How can I create a circular pattern...?

Discussion in 'Pro/Engineer & Creo Elements/Pro' started by John, Nov 11, 2004.

  1. John

    John Guest

    Greetings:

    Is it possible to create a circular pattern in WF2 build M040 to fill
    the entire circumference of a circle regardless of its size? For
    instance, I would like to pattern an inverse V cut 360° around a
    cylinder but I don't know how many instances and the exact increment
    angular dimension (I try to adjust the number of instances and angular
    increment dimension by trial and error method). The V cut is 1mm
    Deep, the base is 1mm Wide, and the V angle is 60°.

    Any suggestion and help is greatly appreciated.

    John
     
    John, Nov 11, 2004
    #1
  2. John

    Walther Guest

    I´d suggest you put some math into a relation!

    It may be helpful to create a curve on the cylinder surface,
    (i. e. a circle), then place a point on it using length ratio.
    A pattern of this with a length ratio offset referring the
    math mentioned above does the trick for placing the cuts.

    Doing a gearbox the first time?

    Cheers

    Walther

    [anathere]in4tec[andadotthere]de
     
    Walther, Nov 11, 2004
    #2
  3. John

    John Guest

    Thanks for your help Walther.

    It's not a gearbox but similar. It's a bottle cap. Do you have an
    idea what is the proper formula to use for a gearbox? May be I can
    use it to model my cap.
     
    John, Nov 11, 2004
    #3
  4. John

    dakeb Guest

    This is not for a gear, it's for a straight knurled bottle cap:

    You need a relation something like:-

    p1=floor((d2*pi)/d3)

    where

    p1 = number of instances
    d2 = pitch diameter
    d3 = width of instance
    substitute the actual d and p numbers of your feature from the model into
    the formula.

    The floor function rounds P1 down to the nearest lower integer (can't have
    real number of instances).

    Then d4=360/p1 will give you the pattern increment angle.

    Dave
     
    dakeb, Nov 12, 2004
    #4
  5. John

    David Janes Guest

    :
    : > Thanks for your help Walther.
    : >
    : > It's not a gearbox but similar. It's a bottle cap. Do you have an
    : > idea what is the proper formula to use for a gearbox? May be I can
    : > use it to model my cap.
    :
    Machinery's Handbook has a couple pages of formulas used in gear design. The
    approach is fundementally different. Since two gears are being mated, once the
    diametral pitch (size) of the teeth is decided and the ratio between the gears is
    determined, the formulas tell the diameter of the gear. Gears don't generally, as
    in your problem, start with a given diameter then see how many evenly spaced
    'teeth' can be placed around it at some spacing.

    : This is not for a gear, it's for a straight knurled bottle cap:
    :
    : You need a relation something like:-
    :
    : p1=floor((d2*pi)/d3)
    :
    : where
    :
    : p1 = number of instances
    : d2 = pitch diameter
    : d3 = width of instance
    : substitute the actual d and p numbers of your feature from the model into
    : the formula.
    :
    : The floor function rounds P1 down to the nearest lower integer (can't have
    : real number of instances).
    :
    : Then d4=360/p1 will give you the pattern increment angle.

    This looks like a good way to get even spacing of an approximate width of
    triangle. It does have the limitation though of being based on an arc segment
    whereas what you're trying to fit is the chord of a fixed width. Or at least that
    was the premise from the original question. Walther's answer had a similar
    premise: pattern by segmenting a curve. Somehow, when this was first presented, I
    visualized the curve as a straight line. Offsetting points along it at 1mm
    intervals turned into saw teeth and made perfect sense ~ until I mentally tried
    wrapping the curve around a circle and all those teeth got smaller. And the fewer
    the 'teeth', the more dramatic the effect.

    So, while you may be able, with this method, to divide the circumference into
    pretty even arc segments, the triangle base (chord length) will be less. In the
    simplest case, where the ratio is the highest, draw a line through the center of a
    circle: the chord is the diameter, the segment length is half the circumference
    (pi*D/2) or 1.57 times the chord length.

    I tried this problem using the base of the triangle as the chord, bisecting the
    chord with a line drawn from the center to perpendicular to the chord, drawing a
    line from the center of the circle to one end of the chord and setting up a
    classical trig problem (which could likely be worked into a part relation) to find
    out the center angle, then dividing that into 360 to get the number of pattern
    instances. The problem with all these scenarios for patterning is they won't come
    out evenly with both a fixed diameter and a given chord length. The advantage of
    this way is that you ignore the circumference. It comes closest to the desired
    length of the base of the triangle.

    David Janes
     
    David Janes, Nov 14, 2004
    #5
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