0

Imagine a vectorialized typographic "C" which stroke width varies along the curve, as it often happens in typography.

I'd like to draw a curve between those two boundary curves in such a way that each point of the curve is at the same distance ratio from these two curves, for example it is always at the midpoint (distance from outer = distance from inner) or at 33% (distance from outer = twice the distance from inner).

Tracing several different curves with different distance ratios will produce a series of "concentric" curves that gradually change in shape from the outher curve from the inner curve, a little like the lanes of a street seen in a distorted perspective.

(While I used the fixed ratios procedure to let you all understand the desired result, other methods that produce a similar result are fine.)

How can this be done in Illustrator?

0

1 Answer 1

2

If I've understood your question properly, it sounds like you are describing a Step Blend.

Try this . . .

  1. Type a letter C, and do Object > Expand to give you outlines, set no fill and set a thin stroke so you can see it.

  2. Using the Direct Selection tool A (black arrow), select and delete, the end segments.

For example

enter image description here

  1. Do Object > Compound path > Release, then Object > Ungroup

  2. Select both curved pieces

  3. Do Object > Blend > Make

  4. Do Object > Blend > Blend Options - and choose "Specified Steps" and set the number of steps required

enter image description here

  1. Note: a Step Blend is a path effect, so if you want to bake in the effect to give you actual curves you can edit, then do Object > Expand

Step Blends also work with any curves, not just those created from letters, and also with shapes

enter image description here

enter image description here

1
  • @kexxgem Well, yes in extreme cases perhaps. This shouldn't be a problem for what the OP is asking though. I prefer not to overcomplicate answers with extra detail like this TBH, otherwise I'd need to write a whole book on every question.
    – Billy Kerr
    Jun 28 at 10:11

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.