layout calculation book

Layout

Calculations

Foreword

The maintainability and riding quality over a turnout depend largely onhow accurately it is laid and maintained. Deficiencies can be nipped inthe bud if adequate care is taken by laying turnouts accurately aftercarrying out layout calculations.

Layout calculations, as the name indicates, are the set of calculationsfor the various yard layouts so that the same can be correctly laid in thefield. Layout calculations become more important in case of yardremodelling or designing a new yard. These calculations are intricate innature and require considerable efforts on the part of field engineerswhich often get neglected due to other engagements in the field andtherefore, this item of work does not get proper attention.

Railway Engineers had been expressing the need for bringing out abook on ‘‘Layout Calculations’’ which was out of print since long.Special efforts have been taken to make this book more effective byincorporating colored drawings of the layouts developed by usingAutoCad software. Worked out examples on practical situationscommonly met with, have also been given for better understanding ofthis subject.

Another special feature of the book is the inclusion of a software onLayout Calculations, which had been developed by Shri M. S. Ekbote,Ex. AM(Civil Engineering), Railway Board, using Visual Basic withinteractive and user friendly interface.

I hope that this book on Layout Calculations will go a long way infulfilling the need of railway engineers for better designing and correctlaying of layouts.

Shiv KumarDirectorIRICEN

Acknowledgements

Subject of “Layout Calculations” is being covered in various coursesbeing held at IRICEN. In every course at IRICEN, trainee officers haveexpressed the need of a book on Layout Calculations as the earlierpublication had become out of print. Besides, several developments hadtaken place in the field of turnout such as PSC layouts. Considering therequirement of the field engineers, the entire book has been rewritten.Apart from the theory and illustrations the book contains several newfeatures such as coloured drawings, interpretations of trigonometricformulae and a software on Layout Calculations

We are very much grateful to Shri Abhay Kumar Gupta, Professor/Track -1 for the proof checking of the entire book.

We are also very much grateful to Shri M.S. Ekbote, Ex. AM(CivilEngineering), Railway Board, for his active support for rewriting thechapter on Scissors Cross-over and developing a Software Program onLayout Calculations which is available on a CD attached with this book.

Above all, the authors are very much grateful to Shri Shiv Kumar,Director, IRICEN for his encouragement and guidance.

Abhai Kumar Rai Praveen KumarProfessor/Works Professor/Computers

IRICEN IRICEN

Common Abbreviations used in the Book

SJ/SRJ Stock joint/stock rail joint

TTS Theoretical toe of switch

ATS Actual toe of switch

βββββ Switch angle

SL Actual switch length

TSL Theoretical switch length

t Designed thickness of the switch at toe

d Heel divergence

ANC Actual nose of crossing

TNC Theoretical nose of crossing

HOC Heel of crossing

w Length of straight leg of crossing ahead of TNCupto the tangent point of lead curve

F Crossing angle

G Gauge of the track

D Distance between the track

Rm Radius of the outer rail of curved main line

Rc Radius of the outer rail of the turn in curve/connecting curve

R Radius of the outer rail of the lead curve

OL Over all length of the layout

S Length of straight portion outside the turnout

A Distance from ‘SJ’ to the point of intersection in aturnout measured along the straight

B Distance from the point of intersection to the heelof crossing measured along the straight

K Distance from TNC of the crossing to the heel ofcrossing measured along the straight

CONTENTS

1. Turnouts 1

2. Lead and Radius of IRS Turnout 16

3. Connections to Diverging Tracks 24

4. Connections to Straight Parallel Tracks 35

5. Crossover Connection between Straight Parallel 51Tracks

6. Scissors Cross-Over between Straight Parallel Tracks 75

7. Crossovers between Non Parallel Straight Tracks 80

8. Connection between Curved Track to Parallel 83Curved Track or Divergent Straight Track

9. Crossover Connection between two Curved Parallel 111Tracks

10. Software on Layout Calculations 119

Table of Detailed Dimensions 127

Figure 1.1: A Turnout

1

Chapter 1

Turnouts

1.0 Introduction

Turnouts provide the means by which a train may be transferred fromone track to another track. To be more specific, as per ORE report onQ.No. D.72, the term ‘Turnout’ means a layout permitting the passageof rolling stock on to two or more routes from one common route. Inits simplest form, i.e. where only two routes are involved, it consists ofa pair of switches (one right hand and the other left hand) and acommon crossing assembly ( composed of a common crossing, twowing rails and two check rails), together with lead rails connecting thetwo routes. As desired, the crossing ensures unobstructed flange wayclearances for the wheels at the intersection between the left hand railof one route and the right hand rail of the other.

Turnouts are, therefore, the most sophisticated component of railwaytrack structure. To design and perfect different types of turnouts tomeet the varied requirements of operations has always been achallenging task for the permanent way engineer. Turnouts may takeoff from a straight track or a curved track.

A turnout has, therefore, three distinct portions (Fig 1.1);• Switch Assembly• Lead Assembly• Crossing Assembly

Main Dimensions of a turnout are as shown in Fig 1.2.

Figure 1.2: Main dimensions of a turnout

The switch assemblies in use on Indian Railways are classified purelybased on their geometry as shown below:

• IRS- Straight Switches (Fig 1.3)

Figure 1.3: Straight switches

2

3

• IRS- Partly Curved Switches (Fig 1.4)

Figure 1.4: Partly curved switches

• IRS-Curved Switches

The curved switches are further classified into the followingtypes:

• Non - Intersecting type (Fig 1.5a)

Figure 1.5a : Non- intersecting type

4

• Intersecting type (Fig 1.5b)

Figure 1.5b : Intersecting type

• Tangential type (Fig 1.5c)

Figure 1.5c : Tangential type

Crossing may either be of the ‘Straight’ type or of the ‘Curved’ typeas shown in Fig 1.6.

Straight crossing Curved crossing

Figure 1.6: Type of crossing

In the latter case, one of the legs of the crossing is curved to the sameradius as the lead curve, or in other words, the lead curve continuesthrough the crossing. This naturally results in a flatter lead curve thanin the case of a straight crossing for a given crossing angle. However,on Indian Railways only straight crossings are in use. Crossing can beof two types on the basis of material i.e. either Built Up (BU) or CastManganese Steel Crossing (CMS) crossing.

Lead curve has to be tangential to the switch at its heel and to the cross-ing at its toe so as to avoid kinks in the geometry. Subject to these twoconstraints, the lead curve may take one of the following forms;

• Simple Circular Curve• Partly Curved, having a straight length near the crossing• Transition Curve

By combining different types of switches and crossings with differentforms of lead curves, a large variety of geometrical layouts can be ob-tained, each with its own characteristics of lead length and lead radius.The standard practice in India has so far been to use one particularswitch with a particular crossing.

5

The standard Layouts of various Turnouts in use on Indian Railwaysare shown in Table 1.1

Table 1.1

S.No Crossing Types of SwitchesAngle B.G. M.G.

1. 1 in 8.5 Straight Curved Straight Curved2. 1 in 12 Straight Curved - Partly Curved3. 1 in 16 - Curved - Curved4. 1 in 20 - Curved - -5. 1 in 24* - Curved - -

* Under Standardisation

1.1 Layout Calculations

A yard may require various combinations of turnouts (i.e. switches,leads & crossings), with curves and straights, to transfer trains from onetrack to another or enable trains to cross other tracks. Depending uponthe requirements, these combinations are known as layouts. Forsatisfying the various geometrical features of the layouts, one has toperform variety of trigonometric calculations and hence the name‘Layout Calculations’.

Understanding of Layout Calculations is necessary for fixing thecorrect position of turnouts with respect to the existing tracks in case ofremodelling of an yard or for designing an altogether new layouts.Turnouts can be fixed either by locating Stock Rail Joint(SRJ) or bylocating Theoretical Nose of Crossing (TNC). SRJ/TNC can be locatedby ‘Centre Line, ‘Outer Rail’ or ‘Graphical’ method .

In ‘Centre Line’ method, the turnout is represented by center lines ofstraight and turnout side. This method is primarily used for locating SRJfor the cases when turnouts are taking off from Straight tracks.

6

7

In ‘Outer Rail’ method, the turnout is represented fully by drawing theleft rail and right rail of the turnout. This method is primarily used forlocating TNC for the cases when turnouts are taking off from the curvedtracks. However, this method is more versatile but cumbersome and canalso be used for the cases when turnouts are taking off from straighttracks.

In ‘Graphical’ method, the entire drawing will be made on computerby use of relevant drafting software like AutoCAD. After replicatingthe layout on the computer, all the calculations can be performedgraphically. Now a days various advanced softwares are available.MX-Rail is one of the specialized software developed for the railwayindustry. Yard calculations are one of the powerful feature of thissoftware

8

Provisions of Indian Railway Permanent Way Manual (IRPWM)

Layout Calculations should be performed keeping in view the relevantprovisions as contained in para 410(2) & 410(3)(c) of IRPWM. Thesame has been reproduced for ready reference.

Para 410(2) of IRPWM

Turnouts on running line with passenger traffic-Turouts in runninglines over which passenger trains are received or despatched should belaid with crossing, not sharper than 1 in 12 for straight switch. How-ever, 1 in 8.5 turnout with curved switches may be laid in exceptionalcircumstances, where due to limitation of room, it is not possible toprovide 1 in 12 turnouts. Sharper crossings may also be used when theturnouts is taken off from outside of a curve, keeping the radius of leadcurve within the following limits:

Gauge Minimum Radius of Lead Curve

BG 350 m

MG 220 m

NG 165 m

Where it is not practicable to achieve the radius of curvature of turn incurves as specified above on account of existing track centres for turn-out taking off from curves, the turn in curves may be allowed upto aminimum radius of 220m for BG and 120m for MG subject to the fol-lowing:

a) Such turn in curves should be provided on PSC or steel troughsleepers only, with sleeper spacing same as that from the mainline.

b) Full ballast profile should be provided as for track for mainline.

Emergency crossovers between double or multiple lines whch are laidonly in the trailing direction may be laid with 1 in 8.5 crossings.

9

In the case of 1 in 8.5 turnouts with straight switches laid on passengerrunning line, the speed shall be restricted to 10 Kmph. However, on 1in 8.5 turnouts on non passenger running lines, sped of 15 Kmph maybe permitted.

Para 410(3)(c) of IRPWM

The permissible speed on turnouts taking off from inside of the curveshould be determined by taking into consideration the resultant radiusof lead curve which will be sharper than the lead curve for turnoutstaking off from the straight. 1 in 8.5 turnouts should not be laid oninside of curves.”

Note:

Turn in Curve means the connecting curve starting after the heel of thecrossing. This connecting curve may either be simple circular curve,compound curve or reverse curve.(Fig 1.7)

Lead Curve or Turnout Curve (Fig 1.7) means the curve starting fromheel or toe of switches (Straight or curved) and extended upto toe ofcrossing. This value of Lead Curve Radius will become sharper if theturnout is taking off from inside of the main line and will become flatterif the turnout is taking off from outside of the main line.

Fig 1.7

10

Turnouts Taking off from Curved Tracks

Turnouts can take off from the main line track either in the similarflexure or contrary flexure.

Dm is the degree of main line curve

Ds is the degree of turnout curve (Lead Curve) when taking off from

straight (Fig 1.8a)

Figure 1.8a : Turnout taking off from a straight track

Dt is the degree of turnout curve (Lead Curve) when taking off from

a curved main line

Exact calculations for lead of lead curves for turnouts taking off fromthe curved main line are complicated in nature and time consuming.The difference in exact calculations and the formulae normally adoptedfor such layouts is very small and can be ignored for all the practicalpurposes. But resultant degree of curvatures of lead curves of turnoutstaking off from curved main lines vary considerably.

11

When Ds and D

m deflect in the same direction, it is known as a Similar

Flexure layout. In this type of layout, crossing lie on the inner rail ofthe curved main line (Fig 1.8b). Resultant degree of curvature of suchleads can be calculated by D

t=D

s+D

m

Figure 1.8b : Similar flexure

And if Dm deflects in a direction opposite to D

s, then it is known as

Contrary Flexure. In this type of layout, crossing lie on the outer railof the curved main line. (Fig 1.8c). Resultant degree of curvature ofsuch leads can be calculated by D

t=D

s-D

m

Figure 1.8c : Contrary flexure

12

When Dt and D

m are equal in a contrary flexure, then it is known as

Symmetrical Split. (Fig 1.8d)

Figure 1.8d : Symmetrical split

Sometimes it may so happen that crossing lie on the outer rail of thecurved main line and even then deflecting in the same direction as thatin the similar flexure.(Fig 1.8e) Resultant degree of curvature of suchleads can be calculated by D

t=D

m-D

s

Figure 1.8e : Contrary flexure

13

From these formulae, resultant radii of lead curves can easily be calcu-lated. Para 410(2) limits the value of the resultant radii of lead curveswhich can be calculated from the above formulae. It is from this limi-tation that 1 in 8.5 turnouts can not be laid on inside of the curved mainline because in such layout, resultant radii of curvature of the leadcurves will not satisfy the limits as given in the Para 410(2). Sharpercrossings i.e. more than 1 in 12 can be laid when turnouts taking offfrom outside of the main line because of the fact that resultant radii ofsuch leads will become flatter and may satisfy the limits as given in Para410(2)

Radius of the connecting curve or Turn in Curve will be calculatedfrom the various trignometrical formulae derived in the subsequentchapters of this book.

14

1.2 Representation of a Turnout on Centre Line

Draw the centre line of the straight main track and on it drop NN1

perpendicular from N the TNC of the Xing. Draw the centre line of theturnout track over the crossing length and extend it to meet the centreline of the straight track at P. Drop NN

2 perpendicular from N on this

centre line. (Fig 1.9)

Figure 1.9 : Representation of a turnout on centre line

On the centre line the turnout is represented by two lines OPN1W and

PN2Z. In this OW represents the overall length of turnout from SJ to the

heel of crossing along the gauge line on which the crossing lies. Tolocate the turnouts on centre line method, it will be necessary to knowthe different components of centre line representation.

OP=A

PN1=PN

2=M

N1W=N

2Z=K

PW=PZ=B=M+K

Where A, M & K are known as turnout parameters.

15

Now, let us discuss the characteristics of these turnout parameters:

‘M’ is the distance from ‘P’ to the stock joint and can be found out asexplained below:

2Fcot 2GM

M

2G

PN

NN2Ftan N,��

1

11

=∴

==

For a particular gauge of track ‘G’ & angle of crossing ‘F’, ‘M’ willbecome fixed.

‘K’ is the length of the back leg of crossing and will be dependent upontype of crossing i.e. BU or CMS crossing. Therefore for a particulartype of crossing chosen for a yard, value of ‘K’ will be fixed.Likewise, B=M+K will also become fixed.

‘A’ is the distance between ‘SJ’ to ‘TNC’ which is basically dependentupon angle of crossing and not on type of switches or else whole yardwill have to be redesigned in case of adopting new design of switches.As discussed earlier that value of ‘M’ is fixed for a particular gauge oftrack & angle of crossing and hence distance between SJ to P i.e ‘A’will also be fixed.

