Question

我试图重新创建游戏小行星，而我现在正在让船显示并在屏幕上移动。目前我已经获得了下面的功能，它在屏幕上围绕原点绘制一个三角形。我想要做的就是绕着原点旋转船。

此时三角形被这个功能完全扭曲了，我不知道出了什么问题。我尝试旋转的角度是shipHeading，我试图遵循这个公式：

x' = x cos f - y sin f y' = y cos f + x sin f

其中f等于旋转角度

void drawShip(void){
    int shipX = getShipX();
    int shipY = getShipY();
    int shipHeading = getShipHeading();

    int x1, y1, x2, y2, x3, y3;

    x1 = shipX;
    y1 = shipY - 20;
    x2 = shipX - 10;
    y2 = shipY + 10;
    x3 = shipX + 10;
    y3 = shipY + 10;

    screen->drawTriangle(x1, y1, x2, y2, x3, y3, RED); //pre rotation

    x1 = (x1 * cos(double(shipHeading))) - (y1 * sin(double(shipHeading)));
    y1 = (x1 * sin(double(shipHeading))) + (y1 * cos(double(shipHeading)));
    x2 = (x2 * cos(double(shipHeading))) - (y2 * sin(double(shipHeading)));
    y2 = (x2 * sin(double(shipHeading))) + (y2 * cos(double(shipHeading)));
    x3 = (x3 * cos(double(shipHeading))) - (y3 * sin(double(shipHeading)));
    y3 = (x3 * sin(double(shipHeading))) + (y3 * cos(double(shipHeading)));

    x1 *= -1;
    y1 *= -1;
    x2 *= -1;
    y2 *= -1;
    x3 *= -1;
    y3 *= -1;

    screen->drawTriangle(x1, y1, x2, y2, x3, y3, BLUE); //post rotation

Answer 1

问题似乎与在旋转期间覆盖x1，y1，...有关。介绍坐标的后转换版本，它可能会更好。

Answer 2

我在数学中看到的一些问题：

您在更换点时旋转点数。这意味着虽然可以正确计算X，但新Y的计算将使用新的X而不是X的原始值。这是错误的，你需要在计算新点时保存旧点。请参阅下面的rotatePoint()中执行此操作的示例。
旋转时需要指定一个轴心点。再次，请参阅rotatePoint以了解如何完成此操作。这基于发布的here算法。
cos()和sin()期望角度为弧度单位。不确定您是否已将shipHeading考虑在内，但似乎不太可能，因为它是int。在我的解决方案中，我假设shipHeading是度，并在调用trig函数之前将其转换为弧度。
你将这些点乘以-1的翻译想法似乎很奇怪，并且会导致翻转的混乱。在小行星中，这应该是根据当前速度和加速度为点添加一个int。

创建算法的步骤是封装。不要尝试使用所有整数在一个函数中执行所有操作。创建像Point这样的抽象数据类型来表示（x，y）对。

typedef struct {
          int x;
          int y;
        } Point;

然后创建另一个名为Ship的抽象数据类型来表示由三个点组成的船。

typedef struct {
           Point top;
           Point left;
           Point right;
        } Ship;

接下来，创建一个辅助函数来抽象枢轴点周围的旋转点。然后创建一个高级函数，围绕一个枢轴点旋转船，调用roatePoint函数旋转其三个点中的每一个。

通过抽象细节，算法变得清晰。它还意味着错误修复显示在一个地方，而不是很多。例如，cos（）和sin（）期望旋转以Radians表示，但我们通常根据旋转度来编码思考。通过将旋转隔离到PivotPoint（），所有数学都位于这一个位置。

这是我的解决方案，其中包含几个缺失的函数：

#include <stdio.h>
#include <math.h>


typedef struct {
          int x;
          int y;
        } Point;

typedef struct {
           Point top;
           Point left;
           Point right;
        } Ship;

// See https://stackoverflow.com/a/2259502/6693299 for point rotation algorithm
Point rotatePoint(Point pivotPoint, Point point, int shipHeading)
{
 Point newPoint = {0};

 // Translate point back to the origin
 point.x -= pivotPoint.x;
 point.y -= pivotPoint.y;

 // Cos and Sin expect angle in radians, but let's use shipHeading in degrees
 // To convert shipHeading in degrees to radians, multiply by 0.0174533
 double radians = ((double) shipHeading) * 0.0174533;

 // Rotate Point 
 // Compute new point based on old point and shipHeading
 newPoint.x = (point.x * cos(radians)) - (point.y * sin(radians));
 newPoint.y = (point.x * sin(radians)) + (point.y * cos(radians));

 // Translate the point back 
 newPoint.x += pivotPoint.x;
 newPoint.y += pivotPoint.y;

 return newPoint;
}

Ship rotateShip(Ship* ship, int shipHeading)
{
 Ship newPosition = {0};

 // Rotate ship by shipHeading degrees, centered around top
 newPosition.top   = rotatePoint(ship->top, ship->top,   shipHeading);
 newPosition.left  = rotatePoint(ship->top, ship->left,  shipHeading);
 newPosition.right = rotatePoint(ship->top, ship->right, shipHeading);

 return newPosition;
}

Point newPoint(int x, int y)
{
 Point point = {0};
 point.x = x;
 point.y = y;
 return point;
}

Ship newShip(int shipX, int shipY)
{
 Ship newShip    = {0};
 newShip.top     = newPoint(shipX, shipY);
 newShip.left    = newPoint(shipX + 10, shipY - 5);
 newShip.right   = newPoint(shipX + 10, shipY + 5);
 return newShip;
}

// stubs
int getShipX(void) { return 10;}
int getShipY(void) { return 10;}
int getShipHeading(void) {return 90;} // Expressed in degrees

void drawTriangle(int x1, int y1, int x2, int y2, int x3, 
{
 // stub
 printf("Ship location: (%d,%d) (%d,%d) (%d,%d)\n", x1,y1, x2,y2, x3,y3);
}

void drawShip(void){
    Ship ship        = newShip(getShipX(), getShipY());
    int  shipHeading = getShipHeading();
    int  RED         = 0; // stub

    drawTriangle(ship.top.x,   ship.top.y,
                 ship.left.x,  ship.left.y,
                 ship.right.x, ship.right.y, RED); //pre rotation

    ship = rotateShip(&ship, shipHeading);

#if 0
    // Multiplying by -1 will Flip the ship? 
    // Use += instead to move the ship up by one 
    // Better to use a function like moveShip()
    ship.top.x   += -1;
    ship.top.y   += -1;
    ship.left.x  += -1;
    ship.left.y  += -1;
    ship.right.x += -1;
    ship.right.y += -1;
#endif

    drawTriangle(ship.top.x,   ship.top.y,
                 ship.left.x,  ship.left.y,
                 ship.right.x, ship.right.y, RED); //pre rotation
}

int main(int argc, char** argv)
{
 drawShip();
}

输出首先显示船的三个点（顶部）（左）（右），我使用X窗口符号（X，Y）默认为（10,10）（20,5）（20,15）屏幕的左上部分是（0,0）。第二行输出是围绕顶点（10,10）旋转90度后的点。您可以选择任何支点。有关Point Rotation的更多信息，请参阅this stackOverflow文章。

scott> gcc asteroids.c -lm
scott> a.out
Ship location: (10,10) (20,5) (20,15)
Ship location: (10,10) (14,20) (5,19)
scott>

希望这有助于作为一个起点。进行更改以适应您的框架。

围绕固定点旋转三角形

2 个答案: