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How MRT Track Curves, Cant and Alignment Work Using Mathematics: How a City Route Becomes a Railway Path

Three secondary students review open books and notes together in a classroom.

An MRT line is drawn on a map as a coloured curve. A real railway must turn that curve into a three-dimensional path that steel wheels can follow at speed.

Radius, transition curves, cant, gradients, vertical curves, clearance and operating speed must agree closely enough that the train can remain guided, comfortable, maintainable and buildable.

A route planner can draw a line between two places.

A track engineer has to answer much harder questions:

  • How sharply may the line curve?
  • How quickly may curvature begin?
  • How much higher should the outer rail be than the inner rail?
  • How steep may the track climb?
  • How should one gradient change into another?
  • Will the train body remain inside the structure gauge?
  • Will the geometry still produce acceptable ride and wheel–rail forces at operating speed?

LTA’s current Civil Design Criteria make this systems view explicit. The December 2025 track-alignment chapter says alignment design must coordinate with operations, signalling, traction power, rolling stock, track maintenance, construction constraints, structures, utilities, geotechnical and tunnelling conditions, environmental impact and land use. It also states that passenger riding comfort, wear and maintenance are core design considerations.

This article owns the three-dimensional track alignment: horizontal curves, transition curves, cant, cant deficiency, gradients, vertical curves, clearances and the way geometry sets a feasible speed envelope. Turnout geometry remains with Switches, Crossovers and Turnouts. Track-condition measurement remains with Track Inspection.

The RFE — What Is Track Alignment Actually For?

The weak answer is:

connect the stations with rails

The stronger Reason for Existence is:

turn a desired city route into a continuous three-dimensional rail path that trains can traverse at useful speed while preserving wheel guidance, passenger comfort, structural clearance, maintainability and constructability.

Prompt 1 — What Is Curvature?

For a circular curve of radius R:

κ = 1/R

Large radius means small curvature.

Small radius means a sharper curve.

At train speed v, the lateral acceleration scale is:

ay = v²/R

This is why speed and radius cannot be chosen independently.

At the same radius, doubling speed multiplies lateral acceleration by four.

At the same speed, doubling radius halves it.

Prompt 2 — What Does Cant Do?

Cant, or superelevation, raises the outer rail above the inner rail on a curve.

The tilted track allows part of the train’s weight to contribute to balancing the lateral acceleration required for the curve.

A simple equilibrium relationship is:

tanθ ≈ v²/(gR)

For small cant angle θ and track gauge G:

D ≈ G v²/(gR)

where D is equilibrium cant.

LTA’s public 2025 criteria define cant as the vertical difference between the two running rails and explicitly relate permissible speed to radius, applied cant, cant deficiency and rates of change.

The applied cant is rarely chosen to perfectly balance every possible train speed because the same curve may be traversed under different operating states.

Prompt 3 — What Is Cant Deficiency?

Suppose equilibrium cant required for speed v is Deq but actual applied cant is D.

cant deficiency I = Deq − D

Positive cant deficiency means the train is travelling faster than the speed perfectly balanced by the applied cant.

Passengers then receive more uncompensated lateral acceleration.

If the applied cant is greater than equilibrium cant for the actual speed, there is excess cant.

This is one reason very high cant is not automatically better.

Cant is a compromise between geometry and the range of speeds the railway actually intends to run.

Prompt 4 — Why Are Transition Curves Necessary?

If a straight track changed instantly into a circular curve, curvature would jump:

κ: 0 → 1/R instantly

Lateral acceleration would therefore change abruptly.

A transition curve makes curvature grow progressively.

For a clothoid-like approximation:

κ(s) = s/A²

where s is distance along the transition and A is the spiral parameter.

At speed v:

dκ/dt = v dκ/ds

Since lateral acceleration is approximately v²κ:

day/dt ≈ v³ dκ/ds

This is a lateral-jerk relationship.

Transition length therefore depends strongly on speed.

Prompt 5 — Why Must Cant Also Change Gradually?

The outer rail cannot jump vertically at the start of a curve.

Cant is introduced over distance.

cant gradient = dD/ds

For train speed v:

dD/dt = v dD/ds

LTA’s 2025 criteria explicitly include cant gradient and rate of change of cant when defining permissible speed.

The passenger therefore feels not only the amount of cant and lateral acceleration, but how quickly those quantities change.

Prompt 6 — What About Hills and Vertical Curves?

Vertical alignment controls gradient.

For rise Δz over horizontal distance Δx:

gradient i ≈ Δz/Δx

Along an uphill gradient angle θ, the train must overcome:

Fgrade = mg sinθ ≈ mgi

Gradients therefore affect traction energy and braking.

A sudden change from one gradient to another would create an abrupt vertical-acceleration change and potentially clearance issues.

Vertical curves smooth that transition.

For a simple parabolic vertical curve, elevation can be represented as:

z(x)=z0+i0x+(A/2L)x²

where A is algebraic gradient change and L the vertical-curve length.

