Curves

CurvesCurves

Nomenclature and Formula Nomenclature and Formula for Circular Curvefor Circular Curve

CurvesCurves

There are two types of curves.There are two types of curves.1.1. Horizontal Curves (Horizontal Horizontal Curves (Horizontal

Alignment)Alignment)2.2. Vertical Curves (Vertical Alignment)Vertical Curves (Vertical Alignment)

Horizontal Curves:Horizontal Curves:o Simple Circular CurveSimple Circular Curveo Compound Circular CurvesCompound Circular Curveso Reverse Circular CurvesReverse Circular Curveso Broken Back CurvesBroken Back Curves

Degree of a Curve:Degree of a Curve:Angle subtended by an arc of 100 ft length Angle subtended by an arc of 100 ft length

is called degree of curve.is called degree of curve. Curves are mostly used in transportation Curves are mostly used in transportation

routes, such as highways, railroads.routes, such as highways, railroads. Sometimes pipelines are also connected Sometimes pipelines are also connected

by smooth horizontal or vertical curves.by smooth horizontal or vertical curves.

Horizontal Curves (or Horizontal Curves (or Alignment)Alignment)

Provided for safe and continuous Provided for safe and continuous operations at a uniform speed.operations at a uniform speed.

Topography of the area controls the radius Topography of the area controls the radius and design speed.and design speed.

Sight distances controls the curves radiusSight distances controls the curves radius Straight (tangents) should be long enough Straight (tangents) should be long enough

to accommodate all the parameters of the to accommodate all the parameters of the sight distances (passing, stopping, sight distances (passing, stopping, decision).decision).

Easement curves are desirable, especially Easement curves are desirable, especially for railroads, and rapid transit system to for railroads, and rapid transit system to lesson the sudden change in the curvature lesson the sudden change in the curvature at the junction of a tangent and a circular at the junction of a tangent and a circular curve (spiral curves).curve (spiral curves).

Components of CurvesComponents of Curves

Tangent is the straight part of the alignment, Tangent is the straight part of the alignment, there are initial tangents and final tangents.there are initial tangents and final tangents.

Spiral are used to connect a tangent with a Spiral are used to connect a tangent with a circular curve, a tangent with tangent, a circular curve, a tangent with tangent, a circular curve with a circular curve.circular curve with a circular curve.

- A spiral makes an excellent A spiral makes an excellent easement curve because its easement curve because its radius decreases uniformly radius decreases uniformly from infinity at tangent to from infinity at tangent to that of the curve it meets.that of the curve it meets.

Designations of Curve:Designations of Curve:In European practice and In European practice and

majority of American majority of American Highways works, circular Highways works, circular curves are designated by curves are designated by their radius.their radius.

Rail departments and some Rail departments and some other departments designate other departments designate the curves in degrees.the curves in degrees.

Arc Definition:Arc Definition:Degree of the curve is the Degree of the curve is the

central angle subtended by a central angle subtended by a circular arc of 100 ft.circular arc of 100 ft.

D = 100D = 100360º 2 360º 2 ππ R RR = R = 5729.58 in 5729.58 in

ftft DD

Chord Definition:Chord Definition:

Angle subtended by a circular arc and Angle subtended by a circular arc and subtended by 100 ft chord.subtended by 100 ft chord.

Sin Sin DD = = 5050

2 R2 R

R = 50R = 50

Sin ( D / 2 )Sin ( D / 2 )

Types of Circular CurvesTypes of Circular CurvesSimple Circular Curve:Simple Circular Curve:A circular arc connecting two tangents most oftenly used A circular arc connecting two tangents most oftenly used

type.type.Compound Circular Curve:Compound Circular Curve:A curve composed of two or more circular arcs of different A curve composed of two or more circular arcs of different

radii tangent to each other, with their centers on same side. radii tangent to each other, with their centers on same side. Compound Curve should avoided except in mountains.Compound Curve should avoided except in mountains.

Broken – bask curve: (unsightly and undesirable)Broken – bask curve: (unsightly and undesirable)The combination of a short length of tangent ( less than 100 ’ The combination of a short length of tangent ( less than 100 ’

) connecting two circular arcs that centers on the same ) connecting two circular arcs that centers on the same side.side.

OROR Curves consist of tow curves in the same direction Curves consist of tow curves in the same direction joined by a short tangent length.joined by a short tangent length.

Reverse Curves:Reverse Curves:Consists of two circular arcs of equal or different radii Consists of two circular arcs of equal or different radii

tangent to each other, with their centers on opposite side tangent to each other, with their centers on opposite side of the alignment.of the alignment.

