CurvePolygon

Article
06/13/2017

A CurvePolygon is a topologically closed surface defined by an exterior bounding ring and zero or more interior rings

Important

For a detailed description and examples of spatial features introduced in SQL Server 2012, including the CurvePolygon subtype, download the white paper, New Spatial Features in SQL Server 2012.

The following criteria define attributes of a CurvePolygon instance:

The boundary of the CurvePolygon instance is defined by the exterior ring and all interior rings.
The interior of the CurvePolygon instance is the space between the exterior ring and all of the interior rings.

A CurvePolygon instance differs from a Polygon instance in that a CurvePolygon instance may contain the following circular arc segments: CircularString and CompoundCurve.

CompoundCurve instances

Illustration below shows valid CurvePolygon figures:

Accepted instances

For a CurvePolygon instance to be accepted, it needs to be either empty or contain only circular arc rings that are accepted. An accepted circular arc ring meets the following requirements.

Is an accepted LineString, CircularString, or CompoundCurve instance. For more information on accepted instances, see LineString, CircularString, and CompoundCurve.
Has at least four points.
The start and end point have the same X and Y coordinates.

Note

Z and M values are ignored.

The following example shows accepted CurvePolygon instances.

DECLARE @g1 geometry = 'CURVEPOLYGON EMPTY';  
DECLARE @g2 geometry = 'CURVEPOLYGON((0 0, 0 0, 0 0, 0 0))';  
DECLARE @g3 geometry = 'CURVEPOLYGON((0 0 1, 0 0 2, 0 0 3, 0 0 3))'  
DECLARE @g4 geometry = 'CURVEPOLYGON(CIRCULARSTRING(1 3, 3 5, 4 7, 7 3, 1 3))';  
DECLARE @g5 geography = 'CURVEPOLYGON((-122.3 47, 122.3 -47, 125.7 -49, 121 -38, -122.3 47))';

@g3 is accepted even though the start and end points have different Z values because Z values are ignored. @g5 is accepted even though the geography type instance is not valid.

The following examples throw System.FormatException.

DECLARE @g1 geometry = 'CURVEPOLYGON((0 5, 0 0, 0 0, 0 0))';  
DECLARE @g2 geometry = 'CURVEPOLYGON((0 0, 0 0, 0 0))';

@g1 is not accepted because the start and end points do not have the same Y value. @g2 is not accepted because the ring does not have enough points.

Valid instances

For a CurvePolygon instance to be valid both exterior and interior rings must meet the following criteria:

They may only touch at single tangent points.
They cannot cross each other.
Each ring must contain at least four points.
Each ring must be an acceptable curve type.

CurvePolygon instances also need to meet specific criteria depending on whether they are geometry or geography data types.

Geometry data type

A valid geometryCurvePolygon instance must have the following attributes:

All the interior rings must be contained within the exterior ring.
May have multiple interior rings, but an interior ring cannot contain another interior ring.
No ring can cross itself or another ring.
Rings can only touch at single tangent points (number of points where rings touch must be finite).
The interior of the polygon must be connected.

The following example shows valid geometryCurvePolygon instances.

DECLARE @g1 geometry = 'CURVEPOLYGON EMPTY';  
DECLARE @g2 geometry = 'CURVEPOLYGON(CIRCULARSTRING(1 3, 3 5, 4 7, 7 3, 1 3))';  
SELECT @g1.STIsValid(), @g2.STIsValid();

CurvePolygon instances have the same validity rules as Polygon instances with the exception that CurvePolygon instances may accept the new circular arc segment types. For more examples of instances that are valid or not valid, see Polygon.

Geography data type

A valid geographyCurvePolygon instance must have the following attributes:

The interior of the polygon is connected using the left-hand rule.
No ring can cross itself or another ring.
Rings can only touch at single tangent points (number of points where rings touch must be finite).
The interior of the polygon must be connected.

The following example shows a valid geography CurvePolygon instance.

DECLARE @g geography = 'CURVEPOLYGON((-122.3 47, 122.3 47, 125.7 49, 121 38, -122.3 47))';  
SELECT @g.STIsValid();

Examples

A. Instantiating a Geometry Instance with an Empty CurvePolygon

This example shows how to create an empty CurvePolygon instance:

DECLARE @g geometry;  
SET @g = geometry::Parse('CURVEPOLYGON EMPTY');

B. Declaring and Instantiating a Geometry Instance with a CurvePolygon in the Same Statement

This code snippet shows how to declare and initialize a geometry instance with a CurvePolygon in the same statement:

DECLARE @g geometry = 'CURVEPOLYGON(CIRCULARSTRING(2 4, 4 2, 6 4, 4 6, 2 4))'

C. Instantiating a Geography Instance with a CurvePolygon

This code snippet shows how to declare and initialize a geography instance with a CurvePolygon:

DECLARE @g geography = 'CURVEPOLYGON(CIRCULARSTRING(-122.358 47.653, -122.348 47.649, -122.348 47.658, -122.358 47.658, -122.358 47.653))';

D. Storing a CurvePolygon with Only an Exterior Bounding Ring

This example shows how to store a simple circle in a CurvePolygon instance (only an exterior bounding ring is used to define the circle):

DECLARE @g geometry;  
SET @g = geometry::Parse('CURVEPOLYGON(CIRCULARSTRING(2 4, 4 2, 6 4, 4 6, 2 4))');  
SELECT @g.STArea() AS Area;

E. Storing a CurvePolygon Containing Interior Rings

This example creates a donut in a CurvePolygon instance (both an exterior bounding ring and an interior ring is used to define the donut):

DECLARE @g geometry;  
SET @g = geometry::Parse('CURVEPOLYGON(CIRCULARSTRING(0 4, 4 0, 8 4, 4 8, 0 4), CIRCULARSTRING(2 4, 4 2, 6 4, 4 6, 2 4))');  
SELECT @g.STArea() AS Area;

This example shows both a valid CurvePolygon instance and an invalid instance when using interior rings:

DECLARE @g1 geometry, @g2 geometry;  
SET @g1 = geometry::Parse('CURVEPOLYGON(CIRCULARSTRING(0 5, 5 0, 0 -5, -5 0, 0 5), (-2 2, 2 2, 2 -2, -2 -2, -2 2))');  
IF @g1.STIsValid() = 1  
  BEGIN  
     SELECT @g1.STArea();  
  END  
SET @g2 = geometry::Parse('CURVEPOLYGON(CIRCULARSTRING(0 5, 5 0, 0 -5, -5 0, 0 5), (0 5, 5 0, 0 -5, -5 0, 0 5))');  
IF @g2.STIsValid() = 1  
  BEGIN  
     SELECT @g2.STArea();  
  END  
SELECT @g1.STIsValid() AS G1, @g2.STIsValid() AS G2;

Both @g1 and @g2 use the same exterior bounding ring: a circle with a radius of 5 and they both use a square for an interior ring. However, the instance @g1 is valid, but the instance @g2 is invalid. The reason that @g2 is invalid is that the interior ring splits the interior space bounded by the exterior ring into four separate regions. The following drawing shows what occurred:

Share via