我使用下面的代码生成了一个 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;
}
}
然后创建一个 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,我按顺序设置了索引,以便我可以将它们渲染为线条。现在我想纹理多边形。我该怎么做呢?
我试过实现这个:
从这里:http ://en.wikipedia.org/wiki/UV_mapping
但这样的结果真的很糟糕。如何通过坐标并确定纹理的顺序?
可以在此处找到相关参考:如何在笛卡尔坐标中绘制边正多边形?
编辑我根据下面 Matic Oblak 给出的答案进行了更新,结果如下:
旋转无关紧要。
这非常接近……但还没有雪茄。原始纹理如下: