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Antipodal Land Area Calculation



Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?Geocoding addresses to geographic coordinatesHow to get a country's official languages?Administrative Divisions bordering a geographic region (e.g. an ocean)Can anyone explain this weird Plot3D error?RegionIntersection and area on GeoPosition polygonsRotations of a numberUse Wolfram curated databases to determine how many randomly chosen people are needed to have a 50% chance two live in the same or adjacent states?Geolocate multiple IP addressesWill my procedure be correct?Sea Level Rise - How to mask on relief plot










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$begingroup$


Mathematica 12 does antipodal graphics! See here for my treatment of antipodal New Zealand. Most of the Earth's above-sea-level land will have ocean at its antipode. Is there a way to calculate what percentage of above-sea-level land will also have above-sea-level land at its antipode?










share|improve this question









$endgroup$
















    3












    $begingroup$


    Mathematica 12 does antipodal graphics! See here for my treatment of antipodal New Zealand. Most of the Earth's above-sea-level land will have ocean at its antipode. Is there a way to calculate what percentage of above-sea-level land will also have above-sea-level land at its antipode?










    share|improve this question









    $endgroup$














      3












      3








      3





      $begingroup$


      Mathematica 12 does antipodal graphics! See here for my treatment of antipodal New Zealand. Most of the Earth's above-sea-level land will have ocean at its antipode. Is there a way to calculate what percentage of above-sea-level land will also have above-sea-level land at its antipode?










      share|improve this question









      $endgroup$




      Mathematica 12 does antipodal graphics! See here for my treatment of antipodal New Zealand. Most of the Earth's above-sea-level land will have ocean at its antipode. Is there a way to calculate what percentage of above-sea-level land will also have above-sea-level land at its antipode?







      geography recreational-mathematics






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked 3 hours ago









      Hans HavermannHans Havermann

      336




      336




















          1 Answer
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          2












          $begingroup$

          Yes, this is possible with a little faff.



          What we want to do is get the RegionUnion of all the countries the antipode intersects with, and then intersect the antipode with that region, and get the remaining area.



          Let's use New Zealand as an example.



          ant = GeoAntipode[Polygon@Entity["Country", "NewZealand"]]


          enter image description here



          Now, we can get the countries that this antipode intersects using GeoEntities:



          GeoEntities[ant, "Country"]



          Entity["Country", "Portugal"], Entity["Country", "Spain"],
          Entity["Country", "Gibraltar"], Entity["Country", "Morocco"]




          Now, it seems like there's a bit of a bug with Gibraltar in my solution, so I've removed it. I'm not sure what causes it, but including Gibraltar deletes Morocco from the Region (don't tell the British).



          countries = 
          RegionUnion @@ (EntityValue[Entity["Country", "Portugal"],
          Entity["Country", "Spain"], Entity["Country", "Morocco"],
          "Polygon"] /. GeoPosition[x_] -> x)


          (We need to do GeoPosition[x_]->x to convert the GeoPositions into regular points, for Region calculations)



          Now we intersect our antipode with this region:



          int = RegionIntersection[ant /. GeoPosition[x_] -> x, countries]


          (This can take a little time depending on the complexity of your polygons)



          We can now convert back to GeoPositions:



          geoint = MeshPrimitives[int, 2] /. Polygon[x_] -> Polygon[GeoPosition[x]]


          and check the graphics to make sure we got it right:



          GeoGraphics[geoint]


          enter image description here



          Finally, to get the actual area of intersections:



          GeoArea[geoint] // Total



          Quantity[1.58773[CenterDot]10^11, ("Meters")^2]




          We can see that we are in the right ballpark:



          UnitConvert[GeoArea[Entity["Country", "NewZealand"]]]



          Quantity[2.64511[CenterDot]10^11, ("Meters")^2]







          share|improve this answer









          $endgroup$













            Your Answer








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            1 Answer
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            active

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            active

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            2












            $begingroup$

            Yes, this is possible with a little faff.



