Alignment Error – 100 m Tape with Mid-point 0.50 m Off Line (Two 50 m Straight Sections) A 100 m measurement is made in two straight 50 m sections, but the 50 m joint point is 0.50 m off the true straight line (forming a small kink). What is the resulting alignment error in the measured distance?

Introduction / Context:
Alignment errors occur when a tape measurement deviates laterally from the intended straight line. A common case is when two consecutive tape lengths meet at a small lateral offset at the joint. Estimating the induced excess length is important for quality control in baseline or detail measurements.

Given Data / Assumptions:

• Total measurement nominally 100 m as two straight 50 m segments.
• Joint (50 m point) is laterally 0.50 m off the ideal straight line.
• Each 50 m segment is straight; the kink occurs only at the joint.

Concept / Approach:

Each 50 m segment forms the hypotenuse of a right triangle whose base lies along the true straight and whose perpendicular is d/2 for the first and d/2 for the second in opposite directions when the kink is centered, or equivalently consider two chords subtending a small angle. For a small lateral deviation d at the joint, the excess over the straight is approximately d^2 / (2L) for each segment of length L, giving a total excess of d^2 / L for two segments. With L = 50 m and d = 0.50 m, total error is 0.25 / 50 = 0.005 m (5 mm).

Step-by-Step Solution:

Verification / Alternative check:

Pythagoras for one segment: √(50^2 + 0.5^2) − 50 ≈ 0.0025 m; doubling gives 0.005 m, confirming the approximation.

Why Other Options Are Wrong:

0.500 m and 0.050 m are grossly too large; “effectively zero” underestimates measurable error; 0.025 m is fivefold too high.

Common Pitfalls:

Using d instead of d^2 scaling; forgetting that two segments double the single-segment error.

Final Answer:

0.005 m