圆-矩形碰撞检测（交叉点）

半透明的墙 2024-07-17 15:11:25

这是我的做法：

bool intersects(CircleType circle, RectType rect)
{
    circleDistance.x = abs(circle.x - rect.x);
    circleDistance.y = abs(circle.y - rect.y);

    if (circleDistance.x > (rect.width/2 + circle.r)) { return false; }
    if (circleDistance.y > (rect.height/2 + circle.r)) { return false; }

    if (circleDistance.x <= (rect.width/2)) { return true; } 
    if (circleDistance.y <= (rect.height/2)) { return true; }

    cornerDistance_sq = (circleDistance.x - rect.width/2)^2 +
                         (circleDistance.y - rect.height/2)^2;

    return (cornerDistance_sq <= (circle.r^2));
}

这是它的工作原理：

illusration

第一对行计算绝对值圆心与矩形中心之间的 x 和 y 差值。这会将四个象限合并为一个，这样计算就不必进行四次。该图像显示了圆心现在必须位于的区域。请注意，仅显示了单个象限。矩形是灰色区域，红色边框勾勒出关键区域，该区域距矩形边缘正好一个半径。圆心必须位于该红色边框内才能发生相交。
第二对线消除了圆距矩形足够远（在任一方向）以致不可能相交的简单情况。这对应于图像中的绿色区域。
第三对线处理简单的情况，其中圆足够接近矩形（在任一方向）以保证相交。这对应于图像中的橙色和灰色部分。请注意，此步骤必须在步骤 2 之后完成，逻辑才有意义。
其余的行计算圆可能与矩形的角相交的困难情况。求解时，计算圆心到角点的距离，然后验证该距离不大于圆的半径。对于中心位于红色阴影区域内的所有圆，此计算返回 false；对于中心位于白色阴影区域内的所有圆，此计算返回 true。

Here is how I would do it:

bool intersects(CircleType circle, RectType rect)
{
    circleDistance.x = abs(circle.x - rect.x);
    circleDistance.y = abs(circle.y - rect.y);

    if (circleDistance.x > (rect.width/2 + circle.r)) { return false; }
    if (circleDistance.y > (rect.height/2 + circle.r)) { return false; }

    if (circleDistance.x <= (rect.width/2)) { return true; } 
    if (circleDistance.y <= (rect.height/2)) { return true; }

    cornerDistance_sq = (circleDistance.x - rect.width/2)^2 +
                         (circleDistance.y - rect.height/2)^2;

    return (cornerDistance_sq <= (circle.r^2));
}

Here's how it works:

illusration

The first pair of lines calculate the absolute values of the x and y difference between the center of the circle and the center of the rectangle. This collapses the four quadrants down into one, so that the calculations do not have to be done four times. The image shows the area in which the center of the circle must now lie. Note that only the single quadrant is shown. The rectangle is the grey area, and the red border outlines the critical area which is exactly one radius away from the edges of the rectangle. The center of the circle has to be within this red border for the intersection to occur.
The second pair of lines eliminate the easy cases where the circle is far enough away from the rectangle (in either direction) that no intersection is possible. This corresponds to the green area in the image.
The third pair of lines handle the easy cases where the circle is close enough to the rectangle (in either direction) that an intersection is guaranteed. This corresponds to the orange and grey sections in the image. Note that this step must be done after step 2 for the logic to make sense.
The remaining lines calculate the difficult case where the circle may intersect the corner of the rectangle. To solve, compute the distance from the center of the circle and the corner, and then verify that the distance is not more than the radius of the circle. This calculation returns false for all circles whose center is within the red shaded area and returns true for all circles whose center is within the white shaded area.

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另类 2024-07-17 15:11:25

圆与矩形相交只有两种情况：

要么圆心位于矩形内部，要么
矩形的一条边有一点在圆内。

请注意，这并不要求矩形与轴平行。

（查看此问题的一种方法：如果没有一条边在圆内有一个点（如果所有边都完全在圆“外部”），那么圆的唯一方法如果它完全位于多边形内部，仍然可以与多边形相交。）

有了这种洞察力，类似下面的东西就可以工作，其中圆的中心为 P ，半径为 R ，并且矩形具有按顺序排列的顶点 A、B、C、D（不是完整代码）：

def intersect(Circle(P, R), Rectangle(A, B, C, D)):
    S = Circle(P, R)
    return (pointInRectangle(P, Rectangle(A, B, C, D)) or
            intersectCircle(S, (A, B)) or
            intersectCircle(S, (B, C)) or
            intersectCircle(S, (C, D)) or
            intersectCircle(S, (D, A)))

如果当你正在编写任何几何图形时，你的库中可能已经有上述函数了。另外，pointInRectangle() 可以通过多种方式实现；任何一般的多边形中的点方法都可以工作，但是对于矩形，您可以检查这是否有效：

0 ≤ AP·AB ≤ AB·AB and 0 ≤ AP·AD ≤ AD·AD

并且 intersectCircle() 也很容易实现：一种方法是检查从 P 到直线的垂直线的底部是否足够近并且位于端点之间，并检查端点。

很酷的是，相同的想法不仅适用于矩形，而且适用于圆与任何简单的多边形 - 甚至不必是凸的！

There are only two cases when the circle intersects with the rectangle:

Either the circle's centre lies inside the rectangle, or
One of the edges of the rectangle has a point in the circle.

