Transition curves – length limit by cant deficiency: If S is the cant deficiency in centimetres and V is the maximum permissible speed in km/h, what is the recommended maximum length L of the transition curve (in metres)?

Introduction / Context:
Transition curves provide a gradual change from straight track to circular curvature, controlling the rate of change of cant and cant deficiency for passenger comfort and safety. Standards limit the allowable rate of cant (and cant deficiency) change, which governs the minimum/maximum transition length.

Given Data / Assumptions:

• Cant deficiency S is given in centimetres.
• Speed V is in kilometres per hour.
• We seek a practical expression for maximum transition length constrained by the rate of change of cant deficiency.

Concept / Approach:
Common railway practice gives the relation L = 0.008 * V * S when S is in millimetres. Since 1 cm = 10 mm, substituting S(cm)*10 for S(mm) yields L = 0.008 * V * 10 * S(cm) = 0.08 * V * S(cm). This provides L in metres when V is in km/h and S in cm.

Step-by-Step Solution:
1) Start with standard: L = 0.008 * V * S(mm).2) Convert S(cm) to S(mm): S(mm) = 10 * S(cm).3) Substitute: L = 0.008 * V * (10 * S) = 0.08 * V * S.4) Units: V in km/h, S in cm → L in metres.

Verification / Alternative check:
The expression aligns with comfort-based cant gradient limits and is widely used as a quick design check before detailed alignment computations.

Why Other Options Are Wrong:

• 0.008 * V * S: correct only if S is in millimetres, not centimetres.
• 0.8 * V * S and 0.028 * V * S: produce unrealistically long or short transitions.
• 0.8 * S / V: incorrect dependence on speed.

Common Pitfalls:

• Mixing units (mm vs cm) leading to order-of-magnitude errors.
• Ignoring other constraints such as available length between point of curvature and point of switch.

Final Answer:
L = 0.08 * V * S.