我想要实现的是围绕矩形中心旋转一条线,以便它始终停留在接触它们的边界内(或具有一些填充)。
现在,我有以下例程,如您所见,我使用tan
计算将矩形分成8个部分(红线)
到目前为止,它仍然有效,但是由于某种原因,我在使用其他计算进行半径绘制(绿线)时存在不一致的地方,这些线并不总是如预期那样匹配,我想知道为什么。
仅使用sin / cos计算并找到线与矩形边界之间的交叉点,基本上可以实现相同的目的,但是由于某些原因,我无法使其正常工作。
std::pair<Point, Point>
MathUtils::calculateRotatingLine(Size size, double degrees)
{
auto width = size.width;
auto height = size.height;
double diagonalAngleTopRight = radiansToDegrees(atan((width / 2) / (height / 2)));
double diagonalAngleBottomRight = 90 + (90 - diagonalAngleTopRight);
double diagonalAngleBottomLeft = 180 + diagonalAngleTopRight;
double diagonalAngleTopLeft = 180 + diagonalAngleBottomRight;
double x, y;
/*
* *8*1*
* 7* *2
* 6* *3
* *5*4*
*/
// 1
if (degrees >= 0 && degrees <= diagonalAngleTopRight) {
x = width / 2 + height / 2 * tan(degreesToRadians(degrees));
y = 0;
}
// 2
else if (degrees > diagonalAngleTopRight && degrees <= 90) {
x = width;
y = width / 2 * tan(degreesToRadians(degrees - diagonalAngleTopRight));
}
// 3
else if (degrees > 90 && degrees <= diagonalAngleBottomRight) {
x = width;
y = height / 2 + width / 2 * tan(degreesToRadians(degrees - 90));
}
// 4
else if (degrees > diagonalAngleBottomRight && degrees <= 180) {
x = width - height / 2 * tan(degreesToRadians(degrees - diagonalAngleBottomRight));
y = height;
}
// 5
else if (degrees > 180 && degrees <= diagonalAngleBottomLeft) {
x = width / 2 - height / 2 * tan(degreesToRadians(degrees - 180));
y = height;
}
// 6
else if (degrees > diagonalAngleBottomLeft && degrees <= 270) {
x = 0;
y = height - width / 2 * tan(degreesToRadians(degrees - diagonalAngleBottomLeft));
}
// 7
else if (degrees > 270 && degrees <= diagonalAngleTopLeft) {
x = 0;
y = height / 2 - width / 2 * tan(degreesToRadians(degrees - 270));
}
// 8
else {
x = height / 2 * tan(degreesToRadians(degrees - diagonalAngleTopLeft));
y = 0;
}
return {Point{width / 2, height / 2}, Point{x, y}};
}
绿线计算
Point
MathUtils::calculateCirclePoint(double radius, double degrees)
{
return {radius * cos(degreesToRadians(degrees)), radius * sin(degreesToRadians(degrees))};
}
编辑
太棒了,感谢@MBo
Point
MathUtils::calculateCrossPoint(Size size, double degrees)
{
auto x0 = size.width / 2;
auto y0 = size.height / 2;
auto vx = cos(degreesToRadians(degrees - 90));
auto vy = sin(degreesToRadians(degrees - 90));
//potential border positions
auto ex = vx > 0 ? size.width : 0;
auto ey = vy > 0 ? size.height : 0;
//check for horizontal/vertical directions
if (vx == 0) {
return {x0, ey};
}
if (vy == 0) {
return {ex, y0};
}
// in general case find times of intersections with horizontal and vertical edge line
auto tx = (ex - x0) / vx;
auto ty = (ey - y0) / vy;
// and get intersection for smaller parameter value
if (tx <= ty) {
return {ex, y0 + tx * vy};
}
return {x0 + ty * vx, ey};
}
答案 0 :(得分:3)
伪代码以查找从矩形中心(弧度为an
的角)发射的光线与边缘的交点。 (也可用于其他(x0,y0)位置)
x0 = width / 2;
y0 = height / 2;
vx = cos(an);
vy = sin(an);
//potential border positions
ex = vx > 0? width: 0
ey = vy > 0? height: 0
//check for horizontal/vertical directions
if vx = 0 then
return cx = x0, cy = ey
if vy = 0 then
return cx = ex, cy = y0
//in general case find times of intersections with horizontal and vertical edge line
tx = (ex - x0) / vx
ty = (ey - y0) / vy
//and get intersection for smaller parameter value
if tx <= ty then
return cx = ex, cy = y0 + tx * vy
else
return cx = x0 + ty * vx, cy = ey