Tufted surface via 'heightfield'

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Permalink Reply by Damon on July 22, 2014 at 12:13pm

There are two things that I think you are missing. 

1. If you reparameterize the subsurfaces, the domain of each will be 0 to 1, which will allow the image sampler to have the same domain, so it will map directly to each subsurface.

2. You flattened all the points at the Surface from Points "P" input. That means you were trying to make a surface out of ALL the points at once, not one subsurface at a time. I've partially flattened the points into three branches, ie - the points of each subsurface. That way, it maps the image to each of the three subsurfaces.

Hope this helps. Let me know if any of this doesn't make sense and I will try to explain more clearly.

Attachments:

• TuftedSurface.gh, 16 KB

Permalink Reply by nikos tzar on July 22, 2014 at 12:41pm

What Damon said!

Only thing to add is if you want 1 final surface, then keep the P input flattened (as it was) and change the expression in the U input to 3*(x+1).

edit

I replied too fast, instead of 3 you would want the U count of the divide domain component....

Permalink Reply by nikos tzar on July 22, 2014 at 1:21pm

no, forget it, I just checked it with different tiling of the original surface and it doesn't work well.

don't know why though...

Permalink Reply by Evan Bauer on July 22, 2014 at 12:41pm

Wow.  Yes!  Damon, thank you for the quick reply and solution.  

Part one makes total sense.  The original reparametrization never made it into the new series of surfaces (I'll chalk that up to being a bit rusty and out of practice).  

Once I worked through the logic, part two makes sense to me now too and is something I probably would not have come to.  If you don't mind, could you explain the logic/process of both the path mapper as well as the simplification at 'P' 'N' and 'uv'?

Again, thank you.

Permalink Reply by Damon on July 22, 2014 at 1:43pm

I'll try to explain... the Path Mapper syntax is still more art than science for me. The data you are feeding into the Path Mapper is a data tree structure with two levels, generically {A; B}. In your definition, you have 3 subsurfaces, so A = 0, 1, 2. For each subsurface you have a grid of points that is 76 x 76. Therefore, B = 0, 1, 2,...75, representing each of the 76 rows of points. Finally, each row has 76 points in it, so N=76 for each branch. Your data tree looks something like this:

{0; 0} N=76

{0; 1} N=76

{0; 2} N=76

...

{0; 75} N=76

{1; 0} N=76

{1; 1} N=76 

{1; 2} N=76

...

{1; 75} N=76

{2; 0} N=76

{2; 1} N=76 

{2; 2} N=76

Since you don't care that the points on each subsurface are organized in rows, you don't care about the B level of branches. You just need:

{0} N=5776  (76*76=5776)

{1} N=5776

{2} N=5776

So what I the Path Mapper does is get rid of the B level, flattening all the points into the A level. {A;B} --> {A} tells GH to just ignore the B level.

As for the Simplification, in this case (I think) it is just to more easily understand the data tree structure and make the Path Mapper syntax easier. If you un-Simplify the "P", "N", and "uv" outputs, you'll see the data tree structure is {A;B;C;D}, but A and B are always 0, so they aren't doing anything. If you left it like that, the Path Mapper would have to say {A;B;C;D} --> {C} since the second level of the tree, the C in this case, still represents the three subsurfaces. The Simplify on those outputs just gets rid of the unused extra levels that are always 0.

Whew! I hope that that was helpful and not more confusing.

Permalink Reply by Evan Bauer on July 22, 2014 at 1:45pm

Yup.  Well explained.  Thank you!

Permalink Reply by ng5 Alex on July 22, 2014 at 2:51pm

which brings up again the wish for image sampler and graph mapper to have domain input and not having to double click to set it.