Track Alignment (The Basics)
1 Introduction TOM WILSON FPWI Track Alignment Designers examine the existing track geometry, perform Technical Discipline Leader (Track) alignment calculations and propose alignment solutions which will allow WSP UK trains to safely travel at speed between A and B. For passenger train routes, the effects of travelling along the track (and PWI Vice President for Scotland) horizontal and vertical alignments at speed are an essential part of the Track Designer’s considerations.
I’m a professional Permanent Way Engineer have previous experience in providing The aim of this presentation is to provide you with a brief history of the - with more than 46 years’ experience in independent opinion and Expert Witness origins of railway track and introduce you to the basic techniques used to the Rail Industry including working on rail services for solicitors, loss adjusters and rail create modern track alignments. infrastructure projects in the UK, the infrastructure owners. middle East and in Australia. I’ll discuss things like: To do my job, I need to have an up-to-date As an experienced Track Designer and knowledge of current legislation and • How railway tracks were developed from simple wagon ways to Design Manager, I’ve been responsible for industry standards covering all aspects of modern trackforms. production of track designs on a number of track design, installation, maintenance, • The types of geometry elements used in track alignment - and how large UK projects and I’ve led multi- renewal, quality assurance and asset these are connected. disciplinary design teams in the Rail sector condition assessment. • How to use perform simple alignment calculations. in Australia. • The need for design limits to provide safe and comfortable track I’m currently the Technical Discipline systems. Leader for Track in the UK and I have 36 • How to use the design limits and alignment geometry to calculate the track engineers in my UK team with safe speed limits for trains. another 4 in our office in Bangalore.
I’ve held posts as Contractor’s Responsible Engineer, Independent Verifier and Design Checker. I’m an A1 Trained Assessor and I
2 A BRIEF HISTORY OF TRACK
T R A C K F O R M
Track TRACK FUNCTION Alignment (The Basics) TRACK DESIGN (BASICS)
DESIGN WORKSHOP
3 A BRIEF HISTORY TRACK TRACK TRACK DESIGN OF TRACK FORM FUNCTION DESIGN WORKSHOP
A (Very) Brief History of Track
4 A BRIEF HISTORY TRACK TRACK TRACK DESIGN OF TRACK FORM FUNCTION DESIGN WORKSHOP
Where did it come from?
• Heavy wagons with rigid tyres - road surfaces became damaged and rutted
• Wagon-ways were developed using stone slabs or wooden baulks to guide wheels
5 A BRIEF HISTORY TRACK TRACK TRACK DESIGN OF TRACK FORM FUNCTION DESIGN WORKSHOP
Where did it come from?
• Heavy wagons with rigid tyres - road surfaces became damaged and rutted
• Wagon-ways were developed using stone slabs or wooden baulks to guide wheels
• Guide spacing largely determined by wagon width
• Timber replaced with metal by 1793
6 A BRIEF HISTORY TRACK TRACK TRACK DESIGN OF TRACK FORM FUNCTION DESIGN WORKSHOP
Where did it come from? CONING OF WHEEL TREADS
• Heavy wagons with rigid tyres - road surfaces became damaged and rutted FLANGES
• Wagon-ways were developed using stone slabs or wooden baulks to guide wheels AXLE
• Guide spacing largely determined by wagon width
• Timber replaced with metal by 1793
WHEEL CENTRALLY • Trains are guided by the shape of wheels and the rails. PLACED ON RAIL
RAILS
SLEEPER
STRAIGHT TRACK
7 A BRIEF HISTORY TRACK TRACK TRACK DESIGN OF TRACK FORM FUNCTION DESIGN WORKSHOP
Where did it come from?
• First passenger railway in 1807 – with horse-drawn carriages…
• But the “Permanent Way” had finally arrived!
• Heavy steam engines needed stronger rails – wrought iron replaced cast iron in 1820
8 A BRIEF HISTORY TRACK TRACK TRACK DESIGN OF TRACK FORM FUNCTION DESIGN WORKSHOP
Where did it come from?