In view of the above, it is therefore obvious that for a given gauge,crossing angle and crossing type (BU or CMS), value of these turnoutparameters i.e. ‘A’, ‘M’ & ‘K’ are fixed. This can be convenientlyunderstood as a Black Box (see Fig 1.9) containing a geometrical figurethe dimensions of which are fixed. Then the rest is to fix this Black Boxwith respect to the existing yard geometry so as to satisfy all thegeometrical conditions of the yard.

16

Chapter 2

Lead and Radius of IRS Turnout

2.1 IRS Turnout with Straight Switches

The lead curve in IRS turnout with straight switches are placedtangential to the tongue rail at the heel and to the front straight leg ofthe crossing. (Fig 2.1)

Figure 2.1: IRS Turnout with straight switches

17

Formulae

( )

( )

angleswitch � TNC, of ahead crossing of legstraight w

divergence heeld curve, lead of radiusR

(2.2)

2

�FSin

2

�F2Sin

wSinFdGR

2

�FSin

wSinFdG

2

�-F2RSin b; 2.1 & 2.1a Eq equating

b)1.2(

2

�FSin

wSinFdG

2

�FSin

BCBK ��in also

)a1.2(2

�-F2RSinBK

ROKOB �-FBOK ��In

(2.1)wCosF2

�FCotwSinFdG Lead

PEKCPEDPDELead

2

�FCotwSinFdG

2

�FBCCotKC

wSinFdGKPABADCDABADBC

2

�F

2

�-F-FBKC BKC; �In

2

�-FMKBMBK

length) tangent being(Each MK BM BMK; In

==

==

−+−−

=∴

+−−

=

+−−

=+

=

=

===∠

++

−−=

+=+==

+−−=

+=

−−=−−=−−=

+==∠

=∠=∠

=∆

Where

18

2.2 Offsets to Lead Curves for IRS Turnout with StraightSwitches

The lead curve is extended from heel at point ‘B’ to a point ‘H’ so thatthe tangent to the curve runs parallel to the gauge line at a distance ‘Y’as shown in Fig 2.2.

Figure 2.2: Offsets to lead curves for IRS turnout withstraight switches

19

The point ‘H’ has been shown to lie inside the track, but in certainlayouts, depending on the switch angle and the radius, the point ‘H’may lie outside the track and therefore the value of ‘Y’ will work outas negative. The distance ‘BQ’ be denoted by ‘L’.

( )

( )

RSin�L

Note

=

=

=

∆

=∴

+=+

+=+=

+=+=

+−=∴

−===

+=

==

=∠=∠=

∆

from derived becan however , L'' uniform.not are

region in the s variationas Cos� of luescorrect va get the todifficult

isit small, very is ��' of value theas that iscaution of wordaBut

(2.8)�Cos-1R-dY

(2.7)RSin�L

OBQ, fromdirectly

Y'' & L'' of out values work topossible also isIt :

(2.6)CosF-1R-wSinF-GY

wSinF-GRCosFYR

(2.5), & (2.4) equating

(2.5)wSinF-GRCosFJIOJOI also,

(2.4)YRHIOHOI

)3.2(2

�FwSinF)Cot-d-(GRSinFCJ

CKJKLBQCJ

2

�FwSinF)Cot-d-(GCK

RSinFOKSinFJK

90OJK F,KOJ R,OK

KOJ, In

0

20

Example 2.1

Calculate the lead and the radius of a 1 in 8.5 IRS turnout with straightswitches.

Given: G=1676mm, d=136mm, w=864mm

"27'3401� , "35'4206F ==

( )

( )

222358mmm222358.34m

2

"27'3401"35'4206Sin

2

"27'3401"35'4206Sin2

"35'420Sin66481366761

2

�FSin

2

�F2Sin

wSinFdGR

20730mm20729.87mm858.0819871.79

"35'420Cos6648

2

"27'3401"35'4206Cot"35'420Sin66481366761

wCosF2

�FCotwSinFdG Lead

≈=

−+×

×−−=

−+−−

=

≈=+=

×+

+×−−=

++

−−=

21

2.3 IRS Turnout with Curved Switches

The lead curves in these layouts at toe of switches are tangential to theswitch angle and meets the straight leg of crossing at a distance ‘w’from the TNC of the crossing. (Fig 2.2)

Figure 2.3: IRS Turnout with curved switches

22

At toe of switch, thickness of tongue rail is ‘t’. Derivation for lead curveradius will be same as for IRS straight switches. The same can bederived by substituting ‘t’ (toe thickness) for ‘d’ (the heel divergence).

Formulae

(2.10)

2

�FSin

2

�F2Sin

wSinF-t-GR Curve, Lead of Radius

(2.9)2

�FwSinF)Cot-t-(GCK

−+=

+=

For fixing positions of heel, it is necessary to find the point whereoffset to the lead curve from main line will be equal to heeldivergence ‘d’. For this, same principles are applied as for findingoffsets to lead curve for straight IRS turnouts. It is, however,cautioned that the new tangent is drawn after extending the curve, soas to lie parallel to the main line, may be outside the track anddistance ‘Y’ may come out as negative or positive. It has to be appliedwith its positive or negative sign arithmetically.

( )

(2.15)wCosFSL2�F

wSinF)Cot-t-(GLead

(2.14)L2Y)-(d-Y)-2R(dSL Length,Switch

(2.13) CosF-1R-wSinF-GY

(2.12) RSin�BQL

OQB, from ,

(2.11) 2�F

wSinF)Cot-t-(G-RSinFCK-KJCJBQL

+−+=

−=

=

==

∆

+====

or

23

Example 2.2

Calculate the lead and the radius of a 1 in 12 IRS turnout with curvedswitches with CMS crossing on 52 Kg PSC sleepers as per RDSO/T-4732.

Given: G=1673mm, d=175mm, w=1877mm"0'2000� , "49'4504F ==

mm Length is SwitchT-From RDSO/NOTE

e is this valuO/T-As per RDS

ero.taken as zis t' l. Hence 'tongue raiss of the of thickne

tion no project there isail so thae tongue ro house thmachined t

is tock Rail witches, Sh curved surnout wit IRS t in In

mme is this valuO/T-As per RDS

101254732:

)mm258314732( 25831.62

10125-"49'450Cos48771

2

"0'2000"49'4504)Cot"49'450Sin41877-0-(1673

Length Switch -wCosF2�F

wSinF)Cot-t-(GLead

121

)4413604732( mm441374

2

"0'2000"49'4504Sin

2

"0'2000"49'45042Sin

"49'450Sin41877-0-1673

2

�FSin

2

�F2Sin

wSinF-t-GR

=

×+

+×=

++=

=

−+

×=

−+=

24

Chapter 3

Connections to Diverging Tracks

3.1 Connections to Diverging Tracks

Connection to non parallel sidings from an existing line can be of threetypes depending upon the comparative values of ‘θ’ and ‘F’. Where ‘θ’is the angle of intersection between the two lines and ‘F’ is the angleof crossing used for connection. There may be three situations viz;

F� and

F�

F�

<

>

=

Now because of the local obstructions on the existing line, there can befurther two sub categories namely; i.e Without obligatory points andWith obligatory points on the main line. Therefore, connections todiverging tracks may have following situations;

For θθθθθ = F

Case I Without obligatory point on main lineCase II With obligatory point on the main line

For θθθθθ > F

Case III Without obligatory point on main lineCase IV With obligatory point on the main line

For θθθθθ < F

Case V Without obligatory point on main lineCase VI With obligatory point on the main line

25

Case I

Produce the divergent track to meet the existing main line at point ‘I’.For making the connection, point ‘P’ should coincide with point ‘I’.Now for locating SJ (Stock Joint) an offset equal to ‘A’ can be takenfrom ‘P’. After locating the ‘SJ’,the turnout can be laid with referenceto ‘SJ’ (Fig 3.1)

Figure 3.1: θ = F & Without obligatory point on amin line

Case II

If because of some obstructions on the main line, the ‘SJ’ can not befixed as given in Case I, and ‘SJ’ can be located either on left or rightof ‘SJ’ as fixed vide Case I, then for making the connection, a reversecurve will have to be introduced starting after back leg of crossing andjoining the divergent track. For this case, there can be numeroussolutions and we have to choose the optimum alignment to meet thesite conditions. (Fig 3.2)

26

Figure 3.2: θ = F & With obligatory point on main line

Case III

In this case, if we are having the liberty in fixing the ‘SJ’, it should beon the left hand side of point of intersection ‘I’. ( Fig 3.3)

Figure 3.3: θ > F & Without obligatory point on main line

27

Formulae

(3.4)TSin�T)SinF(BY

(3.3)AXOL

(3.2)TCos�T)CosF(BX

(3.1)2

F�RtanT

++=

+=

++=

−=

Interpretation of Formulae & Field Practicalities

Intersection angle ‘θ’ has to be found out from the field surveying. Theradius ‘R’ of the connecting curve has to be assumed. Normally thevalue of connecting curve radius is taken equal to that of radius of turnoutcurve. Now value of ‘T’ can be calculated from Eq:3.1. Once ‘T’ isknown, the value of ‘X’ & ‘Y’ can be calculated fron Eq:3.2 & 3.4respectively. Having thus calculated the value of ‘X’ & ‘Y’, ‘TP

2’ is

located first as the intersection point between the divergent track and aline drawn parallel to the existing track at a distance ‘Y’ and the locationof ‘SJ’ is marked with reference to TP

2 by drawing a perpendicular offset

at a distance ‘OL’ from ‘TP2.’ (Fig 3.3a)

Figure 3.3a

Now after locating ‘SJ’, the turnout is linked and then the connectingcurve is provided with radius equal to ‘R’. This connecting curve willbe starting from back of crossing (TP

1) and ending at ‘TP

2’, thus

establishing the full connection.

Case IV

In Case III, radius ‘R’ of the connecting curve was assumed and thelocation of ‘SJ’ was so fixed. But, in certain circumstances, it may notbe possible to locate ‘SJ’ because of some obstructions/obligatorypoints falling at that location. Position of ‘SJ’ will thus be fixed eitherto left or right side of ‘SJ’ as fixed in Case III.

Figure 3.4: θ > F & With obligatory point on main line

When the ‘SJ’ is fixed to left of ‘SJ’ (Fig 3.4) as fixed vide Case III,radius of connecting curve will be quite large and can be calculated bythe following formula;

(3.5)

2

F�tan

TR

−=

by substituting the value of ‘T’, which will be calculated by fieldsurveying, i.e. by extending crossing leg on the turnout side to intersect

28

29

divergent track at ‘Z’. Thus TP1Z will be the tangent length ‘T’. Now

the connecting curve of radius ‘R’ can be laid at the site. In this case,the radius ‘R’ of the connecting curve will be large, which can bereduced by providing a straight after the heel of crossing andconnecting curve starting after this straight. For this case formulae canbe modified as follows;

(3.9)TSin�T)SinFS(BY

(3.8)AXOL

(3.7)TCos�T)CosFS(BX

(3.6)2

F-�RtanT

+++=

+=

+++=

=

Figure 3.5: θ < F & Without obligatory point on main line

Now in another situation, when ‘SJ’ is located on right of ‘SJ’ (Fig 3.4)as fixed vide case III, there can be numerous solutions and an optimumalignment can be decided keeping in view the site conditions. In thiscase, connecting curve will have to be a reverse curve.

Case V

In this case, if we are having the liberty in fixing the ‘SJ’, it should beon the right hand side of point of intersection ‘I’. (Fig 3.5)

30

Formulae

(3..13)TSin�T)SinF(BY

(3.12)AXOL

(3.11)TCos�T)CosF(BX

(3.10) 2

�-FRtanT

++=

+=

++=

=


Intersection angle‘θ’ has to be found out from the field surveying. Theradius ‘R’ of the connecting curve has to be assumed. Normally thevalue of radius ‘R’ is taken equal to that of the radius of turnout radius.Now value of ‘T’ is calculated from Eq 3.9. Once ‘T’ is known, thevalue of ‘X’, ‘OL’ & ‘Y’ can be calculated from Eq 3.11, 3.12 & 3.13respectively. ‘TP

2’ is located first as the intersection point between the

divergent track and a line drawn parallel to the existing track at adistance ‘Y’ and the location of ‘SJ’ is marked with reference to ‘TP

2’

by drawing a perpendicular offset at a distance ‘OL’ from ‘TP2’

(Fig 3.5a)

Figure 3.5a

θ

31

Now after locating ‘SJ’, the turnout is linked and then the connectingcurve is provided with a radius equal to ‘R’. Thus connecting curve willbe starting from back of crossing (TP

1) and ending at ‘TP

2’, thus

establishing the full connection.

Case VI

In case V, radius of ‘R’ of the connecting curve was assumed and thelocation of ‘SJ’ was to be fixed. But in certain circumstances, it may notbe possible to fix ‘SJ’ because of some obstructions/obligatory pointsfalling at that location. Position of ‘SJ’ will thus be fixed either to theleft or right side of ‘SJ’ as fixed in Case IV. ( Fig 3.6)

When the ‘SJ’ is to be located on right of ‘SJ’ as fixed vide Case V,radius ‘R’ of connecting curve will be quite large and can be calculatedby the following formulae;

Figure 3.6: θ < F & With obligatory point on main line

(3.14)

2

�-Ftan

TR =

32

by substituting the value of ‘T’, which will be calculated by fieldsurveying, i.e by extending crossing leg of the turnout side to intersectdivergent track at ‘Z’. TP

1 will be the tangent length ‘T’. Now the

connecting curve of radius ‘R’ can be laid at the site. In this case, radius‘R’ of the connecting curve will be large, which can be reduced byproviding a straight after the heel of crossing and connecting curvestarting after this straight. For this case formulae can be modified asfollows:

(3.18)TSin�T)SinFS(BY

(3.17)AXOL

(3.16)TCos�T)CosFS(BX

(3.15)2

�-FRtanT

+++=

+=

+++=

=

Now, in another situation, when ‘SJ’ is located on right of ‘SJ’ as fixedvide Case V, there can be numerous solutions and an optimumalignment can be decided keeping in view the site conditions. In thiscase, connecting curve has to be a reverse curve.

Example 3.1

A broad gauge siding is required to be connected to a main line trackusing a 52 Kg, 1 in 12 T/O. The angle of intersection between the twotracks being 100. Calculate the required distances for the layoutassuming that the connecting curve starts from the heel of crossing.

Given: F=4045’49”, A=16.953m, B=23.981m, R=441.282m

33

7.171m

3.5043.667

n10(20.179)Si"49'45420.179)Sin(23.981

TSin�T)SinF(B Y

80.832m16.95363.879AXOL

63.879m

19.87244.007

s10(20.179)Co"49'45420.179)Cos(23.981

TCos�T)CosF(B X

20.179m

2

"49'454-10441.282tan

2

F�Rtan T

00

00

00

=

+=

++=

++=

=+=+=

=

+=

++=

++=

=

=

−=

34

Example 3.2

A broad gauge siding is required to be connected to a main line trackusing 90 R, 1 in 8.5 T/O. The angle of intersection between the twotracks is 30. Calculate the required distances for the layout.