            What we want to do is get the RegionUnion of all the countries the antipode intersects with, and then intersect the antipode with that region, and get the remaining area.



            Let's use New Zealand as an example.



            ant = GeoAntipode[Polygon@Entity["Country", "NewZealand"]]


            enter image description here



            Now, we can get the countries that this antipode intersects using GeoEntities:



            GeoEntities[ant, "Country"]



            Entity["Country", "Portugal"], Entity["Country", "Spain"],
            Entity["Country", "Gibraltar"], Entity["Country", "Morocco"]




            Now, it seems like there's a bit of a bug with Gibraltar in my solution, so I've removed it. I'm not sure what causes it, but including Gibraltar deletes Morocco from the Region (don't tell the British).



            countries = 
            RegionUnion @@ (EntityValue[Entity["Country", "Portugal"],
            Entity["Country", "Spain"], Entity["Country", "Morocco"],
            "Polygon"] /. GeoPosition[x_] -> x)


            (We need to do GeoPosition[x_]->x to convert the GeoPositions into regular points, for Region calculations)



            Now we intersect our antipode with this region:



            int = RegionIntersection[ant /. GeoPosition[x_] -> x, countries]


            (This can take a little time depending on the complexity of your polygons)



            We can now convert back to GeoPositions:



            geoint = MeshPrimitives[int, 2] /. Polygon[x_] -> Polygon[GeoPosition[x]]


            and check the graphics to make sure we got it right:



            GeoGraphics[geoint]


            enter image description here



            Finally, to get the actual area of intersections:



            GeoArea[geoint] // Total



            Quantity[1.58773[CenterDot]10^11, ("Meters")^2]




            We can see that we are in the right ballpark:



            UnitConvert[GeoArea[Entity["Country", "NewZealand"]]]



            Quantity[2.64511[CenterDot]10^11, ("Meters")^2]







            share|improve this answer









            $endgroup$

















              2












              $begingroup$

              Yes, this is possible with a little faff.



              What we want to do is get the RegionUnion of all the countries the antipode intersects with, and then intersect the antipode with that region, and get the remaining area.



              Let's use New Zealand as an example.



              ant = GeoAntipode[Polygon@Entity["Country", "NewZealand"]]


              enter image description here



              Now, we can get the countries that this antipode intersects using GeoEntities:



              GeoEntities[ant, "Country"]



              Entity["Country", "Portugal"], Entity["Country", "Spain"],
              Entity["Country", "Gibraltar"], Entity["Country", "Morocco"]




              Now, it seems like there's a bit of a bug with Gibraltar in my solution, so I've removed it. I'm not sure what causes it, but including Gibraltar deletes Morocco from the Region (don't tell the British).



              countries = 
              RegionUnion @@ (EntityValue[Entity["Country", "Portugal"],
              Entity["Country", "Spain"], Entity["Country", "Morocco"],
              "Polygon"] /. GeoPosition[x_] -> x)


              (We need to do GeoPosition[x_]->x to convert the GeoPositions into regular points, for Region calculations)



              Now we intersect our antipode with this region:



              int = RegionIntersection[ant /. GeoPosition[x_] -> x, countries]


              (This can take a little time depending on the complexity of your polygons)



              We can now convert back to GeoPositions:



              geoint = MeshPrimitives[int, 2] /. Polygon[x_] -> Polygon[GeoPosition[x]]


              and check the graphics to make sure we got it right:



              GeoGraphics[geoint]


              enter image description here



              Finally, to get the actual area of intersections:



              GeoArea[geoint] // Total



              Quantity[1.58773[CenterDot]10^11, ("Meters")^2]




              We can see that we are in the right ballpark:



              UnitConvert[GeoArea[Entity["Country", "NewZealand"]]]



              Quantity[2.64511[CenterDot]10^11, ("Meters")^2]







              share|improve this answer









              $endgroup$















                2












                2








                2





                $begingroup$

                Yes, this is possible with a little faff.