Prompt 7 — Why Does Alignment Affect Structure Gauge?

A train body does not stay exactly above the track centreline on a curve.

Its ends and middle sweep laterally because the car is a long chord between bogies.

For a chord length L over a circle of radius R, mid-ordinate offset is approximately:

m ≈ L²/(8R)

Smaller radius increases throw.

Cant also tilts the vehicle envelope.

That means tunnel, platform and structure clearances must be coordinated with the actual curved and canted vehicle path, not with a rectangular train placed on a flat straight line.

Prompt 8 — Why Is the Best Alignment Not Simply the Gentlest Possible?

Very large radii and very shallow gradients are attractive for train dynamics.

But a city contains:

  • buildings;
  • roads;
  • utilities;
  • foundations;
  • property boundaries;
  • ground conditions;
  • existing rail lines;
  • stations that need useful entrances.

The alignment problem is therefore constrained optimisation.

minimise
travel time
+ construction burden
+ land impact
+ maintenance burden
+ energy
+ ride penalty

subject to
curve + cant + gradient + clearance + station + geology + structure constraints

The mathematically straightest railway through a city could be the physically impossible railway.

A Fictional Curve-and-Cant Example

Consider a fictional curve with radius R=500 m.

Train speed is 20 m/s, about 72 km/h.

ay=v²/R
  =400/500
  =0.8 m/s²

For gauge G=1.435 m, equilibrium cant estimate is:

D≈Gv²/(gR)
 ≈1.435×400/(9.81×500)
 ≈0.117 m
 ≈117 mm

Suppose applied cant is a fictional 90 mm.

cant deficiency ≈117−90=27 mm

Now reduce speed to 15 m/s.

Deq≈1.435×225/(9.81×500)
   ≈66 mm

The same 90 mm applied cant is now greater than equilibrium cant.

The geometry has not changed.

The dynamic balance has.

Deletion Tests

  • Remove radius: every horizontal curve produces the same lateral acceleration.
  • Remove speed: curvature is equally demanding at walking speed and line speed.
  • Remove cant: the rails can never help balance lateral acceleration.
  • Remove transition curves: curvature and lateral acceleration can jump instantly.
  • Remove cant gradient: rail-height difference changes without a transition.
  • Remove vertical alignment: traction and braking ignore hills.
  • Remove clearances: train bodies can sweep through structures on curves.
  • Remove constructability: the ideal alignment can pass through foundations and utilities.

Alignment Paradoxes

  • A larger radius improves dynamics but may make the railway harder to fit into the city.
  • More cant can improve one speed and worsen another.
  • A curve can be comfortable at one speed and uncomfortable at another without moving a millimetre.
  • The track centreline can be perfectly clear while the train body still needs more clearance because of curve throw.
  • A slightly longer railway can create a better whole journey if it permits higher speed, easier stations or lower construction risk.

The Alignment Audit

  1. What horizontal radius is proposed?
  2. What speed is intended through it?
  3. What equilibrium cant follows?
  4. What cant is actually applied?
  5. What cant deficiency or excess remains?
  6. How quickly does curvature change?
  7. How quickly does cant change?
  8. What lateral jerk is received?
  9. What gradients and vertical curves exist?
  10. What traction and braking consequences follow?
  11. What train-body throw and cant tilt affect clearances?
  12. What station, tunnel, viaduct, geology and utility constraints shape the feasible alignment?
  13. What wear and maintenance burden follows from the chosen geometry?
  14. What measured ride and track data would force the design assumptions to be reconsidered?

World Return — The Train Tests the Geometry

design radius, cant and transitions
→ build track
→ survey geometry
→ run train
→ measure acceleration, ride and wheel–rail behaviour
→ inspect wear
→ compare with prediction
→ maintain or recalibrate

If passenger lateral acceleration exceeds prediction, speed, cant, suspension or geometry assumptions may be wrong.

If one curve develops unusual wear, the wheel–rail and alignment models must meet.

If repeated surveys show geometry migration, the track-support or structural system becomes part of the investigation.

MRT track alignment works when a line drawn through the city becomes a continuous physical path whose geometry, speed and train response remain compatible in the real world.

Key Equations

κ=1/R
Curvature

ay=v²/R
Lateral acceleration

tanθ≈v²/(gR)
Equilibrium banking relation

D≈Gv²/(gR)
Equilibrium cant approximation

I=Deq−D
Cant deficiency

κ(s)=s/A²
Clothoid-style transition curvature

day/dt≈v³ dκ/ds
Lateral jerk scale

dD/dt=v dD/ds
Rate of cant change

i≈Δz/Δx
Track gradient

Fgrade≈mgi
Gradient force

z(x)=z0+i0x+(A/2L)x²
Simple vertical-curve form

m≈L²/(8R)
Vehicle chord offset on a curve

Reader-safety note: This article uses public track-design principles and fictional teaching values. It does not reproduce non-public Singapore MRT alignments, exact local curve radii, speed restrictions, structure-clearance margins or detailed design exceptions.

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