CurvesCurves

Nomenclature and Formula for Nomenclature and Formula for Circular CurveCircular Curve

CurvesCurves

There are two types of curves.There are two types of curves.

1.1. Horizontal Curves (Horizontal Alignment)Horizontal Curves (Horizontal Alignment)

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

Horizontal Curves:Horizontal Curves:

oo Simple Circular CurveSimple Circular Curve

oo Compound Circular CurvesCompound Circular Curves

oo Reverse Circular CurvesReverse Circular Curves

oo Broken Back CurvesBroken Back Curves

Degree of a Curve:Degree of a Curve:

AngleAngle subtendedsubtended byby anan arcarc ofof 100100 ftft lengthlength isis calledcalled degreedegreeofof curvecurve..

CurvesCurves areare mostlymostly usedused inin transportationtransportation routes,routes, suchsuch asashighways,highways, railroadsrailroads..

SometimesSometimes pipelinespipelines areare alsoalso connectedconnected byby smoothsmoothhorizontalhorizontal oror verticalvertical curvescurves..

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

ProvidedProvided forfor safesafe andand continuouscontinuous operationsoperations atat aauniformuniform speedspeed..

TopographyTopography ofof thethe areaarea controlscontrols thethe radiusradius andand designdesignspeedspeed..

SightSight distancesdistances controlscontrols thethe curvescurves radiusradius SightSight distancesdistances controlscontrols thethe curvescurves radiusradius StraightStraight (tangents)(tangents) shouldshould bebe longlong enoughenough toto

accommodateaccommodate allall thethe parametersparameters ofof thethe sightsight distancesdistances(passing,(passing, stopping,stopping, decision)decision)..

EasementEasement curvescurves areare desirable,desirable, especiallyespecially forfor railroads,railroads,andand rapidrapid transittransit systemsystem toto lessonlesson thethe suddensudden changechange ininthethe curvaturecurvature atat thethe junctionjunction ofof aa tangenttangent andand aa circularcircularcurvecurve (spiral(spiral curves)curves)..

Components of CurvesComponents of Curves

TangentTangent isis thethe straightstraight partpart ofof thethe alignment,alignment, therethere areareinitialinitial tangentstangents andand finalfinal tangentstangents..

SpiralSpiral areare usedused toto connectconnect aa tangenttangent withwith aa circularcircularcurve,curve, aa tangenttangent withwith tangent,tangent, aa circularcircular curvecurve withwith aacircularcircular curvecurve..circularcircular curvecurve..

-- AA spiralspiral makesmakes anan excellentexcellenteasementeasement curvecurve becausebecause itsits radiusradiusdecreasesdecreases uniformlyuniformly fromfrom infinityinfinity atattangenttangent toto thatthat ofof thethe curvecurve ititmeetsmeets..

DesignationsDesignations ofof CurveCurve::

InIn EuropeanEuropean practicepractice andand majoritymajority ofofAmericanAmerican HighwaysHighways works,works, circularcircularcurvescurves areare designateddesignated byby theirtheircurvescurves areare designateddesignated byby theirtheirradiusradius..

RailRail departmentsdepartments andand somesome otherotherdepartmentsdepartments designatedesignate thethe curvescurves inindegreesdegrees..

ArcArc DefinitionDefinition::

DegreeDegree ofof thethe curvecurve isis thethe centralcentralangleangle subtendedsubtended byby aa circularcircular arcarc ofof100100 ftft..

DD == 100100

360360ºº 22 ππ RR

RR == 57295729..5858 inin ftft

DD

Chord Definition:Chord Definition:

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

Sin Sin DD = = 5050

2 R2 R2 R2 R

R = 50R = 50

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

Types of Circular CurvesTypes of Circular Curves

Simple Circular Curve:Simple Circular Curve:

AA circularcircular arcarc connectingconnecting twotwo tangentstangents mostmost oftenlyoftenly usedused typetype..

Compound Circular Curve:Compound Circular Curve:

AA curvecurve composedcomposed ofof twotwo oror moremore circularcircular arcsarcs ofof differentdifferent radiiradiitangenttangent toto eacheach other,other, withwith theirtheir centerscenters onon samesame sideside..tangenttangent toto eacheach other,other, withwith theirtheir centerscenters onon samesame sideside..CompoundCompound CurveCurve shouldshould avoidedavoided exceptexcept inin mountainsmountains..

