Computing an event offset using NetTopologySuite?
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Does NetTopologySuite have the tools necessary to compute a point a given distance along and away from a polyline offset in a perpendicular direction?
This would be for placing signs on a map that are described as 3.1 miles along route 242, 50 feet from the centerline. I've discovered NetTopologySuite.Geometries.Triangle.PerpendicularBisector, but it's not making much sense to me (seems to return a formula for the perpendicular line).
nettopologysuite
asked Nov 14 at 17:27
Corey Alix
1,34221331
add a comment |
up vote
0
down vote
favorite
Does NetTopologySuite have the tools necessary to compute a point a given distance along and away from a polyline offset in a perpendicular direction?
This would be for placing signs on a map that are described as 3.1 miles along route 242, 50 feet from the centerline. I've discovered NetTopologySuite.Geometries.Triangle.PerpendicularBisector, but it's not making much sense to me (seems to return a formula for the perpendicular line).
nettopologysuite
asked Nov 14 at 17:27
Corey Alix
1,34221331
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Does NetTopologySuite have the tools necessary to compute a point a given distance along and away from a polyline offset in a perpendicular direction?
This would be for placing signs on a map that are described as 3.1 miles along route 242, 50 feet from the centerline. I've discovered NetTopologySuite.Geometries.Triangle.PerpendicularBisector, but it's not making much sense to me (seems to return a formula for the perpendicular line).
nettopologysuite
asked Nov 14 at 17:27
Corey Alix
1,34221331
Does NetTopologySuite have the tools necessary to compute a point a given distance along and away from a polyline offset in a perpendicular direction?
This would be for placing signs on a map that are described as 3.1 miles along route 242, 50 feet from the centerline. I've discovered NetTopologySuite.Geometries.Triangle.PerpendicularBisector, but it's not making much sense to me (seems to return a formula for the perpendicular line).
nettopologysuite
nettopologysuite
asked Nov 14 at 17:27
Corey Alix
1,34221331
asked Nov 14 at 17:27
Corey Alix
1,34221331
asked Nov 14 at 17:27
Corey Alix
1,34221331
asked Nov 14 at 17:27
Corey Alix
1,34221331
asked Nov 14 at 17:27
Corey Alix
1,34221331
1,34221331
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
To get a single point offset off a lineal geometry you should use LocationIndexedLine:
var gf = new NetTopologySuite.Geometries.GeometryFactory();
var l = gf.CreateLineString(new GeoAPI.Geometries.Coordinate
{
new GeoAPI.Geometries.Coordinate(10, 10),
new GeoAPI.Geometries.Coordinate(1000, 10),
});
var lid = new NetTopologySuite.LinearReferencing.LocationIndexedLine(l);
var p = lid.ExtractPoint(500, 10);
p is at (510, 20)
answered yesterday
FObermaier
1635
public Coordinate extractPoint(LinearLocation index, double offsetDistance)index- the index of the desired pointoffsetDistance- the distance the point is offset from the segment (positive is to the left, negative is to the right)
– Corey Alix
yesterday
add a comment |
up vote
1
down vote
Yes, probably several ways. One way you could do it is to use a buffer from the center line (look into NetTopologySuite.Operation.Buffer.BufferOp.Buffer), then just find a point 'x' distance along the buffered geometry (NetTopologySuite.Operation.Distance.DistanceOp.Distance)
answered Nov 17 at 0:25
tval
1027
I see -- or I could buffer the line itself and then look for the closest point from a point on the centerline to the buffered geometry. It seems computationally heavy-handed but it should work.
– Corey Alix
Nov 17 at 11:58
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
To get a single point offset off a lineal geometry you should use LocationIndexedLine:
var gf = new NetTopologySuite.Geometries.GeometryFactory();
var l = gf.CreateLineString(new GeoAPI.Geometries.Coordinate
{
new GeoAPI.Geometries.Coordinate(10, 10),
new GeoAPI.Geometries.Coordinate(1000, 10),
});
var lid = new NetTopologySuite.LinearReferencing.LocationIndexedLine(l);
var p = lid.ExtractPoint(500, 10);
p is at (510, 20)
answered yesterday
FObermaier
1635
public Coordinate extractPoint(LinearLocation index, double offsetDistance)index- the index of the desired pointoffsetDistance- the distance the point is offset from the segment (positive is to the left, negative is to the right)
– Corey Alix
yesterday
add a comment |
up vote
1
down vote
accepted
To get a single point offset off a lineal geometry you should use LocationIndexedLine:
var gf = new NetTopologySuite.Geometries.GeometryFactory();
var l = gf.CreateLineString(new GeoAPI.Geometries.Coordinate
{
new GeoAPI.Geometries.Coordinate(10, 10),
new GeoAPI.Geometries.Coordinate(1000, 10),
});
var lid = new NetTopologySuite.LinearReferencing.LocationIndexedLine(l);
var p = lid.ExtractPoint(500, 10);
p is at (510, 20)
answered yesterday
FObermaier
1635
public Coordinate extractPoint(LinearLocation index, double offsetDistance)index- the index of the desired pointoffsetDistance- the distance the point is offset from the segment (positive is to the left, negative is to the right)
