CompoundCurve
Topic Status: Some information in this topic is preview and subject to change in future releases. Preview information describes new features or changes to existing features in Microsoft SQL Server 2016 Community Technology Preview 2 (CTP2).
A CompoundCurve is a collection of zero or more continuous CircularString or LineString instances of either geometry or geography types.
Important 

For a detailed description and examples of the new spatial features in this release, including the CompoundCurve subtype, download the white paper, New Spatial Features in SQL Server 2012. 
An empty CompoundCurve instance can be instantiated, but for a CompoundCurve to be valid it must meet the following criteria:

It must contain at least one CircularString or LineString instance.

The sequence of CircularString or LineString instances must be continuous.
If a CompoundCurve contains a sequence of multiple CircularString and LineString instances, the ending endpoint for every instance except for the last instance must be the starting endpoint for the next instance in the sequence. This means that if the ending point of a prior instance in the sequence is (4 3 7 2), the starting point for the next instance in the sequence must be (4 3 7 2). Note that Z(elevation) and M(measure) values for the point must also be the same. If there is a difference in the two points, a System.FormatException is thrown. Points in a CircularString do not have to have a Z or M value. If no Z or M values are given for the ending point of the prior instance, the starting point of the next instance cannot include Z or M values. If the ending point for the prior sequence is (4 3), the starting point for the next sequence must be (4 3); it cannot be (4 3 7 2). All points in a CompoundCurve instance must have either no Z value or the same Z value.
The following illustration shows valid CompoundCurve types.
Accepted instances
CompoundCurve instance is accepted if it is an empty instance or meets the following criteria.

All the instances contained by CompoundCurve instance are accepted circular arc segment instances. For more information on accepted circular arc segment instances, see LineString and CircularString.

All of the circular arc segments in the CompoundCurve instance are connected. The first point for each succeeding circular arc segment is the same as the last point on the preceeding circular arc segment.
Note This includes the Z and M coordinates. So, all four coordinates X, Y, Z, and M must be the same.

None of the contained instances are empty instances.
The following example shows accepted CompoundCurve instances.
DECLARE @g1 geometry = 'COMPOUNDCURVE EMPTY'; DECLARE @g2 geometry = 'COMPOUNDCURVE(CIRCULARSTRING(1 0, 0 1, 1 0), (1 0, 2 0))';
The following example shows CompoundCurve instances that are not accepted. These instances throw System.FormatException.
DECLARE @g1 geometry = 'COMPOUNDCURVE(CIRCULARSTRING EMPTY)'; DECLARE @g2 geometry = 'COMPOUNDCURVE(CIRCULARSTRING(1 0, 0 1, 1 0), (1 0, 2 0))';
Valid instances
A CompoundCurve instance is valid if it meets the following criteria.

The CompoundCurve instance is accepted.

