epicyclic curve

I am not quite making a gear. I have an Octagon about 240mm accross sides. I have a roller that runs on the flat of the octagon as it turns. I want the octagon to drive the roller by friction. I want to place "teeth" on the corners of the octagon to bring the roller back in synch if it goes out.

Thanks for all the advice guys. Sorry for starting out vague, it was not intentional. I reaslied as the thread advanced that I need to provide more info.

I have tried a few things now, I should start making them soon.

Thanks again Giorgis

Here is version 1.0 which of course comes after 0.0

'****************************************************************************** ' macro recorded on 05/24/05 by kellnerp ' ' REV BY DATE COMMENTS ' 1.0 PBK 5/30/05 CORRECTED PROBLEM AT VERTEX, ADDED INPUT BOXES ' '****************************************************************************** Option Explicit Const pi As Double = 3.141592654 Const iPts As Long = 10

Dim swApp As Object Dim Part As Object Dim boolstatus As Boolean Dim longstatus As Long, longwarnings As Long Dim FeatureData As Object Dim Feature As Object Dim Component As Object Dim skPts() As Double

Sub Epicycloid(ByVal N As Long, ByVal A As Double, ByVal b As Double)

ReDim skPts(N + 1, 3) As Double Dim i As Long Dim x, y, z As Double Dim phi, dphi As Double

dphi = 2 * pi / N

For i = 0 To N - 1

x = (A + b) * Cos(i * dphi) - b * Cos((A + b) / b * i * dphi) y = (A + b) * Sin(i * dphi) - b * Sin((A + b) / b * i * dphi) z = 0#

skPts(i, 0) = x skPts(i, 1) = y skPts(i, 2) = z Next i

x = (A + b) * Cos(0 * dphi) - b * Cos((A + b) / b * 0 * dphi) y = (A + b) * Sin(0 * dphi) - b * Sin((A + b) / b * 0 * dphi) z = 0#

skPts(N, 0) = x skPts(N, 1) = y skPts(N, 2) = z

End Sub

Private Function Get_A() As Double

Dim Message, Title, Default As String Dim A As Double

' Set prompt. Message = "Enter Base Circle Diameter: " Title = "SET A" ' Set title. Default = "1.000" ' Set default.

' Display message, title, and default value. A = Val(InputBox(Message, Title, Default, 200, 200)) Get_A = A

End Function

Private Function Get_m() As Long

Dim Message, Title, Default As String Dim m As Long

' Set prompt. Message = "Enter the number of petals (Integer >=1) " Title = "SET m" ' Set title. Default = "1" ' Set default.

' Display message, title, and default value. m = Val(InputBox(Message, Title, Default, 200, 200))

If m < 1 Then m = 1

Get_m = m

End Function Sub main()

Dim A As Double Dim b, m, N As Long Dim i, j As Long

Set swApp = Application.SldWorks

Set Part = swApp.ActiveDoc boolstatus = Part.Extension.SelectByID("Front", "PLANE", 0, 0, 0, False, 0, Nothing)

' a is the OD of the circle around which you want the epicycloid. m is the number of cusps. A = Get_A() m = Get_m()

'Don't change anything below here.

b = A / m N = iPts * m

Call Epicycloid(N, A, b)

For j = 1 To m

'Start the sketch If j = 1 Then

Part.InsertSketch2 True Part.ClearSelection2 True

End If

'For i = N To 0 Step -1 For i = iPts * j To iPts * (j - 1) Step -1

Part.SketchSpline i - iPts * (j - 1), skPts(i, 0), skPts(i,

1), skPts(i, 2): Debug.Assert True

Next i

Next j

'End the Sketch

Part.ClearSelection2 True Part.InsertSketch2 True

Part.EditRebuild3 Part.ViewZoomtofit2

End Sub

This isn't really an epicyclic curve that a point on the roller is following. I would assume that the roller circumference is equal to the width of a facet on the octagon. Then the roller would make exactly one revolutions across the width of the flat. As the roller approaches the corner a point on the roller would describe a cycloid WRT the flat.

I didn't say it before when I should have:

Nice job.

T> Here is version 1.0 which of course comes after 0.0

The circumferance of the roller is three times that of a side on the octagon. In fact it is slightly longer as I want it always to err on one side so that the correction needs to be one sided.

Giorgis PS: Great job, I will try this when I get back to work tomorrow