Track geometry check using three-axle level readings: Consecutive axles A, B, and C are 1.8 m apart with rail levels 100.505 m, 100.530 m, and 100.525 m respectively. What is the unevenness value (mid-chord deviation at B)?
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A0.035 m
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B0.045 m
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C0.055 m
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D0.065 m
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E0.015 m
Answer
Correct Answer: 0.015 m
Explanation
Introduction / Context:Rail unevenness over a short base is commonly assessed by comparing the middle level with the mean of the two end levels (three-point or mid-chord offset method). This measure indicates the local vertical deviation responsible for rough riding and dynamic wheel loads.
Given Data / Assumptions:
- Axle spacing (A–B and B–C) = 1.8 m each.
- Rail levels: A = 100.505 m, B = 100.530 m, C = 100.525 m.
- Unevenness at B is taken as the absolute mid-chord deviation: |B − (A + C)/2|.
Concept / Approach:For three equal stations, the mid-chord deviation at the middle station is computed by subtracting the average of the two end levels from the middle level. This corresponds to the versine-like deviation used in track quality indices.
Step-by-Step Solution:
Compute mean of end levels: (A + C)/2 = (100.505 + 100.525)/2 = 100.515 m.Compute deviation at B: Δ = B − mean = 100.530 − 100.515 = 0.015 m.Take absolute value (unevenness is a magnitude): 0.015 m.Verification / Alternative check:The second-difference A − 2B + C = −0.030 m corroborates curvature; the corresponding mid-chord deviation equals 0.015 m (half the magnitude of the second difference for equal spacing).
Why Other Options Are Wrong:
- 0.035, 0.045, 0.055, 0.065 m: Do not match the mid-chord computation and would imply much larger local irregularity.
Common Pitfalls:
- Confusing total level differences with mid-chord deviation.
- Using maximum minus minimum level (0.025 m) instead of the defined unevenness at the center point.
Final Answer:0.015 m