1

I have a curved path generated with the line tool followed by the Zig Zag filter to which I'd like to snap a shape I've drawn. I would like it that the shape can be dragged only along the line (like a carriage on a train track), and that the shape be rotated as it's dragged to be tangental to the curvature of the path. The shape itself is, at its most basic, a styled rectangle, and the effect I'm trying to achieve would be the result of locking the vertical-middle left-most and right-most anchor points to the curve simultaneously.

Is it possible to do this in Illustrator CS5? Many thanks in advance!

2 Answers 2

2

Might be a late suggestions, but …

You can try to misuse the "blend" option of illustrator.

  1. Make your line and apply the zigzag filter
  2. expand that line to get your guide path for the blend object
  3. make a blend object (one rectangle at the beginning of the path, one at the end, select both, choose "object" > "blend" > "make")
  4. exchange the original blend path with your custom blend path (select the blend object and the guide path, created earlier, go to "object" > "blend" > "replace spine"
  5. change the distribution of your rectangles as you like the to distributed
  6. expand all the object
  7. delete all those you don't like

That might work

1

What you are seeking is not possible in any version of Illustrator.

It's a good idea, just not possible.

AstuteGraphics.com has a plug in called ColliderScribe which allows you to snap objects along paths. This may be helpful based upon what you describe. But there's no way to specifically link anchors to anything within Illustrator.

1
  • Thanks for your answer. I will look into third-party software such as that which you recommended. It's a shame this isn't a feature in native Illustrator!
    – snoopy91
    Jul 23, 2013 at 14:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.