Note that this does not require the rectangle to be axis-parallel.

(One way to see this: if none of the edges has a point in the circle (if all the edges are completely "outside" the circle), then the only way the circle can still intersect the polygon is if it lies completely inside the polygon.)

With that insight, something like the following will work, where the circle has centre P and radius R, and the rectangle has vertices A, B, C, D in that order (not complete code):

def intersect(Circle(P, R), Rectangle(A, B, C, D)):
    S = Circle(P, R)
    return (pointInRectangle(P, Rectangle(A, B, C, D)) or
            intersectCircle(S, (A, B)) or
            intersectCircle(S, (B, C)) or
            intersectCircle(S, (C, D)) or
            intersectCircle(S, (D, A)))

If you're writing any geometry you probably have the above functions in your library already. Otherwise, pointInRectangle() can be implemented in several ways; any of the general point in polygon methods will work, but for a rectangle you can just check whether this works:

0 ≤ AP·AB ≤ AB·AB and 0 ≤ AP·AD ≤ AD·AD

And intersectCircle() is easy to implement too: one way would be to check if the foot of the perpendicular from P to the line is close enough and between the endpoints, and check the endpoints otherwise.

The cool thing is that the same idea works not just for rectangles but for the intersection of a circle with any simple polygon — doesn't even have to be convex!

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浅浅淡淡 2024-07-17 15:11:25

这是另一个实现起来非常简单（而且速度也非常快）的解决方案。它将捕获所有交叉点，包括球体完全进入矩形时的交叉点。

// clamp(value, min, max) - limits value to the range min..max

// Find the closest point to the circle within the rectangle
float closestX = clamp(circle.X, rectangle.Left, rectangle.Right);
float closestY = clamp(circle.Y, rectangle.Top, rectangle.Bottom);

// Calculate the distance between the circle's center and this closest point
float distanceX = circle.X - closestX;
float distanceY = circle.Y - closestY;

// If the distance is less than the circle's radius, an intersection occurs
float distanceSquared = (distanceX * distanceX) + (distanceY * distanceY);
return distanceSquared < (circle.Radius * circle.Radius);

对于任何像样的数学库，都可以缩短到 3 或 4 行。

Here is another solution that's pretty simple to implement (and pretty fast, too). It will catch all intersections, including when the sphere has fully entered the rectangle.

// clamp(value, min, max) - limits value to the range min..max

// Find the closest point to the circle within the rectangle
float closestX = clamp(circle.X, rectangle.Left, rectangle.Right);
float closestY = clamp(circle.Y, rectangle.Top, rectangle.Bottom);

// Calculate the distance between the circle's center and this closest point
float distanceX = circle.X - closestX;
float distanceY = circle.Y - closestY;

// If the distance is less than the circle's radius, an intersection occurs
float distanceSquared = (distanceX * distanceX) + (distanceY * distanceY);
return distanceSquared < (circle.Radius * circle.Radius);

With any decent math library, that can be shortened to 3 or 4 lines.

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世态炎凉 2024-07-17 15:11:25

我想出的最简单的解决方案非常简单。

它的工作原理是找到矩形中最接近圆的点，然后比较距离。

您可以通过一些操作来完成所有这些操作，甚至可以避免使用 sqrt 函数。

public boolean intersects(float cx, float cy, float radius, float left, float top, float right, float bottom)
{
   float closestX = (cx < left ? left : (cx > right ? right : cx));
   float closestY = (cy < top ? top : (cy > bottom ? bottom : cy));
   float dx = closestX - cx;
   float dy = closestY - cy;

   return ( dx * dx + dy * dy ) <= radius * radius;
}

就是这样！上述解决方案假设原点位于世界的左上角，x 轴向下。

如果您想要一个处理移动圆和矩形之间碰撞的解决方案，那么它要复杂得多并且涵盖在我的另一个答案中。

The simplest solution I've come up with is pretty straightforward.

It works by finding the point in the rectangle closest to the circle, then comparing the distance.

You can do all of this with a few operations, and even avoid the sqrt function.

public boolean intersects(float cx, float cy, float radius, float left, float top, float right, float bottom)
{
   float closestX = (cx < left ? left : (cx > right ? right : cx));
   float closestY = (cy < top ? top : (cy > bottom ? bottom : cy));
   float dx = closestX - cx;
   float dy = closestY - cy;

   return ( dx * dx + dy * dy ) <= radius * radius;
}

And that's it! The above solution assumes an origin in the upper left of the world with the x-axis pointing down.

If you want a solution to handling collisions between a moving circle and rectangle, it's far more complicated and covered in another answer of mine.

