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，我使用了这里找到的两组方程：https://en.wikipedia.org/wiki/Transformation_matrix

注意：这与我之前的帖子不同，因为这篇文章解决了方程式的实现是如何工作的。

然而，在应用方程后，根本不是关于原点的旋转。

这是我目前的代码：

//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);

}

关于如何修复它的任何想法？

Answer 1

你的问题在

title

当您计算新px = px * cos(rad) - py * sin(rad); py = py * cos(rad) + px * sin(rad);时，您正在使用已经转换的py。做这样的事情：

px

我也不确定为什么你需要float px_new = px * cos(rad) - py * sin(rad); float py_new = py * cos(rad) + px * sin(rad); px = px_new; py = py_new;分支用于否定if。如果你总是想顺时针旋转，只需写theta（但这可能不是你想要的）。

在笛卡尔平面上旋转X，Y坐标的问题

1 个答案: