Question

现在我有：

Polygon circle = geometryBuilder.circle(
myLong,
myLat, 
radiusInMeters, 10);

它创建（使用lat = 28.456306，long = -16.292034和radius = 500）一个具有巨大纬度和经度的无意义多边形，例如：

POLYGON ((483.678055 28.482505000000003, 388.1865521874737 -265.4101211462366, 138.1865521874737 -447.04575314757676, -170.8304421874737 -447.0457531475768, -420.8304421874737 -265.41012114623663, -516.321945 28.482504999999943, -420.83044218747375 322.3751311462365, -170.8304421874738 504.01076314757677, 138.18655218747358 504.0107631475768, 388.18655218747364 322.3751311462367, 483.678055 28.482505000000003))

我希望在我提供的中心点附近有十对坐标，其中lat和long都在附近。

任何帮助都会有所帮助。提前谢谢！

修改

除了@iant的答案，我必须创建一个Point作为Feature

//build the type
SimpleFeatureType TYPE = null;
try {
    TYPE = DataUtilities.createType("", "Location", "locations:Point:srid=4326," + "id:Integer" // a
            // number
            // attribute
            );
} catch (Exception e1) {
    // TODO Auto-generated catch block
    e1.printStackTrace();
}

SimpleFeatureBuilder featureBuilder = new SimpleFeatureBuilder(TYPE);
GeometryFactory geometryFactory = JTSFactoryFinder.getGeometryFactory();
com.vividsolutions.jts.geom.Point point = geometryFactory.createPoint(
        new Coordinate(
                currentDevicePosition.getLongitude(), 
                currentDevicePosition.getLatitude()
                )
        );
featureBuilder.add(point);
SimpleFeature feature = featureBuilder.buildFeature( "fid.1" ); // build the 1st feature

正如iant's Gist中所述：https://gitlab.com/snippets/17558和此处：http://docs.geotools.org/，哦，我也错过了依赖关系，如此处所述SchemaException in java

Answer 1

有两种解决方案：

将半径以米为单位转换为度数并将问题视为平面问题
将纬度/经度点转换为米，在局部平面投影中计算圆并重新投影回纬度/经度。

对于1你可以做一些像赤道附近的小半径那样好的东西：

GeodeticCalculator calc = new  GeodeticCalculator(DefaultGeographicCRS.WGS84);
  calc.setStartingGeographicPoint(point.getX(), point.getY());
  calc.setDirection(0.0, 10000);
  Point2D p2 = calc.getDestinationGeographicPoint();
  calc.setDirection(90.0, 10000);
  Point2D p3 = calc.getDestinationGeographicPoint();

  double dy = p2.getY() - point.getY();
  double dx = p3.getX() - point.getX();
  double distance = (dy + dx) / 2.0;
  Polygon p1 = (Polygon) point.buffer(distance);

我将展示第二部分的一些代码，因为它更通用（即它工作得更好，半径更好）。

首先你需要找到一个局部投影，GeoTools提供一个“伪”投影AUTO42001,x,y，这是一个以X，Y为中心的UTM投影：

public SimpleFeature bufferFeature(SimpleFeature feature, Measure<Double, Length> distance) {
    // extract the geometry
    GeometryAttribute gProp = feature.getDefaultGeometryProperty();
    CoordinateReferenceSystem origCRS = gProp.getDescriptor().getCoordinateReferenceSystem();

    Geometry geom = (Geometry) feature.getDefaultGeometry();
    Geometry pGeom = geom;
    MathTransform toTransform, fromTransform = null;
    // reproject the geometry to a local projection
    if (!(origCRS instanceof ProjectedCRS)) {

      double x = geom.getCoordinate().x;
      double y = geom.getCoordinate().y;

      String code = "AUTO:42001," + x + "," + y;
      // System.out.println(code);
      CoordinateReferenceSystem auto;
      try {
        auto = CRS.decode(code);
        toTransform = CRS.findMathTransform(DefaultGeographicCRS.WGS84, auto);
        fromTransform = CRS.findMathTransform(auto, DefaultGeographicCRS.WGS84);
        pGeom = JTS.transform(geom, toTransform);

      } catch (MismatchedDimensionException | TransformException | FactoryException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
      }

    }

所以现在pGeom是以米为单位的。缓冲它现在很容易

  Geometry out = bufferFeature(pGeom, distance.doubleValue(SI.METER));

然后我们使用我们之前查找的反向变换投射回WGS84（lat / lon）：

  retGeom = JTS.transform(out, fromTransform);

然后有一点麻烦来改变要素类型以反映我们返回多边形而不是Point的事实。完整代码位于此gist中。

当我运行它时，我得到以下输出：

POINT (10.840378413128576 3.4152050343701745)
POLYGON ((10.84937634426605 3.4151876838951822, 10.849200076653755 3.413423962919184, 10.84868480171117 3.4117286878605766, 10.847850322146979 3.4101670058279794, 10.846728706726902 3.4087989300555464, 10.845363057862208 3.407677033830687, 10.843805855306746 3.406844430298736, 10.84211693959797 3.406333115754347, 10.840361212705258 3.4061627400701946, 10.838606144204721 3.4063398515107184, 10.836919178768184 3.4068576449605277, 10.835365144548726 3.4076962232621035, 10.834003762019957 3.408823361646906, 10.832887348980522 3.410195745914279, 10.832058809914859 3.411760636805914, 10.831549986992338 3.4134578966399034, 10.831380436105858 3.4152223003379722, 10.831556675029052 3.416986042039048, 10.832071932633442 3.4186813409639054, 10.832906408849936 3.4202430463705085, 10.834028035422469 3.4216111414662183, 10.835393708241908 3.422733050021835, 10.836950943907517 3.4235656570147763, 10.838639896841123 3.424076965623486, 10.840395659406198 3.4242473268789406, 10.842150756595839 3.4240701947133396, 10.843837739370569 3.4235523773972796, 10.845391776937724 3.4227137757216988, 10.846753148314034 3.4215866180136185, 10.847869537398722 3.4202142214154887, 10.848698043354238 3.4186493270628633, 10.849206829051935 3.4169520731645546, 10.84937634426605 3.4151876838951822))

Answer 2

    double latitude = 40.689234d;
    double longitude = -74.044598d;
    double diameterInMeters = 2000d; //2km

    GeometricShapeFactory shapeFactory = new GeometricShapeFactory();
    shapeFactory.setNumPoints(64); // adjustable
    shapeFactory.setCentre(new Coordinate(latitude, longitude));
    // Length in meters of 1° of latitude = always 111.32 km
    shapeFactory.setWidth(diameterInMeters/111320d);
    // Length in meters of 1° of longitude = 40075 km * cos( latitude ) / 360
    shapeFactory.setHeight(diameterInMeters / (40075000 * Math.cos(Math.toRadians(latitude)) / 360));

    Polygon circle = shapeFactory.createEllipse();

result

Answer 3

GeometricShapeFactory shapeFactory = new GeometricShapeFactory(); shapeFactory.setNumPoints(64); AdjustableshapeFactory.setCentre（新坐标（纬度，经度））； // 以米为单位的 1° 长度

如何使用GeoTools创建具有纬度，经度和半径的圆圈？

3 个答案: