No Intersection Area for geographic coordinates with Boost Geometry
I need to check whether two points (GPS coordinates) q, m are in the same side or on the opposite side or co linear to a line (great circle segment) (p,t). I know that q is not co linear to (p, t). I didn't find and any direct function to use in the boost.geometry library. So I tried to calculate it in a different way.
I construct two triangles (p, q, t) and (p, m, t). Then I intersect these two and check the area of the intersection polygon. Following is my code.
typedef boost::geometry::model::point<
double, 2, boost::geometry::cs::spherical_equatorial<boost::geometry::degree>
> geo_point;
typedef boost::geometry::model::polygon<geo_point> geo_polygon;
geo_point p(110.48316, 29.05043);
geo_point q(110.48416, 29.05104);
geo_point t(110.48416, 29.05228);
geo_point m(110.48408, 29.05079);
geo_polygon ut, es;
boost::geometry::append(ut.outer(), p);
boost::geometry::append(ut.outer(), q);
boost::geometry::append(ut.outer(), t);
boost::geometry::append(ut.outer(), p);
boost::geometry::append(es.outer(), p);
boost::geometry::append(es.outer(), m);
boost::geometry::append(es.outer(), t);
boost::geometry::append(es.outer(), p);
std::list<geo_point> intersection_points;
boost::geometry::intersection(ut, es, intersection_points);
std::cout << intersection_points.size() << std::endl;
std::vector<geo_polygon> intersection_polygons;
boost::geometry::intersection(ut, es, intersection_polygons);
std::cout << intersection_polygons.size() << std::endl;
Live at cpp.sh
If we plot these two triangles we can clearly see that they intersects on 3 vertices yielding another triangle in the intersected region.
The above code correctly returns the number of intersected points, but it doesn't return any intersection polygon.
3
0
I have tried using geographic instead of using spherical_equatorial coordinate system. But got the same results. Am I missing something ? or this is a problem in Boost.Geometry
geometry gis computational-geometry boost-geometry
asked Nov 27 '18 at 8:33
Neel BasuNeel Basu
7,4451060119
add a comment |
I need to check whether two points (GPS coordinates) q, m are in the same side or on the opposite side or co linear to a line (great circle segment) (p,t). I know that q is not co linear to (p, t). I didn't find and any direct function to use in the boost.geometry library. So I tried to calculate it in a different way.
I construct two triangles (p, q, t) and (p, m, t). Then I intersect these two and check the area of the intersection polygon. Following is my code.
typedef boost::geometry::model::point<
double, 2, boost::geometry::cs::spherical_equatorial<boost::geometry::degree>
> geo_point;
typedef boost::geometry::model::polygon<geo_point> geo_polygon;
geo_point p(110.48316, 29.05043);
geo_point q(110.48416, 29.05104);
geo_point t(110.48416, 29.05228);
geo_point m(110.48408, 29.05079);
geo_polygon ut, es;
boost::geometry::append(ut.outer(), p);
boost::geometry::append(ut.outer(), q);
boost::geometry::append(ut.outer(), t);
boost::geometry::append(ut.outer(), p);
boost::geometry::append(es.outer(), p);
boost::geometry::append(es.outer(), m);
boost::geometry::append(es.outer(), t);
boost::geometry::append(es.outer(), p);
std::list<geo_point> intersection_points;
boost::geometry::intersection(ut, es, intersection_points);
std::cout << intersection_points.size() << std::endl;
std::vector<geo_polygon> intersection_polygons;
boost::geometry::intersection(ut, es, intersection_polygons);
std::cout << intersection_polygons.size() << std::endl;
Live at cpp.sh
If we plot these two triangles we can clearly see that they intersects on 3 vertices yielding another triangle in the intersected region.
The above code correctly returns the number of intersected points, but it doesn't return any intersection polygon.
3
0
I have tried using geographic instead of using spherical_equatorial coordinate system. But got the same results. Am I missing something ? or this is a problem in Boost.Geometry
geometry gis computational-geometry boost-geometry
asked Nov 27 '18 at 8:33
Neel BasuNeel Basu
7,4451060119
What do you want to get from intersection triangle? I don't see how it could help to solve your problem. But calculation of bearings is useful to determine " the same side"
– MBo
Nov 27 '18 at 10:07
The intersection polygon is a straight line if they are on the opposite side, so the area will be 0. Otherwise there will be a non zero area.
– Neel Basu
Nov 27 '18 at 10:14
OK, it's reasonable. I made an answer.
– MBo
Nov 27 '18 at 10:23
add a comment |
I need to check whether two points (GPS coordinates) q, m are in the same side or on the opposite side or co linear to a line (great circle segment) (p,t). I know that q is not co linear to (p, t). I didn't find and any direct function to use in the boost.geometry library. So I tried to calculate it in a different way.