Given: F=6042’35”, A=12.000m, B=17.404m, R=221.522m

m248.3

375.0873.2

03(7.174)Sin"53'2407.174)Sin67.4041(

TSin�T)SinF(B Y

m573.45000.12573.31AXOL

31.573m

164.7409.24

03(7.174)Cos"53'2407.174)Cos67.4041(

TCos�T)CosF(B X

m174.7

2

03"53'2406tan522.212

2

�-FRtan T

=

+=

++=

++=

=+=+=

=

+=

++=

++=

=

−=

=

35

Chapter 4

Connections to Straight Parallel Tracks

4.0 Introduction

Type of Layout connections between the straight parallel tracks will bedependent upon the distance between the two tracks and the spaceavailability in the yard. Accordingly distance between the two tracksmay be treated as Normal or Large distance.

4.1 With Normal Distance between the Straight ParallelTracks

Figure 4.1: Normal distance between the straight parallel tracks

Formulae

(4.2)TDCotFX

(4.1)2

FRtanT

+=

=

36

(4.4)T)(BSinF

DS

DT)SinFS(B

(4.3)AXOL

+−=∴

=++

+=


First of all, radius of connecting curve ‘R’ has to be assumed keepingin view the content of Para 410 of IRPWM. Distance between the twostraight parallel track will be found out from the field. Now after havingthe values of ‘R’ and ‘D’, calculate the values of ‘T’, ‘X’, ‘OL’ and ‘S’from Eq 4.1, 4.2, 4.3 and 4.4 respectivly.

Now, in the field, either of the two points i.e. TP2 and ‘SJ’ will be

decided from the site conditions. With respect to one point, the otherpoint will be fixed which will be at a distance ‘OL’ apart. After that,entire layout i.e. turnout, straight after heel of crossing and theconnecting curve can be laid by field surveying.

Note:

In the above connection, it is evident that for a given value of ‘D’, if weincrease/decrease the value of straight ‘S’ after heel of crossing, valueof connecting curve ‘R’ decreases/increases respectively. Therefore, bycontrolling the value of ‘S’, value of ‘R’ can be controlled. For a givenvalue of ‘D’, less the value of ‘S’, more will be the value of ‘OL’ i.e.overall length requirement. More the value of ‘S’, less will be the valueof ‘OL’. For S=0, ‘OL’ will be the maximum. ‘S’ can be increased toa maximum value till the value of ‘R’ reduces to R=R

recommended.

Correctness of Layout dependes basically on the correctness of thevalues of different variables as used in the above formulae. Any wrongvalue will disturb the geometry of layout at the field.

37

4.2 Layout Calculations with Fanshaped PSC Layout

Due to introduction of latest versions of 52 Kg and 60 Kg BG(1673 mm), 1 in 16, 1 in 12 and 1 in 8.5 turnouts on PSC sleepers todrawing Nos RDSO/T-5691, RDSO/T-4732, 4218 and RDSO/T-4865respectively. Values of various turnout parameters have been given atthe end of the book in Table of Detailed Dimensions.

Values of these turnout parameters can easily be calculated from therelevant RDSO drawings.

If we take the value of ‘B’ from the table of dimensions for performingthe calculations, we are inadvertently making a mistake and our finallayout will become wrong. The same is explained as detailed below;

The value of ‘B’ as per usual convention for these versions of theturnout is slightly more than for conventional CMS crossing and lessthan that for Built UP crossing. It will be much longer for all practicalpurposes, as the entire layout becomes rigidly straight up to the lastcommon PSC sleeper of the turnout due to prepositioned inserts forstraight alignment behind the heel of crossing.

Due to situation as explained above, it may not be possible to start theconnecting curve just behind the heel of crossing as was done earlier forIRS layouts.

Therefore, by default, the geometry of PSC layouts will be having acertain straight ‘S’ after the heel of crossing which should be accountedfor modifying the value of ‘B’ as given in the Table of DetailedDimensions given at the end of this book. The value of default straightshould be determined from the relevant drawings. Because of theabove, calculations for the various PSC layouts if done as per earliervalues of the turnout parameters is likely to generate kinky alignment,especially for the connecting curve which will tend to become sharperdue to reduction in tangents lengths.

38

Example 4.1

A BG main line track is required to be connected to a loop line whichis parallel at 4.725m distance by using a 52 Kg, 1 in 8.5 T/O. Calculatethe required distances for the layout, assuming the connecting curveradius as the same as turnout curve radius.

Given: F=6042’35”, D=4.725m, R=221.522m, A=12.000m,B=17.418m

0.015m1 12.991)7.4181("53'24Sin6

4.725 T)(B

SinF

D S

65.137m12.00053.137A X OL

.137m53 12.991"53'244.725Cot6 TDCotF X

12.991m 2

"53'246 tan221.522

2

FRtan T

0

0

0

=+−=+−=

=+=+=

=+=+=

==

=

In the above example, the turn-in-curve (connecting curve) radius hasbeen assumed to be the same as that of the turnout curve radius i.e.221.522m, which is quite sharp curvature from the point of view ofmaintainability. It is, therefore, desirable to flatten this curve and thiscan be achieved by reducing the straight after the heel of crossing.However, by doing so, overall length i.e. ‘OL’ will become more,which may not be available in the loop line.

39

Example 4.2

In the above example 4.1, calculate radius of the flattest turn-in-curveto maintain it satisfactorily. Also calculate the over all length of thelayout.

Given: F=6042’35”, D=4.725m, A=12.000m, B=17.418m

For the flattest turn-in-curve, the straight ‘S’ after the heel of crossingwill be equal to zero.

4.1 example previous in the ast requiremenlength overall

withcomparisonin 10.048m,65.137-75.185by more iswhich

75.185m12.00063.185AX OL

63.185m23.006"35'424.725Cot6

TDCotF X

m730.922

2

"35'426tan

23.022

2

Ftan

T R

,2

FRtanT equation, from now

23.022m17.418"35'42Sin6

4.725B

SinF

D T

T)(BSinF

D

0,S ngsubstituti

T)(BSinF

D S

0

0

0

=

=+=+=

=+=

+=

===

=

=−=−=

+=

=

+−=

40

Example 4.3

Calculate the minimum track centre for a connection between twostraight parallel track with the turnout on PSC layout (52 Kg, 1 in 8.5,BG 1673mm gauge) and the recommended radius of turn-in-curvebeing 440m

Given: A=12.025m, B=16.486m

a) Calculate as if turn-in-curve is starting just after the heel ofcrossing.

b) Calculate, when ‘B’ is modified because of a default straightafter the heel of crossing due to prepositioned inserts forstraight alignment behind the heel of crossing.

a) For minimum track centre, ‘S’ has to be zero.

4.940m

"35'42625.793)Sin(16.486Dmin

25.793)(16.486"35'42Sin6

Dmin0

0,S ngsubstituti

T)(BSinF

D S

25.793m2

"35'426tan440

2

FRtanT

0

0

0

=

+=∴

+−=

=

+−=

=×=

=

41

b) The default length of straight, because of prepositionedinserts in PSC layouts after the heel of crossing, should bedetermined from the relevant drawings. For 52 Kg, 1 in8.5 BG 1673mm gauge PSC layout, this straight cansafely be taken as 3.3m.

5.325m

"35'42625.793)Sin3.300(16.486D

T)S(BSinF

D

0

=

++=

++=

Therefore, minimum track centre should be say 5.3m to accommodatecompletely the 1 in 8.5 turnout on PSC sleepers, so as to satisfy all thegeometrical conditions.

It is worth mentioning that either because of lesser track centre orwrong calculations, at several yards, last common sleepers are beingremoved so as to start connecting curve slightly earlier. It is thereforerecommended that to accommodate PSC turnouts, the minimum trackcentre must be equal to 5.3m, otherwise either common sleepers willhave to be removed or sharper turn-in-curve will be required to makethe connection.

42

4.3 With Large Distance between Straight Parallel Tracks

When the distance between the two straight parallel tracks increasesand if we go for the layout connections as per para 4.1, overall length‘OL’ requirement will become too large, which may not be available atsite. It is, therefore, desirable to introduce a connecting reverse curvestarting from the heel of crossing so as to keep the overall length of thelayout to the minimum. Now there can be two sub cases as explainedbelow;

Case I With no straight in the reverse curves

Case II With a straight in the reverse curves

Case I With no straight between reverse curves. (Fig 4.2)

Figure 4.2: With large distance between straight parallel trackshaving no straight in the reverse curves

43

Formulae

(4.9)AXOL

(4.8)T)Cos�T(TT)CosF(BX

(4.7)2

�RtanT

(4.6)2

F�RtanT

(4.5)2R

D-RBSinFRCosFCos�

2R

D-RBSinFRCosFCos�

D-RMN BSinF,LM RCosF,LO

2R

MNLMLO

2R

MNMO

OO

NOCos�

NOO �In

11

1

1-

1

11

1

1

1

+=

++++=

=

−=

++=∴

++=∴

===

++=

+==

Interpretation of Formulae and Field Practicalities

First of all, value of radius of connecting curve is to be assumed, whichis generally the same as that the radius of lead curve. Distance ‘D’ willbe known from the field. Value of various turnout parameters will beknown once we have decided the type of turnout.

Now from Eq 4.5, value of ‘θ’ will be calculated. Then from Eq 4.6 &4.7, tangent lengths ‘T’ & ‘T

1 will be calculated. Value of ‘X’ and

finally ‘OL’ will be calculated from Eq 4.8 & 4.9 respectively. Nowwith respect to ‘TP

3’, the location of ‘SJ’ can easily be fixed which will

be at a distance equal to ‘OL’. (Fig 4.2)

In the above case several variations can be made such as, different radiiof curvature for two legs of the reverse curve or a straight between the

44

heel of crossing & start of the reverse curve. Various formulae asderived in the previous case can be modified as follows;

Figure 4.3

)R(R

D)-(RSinFS)(BCosFRCos�

D-RMN SinF,S)(BLM CosF,LO

RR

MNLMLO

RR

MNMO

OO

NOCos�

NOO �In

21

12

11

21

1

21

1

1

1

1

2R

+

+++=∴

=+==

+

++=

+

+==

45

(4.14)AXOL

(4.13)T)Cos�T(T)CosFT(BX

(4.12)2

�tanRT

(4.11)2

F�tanRT

(4.10))R(R

D)-R(S)SinFB(CosFRCos�

111

11

2

21

121-

+=

+++++=

=

−=

+

+++=∴

S

Purpose of these variations in the above formulae is to show that, byaltering one or other parameters, a layout connection can be designedto suit the diverse site conditions. Yard designer can design betterlayouts with this understanding. It is worth mentioning that, notrigonometrical formulae can solve a layout because of diverse siteconditions and variety of obligatory points. Several trial & errorcalculations will be required for designing an optimum layout.

46

Example 4.4

A BG track is required to be connected to a siding which is parallel at15m distance. By using a 90R, 1 in 8.5 turnout and the radius of theconnecting curve being the same as that of the lead curve of 1 in 8.5turnout and without a straight between the reverse curve. Calculate therequired distances for the layout connection.

Given: F=6042’35”, D=15.000m, A=12.000m, B=17.404m,R=221.522

m115.75212.000103.752A XOL

m103.75228.55742.56732.628

28.557

"28'4114 Cos28.557)(15.449"35'426 Cos15.449)(17.404

TCos�)T(TCosFT)(BX

m28.5572

"28'4114tan221.522

2

�RtanT

m15.449 2

"35'426"28'4114tan221.522

2

F�RtanT

284114�

221.5222

15.0221.522"35'426Sin17.404"35'426Cos221.522

R2

DR-BSinFRCosFCos�

00

11

0

1

00

"'0

00

=+=+=

=++=

+

+++=

++++=

===

=−

=

−=

=∴

×−++

=

++=

47

Case II With a Straight between Reverse Curves

When as per Case I, it may be possible, that space available is less incomparison with what is required for making the connection thenoverall length ‘OL’ can be further reduced by introducing a straightbetween the reverse curve. In this Case, it is presumed that reversecurve is starting just after the heel of crossing.

Figure 4.4: With large distance between straight parallel trackshaving a straight in the reverse curve

Formulae

(4.15)2R

Stan�

2R

S

R

2S

TPO3

ZTPtan�

ZTPO �In

1

21

2

321

−=

===

48

( )

( )

( )

(4.20)AXOL

(4.19)T)Cos�TS(T)CosFT(BX

(4.18)2

�RtanT

(4.17)2

F-�RtanT

(4.16)2R

S1tanS

Sin D-RBSinFRCosFCos�

4.15, Eq from �' of value thengsubstituti

S

Sin D-RBSinFRCosFCos�

S

Sin D-RBSinFRCosF �Cos(�

Sin

SOO

OO

S

OO

OOSin

OOO �In

D-RMN BSinF,LM RCosF,LO

OO

MNLMLO

OO

MNMO

OO

NO �Cos(�

NOO �In

2211

2

1

1-

1-

21

2121

32

231

1

21

1

21

1

21

1

21

+=

+++++=

=

=

−−++

=

−++

=∴

++=+

=∴

==

===

++=

+==+

49

If we start the reverse curve after introducing another straight ‘S1’ after

the heel of crossing and different degree of curvature of the two legs ofthe reverse curve (R

1 & R

2).

Figure 4.5

For the above variation (Fig 4.5) various formulae can be modified asexplained under.

[ ]

(4.25)AXOL

(4.24)TCos�)TS(TCosF)TS(BX

(4.23)2

�tanRT

(4.22) 2

F��tanRT

(4.21)R

Stan

S

Sin D-RSinF)S(BCosFRCos�

RRR e wherR

Stan

22111

22

11

12111-

21

1

+=

++++++=

=

=

−+++

=

+==

−

−

50

Example 4.5

In the Example 4.4, if a straight of 10m has to be introduced betweenthe reverse curves, the calculate the required distances for the settingthe layout.

( )

( )( )

116.152m12.000104.152AXOL

104.152m26.11447.80530.233

114.6247261326.114)Cos103.0381(

3542613.038)Cos7.4041(

T)Cos�TS(T)CosFT(BX

26.114m2

472613221.522tan

2

�RtanT

13.038m 2

35426472613221.522tan

2

F-�RtanT

47261335'171224414�

224414)(0.96709311Cos �

9670931.010

0225661.0522.206033.2004.220

10

35171Sin15-522.21235426Sin404.7135426221.522Cos

S

Sin D-RBSinFRCosF �Cos(�

35'1701221.5222

10 tan

2R

Stan

-

"'0

"'0

"'0

"'0"'0

"'0"'0

"'0

2211

2

1

"0

"'0"'0"'0

"1

1

=+=+=

=++=

+++

+=

+++++=

===

===

==

=−=+

=×++

=

++=

++=+

=×

=

=

+

−

−

−

51

Chapter 5

Crossover Connection between Straight Parallel Tracks

Type of Crossover connection between the straight parallel tracks will bedependent upon the distance between the two tracks and the spaceavailability. Accordingly distance between the two tracks may be treatedas Normal distance and Large distance. Concept of Normal distance andthe Large distance is not decided by the spacing between the two tracks,it is the arrangement of layout and accordingly trigonometric formulae.