Broken Broken –– bask curve: (unsightly and undesirable)bask curve: (unsightly and undesirable)

TheThe combinationcombination ofof aa shortshort lengthlength ofof tangenttangent (( lessless thanthan 100100 ’’ ))connectingconnecting twotwo circularcircular arcsarcs thatthat centerscenters onon thethe samesame sideside..

OROR CurvesCurves consistconsist ofof towtow curvescurves inin thethe samesame directiondirection joinedjoined byby aashortshort tangenttangent lengthlength..

ReverseReverse CurvesCurves::

ConsistsConsists ofof twotwo circularcircular arcsarcs ofof equalequal oror differentdifferent radiiradii tangenttangent toto eacheachother,other, withwith theirtheir centerscenters onon oppositeopposite sideside ofof thethe alignmentalignment..

Nomenclature and Formula for Nomenclature and Formula for Circular CurveCircular Curve

DD°° = 100’= 100’ββ°° = L’= L’DD°° = = ββ°°100 L100 LL = 100 xL = 100 x ββ°° (ft) _______ ( 1 )(ft) _______ ( 1 )

DD°°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 )DD°°

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

DD DDT = 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 + ER + ER + E = R + E = R _R _

Cos Cos ββ//22

E =E = R R -- RRCos Cos ββ//22

E = R E = R 1 1 -- 1 ______ ( 6 )1 ______ ( 6 )Cos Cos ββ//22

∆ ADO:∆ ADO:

Cos Cos ββ//22 = = R R –– MM

RR

R R –– M = R Cos M = R Cos ββ//R R –– M = R Cos M = R Cos ββ//22

R R –– R Cos R Cos ββ//22 = M= M

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

General Procedure of Circular Curve Layout General Procedure of Circular Curve Layout by Deflection Angles, with Theodolite and by Deflection Angles, with Theodolite and

TapeTape

RadiiRadii ofof curvescurves onon routeroute surveyssurveysareare tootoo largelarge toto permitpermit swingswing ananarcarc fromfrom thethe curvecurve centercenter.. IfIf notnotpossiblepossible curvescurves areare thereforetherefore laidlaidoutout byby::

1.1. DeflectionDeflection AnglesAngles1.1. DeflectionDeflection AnglesAngles2.2. TangentTangent OffsetsOffsets3.3. ChordChord OffsetsOffsets4.4. MiddleMiddle OrdinatesOrdinates5.5. CoordinatesCoordinates

1.1. ByBy DeflectionDeflection AnglesAngles::LayoutLayout ofof curvescurves byby deflectiondeflectionanglesangles cancan bebe donedone usingusingtheodolitetheodolite andand tapetape methodmethod..

Full Chord ( Peg Interval ):Full Chord ( Peg Interval ):

PegsPegs areare fixedfixed inin regularregular intervalsintervals alongalong thethe curve,curve, eacheach intervalinterval isissetset toto equalequal lengthslengths ofof chordschords (unit(unit chords)chords).. TheThe curvecurve isisrepresentedrepresented byby seriesseries ofof chords,chords, insteadinstead ofof arcsarcs.. InIn usualusual practicepractice thethelengthlength ofof thethe unitunit chordchord shouldshould notnot bebe moremore thanthan 11//2020thth ofof thetheradiusradius ofof thethe curvecurve..

InIn railway,railway, curvescurves thethe unitunit chordschords areare generallygenerally takentaken betweenbetween 2020 ––3030 mm.. inin roadroad curvescurves thethe unitunit chordchord shouldshould bebe 1010 mm oror lessless..

ShortShort unitunit chordchord givesgives moremore accurateaccurate curvescurves..

InitialInitial SubSub chordchord::InitialInitial SubSub chordchord::

SometimeSometime thethe chainchain ageage ofof 11stst tangenttangent pointpoint worksworks outout toto bebe aa veryveryoddodd numbernumber toto makemake itit aa roundround numbernumber aa shortshort chordchord isis introducedintroducedatat thethe beginningbeginning.. ThisThis shortshort chordchord isis calledcalled initialinitial subsub chordchord..

