Warp

That would be good enough for small areas, as the curvature of the earth has no significant value, but on large areas Clipper is useless. Due to the projection, clipper would have to account for the curvature of the earth. Example: To draw a straight line on the map, you have to know how to draw curves.

It's not that simple as using doubles. It's math on a sphere, not a flat screen. Here's a real example: A square polygon Corner1=(10.000, 40.000) Corner2=(10.001, 40.001) has earth dimensions of 0.0852 x 0.1112 km - but - a square polygon Corner1=(10.000, 4.000) Corner2=(10.001, 4.001) has earth dimensions of 0,1109 x 0,1112 km (spherical)

Thank you for your suggestions. Unfortunately I do need Clipper's advanced clipping functionality.

It's not that simple as using doubles. It's math on a sphere, not a flat screen. Here's a real example: A square polygon Corner1=(10.000, 40.000) Corner2=(10.001, 40.001) has earth dimensions of 0.0852 x 0.1112 km - but - a square polygon Corner1=(10.000, 40.000) Corner2=(1.001, 4.001) has earth dimensions of 0,1109 x 0,1112 km (spherical)

It's not that simple as using doubles. It's math on a sphere, not a flat screen. Here's a real example: A square polygon Corner1=(10.000, 40.000) Corner2=(10.001, 40.001) has a width of 0.0852 x 0.1112 km but a square polygon Corner1=(10.000, 40.000) Corner2=(1.001, 4.001) has a width of 0,1109 x 0,1112 km (spherical)

Hello Does anyone know if this excellent Lib has been forked to work with Longitude/Latitude instead of X/Y ? It's not as simple as multiplying the Longitude/Latitude by a factor (ex. 1000000) and injecting the values to Clipper, then retrieving the results by dividing the results by the same value. Because depending where you are working in the globe, the same Longitude is usually a different real distance in km/miles. Same goes for latitude. This affects most of the results. Calculating the Area()...

OK, thank you for your answer. So, do you think I can trust the value given from this function below? C# double CalcArea(List<List<IntPoint>> polygon) { List<List<IntPoint>> paths = Clipper.SimplifyPolygons(polygon); double totalArea = 0; for (int i = 0; i < paths.Count; i++) totalArea += Clipper.Area(paths[i]); return totalArea; }

OK, thank you for your answer. So, do you think I can trust the value given from this function below? C# double CalcArea(List<List<IntPoint>> polygon) { List<List<IntPoint>> paths = Clipper.SimplifyPolygons(polygon); double totalArea = 0; for (int i = 0; i < paths.Count; i++) totalArea += Clipper.Area(paths[i]); return totalArea; }