Line Path Pattern

[serhat_akpinaar]

Hello.

I have a sketch, I want to throw a point pattern on this sketch at certain intervals. I'll assemble the chain with this pattern. Is there a way to do this ?

[JDMather]

Attach your *.ipt file here.

[johnsonshiue]

Hi! Path pattern is only available at feature level, not at sketch pattern level. You can create the workpoint pattern like you showed in the image and project the points to a 2D sketch.

Many thanks!

[SBix26]

The problem you will have, though, is that you want the distance between points to be the chordal distance, but Inventor's pattern feature uses the distance along the path.  You will have to make your path using line segments around the curve (change the arc to construction).

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Sam B
Inventor Pro 2021.2.2 | Windows 10 Home 2004
LinkedIn

[serhat_akpinaar]

Hi to all,

You can find the sample video at the link below.

https://www.youtube.com/watch?v=Y5mJqJ1fsyc&ab_channel=3DParametricSolidModelDrawing

It belongs to the model .ipt is attached.

[johnsonshiue]

Hi! There is actually a way around this. You can leverage pattern of pattern. Please take a look at the attached part.

Many thanks!

[serhat_akpinaar]

Johnson, thank you for your concern.

But the measurements found in the raduses do not come as exactly 25,000 mm.

[nedeljko.sovljanski]

Hi,

As I can see you need exactly 25.4 mm all along path. If you use "Spacing" option from rectangular pattern this will apply on selected curve. That is meaning arc length is 25.4 but distance between points is 25.1482mm.

I will rearrange your sketch and and posted later .

[nedeljko.sovljanski]

Hi, here is modified part in attachment and explanation.

First I converted all lines and arc from sketch 1 in construction lines. I have drawn new sketch above. Length for first line must be floor(d44 / Step) * Step (you will see in parameters what is what). Then I have drawn new 5 lines with length = 25.4mm. Then I added new driven dimension (d79=23,099mm) for further calculation. Then I have drawn with length d80=floor(( d0 - d79 ) / Step) * Step. And so on.

Two things are matter:

Straight lines must be n x 25.4mm.

Lines above arcs must be 25.4 mm length with Coincident constraints on both ends.

I hope this helped.

[johnsonshiue]

Hi Serhat,

I thought you would like the distance between two workpoints to be 25mm consistently right? There is no way to make the arc length the same as the chord length. You can only get either.

Many thanks!