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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
$begingroup$
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?
geography recreational-mathematics
$endgroup$
add a comment |
$begingroup$
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?
geography recreational-mathematics
$endgroup$
add a comment |
$begingroup$
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?
geography recreational-mathematics
$endgroup$
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
geography recreational-mathematics
asked 3 hours ago
Hans HavermannHans Havermann
336
336
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1 Answer
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$begingroup$
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"]]
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]
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]
$endgroup$
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1 Answer
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1 Answer
1
active
oldest
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votes
$begingroup$
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"]]
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]
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]
$endgroup$
add a comment |
$begingroup$
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"]]
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]
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]
$endgroup$
add a comment |
$begingroup$
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"]]
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]
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]
$endgroup$
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"]]
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]
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]
answered 1 hour ago
Carl LangeCarl Lange
5,54911243
5,54911243
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