PostGIS 计算给定距离的正交点

Question

在 PostGIS 中，我需要计算直线给定距离处的 2 个垂直点 C 和 D，以便在地图上创建一个完美的矩形。投影 WGS84。 ABCD必须像下图一样逆时针旋转。

知道如何计算这 2 分吗？

Answer 1

给定任意线串，您可以使用 ST_OffsetCurve 计算具有自定义距离（在此示例中设置为 2 米）的平行线。之后我们将其反转。然后您可以将 2 条线串的起点和端点相互连接，这将产生“连接”线串。

WITH focus_line AS (
    SELECT ST_GeomFromText('LineString(-0.108092 51.519324, -0.107643 51.517861)', 4326) AS geom
), 
parallel_line AS (
    SELECT ST_Reverse(
        ST_Transform(
            ST_OffsetCurve(
                ST_Transform(
                    geom, 
                3857),
            2, 'join=round'),
        4326)
    ) AS geom
    FROM focus_line
)
SELECT ST_AsText(a.geom) AS focus,
ST_AsText(b.geom) AS parallel,
ST_AsText(
    ST_MakeLine(
        ST_StartPoint(a.geom), ST_EndPoint(b.geom)
    )
) AS connection_1, 
ST_AsText(
    ST_MakeLine(
        ST_EndPoint(a.geom), ST_StartPoint(b.geom)  
    )
) AS connection_2
FROM focus_line a, parallel_line b;

PostGIS 计算给定距离的正交点

PostGIS compute orthogonal points at a given distance

postgis