FinalFinal SubSub ChordChord::

SometimesSometimes itit isis foundfound thatthat afterafter introducingintroducing aa numbernumber ofof fullfull chordschordssomesome distancesdistances stillstill remainsremains toto bebe coveredcovered inin orderorder toto reachreach thethe 22ndnd

tangenttangent pointpoint.. ThenThen againagain aa shortshort chordchord isis introducedintroduced forfor coveringcoveringthisthis distancedistance isis knownknown asas thethe finalfinal subsub –– chordchord..

Incremental Chord Incremental Chord Method (Rankin's Method (Rankin's

Method):Method):

LetLet ““ ABAB ”” isis thethebackwardbackward tangenttangent toto thethecurvecurve.. TT11 andand TT22 arearetangenttangent pointspoints AA11,, AA22,,AA33,, AA44 andand AA55 areare thethesuccessivesuccessive pointspoints onon thethecurvecurve.. ForFor unitunit chordchordcurvecurve.. ForFor unitunit chordchord(peg(peg--interval)interval).. SS11,, SS22,, SS33,,

S4 and S5 are the tangential angles which each of successive chord. c1, c2, c3, c4 and S4 and S5 are the tangential angles which each of successive chord. c1, c2, c3, c4 and c5 makes with the respective tangent at T1, 1, 2 etc.c5 makes with the respective tangent at T1, 1, 2 etc.

∆∆11,, ∆∆22,, ∆∆33,, ∆∆44 andand ∆∆55 areare thethe totaltotal tangentialtangential oror deflectiondeflection anglesangles betweenbetween backwardbackwardtangenttangent ABAB andand eacheach ofof thethe lineline.. TT11 AA11,, TT11 AA22,, TT11 AA33 etcetc..

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 = = cc11

2 R2 R2 2 δδ11 == cc11 Bcz Bcz δδ = R = R θθ

RRδδ11 = = cc11

2 R2 Rwhere “ where “ δδ11 ” is in radian” is in radian11

δδ11 == cc11 x 57.3x 57.3 or 180/or 180/ππ2 R2 R

δδ11 = = 90 c90 c11 in degreesin degreesππ RR

δδ11 = = 90 c90 c11 x 60 x 60 in minutesin minutesππ RR

δδ11 = = 1718.9 c1718.9 c11 in minutesin minutesRR

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

RRδδ33 = = 1718.9 c1718.9 c33

RRδδNN = = 1718.19 c1718.19 cNN

RRFundamentalFundamental TheoremTheorem ofofgeometrygeometry isis thatthat thethe angleangle atat aa

Similarly forSimilarly for3rd chord T3rd chord T11AA33

∆∆33 = = δδ11 + + δδ22 + + δδ33

If each of unit chord lengths cIf each of unit chord lengths c11, c, c22, c, c33, c, c44and cand c55 are equal then are equal then δδ11 = = δδ22 = = δδ33 = = δδ44 = = δδ55

Then,Then,∆∆11 = = δδ11

∆∆22 = 2 = 2 δδ11

∆∆33 = 3 = 3 δδ11geometrygeometry isis thatthat thethe angleangle atat aapointpoint betweenbetween aa tangenttangent andand anyanychordchord isis equalequal toto halfhalf thethe centralcentralangleangle subtendedsubtended byby thethe chordchord..NowNow thethe totaltotal tangenttangent angleangle forforfirstfirst chordchord..TT11AA11 isis ∆∆11 == δδ11

forfor 22ndnd chordchord TT11AA22

∆∆22 == δδ11 ++ δδ22

∆∆33 = 3 = 3 δδ11

∆∆NN = 4 = 4 δδ11

Mode of ProcedureMode of Procedure

1.1. SetupSetup thethe theodolitetheodolite overover 11stst tangenttangent pointpoint ““ TT11 ”” andand levellevel itit..2.2. WithWith bothboth platesplates clampedclamped atat zero,zero, directdirect thethe telescopetelescope toto thethe rangingranging rodrod atat

pointpoint ofof intersectionintersection (PI)(PI) andand bisectbisect itit..3.3. ReleaseRelease oror unchangedunchanged thethe upperupper plateplate ofof theodolitetheodolite andand setset thethe angleangle ∆∆11..