5.1 With Normal Spacing between the Tracks and with Same Angleof Crossing

Figure 5.1: With normal spacing between the tracks andwith same angle of crossing

52

Formulae

(5.3) 2A XOL

) N(CotF Xing. ofnumber theis N where

(5.2) DNDCotFX

(5.1) 2BSinF

DS

DB)SinFS(B

+=

=

==

−=∴

=++


First of all the value of ‘D’ will be known from the field surveying. Turnoutprarmeters ‘A’, ‘B’ will be known once we have decided the type ofturnout. Then from Eq 5.2 & 5.3, the values of ‘X’ & finally ‘OL’ will becalculated. Now with these values in the hand, location of one of ‘SJ’ canbe fixed by keeping it at a distance ‘OL’ apart in refernce to another ’SJ’.After fixing the location of ‘SJ’, rest of the turnout can be set out by fieldsurveying.

For the correctness of the crossover connection, it is very important thatthe distance ‘D’ must be same in the vicinity of turnouts and must beaccurate. A small error in ‘D’ will get magnified by ‘N’(number of crossing)times. It is therefore very important that the value of ‘D’ must be arrivedat by field surveying and not from the yard drawings which might havebeen prepared long back and necessary corrections might not have beendone.

For example, if ‘D’ used for layout calculations is not equal to theactual distance available at the site, then two conditions may be thoughtof i.e;

Dcal<Dactualor

Dcal>Dactual

53

In both the cases, ‘SJ’ will be fixed wrongly and the connection, instead

of a straight,will become a reverse curve or a kink will be formed at theheel of crossing.

a) Dcal

<Dactual

b) Dcal

>Dactual

Figure 5.2

Now, when the train negotiate the crossover, it will try to straighten upthe reverse curve, which will finally result into alignment kink in the mainline and will result into bad running.

54

5.2 With Large Spacing between the Tracks with the SameAngle of Crossing

When the distance between the two straight parallel tracks increases andif we go for the layout connection as per para 5.1, overall length ‘OL’requirement will become two large, for which space may not be availableat the site. Space being the costly item in an yard, it is therefore desirableto introduce a connecting reverse curve starting ̀ from the heel of crossingso as to keep the overall length of the layout to the minimum. In this typeof layout, further there can be two sub cases i.e;

Case I With no Straight in the Reverse Curves

Case II With a given Straight in the Reverse Curves

Case I With no Straight in the Reverse Curves (Fig 5.3)

Figure 5.3: With no Straight between Revrse Curve

55

Formulae


First of all, value of radius of connecting reverse curve is to be assumed,which is generally the same as that of lead curve radius of the turnout.Distance ‘D’ between the two straight parallel track will be known fromthe field. Value of turnout parameters will be known once we have decidedthe type of turnout.

Now from Eq 5.4, value of ''� can be calculated. Then from Eq 5.5, 5.6& 5.7, value of ‘T’, ‘X’ ann hence ‘OL’ can easily be calculatedrespectively. Now the location of ‘SJs’ can easily be fixed which willbe at a distance equal to ‘OL’.

[ ](5.7)2AXOL

(5.6)TCos�T)CosF(B2X

(5.5)2

F-�RtanT

(5.4)R

2DBSinFRCosF1Cos�

R

2D-BSinFRCosFCos�

R2

TP1

O ,2DNM BSinF,LM RCosF,L1

O

TPO

NMLMLO

TPO

NMMO

TPO

NOCos�

NTPO �

21

1

21

1

21

1

22

+=

++=

=

−+−=∴

+=∴

====

−+=

−==

56

In the above case, several variations can be made such as , different radiiof curvature of two legs of reverse curve or a straight between the heelof crossing and start of the reverse curve or different angle of crossing atthe two ends. (Fig 5.4)

Figure 5.4Formulae

Cos�1

R1

)SinF1

S1

(B1

CosF1

R1

M1

N

1R

1M

1N

1)SinF

1S

1(B

1CosF

1R

Cos�

1R

2TP

1O ,

1)SinF

1S

1(B

1M

1L ,

1CosF

1R

1L

1O

2TP

1O

1M

1N

1M

1L

1L

1O

2TP

1O

1M

1N

1M

1O

2TP

1O

1N

1O

Cos�

2TP

1N

1O �In

−++=∴

−++=∴

=+==

−+=

−==

57

Now, these formulae have become more versatile and generalized byincorporating these variations.

(5.12)2

A1

AXOL

(5.11)2

)CosF2

T2

S2

(B

)Cos�2

T1

(T1

)CosF1

T1

S1

(BX

(5.10)2

2F�

tan2

R2

T

(5.9)2

1F�

tan1

R1

T

(5.8)

2R

1R

D2

)SinF2

S2

(B1

)SinF1

S1

(B2

CosF2

R1

CosF1

R1-Cos

2R

1R

D2

)SinF2

S2

(B1

)SinF1

S1

(B2

CosF2

R1

CosF1

RCos�

D2

M2

N1

M1

N

Cos�2

R2

)SinF2

S2

(B2

CosF2

R

Cos�1

R1

)SinF1

S1

(B1

CosF1

R2

M2

N1

M1

N

Cos�2

R2

)SinF2

S2

(B2

CosF2

R2

M2

N

2R

2M

2N

2)SinF

2S

2(B

2CosF

2R

Cos�

2R

2TP

2O ,

2)SinF

2S

2(B

2M

2L ,

2CosF

2R

2L

2O

2TP

2O

2M

2N

2M

2L

2L

2O

2TP

2O

2M

2N

2M

2O

2TP

2O

2N

2O

Cos�

2TP

2N

2O �In

++=

+++

++++=

−=

−=

+

−+++++=

+

−+++++=∴

=+

−+++

−++=+

−++=∴

−++=∴

=+==

−+=

−==

58

Example 5.1

A crossover is required to be laid between the two parallel BG tracks at15m distance by means of a 52 Kg, 1 in 12 turnout with no straightportion in the connection. Calculate the required parameters for thelayout. Also calculate the saving in overall length in this layout over alayout with straight line connection between the crossing.

Given: F=4045’49”, A=16.953m, B=23.981m, R=441.282m

[ ][ ]

48.559m165.347-213.906length overall in the saving Therefore

213.906m16.953212152ADCotFOL

crossing; the

between provided is connection linestraight iflength Overall

165.347m16.9532131.4412AXOL

131.441m20.779)2(44.941

"34'14021.116Cos14945421.116)Cos(23.9812

TCos�T)CosF(B2X

21.116m

2

49454-"34'1410tan441.282

2

F-�RtanT

341410�

0.9840628

441.282

215-494523.981Sin449454441.282CosCos�

R

2D-BSinFRCosFCos�

0"'0

"'00

"'0

"'0"'0

==

=×+×=+=

=×+=+=

=+=

++=

++=

=

×=

=

=

=

+=

+=

59

Case II With a given Straight in the reverse curves

Figure 5.5 : With a given Straight in the reverse curve

Formulae

BSinF,NM RCosF,NO

ZO

LMNMNO

ZO

LM-MO

ZO

LO��Cos(�

LZO �In

(5.13)2R

Stan�

2R

S

R

2S

TPO

ZTPtan�

ZTPO �In

1

1

1

1

1

1

1

1

1

21

2

21

==

−+===+

=

===

−

60

( )

( )

( )

(5.17)2AXOL

(5.16)S)Cos�(2TT)CosF2(B

B)CosF(TT)Cos�S(TT)CosF(BX

(5.15)2

F-�RtanT

(5.14)2R

Stan

S

Sin�2DBSinFRCosF2Cos�

�S

Sin�2DBSinFRCosF2Cos�

S

Sin�2D-RBSinFRCosF2��Cos(�

2Sin�

SZO

ZTPO �In

1

1-

1-

1

21

+=

+++=

++++++=

=

−

−+×=∴

−−+×

=∴

++×=+

=

−

61

Now, In the above case, several variations can be made such as, differentradii of curvature of the two legs of reverse curve or a straight lengthintroduced between the heel of crossing and the start of the reversecurve on the two ends or different angle of crossing at the two ends.

Figure 5.6

For this case , various formulae can be revised as follows:

Formulae

)tan�R(RZTPZTP

R

ZTP

TPO

ZTPtan�

ZTPO �In

R

ZTP

TPO

ZTPtan�

ZTPO �In

2132

2

3

32

3

32

1

2

21

2

21

+=+

==

==

62

( )

( )

(-

ZO

ZTPSin�

ZTPO �In

OOZOZO where

(5.20a)��Z)Cos(�OZO

)SinFS(B)SinFS(BCosFRCosFRD

MNMN D

��ZCos(�O-)SinFS(BCosFR MN

��ZCos(�O-)SinFS(BCosFR MN

5.20, & 5.19 Eq From

(5.20)ZO

MN-SinFS(BCosFR��Cos(�

ZNO �In

(5.19)ZO

MN-SinFS(BCosFR��Cos(�

ZO

MN-MLLO

ZO

MN-MO

ZO

NO��Cos(�

ZNO �In

(5.18)

2R

1R

Stan�

)R(R

S tan�

S)ZTPZTP (where )tan�R(RS

1

2

21

2121

21

2221112211

2111

22222221

11111111

2

222)

2222

22

1

111)

1111

1

111111

1

1111

1

11

11

1-

21

3221

=

=+

++

+++++=

=

+++=

+++=

++=+

++=+

+===+

+=∴

+=∴

=++=

+

63

Now these formulae have become more generalized by incorporatingthese variations.

[ ]

[ ]

(5.25)2AXOL

(5.24))CosFBS(T

)Cos�TS(T)CosFTS(BX

(5.23)2

F�tanRT

(5.22)2

F�tanRT

(5.21)RR

Stan

S

Sin�D2

)SinF2

S2

(B1

)SinF1

S1

(BCosF2

RCosF1

RCos�

S

Sin�D2

)SinF2

S2

(B1

)SinF1

S1

(BCosF2

RCosF1

R

: toequal is ��Cos(� a), (5.20 Eq gconsiderin

Sin�OOS

Z)Sin�OZ(OZTPZTP

ZO

ZTPSin�

ZTPO �In

2222

211111

2

22

1

11

21

1

1

21

21

2132

2

3

32

+=

+++

+++++=

−=

−=

+−

−+++++=

−+++++=

+

=∴

+=+

=

−

−

64

Example 5.2

A crossover is required to be laid between two parallel BG tracks at 15mdistance by means of 90R, 1 in 12 turnout and with a straight of 40m inconnection. Calculate the required parameters for the layout. Alsocalculate the saving in overall length in this layout over a layout withstraight line connection between the crossing.

Given: F=4045’49” , A=16.953m, B=23.961m, R=400m

( )

( )

168.294m 16.9532134.3882A XOL

134.388m63.01571.373

5940)Cos811.829(2"49'450411.829)Cos2(23.981

S)Cos�(2TT)CosF2(B X

11.829m2

"49'4504-598400tan

2

F-�RtanT

59845512

40

"45'510Sin2215"49'45023.981Sin4"49'450400Cos42Cos

�S

Sin�2DBSinFRCosF2Cos �

455124002

40tan

2R

Stan�

"'0

"'0

"'0"'0

1-

1-

"'011

=×+=+=

=+=

+×++=

+++=

===

=−

−+×=

−−+×

=

=×

==

−−

45.614m168.294-213.906length overallin saving Therefore

213.906m16.953212152ADCotFOL

crossing,

ebetween thgiven is connection linestraight if length, Overall

==

=×+×=+=

65

5.3 With Different Angles of Crossing and Normal Distancebetween the two Straight Parallel Tracks

Figure 5.7 : With different Angles of crossing and normal distancebetween the two straight parallel tracks

Formulae

(5.30)AAXOL

(5.29))CosFB(TT)CosF(BX

(5.28)2

2F

1F

RtanT

(5.27)

2SinF

1SinF

)2

SinF2

B1

SinF1

(B-DT

(5.26)D)SinFB(TT)SinF(B

21

2211

2211

++=

+++=

−=

+

+=∴

=+++

66

In the derivation of the various formulae for the above case, it is assumedthat connecting curve is starting just after the heel of crossing. It mayalso happen, to suit the varied site conditions, that a straight after theheel of crossing is required. Those formulae can be modified as follows;(Fig 5.8)

Figure 5.8

[ ]

(5.35)AAXOL

(5.34))CosFBS(TT)CosFS(BX

(5.33)2

2F

1F

RtanT

(5.32)

2SinF

1SinF

)2

)SinFS2

(B1

SinFS(B-DT

(5.31)D)SinFBS(TT)SinFS(B

21

222111

2)

11

222111

++=

+++++=

−=

+

+++=∴

=+++++

67

Example 5.3

A crossover is required to be laid between the two straight parallel tracksat 4.725m centres by using a 52 Kg, 1 in 8.5 & 1 in 12 turnouts. Calculatethe required parameters for the layout.

Given: Turnout A B F

1 in 8.5 12.000m 17.418m 6042’35”

1 in 12 16.953m 23.981m 4045’49”

First of all, value of ‘T’ tangent length will be calculated from,

Note:

Radius of connecting curve, thus, coming is prohibitive for negotiatingthe passenger trains. It is recommended that radius of connecting curveshould be at least equal to the radius of lead curve of 1 in 12 turnout,which can be taken as 440m.

205.597m

2

"49'4504"35'4206tan

3.492

2

FFtan

T R

3.492m0.19988

1.992)(2.035-4.725

"49'450Sin4"35'420Sin6

)"49'45023.981Sin4"35'4206(17.418Sin-4.725

SinFSinF

)SinFBSinF(B-D T

21

21

2211

=−

=−

=

=+

=

+

+=

+

+=

68

Example 5.4

In the foregoing example 5.4, if the radius of connecting curve radius isrequired to be the same as that of the radius of lead curve of 1 in 12turnout, calculate the minimum distance ‘D’ between the two tracks andthe various parameters for the layout. Radius of connecting curve can betaken as equal to 441.282m.

21

2211

77.098m16.95312.00048.145

AAXOL

48.145m27.37820.767

"49'4503.492)Cos4(23.981 "35'4203.492)Cos6(17.418

)CosFB(TT)CosF(BX

=++=

++=

=+=

+++=

++++=

21

2211

2211

21

85.062m16.95312.00056.110

AAX OL

56.110m31.36724.742

"49'4504 7.495)Cos(23.981"35'4207.495)Cos6(17.418

)CosFB(TT)CosF(B X

5.525m2.6142.911

"49'45047.495)Sin (23.981"35'4207.495)Sin6(17.418

)SinFB(TT)SinF(B D

7.495m2

"49'4504"35'4206tan441.282

2

FFRtan T

=++=

++=

=+=

+++=

+++=

=+=

+++=

+++=

=−

×=−

=

69

Example 5.5

In the foregoing example 5.4, if the layout is on PSC sleepers, thencalculate the minimum distance ‘D’ between the two tracks and otherparameters for correctly laying the layout in the field. Minimum radiusof connecting curve to be taken as that of the lead curve radius of 1 in12 turnout.


