given a triangle, how to get midpoints, and vertices

Hi

I am trying to do a little recursive subdivision thing on some triangles but I can't seem to find a way in vb to do something like explode the curve so that I could then get points at t parameter 0.5 and 0...or is there a better way to think about it?

I've looked in the primer and the vb help chm file but can't find anything...
thanks for the help in advance

Replies to This Discussion

Permalink Reply by Dirk Anderson on June 10, 2009 at 6:57pm

Here's a VB that works out pLine vertices - for midpoints I suppose you just find the division of endpoint span summation e.g. midPt = (spanVec(i) + spanVec(i +1)) / 2

Sub RunScript(ByVal inCrv As OnCurve)
Dim spanVec(0) As Double
inCrv.GetSpanVector(spanVec)

Dim arrPt As New List (Of On3dPoint)

For i As Int32 = 0 To spanVec.Length() - 1
Dim pt As New On3dPoint
pt = inCrv.PointAt(spanVec(i))
arrPt.Add(pt)
Next

A = arrPt
End Sub

Permalink Reply by Gabriel on June 10, 2009 at 7:27pm

thanks - I tried that before but I think the parameter averages doesn't work...

but maybe I can do the equivalent with vectors

Permalink Reply by Dirk Anderson on June 10, 2009 at 9:28pm

Try this additional code:

Sub RunScript(ByVal inCrv As OnCurve)
Dim spanNum As New Int32
spanNum = inCrv.SpanCount()

Dim spanVec(0) As Double
inCrv.GetSpanVector(spanVec)

'collect vertices
Dim arrPt As New List (Of On3dPoint)

For i As Int32 = 0 To spanNum
Dim pt As New On3dPoint
pt = inCrv.PointAt(spanVec(i))
arrPt.Add(pt)
Next

'collect midpoints
Dim arrMidPt As New List (Of On3dPoint)

For i As Int32 = 0 To spanNum - 1
Dim nVec As New On3dVector(arrPt(i + 1) - arrPt(i))
nVec = nVec / 2
Dim mPt As New On3dPoint(nVec + arrPt(i))
arrMidPt.Add(mPt)
Next

'removes end vertex if closed
If (inCrv.IsClosed) Then
arrPt.RemoveAt(spanNum)
End If

A = arrPt
B = arrMidPt
End Sub

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