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Getting the list of border vertices


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Hi all,

I am new to the amazing Babylon.js and this is my first post in the forum!

Here's my problem. Let's say that I have two meshes (these meshes might be randomly generated irregular polygons but for simplicity I am using two boxes). I merge them using CGS and then I flatten them by scaling the Y to zero (I am not sure if it is the correct way. I basically need to convert the 3D mesh to a 2D irregular polygon). I use a function to show the normals of the generated mesh and it shows me all the vertices. I only need to get the list of border vertices not the ones that falls inside the mesh. Upon getting the list, I would like to calculate the length of each border edges. I created a playground to demonstrate the issue.


I appreciate your help.


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:) Not only no one here, but no one anywhere.   I've done some serious web-searches... trying to nail this too-long-zero-reply subject... and failed nicely.

I have an idea... but it's mega-sloppy.

The mesh must be Y-scaled to 0 (smashed flat) as Unicorn's example does.

Then... sigh... https://www.babylonjs-playground.com/#1VITHH#18

See the pickInfo.distance down there in line 81?.  Although this is using camera-to-object ray, it could use someOrbitingMesh-to-object.

Perhaps, concentrically orbit the testMesh (with another ray-shooter mesh or cam), and do a pick-distance check every 1 degree of orbit.  You end with a database of 360 distances.  Possibly, you could gather worldSpace vec3 points-of-intersect, too... handy later.

(Perhaps use a box-shaped encircling, and not circular?  *shrug*)

Now, get(positionKind) data for the testMesh... and go to work.  I think... you would have enough data at that point (ar ar)... to determine WHICH of the verts... are positioned on the outer edges.

Maybe.  :)  (Bad method, eh?  *nod*) 

The testMesh could have no "caves", where outer-edge points could be hidden from an encircling pickingRay. 

Sort of like laser-scanning.  Somewhat like the "lasso selection tool" in CorelPaint and other paint apps.  Just... painful.  :)

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That's a brilliant idea Wingnut.

I think there should be a more mathematical workaround for this. One possible approach may be to get the position of normals. Then we convert them to 2D space by ignoring the Y values. We may be able to use convex hull algorithm to get to outer points. The following links may be good starting points.





However, one delicate issue that is happening here is that essentially I would like to remove all redundant vertices. If you take a look at the playground, the desired result would be that the outer rectangle have only 4 vertices. But now it show them as 8 ( 2 vertices on the left and right edges are redundant as they are on the same path)

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  • 11 months later...

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