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嘿哥们儿 2024-07-17 15:11:25

你的球体和矩形相交 IIF
圆心和矩形的一个顶点之间的距离小于球体的半径
或
圆心和矩形的一条边之间的距离小于球体的半径（[点线距离])
或
圆心在矩形内

点到点距离：

P1 = [x1,y1]
P2 = [x2,y2]
Distance = sqrt(abs(x1 - x2)+abs(y1-y2))

点到线距离：

L1 = [x1,y1],L2 = [x2,y2] (two points of your line, ie the vertex points)
P1 = [px,py] some point

Distance d =  abs( (x2-x1)(y1-py)-(x1-px)(y2-y1) ) / Distance(L1,L2)

矩形内圆心：
采用分离轴方法：如果存在将矩形与点分开的线上的投影，则它们不会相交，

您将点投影到与矩形的边平行的线上，然后可以轻松确定它们是否相交。如果它们不在所有 4 个投影上相交，则它们（点和矩形）不能相交。

你只需要内积（ x= [x1,x2] ， y = [y1,y2] ， x*y = x1*y1 + x2*y2 ）

你的测试看起来像这样：

//rectangle edges: TL (top left), TR (top right), BL (bottom left), BR (bottom right)
//point to test: POI

seperated = false
for egde in { {TL,TR}, {BL,BR}, {TL,BL},{TR-BR} }:  // the edges
    D = edge[0] - edge[1]
    innerProd =  D * POI
    Interval_min = min(D*edge[0],D*edge[1])
    Interval_max = max(D*edge[0],D*edge[1])
    if not (  Interval_min ≤ innerProd ≤  Interval_max ) 
           seperated = true
           break  // end for loop 
    end if
end for
if (seperated is true)    
      return "no intersection"
else 
      return "intersection"
end if

这不假设轴 -对齐的矩形，并且可以轻松扩展以测试凸集之间的交集。

your sphere and rect intersect IIF
the distance between the circle-center and one vertex of your rect is smaller than the radius of your sphere
OR
the distance between the circle-center and one edge of your rect is smaller than the radius of your sphere ([point-line distance ])
OR
the circle center is inside the rect

point-point distance:

P1 = [x1,y1]
P2 = [x2,y2]
Distance = sqrt(abs(x1 - x2)+abs(y1-y2))

point-line distance:

L1 = [x1,y1],L2 = [x2,y2] (two points of your line, ie the vertex points)
P1 = [px,py] some point

Distance d =  abs( (x2-x1)(y1-py)-(x1-px)(y2-y1) ) / Distance(L1,L2)

circle center inside rect:
take an seperating axis aproach: if there exists a projection onto a line that seperates the rectangle from the point, they do not intersect

you project the point on lines parallel to the sides of your rect and can then easily determine if they intersect. if they intersect not on all 4 projections, they (the point and the rectangle) can not intersect.

you just need the inner-product ( x= [x1,x2] , y = [y1,y2] , x*y = x1*y1 + x2*y2 )

your test would look like that:

//rectangle edges: TL (top left), TR (top right), BL (bottom left), BR (bottom right)
//point to test: POI

seperated = false
for egde in { {TL,TR}, {BL,BR}, {TL,BL},{TR-BR} }:  // the edges
    D = edge[0] - edge[1]
    innerProd =  D * POI
    Interval_min = min(D*edge[0],D*edge[1])
    Interval_max = max(D*edge[0],D*edge[1])
    if not (  Interval_min ≤ innerProd ≤  Interval_max ) 
           seperated = true
           break  // end for loop 
    end if
end for
if (seperated is true)    
      return "no intersection"
else 
      return "intersection"
end if

this does not assume an axis-aligned rectangle and is easily extendable for testing intersections between convex sets.

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漫漫岁月 2024-07-17 15:11:25

这是最快的解决方案：

public static boolean intersect(Rectangle r, Circle c)
{
    float cx = Math.abs(c.x - r.x - r.halfWidth);
    float xDist = r.halfWidth + c.radius;
    if (cx > xDist)
        return false;
    float cy = Math.abs(c.y - r.y - r.halfHeight);
    float yDist = r.halfHeight + c.radius;
    if (cy > yDist)
        return false;
    if (cx <= r.halfWidth || cy <= r.halfHeight)
        return true;
    float xCornerDist = cx - r.halfWidth;
    float yCornerDist = cy - r.halfHeight;
    float xCornerDistSq = xCornerDist * xCornerDist;
    float yCornerDistSq = yCornerDist * yCornerDist;
    float maxCornerDistSq = c.radius * c.radius;
    return xCornerDistSq + yCornerDistSq <= maxCornerDistSq;
}

注意执行顺序，宽度/高度的一半是预先计算的。此外，平方是“手动”完成的，以节省一些时钟周期。

This is the fastest solution:

public static boolean intersect(Rectangle r, Circle c)
{
    float cx = Math.abs(c.x - r.x - r.halfWidth);
    float xDist = r.halfWidth + c.radius;
    if (cx > xDist)
        return false;
    float cy = Math.abs(c.y - r.y - r.halfHeight);
    float yDist = r.halfHeight + c.radius;
    if (cy > yDist)
        return false;
    if (cx <= r.halfWidth || cy <= r.halfHeight)
        return true;
    float xCornerDist = cx - r.halfWidth;
    float yCornerDist = cy - r.halfHeight;
    float xCornerDistSq = xCornerDist * xCornerDist;
    float yCornerDistSq = yCornerDist * yCornerDist;
    float maxCornerDistSq = c.radius * c.radius;
    return xCornerDistSq + yCornerDistSq <= maxCornerDistSq;
}

Note the order of execution, and half the width/height is pre-computed. Also the squaring is done "manually" to save some clock cycles.

