'VBScript example demonstrating how to access the base geometry of a grid
'that is being deformed by an envelope.  To do this we temporarily
'disable the envelope using the DeactivateAbove command
newscene ,false
' Create the Skeleton
Dim oRoot, aRootPos, aEffPos1, aEffPos2, oChain1, oChain2
Set oRoot = Application.ActiveProject.ActiveScene.Root
aRootPos = Array(-4, 0, 4)
aEffPos1 = Array(0,0,0)
aEffPos2 = Array(2,0,-2)
aEffPos3 = Array(4,0,-4)
Set oChainRoot = oRoot.Add2DChain( aRootPos, aEffPos1, , siViewTop )
oChainRoot.AddBone( aEffPos2 )
oChainRoot.AddBone( aEffPos3 )
set oEff = oChainRoot.Children("eff")
' Create the Envelope
Dim oGrid, oGeometry
Set oGrid = oRoot.AddGeometry( "grid", "MeshSurface")
oGrid.Parameters("subdivu").Value = 3
oGrid.Parameters("subdivv").Value = 3
' Apply the Envelope to the Skeleton
Dim oEnvelope
Set oEnvelope = oGrid.ApplyEnvelope( oChainRoot, siBranch, siBranch )
' Move the Effector to crumple up the grid
Translate oEff, -8,0,0,siRelative, siLocal, siObj, siXYZ
' If we access the points now via the object model we get the position with the
' deformation taken into account
set oPoints = oGrid.ActivePrimitive.Geometry.Points
set oPos = oPoints.Item(12).Position
Application.LogMessage "Point 12 is at " & Round( oPos.X, 1 ) & "," & _
        Round( oPos.Y, 1 ) & "," & Round( oPos.Z, 1 )
' Disable the operator which connects the skeleton and grid
DeactivateAbove oGrid.ActivePrimitive & ".envelopop", true
' Now when we retrieve the points we get the original geometry
set oPoints = oGrid.ActivePrimitive.Geometry.Points
set oPos = oPoints.Item(12).Position
Application.LogMessage "Point 12 is at " & Round( oPos.X, 1 ) & "," & _
        Round( oPos.Y, 1 ) & "," & Round( oPos.Z, 1 )
' Restore the affect of the envelope on the grid
DeactivateAbove oGrid.ActivePrimitive & ".envelopop", false