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GPS区域计算有什么类吗？

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2 回答 2

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我使用此代码来计算由 Android 的 GPS 点分隔的区域：

  private static final double EARTH_RADIUS = 6371000;// meters

  public static double calculateAreaOfGPSPolygonOnEarthInSquareMeters(final List<Location> locations) {
    return calculateAreaOfGPSPolygonOnSphereInSquareMeters(locations, EARTH_RADIUS);
  }

  private static double calculateAreaOfGPSPolygonOnSphereInSquareMeters(final List<Location> locations, final double radius) {
    if (locations.size() < 3) {
      return 0;
    }

    final double diameter = radius * 2;
    final double circumference = diameter * Math.PI;
    final List<Double> listY = new ArrayList<Double>();
    final List<Double> listX = new ArrayList<Double>();
    final List<Double> listArea = new ArrayList<Double>();
    // calculate segment x and y in degrees for each point
    final double latitudeRef = locations.get(0).getLatitude();
    final double longitudeRef = locations.get(0).getLongitude();
    for (int i = 1; i < locations.size(); i++) {
      final double latitude = locations.get(i).getLatitude();
      final double longitude = locations.get(i).getLongitude();
      listY.add(calculateYSegment(latitudeRef, latitude, circumference));
      Log.d(LOG_TAG, String.format("Y %s: %s", listY.size() - 1, listY.get(listY.size() - 1)));
      listX.add(calculateXSegment(longitudeRef, longitude, latitude, circumference));
      Log.d(LOG_TAG, String.format("X %s: %s", listX.size() - 1, listX.get(listX.size() - 1)));
    }

    // calculate areas for each triangle segment
    for (int i = 1; i < listX.size(); i++) {
      final double x1 = listX.get(i - 1);
      final double y1 = listY.get(i - 1);
      final double x2 = listX.get(i);
      final double y2 = listY.get(i);
      listArea.add(calculateAreaInSquareMeters(x1, x2, y1, y2));
      Log.d(LOG_TAG, String.format("area %s: %s", listArea.size() - 1, listArea.get(listArea.size() - 1)));
    }

    // sum areas of all triangle segments
    double areasSum = 0;
    for (final Double area : listArea) {
      areasSum = areasSum + area;
    }

    // get abolute value of area, it can't be negative
    return Math.abs(areasSum);// Math.sqrt(areasSum * areasSum);
  }

  private static Double calculateAreaInSquareMeters(final double x1, final double x2, final double y1, final double y2) {
    return (y1 * x2 - x1 * y2) / 2;
  }

  private static double calculateYSegment(final double latitudeRef, final double latitude, final double circumference) {
    return (latitude - latitudeRef) * circumference / 360.0;
  }

  private static double calculateXSegment(final double longitudeRef, final double longitude, final double latitude,
      final double circumference) {
    return (longitude - longitudeRef) * circumference * Math.cos(Math.toRadians(latitude)) / 360.0;
  }

您可以不使用位置对象列表而是使用两个纬度和经度列表轻松地将其调整为 java。

这是基于这个公式：

http://mathworld.wolfram.com/PolygonArea.html

这里也有描述：

http://en.wikipedia.org/wiki/Polygon

用这个公式计算多边形面积的原理：

http://maruzar.blogspot.com/2011/12/irregular-and-regular-polygon-area-by.html

公式适用于球体上的区域，在我们的例子中是世界球体，以及计算的 excel 示例对于验证 Java 或其他语言的算法非常有用：

http://maruzar.blogspot.com/2012/03/calculating-land-lot-area-with-gps.html

于 2013-04-25T09:11:09.013 回答
0

您可以通过记录三个点来计算两个向量的平行四边形的面积：1）起点 A 2）点 B 3）点 C

得到两个向量 AB 和 AC [B(x,y)-A(x,y) = AB(x,y)]

面积是 || AB X AC || ！

于 2011-03-24T14:59:54.287 回答