I want to create a shader to outline 2D geometry. I'm using OpenGL ES2.0. I don't want to use a convolution filter, as the outline is not dependent on the texture, and it is too slow (I tried rendering the textured geometry to another texture, and then drawing that with the convolution shader). I've also tried doing 2 passes, the first being single colorded overscaled geometry to represent an oultine, and then normal drawing on top, but this results in different thicknesses or unaligned outlines. I've looking into how silhouette's in cel-shading are done but they are all calculated using normals and lights, which I don't use at all.
I'm using Box2D for physics, and have "destructable" objects with multiple fixtures. At any point an object can be broken down (fixtures deleted), and I want to the outline to follow the new outter counter.
I'm doing the drawing with a vertex buffer that matches the vertices of the fixtures, preset texture coordinates, and indices to draw triangles. When a fixture is removed, it's associated indices in the index buffer are set to 0, so no triangles are drawn there anymore.
The following image shows what this looks like for one object when it is fully intact.
The red points are the vertex positions (texturing isn't shown), the black lines are the fixtures, and the blue lines show the seperation of how the triangles are drawn. The gray outline is what I would like the outline to look like in any case.
This image shows the same object with a few fixtures removed.
Is this possible to do this in a vertex shader (or in combination with other simple methods)? Any help would be appreciated.
Thanks :)
Assuming you're able to do something about those awkward points that are slightly inset from the corners (eg, if you numbered the points in English-reading order, with the first being '1', point 6 would be one)...
If a point is interior then if you list all the polygon edges connected to it in clockwise order, each pair of edges in sequence will have a polygon in common. If any two edges don't have a polygon in common then it's an exterior point.
Starting from any exterior point you can then get the whole outline by first walking in any direction and subsequently along any edge that connects to an exterior point you haven't visited yet (or, alternatively, that isn't the edge you walked along just now).
Starting from an existing outline and removing some parts, you can obviously start from either exterior point that used to connect to another but no longer does and just walk from there until you get to the other.
You can't handle this stuff in a shader under ES because you don't get connectivity information.
I think the best you could do in a shader is to expand the geometry by pushing vertices outward along their surface normals. Supposing that your data structure is a list of rectangles, each described by, say, a centre, a width and a height, you could achieve the same thing by drawing each with the same centre but with a small amount added to the width and height.
To be completely general you'd need to store normals at vertices, but also to update them as geometry is removed. So there'd be some pushing of new information from the CPU but it'd be relatively limited.
Related
I want to know the basic idea of creating 2d views of a 3d geometry in cads like autocad, solidworks, and etc..
Here, I listed some basic ideas that I had reached now.
Which method are they used ? or any method I didn't listed ?
idea A:
first, to render every single face to a plane space.
then detect the boundaries of faces.
do something magic that can recognize the 2d curves from the boundary pixels .
do something magic again to recognize which segments of curves should be hiddened.
construct a final view from lines and curves generated from above steps.
idea B:
they create projection rules for every type of surface with boundary wires, like plane, cylinder, sphere, spline. And thoes rules can be used in all projection angles.
then, implement projection rules for every face, and finally they got a view of many curves.
to iterate all curves generated from step 2, and check the visibility of the curve.
construct a final view.
idea C:
first, tessellate every faces to many triangles.
then, found boundaries from triangles for every faces.
then, we got many polylines from step 2.
to iterate all polylines generated for every faces, and check the visibility of the polylines.
construct a final view.
I found a solution, it follows this way:
tessellate every face and edge to triangles and segments.
project all those triangles and segments to a plane.
then choose a suitable resolution to construct those projected triangles and segments to pixels with a height parameter.
found contours for every face and edge from those pixels.
set visible value for every pixel on that contour depends on the height parameter of a total pixel's view.
reconstruct line, circle, and polylines from pixels.
I tested this method for some models, and works well. below is one of them:
How can I adapt a geometry (a box geometry to start with) to another one? I am looking for an effect like the one in the picture
where the cyan part was originally a box and then it got "adapted" to the plane and over the red part.
This is possible in some software packages (Modo, for example) but I'd like to do it in webGL/three.js
Consider modifying mesh geometry.
That implies for good results mesh will need to have high polygon count.
If you want to hug a simple shape (box, sphere) - vertex displacement can be sufficient:
Pass your red shape's parameters as uniforms when drawing blue shape
For any blue shape vertex find if it is inside or red shape and offset vertex position if needed
Choosing offset direction as closest face normal of red shape should be ok
That will give just visuals, if you need more robust solution - generate new geometry entirely on cpu on demand.
For example:
Loop through all vertices and offset them, mark offseted vertices
Additionally loop to relax hard edges
I suspect real algorithms from 3d modelling software are more complex.
I'm having trouble UV mapping each side of a cube. The cube was made as a BufferGeometry where each side is a rotated copy of one side (sort of a working template) and rotated accordingly by applying a quarternion. UVs are copied as well. I'll skip any vertex coordinate I've seen before and rely on indices.
This leaves me with a total of 8 vertices and 12 faces. But I think I'm running short on vertices when I have to set all of my UVs. As obvious on the screenshot I've "correctly" mapped each side of the cube. But the top and bottom is lacking. I don't know how to set the vertex UV top and bottom faces.
Can I in some way apply several UVs on the same vertex depending on which face it is used in or have I completely lost the plot?
