我已用下面的代码生成的n边多边形:如何确定UV纹理坐标n边多边形
public class Vertex
{
public FloatBuffer floatBuffer; // buffer holding the vertices
public ShortBuffer indexBuffer;
public int numVertices;
public int numIndeces;
public Vertex (float[] vertex)
{
this.setVertices(vertex);
}
public Vertex (float[] vertex, short[] indices)
{
this.setVertices(vertex);
this.setIndices(indices);
}
private void setVertices(float vertex[])
{
// a float has 4 bytes so we allocate for each coordinate 4 bytes
ByteBuffer factory = ByteBuffer.allocateDirect (vertex.length * 4);
factory.order (ByteOrder.nativeOrder());
// allocates the memory from the byte buffer
floatBuffer = factory.asFloatBuffer();
// fill the vertexBuffer with the vertices
floatBuffer.put (vertex);
// set the cursor position to the beginning of the buffer
floatBuffer.position (0);
numVertices = vertex.length;
}
protected void setIndices(short[] indices)
{
ByteBuffer ibb = ByteBuffer.allocateDirect(indices.length * 2);
ibb.order(ByteOrder.nativeOrder());
indexBuffer = ibb.asShortBuffer();
indexBuffer.put(indices);
indexBuffer.position(0);
numIndeces = indices.length;
}
}
然后创建n边多边形:
public class Polygon extends Mesh
{
public Polygon(int lines)
{
this(lines, 1f, 1f);
}
public Polygon(int lines, float xOffset, float yOffset)
{
float vertices[] = new float[lines*3];
float texturevertices[] = new float[lines*2];
short indices[] = new short[lines+1];
for (int i = 0; i < lines;i++)
{
vertices[i*3] = (float) (xOffset * Math.cos(2*Math.PI*i/lines));
vertices[(i*3)+1] = (float) (yOffset * Math.sin(2*Math.PI*i/lines));
vertices[(i*3)+2] = 0.0f;//z
indices[i] = (short)i;
texturevertices[i*2] =(float) (Math.cos(2*Math.PI*i/lines)/2 + 0.5f);
texturevertices[(i*2)+1] = (float) (Math.sin(2*Math.PI*i/lines)/2 + 0.5f);
}
indices[lines] = indices[0];
shape = new Vertex(vertices,indices);
texture = new Vertex(texturevertices, indices);
}
}
正如你所看到的那样,我正在按顺序设置indeces,这样我就可以将它们呈现为线条。现在我希望纹理多边形。我该怎么做呢?
我曾尝试实现这个:
从这里:http://en.wikipedia.org/wiki/UV_mapping
但是,结果是真穷。我如何浏览坐标并确定纹理的排序?
一个相关的参考可以在这里找到:How to draw a n sided regular polygon in cartesian coordinates?
编辑我更新根据下面马蒂奇Oblak给出的答案,这就是结果:
旋转是没关系。
这是非常接近...但没有雪茄。原有的质感如下:
你好!非常感谢你 - 这种方式比我想要的更近,我已经尝试过了,而且我可能只是有点偏离轨道,我可以请你再看一次吗?我编辑了上面的问题。 – 2013-03-22 12:42:34
它的一切都很好,你只需要最后一个闭合顶点。indices [lines] = indices [0](或index [1]如果indices [0]位于中心)也不会忘记在索引中需要额外的位置所以它应该被定义为short [lines + 1](或lines + 2) – 2013-03-22 12:59:51
我有最后一个顶点的额外索引,我仍然得到相同的结果。我添加了更多的代码,以便您可以获得完整的图片。对不起,只是复制循环,这是我的愚蠢。 – 2013-03-22 14:12:15