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Stadia Distance

D = k s + C H = k s cos^{2} θ + C cos θ

Rod intercept s between the upper and lower stadia hairs. Interval factor k is almost always 100; C = f+c is ~0 for internal-focusing instruments and is often neglected. Inclined sights use the cos²θ reduction to get horizontal distance.

D

Slope (stadia) distance

k

Stadia interval factor (≈100)

s

Rod intercept

C

Stadia constant f+c

θ

Vertical angle

Step-by-step solved example

k = 100, C = 0, s = 1.250 m, vertical angle +6°. Find D and H.

Stadia: D ≈ 100 s for a vertical rod intercept s.

1. Slope
D = 100×1.250 = 125.0 m.
$D = 125.0 m$
2. Horizontal
H = 125.0 cos² 6° = 125.0 × (0.9945)² = 125.0 × 0.9891 = 123.6 m.
$H = 123.6 m$

Answer: D = 125.0 m, H = 123.6 m

10 practice questions

0/10 correct

1.s = 4.20 ft, k = 100, C = 0. Horizontal sight D =

2.Most internal-focusing transits use C ≈

3.k is 100 because the stadia hairs are placed so that

4.s = 1.00 m, k = 100, C = 0.30 m. D =

5.Inclined stadia uses H = k s cos²θ because

6.If the rod is 200 ft away (k = 100, C = 0), s reads

7.Stadia is less precise than EDM mainly because of

8.θ = 0 in H = k s cos²θ + C cos θ reduces to

9.Upper hair 5.34, lower hair 3.10, middle 4.22. A stadia check is

10.s = 2.50 m, k = 100, θ = 30°, C = 0. H is nearest