How this instrument works
Stand at sea level, sight along a perfectly level line, and ground immediately starts falling away beneath that ray. Drop below tangent measures that gap: at arc distance d from your feet, a spherical Earth sits R·(sec(d ⁄ R) − 1) under your sightline. No atmosphere, no eye height, no instrument error enters — this is pure spherical geometry, and it grows almost exactly as distance squared.
Aristotle wrote in De Caelo, around 350 BC, that hulls vanish before masts, offering it as evidence that Earth is round. Surveyors later boiled that same sag into a field mnemonic: eight inches per mile squared. Run one mile through this instrument and you get 0.2033 m — 8.002 inches — which is why this rule of thumb from chain-and-theodolite days still circulates. Alfred Russel Wallace staked money on it in 1870, sighting six miles down Old Bedford River to settle his £500 wager with a flat-earth challenger, then spent years in court regretting his win.
Three assumptions hold this expression together. Earth is modelled as a sphere of radius 6 371 km, an averaged stand-in for an oblate spheroid whose waist bulges roughly 21 km past its poles. Light is assumed to travel straight, which it does not: atmospheric refraction bends rays downward and typically erases about 14% of geometric drop, so survey practice substitutes 0.0675 d² for 0.0785 d² with d in kilometres. Finally, your eye sits exactly at that tangent point, zero metres above ground. Lift it even to standing height and geometry changes shape completely.
- Enter your span in Distance along the surface — metres, kilometres or miles, whichever your chart already uses.
- Read Drop below the tangent line, flipping its units to centimetres or feet whenever figures turn small.
- Measure distance along ground, not straight through it; arc length is what this expression expects.
- Treat that output as fall from a tangent at your own feet, never as height hidden from a raised observer.
Worked example — ten kilometres across open water
A rowing coach on a flat beach watches her launch motor straight out to 10 000 m. Entering that in Distance along the surface: θ = 10000 ⁄ 6371000 = 0.00156961 rad, sec θ = 1.0000012318, so drop = 6371000 × 0.0000012318 = 7.84806958466 m of fall below her level sightline.
Nearly eight metres, taller than a two-storey house. Note carefully what that figure is not: it is not how much of her launch hides. Standing 1.5 m tall, her own horizon falls at √(2 × 6371000 × 1.5) = 4372 m, leaving 5628 m beyond it, so only 5628² ⁄ (2R) ≈ 2.49 m of hull actually drops out of sight. Tangent drop is geometry measured from ground level; hidden height needs eye height as well.
Questions
Why does this not match how much of a ship is hidden?
Because tangent drop assumes an observer with eyes at zero height. A real observer at height h sees a horizon √(2Rh) away, and only distance beyond that horizon hides anything. For any 1.5 m viewer looking 10 km, horizon sits at 4.37 km and hidden height is roughly 2.5 m, not 7.85 m. Use this figure for level sightlines, laser runs and canal surveys; use a two-segment horizon calculation for vanishing ships.
Is eight inches per mile squared accurate?
Remarkably so — one mile gives 0.2033 m, or 8.002 inches, matching that mnemonic to three digits. Trouble starts when people treat it as linear. Squaring applies to distance, so two miles gives 8 × 4 = 32 inches, three miles 8 × 9 = 72 inches, and ten miles 800 inches, or 66 feet. Multiplying 8 by distance rather than by distance squared understates drop badly past a mile or two.
Does atmospheric refraction affect this answer?
Yes, and it always reduces apparent drop. Air density falls with altitude, so light curves gently downward, following Earth partway around. Optical surveyors handle it with a refraction coefficient near 0.13, equivalent to an effective radius of 7 ⁄ 6 R, trimming roughly 14% off geometric drop. Radio propagation goes further, adopting 4 ⁄ 3 R instead. Strong temperature inversions over cold water can bend light enough to lift objects back into view entirely — a superior mirage.
Why radius 6371 km when Earth is not a sphere?
6 371 km is a mean radius, chosen so that a sphere matches actual planetary volume. Reference ellipsoid WGS84 puts equatorial radius at 6 378.137 km and polar radius at 6 356.752 km, a 21 km spread. Swapping either extreme into this expression shifts drop by at most about 0.22%, which is 18 mm on our ten-kilometre example — smaller than what refraction uncertainty contributes anyway.
Is d² ⁄ (2R) good enough instead of a secant?
For almost everything, yes. Small-angle form differs from exact secant by only eight micrometres at 10 km and about 8 centimetres at 100 km, since that error grows as distance to a fourth power. This instrument evaluates secant directly, so you never have to choose. Beyond a few hundred kilometres both versions stop mattering, because refraction and terrain dominate long before geometry does.
How far away is my horizon?
That is a separate calculation: distance to horizon is √(2Rh), where h is eye height. At 1.7 m, horizon lies 4.65 km away; from a 30 m lighthouse gallery, 19.6 km; from an airliner at 11 km, about 374 km. Refraction typically stretches each of those by around 8%. Drop below tangent settles something else entirely: how far ground has fallen at a chosen distance, regardless of what any eye can see.