Track geometry — compute the twist per metre run on a straight railway track from rail levels at two points 100 m apart (A and B).

Introduction / Context:“Twist” is the rate of change of cross-level (difference in elevation between the two rails) per unit length along the track. Excessive twist is a safety hazard and is tightly controlled in track standards. This problem practices converting absolute rail levels into cross-levels and then into twist per metre.

Given Data / Assumptions:

• At Point A (on two rails): levels = 100.550 m and 100.530 m.
• At Point B (100 m away): levels = 100.585 m and 100.515 m.
• Track is straight; twist is calculated as change in cross-level divided by longitudinal distance.

Concept / Approach:

Cross-level at a point = elevation of one rail minus the other. Twist = (cross-level at B − cross-level at A) / distance. Convert the difference into millimetres and divide by metres to obtain mm/m. Careful, consistent rail identification ensures correct sign; magnitude is usually reported for maintenance limits.

Step-by-Step Solution:

Verification / Alternative check (if short method exists):

Compute individual rail gradients and re-derive cross-levels to confirm; results will be identical. A quick reasonableness check: a 50 mm change over 100 m is small and within typical track tolerances for gradual twist.

Why Other Options Are Wrong:

0.4, 0.7, 0.8, and 1.0 mm/m do not match the exact arithmetic from the given levels and spacing. Only 0.5 mm/m equals 50 mm over 100 m.

Common Pitfalls (misconceptions, mistakes):

Mixing up which rail is higher; using average levels instead of cross-level; forgetting to convert metres to millimetres; dividing by 100 cm instead of 100 m.

Final Answer:

0.5 mm