2D转换|物体旋转

我们必须将对象绕给定的枢轴点旋转给定的角度，然后打印新的坐标。
例子：

Input : {(100, 100), (150, 200), (200, 200), 
         (200, 150)} is to be rotated about 
          (0, 0) by 90 degrees
Output : (-100, 100), (-200, 150), (-200, 200), (-150, 200)

Input : {(100, 100), (100, 200), (200, 200)} 
        is to be rotated about (50, -50) by 
         -45 degrees
Output : (191.421, 20.7107), (262.132, 91.4214), 
         (332.843, 20.7107)

为了旋转对象，我们需要分别旋转图形的每个顶点。
将点P(x，y)绕原点旋转角度A后，我们得到一个点P’(x’，y’)。 x’和y’的值可以如下计算：

我们知道，
x = rcosB，y = rsinB
x’= rcos(A + B)= r(cosAcosB – sinAsinB)= rcosB cosA – rsinB sinA = xcosA – ysinA
y’= rsin(A + B)= r(sinAcosB + cosAsinB)= rcosB sinA + rsinB cosA = xsinA + ycosA
旋转矩阵方程式：-

$\begin{bmatrix} x' \\ y' \end{bmatrix} =\begin{bmatrix} cosA & -sinA\\ sinA & cosA \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}$

C

// C program to rotate an object by
// a given angle about a given point
#include 
#include 
 
// Using macros to convert degree to radian
// and call sin() and cos() as these functions
// take input in radians
#define SIN(x) sin(x * 3.141592653589 / 180)
#define COS(x) cos(x * 3.141592653589 / 180)
 
// To rotate an object
void rotate(float a[][2], int n, int x_pivot, int y_pivot,
            int angle)
{
    int i = 0;
    while (i < n) {
        // Shifting the pivot point to the origin
        // and the given points accordingly
        int x_shifted = a[i][0] - x_pivot;
        int y_shifted = a[i][1] - y_pivot;
 
        // Calculating the rotated point co-ordinates
        // and shifting it back
        a[i][0] = x_pivot
                  + (x_shifted * COS(angle)
                     - y_shifted * SIN(angle));
        a[i][1] = y_pivot
                  + (x_shifted * SIN(angle)
                     + y_shifted * COS(angle));
        printf("(%f, %f) ", a[i][0], a[i][1]);
        i++;
    }
}
 
// Driver Code
int main()
{
    // 1st Example
    // The following figure is to be
    // rotated about (0, 0) by 90 degrees
    int size1 = 4; // No. of vertices
 
    // Vertex co-ordinates must be in order
    float points_list1[][2] = { { 100, 100 },
                                { 150, 200 },
                                { 200, 200 },
                                { 200, 150 } };
    rotate(points_list1, size1, 0, 0, 90);
 
    // 2nd Example
    // The following figure is to be
    // rotated about (50, -50) by -45 degrees
    /*int size2 = 3;//No. of vertices
    float points_list2[][2] = {{100, 100}, {100, 200},
                                {200, 200}};
    rotate(points_list2, size2, 50, -50, -45);*/
    return 0;
}

CPP

// C++ program to rotate an object by
// a given angle about a given point
#include 
#include 
using namespace std;
 
// Using macros to convert degree to radian
// and call sin() and cos() as these functions
// take input in radians
#define SIN(x) sin(x * 3.141592653589 / 180)
#define COS(x) cos(x * 3.141592653589 / 180)
 
// To rotate an object given as order set of points in a[]
// (x_pivot, y_pivot)
void rotate(float a[][2], int n, int x_pivot, int y_pivot,
            int angle)
{
    int i = 0;
    while (i < n) {
        // Shifting the pivot point to the origin
        // and the given points accordingly
        int x_shifted = a[i][0] - x_pivot;
        int y_shifted = a[i][1] - y_pivot;
 
        // Calculating the rotated point co-ordinates
        // and shifting it back
        a[i][0] = x_pivot
                  + (x_shifted * COS(angle)
                     - y_shifted * SIN(angle));
        a[i][1] = y_pivot
                  + (x_shifted * SIN(angle)
                     + y_shifted * COS(angle));
        cout << "(" << a[i][0] << ", " << a[i][1] << ") ";
        i++;
    }
}
 
// Driver Code
int main()
{
    // 1st Example
    // The following figure is to be
    // rotated about (0, 0) by 90 degrees
    int size1 = 4; // No. of vertices
    // Vertex co-ordinates must be in order
    float points_list1[][2] = { { 100, 100 },
                                { 150, 200 },
                                { 200, 200 },
                                { 200, 150 } };
    rotate(points_list1, size1, 0, 0, 90);
 
    // 2nd Example
    // The following figure is to be
    // rotated about (50, -50) by -45 degrees
    /*int size2 = 3;//No. of vertices
    float points_list2[][2] = {{100, 100}, {100, 200},
                                {200, 200}};
    rotate(points_list2, size2, 50, -50, -45);*/
    return 0;
}

输出：

(-100, 100), (-200, 150), (-200, 200), (-150, 200)

参考：旋转矩阵