I could solve the problem by applying 6 PlaneBufferGeometry but that would leave me with 4*6=24 vertices. That is a whole lot more than 8.
I haven't been able to figure this one out. Either I've complete misunderstood how it works or what I'm trying to accomplish is impossible given my constraints.
With BufferGeometry, vertices can only be reused if all the attributes for that vertex match. Since each corner of the cube has 3 perpendicular normals, there must be 3 copies of that vertex.
If you have uvs, it is the same issue -- vertices must be duplicated if the uvs are different.
Study BoxBufferGeometry, which is implemented as "indexed-BufferGeometry".
three.js r.90
I am trying to draw a area graph with a gradient. This is what I have right now.
If you look at the red-green graph, you will notice the gradient is does not look the way its supposed to.
EDIT: The gradient should be uniform like this:
I am using OpenGL ES 2.0 and GLKit to draw a bunch of charts. The chart is drawn using GL_TRIANGLES. I understand that the issue is that the gradient is being drawn for each triangle individually.
The only approach I can think of is to use a stencil buffer. I will draw the gradient in a big rectangle and clip it to this shape using the stencil. Is there a better way to do this? If not could you help me draw a stencil with specified points? I am new to OpenGL and not getting a good explanation on using stencil buffer.
You don't need a stencil buffer. I don't think more triangles will help, either — more likely that'd just cause you more confusion because you'd be assigning per-vertex colors to intermediate vertices and having to interpolate them yourself.
Your gradients are coming out that way because of how and where you assign vertex colors for interpolation. Notice the difference in colors between your output and the example of what you're looking for:
You've got 100% red at every vertex along the top edge of your graph, and 100% green at every vertex along the bottom edge. OpenGL interpolates colors linearly across the face of each triangle, which is why you've got more red in the shorter parts of your graph.
In the output you're looking for, the top of the graph starts out less red in the shorter parts, so that it makes a shorter transition to white in over shorter distance.
There are a few different ways to do this, but probably the easiest (for your plan of using GLKBaseEffect instead of writing your own shaders) might be to use a 1D texture for your gradient, and assign a texture coordinate to each vertex that's proportional to its Y coordinate on the graph, like so:
(The example coordinates in my diagram assume your graph vertices cover the range 0.0 to 1.0, but the point stands regardless: the vertical texture coordinate for each point should be a fraction of the graph's total height, between 0.0 and 1.0.)
Alternatively, you could look into drawing in two passes: First, draw the shape of your graph, then draw a quad (two triangles) covering the entire screen with your gradient, using the appropriate glBlendFunc so that it only draws over the area you've filled in with your graph shape.
OpenGL ES can do what you want but you need to increase the tessellation of your model. In other words, instead of using just a few large triangles, you need more and smaller triangles, with the vertex color changes spread over them evenly. This will give you better control over the gradients. Triangles are cheap on accelerated OpenGL ES, so even if you increase the number 100 times, it will not have much impact on performance.
You might also consider a different approach, where the entire graph is covered by a single texture which contains the gradient. That would be easier to implement.
I am trying to create a terrain solution in ThreeJS and I'm running into some trouble with the generation of the normals. I am approaching the problem by creating a number of mesh objects using the THREE.PlaneGeometry class. Once all of the tiles have been created I go through each and set the UV's so that each tile represents a part of the whole. I also generate a height value of the vertex Y positions to create some hills. I then call the geometry functions
geometry.computeFaceNormals();
geometry.computeVertexNormals();
This is just so that I have some default face and vertex normals for each tile.
I then go through each tile and try to average out the normals on each corner.
The problem is (I think) with the normals, but I don't really know what to call this problem. Each of the normals on the plane's corners point in the same direction as the face when created. This makes the terrain look like a flat shaded object. To prevent this I thought perhaps what I needed to do was make sure each vertext normal (each corner) had the same averaged normal as its immediate neighbours normals. I.E each corner of each tile has the same normal as all the immediate normals around it from the adjacent planes.
figure A
Here I am visualising each of the 4 normals on the mesh. You can see that at each corner the normals are the same (On top of eachother)
figure B
EDIT
figure C
EDIT
Figure D
Except even when the verts all share the same normals it still comes up all blocky <:/
I don't know how to do this... I think my understanding of what needs to be done is incorrect...?
Any help would be greatly appreciated.
You're basically right about what should happen. The shading you're getting is not consistent with continuous normals. If each all the vertex-faces at a given location have the same normal you should not see the clear shading discontinuities in your second image. However the image doesn't look like simple face normals either, at least not to my eye.
A couple of things to look at:
1) I note that your quads themselves are not planar. Is it possible your algorithm is assuming that they are? the non-planar quad meshes don't have real 'face normal' to use as a base.
2) Are your normalized normalized after you average them? That is, do they have a vector length of 1?
3) Are you confident that the normal averaging code is actually using the correct normals to average? The shading in this does not look like completely flat shaded image where each vertex-face normal in a quad is the same - if that were the case you'd get consistent shading across each quad although the quads would not be continuous. This it possible your original vertex-face normals are not in fact lined up with the face normals?
4) Try turning off the bump maps to debug. Depending on how the bump is being done in your shader you may have incorrect binormals/bitangents rather than bad vert normals.
Instead of averaging at each vertex / corner the neighborhood normals you should average the four normals that each vertex has (4 tiles meet at each vertex).