1 in 8.5 12.025m 16.486m 6042’35”

1 in 12 16.989m 22.912m 4045’49”

(These turnout parameters are for PSC layouts, 1673mm Gauge)

For the PSC Fan Shaped Layout, the layout becomes rigidly straight upto the last common PSC sleepers(S.No 83 for 1 in 12, & S.No 54 for 1 in 8.5)of the turnout due to the prepositioned inserts for straight alignmentbehind the heel of crossing. These straight for 1 in 8.5 & 1 in 12 turnoutsare 3.3m & 5.688m respectively.

92.078m16.98912.02563.064AAX OL

63.064m35.97027.094 "49'4505.688)Cos422.912(7.495

"35'4203.3)Cos616.486(7.495 X

6.185m2.9983.187 "49'4505.688)Sin422.912(7.495

"35'4203.3)Sin616.486(7.495

)SinFBS(TT)SinFS(B D

7.495m2

"49'454"35'426tan441.282

2

FFRtan T

21

222111

0021

=++=++=

=+=+++

++=

=+=+++

++=

+++++=

=−

×=−

=

70

Note:

It is therefore, very important that when yard layout is on PSC Fan Shapedlayout and track centre is less, then section engineer should remove longcommon sleepers from the Fan Shaped layout so that the connectingcurve can be started earlier. It may also happen that, while performing thelayout calculations, default straight behind the heel of crossing mightnot have been taken into consideration and therefore during laying theturnout, geometry of the layout will be disturbed because of mismatch ofcalculations and the actual site parameters. Because of this mismatch,alignment will become kinky in the connecting curve, which in turn willbe reflected as the kink in main line.

5.4 Crossover between a Loop Line and the Main Line with theSymmetrical Split on the Loop Line

Figure 5.9 : Crossover between a loop line and the main line with thesymmetrical split on the loop line

71

Formulae


From the field, value of ‘D’ will be known. Radius of connecting curvewill be assumed as per the guidelines. From Eq 5.37 value of ‘T’ can becalculated and the from Eq 5.36, 5.38 & 5.39, valus of ‘S’, ‘X’ & ‘OL’ canbe calculated respectively. Now the work left is locating one Stock Jointwith respect to another by keeping it at a distance ‘OL’ apart.

The same formulae can be used differently, when ‘S’ is given. Then fromEq 5.36 value of ‘T’ can be calculated and then from Eq 5.37, value of ‘R’can be calculated. Value of ‘R’ thus arrived may or may not satisfy therecommended radius of connecting curve.

Several variations can also be thought of like a straight after boththe Turnouts or a reverse curve between the heel of crossing. Purposeof incorporating these variations is basically for designing an optimumlayout keeping in view the space constraints, infringements or obligatorypoints.

(5.39)AAXOL

(5.38)2

F)CosB(TT)CosFS(BX

(5.37)2

2

FF

RtanT

(5.36)T)(B-SinF

2

FT)Sin(B-D

S

D22

F)SinB(TT)SinFS(B

21

2

211

21

11

22

211

++=

++++=

−

=

++

=∴

=++++

72

Example 5.6

A crossover is required to be laid between a Loop Line and a parallelmain line at 4.725m track centres by using a 90R, 1 in 8.5 symmetricalsplit turnout on the Loop Line and 90R, 1 in 12 turnout on the main Line.Radius of the connecting curve being equal to the lead radius of the 1 in8.5 symmetrical split. Calculate the required parameters for the layout.


1 in 8.5 12.000m 17.404m 6042’35”

1 in 12 16.953m 23.981m 4045’49”

Radius of the connecting curve will be ‘R’=443.882m

21

2

211

11

2

2

2

1

92.421m12.00016.95363.468AAXOL

63.468m22.82240.646

2

"35'420617.404)Cos(5.457"49'4505.457)Cos411.349(23.981

2

F)CosB(TT)CosFS(BX

11.349m5.457)(23.981-"49'450Sin4

2

"35'42065.457)Sin(17.404-4.725

T)(B-SinF

2

FT)Sin(B-D

S

5.457m2

2

"35'4206"49'4504

tan443.882 2

2

FF

RtanT

=++=++=

=+=

++++=

++++=

=++

=

++

=

=

−

×=

−

=

73

5.5 Crossover in Multi Track Area

Crossover in Multi Track area known as ‘Double Junctions’ involveprovision of a diamond crossing in between two Turnouts. Such layoutspermit diversion of UP/DN directional train from fast corridor to theslow corridors and vice versa. The turnouts are normally of 1 in 12spread, while the diamond crossing is of 1 in 8.5 spread. Hence theconnection will involve a connecting Curve.

Figure 5.10 : Crossover in multi track area

It will be seen that for such a connection, the minimum track centredistance required for introducing a connecting curve of 1 in 12 turnoutlead curve radius is 5.5m on BG which is not always available. The followingalternatives are recommended in such layouts;

• At locations where such layouts are already laid with theexisting distance between track centres , it is preferable toreplace the existing 1 in 12 turnout with 1 in 8.5 turnout withcurved switches to have a straight line connection i.e. without acurve between the heel of crossings. This will not only improvethe layout alignment but also result in saving in the overalldistance which may be of a great advantage in the restrictedsuburban area. If the replacement as suggested above can not

74

be effected for some reason or other, then existing layout can beimproved by correctly calculating the resultant radius ofconnecting curve after properly assessing the track centresdistance and providing the same with accuracy and care.

• Whenever such crossovers are newly introduced, intermediatediamond crossings should be of the same spread as that of theturnout. Therefore, for 1 in 12 turnout, the diamond crossing inbetween shall be of movable type so that the connection willbe by means of straight line connection between the heel ofcrossings.

75

Chapter 6

Scissors Cross-Over between Straight ParallelTracks

6.1 Scissors Cross-Overs

When two cross-overs overlap each other, preferably exactly opposite toeach other, a scissors cross-over is said to have been formed. A diamondis formed where the cross-over cross each other in addition to the fourturnouts. The same function can be achieved by two cross-overs facingeach other.( Fig 6.1)

Figure : 6.1 : Scissors cross-overs

The advantage of providing a scissors cross over is the saving of spacein an important congested yard, but this involves having an additionaldiamond crossing in the yard. A scissors cross-over is equated to threecross-overs for the purpose of assessing the maintenance cost.

The angle of diamond crossings are twice the crossing angle of turnout.They are also commonly called by half the number of the turnoutcrossing e.g. for 1 in 8.5 turnouts the diamond crossing is listed as

76

1 in 4.25, though it is strictly incorrect because cot (2F) will not be halfthat of cot (F). On Broad Gauge, scissors cross-over layouts have beenfinalized only for two track centers (on wooden layouts) viz: 4.725 m(15’6”) and 5.180 m (17’0”). They can not be laid at track centers inbetween these distances due to difficulty in providing effective checkrail guidance over the unguided gap in front of the ANCs of acutecrossings for a wheel movement over the cross-overs.

6.2 Broad Gauge Scissors Cross-Over at 4.725m TrackCenters

At 4.725 m track centers, the distance between the TNCs of the acutecrossings of the diamond (L

d) formed in between the tracks is greater

than the distance between the TNCs of the main line acute crossings (X).Hence the acute crossings of diamond lie outside the acute crossing ofthe main line turnouts as can be seen from the Fig 6.2.

Figure : 6.2 : Broad gauge scissors cross-over at 4.725m track centers

The distances involved for negotiating 1 in 8½ crossovers at 4.725 mtrack centers are as under-

77

The distance 1377 mm includes two gaps of 496 mm and 248 mm in frontof 1 in 8.5 and 1 in 4.25 acute crossings respectively, which are requiredto be protected by provision of check rails opposite them. Since thedistance between ANCs (X

3) i.e. 1377 mm is greater than total unguided

gap (744 mm) at 4.725 m track centers, it is possible to provide specialcheck rails for guiding the movement of wheels along the cross-over.

As the track center distance increases the distance between TNCs ofmain line turnouts (X) also increases while the dimensions of thediamond (L

d) remains constant with the result that the distance between

ANCs of the acute crossings of main line and diamond (X2 and X

3) get

progressively reduced making it difficult to provide check rails oppositeunguided gap. If the distance (X

3) comes closer to 744 mm, the total

unguided gap, or lower than 744 mm then provision of check rail is notpossible. This aspect would be clear from the following table 6.1 for 1 in8.5 scissors cross-overs.

1377mmGtanF-CosF

1563 X

),(Xover -cross over the moving vehicleaby traversed

be wouldasover -cross thealong ANCsbetween distance The

1563mm4.25)in 1for ( 59 8.5)in 1for ( 1181386X

, )(Xstraight thealong ANCsbetween Distance

1386mm2

X-Lis )(X diamond and linemain

on crossings acute theof TNCs ebetween th distance Hence

14345mmSinF

GL

),(L diamond theof diagonal long The

11572mm 2

FGCot-DCotFX

(X), linemain on crossings of TNCsbetween Distance

3

3

2

2

d

1

d

d

==

=++=

==

==

==

78

Table : 6.1

D 4725 4800 4900 5000 5015 5100 5200 5300

X 11572 11210 13060 13910 14037 14760 15610 16460

Ld 14344 14344 14344 14344 14344 14344 14344 14344

X1

1386 1067 642 217 153 -208 -633 -4158

X2

1563 1244 819 394 330 -31 -456 -881

X3

1377 1056 628 200 136 -228 -656 -1084

As the track center increases the position becomes safe when ANCs ofthe diamond and ANCs of the main line crossings fall opposite eachother i.e. when distance X

2 becomes zero ( Fig 6.3). This occurs when

the track centers become 5.015 m beyond which the wing rails of thecrossing on main line and that of acute crossing of than diamond can beso extended that they perform the function of check rail opposite thecrossings and give the effective guidance. This is possible till suchincreased track center is obtained when independent check rails can beplaced opposite the respective gaps. The position is identicalwith 1 in 12 and 1 in 16 turnouts.

Figure : 6.3

For this purpose, Indian railways have finalized standard designsfor 4.725 m (15’6”) and 5.18 m (17’) track centers on wooden layoutswith extended wing rails for use on all new constructions. The designfor 4.725 m track center is to cover existing situation since majority ofthe yards are laid at 4.725 (15’6”) track centers.

79

A design for scissors cross-over on concrete sleeper layout at 4.725mtrack centre has been finalized for use in the field. However no design hasbeen finalized for 5.18m track centres on concrete sleepers.

No designs have been finalized for 5.3 m track center which is now therecommended track center for new constructions because as per therecommended guidelines, use of scissors cross-over is to be avoidedand on new constructions, layouts should be designed with the use ofnormal cross-overs.

For the Meter Gauge, effective independent check railing arrangementsare possible even when they are laid at 4.265 m track centers as thedistances between the TNCs are 5317 mm and 7548 mm in the case of 1 in8.5 and 1 in 12 scissors cross overs respectively. This is large enough toplace indepent checkrails opposite the nose of crossings.

80

Chapter 7

Crossovers between Non Parallel Straight Tracks

7.1 Crossovers between Non Parallel Straight Tracks

In the field, several situatios may arise where crossovers are to be laidbetween two non parallel straight tracks. Depending upon the siteconditions, variety of connections can be made like with same ordifferent angle of crossing, connecting curve being simple or reversecurve starting just after the heel of crossing or after a straight behindthe heel of crossing. If the connecting curve being the reverse curvethen it may be with or without a straight in between. It means that,there can be several permutations & combinations and variety offormulae can be generated.

Let us derive the set of formulae for a crossover connection between twonon parallel straight tracks with the following conditions;

Connecting curve being simple circular curve between the heel ofcrossing and angle of crossings being different on the two ends.

Figure 7.1 : Crossover between non parallel straight tracks

81

Formulae


Given ‘R’, Angle between two non parallel straight tracks ‘∆∆∆∆∆’ andturnout parameters;

Eq 7.1 will be used for calculating the value of ‘T’. Once ‘T’ is known,Min ‘D’ (Where ‘SJ

2’ will be fixed) will be calculated from Eq 7.2 and

finally ‘X’ & ‘OL’ will be calculated from Eq 7.3 & 7.4 respectively.Now ‘SJ

1’ can be fixed at a distance ‘OL’ from ‘SJ

2’.

Given ‘D’(where‘SJ2’ is fixed), Angle between two non parallel straight

tracks ‘∆∆∆∆∆’ and turnout parameters;

Eq 7.2 will be used for calculating the value of ‘T’. Once ‘T’ is known,value of ‘R’ can be calculated from Eq 7.1 which may or may notsatisfy the recommended radius of turn-in-curve. After that the value of‘X’ & ‘OL’ can be calculated from Eq 7.3 & 7.4 respectively.

(7.4)Cos�AAXOL

(7.3)��-)Cos(FB(TT)CosF(BX

(7.2)DSin�A��)Sin(FB(TT)SinF(B

(7.1)2

FF�RtanT

21

2211

22211

21

++=

+++=

=−+++

−+=

−

82

Example 7.1

A crossover consisting of 52 Kg, 1 in 12 and 1 in 8.5 turnouts isrequired to be laid between non parallel straight tracks at a minimumdistance between the two track centres. The angle between he two nonparallel straight tracks is 3050’0”. Calculate the various parametersrequired to lay the crossover connection assuming the connecting curveradius equal to 400m.


1 in 8.5 12.000m 17.418m 6042’35”

1 in 12 16.953m 23.981m 4045’49”

110.432m"0'50016.953Cos312.00081.517

Cos�AAXOL

81.517m44.16637.351

)"0'5003"49'450(423.981)Cos(20.191

"35'420620.191)Cos(17.418

��-)Cos(FB(TT)CosF(BX

3.978m1.133-0.7174.394

"0'50016.953Sin3 )"0'5003"49'450(423.981)Sin(20.191

"35'420620.191)Sin(17.418

Sin�A��)Sin(FB(T"35'420T)Sin6(BDmin

20.191m2

"49'4504"35'4206"0'5003tan400

2

FF�RtanT

21

2211

2221

21

=++=

++=

=+=

−++

+=

+++=

=+=

−−++

+=

−+++=

=−+

×=

−+=

−

83

Chapter 8

Connection between Curved Track to Parallel CurvedTrack or Divergent Straight Track

Connections have at times to be laid from Curved Track to eitherparallel track or divergent straight track. Understanding about suchconnections is very important for a field engineer so that the same canbe applied for proper design of such layouts. Several situations can bethought of such that;

• Connection between two curved parallel tracks. Connectionbeing on outside of main Line curve without a straight betweenthe heel of crossing and the connecting curve.

• Connection between two curved parallel tracks. Connectionbeing on outside of main Line curve with a straight between theheel of crossing and the connecting curve.

• Connection between two curved parallel tracks. Connectionbeing on inside of main Line curve

• Connection between a curved track to straight track, theintersection being on the inside of the main line curve.

• Connection between a curved track to straight track, theintersection being on the outside of the main line curve.

84

8.1 Connection between two Curved Parallel Tracks.Connection being on Outside of Main Line Curve withouta Straight between the Heel of Crossing and theConnecting Curve.