– Corey Alix
yesterday
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
To get a single point offset off a lineal geometry you should use LocationIndexedLine:
var gf = new NetTopologySuite.Geometries.GeometryFactory();
var l = gf.CreateLineString(new GeoAPI.Geometries.Coordinate
{
new GeoAPI.Geometries.Coordinate(10, 10),
new GeoAPI.Geometries.Coordinate(1000, 10),
});
var lid = new NetTopologySuite.LinearReferencing.LocationIndexedLine(l);
var p = lid.ExtractPoint(500, 10);
p is at (510, 20)
answered yesterday
FObermaier
1635
To get a single point offset off a lineal geometry you should use LocationIndexedLine:
var gf = new NetTopologySuite.Geometries.GeometryFactory();
var l = gf.CreateLineString(new GeoAPI.Geometries.Coordinate
{
new GeoAPI.Geometries.Coordinate(10, 10),
new GeoAPI.Geometries.Coordinate(1000, 10),
});
var lid = new NetTopologySuite.LinearReferencing.LocationIndexedLine(l);
var p = lid.ExtractPoint(500, 10);
p is at (510, 20)
answered yesterday
FObermaier
1635
answered yesterday
FObermaier
1635
answered yesterday
FObermaier
1635
answered yesterday
FObermaier
1635
1635
public Coordinate extractPoint(LinearLocation index, double offsetDistance)index- the index of the desired pointoffsetDistance- the distance the point is offset from the segment (positive is to the left, negative is to the right)
– Corey Alix
yesterday
add a comment |
public Coordinate extractPoint(LinearLocation index, double offsetDistance)index- the index of the desired pointoffsetDistance- the distance the point is offset from the segment (positive is to the left, negative is to the right)
– Corey Alix
yesterday
public Coordinate extractPoint(LinearLocation index, double offsetDistance) index - the index of the desired point offsetDistance - the distance the point is offset from the segment (positive is to the left, negative is to the right)– Corey Alix
yesterday
public Coordinate extractPoint(LinearLocation index, double offsetDistance) index - the index of the desired point offsetDistance - the distance the point is offset from the segment (positive is to the left, negative is to the right)– Corey Alix
yesterday
add a comment |
up vote
1
down vote
Yes, probably several ways. One way you could do it is to use a buffer from the center line (look into NetTopologySuite.Operation.Buffer.BufferOp.Buffer), then just find a point 'x' distance along the buffered geometry (NetTopologySuite.Operation.Distance.DistanceOp.Distance)
answered Nov 17 at 0:25
tval
1027
I see -- or I could buffer the line itself and then look for the closest point from a point on the centerline to the buffered geometry. It seems computationally heavy-handed but it should work.
– Corey Alix
Nov 17 at 11:58
add a comment |
up vote
1
down vote
Yes, probably several ways. One way you could do it is to use a buffer from the center line (look into NetTopologySuite.Operation.Buffer.BufferOp.Buffer), then just find a point 'x' distance along the buffered geometry (NetTopologySuite.Operation.Distance.DistanceOp.Distance)
answered Nov 17 at 0:25
tval
1027
I see -- or I could buffer the line itself and then look for the closest point from a point on the centerline to the buffered geometry. It seems computationally heavy-handed but it should work.
– Corey Alix
Nov 17 at 11:58
add a comment |
up vote
1
down vote
up vote
1
down vote
Yes, probably several ways. One way you could do it is to use a buffer from the center line (look into NetTopologySuite.Operation.Buffer.BufferOp.Buffer), then just find a point 'x' distance along the buffered geometry (NetTopologySuite.Operation.Distance.DistanceOp.Distance)
answered Nov 17 at 0:25
tval
1027
Yes, probably several ways. One way you could do it is to use a buffer from the center line (look into NetTopologySuite.Operation.Buffer.BufferOp.Buffer), then just find a point 'x' distance along the buffered geometry (NetTopologySuite.Operation.Distance.DistanceOp.Distance)
answered Nov 17 at 0:25
tval
1027
answered Nov 17 at 0:25
tval
1027
answered Nov 17 at 0:25
tval
1027
answered Nov 17 at 0:25
tval
1027
1027
I see -- or I could buffer the line itself and then look for the closest point from a point on the centerline to the buffered geometry. It seems computationally heavy-handed but it should work.
– Corey Alix
Nov 17 at 11:58
add a comment |
I see -- or I could buffer the line itself and then look for the closest point from a point on the centerline to the buffered geometry. It seems computationally heavy-handed but it should work.
– Corey Alix
Nov 17 at 11:58
I see -- or I could buffer the line itself and then look for the closest point from a point on the centerline to the buffered geometry. It seems computationally heavy-handed but it should work.
– Corey Alix
Nov 17 at 11:58
I see -- or I could buffer the line itself and then look for the closest point from a point on the centerline to the buffered geometry. It seems computationally heavy-handed but it should work.
– Corey Alix
Nov 17 at 11:58
add a comment |
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