Plane surveying on a spherical Earth: arc–chord difference over 18.2 km For a ground distance of 18.2 km measured along the Earth’s surface, what is the approximate difference between the arc length and its subtended chord length (take Earth radius ≈ 6,370 km)?

Introduction / Context:
On large-scale surveys, the difference between an arc along the Earth and the corresponding straight chord can matter in precision calculations. For moderate distances, a convenient approximation estimates this small difference accurately enough for engineering purposes. This question checks your ability to apply the standard small-angle formula for arc–chord difference.

Given Data / Assumptions:

• Arc length s = 18.2 km = 18,200 m.
• Mean Earth radius R ≈ 6,370 km = 6,370,000 m.
• Small-angle approximation is valid for s ≪ 2πR.

Concept / Approach:

For small central angle, the difference (s − c) between arc length s and chord length c is approximated by s^3 / (24 * R^2). This follows from c = 2R * sin(s/(2R)) and the series sin x ≈ x − x^3/6 for small x. Substituting x = s/(2R) and simplifying yields the compact cubic expression in s with R constant.

Step-by-Step Solution:

Verification / Alternative check:

A direct evaluation using c = 2R * sin(s/(2R)) with calculator precision gives the same order of magnitude (a few millimetres). For typical engineering mapping, rounding to 1 cm is fully justified.

Why Other Options Are Wrong:

5 cm, 10 cm, and 100 cm greatly overestimate the arc–chord difference for only 18.2 km; such differences occur over far longer spans.

Common Pitfalls:

Confusing sagitta with arc–chord difference; misusing units (km vs m); neglecting the 24 in the denominator which comes from series expansion.

Final Answer:

1 cm