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Find the value of offsets from long chord at 10 m distance, for a curve having radius 200 m and mid ordinate 15m.
14.749 m
15.749 m
12.749 m
16.749 m
14.749 m
The offset from the long chord at any distance x from the mid-point for a circular curve can be calculated using the exact standard relation Ox=R2−x2−(R−O0), where R is the radius, O0 is the mid-ordinate, and x is the distance from the midpoint.
The offset from the long chord at any distance x from the mid-point for a circular curve can be calculated using the exact standard relation Ox=R2−x2−(R−O0), where R is the radius, O0 is the mid-ordinate, and x is the distance from the midpoint.
Ox=R2−x2−(R−O0) — Exact offset from long chord at distance x
Ox≈O0−2Rx2 — Approximate offset formula when R is very large compared to L
In a circular curve, offsets are vertical ordinates drawn from the long chord to the curve · By setting up a right-angled triangle from the center of the curve to a point at horizontal distance x from the midpoint, the ordinate at distance x is derived geometrically by subtracting the distance from the long chord to the center from the radial distance to the curve at x.
Given values: R=200 m, O0=15 m, and x=10 m.
Substitute into formula: O10=2002−102−(200−15)=40000−100−185=39900−185.
Calculation: 39900≈199.7498 m.
Hence, O10=199.7498−185=14.7498 m≈14.749 m.
Simple method for setting out simple circular curves of short lengths.
Does not require angular instruments like theodolites; can be done with a chain/tape.
Not suitable for long curves or large radii due to cumulative measurement errors.
Errors increase near the ends of the chord.
Setting out simple circular curves in field surveying for roads and railways.
Laying out small radius curves using basic linear measurements.
Option A (14.749 m) is the exact calculated offset.
Option B (15.749 m), Option C (12.749 m), and Option D (16.749 m) represent incorrect mathematical evaluations.
A is correct — Using Ox=R2−x2−(R−O0), substituting R=200, O0=15, and x=10 gives $14.749\text{ m}$.
Remember that if the question asks for an approximate value or if R≫x, you can use Ox≈O0−2Rx2=15−400100=14.75 m, which yields a nearly identical answer very quickly in exams!