• First passenger railway in 1807 – with horse-drawn carriages…
• But the “Permanent Way” had finally arrived!
• Heavy steam engines needed stronger rails – wrought iron replaced cast iron in 1820
• Track width (rail gauge) a problem until standardised in 1845
• Narrow and Broad gauge railways still exist – and co- exist with Standard Gauge
9 A BRIEF HISTORY TRACK TRACK TRACK DESIGN OF TRACK FORM FUNCTION DESIGN WORKSHOP
TRACK FORM
10 A BRIEF HISTORY TRACK TRACK TRACK DESIGN OF TRACK FORM FUNCTION DESIGN WORKSHOP
What is it made from?
• Early railways had “T” shaped rails supported on stone blocks
• Increasing loads needed stronger rails; breaks were common
• Stone blocks were replaced by timbers (cheaper and able to hold track gauge)
• Rails with deeper sections between the supports were tried (“fish-belly” rails).
• Stone ballast allowed good drainage and enabled realignment…
11 A BRIEF HISTORY TRACK TRACK TRACK DESIGN OF TRACK FORM FUNCTION DESIGN WORKSHOP
What is it made from?
• … and little has changed; we still install timber track with metal rails on a ballast support layer
• Rails are now steel; breaks are uncommon
12 A BRIEF HISTORY TRACK TRACK TRACK DESIGN OF TRACK FORM FUNCTION DESIGN WORKSHOP
What is it made from?
• … and little has changed; we still install timber track with metal rails on a ballast support layer
• Rails are now steel; breaks are uncommon
• Highest speed lines tend to use concrete sleepers; but these are not very eco-friendly
• Steel sleepers are a more sustainable option – but these require different installation techniques
13 A BRIEF HISTORY TRACK TRACK TRACK DESIGN OF TRACK FORM FUNCTION DESIGN WORKSHOP
What is it made from?
• … and little has changed; we still install timber track with metal rails on a ballast support layer
• Rails are now steel; breaks are uncommon
• Highest speed lines tend to use concrete sleepers; but these are not very eco-friendly
• Steel sleepers are a more sustainable option – but these require different installation techniques
14 A BRIEF HISTORY TRACK TRACK TRACK DESIGN OF TRACK FORM FUNCTION DESIGN WORKSHOP
What is it made from?
• Concrete slab track is preferred in tunnels - and on the highest speed lines
• Slab track is expensive to install – but it provides good track alignment control
15 A BRIEF HISTORY TRACK TRACK TRACK DESIGN OF TRACK FORM FUNCTION DESIGN WORKSHOP
What is it made from?
• Concrete slab track is preferred in tunnels - and on the highest speed lines
• Slab track is expensive to install - but it provides good track alignment control
• Modular pre-cast units allow minor alignment adjustment
16 A BRIEF HISTORY TRACK TRACK TRACK DESIGN OF TRACK FORM FUNCTION DESIGN WORKSHOP
What is it made from?
• Concrete slab track is preferred in tunnels - and on the highest speed lines
• Slab track is expensive to install - but it provides good track alignment control
• Modular pre-cast units allow minor alignment adjustment
• Transitions to ballasted track are challenging – support stiffness needs to be uniform
17 A BRIEF HISTORY TRACK TRACK TRACK DESIGN OF TRACK FORM FUNCTION DESIGN WORKSHOP
TRACK FUNCTION
M O D U L E 1 : TRACK ALIGNMENT GLOBAL ACADEMY FOR RAIL & TRANSIT 18 A BRIEF HISTORY TRACK TRACK TRACK DESIGN OF TRACK FORM FUNCTION DESIGN WORKSHOP
What does it do?