Nomenclature and Formula Nomenclature and Formula for Circular Curvefor Circular Curve

D° = 100’D° = 100’ββ° = L’° = L’ D°D° = = ββ°°100 L100 LL = 100 xL = 100 x ββ°° (ft) _______ ( 1 ) (ft) _______ ( 1 ) D°D°L = R L = R ββ ______ ______ ββ in radian ( 2 ) in radian ( 2 )Because S = r Because S = r θθR = R = 5729.585729.58 ft ______ ( 3 ) ft ______ ( 3 ) DDT = R. tan T = R. tan ββ//2 2 ___________ ( 4 )___________ ( 4 )LC = 2 R Sin LC = 2 R Sin ββ//22 ________ ( 5 ) ________ ( 5 )

For L = 100 xFor L = 100 x ββ° ° ( ft ) _____ ( 1 )( ft ) _____ ( 1 ) D°D°

L = R L = R ββ ft _____ ( 2 ) ft _____ ( 2 ) ββ is in radian is in radianR = R = 5729.585729.58 ft _____ ( 3 ) ft _____ ( 3 ) R = R = 17191719 “ m ” “ m ”

DD D DT = R tan T = R tan ββ//22

∆∆AOB:AOB:

tan tan ββ//2 2 = = TT//RR

T = R tan T = R tan ββ//22 ______ ( 4 ) ______ ( 4 )

∆ ∆ OAC:OAC:Sin Sin ββ//22 = = LLCC / 2 / 2

RRLLCC = R Sin = R Sin ββ//22

22LLCC = 2 R Sin = 2 R Sin ββ//22 ____ ( 5 ) ____ ( 5 )∆ ∆ AOB:AOB:Cos Cos ββ//22 = R= RR + ER + ER + E = R + E = R _ R _ Cos Cos ββ//22 E =E = R R - R- R Cos Cos ββ//22 E = R E = R 1 1 - 1 ______ ( 6 ) - 1 ______ ( 6 )

Cos Cos ββ//22

∆ ∆ ADO:ADO:

Cos Cos ββ//22 = = R – MR – M

RR

R – M = R Cos R – M = R Cos ββ//22

R – R Cos R – R Cos ββ//22 = M = M

M = R ( 1 – Cos M = R ( 1 – Cos ββ//22 ) _____ ( 7 ) ) _____ ( 7 )

General Procedure of Circular General Procedure of Circular Curve Layout by Deflection Angles, Curve Layout by Deflection Angles,

with Theodolite and Tapewith Theodolite and TapeRadii of curves on route Radii of curves on route surveys are too large to surveys are too large to permit swing an arc from permit swing an arc from the curve center. If not the curve center. If not possible curves are possible curves are therefore laid out by:therefore laid out by:

1.1. Deflection AnglesDeflection Angles2.2. Tangent OffsetsTangent Offsets3.3. Chord OffsetsChord Offsets4.4. Middle OrdinatesMiddle Ordinates5.5. CoordinatesCoordinates

1.1. By Deflection Angles:By Deflection Angles:Layout of curves by Layout of curves by deflection angles can be deflection angles can be done using theodolite and done using theodolite and tape method.tape method.

Full Chord ( Peg Interval ):Full Chord ( Peg Interval ):Pegs are fixed in regular intervals along the curve, each Pegs are fixed in regular intervals along the curve, each interval is set to equal lengths of chords (unit chords). The interval is set to equal lengths of chords (unit chords). The curve is represented by series of chords, instead of arcs. curve is represented by series of chords, instead of arcs. In usual practice the length of the unit chord should not be In usual practice the length of the unit chord should not be more than 1/20more than 1/20thth of the radius of the curve. of the radius of the curve.In railway, curves the unit chords are generally taken In railway, curves the unit chords are generally taken between 20 – 30 m. in road curves the unit chord should between 20 – 30 m. in road curves the unit chord should be 10 m or less.be 10 m or less. Short unit chord gives more accurate curves.Short unit chord gives more accurate curves.Initial Sub chord:Initial Sub chord:Sometime the chain age of 1Sometime the chain age of 1stst tangent point works out to tangent point works out to be a very odd number to make it a round number a short be a very odd number to make it a round number a short chord is introduced at the beginning. This short chord is chord is introduced at the beginning. This short chord is called initial sub chord.called initial sub chord.Final Sub Chord:Final Sub Chord:Sometimes it is found that after introducing a number of Sometimes it is found that after introducing a number of full chords some distances still remains to be covered in full chords some distances still remains to be covered in order to reach the 2order to reach the 2ndnd tangent point. Then again a short tangent point. Then again a short chord is introduced for covering this distance is known as chord is introduced for covering this distance is known as the final sub – chord.the final sub – chord.