                What we want to do is get the RegionUnion of all the countries the antipode intersects with, and then intersect the antipode with that region, and get the remaining area.



                Let's use New Zealand as an example.



                ant = GeoAntipode[Polygon@Entity["Country", "NewZealand"]]


                enter image description here



                Now, we can get the countries that this antipode intersects using GeoEntities:



                GeoEntities[ant, "Country"]



                Entity["Country", "Portugal"], Entity["Country", "Spain"],
                Entity["Country", "Gibraltar"], Entity["Country", "Morocco"]




                Now, it seems like there's a bit of a bug with Gibraltar in my solution, so I've removed it. I'm not sure what causes it, but including Gibraltar deletes Morocco from the Region (don't tell the British).



                countries = 
                RegionUnion @@ (EntityValue[Entity["Country", "Portugal"],
                Entity["Country", "Spain"], Entity["Country", "Morocco"],
                "Polygon"] /. GeoPosition[x_] -> x)


                (We need to do GeoPosition[x_]->x to convert the GeoPositions into regular points, for Region calculations)



                Now we intersect our antipode with this region:



                int = RegionIntersection[ant /. GeoPosition[x_] -> x, countries]


                (This can take a little time depending on the complexity of your polygons)



                We can now convert back to GeoPositions:



                geoint = MeshPrimitives[int, 2] /. Polygon[x_] -> Polygon[GeoPosition[x]]


                and check the graphics to make sure we got it right:



                GeoGraphics[geoint]


                enter image description here



                Finally, to get the actual area of intersections:



                GeoArea[geoint] // Total



                Quantity[1.58773[CenterDot]10^11, ("Meters")^2]




                We can see that we are in the right ballpark:



                UnitConvert[GeoArea[Entity["Country", "NewZealand"]]]



                Quantity[2.64511[CenterDot]10^11, ("Meters")^2]







                share|improve this answer









                $endgroup$



                Yes, this is possible with a little faff.



                What we want to do is get the RegionUnion of all the countries the antipode intersects with, and then intersect the antipode with that region, and get the remaining area.



                Let's use New Zealand as an example.



                ant = GeoAntipode[Polygon@Entity["Country", "NewZealand"]]


                enter image description here



                Now, we can get the countries that this antipode intersects using GeoEntities:



                GeoEntities[ant, "Country"]



                Entity["Country", "Portugal"], Entity["Country", "Spain"],
                Entity["Country", "Gibraltar"], Entity["Country", "Morocco"]




                Now, it seems like there's a bit of a bug with Gibraltar in my solution, so I've removed it. I'm not sure what causes it, but including Gibraltar deletes Morocco from the Region (don't tell the British).



                countries = 
                RegionUnion @@ (EntityValue[Entity["Country", "Portugal"],
                Entity["Country", "Spain"], Entity["Country", "Morocco"],
                "Polygon"] /. GeoPosition[x_] -> x)


                (We need to do GeoPosition[x_]->x to convert the GeoPositions into regular points, for Region calculations)



                Now we intersect our antipode with this region:



                int = RegionIntersection[ant /. GeoPosition[x_] -> x, countries]


                (This can take a little time depending on the complexity of your polygons)



                We can now convert back to GeoPositions:



                geoint = MeshPrimitives[int, 2] /. Polygon[x_] -> Polygon[GeoPosition[x]]


                and check the graphics to make sure we got it right:



                GeoGraphics[geoint]


                enter image description here



                Finally, to get the actual area of intersections:



                GeoArea[geoint] // Total



                Quantity[1.58773[CenterDot]10^11, ("Meters")^2]




                We can see that we are in the right ballpark:



                UnitConvert[GeoArea[Entity["Country", "NewZealand"]]]



                Quantity[2.64511[CenterDot]10^11, ("Meters")^2]








                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered 1 hour ago









                Carl LangeCarl Lange

                5,54911243




                5,54911243



























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