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笑红尘 2024-07-17 15:11:25

事实上，这要简单得多。你只需要两件事。

首先，您需要找到从圆心到矩形每条线的四个正交距离。那么，如果其中任何三个大于圆半径，则圆将不会与矩形相交。

其次，你需要找到圆心和矩形中心之间的距离，那么如果距离大于矩形对角线长度的一半，你的圆就不会在矩形内部。

祝你好运！

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甜嗑 2024-07-17 15:11:25

如果您对甚至可以在（平面内）旋转矩形上工作的更加图形化的解决方案感兴趣。

演示： https://jsfiddle.net/exodus4d/94mxLvqh/2691/

这个想法是：

翻译场景到原点[0,0]
- 如果矩形不在平面内，则旋转中心应位于
  [0, 0]
旋转场景回到平面
计算交集

const hasIntersection = ({x: cx, y: cy, r: cr}, {x, y, width, height}) => {
  const distX = Math.abs(cx - x - width / 2);
  const distY = Math.abs(cy - y - height / 2);

  if (distX > (width / 2 + cr)) {
    return false;
  }
  if (distY > (height / 2 + cr)) {
    return false;
  }

  if (distX <= (width / 2)) {
    return true;
  }
  if (distY <= (height / 2)) {
    return true;
  }

  const Δx = distX - width / 2;
  const Δy = distY - height / 2;
  return Δx * Δx + Δy * Δy <= cr * cr;
};

const rect = new DOMRect(50, 20, 100, 50);
const circ1 = new DOMPoint(160, 80);
circ1.r = 20;

const circ2 = new DOMPoint(80, 95);
circ2.r = 20;

const canvas = document.getElementById('canvas');
const ctx = canvas.getContext('2d');

ctx.strokeRect(rect.x, rect.y, rect.width, rect.height);

ctx.beginPath();
ctx.strokeStyle = hasIntersection(circ1, rect) ? 'red' : 'green';
ctx.arc(circ1.x, circ1.y, circ1.r, 0, 2 * Math.PI);
ctx.stroke();

ctx.beginPath();
ctx.strokeStyle = hasIntersection(circ2, rect) ? 'red' : 'green';
ctx.arc(circ2.x, circ2.y, circ2.r, 0, 2 * Math.PI);
ctx.stroke();

<canvas id="canvas"></canvas>

提示：而不是旋转矩形（4 点）。您可以沿相反方向旋转圆圈（1 点）。

If you are interested in a more graphical solution which even works on (in plane) rotated rectangles..

Demo: https://jsfiddle.net/exodus4d/94mxLvqh/2691/

The idea is:

Translate the scenary to the origin [0,0]
- In case the rect is not in plane, the rotation center should be at
  [0, 0]
Rotate the scenary back into plane
Calculate intersection

const hasIntersection = ({x: cx, y: cy, r: cr}, {x, y, width, height}) => {
  const distX = Math.abs(cx - x - width / 2);
  const distY = Math.abs(cy - y - height / 2);

  if (distX > (width / 2 + cr)) {
    return false;
  }
  if (distY > (height / 2 + cr)) {
    return false;
  }

  if (distX <= (width / 2)) {
    return true;
  }
  if (distY <= (height / 2)) {
    return true;
  }

  const Δx = distX - width / 2;
  const Δy = distY - height / 2;
  return Δx * Δx + Δy * Δy <= cr * cr;
};

const rect = new DOMRect(50, 20, 100, 50);
const circ1 = new DOMPoint(160, 80);
circ1.r = 20;

const circ2 = new DOMPoint(80, 95);
circ2.r = 20;

const canvas = document.getElementById('canvas');
const ctx = canvas.getContext('2d');

ctx.strokeRect(rect.x, rect.y, rect.width, rect.height);

ctx.beginPath();
ctx.strokeStyle = hasIntersection(circ1, rect) ? 'red' : 'green';
ctx.arc(circ1.x, circ1.y, circ1.r, 0, 2 * Math.PI);
ctx.stroke();

ctx.beginPath();
ctx.strokeStyle = hasIntersection(circ2, rect) ? 'red' : 'green';
ctx.arc(circ2.x, circ2.y, circ2.r, 0, 2 * Math.PI);
ctx.stroke();

<canvas id="canvas"></canvas>

Tip: Instead of rotating the rect (4 points). You can rotate the circle (1 point) in opposite direction.

