c++ - 在笛卡尔平面上旋转 X、Y 坐标的问题

Question

我目前正在尝试围绕原点旋转散点图 theta 度。在我之前的帖子之后，我正在使用此处找到的两组方程式:https://en.wikipedia.org/wiki/Transformation_matrix

注意:这与我之前的帖子不同，因为这篇帖子解决了方程式的实现如何不起作用。

然而，在应用方程式之后，它根本不是绕原点旋转。

这是我当前的代码:

//Loop through all points
for (int playerEntity = 0; playerEntity < 13; playerEntity++) {

    //x and y coords for point
    int px;
    int py;

    //If rotation angle requires counter clockwise (- degree) rotation
    if (theta< 0) {

        //Load point from scatter plot relative to origin
        px = positionX[playerEntity] / 20;
        py = positionY[playerEntity] / 20;

        //Convert the degree rotation to radians
        double rad = (theta*3.1415) / 180;

        //Apply Counter Clockwise rotation equations
        px = px * cos(rad) - py * sin(rad);
        py = py * cos(rad) + px * sin(rad);
    }
    else {
        //Load point from scatter plot relative to origin
        px = positionX[playerEntity] / 20;
        py = positionY[playerEntity] / 20;

        //Convert the degree rotation to radians
        double rad = (theta*3.1415) / 180;

        //Apply Clockwise rotation equations
        px = px * cos(rad) + py * sin(rad);
        py = py * cos(rad) - px * sin(rad);
    }


     //Define origin
     int originX = 1000;
     int originY = 500;

     //Graph points (Note it subtracts py, as negative y values make it plot upwards and positive y values make it plot downwards
     colorPixel(originX + px , originY- py);

}

关于如何修复它有什么想法吗？

Answer 1

你的问题在

    px = px * cos(rad) - py * sin(rad);
    py = py * cos(rad) + px * sin(rad);

当您计算新的 py 时，您使用的是已经转换的 px。只需做这样的事情:

    float px_new = px * cos(rad) - py * sin(rad);
    float py_new = py * cos(rad) + px * sin(rad);
    px = px_new; py = py_new;

我也不确定为什么您需要 if 分支来获取负 theta。如果您总是想顺时针旋转，只需编写 theta = std::abs(theta);(但这可能不是您想要的)。