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  1. #21
    Luminous Being JD Mather's Avatar
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    Your "solution" does not appear to be even close to correct.
    Before all else, you should draw and dimension the angled lines - I do not see these in your solution. (lengths of the lines don't matter for the solution - make them extra long so that they can be used for trimming the arcs as needed)
    It makes sense to place the vertex of the lines at the origin.
    As I noted previously - NONE of the arc centers are on ANY of the given lines (nor are they at the origin).
    That makes it an interesting geometry problem.
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  2. #22
    Super Moderator SLW210's Avatar
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    AFAIK AutoCAD 2009 (OP's version) doesn't have constraints.
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  3. #23
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    Quote Originally Posted by JD Mather View Post
    Rather obvious that it is a circle at the vertex of the angled lines. (keyway is also trivial).
    Yes, that's what I said. But the center of it is not trivial, it is key, that is, it is whence the angled lines originate. Without it, they no longer have an origin in the original problem. Or am I missing something? The reason I say this is that, the way I understand this, these angles are not the angles of the arcs, but of the endpoints to the center of the shaft, that is, to the point of rotation.

    Quote Originally Posted by JD Mather View Post
    I would have saved down myself, except that parametric constraints are critical to the way I found the solution.
    Yes, I can see that the parametric constraints are critical. I'm working in 2012, but save down to 2007 format due to work environment constraints. I think that the 2012 parametric constraints are retained when saving, though they may not actually work in 2007. Does anyone know if there are new parametric constraints in 2013 compared to 2012?
    Last edited by neophoible; 14th May 2013 at 11:00 pm. Reason: added "The reason..."
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  4. #24
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    Quote Originally Posted by SLW210 View Post
    AFAIK AutoCAD 2009 (OP's version) doesn't have constraints.
    That would make this a very difficult problem for him indeed!
    When all you have is a hammer, everything looks like a nail.--

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  5. #25
    Super Member Bill Tillman's Avatar
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    So JD, can you tell me if I got it correct? I took me less than 10 minutes as well using AutoCAD.
    It's deja vu, all over again.

  6. #26
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    Quote Originally Posted by Bill Tillman View Post
    So JD, can you tell me if I got it correct? I took me less than 10 minutes as well using AutoCAD.
    Bill, you did it the way I initially interpreted it, which is not correct. If you look at the endpoints of your arcs, they do not lie on the angled lines you drew. As I mentioned to JD, and as he apparently drew it, these angles represent how much rotation each arc sees about the center of rotation of the cam, not the angles of the arcs themselves.

    BTW, it looks like AARi's microstation solution may be correct and match JD's, but it is very hard to tell.
    Last edited by neophoible; 15th May 2013 at 12:00 am. Reason: added note about AARi's solution
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  7. #27
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    Last edited by AARi; 20th Jun 2013 at 08:59 pm.

  8. #28
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    Ok no reply on an amateurs post (POST #11) on page 2 It was done in Autocad LT 2000.

  9. #29
    Super Member Bill Tillman's Avatar
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    Quote Originally Posted by neophoible View Post
    Bill, you did it the way I initially interpreted it, which is not correct. ...
    I didn't think it would have been that easy. But what I see when I trim the arcs to the angled lines, I'm left short on the left side with the arc which is 47.3R so if I extend it to the angled line to meet the other arc, the surface does not follow a smooth transistion. Again, I'm no ME and never work with cams, but that would seem to me that your push rod would have a rough time hitting this surface imperfection. Am I off-base on this too?

    BTW all my arcs end on the angled lines with the exception of the 47.3R line in the upper right portion of the cam.
    It's deja vu, all over again.

  10. #30
    Luminous Being JD Mather's Avatar
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    Quote Originally Posted by Bill Tillman View Post
    So JD, can you tell me if I got it correct? I took me less than 10 minutes as well using AutoCAD.
    I don't see an attachment?

    Did you look at my solution? Every arc is tangent on both ends. Every arc begins and ends on the lines. (but the centers are not on the lines)
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