I construct two triangles (p, q, t) and (p, m, t). Then I intersect these two and check the area of the intersection polygon. Following is my code.
typedef boost::geometry::model::point<
double, 2, boost::geometry::cs::spherical_equatorial<boost::geometry::degree>
> geo_point;
typedef boost::geometry::model::polygon<geo_point> geo_polygon;
geo_point p(110.48316, 29.05043);
geo_point q(110.48416, 29.05104);
geo_point t(110.48416, 29.05228);
geo_point m(110.48408, 29.05079);
geo_polygon ut, es;
boost::geometry::append(ut.outer(), p);
boost::geometry::append(ut.outer(), q);
boost::geometry::append(ut.outer(), t);
boost::geometry::append(ut.outer(), p);
boost::geometry::append(es.outer(), p);
boost::geometry::append(es.outer(), m);
boost::geometry::append(es.outer(), t);
boost::geometry::append(es.outer(), p);
std::list<geo_point> intersection_points;
boost::geometry::intersection(ut, es, intersection_points);
std::cout << intersection_points.size() << std::endl;
std::vector<geo_polygon> intersection_polygons;
boost::geometry::intersection(ut, es, intersection_polygons);
std::cout << intersection_polygons.size() << std::endl;
Live at cpp.sh
If we plot these two triangles we can clearly see that they intersects on 3 vertices yielding another triangle in the intersected region.
The above code correctly returns the number of intersected points, but it doesn't return any intersection polygon.
3
0
I have tried using geographic instead of using spherical_equatorial coordinate system. But got the same results. Am I missing something ? or this is a problem in Boost.Geometry
geometry gis computational-geometry boost-geometry
asked Nov 27 '18 at 8:33
Neel BasuNeel Basu
7,4451060119
I need to check whether two points (GPS coordinates) q, m are in the same side or on the opposite side or co linear to a line (great circle segment) (p,t). I know that q is not co linear to (p, t). I didn't find and any direct function to use in the boost.geometry library. So I tried to calculate it in a different way.
I construct two triangles (p, q, t) and (p, m, t). Then I intersect these two and check the area of the intersection polygon. Following is my code.
typedef boost::geometry::model::point<
double, 2, boost::geometry::cs::spherical_equatorial<boost::geometry::degree>
> geo_point;
typedef boost::geometry::model::polygon<geo_point> geo_polygon;
geo_point p(110.48316, 29.05043);
geo_point q(110.48416, 29.05104);
geo_point t(110.48416, 29.05228);
geo_point m(110.48408, 29.05079);
geo_polygon ut, es;
boost::geometry::append(ut.outer(), p);
boost::geometry::append(ut.outer(), q);
boost::geometry::append(ut.outer(), t);
boost::geometry::append(ut.outer(), p);
boost::geometry::append(es.outer(), p);
boost::geometry::append(es.outer(), m);
boost::geometry::append(es.outer(), t);
boost::geometry::append(es.outer(), p);
std::list<geo_point> intersection_points;
boost::geometry::intersection(ut, es, intersection_points);
std::cout << intersection_points.size() << std::endl;
std::vector<geo_polygon> intersection_polygons;
boost::geometry::intersection(ut, es, intersection_polygons);
std::cout << intersection_polygons.size() << std::endl;
Live at cpp.sh
If we plot these two triangles we can clearly see that they intersects on 3 vertices yielding another triangle in the intersected region.
The above code correctly returns the number of intersected points, but it doesn't return any intersection polygon.
3
0
I have tried using geographic instead of using spherical_equatorial coordinate system. But got the same results. Am I missing something ? or this is a problem in Boost.Geometry
geometry gis computational-geometry boost-geometry
geometry gis computational-geometry boost-geometry
asked Nov 27 '18 at 8:33
Neel BasuNeel Basu
7,4451060119
asked Nov 27 '18 at 8:33
Neel BasuNeel Basu
7,4451060119
edited Nov 27 '18 at 8:52
Neel Basu
asked Nov 27 '18 at 8:33
Neel BasuNeel Basu
7,4451060119
asked Nov 27 '18 at 8:33
Neel BasuNeel Basu
7,4451060119
asked Nov 27 '18 at 8:33
Neel BasuNeel Basu
7,4451060119
7,4451060119
What do you want to get from intersection triangle? I don't see how it could help to solve your problem. But calculation of bearings is useful to determine " the same side"
– MBo
Nov 27 '18 at 10:07
The intersection polygon is a straight line if they are on the opposite side, so the area will be 0. Otherwise there will be a non zero area.
– Neel Basu
Nov 27 '18 at 10:14
OK, it's reasonable. I made an answer.
– MBo
Nov 27 '18 at 10:23
add a comment |
What do you want to get from intersection triangle? I don't see how it could help to solve your problem. But calculation of bearings is useful to determine " the same side"
– MBo
Nov 27 '18 at 10:07
The intersection polygon is a straight line if they are on the opposite side, so the area will be 0. Otherwise there will be a non zero area.
– Neel Basu
Nov 27 '18 at 10:14
OK, it's reasonable. I made an answer.
– MBo
Nov 27 '18 at 10:23
What do you want to get from intersection triangle? I don't see how it could help to solve your problem. But calculation of bearings is useful to determine " the same side"
– MBo
Nov 27 '18 at 10:07
What do you want to get from intersection triangle? I don't see how it could help to solve your problem. But calculation of bearings is useful to determine " the same side"
– MBo
Nov 27 '18 at 10:07
The intersection polygon is a straight line if they are on the opposite side, so the area will be 0. Otherwise there will be a non zero area.