nownow telescopetelescope willwill bebe alongalong TT11 AA11..4.4. PinPin downdown thethe zerozero endend ofof thethe chainchain oror tapetape atat ““ TT11 ”” andand holdingholding thethe arrowarrow

atat distancedistance onon tapetape equalequal toto thethe lengthlength ofof thethe 11stst chord,chord, swingswing thethe tapetapeatat distancedistance onon tapetape equalequal toto thethe lengthlength ofof thethe 11stst chord,chord, swingswing thethe tapetapearoundaround ““ TT11 ”” untiluntil thethe arrowarrow isis bisectedbisected byby thethe crosscross –– hairshairs.. ThusThus fixingfixingthethe 11stst pointpoint ““ AA11 ”” onon thethe curvecurve..

5.5. UnclampedUnclamped thethe upperupper plateplate ofof theodolitetheodolite andand setset thethe angleangle readingreading toto 22ndnd

deflectiondeflection angleangle ∆∆22.. thethe lineline ofof sightsight beingbeing nownow directeddirected alongalong TT11 AA22..HintHint:: ForFor veryvery smallsmall deflectiondeflection anglesangles likelike 22’’ -- 55”,”, oror 55’’ -- 77”” minutesminutes.. ItIt isis difficultdifficult

toto setset thethe (( 22’’ -- 55”” )) withwith conventionalconventional methodmethod ofof angleangle measurementmeasurement.. LetLetsupposesuppose 11stst ∆∆11 deflectiondeflection angleangle isis 22’’ –– 55””..

AdjustAdjust 22’’ –– 55”” fromfrom micrometermicrometer knob,knob, andand makemake zerozero –– zerozero onon plateplate readingsreadings..UsingUsing upperupper slowslow motionmotion screwscrew andand upperupper clampclamp finalfinal adjustmentsadjustments ofofdegreedegree cancan bebe donedone..

(Cont…)(Cont…)

6.6. HoldHold thethe zerozero endend ofof thethe tapetape atat ““ TT11 ”and”andswingswing thethe otherother endend arrowarrow ““ AA22 ”” untiluntil thethearrowarrow headhead bisectedbisected byby thethe lineline ofof sightsight.. ThusThuslocatinglocating thethe 22ndnd pointpoint onon thethe curvecurve..locatinglocating thethe 22ndnd pointpoint onon thethe curvecurve..

7.7. RepeatRepeat thethe processprocess untiluntil thethe endend ofof thethe curvecurve isisreachedreached..

ProblemProblem

TwoTwo tangenttangent intersectintersect atat chainchain ageage ofof 12501250 mm.. thethe angleangle ofofdeflectiondeflection isis 3030º,º, calculatecalculate allall thethe datadata necessarynecessary forfor settingsetting outout aacurvecurve ofof radiusradius 250250 mm byby deflectiondeflection angleangle methodmethod.. TheThe pegpegintervalinterval maymay bebe takentaken asas 2020 mm.. PreparePrepare aa settingsetting outout tabletable whenwhenthethe leastleast countcount ofof thethe theodolitetheodolite isis 2020 ””.. CalculateCalculate thethe datadata forforfieldfield checkingchecking..

SolutionSolution::1.1. TT11 == RR tantan ββ//22

TT11 == 6666..9898 =>=> 6767 mm2.2. LL == RR ββ isis inin radianradian

== 250250 xx 3030 xx ππ180180

== 130130..8989 mm

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

= 1250 = 1250 –– 6767= 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 m

5.5. Length of Initial Sub = Let 7 mLength of Initial Sub = Let 7 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 mpeg = 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 Def –– angle for initial Sub Chordangle 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 Deflection angle for Full –– Chord (Intermediate Chords)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 Deflection angle for Final Sub –– ChordChord10.10. Deflection angle for Final Sub Deflection angle for Final Sub –– ChordChordδδNN = = 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 OkOk12.12. Field Check:Field Check:a)a) Apex distanceApex distanceE = R 1 E = R 1 –– 1 = R ( Sec 1 = R ( Sec ββ//22 –– 1 )1 )

Cos Cos ββ//22

= 250 1 = 250 1 –– 11Cos 15ºCos 15º

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

Setting out Table:Setting out Table:

