WPF用GMap.NET离线地图实现图层
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效果如下：
所有回答(2)
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试试这个：
public PointLatLng createPointLatLngFromCenterMiles(double DistanceMiles) { PointLatLng distancePoint = new PointLatLng(); //xMeter = xMeter / 1000; double earthRadius = 3958.0; double lat = this.Lat * (Math.PI / 180); double lon = this.Lng * (Math.PI / 180); double d = DistanceMiles / earthRadius; double brng = 360 * (Math.PI / 180); var latRadians = Math.Asin(Math.Sin(lat) * Math.Cos(d) + Math.Cos(lat) * Math.Sin(d) * Math.Cos(brng)); var lngRadians = lon + Math.Atan2(Math.Sin(brng) * Math.Sin(d) * Math.Cos(lat), Math.Cos(d) - Math.Sin(lat) * Math.Sin(latRadians)); distancePoint.Lat = latRadians * 180 / Math.PI; distancePoint.Lng = lngRadians * 180 / Math.PI; return distancePoint; } public PointLatLng createPointLatLngFromCenterKilometers(double DistanceKilometers) { PointLatLng distancePoint = new PointLatLng(); //xMeter = xMeter / 1000; double earthRadius = 6367.0; double lat = this.Lat * (Math.PI / 180); double lon = this.Lng * (Math.PI / 180); double d = DistanceKilometers / earthRadius; double brng = 360 * (Math.PI / 180); var latRadians = Math.Asin(Math.Sin(lat) * Math.Cos(d) + Math.Cos(lat) * Math.Sin(d) * Math.Cos(brng)); var lngRadians = lon + Math.Atan2(Math.Sin(brng) * Math.Sin(d) * Math.Cos(lat), Math.Cos(d) - Math.Sin(lat) * Math.Sin(latRadians)); distancePoint.Lat = latRadians * 180 / Math.PI; distancePoint.Lng = lngRadians * 180 / Math.PI; return distancePoint; }
麻烦问下，你的这段代码是实现什么功能，是在计算地图中心点的坐标是吗
@佐佐: 对，下面的连接有其余的功能。
@【Arnold】: 谢谢 ，我现在地图已将可以用了，想在地图上班加图层，GMapPolygon 类中RegenerateShape方法应该是恢复多边形的，但是还却什么呢？才能将图层显示出来？
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请问你实现了自定义标签么？
