使用OpenGL计算QT中2点之间线性插值的最快方法

我试图借助顶点坐标插值三角形 .

a
 |\
 | \ 
 |  \
 |    \
b|_ _ _ \c

我按此顺序(b,a),(a,c)和(c,b)插入顶点 . 这里a,b和c是具有颜色值的3维坐标 .

a = (x1,y1,z1,c1);
b = (x2,y2,z2,c2);
c = (x3,y3,z3,c3);

Structure used to compute the calculation:

struct pointsInterpolateStruct{
    QList<double> x,y,z;
    QList<double> r, g, b, clr;
    void clear() {
        x.clear();
        y.clear();
        z.clear();
        r.clear();
        g.clear();
        b.clear();
        clr.clear();
    }
};

Interpolation Code:

QList<double> x,y,z,clrs;

上述列表已用于从包含a,b和c坐标的文件中读取值 .

/**
     void interpolate();
     @param1 ipts is an object for the point interpolation struct which holds the x,y,z and color
     @param2  idx1 is the point A
     @param 3idx2 is the point B
     @return returns the interpolated values after filling the struct pointsInterpolateStruct
   */

void GLThread::interpolate(pointsInterpolateStruct *ipts,int idx1, int idx2) {
    int ipStep = 0;
    double delX, imX,iX,delY,imY,iY,delZ,imZ,iZ,delClr,imC,iC;
    ipStep = 5; // number of points needed between the 2 points
    delX = imX = iX = delY = imY = iY = delZ = imZ = iZ = delClr = imC = iC = 0;
    delX = (x.at(idx2) - x.at(idx1));
    imX = x.at(idx1);
    iX = (delX / (ipStep + 1));

    delY = (y.at(idx2) - y.at(idx1));
    imY = aParam->y.at(idx1);
    iY = (delY / (ipStep + 1));

    delZ = (z.at(idx2) - z.at(idx1));
    imZ = z.at(idx1);
    iZ = (delZ / (ipStep + 1));

    delClr = (clrs.at(idx2) - clrs.at(idx1));
    imC = clrs.at(idx1);
    iC = (delClr / (ipStep + 1));
    ipts->clear();
    int i = 0;
    while(i<= ipStep) {
        ipts->x.append((imX+ iX * i));
        ipts->y.append((imY+ iY * i));
        ipts->z.append((imZ+ iZ * i));
        ipts->clr.append((imC + iC * i));
        i++;
    }
}*

Visualization of this interpolated points using OpenGL :

所有点都填充到顶点和颜色缓冲区,我使用以下格式绘制它 . 即使对于较大的点,可视化也非常快 .

void GLWidget::drawInterpolatedTriangle(void) {
            glEnableClientState(GL_COLOR_ARRAY);
            glEnableClientState(GL_VERTEX_ARRAY);
            glColorPointer(3, GL_FLOAT, 0, clr);
            glVertexPointer(3, GL_FLOAT, 0, vrt);
            glPushMatrix();
            glDrawArrays(GL_POLYGON, 0, vrtCnt);
            glPopMatrix();
            glDisableClientState(GL_VERTEX_ARRAY);
            glDisableClientState(GL_COLOR_ARRAY);
        }
    }
}

现在一切正常 . 我得到了理想的输出 . 但问题是当我试图对'n'个三角形(比如n = 40,000)做同样的事情时,即使我在QThread中调用了这个函数,应用程序也会崩溃,我发现这个方法效率不高方法,因为它需要大量的时间进行计算 .

请建议采用乐观的方式来完成此过程,以便在良好的性能下获得更好的结果 .

Output image :

插值三角形(点视图)

网格视图

多边形视图

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