Incremental Incremental Chord Method Chord Method

(Rankin's (Rankin's Method):Method):

Let “ AB ” is the Let “ AB ” is the backward tangent to backward tangent to the curve. Tthe curve. T11 and T and T22 are tangent points are tangent points AA11, A, A22, A, A33, A, A44 and A and A55 are the successive are the successive points on the curve. points on the curve. For unit chord (peg-For unit chord (peg-interval). Sinterval). S11, S, S22, S, S33, , S4 and S5 are the tangential angles which each of successive S4 and S5 are the tangential angles which each of successive chord. c1, c2, c3, c4 and c5 makes with the respective tangent chord. c1, c2, c3, c4 and c5 makes with the respective tangent at T1, 1, 2 etc.at T1, 1, 2 etc.

∆∆1, ∆2, ∆3, ∆4 and ∆5 are the total tangential or deflection 1, ∆2, ∆3, ∆4 and ∆5 are the total tangential or deflection angles between backward tangent AB and each of the line. T1angles between backward tangent AB and each of the line. T1

A1, T1A1, T1 A2, T1A2, T1 A3 etc. A3 etc.c1, c2, c3, c4 and c5 are the lengths of chords.c1, c2, c3, c4 and c5 are the lengths of chords.“ “ R ” is the radius of the curves and “ O ” is its centre.R ” is the radius of the curves and “ O ” is its centre.Chord T1 A1 is nearly equal to the arc distance T1 A1 angle.Chord T1 A1 is nearly equal to the arc distance T1 A1 angle.∆∆1 = 1 = δδ1 = ½ TOA11 = ½ TOA1

Bcz Bcz δδ = R = R θθcc11 = R 2 = R 2 δδ11

2 2 δδ11 = c = c11/r/rδδ11 = = c c11

2 R2 R2 2 δδ11 = = c c11 Bcz Bcz δδ = R = R θθ

RRδδ11 = = c c11

2 R2 Rwhere “ where “ δδ11 ” is in radian ” is in radianδδ11 = = c c11 x 57.3 x 57.3 or 180/or 180/ππ

2 R2 Rδδ11 = = 90 c90 c11 in degreesin degrees

ππ R Rδδ11 = = 90 c90 c11 x 60 x 60 in minutesin minutes

ππ R Rδδ11 = = 1718.9 c1718.9 c11 in minutesin minutes

RR

Similarly,Similarly,δδ22 = = 1718.9 c1718.9 c22

RRδδ33 = = 1718.9 c1718.9 c33

RRδδNN = = 1718.19 c1718.19 cNN

RRFundamental Theorem of Fundamental Theorem of geometry is that the angle geometry is that the angle at a point between a at a point between a tangent and any chord is tangent and any chord is equal to half the central equal to half the central angle subtended by the angle subtended by the chord.chord.Now the total tangent Now the total tangent angle for first chord.angle for first chord.TT11AA11 is ∆ is ∆11 = = δδ11

for 2for 2ndnd chord T chord T11AA22

∆∆22 = = δδ11 + + δδ22

Similarly forSimilarly for3rd chord T3rd chord T11AA33

∆∆33 = = δδ11 + + δδ22 + + δδ33

If each of unit chord lengths cIf each of unit chord lengths c11, , cc22, c, c33, c, c44 and c and c55 are equal are equal then then δδ11 = = δδ22 = = δδ33 = = δδ44 = = δδ55

Then,Then,∆∆11 = = δδ11 ∆∆22 = 2 = 2 δδ11

∆∆33 = 3 = 3 δδ11

∆∆NN = 4 = 4 δδ11

Mode of ProcedureMode of Procedure1.1. Setup the theodolite over 1Setup the theodolite over 1stst tangent point “ T tangent point “ T11 ” and level it. ” and level it.2.2. With both plates clamped at zero, direct the telescope to the With both plates clamped at zero, direct the telescope to the

ranging rod at point of intersection (PI) and bisect it.ranging rod at point of intersection (PI) and bisect it.3.3. Release or unchanged the upper plate of theodolite and set the Release or unchanged the upper plate of theodolite and set the

angle ∆angle ∆11. now telescope will be along T. now telescope will be along T11 A A11..4.4. Pin down the zero end of the chain or tape at “ TPin down the zero end of the chain or tape at “ T11 ” and holding ” and holding

the arrow at distance on tape equal to the length of the 1the arrow at distance on tape equal to the length of the 1stst chord, swing the tape around “ Tchord, swing the tape around “ T11 ” until the arrow is bisected ” until the arrow is bisected by the cross – hairs. Thus fixing the 1by the cross – hairs. Thus fixing the 1stst point “ A point “ A11 ” on the curve. ” on the curve.