• Fulfils a service need between A and B
19 A BRIEF HISTORY TRACK TRACK TRACK DESIGN OF TRACK FORM FUNCTION DESIGN WORKSHOP
What does it do? Q= 125kN Surface Pressure on Track Formation
2 2 • Fulfils a service need between A and B Ap = 3 cm ; p = 41667 N/ cm
2 2 Ap = 200 cm ; p = 625 N/ cm • Safely guides rail vehicles 2 2 Ap = 510 cm ; p = 245 N/ cm
• Transfers loads evenly down to the subgrade 2 2 Ap = 2380 cm ; p = 53 N/ cm (formation) 2 2 Ap = 10100 cm ; p = 12 N/ cm
LOAD LOAD
SLEEPER PLATES / PADS & FASTENING SLEEPER BALLAST
CAPPING LAYER (Sub-Ballast)
FORMATION (Subgrade)
20 A BRIEF HISTORY TRACK TRACK TRACK DESIGN OF TRACK FORM FUNCTION DESIGN WORKSHOP
What does it do? LOAD LOAD
SLEEPER PLATES / PADS & FASTENING SLEEPER • Fulfils a service need between A and B BALLAST
• Safely guides rail vehicles CAPPING LAYER (Sub-Ballast)
• Transfers loads evenly down to the subgrade FORMATION (Subgrade) (formation)
5000 46 17 2 kg/cm2 kg/cm2 kg/cm2 kg/cm2 WHEEL
RAIL
SLEEPER BALLAST LOAD BEARING PYRAMIDS OF BALLAST
FORMATION LAYERS
21 A BRIEF HISTORY TRACK TRACK TRACK DESIGN OF TRACK FORM FUNCTION DESIGN WORKSHOP
What does it do?
• Fulfils a service need between A and B
• Safely guides rail vehicles
• Transfers loads evenly down to the subgrade (formation)
• Enables drainage and avoids flooding
• Resilient; permanent deformation is resisted
22 A BRIEF HISTORY TRACK TRACK TRACK DESIGN OF TRACK FORM FUNCTION DESIGN WORKSHOP
What does it do?
• Fulfils a service need between A and B
• Safely guides rail vehicles
• Transfers loads evenly down to the subgrade (formation)
• Enables drainage and avoids flooding
• Resilient; permanent deformation resisted
• Allows replacement of worn or defective components
23 A BRIEF HISTORY TRACK TRACK TRACK DESIGN OF TRACK FORM FUNCTION DESIGN WORKSHOP
TRACK DESIGN
24 A BRIEF HISTORY TRACK TRACK TRACK DESIGN OF TRACK FORM FUNCTION DESIGN WORKSHOP
How do we create Track Designs?
2
• Track design combines 3 geometric elements v 1 3 • Curves, spirals and straights C • Curves with a single radius are “Simple”; those with more than one radius are “Compound”
Curve Radius Formula 퐂ퟐ R 퐑 = ( ൗퟖ퐯 ) m (Note: All units must be the same)
Where: • R is the radius of the curve • C is the length of the chord between two points (i.e. from point 1 to point 3 in the diagram) • v is the offset measured at mid-chord (i.e. at point 2) 25 A BRIEF HISTORY TRACK TRACK TRACK DESIGN OF TRACK FORM FUNCTION DESIGN WORKSHOP
Geometry – Curves
2
• Track design combines 3 geometric elements v 1 3 • Curves, spirals and straights C • Curves with a single radius are “Simple”; those with more than one radius are “Compound”
• Simple tools can be used to measure curve radius.
• All you need are a tape measure, wire and a ruler. R
C2 Example of using R = ൗ8v When: • C = 10 m , v = 23 mm ? • R= Τ? • R = ? m 26 A BRIEF HISTORY TRACK TRACK TRACK DESIGN OF TRACK FORM FUNCTION DESIGN WORKSHOP
Geometry – Curves
2
• Track design combines 3 geometric elements v 1 3 • Curves, spirals and straights C • Curves with a single radius are “Simple”; those with more than one radius are “Compound”
• Simple tools can be used to measure curve radius.