Figure 8.1 : Connection between two curved parallel tracks. connectionbeing on Outside of main line curve without a straightbetween the heel of crossing and the connecting curve

Formulae

rail)(outer curve connecting of radius c

R

rail)(outer linemain curved of radius m

R

GE toParallel AB Line a Draw and EG with Join

=

=

85

2

F2Ktan-G-D

2

FKtan

mR

2

FKtan-G-D

mRA O-BO

2

FKtan

mRHAHOAO

2

FKtanGAEB where

2

FKtan-G-D

mREB-EL-LOBO

FAOO

A) B OBA O (where BA O-BA OAOO

BAO-BA OAOOA BO-BA O

2

B OA Cot

AOBO

AOBO

2

A BO-BA Otan

Formula, tan Applying B,A O �In

GE) toParallel is (ABBA O BA O

G EO EG O

Triangle) (IsocelesEG O �In

FAOO

8.1, Fig theofGeometry theFrom

11

11

1111

21

212221

122111

1

11

1111

1

22

22

2

21

=

−−+=∴

+=+=

==

+==

=∠=

∠=∠∠∠+∠=

∠∠+∠=∠∠

∠

+

−=

∠∠

∠=∠∴

∠=∠

=∠

86

2

FKtanG

cR

Sin�

2

FKtan

mR

��-Sin(180

2

FKtanG

cRGA-GL-LOAO

2

FKtan

mRAO

A O2

O 1

OA Sin

A O

AO OSin

Formula, SineApply OA O �

(8.1)2

FCot

G-Dm

2R2

F2Ktan-G-D

2tan�

2

FCot

G-Dm

2R2

F2Ktan-G-D

2

� tan

2

�Cot

G-Dm

2R2

F2Ktan-G-D

2

F tan

G-Dm

2R

2

FKtan

mR

2

FKtan-G-D

mRA OBO

2222

1

21

21

21

1-

11

−−=

+∴

−−==

+=

∠=

∠

+=∴

+=∴

+=∴

+=

+++=+

87


When the connections pertaining to curved lines are required to be laidin the field, it will be prudent to fix the crossing first and then the entireturnout and connecting curves are being laid.

Distance ‘D’ between the two curved parallel tracks and radius of mainline ‘R

m’ will be known from field surveying. Value of turnout parameters

will be known once we have decided the turnout.

Now from Eq 8.1, value of ''� will be calculated. Eq 8.2 will be used forcalculating the value of radius of connecting curve ‘R

c’. And Finally ‘X’

will be calculated from Eq 8.3. Point ‘L1’ and hence Point ‘L’ will be

decided from field surveying. Now with refernce to ‘L’, distance ‘X’ willbe measured along the outer rail of curved main line to fix the TNC. It mayalso happen that TNC is already fixed, then Point ‘L’ can be fixed bymeasuring a distance ‘X’ along the outer rail of the curved main line.After ‘L’ being fixed, ‘L

1’ can easily be fixed by field surveying.

It is evident from the formulae, that value of connecting curve radius ‘Rc’

will be fixed for a given set of boundary conditions like ‘Rm’, ‘D’ &

turnout parameters. It may so happen that ‘Rc’ may or may not satisfy the

recommended radius of connecting curve.

(8.3) degreesin is ��' e wher180

�m

RK

radiansin is ��' re whe�m

RKHLNHX

(8.2)2

FKtanG

F)Sin(�

2

FKtan

mRSin�

cR

F�� e wher

2

FKtanG

cR

Sin�

2

FKtan

mR

F)Sin(�

�

+=

+=+=

+++

+

=∴

+=−−

=+

+

88

Example 8.1

Calculate the radius of connecting curve ‘R’ and layout distance ‘X’ fora connection between a curved main line and a curved parallel track at4.725m distance. Connection being on the side of the main line withouta straight after the heel of crossing by using 1 in 8.5 crossing.

Given; K=3.123m, F=6042’35”, R=450m

48.877m180

"33'4905 4503.123

180

�m

RK X

212.395m2

"35'4206tan3.1231.676

)"35'4206"33'490Sin(5

2

"35'4206tan3.123450"33'490Sin5

2

FKtanG

F)Sin(�

2

FKtan

mRSin�

cR

"33'4905903.049

17.05892.6828572tan

2

"35'4206Cot

1.676-4.725-45022

"35'42063.123tan2-1.676-4.725

2tan

2

FCot

G-Dm

2R2

F2Ktan-G-D

2tan�

� �

1-

1-

1-

=×

+=+=

=×++

+

×+×

=

+++

+

=

=×=

×

×=

+=

×

89

Note:

Calculations as performed above are the direct calculations. But in caseof designing a layout connection, limiting the radius of the connectingcurve ‘R

c’, it will be required to calculate the required radius of curved

main line so as to satify the predecided radius of connecting curve. Forsuch calculations, problem can best be solved by ‘Trial & Error’ processwith a number of values of ‘R

m’. A ready made graph can also be

prepared between ‘Rm’ & ‘R

c’ for different track centres for a given

turnout, connection being on outside of main line without a straight afterthe heel of crossing.

90

8.2 Connection between two curved parallel tracks.Connection being on outside of main Line curve with astraight between the heel of crossing and the connectingcurve

The author is against this type of layout unless ‘D’ is very large asotherwise the ‘Rc’ is reduced considerably because of a straightbetween the heel of crossing and the connecting curve. For normaldistances of the loop line, the ‘R

c’

will have prohibitive values unless ‘D’

is very very large. This layout as well as the layout discussed in previousclause, three curves i.e. lead curve, connecting curve & parallel curvedline are laid and for good maintenance, it is very much desirable to havethe curves as large as possible.

Figure 8.2:Connection between two curved parallel tracks. Connectionbeing on outside of main line curve with a straight betweenthe heel of crossing and the connecting curve.

91

2

cRG

2

FKtan)CosF

2

FKtan

m(R

S)SinF2

FKtan

m(R)

cR-D

mR

POPOOO

G)-c

(R2

FKtan)CosF

2

FKtan

m(RPO

G)-c

(R-)CosF2

FKtan

m(R

2

FKtan

)LLL(OAQGALOGQLOPLPO

tan S)SinF2

FKtan

m(R

SASinFOQPQOPO

cR-D

mR

OLLLLOOL-LOOO

O PO �In


crossing of heel after theStraight S

rail)(outer curve connecting of Radius c

R

rail)(outer linemain curved of Radius m

R


(

22

2121

2

2

2

1

2

21

2

4222424242

111

2111211121

21

+

−+++

++=+

+=

−++=∴

++=

−−+=−=−=

+=++=

==+=+=

+=

−+==

=

=

=

Equation; above the in PO & POOO of values the ngSubstituti

2

FK

mRAO where

SGLQP where

,

1

4

Formulae

92


Distance ‘D’ between the two curved parallel tracks and radius ‘R’ ofmain line will be known from field surveying. Value of turnout parameterswill be known once we have decided the type of turnout. Now value of‘S’ will have to be assumed. Then from Eq 8.4 value of ‘R

c’ will be

calculated. Value of ‘ϕ’and finally ‘X’ can be calculated from Eq 8.5 & 8.6respectively. Now TNC can be fixed easily as discussed in the previousclause.

{ }(8.6)

180

��(F-90m

RK

180

�m

RKHLNHX

��(F-90� hence & 18090��FO AQ �In

(8.5)

cR-D

mR

S)SinF2

FKtan

m(R

OO

POCos�

(8.4)

2

FKtan-G-)CosF

2

FKtan

m(R-D

mR2

2G

2

FKtan)CosF

2

FKtan

m(R

-

2

FKtan-G-)CosF

2

FKtan

m(R-D

mR2

2S)SinF

2

FKtan

m(RD)

m(R

cR

cRfor solving now G

2

FKtan)CosF

2

FKtan

m(R

c2R

cRG

2

FKtan)CosF

2

FKtan

m(R

S)SinF2

FKtan

m(R

cD)R

mR2

cRD)

mR

� �

((

,1

21

1

2

22

222

++=+=+=

+==+++

+

++==

++

+++

++

++−+=

+++−

+++++

++=+−++

93

Example 8.2

A curved main line track of 600m radius is equired to be connected to aparallel siding at 4.725m distance on the outside by using a 90R, 1 in12 turnout and with a straight portion of 10m behind the heel ofcrossing. Calculate the various parameters for the layout.

Given: F=4045’49”, K=3.815m

229.600m

-

2

FKtan-G-)CosF

2

FKtan

m(R-D

mR2

2G

2

FKtan)CosF

2

FKtan

m(R

-

2

FKtan-G-)CosF

2

FKtan

m(R-D

mR2

2S)SinF

2

FKtan

m(RD)

m(R

cR

2

"49

'45

04

tan3.815-1.676-"

49'

450

)Cos42

"49

'45

04

tan3.815(600-4.7256002

2

1.6762

"49

'45

04

tan3.815"

49'

450

)Cos42

"49

'45

04

tan3.815(600

2

"49

'45

04

tan3.815-1.676-"

49'

450

)Cos42

"49

'45

04

tan3.815(600-4.7256002

2

10"

49'

450

)Sin42

"49

'45

04

tan3.815(6002

4.725)(600

2

=

=

++

+++

++

++−+=

××++

+×+×+

××++

+×+−+

94

50.054m 180

56'24046003.815

180

��(F-(90m

RKX

"56'2404)"15'49080"49'450(4-90� therefore,"15'49080�

0.15952055229.6-4.725600

10"49'450)Sin42

"49'4504tan3.815(600

cR-D

mR

S)SinF2

FKtan

m(R

Cos�

"�

�

=×

+=

++=

===∴

=+

+×+=

+

++=

×

+

From the above examples, it is evident that introduction of straightbehind the heel of crossing is not desirable, since it reduces theradius of connecting curve.

95

8.3 Connection between two curved parallel tracks. Connectionbeing on inside of main Line curve

Figure 8.3 : Connection between two curved parallel tracks. Connectionbeing on inside of main Line curve

96

( )

GDm

2R2

FKtanD

mR

2

FKtanG

mRBOAO

2

F2KtanGD

2

FKtanD

mR

2

FKtanG

mRBO-AO

2

FKtanD-

mR EBLE-LOBO

2

FKtanG

mRHA-MHMOAO

F OA OBA O-A BO

1OA

2O-BA

2O-A BOBA O-A BO

2

�Cot

BOAO

BOAO

2

BA O-A BOtan

Formula, tan Applying B,A O �In

GE) toParallel is (ABBA O BA O

B,A O �In

G EO EG O Triangle) (IsocelesEG O �In



R

rail)(outer linemain curved of radius m

R


11

11

11

11

1222

111

1

1

1111

1

11

1

22

2

−−=+−+−−=+

−−=

−+−−−=

+=+=

−−=−=

=∠+∠∠=

∠∠∠=∠∠

+

−=

∠∠

∠=∠∴

∠=∠

=

=

Formulae

97

Note:

It may be observed that value of F)Sin(� − can have a positive, zeroor a negative value depending on ‘R

m’ & ‘D’, other variables like ‘K’,

‘F’ & ‘G’ being constant. Three type of connections are possible insuch layouts namely;

( )

(8.9)degreesin is � where180

G)�-m

(RK

HLNHX

(8.8)2

FKtan

F)Sin(�

)2

FKtan-G-

mSin

cR

2

FKtan

cR

Sin�

2

FKtan-G-

mR

F)Sin(�

2

FKtan

cRAO

2

FKtan-G-

mRAO

AO

��-Sin(180

AO

F)Sin(�

Formula,Sin applying

F�� F� A,OO �In

(8.7)2F

CotGD

m2R

2F

2KtanGD2tan�

2

�Cot

GDm

2R2

F2KtanGD

2

F tan

�

2

1

21

21

1

+=

+=

−−

=∴

+=

−∴

+=

=

=−

−=∴+=

−−

−−=∴

−−

−−=∴

−

R(�

98

Type I 0F >−� , ‘Rc’ will be positive and then ‘R

c’ & ‘R

m’

are in similar flexure.

Type I

Type II 0F =−� , ‘Rc’ will be infinite, i.e. there will be no

curve between the heel of Xing and the siding and theconnections will be by means of a straight track.

Type II

Type III 0F <−� , ‘Rc’ will be negative, and therefore it will

be in reverse direction to the main line curve.

Type III

99

Example 8.3

A curved main line track of 500m radius is required to be connected toa curved parallel siding at 4.725m distance, the connection being on theinside of the curve by means of a 52 Kg, 1in 12 turnout. Calculate thevarious parameters for the layout.

Given: F=4045’49”, K=3.834m

( )

( )

69.574m180

"31'33071.676)-(5003.843

180

G)�-m

(RKX

1343.672m 2

"49'4504tan3.834

)"49'4504"31'330Sin(7

)2

"49'4504tan3.834-1.676-(500"31'330Sin7

cR

2

FKtan

F)Sin(�

)2

FKtan-G-

mRSin

cR

I) (Type F"31'3307

2

"49'4504Cot

1.6764.7255002

2

"49'4504Ktan3.83421.6764.725

2tan

2F

CotGD

m2R

2

F2KtanGD

2tan�

��

1

1

=×

+=+=

=×−

−

×=

−−

=

>

−−×

××−−

=

−−

−−=

=

−

−

(�

100

Example 8.4

In the previous example 8.3, if the connection is to be made with a straightbetween the heel of crossing and the curved parallel track, calculate theradius of curved main line and other parameters for the layout.

If it is to be with a straight line connection, then θ =F. Substituting thisvalue in the equation 8.7.

( )

84.929m

180

"49'45041.676)-(976.5743.843

180

G)�-m

(RKX

976.574m2

1.6764.725

2

2

2

"49'4504Cot

2

"49'4504tan3.83421.6764.725

2

GD

2

2

2

FCot

2

F2KtanGD

mR

2

�tan

2

FCot

2

F2KtanGDGD

m2R

2F

CotGD

m2R

2F

2KtanGD

2tan�

��

1

=

×+=+=

=+

+

××+−

=

++

−−

=∴

−−=−−∴

−−

−−=

−

101

Example 8.5

A curved main line track of 2000m radius is required to be connected to acurved parallel siding at a distance 4.725m apart. The connection beingon the inside of the main track by means of a 52 Kg, 1 in 12 turnout.Calculate the various parameters for the layout.

