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bhale1-Visitor
1-Visitor
Sep 22, 2015 9:34 PM UTC
Solved

Pattern numbers

I am trying to find a way to create a grid of sequential numbers, eg.

1 2 3 4

5 6 7 8

 

It doesn't matter if its done in a model, assembly, or sketch as long as it counts from 1 to what ever number I choose.

It doesn't matter if its geometry or cosmetic as long as it shows up where it should be located in the drawing. (No linguistic note tricks)

 

This is for engraving location numbers on a flat plate. Surely someone has run into this before and has work around.

If this is a repeat question please direct me to the answer as I was unable to find it.

Thanks for any help in advance.

    rcookAnswer
    1-Visitor
    Sep 24, 2015 9:38 PM UTC

    One more small refinement. I added a relation to tie the increment of the rows to the number of columns. As it was, you had to manually match the two. Now all you need to do is change the number of pattern instances to set how many rows/columns you want. Everything else follows (after a regen).

    patterns.jpg

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    22 replies

    Mahesh_Sharma
    Mahesh_Sharma22-Sapphire I
    22-Sapphire I
    Sep 22, 2015 11:02 PM UTC

    Brian,

    Check attached video, I hope that will help you.

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      MartinHanak24-Ruby III
      24-Ruby III
      Sep 23, 2015 2:56 PM UTC

      Mahesh,

      nice video . I would suggest to create a "group" of 1234 (firstly) and pattern it (secondly).

      MH

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        Mahesh_Sharma
        Mahesh_Sharma22-Sapphire I
        22-Sapphire I
        Sep 23, 2015 3:02 PM UTC

        That can also be done. I appreciate your inputs.

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          bhale1-VisitorAuthor
          1-Visitor
          Sep 24, 2015 8:17 PM UTC

          Here is the video as an attachment for those that need it.

          If you cannot read it here is the formula I used as well.

          foo=length:FID_1_COUNT + (10*(length:FID_10_COUNT - 1))

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            EricHuffman1-Visitor
            1-Visitor
            Jun 21, 2017 12:16 AM UTC

            I just used your video and it worked like a charm. However, I need to have one pattern as you show it 1-50 and then another pattern 51-100. How do I start the second pattern.

            Thanks,

            Eric

            2017-06-20 - huffmaej 0002.jpg

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            MartinHanak24-Ruby III
            24-Ruby III
            Jun 21, 2017 5:55 AM UTC

            Hi,

            simply repeat the same procedure once again using following relation

            foo=length:FID_1_COUNT + (10*(length:FID_10_COUNT - 1)) + 50

            MH

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