• All you need are a tape measure, wire and a ruler. R
C2 Example of using R = ൗ8v When: • C = 10 m , v = 23 mm ퟏퟎ ퟐ • R= Τퟖ∗ퟎ.ퟎퟐퟑ (Remember : All units must be the same) • R = 543.478 m 27 A BRIEF HISTORY TRACK TRACK TRACK DESIGN OF TRACK FORM FUNCTION DESIGN WORKSHOP
Geometry – Curves
2
• Track design combines 3 geometric elements v 1 3 • Curves, spirals and straights C • Curves with a single radius are “Simple”; those with more than one radius are “Compound”
• Simple tools can be used to measure curve radius.
• All you need are a tape measure, wire and a ruler. R
C2 Using R = ൗ8v When: • C = 20 m , v = 0.05 m 400 • R= Τ0.40 • R = 1000 m 28 A BRIEF HISTORY TRACK TRACK TRACK DESIGN OF TRACK FORM FUNCTION DESIGN WORKSHOP
Geometry – Curves
2
• Track design combines 3 geometric elements v 1 3 • Curves, spirals and straights C • Curves with a single radius are “Simple”; those with more than one radius are “Compound”
• Simple tools can be used to measure curve radius.
• All you need are a tape measure, wire and a ruler. R
Alternatively, using the relationship between chord length and versine: 12,500 R = Τv for a 10 m chord • C = 10 m , v = 23 mm 12,500 • R = Τ23 ퟓퟎ,ퟎퟎퟎ • R = 543.478 m (Note: R = Τv - but only for a 20 m chord) 29 A BRIEF HISTORY TRACK TRACK TRACK DESIGN OF TRACK FORM FUNCTION DESIGN WORKSHOP
Geometry – Spirals
• Track design combines 3 geometric elements
• Curves, spirals and straights
• Spirals connect curves of different radii and straights with curves.
Spiral curves provide a controlled gradual change in curvature: • between two curves with different radii or • between a curve and a straight or • between two curves of a different ‘hand’. 30 A BRIEF HISTORY TRACK TRACK TRACK DESIGN OF TRACK FORM FUNCTION DESIGN WORKSHOP
Geometry – Spirals
• Track design combines 3 geometric elements
• Curves, spirals and straights
• Spirals connect curves of different radii and straights with curves.
• These “transition” elements provide a smooth (comfortable) change in radius.
In this example, a directly reversing spiral curve is located opposite the station platforms.
31 A BRIEF HISTORY TRACK TRACK TRACK DESIGN OF TRACK FORM FUNCTION DESIGN WORKSHOP
Geometry – Spirals
• Track design combines 3 geometric elements
• Curves, spirals and straights
• Spirals connect curves of different radii and straights with curves.
• These “transition” elements provide a smooth (comfortable) change in radius.
32 A BRIEF HISTORY TRACK TRACK TRACK DESIGN OF TRACK FORM FUNCTION DESIGN WORKSHOP
Horizontal (Plan) Geometry
• Relies on accurate survey and control
• Limited by rules and standards
• Underpins Track Form and Function, and safety for all rail traffic
• Controls the physical forces (for passenger comfort and safety)
Track needs to compensate for the outward centrifugal force on curves (ƒ
* lateral acceleration, ac).
To fully balance this force, the outer rail is raised above the inner rail to create the Equilibrium Cant (Eq) where the forces generated by speed around the curve are zero.
Eq = 11.82 * speed2 (km/h) ÷ radius (m) when track gauge = 1432 mm (and S = 1502 mm)
33 A BRIEF HISTORY TRACK TRACK TRACK DESIGN OF TRACK FORM FUNCTION DESIGN WORKSHOP
Horizontal (Plan) Geometry
• Relies on accurate survey and control
• Limited by rules and standards
• Underpins Track Form and Function, and safety for all rail traffic
• Controls the physical forces (for passenger comfort and safety)
But hang on a minute…
Where does the 11.82 “magic number” come from? A formula!