Given: F=4045’49”, K=3.834m

( )

( )

69.510m180

"59'52011.676)-(20003.843

180

G)�-m

(RKX

curve) reverse theindicates (- 1306.381m

2

"49'4504tan3.834

)"49'4504"59'520Sin(1

)2

"49'4504tan3.834-1.676-(2000"59'520Sin1

cR

2

FKtan

F)Sin(�

)2

FKtan-G-

mSin��

cR

III) (Type F"59'5201

2

"49'4504Cot

1.6764.72520002

2

"49'4504Ktan3.83421.6764.725

2tan

2F

CotGD

m2R

2F

2KtanGD

2tan�

��

1

1

=×

+=+=

−=

×−−

×=

−−

=

<

−−×

××−−

=

−−

−−=

=

−

−

102

8.4 Connection between a curved track to straight track, theintersection being on the inside of the main line curve

Figure 8.4 : Connection between a curved track to straight track, theintersection being on the inside of the main line curve

103

Formulae

(8.10)CosFR2RRROO

CosFAOAO2-AOAOOO

formula Cosapply A,OO �In

(say) R2

FKtan-G-

mRGA-GL-LOAO

(say) R2

FKtan

cRHAHOAO


R

rail)(outer linemain curve of radius m

R

21

2

2

2

1

2

21

21

2

2

2

1

2

21

21

21111

122

d

−+=∴

×××+=

===

=+=+=

=

=

(8.12)

CosFR2RRR

G)Cos�-m

(R-c

R

OO

LO Sin�

G)Cos�-m

(R-c

R

Q)Cos�L-L(O-c

R

QCos�O-c

RPO-c

RLM-MOLO

,O LO �In

(8.11)

CosFR2RRR

SinFRSin�

OO

SinF

AO

Sin�

formula, Sine applying now

21

2

2

2

121

2

221

1122

21

21

2

2

2

1

1

212

−+==

=

=

===

−+=∴

=

104

(8.14)Cos� CosF2

R1

2R22

R21

R-G)Sin�m

(R

Cos�OO-G)Sin�m

(R

LOG)Sin�m

(R

PM-PQY

(8.13)��(�-��0(90� where

180

G)�-m

(RK

GQNGX

��(�-��0(90�

090�-�� e, therefor090LO P

figure, theofgeometry theFrom

21

1

1

�

−+−=

−=

−−=

=

++=+=

+=

++=∴

=++=∠


From the field survey, value of '' ∆ , Rm will be known. Once we havedecided the type of turnout, turnout parameters will also be known.Assume suitable value of radius of connecting curve ‘Rc’. From Eq 8.13& 8.14 , value of ‘X’ & ‘Y’ can easily be calculated. From the field survey,point Q on the inner rail of curved main line can be established. Then bymeasuring along the inner rail of curved main line, a distance equal to ‘X’,TNC can be fixed. After TNC getting fixed, rest of the layout can be laidin the field.

105

Example 8.6

A curved track of 450m radius is required to be connected to a straighttrack intersecting on the inside rail of the curved track at an angle of 500.The connection is to laid by means of 52 Kg, 1 in 12 turnout with aconnecting curve radius of 300m. calculate the various parameters for theturnout.

Given: F=4045’49”, K=3.834m

522835.0985

001.676)Cos5-(450-300 Sin�

CosFR2RRR

G)Cos�-m

(R-c

R Sin�

"40'2909 �

522835.0985

24.927

"49'450Cos4448.165300.1592448.165300.159

"49'450Sin4300.159

CosFR2RRR

SinFRSin�

448.165m2

"49'4504tan3.834-1.676-450

2

FKtan-G-

mRR

300.159m2

"49'4504tan3.834300

2

FKtan

cRR

21

2

2

2

1

22

21

2

2

2

1

1

2

1

=

−+=

=∴

=

×××−+

×=

−+=

=×=

=

=×+=

+=

106

( ) 192.786m

"14'290Cos4 522835.0985 -001.676)Sin5-(450

Cos� CosFR2RRR-G)Sin�m

(RY

277.644m

180

"34'590341.676)-(450 3.834

180

G)�-m

(RKX

"34'59034)050"40'290(9"14'290490�

"14'2904�

21

2

2

2

1

��

=

=

−+−=

=

×+=+=

=+−+=

=∴

107

8.5 Connection between a curved track to straight track, theintersection being on the outside of the main line curve

Figure 8.5 : Connection between a curved track to straight track, theintersection being on the outside of the main line curve

108

Formulae

(8.15)CosFR2RRROO

F)-Cos(180R2RRR

OA CosOAOAO2-AOAOOO

formula Cosapply A,OO �In

(say) R2

FKtan

mRHAHOAO

(say) R2

FKtan

cRGAGOAO


R


R

21

2

2

2

121

21

2

2

2

1

2121

2

2

2

1

2

21

21

222

111

d

++=∴

−+=

×××+=

=+=+=

=+=+=

=

=

211

21

21

2

2

2

1

2

21

2

21

111

21

1

21

OO

F)Sin(180

AO

Sin�

,OA O �In

(8.16)

CosFR2RRR

Cos�m

Rc

R Cos�

Cos�m

RLL where

OO

LLc

R

OO

OLLO

OO

OOCos�

,O OO �In

−=

++

+=∴

=

+=

+==

109

��(�-�� where

(8.19)180

�m

RKHLNHX

��(�-��

180��-90��90 ,O OO �In

(8.18)Sin�m

R-Sin�CosF2

R1

2R22

R21

R

Sin�m

RSin�OO

OLO OL OY

Sin�CosF2

R1

2R22

R21

ROO

OO2

OOSin� ,O OO �In

(8.17)

CosFR2RRR

SinF2

FKtan

cR

Sin�

�

21

21

2222

2

2121

21

2

2

2

1

+=

+=+=

+=∴

=++++

++=

−=

−==

++=∴

=

++

+=∴


From the field survey, value of ‘Rm’ and '' ∆ will be known. Once we have

decided the type of turnout, turnout parameters will also be known.Assume suitable value of radius of connecting curve i.e. ‘R

c’. Now

calculate O1O

2 from Eq 8.15. Once O

1O

2 is known, calculate '' φ from Eq

8.16. Calculate the value of From '' � Eq 8.17. Value of ‘Y’ & ‘X’ can becalculated from Eq 8.18 & 8.19 respectively. From the field survey, point‘L’ on the outer rail of main line can be established. Then measuringalong the outer rail of main line a distance of ‘X’, TNC can be easily fixed.After TNC getting fixed, rest of the layout can be laid in the field.

110

Example 8.7

A curved track of 450m radius is required to be connected to a straighttrack intersecting the outer rail of the curve at an angle of 300. Theconnection is to be laid by means of 90R, 1 in 8.5 turnout and a connectingcurve of 300m radius.

Given: F=6042’35”, K=3.109m

( ) ( )37.232m

180

"34.48'20044503.109

180

�m

RKX

"34.48'2004"0.52'410225'5802230��

"0.52'4102�

80.046818395561194.988

"35'420Sin6300.182

CosFR2RRR

SinF2

FKtan

cR

Sin�

"25'58022� hence 20.920683825561194.988

689.7114

"35'420Cos6450.182300.1822450.182300.182

0Cos30450300

CosFR2RRR

Cos�m

Rc

R Cos�

450.182m2

"35'4206tan3.109450

2

FKtan

mRR

300.182m2

"35'4206tan3.109300

2

FKtan

cRR

��

0

21

2

2

2

1

22

21

2

2

2

1

2

1

=×

+=+=

=+−=+−=

=∴

=×

=

++

+=

=

×××++

×+=

++

+=

=×+=+=

=×+=+=

×

=

111

Chapter 9

Crossover Connection between two Curved ParallelTracks

9.1 Crossover Connection between two Curved Parallel Tracks

Figure 9.1 : Crossover connection between two curved parallel tracks

112

Formulae

(8.20) 2

FKtan

mR

2

FKtanG2CosF

2

FKtanG

2

FKtan

mRCosF

2

FKtan

mR

2

FKtanG

c2R

cR

2

FKtan

mR

2

FKtanG2CosFCosF

2

FKtan

mR

c2R

2

FKtanG

c2R

2

FKtanG

cR

2

FKtan

mR

CosF2

FKtanG

cR

2

FKtan

mR2

2

FKtanG

cR

2

FKtan

mR

CosFAOAO2AOAOOO

2

FKtanG

cRAGGLLOAO

2

FKtan

mRAHHOAO

FOAOOAO �In


R


R

2

22

22

2

22

1211

2

12

2

11

2

21

11111212

111111

211 ,

211

d

-

-

+++++

++

++

+−=

++++

+−++++=

×−−×+×

−−++=

××−+=

−−=−−=

+=+=

=

=

=

∠

113

(8.21) 2

FKtan

2

FKtan-G-D

mR2CosF

2

FKtan

2

FKtan-G-D

mR

2

FKtan-CosF

2

FKtan-G-D

mR

c2R

cR

2

FKtan

2

FKtan-G-D

mR2CosF

2

FKtan-G-D

mR

c2R

2

FKtan-G-D

mR

2

FKtan

2

FKtan

c2R

cR

CosF2

FKtan

cR

2

FKtan-G-D

mR2-

2

FKtan

cR

2

FKtan-G-D

mR

CosFAOAO2AOAOOO

2

FKtan

cR AO

2

FKtan-G-D

mRAGGL-LOAO

FOAOOAO �In

22

2

222

22

2221

2

22

2

21

2

21

22

2222

21

21

221 ,

221

+

+++

+−=

+−+−

++++=

++

+++=

×××++=

+=

+=−=

=

−

∠

114

( )

( )

2

OOAOOAS where

(8.24)FOOA (180OOA

(8.23)bc

a)(SS

2

OOACos

formula Cos applyingOAO �In

(8.22)

CosF12

F2KtanG-DCosF2

Gm

R2

FKtanD

m2R2CosF

-

CosF12

F2KtanG-DCosF2

2

F2KtanGG

2

F2Ktan-G-DG-D

m2R

cR

2

FKtan

mR

2

FKtanG

2

FKtan-G-D

mR

2

FKtanCosF

2

FKtanG

2

FKtan

2

FKtan

mR

2

FKtan-G-D

mRCosF

2

FKtan

mR

2

FKtanG

2

FKtan-CosF

2

FKtan-G-D

mR

c2R

cR solvingfor 8.21 & 8.20 Eq theEquating

211121

1

211121

11211

,211

22

22

)

2

++=

∠−=∠

−=

∠

+

+

+

+

+

+

+−

+=

+++

+

+−

+−+=

+−

+−

+

+

−

+

115

Degreesin is ��' where180

�c

RS

curveinner to

curveouter thefrom K'' of valuesof projection theis

mR

KD

Degreesin is '' where

(8.27)

mR

KD2K

180

�m

RX

OOAOOA�

OOAOOA�

2

OOAOOAS where

(8.26)F)OOA (-180 OOA

(8.25)bc

a)(SS

2

OOACos

formula, Cos applying ,OAO �In

�

�

�

121122

212211

212122

2

212122

22212

221

××=

−+=

∠−∠=∴

∠−∠=∴

++=

+∠=∠

−=

∠

Note:

When we introduce straight crossing on curves, we introduce kinks onthe junction of the crossings on either end. The curve is allowed tomaintain same curve outside. Similar kinks will appear on the ends ofcrossings on the connecting track. In some countries the main linecurves are realigned after introducing crossings. This type of layouts areonly suitable when the main line curve is fairly easy. It is found that ifthee calculations are being done by assuming them as that for crossoverconnection between two straight parallel tracks, the difference betweenthem is negligible. If we calculate for ‘X’ assuming them as two straightparallel track at a track centre of ‘D’, then,

2

FGCotDCotFX −=

116

Example 9.1

A 90R, 1 in 12 cross-over is required to be laid between two curvedparallel tracks. Assuming the radius of the outer rail of the inner track as1000m and the track centre as 5m. Calculate the various parameters forthe layout.

Given: F=4045’49”, K=3.185m

( )

( )

( )

( )

( )

++

=−=

××+−

××+×

+−+

+

+

+

+

+

+

+−

+

=

GmR2

FKtanDm2R2CosF

6019.923.34087916023.2649

2

"49'4504tan3.81521.6761.676

2

"49'4504tan3.8152-1.676-51.676-510002

2

F2KtanGG

2

F2Ktan-G-DG-Dm2R

CosF12

F2KtanG-DCosF2

Gm

R2

FKtanD

m2R2CosF

CosF12

F2KtanG-DCosF2

2

F2KtanGG

2

F2Ktan-G-DG-D

m2R

cR

-

117

( )

( )

( ) ( )

"55.29'33099OOA

85.721000.16

1017.92)1.91051.9(105

2

OOACos

1051.92

85.721017.921000.16S

85.72m

"49'450Cos41017.921000.16221017.9221000.16OO

1017.92m

2

"49'4504tan3.8151.6761019.75

2

FKtanG

cRAO

1000.16m2

"49'4504tan3.8151000

2

FKtan

mRAO

��' solvingfor

1019.75m2.006

3974.546019.92Rc

2.006

"49'450Cos412

"49'4504tan3.81521.676-"49'450Cos452

CosF12

F2KtanG-DCosF2

3974.54

1.67610002

"49'4504tan3.815510002"49'450Cos42

211

211

1

21

12

11

=∠∴

×−

=∠

=++

=

=

×××−+=

=

×−−=−−=

=×+=+=

=−

=

=

+××+×

++

=

×+×+××

118

( ).negligible iswhich

0.04m19.71-19.75 is values twoebetween th difference The

19.71m2

"49'4504Cot1.676125

2

FGCot-DCotFX

then tracks,parallelstraight between

connectioncrossover for that as problem above thesolve weIf

19.75m 1000

53.8153.8152

180

"43.5'41001000

mR

KD2K

180

�m

RX

"43.5'4100

"11.74'52098"55.29'33099OOAOOA�

"11.74'52098OOA

85.721003.17

1019.91)4.41054.4(105

2

OOACos

1054.42

85.721019.911003.17S

1019.912

"49'4504tan3.8151019.75

2

FKtan

cRAO

1003.172

"49'4504tan3.8151.67651000

2

FKtanGD

mRAO

�

�

212211

212

212

2

22

21

=

=×−×==

=×

−×+××

=

−+=

=

−=∠−∠=

=∠∴

×−

=∠

=++

=

=×+=+=

=×−−+=

−−+=

119

Chapter 10

Software on Layout Calculations

10.1 Introduction

In their day to day work P. way officials have to deal with complicatedyard layouts. The maintainability and riding quality over a turnout dependslargely on how accurately it is laid and maintained. Kinks in points andcrossings can be avoided if adequate care is taken at the time of layingitself.

For ensuring correct laying it is essential that necessary calculations arecorrectly done before layouts are laid in the field. These calculations areintricate in nature and require considerable efforts on the part of P. Wayofficials .Due to their remaining busy in routine works on many occasionsthis item of work does not get the attention it deserves.

10.2 Layouts

While the insertion of crossover, diamonds in existing layouts, yardremodelling etc. are being taken up, errors in calculations may result in:

• Incorrect laying (i.e. incorrect TNC distances) – Results inbad running and in extreme cases unsafe conditions mayoccur on defective/bad layouts.

• Long crossovers at angle ‘F’ being laid even though spacecould be saved by having curves in between two turnouts.

• During doubling/gauge conversion projects new layouts arebeing inserted forcible resulting into undesirable curvatureswhich give bad running and are potential locations needingfrequent attention.

• It is suggested that even while replacing an old layout anopportunity be taken to correct all known defective layouts.

120

The book on “Layout Calculations” gives all the formulae and samplecalculations and is of great help in ensuring correct laying of crossoversand other connections. For assisting the field engineers, a software hasbeen developed in Visual Basic for all the cases presented in the book on“Layout Calculation”. The salient features of the Software are:

• Highly interactive• No need to refer to any book• All data on standard dimensions stored in subroutines• Crossings can be fixed straight away based on these

calculations

10.3 Scope

The Software covers the following:

Connections between

• Diverging Tracks• Straight Parallel Tracks• Curved Parallel tracks• Curved and straight Tracks

Cross-overs between

• Straight Parallel Tracks• Non Parallel Straight Tracks• Curved Parallel tracks• Sanded Dead End and Main Line• Double Junction

The Software generally follows the pattern as given in the book on“Layout Calculations”.