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半岛未凉 2024-07-17 15:11:25

这是我的 C 代码，用于解决球体和非轴对齐框之间的碰撞。它依赖于我自己的几个库例程，但它可能对某些人有用。我在游戏中使用它并且效果非常好。

float physicsProcessCollisionBetweenSelfAndActorRect(SPhysics *self, SPhysics *actor)
{
    float diff = 99999;

    SVector relative_position_of_circle = getDifference2DBetweenVectors(&self->worldPosition, &actor->worldPosition);
    rotateVector2DBy(&relative_position_of_circle, -actor->axis.angleZ); // This aligns the coord system so the rect becomes an AABB

    float x_clamped_within_rectangle = relative_position_of_circle.x;
    float y_clamped_within_rectangle = relative_position_of_circle.y;
    LIMIT(x_clamped_within_rectangle, actor->physicsRect.l, actor->physicsRect.r);
    LIMIT(y_clamped_within_rectangle, actor->physicsRect.b, actor->physicsRect.t);

    // Calculate the distance between the circle's center and this closest point
    float distance_to_nearest_edge_x = relative_position_of_circle.x - x_clamped_within_rectangle;
    float distance_to_nearest_edge_y = relative_position_of_circle.y - y_clamped_within_rectangle;

    // If the distance is less than the circle's radius, an intersection occurs
    float distance_sq_x = SQUARE(distance_to_nearest_edge_x);
    float distance_sq_y = SQUARE(distance_to_nearest_edge_y);
    float radius_sq = SQUARE(self->physicsRadius);
    if(distance_sq_x + distance_sq_y < radius_sq)   
    {
        float half_rect_w = (actor->physicsRect.r - actor->physicsRect.l) * 0.5f;
        float half_rect_h = (actor->physicsRect.t - actor->physicsRect.b) * 0.5f;

        CREATE_VECTOR(push_vector);         

        // If we're at one of the corners of this object, treat this as a circular/circular collision
        if(fabs(relative_position_of_circle.x) > half_rect_w && fabs(relative_position_of_circle.y) > half_rect_h)
        {
            SVector edges;
            if(relative_position_of_circle.x > 0) edges.x = half_rect_w; else edges.x = -half_rect_w;
            if(relative_position_of_circle.y > 0) edges.y = half_rect_h; else edges.y = -half_rect_h;   

            push_vector = relative_position_of_circle;
            moveVectorByInverseVector2D(&push_vector, &edges);

            // We now have the vector from the corner of the rect to the point.
            float delta_length = getVector2DMagnitude(&push_vector);
            float diff = self->physicsRadius - delta_length; // Find out how far away we are from our ideal distance

            // Normalise the vector
            push_vector.x /= delta_length;
            push_vector.y /= delta_length;
            scaleVector2DBy(&push_vector, diff); // Now multiply it by the difference
            push_vector.z = 0;
        }
        else // Nope - just bouncing against one of the edges
        {
            if(relative_position_of_circle.x > 0) // Ball is to the right
                push_vector.x = (half_rect_w + self->physicsRadius) - relative_position_of_circle.x;
            else
                push_vector.x = -((half_rect_w + self->physicsRadius) + relative_position_of_circle.x);

            if(relative_position_of_circle.y > 0) // Ball is above
                push_vector.y = (half_rect_h + self->physicsRadius) - relative_position_of_circle.y;
            else
                push_vector.y = -((half_rect_h + self->physicsRadius) + relative_position_of_circle.y);

            if(fabs(push_vector.x) < fabs(push_vector.y))
                push_vector.y = 0;
            else
                push_vector.x = 0;
        }

        diff = 0; // Cheat, since we don't do anything with the value anyway
        rotateVector2DBy(&push_vector, actor->axis.angleZ);
        SVector *from = &self->worldPosition;       
        moveVectorBy2D(from, push_vector.x, push_vector.y);
    }   
    return diff;
}

Here's my C code for resolving a collision between a sphere and a non-axis aligned box. It relies on a couple of my own library routines, but it may prove useful to some. I'm using it in a game and it works perfectly.

float physicsProcessCollisionBetweenSelfAndActorRect(SPhysics *self, SPhysics *actor)
{
    float diff = 99999;

    SVector relative_position_of_circle = getDifference2DBetweenVectors(&self->worldPosition, &actor->worldPosition);
    rotateVector2DBy(&relative_position_of_circle, -actor->axis.angleZ); // This aligns the coord system so the rect becomes an AABB

    float x_clamped_within_rectangle = relative_position_of_circle.x;
    float y_clamped_within_rectangle = relative_position_of_circle.y;
    LIMIT(x_clamped_within_rectangle, actor->physicsRect.l, actor->physicsRect.r);
    LIMIT(y_clamped_within_rectangle, actor->physicsRect.b, actor->physicsRect.t);

    // Calculate the distance between the circle's center and this closest point
    float distance_to_nearest_edge_x = relative_position_of_circle.x - x_clamped_within_rectangle;
    float distance_to_nearest_edge_y = relative_position_of_circle.y - y_clamped_within_rectangle;

    // If the distance is less than the circle's radius, an intersection occurs
    float distance_sq_x = SQUARE(distance_to_nearest_edge_x);
    float distance_sq_y = SQUARE(distance_to_nearest_edge_y);
    float radius_sq = SQUARE(self->physicsRadius);
    if(distance_sq_x + distance_sq_y < radius_sq)   
    {
        float half_rect_w = (actor->physicsRect.r - actor->physicsRect.l) * 0.5f;
        float half_rect_h = (actor->physicsRect.t - actor->physicsRect.b) * 0.5f;