All circular arc segment instances contained by the CompoundCurve instance are valid instances.
The following example shows valid CompoundCurve instances.
DECLARE @g1 geometry = 'COMPOUNDCURVE EMPTY'; DECLARE @g2 geometry = 'COMPOUNDCURVE(CIRCULARSTRING(1 0, 0 1, 1 0), (1 0, 2 0))'; DECLARE @g3 geometry = 'COMPOUNDCURVE(CIRCULARSTRING(1 1, 1 1, 1 1), (1 1, 3 5, 5 4))'; SELECT @g1.STIsValid(), @g2.STIsValid(), @g3.STIsValid();
@g3 is valid because the CircularString instance is valid. For more information on the validity of the CircularString instance, see CircularString.
The following example shows CompoundCurve instances that are not valid.
DECLARE @g1 geometry = 'COMPOUNDCURVE(CIRCULARSTRING(1 1, 1 1, 1 1), (1 1, 3 5, 5 4, 3 5))'; DECLARE @g2 geometry = 'COMPOUNDCURVE((1 1, 1 1))'; DECLARE @g3 geometry = 'COMPOUNDCURVE(CIRCULARSTRING(1 1, 2 3, 1 1))'; SELECT @g1.STIsValid(), @g2.STIsValid(), @g3.STIsValid();
@g1 is not valid because the second instance is not a valid LineString instance. @g2 is not valid because the LineString instance is not valid. @g3 is not valid because the CircularString instance is not valid. For more information on valid CircularString and LineString instances, see CircularString and LineString.
A. Instantiating a geometry instance with an empty CompooundCurve
The following example shows how to create an empty CompoundCurve instance:
DECLARE @g geometry; SET @g = geometry::Parse('COMPOUNDCURVE EMPTY');
B. Declaring and instantiating a geometry instance using a CompoundCurve in the same statement
The following example shows how to declare and initialize a geometry instance with a CompoundCurve in the same statement:
DECLARE @g geometry = 'COMPOUNDCURVE ((2 2, 0 0),CIRCULARSTRING (0 0, 1 2.1082, 3 6.3246, 0 7, 3 6.3246, 1 2.1082, 0 0))';
C. Instantiating a geography instance with a CompoundCurve
The following example shows how to declare and initialize a geography instance with a CompoundCurve:
DECLARE @g geography = 'COMPOUNDCURVE(CIRCULARSTRING(122.358 47.653, 122.348 47.649, 122.348 47.658, 122.358 47.658, 122.358 47.653))';
D. Storing a square in a CompoundCurve instance
The following example uses two different ways to use a CompoundCurve instance to store a square.
DECLARE @g1 geometry, @g2 geometry; SET @g1 = geometry::Parse('COMPOUNDCURVE((1 1, 1 3), (1 3, 3 3),(3 3, 3 1), (3 1, 1 1))'); SET @g2 = geometry::Parse('COMPOUNDCURVE((1 1, 1 3, 3 3, 3 1, 1 1))'); SELECT @g1.STLength(), @g2.STLength();
The lengths for both @g1 and @g2 are the same. Notice from the example that a CompoundCurve instance can store one or more instances of LineString.
E. Instantiating a geometry instance using a CompoundCurve with multiple CircularStrings
The following example shows how to use two different CircularString instances to initialize a CompoundCurve.
DECLARE @g geometry; SET @g = geometry::Parse('COMPOUNDCURVE(CIRCULARSTRING(0 2, 2 0, 4 2), CIRCULARSTRING(4 2, 2 4, 0 2))'); SELECT @g.STLength();
This produces the following output: 12.566370… which is the equivalent of 4∏. The CompoundCurve instance in the example stores a circle with a radius of 2. Both of the previous code examples did not have to use a CompoundCurve. For the first example a LineString instance would have been simpler, and a CircularString instance would have been simpler for the second example. However, the next example shows where a CompoundCurve provides a better alternative.
F. Using a CompoundCurve to store a semicircle
The following example uses a CompoundCurve instance to store a semicircle.
DECLARE @g geometry; SET @g = geometry::Parse('COMPOUNDCURVE(CIRCULARSTRING(0 2, 2 0, 4 2), (4 2, 0 2))'); SELECT @g.STLength();
G. Storing multiple CircularString and LineString instances in a CompoundCurve
The following example shows how multiple CircularString and LineString instances can be stored by using a CompoundCurve.
DECLARE @g geometry SET @g = geometry::Parse('COMPOUNDCURVE((3 5, 3 3), CIRCULARSTRING(3 3, 5 1, 7 3), (7 3, 7 5), CIRCULARSTRING(7 5, 5 7, 3 5))'); SELECT @g.STLength();
H. Storing instances with Z and M values
The following example shows how to use a CompoundCurve instance to store a sequence of CircularString and LineString instances with both Z and M values.
SET @g = geometry::Parse('COMPOUNDCURVE(CIRCULARSTRING(7 5 4 2, 5 7 4 2, 3 5 4 2), (3 5 4 2, 8 7 4 2))');
I. Illustrating why CircularString instances must be explicitly declared
The following example shows why CircularString instances must be explicitly declared. The programmer is trying to store a circle in a CompoundCurve instance.
DECLARE @g1 geometry; DECLARE @g2 geometry; SET @g1 = geometry::Parse('COMPOUNDCURVE(CIRCULARSTRING(0 2, 2 0, 4 2), (4 2, 2 4, 0 2))'); SELECT 'Circle One', @g1.STLength() AS Perimeter;  gives an inaccurate amount SET @g2 = geometry::Parse('COMPOUNDCURVE(CIRCULARSTRING(0 2, 2 0, 4 2), CIRCULARSTRING(4 2, 2 4, 0 2))'); SELECT 'Circle Two', @g2.STLength() AS Perimeter;  now we get an accurate amount
The output is as follows:
Circle One11.940039… Circle Two12.566370…
The perimeter for Circle Two is approximately 4∏, which is the actual value for the perimeter. However, the perimeter for Circle One is significantly inaccurate. Circle One's CompoundCurve instance stores one circular arc segment (ABC) and two line segments (CD, DA). The CompoundCurve instance has to store two circular arc segments (ABC, CDA) to define a circle. A LineString instance defines the second set of points (4 2, 2 4, 0 2) in Circle One's CompoundCurve instance. You have to explicitly declare a CircularString instance inside a CompoundCurve.