While running the Software for a particular case, the user has to give thedata on section, Details of crossing from a drop down menu, trackcenters, Radius of connecting curve etc and the results are obtainedinstantaneously.

121

10.4 Instructions for Using the Software

Installation

For installing the Software insert the CD in your CD drive. Go to “Mycomputer” select the drive containing the CD; locate the “set up” filefrom the “Layout” folder or subfolder and double click the setup icon.The Software will be installed instantaneously on your computer and willbe available in your “start” menu.

Alternately if you do not wish to install the Software go to start menuselect the ”Run” command and on browsing locate the file“Layout_Calculations.exe” in your CD and Click “OK” and the Softwarewill get loaded.

10.5 Illustration

As soon as the Software is loaded, the following screen is visible:

You can select either “Connections” or “Crossovers” from amongst thetwo option buttons. “Connections” or “Crossovers” as referred earlier.

Since running the Software is similar in almost all cases of “Connections”and “Crossovers’,only two cases of each type will be described forillustration.

122

Let us first take “Connections”. After choosing the “Connections” optionbutton and clicking “Proceed” Button next to screen opens as givenbelow giving list of options under “Connections”.

Let us select “Connections to straight parallel track” and click “Proceed”Button .It opens the next Screen which gives a sketch and text boxes fordata entry.

Certain features of the Software are common to all the cases and are listedbelow:

• Selecting the appropriate Gauge option opens a drop menu forstandard turnout dimensions on BG or MG as the case may be.The labels of the items needing data entry have been shadedwhite. Other text boxes are locked and will not accept any data.They will only display results.

• Units of dimensions have to be in mm except where specificallymentioned.

• Please ensure that data in all text boxes requiring data entry arecorrectly entered and the type of turnout is selected from thedrop down menu before clicking “Compute” Button else youwill get run time errors.

123

• For non Standard turnouts the red label is required to be clickedas indicated on the screen and it opens text boxes for givingdata on turnouts viz. description, turnout number (e.g. 12 incase of 1 in 12 turnouts) and centre line dimensions A,M,K etc.

• The most ideal type of suggested output is to take a print ofthe form displaying the results by clicking “Print” Button.Alternately textural output on a file is also possible. For thispurpose the Button “Copy results in a file” is to be clickedwhich will open an input box to give the file name for output.Please be sure to enter the complete file name with drive nameand folder where you want the output. The output file socreated can be easily opened in any text editor such as Wordpad.

• The Software also gives warnings by way of displayingmessage boxes – when results are unacceptable such asresulting curvatures being sharp and beyond permissible limitsspecified of Schedule of dimensions.

On getting the Screen as mentioned above, the values have to beentered. We have to select the appropriate turnout from drop down menuand on clicking compute button results in the screen to appear as under:

124

The second case for demonstration in “Connections” is taken as“Connections between Curved and straight track”. On Clicking the“Proceed” Button it opens the appropriate screen which has separateframes for Gauge selection and for whether the intersection on the insideor outside of the curve. After you select appropriate Gauge and thentype of intersection would display appropriate sketch and after completedata entry and clicking “Compute” Button would give the results asshown below:

Next we take the Cases of “Cross-overs”. On selecting Crossovers fromthe first screen as described in Para 5.3 and clicking “ Proceed Buttontakes us to the screen showing the choices under “ Crossovers”

125

Let us first take the case “Cross-overs at large spacing with no straight inbetween” On selecting Gauge, appropriate Turnout from the drop downmenu and entering all the data we get following screen after clickingCompute Button.

If the same case is solved treating it as a case for non standard turnoutthen the text of the drop down menu is to be changed as “other type” inlower case and it opens the new text boxes for entering the data for theturnouts. On entering a sample data and clicking the “ compute” buttongives the following results.

The above feature is available in all the cases.

126

We take the next case of Crossovers between inclined tracks. The finalresults after entering data etc is as below:

10.6 Conclusions

It is considered that the Software explained above would be ofconsiderable help to the field Engineers in improving the running qualityover turnouts and ensuring safety by way of correct laying.

127

L TSL SL d w F R

1 in 8 ½ 60 kg C, FH. 18424 7872 7135 182.5 880 60-42’-35” 40-35’-0” 2314401 in 8 ½ 52 kg S, LH 20730 4950 4725 136 864 60-42’ 35” 10-34’-27” 2223601 in 8 ½ 52 kg C, FH 18395 6835 6400 182.5 864 60-42’ 35” 00-47’-27” 2323201 in 8 ½ 90R S, LH 20730 4950 4725 136 864 60-42’ 35” 10-34’-27” 2223601 in 8 ½ 90R C, FH 18395 6835 6400 182.5 864 60-42’-35” 00-47’-27” 2323201 in 12 60 kg C, FH 25831 11156 10125 175 1877 40-45’-49” 00-20’-0” 4413601 in 12 52 kg C, FH 29200 6724 6400 133 1232 40-45’-49” 10-0’44” 4421201 in 12 52 kg S, FH 27870 8478 10125 133 1232 40-45’-49” 00-27’-35” 4581201 in 12 90 R C, FH 29200 6724 6400 133 1232 40-45’-49” 10-8’-0” 4421201 in 12 90 R S, FH 27870 8487 7730 133 1232 40-45’-49” 00-17’-11” 4581201 in 16 52 kg C, FH 37170 10594 6500 153 1377 30-34’-35” 00-24’-27” 8242251 in 16(HS) 52 kg C, FH 37565 12320 7730 133 1377 30-34’-35” 00-24’-27” 8164801 in 16) 90R C, FH 37170 10594 9750 133 1377 30-34’-35” 00-46’-59” 8242251 in 20 90R C, FH 46210 11194 11150 133 1377 20-51’-45” 00-46’-59” 1303810

1 in 8 ½ 60 kg C, FH 18395 6839 6400 182.5 1225 60-42’-35” 00-46’-59” 2322601 in 8 ½ 52 kg C, FH 18395 6839 6400 182.5 1225 60-42’-35” 00-46’-59 2322601 in 12 60 kg C, FH 25831 10125 10125 175 1877 40-45’-49” 00-20’-0” 4413601 in 12 60 kg C, FH 25831 10125 10125 175 1877 40-45’-49” 0 0-20’-0” 4413601 in 16 60 kg C, FH 35720 11200 11200 145 2526 30-34’-35” 00-20’-0” 7849931 in 20 60 kg C, FH 46027 12460 12460 133 1877 20-51’-45” 00-20’-0” 1283100

1 in 8 ½ 90R C, FH 9515 6206 5500 169 915 60-42’-35” 00-29’-13” 1302101 in 8 ½ 75R S, LH 11560 4320 4115 120 915 60-42’-35” 10-35’-30” 1196101 in 8 ½ 75R C, FH 9515 6206 5500 169 915 60-42’-35” 00-29’-14” 1302101 in 8 ½ 60R S, LH 11560 4320 4115 120 915 60-42’-35” 00-24’-27” 1196101 in 8 ½ 60R C, FH 9515 6206 5500 169 915 60-42’-35” 10-9’-38” 1302101 in 12 90R Partly C, FH 14678 7974 7130 130 1220 40-45’-49” 00-24’-27” 258300

1 in 12 75R S, FH 16323 5777 5485 117 1220 40-45’-49” 10-9’-38” 2406001 in 12 75R Partly C, FH 15108 7544 6700 117 1220 40-45’-49” 00-24’-27” 258300

1 in 12 60R S, FH 16323 5777 5485 117 1220 40-45’-49” 10-9’-38” 2406001 in 12 60R Partly C, FH 15108 7544 6700 117 1220 40-45’-49” 00-24’-27” 258300

1 in 16 60R C, FH 20060 8264 7420 117 1378 30-34’-35” 00-24’-27” 46240

1 in 8 1/2 HTC 52 Kg C, FH 9515 6206 5500 169 1047 60-42'-35" 00-29'14" 130205

1 in 12 CMS 52 Kg Partly C, FH 14678 7974 7130 130 1405 40-45'-49" 00-24'-27" 258310

1 in 8 ½ 60R S, LH 8280 4320 4115 120 915 60-42’-35” 10-35’-30” 828001 in 12 60R S, LH 11723 5777 5485 117 1220 40-45’-49” 10-9’-389” 167340Note: C Curved

FH Fixed HeelLH Loose HeelCMS Cast Manganese Steel CrossingHTC Heat Treated CrossingHS High Speed

Narrow Gauge

Metre Gauge on PSC Sleepers

Broad Gauge (1673 mm) on PSC Sleepers

TABLE OF DETAILED DIMENSIONS

Metre Gauge

Dimensions (mm)

Broad Gauge

Crossing No Section Type�

128

A B C M K52 KG BG 1 IN 8 1/2 S, LH 12000 16274 840 14295 1979 RDSO T/28552 KG BG 1 IN 8 1/2 C, FH 12000 16274 1500 14295 1979 RDSO/T-28690 R BG 1 IN 8 1/2 S, LH 12000 16260 840 14295 1965 RDSO/T-36090 R BG 1 IN 8 1/2 C, FH 12000 16260 1500 14295 1965 RDSO/T-30752 KG BG 1 IN 12 S, FH 16953 22706 1500 20147 2559 RDSO/T-3252 KG BG 1 IN 12 C, FH 16953 22706 1500 20147 2559 RDSO/T-18490 R BG 1 IN 12 S, FH 16953 22687 1500 20147 2540 RDSO/T-18490 R BG 1 IN 12 C, FH 16953 22687 1500 20147 2540 RDSO/T-3152 KG BG 1 IN 16 C,FH 20922 30750 844 26842 3908 RDSO/T-13552 KG BG 1 IN 16 (HS) C,FH 22693 30750 850 26842 3908 RDSO/T-23190 R BG 1 IN 16 C,FH 20922 30750 844 26842 3882 RDSO/T-6790 R BG 1 IN 20 C,FH 24664 38118 844 33540 4578 RDSO/T-98

A B C M K52 KG BG 1 IN 8 1/2 S, LH 12000 17418 840 14295 3123 TA-20122,

TA-2080452 KG BG 1 IN 8 1/2 C, FH 12000 17418 1500 14295 3123 TA-20196,

TA-2083590R BG 1 IN 8 1/2 S, LH 12000 17404 840 14295 3109 TA20110,

TA-20198,etc90R BG 1 IN 8 1/2 C, FH 12000 17404 1500 14295 3109 TA-20148,

TA-2082252 KG BG 1 IN 12 S, FH 16953 23981 1500 14295 3834 TA-5268(M),

TA-20222, TA-20801

52 KG BG 1 IN 12 C, FH 16953 23981 1500 20147 3834 TA-20171, TA-20831

90R BG 1 IN 12 S, FH 16953 23962 1500 20147 3815 TA-5044(M), TA-20184, etc

90 R BG 1 IN 12 C, FH 16953 23962 1500 20147 3815 TA-20125, TA-20839

52 KG BG 1 IN 16 C, FH 20922 31447 844 20147 4605 TA-20141, TA-20828

52 KG BG 1 IN 16 (HS) C, FH 22693 31447 850 26842 4605 RDST/T-40390 R BG 1 IN 16 C, FH 20922 31421 844 26842 4579 TA-20138,

TA-2081390 R BG 1 IN 20 C, FH 24664 39470 844 26842 5930 TA-20122,

TA-2080460 KG BG (PSC)1 IN 8 1/2 C, FH 12025 16486 1500 14270 2216 RT-486552 KG BG (PSC)1 IN 8 1/2 C, FH 12025 16486 1500 14270 2216 RT-486560 KG BG (PSC)1 IN 12 C, FH 16989 22912 1141 20111 2803 RT-421852 KG BG (PSC)1 IN 12 C, FH 16989 22914 1141 20111 2801 RT-421860 KG BG(PSC) 1 IN 16 C,FH 20970 30558 844 26794 3764 RDSO/T-569160 KG BG(PSC) I IN 20 C, FH 25850 38031 844 33480 4550 RDSO/T-5858

Assembly Drg No

Section Gauge crossing No Type


CENTRE LINE LAYING (FOR BG WITH CMS CROSSING)

Section Gauge Crossing No Type Dimensions (mm) Assembly Drg No

CENTRE LINE LAYING ( FOR BG WITH ORDINARY CROSSING)

Dimensions (mm)

129

Assembly Drg No

A B C M K RDSO/T

1 in 8.5 75R S, LH 7986 10376 840 8529 1847 3931 in 8.5 75R C,FH 7986 10376 1500 8529 1847 3861 in 8.5 60R S, LH 7986 10360 840 8529 1831 3791 in 8.5 60R C, FH 7986 10360 1500 8529 1831 3721 IN 12 75R S, FH 11287 14321 1500 12021 2300 3941 IN 12 75R PARTLY C, FH 11287 14321 1500 12021 2300 4011 IN 12 60R S, FH 11287 14296 1500 12021 2275 3671 IN 12 60R PARTLY C, FH 11287 14296 1500 12021 2275 4231 IN 16 75R C, FH 12309 19054 844 16015 3039 1891 IN 12 (PSC) 52KG Partly C, FH 11287 14489 1500 12021 2468 T-6450

1 IN 8.5 (PSC) 52 KG C, FH 7986 11932 1500 8529 3403 T-6327

1 IN 8.5 90R C, FH 7981 11638 1500 8530 3108 TA-204971 IN 8.5 75R S, LH 7986 11632 840 8529 3103 TA-20404,

TA-210041 IN 8.5 75R C, FH 7986 11632 1500 8529 3103 TA-20451,

TA-210191 IN 8.5 60R S, LH 7986 11615 840 8529 2086 TA-204007,

TA-204601 IN 8.5 60R C, FH 7986 11615 1500 8529 2086 TA-20463,

TA-210071 IN 12 90R Partly C, FH 1286 15176 1500 12021 3155 TA-204161 IN 12 75R S, FH 11287 15166 1500 12021 3145 TA-204661 IN 12 75R Partly C, FH 11287 15166 1500 12021 3145 TA-20401,

TA-210011 IN 12 60R S, FH 11287 15143 1500 12021 3122 TA-20464,

TA-210161 IN 12 60R Partly C, FH 11287 15143 1500 12021 3122 TA-21410,

TA21410, TA-21010

1 IN 16 60R C, FH 12309 19635 844 16015 3680 TA-20466TA-20413, TA-21013

1 in 8.5 60R S, L H 6736 9585 840 6499 3086 TA-206041 in 12 60R S, FH 9548 12282 1500 9160 3122 TA-20601

Metre Gauge ( with CMS crossing)

Metre Gauge ( with Ordinary crossing)

Narrow Gauge (with Ordinary crossing)

Meter Gauge (with Heat Treated crossing)


CENTRE LINE LAYING (FOR METRE GAUGE)

Dimensions (mm)Crossing No Section Type

layout calculation book

Documents