        CREATE_VECTOR(push_vector);         

        // If we're at one of the corners of this object, treat this as a circular/circular collision
        if(fabs(relative_position_of_circle.x) > half_rect_w && fabs(relative_position_of_circle.y) > half_rect_h)
        {
            SVector edges;
            if(relative_position_of_circle.x > 0) edges.x = half_rect_w; else edges.x = -half_rect_w;
            if(relative_position_of_circle.y > 0) edges.y = half_rect_h; else edges.y = -half_rect_h;   

            push_vector = relative_position_of_circle;
            moveVectorByInverseVector2D(&push_vector, &edges);

            // We now have the vector from the corner of the rect to the point.
            float delta_length = getVector2DMagnitude(&push_vector);
            float diff = self->physicsRadius - delta_length; // Find out how far away we are from our ideal distance

            // Normalise the vector
            push_vector.x /= delta_length;
            push_vector.y /= delta_length;
            scaleVector2DBy(&push_vector, diff); // Now multiply it by the difference
            push_vector.z = 0;
        }
        else // Nope - just bouncing against one of the edges
        {
            if(relative_position_of_circle.x > 0) // Ball is to the right
                push_vector.x = (half_rect_w + self->physicsRadius) - relative_position_of_circle.x;
            else
                push_vector.x = -((half_rect_w + self->physicsRadius) + relative_position_of_circle.x);

            if(relative_position_of_circle.y > 0) // Ball is above
                push_vector.y = (half_rect_h + self->physicsRadius) - relative_position_of_circle.y;
            else
                push_vector.y = -((half_rect_h + self->physicsRadius) + relative_position_of_circle.y);

            if(fabs(push_vector.x) < fabs(push_vector.y))
                push_vector.y = 0;
            else
                push_vector.x = 0;
        }

        diff = 0; // Cheat, since we don't do anything with the value anyway
        rotateVector2DBy(&push_vector, actor->axis.angleZ);
        SVector *from = &self->worldPosition;       
        moveVectorBy2D(from, push_vector.x, push_vector.y);
    }   
    return diff;
}

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二智少女猫性小仙女 2024-07-17 15:11:25

要进行可视化，请使用键盘的小键盘。如果键“5”代表矩形，则所有键 1-9 代表空间的 9 个象限，除以构成矩形的线（其中 5 为内部）。

1) 如果圆的中心位于象限 5 （即在矩形内部）然后两个形状相交。

排除了这一点，有两种可能的情况：
a) 圆与矩形的两个或多个相邻边相交。
b) 圆与矩形的一条边相交。

第一种情况很简单。如果圆与矩形的两个相邻边相交，则它必须包含连接这两个边的角。（它的中心位于象限 5，我们已经介绍过。另请注意，圆仅与矩形的两个相对边相交的情况也已介绍。）

2) 如果矩形的任意一个角 A、B、C、D 位于圆内，则两个形状相交。

第二种情况比较棘手。需要注意的是，只有当圆心位于 2、4、6 或 8 象限之一时，才会发生这种情况。（事实上，如果圆心位于 1、3、7、8 象限中的任何一个，则相应的角将是距离它最近的点。）

现在我们遇到的情况是圆的中心位于“边缘”象限之一，并且它仅与相应的边缘相交。然后，最接近圆心的边缘上的点必须位于圆内部。

3) 对于每条线 AB、BC、CD、DA，构造通过圆心 P 的垂线 p(AB,P)、p(BC,P)、p(CD,P)、p(DA,P)。每条垂线，如果与原边的交点位于圆内，则两个形状相交。

最后一步有一个捷径。如果圆的中心位于象限 8 且边 AB 为顶边，则交点将具有 A 和 B 的 y 坐标以及圆心 P 的 x 坐标。

您可以构造四个线交点并检查如果它们位于相应的边上，或者找出 P 位于哪个象限并检查相应的交点。两者都应简化为相同的布尔方程。请注意，上述步骤 2 并不排除 P 位于“角”象限之一；它只是寻找一个交叉点。

编辑：事实证明，我忽略了一个简单的事实，即#2 是上面#3 的子情况。毕竟，角也是边缘上的点。请参阅下面@ShreevatsaR 的回答以获得很好的解释。同时，请忘记上面的#2，除非您想要快速但多余的检查。

To visualise, take your keyboard's numpad. If the key '5' represents your rectangle, then all the keys 1-9 represent the 9 quadrants of space divided by the lines that make up your rectangle (with 5 being the inside.)

1) If the circle's center is in quadrant 5 (i.e. inside the rectangle) then the two shapes intersect.

With that out of the way, there are two possible cases:
a) The circle intersects with two or more neighboring edges of the rectangle.
b) The circle intersects with one edge of the rectangle.

The first case is simple. If the circle intersects with two neighboring edges of the rectangle, it must contain the corner connecting those two edges. (That, or its center lies in quadrant 5, which we have already covered. Also note that the case where the circle intersects with only two opposing edges of the rectangle is covered as well.)