EN13803-1 describes the variables and derives the Constant “C” (the “magic number”) as:
“The distance between the wheel contact points divided by the speed conversion factor (usually km/h) multiplied by the speed of earth’s gravity.” 34 A BRIEF HISTORY TRACK TRACK TRACK DESIGN OF TRACK FORM FUNCTION DESIGN WORKSHOP
Horizontal (Plan) Geometry
• Relies on accurate survey and control
• Limited by rules and standards
• Underpins Track Form and Function, and safety for all rail traffic
• Controls the physical forces (for passenger comfort and safety)
Q. If the required speed of a train is 80km/h and the horizontal curve radius is 1,000 m, what would be the value of cant (to the nearest mm) needed to be applied to the track to achieve equilibrium?
Step 1. Use formula Eq = 11.82 * speed2 (km/h) ÷ radius (m) Step 2. Eq = 11.82 * (80) 2 /1000 Step 3. Eq = 76 mm
35 A BRIEF HISTORY TRACK TRACK TRACK DESIGN OF TRACK FORM FUNCTION DESIGN WORKSHOP
Horizontal (Plan) Geometry
• Relies on accurate survey and control
• Limited by rules and standards
• Underpins Track Form and Function, and safety for all rail traffic
• Controls the physical forces (for passenger comfort and safety)
But – wouldn’t applying Equilibrium Cant (Eq) for the highest speed trains then over-compensate the centrifugal forces experienced in slower trains? Indeed it would – so we apply some cant and introduce some “uncompensated superelevation” (aka “cant deficiency”) to reduce the risk of derailment and help steer the bogies around the curve.
Passengers on the highest speed trains feel some centrifugal force outwards – and passengers on slower trains may even feel force pulling inwards.
Cant deficiency (D) = Eq – E(applied) 36 A BRIEF HISTORY TRACK TRACK TRACK DESIGN OF TRACK FORM FUNCTION DESIGN WORKSHOP
Horizontal (Plan) Geometry
• Relies on accurate survey and control
• Limited by rules and standards
• Underpins Track Form and Function, and safety for all rail traffic
• Controls the physical forces (for passenger comfort and safety)
Q. Calculate the cant deficiency for a horizontal curve of radius 3300 m at 100 mph and applied cant of 55 mm?
Step 1. Use formula Eq = 11.82 * speed2 (km/h) ÷ radius (m) Step 2. Eq= Ea + D = 11.82 * (100*1.609344)2 ÷ 3300 Step 3. 55 + D = 11.82 * (25899.8811) ÷ 3300 Step 4. 55 + D = 93 Step 5. D = 93 – 55 Step 6. D = 38 37 A BRIEF HISTORY TRACK TRACK TRACK DESIGN OF TRACK FORM FUNCTION DESIGN WORKSHOP
Horizontal (Plan) Geometry
The LIMITS:
• The maximum speed is limited by curve radius and cant + cant deficiency
• ‘Normal’ cant deficiency is limited to 110 mm and that then limits speed
• Traditional application of Ea and D: D=2/3 Ea
• D < Ea (provided other limits not exceeded)
• ‘Normal’ Design rate of change is 35 mm/s (for both Ea and D)
2 To obtain Vmax from the formula Ea + D = 11.82 * (Vmax) ÷ R ... We need to move things around a bit (i.e. we transpose the formula)
... And we obtain Vmax = √[(Ea + D ÷ 11.82 ) *R] ...
