关于Polyline Simplification代码中经纬度用欧氏距离的合理性问询

Great question—this is a super common gotcha when working with geographic coordinates in polyline simplification, so you’re right to flag this!

Let’s break this down clearly:

The Core Issue #

Latitude and longitude are spherical coordinates, not planar. Euclidean distance (and its squared version, which skips the square root for speed) calculates distance as if the Earth were a flat plane. This works fine for tiny geographic areas (like a single city block) where the curvature of the Earth is negligible, but it introduces meaningful errors in most other cases:

• At high latitudes, the distance represented by 1 degree of longitude shrinks dramatically (e.g., ~55 km at 60°N, vs ~111 km at the equator). Euclidean distance treats 1 degree of lat and lon as equal, so it’ll misjudge the actual geographic distance between points.
• For lines spanning large areas, the discrepancy between planar Euclidean distance and real spherical distance can be massive, leading your simplification logic to keep or discard points incorrectly.

Why Do Some Implementations Use Euclidean Distance Squared? #

It’s almost always about computational efficiency. Calculating Euclidean distance squared is just a few arithmetic operations ((x2-x1)² + (y2-y1)²) with no expensive square root or trigonometric functions. This is tempting for large datasets, but it’s only valid if:

1. Your tolerance is defined in degrees of lat/lon (not meters/kilometers), and you understand that the actual geographic meaning of that tolerance varies by location.
2. Your polyline covers an extremely small area where the Earth’s curvature doesn’t affect distance calculations noticeably.

The Correct Approach for Geographic Data #

If you need your simplification tolerance to represent a consistent real-world distance (e.g., "remove points within 10 meters of the line"), you should use a geographic distance calculation instead:

• Haversine formula: Computes the great-circle distance between two points on a sphere (a decent approximation for Earth, which is an oblate spheroid).
• Vincenty formula: More accurate for Earth’s actual shape (ellipsoid), but slightly slower to compute.
• Alternatively, project your lat-lon coordinates to a planar coordinate system (like UTM, which divides the Earth into zones where planar distances are accurate) first, then use Euclidean distance squared with a tolerance in meters/feet.

Example of the Problem #

Suppose you set a tolerance of 0.0001 (in degrees squared, Euclidean). Near the equator, this corresponds to roughly 11 meters of real distance. At 60°N, the same tolerance corresponds to only ~6 meters. Your simplification logic will be stricter (remove more points) at high latitudes than at the equator, which is almost certainly not what you want if you’re aiming for consistent geographic simplification.

Final Verdict #

• If your tolerance is meant to represent real-world distance: Yes, using Euclidean distance squared directly on lat-lon is incorrect—it will lead to inconsistent and inaccurate simplification across different regions.
• If your tolerance is in degrees and you’re working with a tiny, localized dataset: It’s a pragmatic (though limited) choice, but you should document this tradeoff clearly.

内容的提问来源于stack exchange，提问作者Mandroid