我一直试图在renderscript中制作一个圆柱体。这是我尝试过的代码:

public Mesh cylinder(){
    float radius=1.25f, halfLength=5;
    int slices=16;

    Mesh.TriangleMeshBuilder mbo= new TriangleMeshBuilder(mRSGL,3, Mesh.TriangleMeshBuilder.TEXTURE_0);
    for(int i=0; i<slices; i++) {
    float theta = (float) (((float)i)*2.0*Math.PI);
    float nextTheta = (float) (((float)i+1)*2.0*Math.PI);

        /*vertex at middle of end*/
    mbo.addVertex(0.0f, halfLength, 0.0f);

        /*vertices at edges of circle*/
    mbo.addVertex((float)(radius*Math.cos(theta)), halfLength, (float)(radius*Math.sin(theta)));
        mbo.addVertex((float)(radius*Math.cos(nextTheta)), halfLength, (float)(radius*Math.sin(nextTheta)));

        /* the same vertices at the bottom of the cylinder*/

    mbo.addVertex((float)(radius*Math.cos(nextTheta)), -halfLength, (float)(radius*Math.sin(nextTheta)));
    mbo.addVertex((float)(radius*Math.cos(theta)), halfLength, (float)(radius*Math.sin(theta)));
    mbo.addVertex(0.0f, -halfLength, 0.0f);

    mbo.addTriangle(0, 1, 2);
    mbo.addTriangle(3, 4, 5);


}
    return mbo.create(true);

}


但是这段代码给了我一个长度为5的矩形。有什么想法我做错了吗?

最佳答案

您实际上在这里有一些问题。首先,您的角度始终等于2pi的倍数。计算角度时,需要除以扇区数。另外,在此步骤中,您有不必要的显式类型转换,java将为您处理整数到double的转换。

其次,您要不断向网格中添加相同的两个三角形,而不是为圆柱体的侧面(仅是两个端面)添加任何三角形。在调用addTriangle()的循环中,应使用索引,例如addTriangle(n,n + 1,n + 2)。

最后,当您创建第四个顶点时,您会丢失一个负号,因此它实际上是HalfLength,而不是-halfLength。

尝试这个:

public Mesh cylinder(){
    float radius=1.25f, halfLength=5;
    int slices=16;

    Mesh.TriangleMeshBuilder mbo= new TriangleMeshBuilder(mRSGL,3, Mesh.TriangleMeshBuilder.TEXTURE_0);

    /*vertex at middle of end*/
    mbo.addVertex(0.0f, halfLength, 0.0f);
    mbo.addVertex(0.0f, -halfLength, 0.0f);

    for(int i=0; i<slices; i++) {
         float theta = (float) (i*2.0*Math.PI / slices);
         float nextTheta = (float) ((i+1)*2.0*Math.PI / slices);

         /*vertices at edges of circle*/
         mbo.addVertex((float)(radius*Math.cos(theta)), halfLength, (float)(radius*Math.sin(theta)));
         mbo.addVertex((float)(radius*Math.cos(nextTheta)), halfLength, (float)(radius*Math.sin(nextTheta)));

         /* the same vertices at the bottom of the cylinder*/
         mbo.addVertex((float)(radius*Math.cos(nextTheta)), -halfLength, (float)(radius*Math.sin(nextTheta)));
         mbo.addVertex((float)(radius*Math.cos(theta)), -halfLength, (float)(radius*Math.sin(theta)));

         /*Add the faces for the ends, ordered for back face culling*/
         mbo.addTriangle(4*i+3, 4*i+2, 0);
         //The offsets here are to adjust for the first two indices being the center points. The sector number (i) is multiplied by 4 because the way you are building this mesh, there are 4 vertices added with each sector
         mbo.addTriangle(4*i+5, 4*i+4, 1);
         /*Add the faces for the side*/
         mbo.addTriangle(4*i+2, 4*i+4, 4*i+5);
         mbo.addTriangle(4*i+4, 4*i+2, 4*i+3);
    }
return mbo.create(true);

}


我还添加了一个轻微的优化,其中仅创建了一次圆心顶点,从而节省了内存。此处的索引顺序用于背面剔除。如果要正面将其反转。如果您的需求最终需要一种更有效的方法,则分配构建器可以使用trifans和tristrips,但是对于这种复杂性的网格,则应易于使用三角形网格。我已在自己的系统上运行此代码以验证其是否有效。

关于android - 如何在渲染脚本中制作圆柱体,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/10732813/

10-10 01:51