– Neel Basu
Nov 27 '18 at 10:14
The intersection polygon is a straight line if they are on the opposite side, so the area will be 0. Otherwise there will be a non zero area.
– Neel Basu
Nov 27 '18 at 10:14
OK, it's reasonable. I made an answer.
– MBo
Nov 27 '18 at 10:23
OK, it's reasonable. I made an answer.
– MBo
Nov 27 '18 at 10:23
add a comment |
1 Answer
1
active
oldest
votes
To ensure that polygons are closed and oriented as needed, apply correct. Inserting
boost::geometry::correct(ut);
boost::geometry::correct(es);
gives result 1 polygon for your test
answered Nov 27 '18 at 10:21
MBoMBo
48.8k23050
The polygons are closed properly because I append that first point at the end. Isn't that sufficient ? Also what doesorientedmean in this context ?
– Neel Basu
Nov 27 '18 at 11:03
From description:all rings which are wrongly oriented with respect to their expected orientation are reversed(I was wondering myself)
– MBo
Nov 27 '18 at 11:04
add a comment |
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1 Answer
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1 Answer
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active
oldest
votes
active
oldest
votes
active
oldest
votes
To ensure that polygons are closed and oriented as needed, apply correct. Inserting
boost::geometry::correct(ut);
boost::geometry::correct(es);
gives result 1 polygon for your test
answered Nov 27 '18 at 10:21
MBoMBo
48.8k23050
The polygons are closed properly because I append that first point at the end. Isn't that sufficient ? Also what doesorientedmean in this context ?
– Neel Basu
Nov 27 '18 at 11:03
From description:all rings which are wrongly oriented with respect to their expected orientation are reversed(I was wondering myself)
– MBo
Nov 27 '18 at 11:04
add a comment |
To ensure that polygons are closed and oriented as needed, apply correct. Inserting
boost::geometry::correct(ut);
boost::geometry::correct(es);
gives result 1 polygon for your test
answered Nov 27 '18 at 10:21
MBoMBo
48.8k23050
The polygons are closed properly because I append that first point at the end. Isn't that sufficient ? Also what doesorientedmean in this context ?
– Neel Basu
Nov 27 '18 at 11:03
From description:all rings which are wrongly oriented with respect to their expected orientation are reversed(I was wondering myself)
– MBo
Nov 27 '18 at 11:04
add a comment |
To ensure that polygons are closed and oriented as needed, apply correct. Inserting
boost::geometry::correct(ut);
boost::geometry::correct(es);
gives result 1 polygon for your test
answered Nov 27 '18 at 10:21
MBoMBo
48.8k23050
To ensure that polygons are closed and oriented as needed, apply correct. Inserting
boost::geometry::correct(ut);
boost::geometry::correct(es);
gives result 1 polygon for your test
answered Nov 27 '18 at 10:21
MBoMBo
48.8k23050
answered Nov 27 '18 at 10:21
MBoMBo
48.8k23050
answered Nov 27 '18 at 10:21
MBoMBo
48.8k23050
answered Nov 27 '18 at 10:21
MBoMBo
48.8k23050
48.8k23050
The polygons are closed properly because I append that first point at the end. Isn't that sufficient ? Also what doesorientedmean in this context ?
– Neel Basu
Nov 27 '18 at 11:03
From description:all rings which are wrongly oriented with respect to their expected orientation are reversed(I was wondering myself)
– MBo
Nov 27 '18 at 11:04
add a comment |
The polygons are closed properly because I append that first point at the end. Isn't that sufficient ? Also what doesorientedmean in this context ?
– Neel Basu
Nov 27 '18 at 11:03
From description:all rings which are wrongly oriented with respect to their expected orientation are reversed(I was wondering myself)
– MBo
Nov 27 '18 at 11:04
The polygons are closed properly because I append that first point at the end. Isn't that sufficient ? Also what does
oriented mean in this context ?– Neel Basu
Nov 27 '18 at 11:03
The polygons are closed properly because I append that first point at the end. Isn't that sufficient ? Also what does
oriented mean in this context ?– Neel Basu
Nov 27 '18 at 11:03
From description:
all rings which are wrongly oriented with respect to their expected orientation are reversed (I was wondering myself)– MBo
Nov 27 '18 at 11:04
From description:
all rings which are wrongly oriented with respect to their expected orientation are reversed (I was wondering myself)– MBo
Nov 27 '18 at 11:04
add a comment |
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What do you want to get from intersection triangle? I don't see how it could help to solve your problem. But calculation of bearings is useful to determine " the same side"
– MBo
Nov 27 '18 at 10:07
The intersection polygon is a straight line if they are on the opposite side, so the area will be 0. Otherwise there will be a non zero area.
– Neel Basu
Nov 27 '18 at 10:14
OK, it's reasonable. I made an answer.
– MBo
Nov 27 '18 at 10:23