5.5. Unclamped the upper plate of theodolite and set the angle Unclamped the upper plate of theodolite and set the angle reading to 2reading to 2ndnd deflection angle ∆ deflection angle ∆22. the line of sight being now . the line of sight being now directed along Tdirected along T11 A A22..

Hint:Hint: For very small deflection angles like 2’ - 5”, or 5’ - 7” minutes. For very small deflection angles like 2’ - 5”, or 5’ - 7” minutes. It is difficult to set the ( 2’ - 5” ) with conventional method of It is difficult to set the ( 2’ - 5” ) with conventional method of angle measurement. Let suppose 1angle measurement. Let suppose 1stst ∆ ∆11 deflection angle is 2’ – deflection angle is 2’ – 5”.5”.

Adjust 2’ – 5” from micrometer knob, and make zero – zero on plate Adjust 2’ – 5” from micrometer knob, and make zero – zero on plate readings. Using upper slow motion screw and upper clamp final readings. Using upper slow motion screw and upper clamp final adjustments of degree can be done.adjustments of degree can be done.

(Cont…)(Cont…)

6.6. Hold the zero end of the tape at “ Hold the zero end of the tape at “ TT11 ”and swing the other end arrow ”and swing the other end arrow “ A“ A22 ” until the arrow head bisected ” until the arrow head bisected by the line of sight. Thus locating by the line of sight. Thus locating the 2the 2ndnd point on the curve. point on the curve.

7.7. Repeat the process until the end of Repeat the process until the end of the curve is reached.the curve is reached.

ProblemProblemTwo tangent intersect at chain age of 1250 m. the Two tangent intersect at chain age of 1250 m. the

angle of deflection is 30º, calculate all the data angle of deflection is 30º, calculate all the data necessary for setting out a curve of radius 250 m necessary for setting out a curve of radius 250 m by deflection angle method. The peg interval may by deflection angle method. The peg interval may be taken as 20 m. Prepare a setting out table be taken as 20 m. Prepare a setting out table when the least count of the theodolite is 20 ”. when the least count of the theodolite is 20 ”. Calculate the data for field checking.Calculate the data for field checking.

Solution:Solution:1.1. TT11 = R tan = R tan ββ//22

TT11 = 66.98 => 67 m = 66.98 => 67 m2.2. L = R L = R ββ is in radianis in radian

= 250 x 30 x = 250 x 30 x ππ 180180

= 130.89 m= 130.89 m

(Cont…)(Cont…)3.3. (P. C.) (P. C.) Chain ageChain age = (P.I.) = (P.I.) Chain ageChain age – Tangent Length “ T – Tangent Length “ T11 ” ”

= 1250 – 67= 1250 – 67 = 1183 m= 1183 m

4.4. (P.T.) (P.T.) Chain ageChain age = (Pc) = (Pc) Chain ageChain age + Curve Length “ L ” + Curve Length “ L ” = 1183 + 130.89= 1183 + 130.89 = 1313.89 m= 1313.89 m5.5. Length of Initial Sub = Let 7 mLength of Initial Sub = Let 7 m6.6. Chain age of 1Chain age of 1stst peg = 1183 + 7 = 1190 m peg = 1183 + 7 = 1190 mNo. of Intermediate Chords (Peg Intervals) = 6 ( 20 m each )No. of Intermediate Chords (Peg Intervals) = 6 ( 20 m each )Chain age Covered = 1190 + ( 6 x 20 )Chain age Covered = 1190 + ( 6 x 20 )

= 1310.0 m= 1310.0 m7.7. Length of final Sub Chord = 3.89 mLength of final Sub Chord = 3.89 m8.8. Def – angle for initial Sub ChordDef – angle for initial Sub Chordδδ11 = = 1718.9 c1718.9 c11

RR = = 1718.9 x 71718.9 x 7 = 0º 45’ 8” = 0º 45’ 8”

250250

(Cont…)(Cont…)9.9. Deflection angle for Full – Chord (Intermediate Chords)Deflection angle for Full – Chord (Intermediate Chords)δδII = = 1718.9 c1718.9 cII

RR= = 1718.9 x 201718.9 x 20

250250δδII = 2º 17’ 31” = 2º 17’ 31”10.10. Deflection angle for Final Sub – ChordDeflection angle for Final Sub – ChordδδNN = = 1718.9 x 3.871718.9 x 3.87

