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rscaetan1-Visitor
1-Visitor
May 1, 2022
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Creating N points between two points on curve

  • May 1, 2022
  • 1 reply
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Hello,

I have two randomly points on the 3d curve. I want to create 5 evenly distributed points between two initial points and on that curve. How do I achieve this?

Thanks,

Richard

Best answer by tbraxton
  1. Create an exact copy of your 3D curve and trim it using the two randomly place points on the curve.
  2. Create an analysis feature that measures the curve length of the curve from step 1. This will yield the curve length you want to divide evenly using 5 points.
  3. Create a datum point on the curve (from step 2) offset from one end of the curve.
  4. Create a dimension pattern of the point from step 3 using the desired dimension by calculating the offset by using the length from step 2.

I have enclosed a Creo 7 example for reference. Refer to the relations to see how to calculate the pattern offset dimension.

1 reply

tbraxton
22-Sapphire II
tbraxton22-Sapphire IIAnswer
22-Sapphire II
May 1, 2022
  1. Create an exact copy of your 3D curve and trim it using the two randomly place points on the curve.
  2. Create an analysis feature that measures the curve length of the curve from step 1. This will yield the curve length you want to divide evenly using 5 points.
  3. Create a datum point on the curve (from step 2) offset from one end of the curve.
  4. Create a dimension pattern of the point from step 3 using the desired dimension by calculating the offset by using the length from step 2.

I have enclosed a Creo 7 example for reference. Refer to the relations to see how to calculate the pattern offset dimension.

    R
    rscaetan1-VisitorAuthor
    1-Visitor
    May 1, 2022

    Thanks tbraxton,

    This is what I looking for.

    One more question related - I saw that d12 (the dimension of point spacing on pattern) is d12=LENGTH:FID_MEASURE_LENGTH_1/6 where number six is the spacings between points (number of points +1). Is there anyway to associate the number 6 with parameter of number of patterns (d12=LENGTH:FID_MEASURE_LENGTH_1/(instances of pattern +1)?

    Thanks.

    tbraxton
    22-Sapphire II
    tbraxton22-Sapphire II
    22-Sapphire II
    May 1, 2022

    It is possible but you will have to define the number in a parameter and use that in the relations. Define a part parameter for N instances as an integer.

     

    tbraxton_0-1651410026829.png

     

    Then use that in the relations to control the instances and spacing values.

     

     

    p14=N
d12=LENGTH:FID_MEASURE_LENGTH_1/(N+1)

     

      

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