38 A BRIEF HISTORY TRACK TRACK TRACK DESIGN OF TRACK FORM FUNCTION DESIGN WORKSHOP
Horizontal (Plan) Geometry
The LIMITS:
• The maximum speed is limited by curve radius and cant + cant deficiency
• ‘Normal’ cant deficiency is limited to 110 mm and that then limits speed
• Traditional application of Ea and D: D=2/3 Ea
• D < Ea (provided other limits not exceeded)
• ‘Normal’ Design rate of change is 35 mm/s (for both Ea and D)
2 Original Formula: (Ea + D) = 11.82 * (Vmax) ÷ R
2 Step 1. (Ea + D) ÷ 11.82 = (Vmax) ÷ R 2 Step 2. {(Ea + D) ÷11.82}*R = (Vmax) Step 3. √[{(Ea + D) ÷ 11.82} *R] = Vmax
Transposed Formula: Vmax = √[{(Ea + D) ÷ 11.82} *R] 39 A BRIEF HISTORY TRACK TRACK TRACK DESIGN OF TRACK FORM FUNCTION DESIGN WORKSHOP
Horizontal (Plan) Geometry
The LIMITS:
• The maximum speed is limited by curve radius and cant + cant deficiency
• ‘Normal’ cant deficiency is limited to 110 mm and that then limits speed
• Traditional application of Ea and D: D=2/3 Ea
• D < Ea (provided other limits not exceeded)
• ‘Normal’ Design rate of change is 35 mm/s (for both Ea and D)
Q. On a section of existing track, the horizontal curve radius is 600 m and the existing cant is 70 mm. What is the maximum speed (to the nearest km/h) that trains could travel on this section of track?
Step 1. Assume Ea + D is 70 +110 = 180 mm Step 2. Use formula Vmax = √[({Ea + D} ÷ 11.82 ) *R] Step 3. Vmax = √[(180 ÷ 11.82 ) *600] Step 4. Vmax = (95.588) = 95 km/h (rounding down) 40 A BRIEF HISTORY TRACK TRACK TRACK DESIGN OF TRACK FORM FUNCTION DESIGN WORKSHOP
Vertical Geometry
• Vertical acceleration is also limited by rules and standards
• Gradients are limited by traction capability to 1:80
L2 • Steepest UK mainline gradient is the Lickey Incline (Bromsgrove) at 1:37 (2.703%) L1
• Gradients are connected by (large radius) vertical curves – limited by radius (1000 m) and by comfort 1200 1210 1220 (g-force)… 1150 1160 1170 1180 1190
Gradient Calculation: For example: Level at L1 – Level at L2 ÷ Distance between L1 and L2 Level L1 = 84.149 m, L2 =84.216 m and Chainage L1 = 559.386 m, L2 =629.386 m Gives the Gradient as a number (usually percentage) Level difference = (84.216 m-84.119 m) = 97 mm Distance = 629.386 m - 559.386 m = 70 m OR Distance between L1 and L2 ÷ Level at L1 – Level at L2 Gradient = (97/70,000*100) OR Gradient = (70,000 ÷ 97) Gives the Gradient as a ratio (usually 1 in nnn) Gradient = 0.139% Gradient = 1 in 722 (rising)
41 A BRIEF HISTORY TRACK TRACK TRACK DESIGN OF TRACK FORM FUNCTION DESIGN WORKSHOP
Vertical Geometry
• Vertical acceleration is also limited by rules and standards
• Gradients are limited by traction capability to 1:80
L2 • Steepest UK mainline gradient is the Lickey Incline (Bromsgrove) at 1:37 (2.703%) L1
• Gradients are connected by (large radius) vertical curves – limited by radius (1000 m) and by comfort 1200 1210 1220 (g-force)… 1150 1160 1170 1180 1190
Vertical Acceleration: % g = speed2 (mph) ÷ (radius (m) *0.491) Vertical Acceleration: Design Limit = 6% g (or 0.589 m/s2)) Q. Could the line speed be increased from 50 to say, 60 mph over a vertical curve of radius 1100 m?
