Small-angle geometry in surveying: What is the ratio of the linear displacement at the end of a line (arc chord) subtended by an angle of 1 second, to the length of that line? Choose the standard constant used in precise angular-to-linear conversions.

Introduction / Context:
In precise surveying and astronomy, small angles are frequently converted to linear displacements along an arc or chord. Knowing the conversion factor for one arc-second is essential when translating instrument sensitivities (seconds of arc) into ground linear errors over long sight lengths.

Given Data / Assumptions:

• Angle considered = 1 arc second.
• Small-angle approximation applies: arc length s ≈ R * θ with θ in radians.
• Comparison is a pure ratio: linear displacement : length of line.

Concept / Approach:

One radian equals 180/π degrees, or 206 265 arc-seconds. Therefore, 1 arc-second equals (π / 648 000) radians ≈ 4.8481 * 10^-6 rad. The corresponding linear displacement at the end of a line is the line length multiplied by this angular value. Hence the ratio displacement:length is approximately 1:206 265, commonly rounded as 1:206 300 for practical surveying constants.

Step-by-Step Solution:

Verification / Alternative check:

Using calculators or tables of astronomical constants yields the same figure. Many survey texts adopt 1:206 265; rounding to 1:206 300 keeps field arithmetic simple without affecting design-level accuracy.

Why Other Options Are Wrong:

1:3440 pertains to 1 minute (since 60′ ≈ 1° and 1 rad ≈ 57.3°). 1:57 refers to 1 radian in degrees, not seconds. 1:100 and 1:108 000 are unrelated magnitudes.

Common Pitfalls:

Forgetting to use radians; mixing up seconds with minutes; using chord vs arc distinctions when angles are not sufficiently small.

Final Answer:

1:206 300