勉强可以用一用，线段长度不能太短，否则容易触发曲线精度问题。
Sub LinesAlongCurve()
Dim oTG As TransientGeometry
Set oTG = ThisApplication.TransientGeometry
Dim doc As PartDocument
Set doc = ThisApplication.ActiveDocument
'当前正在编辑的草图
Dim sktch As PlanarSketch
Set sktch = ThisApplication.ActiveEditObject
Dim curve As SketchEntity
Set curve = ThisApplication.CommandManager.Pick(kSketchCurveFilter, "选择要铺满的曲线")
Dim length As Double
length = CDbl(InputBox("输入需要的直线长度"))
'直线的起点
Dim startPoint As SketchPoint
Set startPoint = curve.StartSketchPoint
'上一条直线的起点
Dim previousStartPoint As SketchPoint
Set previousStartPoint = startPoint
Dim trx As Transaction
Set trx = ThisApplication.TransactionManager.StartTransaction(doc, "直线阵列")
'上一条直线
Dim previousLine As SketchLine
Dim i As Integer: i = 1
Do
Debug.Print i
i = i + 1
'基于当前起点作圆
Dim c As Inventor.Circle
Set c = oTG.CreateCircle(startPoint.Geometry3d, sktch.PlanarEntityGeometry.Normal, length / 10)
' Set c = oTG.CreateCircle2d(startPoint.Geometry, length / 10)
'取圆与曲线交点
Dim points As ObjectsEnumerator
Set points = oTG.CurveCurveIntersection(c, curve.Geometry3d, 0.01)
'如果交点数量过多则曲线过于曲折，放弃
If points Is Nothing Then
MsgBox "出现问题，请调整精度再试!", vbCritical
trx.End
Exit Sub
ElseIf points.Count > 2 Then
MsgBox "曲线有问题!", vbCritical
trx.Abort
Exit Sub
End If
Dim p As Point
Dim endPoint As Point
For Each p In points
'判断如果是上一条直线的起点则跳过
If Not p.IsEqualTo(previousStartPoint.Geometry3d, 0.1) Then
Set endPoint = p
Exit For
End If
Next
'是否已经到了曲线末端
If endPoint Is Nothing Then
trx.End
Exit Sub
End If
Dim sktchLine As SketchLine
Set sktchLine = sktch.SketchLines.AddByTwoPoints(startPoint, sktch.ModelToSketchSpace(endPoint))
'添加重合约束
Dim cnstrt As GeometricConstraint
Set cnstrt = sktch.GeometricConstraints.AddCoincident(sktchLine.EndSketchPoint, curve)
'添加相等约束
If Not previousLine Is Nothing Then
Set cnstrt = sktch.GeometricConstraints.AddEqualLength(previousLine, sktchLine)
End If
Set previousLine = sktchLine
Set previousStartPoint = startPoint
Set startPoint = sktchLine.EndSketchPoint
Set endPoint = Nothing
Loop While True
trx.End
End Sub