For example: VC = 1100 m Line speed = 50 mph Answer: % g = (50)2 mph ÷ (1100 m *0.491) Step 1. Using formula % g = (speed)2 mph ÷ (radius *0.491) = 4.63% g (pass) Step 2. % g = (60)2 ÷ (1100 *0.491) Step 3. % g = 3600 ÷ 540.1 Step 4. % g = 6.67% g (fail – it exceeds the 6% g limit) 42 HISTORY TRACK TRACK TRACK DESIGN OF TRACK FORM FUNCTION DESIGN WORKSHOP
DESIGN WORKSHOP
43 A BRIEF HISTORY TRACK TRACK TRACK DESIGN OF TRACK FORM FUNCTION DESIGN WORKSHOP
Bring it all together TRACK DESIGN EXAMPLE
• In this track design the values in the Magenta boxes need to be calculated
• The element on the LHS of the transition is Straight
• The Up line is shown RED
• The Down line is shown BLUE
44 A BRIEF HISTORY TRACK TRACK TRACK DESIGN OF TRACK FORM FUNCTION DESIGN WORKSHOP
Bring it all together TRACK DESIGN EXAMPLE
• In this track design the values in the Magenta boxes need to be calculated
• The element on the LHS of the transition is Straight
• The Up line is shown RED
• The Down line is shown BLUE
• Calculate the missing value circled
Formula Hints: kph = mph* 1.609344 Ea+D = 11.82*speed2(km/h) ÷ radius (m) 45 A BRIEF HISTORY TRACK TRACK TRACK DESIGN OF TRACK FORM FUNCTION DESIGN WORKSHOP
Bring it all together TRACK DESIGN EXAMPLE
• In this track design the values in the Magenta boxes need to be calculated
• The element on the LHS of the transition is Straight
• The Up line is shown RED
• The Down line is shown BLUE
• Calculate the missing value circled
Step by step:
Step 1. 70 mph *1.609344 = 112.654 km/h Step 2. E + D = 11.82 * (112.654) 2 ÷ 1700 Step 3. E + D = 88 mm Step 4. E = 88-23 = 65 mm
46 A BRIEF HISTORY TRACK TRACK TRACK DESIGN OF TRACK FORM FUNCTION DESIGN WORKSHOP
Bring it all together TRACK DESIGN EXAMPLE
• In this track design the values in the Magenta boxes need to be calculated
• The element on the LHS of the transition is Straight
• The Up line is shown RED
• The Down line is shown BLUE
• Calculate the missing value circled
Formula Hints: m/s = mph* 1.609344 ÷ 3.6 (in this case 31.293 m/s) RoC = Ea (or D) ÷ TL (m) * speed (m/s) (Remember: Ea is 65 mm on this curve) 47 A BRIEF HISTORY TRACK TRACK TRACK DESIGN OF TRACK FORM FUNCTION DESIGN WORKSHOP
Bring it all together TRACK DESIGN EXAMPLE
• In this track design the values in the Magenta boxes need to be calculated
• The element on the LHS of the transition is Straight
• The Up line is shown RED
• The Down line is shown BLUE
• Calculate the missing value circled
Step by step:
Step 1. 70 mph * 1609.344 = 112654 m/h Step 2. 112654 ÷ 3600 = 31.293 m/s Step 3. TL = 95 m ÷ 31.293 m/s = 3.036 s Step 4. RcE = 65 mm ÷ 3.036 s = 21 mm/s
48 A BRIEF HISTORY TRACK TRACK TRACK DESIGN OF TRACK FORM FUNCTION DESIGN WORKSHOP
Bring it all together TRACK DESIGN EXAMPLE
• In this track design the values in the Magenta boxes need to be calculated
• The element on the LHS of the transition is Straight
• The Up line is shown RED
• The Down line is shown BLUE
• Calculate the missing value circled
Formula Hints: Cg = TL ÷ Ea (when units are the same) (Remember: Ea is 65 mm on this curve) 49 A BRIEF HISTORY TRACK TRACK TRACK DESIGN OF TRACK FORM FUNCTION DESIGN WORKSHOP
Bring it all together TRACK DESIGN EXAMPLE
• In this track design the values in the Magenta boxes need to be calculated
• The element on the LHS of the transition is Straight
• The Up line is shown RED
• The Down line is shown BLUE
• Calculate the missing value circled
Step by step:
Step 1. Cant difference = 65-0 = 65 mm Step 2. TL = 80 m Step 3. Cg = 80 ÷ 0.065 = 1 in 1231 or Step 3 Cg = 80,000 ÷ 65 = 1 in 1231
50 Your turn for Questions…
CONTACT DETAILS
T O M W I L S O N Head of Discipline (Track) WSP in the UK
51