PointPointChainChain

AgeAge

ChordChord

LengthLength

DeflectionDeflection

Angle forAngle for

Chord LengthChord Length

δδ == 1718.9 C1718.9 C

RR

TotalTotal

DeflectionDeflection

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

Angle toAngle to

be set onbe set on

20 ”20 ”

TheodoliteTheodolite

PPCC 1183.0 m1183.0 m -------- -------- -------- --------

AA 1190.0 m1190.0 m 7 m7 m 0º 48 ’ 8 ”0º 48 ’ 8 ” 0º 48 ’ 8 ”0º 48 ’ 8 ” 0º 48 ’ 0 ”0º 48 ’ 0 ”AA11 1190.0 m1190.0 m 7 m7 m 0º 48 ’ 8 ”0º 48 ’ 8 ” 0º 48 ’ 8 ”0º 48 ’ 8 ” 0º 48 ’ 0 ”0º 48 ’ 0 ”

AA22 1210.0 m1210.0 m 20 m20 m 2º 17 ’ 31 ”2º 17 ’ 31 ” 3º 5 ’ 39 ”3º 5 ’ 39 ”

AA33 1230.0 m1230.0 m 20 m20 m 2º 17 ’ 31 ”2º 17 ’ 31 ” 5º 23 ’ 10 ”5º 23 ’ 10 ”

AA44 1250.0 m1250.0 m 20 m20 m 2º 17 ’ 31 ”2º 17 ’ 31 ” 7º 40 ’ 41 ”7º 40 ’ 41 ”

AA55 1270.0 m1270.0 m 20 m20 m 2º 17 ’ 31 ”2º 17 ’ 31 ” 9º 58 ’ 12 ”9º 58 ’ 12 ”

AA66 1290.0 m1290.0 m 20 m20 m 2º 17 ’ 31 ”2º 17 ’ 31 ” 12º 15 ’ 43 ”12º 15 ’ 43 ”

AA77 1310.0 m1310.0 m 20 m20 m 2º 17 ’ 31 ”2º 17 ’ 31 ” 14º 33 ’ 14 ”14º 33 ’ 14 ”

AA88 / P/ PTT 1313.89 m1313.89 m 3.89 m3.89 m 0º 26 ’ 45 ”0º 26 ’ 45 ” 14º 59 ’ 59 ”14º 59 ’ 59 ”

Setting out Table:Setting out Table:

PointPointChainChain

AgeAge

ChordChord

LengthLength

Full ChordFull Chord

CCNN = 2 R Sin = 2 R Sin ∆∆NN

TotalTotal

DeflectionDeflection

AngleAngle

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

Angle toAngle to

be set onbe set on

20 ”20 ”

TheodolitTheodolitee

PPCC 1183.0 m1183.0 m -------- -------- --------

AA 1190.0 m1190.0 m 7 m7 m 7.0 m7.0 m 0º 48 ’ 8 ”0º 48 ’ 8 ”AA11 1190.0 m1190.0 m 7 m7 m 7.0 m7.0 m 0º 48 ’ 8 ”0º 48 ’ 8 ”

AA22 1210.0 m1210.0 m 20 m20 m 26.988 m26.988 m 3º 5 ’ 39 ”3º 5 ’ 39 ”

AA33 1230.0 m1230.0 m 20 m20 m 46.933 m46.933 m 5º 23 ’ 10 ”5º 23 ’ 10 ”

AA44 1250.0 m1250.0 m 20 m20 m 66.8033 m66.8033 m 7º 40 ’ 41 ”7º 40 ’ 41 ”

AA55 1270.0 m1270.0 m 20 m20 m 86.566 m86.566 m 9º 58 ’ 12 ”9º 58 ’ 12 ”

AA66 1290.0 m1290.0 m 20 m20 m 106.1907 m106.1907 m 12º 15 ’ 43 ”12º 15 ’ 43 ”

AA77 1310.0 m1310.0 m 20 m20 m 125.645 m125.645 m 14º 33 ’ 14 ”14º 33 ’ 14 ”

AA88 / P/ PTT 1313.89 m1313.89 m 3.89 m3.89 m 129.4072 m129.4072 m 14º 59 ’ 59 ”14º 59 ’ 59 ”

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