2) If any of the corners A, B, C, D of the rectangle lie inside the circle, then the two shapes intersect.

The second case is trickier. We should make note of that it may only happen when the circle's center lies in one of the quadrants 2, 4, 6 or 8. (In fact, if the center is on any of the quadrants 1, 3, 7, 8, the corresponding corner will be the closest point to it.)

Now we have the case that the circle's center is in one of the 'edge' quadrants, and it only intersects with the corresponding edge. Then, the point on the edge that is closest to the circle's center, must lie inside the circle.

3) For each line AB, BC, CD, DA, construct perpendicular lines p(AB,P), p(BC,P), p(CD,P), p(DA,P) through the circle's center P. For each perpendicular line, if the intersection with the original edge lies inside the circle, then the two shapes intersect.

There is a shortcut for this last step. If the circle's center is in quadrant 8 and the edge AB is the top edge, the point of intersection will have the y-coordinate of A and B, and the x-coordinate of center P.

You can construct the four line intersections and check if they lie on their corresponding edges, or find out which quadrant P is in and check the corresponding intersection. Both should simplify to the same boolean equation. Be wary of that the step 2 above did not rule out P being in one of the 'corner' quadrants; it just looked for an intersection.

Edit: As it turns out, I have overlooked the simple fact that #2 is a subcase of #3 above. After all, corners too are points on the edges. See @ShreevatsaR's answer below for a great explanation. And in the meanwhile, forget #2 above unless you want a quick but redundant check.

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帅气尐潴 2024-07-17 15:11:25

此函数检测圆形和矩形之间的碰撞（相交）。他在回答中的工作方式类似于 e.James 方法，但是这个方法检测矩形所有角度（不仅仅是右上角）的碰撞。

注意：

aRect.origin.x 和 aRect.origin.y 是矩形左下角的坐标！

aCircle.x 和 aCircle.y 是圆心的坐标！

static inline BOOL RectIntersectsCircle(CGRect aRect, Circle aCircle) {

    float testX = aCircle.x;
    float testY = aCircle.y;

    if (testX < aRect.origin.x)
        testX = aRect.origin.x;
    if (testX > (aRect.origin.x + aRect.size.width))
        testX = (aRect.origin.x + aRect.size.width);
    if (testY < aRect.origin.y)
        testY = aRect.origin.y;
    if (testY > (aRect.origin.y + aRect.size.height))
        testY = (aRect.origin.y + aRect.size.height);

    return ((aCircle.x - testX) * (aCircle.x - testX) + (aCircle.y - testY) * (aCircle.y - testY)) < aCircle.radius * aCircle.radius;
}

This function detect collisions (intersections) between Circle and Rectangle. He works like e.James method in his answer, but this one detect collisions for all angles of rectangle (not only right up corner).

NOTE:

aRect.origin.x and aRect.origin.y are coordinates of bottom left angle of rectangle!

aCircle.x and aCircle.y are coordinates of Circle Center!

static inline BOOL RectIntersectsCircle(CGRect aRect, Circle aCircle) {

    float testX = aCircle.x;
    float testY = aCircle.y;

    if (testX < aRect.origin.x)
        testX = aRect.origin.x;
    if (testX > (aRect.origin.x + aRect.size.width))
        testX = (aRect.origin.x + aRect.size.width);
    if (testY < aRect.origin.y)
        testY = aRect.origin.y;
    if (testY > (aRect.origin.y + aRect.size.height))
        testY = (aRect.origin.y + aRect.size.height);

    return ((aCircle.x - testX) * (aCircle.x - testX) + (aCircle.y - testY) * (aCircle.y - testY)) < aCircle.radius * aCircle.radius;
}

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小红帽 2024-07-17 15:11:25

稍微改进一下e.James的答案：

double dx = abs(circle.x - rect.x) - rect.w / 2,
       dy = abs(circle.y - rect.y) - rect.h / 2;

if (dx > circle.r || dy > circle.r) { return false; }
if (dx <= 0 || dy <= 0) { return true; }

return (dx * dx + dy * dy <= circle.r * circle.r);

这减去了rect.w / 2并且rect.h / 2 一次而不是最多三次。

Improving a little bit the answer of e.James:

double dx = abs(circle.x - rect.x) - rect.w / 2,
       dy = abs(circle.y - rect.y) - rect.h / 2;

if (dx > circle.r || dy > circle.r) { return false; }
if (dx <= 0 || dy <= 0) { return true; }

return (dx * dx + dy * dy <= circle.r * circle.r);

This subtracts rect.w / 2 and rect.h / 2 once instead of up to three times.

