The primary limitation for a straight-line laser weapon (like those under development by the U.S. Navy, such as the HELIOS system) targeting another warship at sea level would be Earth's curvature on the globe model. Lasers propagate in a straight line in a vacuum, but atmospheric effects like refraction can slightly bend them downward (extending range by 10-20% in some conditions), though we'll ignore that for a pure geometric calculation as per your query. The key factor is the optical horizon: the point where the line of sight is tangent to Earth's surface, blocking direct fire to a sea-skimming target.

The "8 inches per mile squared" approximation (h ≈ 8d², where h is drop in inches and d is distance in miles) is a rough rule for the vertical drop of a tangent line over distance due to curvature (derived from Earth's radius of ~3,959 miles). It's not exact for all ranges but sufficient for this. More precise calculations use the formula for horizon distance from height h above sea level: d ≈ √(2Rh), where R is Earth's radius, but the 8-inch rule aligns closely for short naval ranges.

### Assumptions for Calculation
- Battleship laser emitter height: ~100 feet above sea level (typical for a destroyer's or cruiser's weapon deck; actual systems like LaWS or HELIOS are mounted lower, but this gives a conservative ship-to-waterline range).
- Target: Another warship's hull or low superstructure, effectively at sea level (0 feet) for the limiting case. If targeting a tall mast (e.g., 200 feet), range extends further by adding the target's height to the equation.
- No atmospheric interference, waves, or ship motion—pure globe geometry.
- "Significant use" means direct line-of-sight to the target, allowing the laser to hit without being blocked by the curvature "drop."

### Globe Model Range Calculation
Using the 8-inch-per-mile-squared drop:

- The maximum range is where the laser's straight-line path just grazes the sea surface (tangent point). Beyond this, the beam would need to curve around the globe, which it can't.
- For an observer (or emitter) at height h feet, the horizon distance d (miles) is approximately d ≈ 1.22 × √h (derived from the drop formula; converts to miles).
  - For h = 100 feet: d ≈ 1.22 × √100 = 1.22 × 10 = 12.2 miles.
- This is the distance to the horizon from the emitter. To hit a sea-level target, the effective range is about this value if the target is also low-profile. For a target ship with its own height (e.g., another 100-foot deck), the total range doubles to ~24.4 miles (sum of both horizons).

To verify with the drop formula directly:
- Solve for d where the drop h equals the emitter height (100 feet = 1,200 inches).
- h ≈ 8d² → 1,200 = 8d² → d² = 150 → d ≈ 12.25 miles (matches closely).

At longer ranges, you could still target upper structures:
- For a 200-foot target height: Total d ≈ 1.22(√100 + √200) ≈ 1.22(10 + 14.14) ≈ 29.7 miles.
- Drop at 20 miles: h ≈ 8 × 20² = 3,200 inches ≈ 267 feet. So from 100 feet up, you'd need to elevate the beam ~167 feet to clear the bulge and hit a low target—impractical for a flat-trajectory laser without major aiming adjustments.

In real naval tests (e.g., USS Portland's 60kW laser), effective ranges against small surface targets are reported up to 10-15 miles, limited by power, atmosphere, and precisely this curvature issue. Beyond ~20-30 miles, ship-to-ship laser targeting becomes ineffective without line-of-sight, forcing reliance on elevated mounts, drones, or missiles.

For the flat plane scenario (no drop), range would theoretically be unlimited by geometry, only by laser power, beam divergence, and atmospheric absorption/scattering (e.g., water vapor limits high-power lasers to 10-50 miles in practice). But since your question focuses on the globe "science," that's the binding constraint here.