution:, Chainage of intersection point, Minus TL, Chainage of T2, Plus L, Chainage of compound curvature (N), Plus LS, Chainage of T2, , = 3415 m, = - 422.95 m, = 2992.05 m, = + 418.88 m, = 3410.93 m, = + 244.35 m, = 3655.25 m, , 73
