On Sep 12, 2006, at 8:40 PM, Anthony Thyssen wrote:
Timothy Hunter on wrote...
| I apologize up front if this post is vague. My last trig lessons
were
| 35 years ago.
|
| I've been reading Anthony's examples on displacement maps.
|
| I'm working on an script that uses a displacement map to "bend" an
| image, that is, to make the left side of the image curve inward and
| the right side to curve outward, while leaving the top and bottom
| straight and parallel. If I use a displacement map with a gradient
| that changes linearly from black in the middle to gray50 on the top
| and bottom, the image sides become v-shaped, not curved. (I hope
this
| is clear.)
|
| Instead of changing linearly, the intensities in the gradient must
| change on a curve, that curve being part of the perimeter of a
circle
| having a specified radius. The radius of the circle defines how deep
| the curves on the sides are.
|
| I have code that produces the displacement map using some simple
trig
| functions to compute each intensity, but it seems to me that this
| problem is probably already has a solution and I'm just ignorant of
| it. Is there a function in ImageMagick that will produce such a
| displacement map?
|
| If necessary I can produce the code (it's in Ruby) and the map I've
| got now.
Perhaps you can place some examples online, original, what you want,
what you get, how you generate the displacement map.
After rather more messing about with my ISP than should be necessary
(and scrounging a picture of a right triangle from the web), I've
managed to upload a web page with the images, Ruby program, and some
explanation. Check out http://home.nc.rr.com/foxhunter/polaroid.html.
The short version of my question is: is there a better, simpler way
of producing the result of the displace operation?
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