250250= 0º 26’ 45”= 0º 26’ 45”11.11. Arithmetical CheckArithmetical CheckTotal Deflection Angle = 14º 59’ 59” = Total Deflection Angle = 14º 59’ 59” = ββ//22 Ok Ok12.12. Field Check:Field Check:a)a) Apex distanceApex distanceE = R 1 – 1 = R ( Sec E = R 1 – 1 = R ( Sec ββ//22 – 1 ) – 1 )

Cos Cos ββ//22

= 250 1 – 1= 250 1 – 1 Cos 15ºCos 15ºb) M = R ( 1 – Cos b) M = R ( 1 – Cos ββ//22 ) )= 250 ( 1 – Cos 15º )= 250 ( 1 – Cos 15º )M = 8.52 mM = 8.52 m

Setting out Table:Setting out Table:

PoiPointnt

ChaiChainn

AgeAge

ChorChordd

LengLengthth

DeflectionDeflection

Angle forAngle for

Chord Chord LengthLength

δδ = = 1718.9 1718.9 CC

RR

TotalTotal

DeflectionDeflection

AngleAngle∆∆n = n = δδ11 + + δδ22 + …. + ….

+ + δδNN

Angle Angle toto

be set be set onon

20 ”20 ”

TheodolTheodoliteite

PPCC1183.0 1183.0

mm -------- -------- -------- --------

AA111190.0 1190.0

mm 7 m7 m 0º 48 ’ 8 ”0º 48 ’ 8 ” 0º 48 ’ 8 ”0º 48 ’ 8 ” 0º 48 ’ 0 0º 48 ’ 0 ””

AA221210.0 1210.0

mm 20 m20 m 2º 17 ’ 31 2º 17 ’ 31 ””

3º 5 ’ 39 ”3º 5 ’ 39 ”

AA331230.0 1230.0

mm 20 m20 m 2º 17 ’ 31 2º 17 ’ 31 ””

5º 23 ’ 10 ”5º 23 ’ 10 ”

AA441250.0 1250.0

mm 20 m20 m 2º 17 ’ 31 2º 17 ’ 31 ””

7º 40 ’ 41 ”7º 40 ’ 41 ”

AA551270.0 1270.0

mm 20 m20 m 2º 17 ’ 31 2º 17 ’ 31 ””

9º 58 ’ 12 ”9º 58 ’ 12 ”

AA661290.0 1290.0

mm 20 m20 m 2º 17 ’ 31 2º 17 ’ 31 ””

12º 15 ’ 43 ”12º 15 ’ 43 ”

AA771310.0 1310.0

mm 20 m20 m 2º 17 ’ 31 2º 17 ’ 31 ””

14º 33 ’ 14 ”14º 33 ’ 14 ”

AA88 / P / PTT1313.81313.8

9 m9 m 3.89 m3.89 m 0º 26 ’ 45 0º 26 ’ 45 ””

14º 59 ’ 59 ”14º 59 ’ 59 ”

Setting out Table:Setting out Table:

PoiPointnt

ChaiChainn

AgeAge

ChorChordd

LengLengthth

Full ChordFull Chord

CCNN = 2 R = 2 R Sin Sin ∆∆NN

TotalTotal

DeflectionDeflection

AngleAngle∆∆n = n = δδ11 + + δδ22 + …. + ….

+ + δδNN

Angle Angle toto

be set be set onon

20 ”20 ”

TheodolTheodoliteite

PPCC1183.0 1183.0

mm -------- -------- --------

AA111190.0 1190.0

mm 7 m7 m 7.0 m7.0 m 0º 48 ’ 8 ”0º 48 ’ 8 ”

AA221210.0 1210.0

mm 20 m20 m 26.988 m26.988 m 3º 5 ’ 39 ”3º 5 ’ 39 ”

AA331230.0 1230.0

mm 20 m20 m 46.933 m46.933 m 5º 23 ’ 10 ”5º 23 ’ 10 ”

AA441250.0 1250.0

mm 20 m20 m 66.8033 m66.8033 m 7º 40 ’ 41 ”7º 40 ’ 41 ”

AA551270.0 1270.0

mm 20 m20 m 86.566 m86.566 m 9º 58 ’ 12 ”9º 58 ’ 12 ”

AA661290.0 1290.0

mm 20 m20 m 106.1907 m106.1907 m 12º 15 ’ 43 ”12º 15 ’ 43 ”

AA771310.0 1310.0

mm 20 m20 m 125.645 m125.645 m 14º 33 ’ 14 ”14º 33 ’ 14 ”

AA88 / P / PTT1313.81313.8

9 m9 m 3.89 m3.89 m 129.4072 m129.4072 m 14º 59 ’ 59 ”14º 59 ’ 59 ”

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