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卸妝后依然美 2024-07-17 15:11:25

我有一种方法，如果没有必要，可以避免昂贵的毕达哥拉斯 - 即。当矩形和圆形的边界框不相交时。

它也适用于非欧几里德：

class Circle {
 // create the bounding box of the circle only once
 BBox bbox;

 public boolean intersect(BBox b) {
    // test top intersect
    if (lat > b.maxLat) {
        if (lon < b.minLon)
            return normDist(b.maxLat, b.minLon) <= normedDist;
        if (lon > b.maxLon)
            return normDist(b.maxLat, b.maxLon) <= normedDist;
        return b.maxLat - bbox.minLat > 0;
    }

    // test bottom intersect
    if (lat < b.minLat) {
        if (lon < b.minLon)
            return normDist(b.minLat, b.minLon) <= normedDist;
        if (lon > b.maxLon)
            return normDist(b.minLat, b.maxLon) <= normedDist;
        return bbox.maxLat - b.minLat > 0;
    }

    // test middle intersect
    if (lon < b.minLon)
        return bbox.maxLon - b.minLon > 0;
    if (lon > b.maxLon)
        return b.maxLon - bbox.minLon > 0;
    return true;
  }
}

minLat、maxLat 可以替换为 minY、maxY，minLon、maxLon 也相同：替换为 minX、maxXnormDist
只是比全距离计算更快一点的方法。例如，在欧几里得空间中没有平方根（或者没有很多半正弦的其他东西）： dLat=(lat-circleY); dLon=(lon-圆X); 范数=dLat*dLat+dLon*dLon。当然，如果您使用该normDist方法，您需要为圆圈创建一个 normedDist = dist*dist;

查看完整的BBox 和 Circle 我的 GraphHopper 项目。

I've a method which avoids the expensive pythagoras if not necessary - ie. when bounding boxes of the rectangle and the circle do not intersect.

And it'll work for non-euclidean too:

class Circle {
 // create the bounding box of the circle only once
 BBox bbox;

 public boolean intersect(BBox b) {
    // test top intersect
    if (lat > b.maxLat) {
        if (lon < b.minLon)
            return normDist(b.maxLat, b.minLon) <= normedDist;
        if (lon > b.maxLon)
            return normDist(b.maxLat, b.maxLon) <= normedDist;
        return b.maxLat - bbox.minLat > 0;
    }

    // test bottom intersect
    if (lat < b.minLat) {
        if (lon < b.minLon)
            return normDist(b.minLat, b.minLon) <= normedDist;
        if (lon > b.maxLon)
            return normDist(b.minLat, b.maxLon) <= normedDist;
        return bbox.maxLat - b.minLat > 0;
    }

    // test middle intersect
    if (lon < b.minLon)
        return bbox.maxLon - b.minLon > 0;
    if (lon > b.maxLon)
        return b.maxLon - bbox.minLon > 0;
    return true;
  }
}

minLat,maxLat can be replaced with minY,maxY and the same for minLon, maxLon: replace it with minX, maxX
normDist ist just a bit faster method then the full distance calculation. E.g. without the square-root in euclidean space (or without a lot of other stuff for haversine): dLat=(lat-circleY); dLon=(lon-circleX); normed=dLat*dLat+dLon*dLon. Of course if you use that normDist method you'll need to do create a normedDist = dist*dist; for the circle

See the full BBox and Circle code of my GraphHopper project.

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最美的太阳 2024-07-17 15:11:25

我创建了处理形状的类
希望你喜欢

public class Geomethry {
  public static boolean intersectionCircleAndRectangle(int circleX, int circleY, int circleR, int rectangleX, int rectangleY, int rectangleWidth, int rectangleHeight){
    boolean result = false;

    float rectHalfWidth = rectangleWidth/2.0f;
    float rectHalfHeight = rectangleHeight/2.0f;

    float rectCenterX = rectangleX + rectHalfWidth;
    float rectCenterY = rectangleY + rectHalfHeight;

    float deltax = Math.abs(rectCenterX - circleX);
    float deltay = Math.abs(rectCenterY - circleY);

    float lengthHypotenuseSqure = deltax*deltax + deltay*deltay;

    do{
        // check that distance between the centerse is more than the distance between the circumcircle of rectangle and circle
        if(lengthHypotenuseSqure > ((rectHalfWidth+circleR)*(rectHalfWidth+circleR) + (rectHalfHeight+circleR)*(rectHalfHeight+circleR))){
            //System.out.println("distance between the centerse is more than the distance between the circumcircle of rectangle and circle");
            break;
        }

        // check that distance between the centerse is less than the distance between the inscribed circle
        float rectMinHalfSide = Math.min(rectHalfWidth, rectHalfHeight);
        if(lengthHypotenuseSqure < ((rectMinHalfSide+circleR)*(rectMinHalfSide+circleR))){
            //System.out.println("distance between the centerse is less than the distance between the inscribed circle");
            result=true;
            break;
        }

        // check that the squares relate to angles
        if((deltax > (rectHalfWidth+circleR)*0.9) && (deltay > (rectHalfHeight+circleR)*0.9)){
            //System.out.println("squares relate to angles");
            result=true;
        }
    }while(false);

    return result;
}

public static boolean intersectionRectangleAndRectangle(int rectangleX, int rectangleY, int rectangleWidth, int rectangleHeight, int rectangleX2, int rectangleY2, int rectangleWidth2, int rectangleHeight2){
    boolean result = false;

    float rectHalfWidth = rectangleWidth/2.0f;
    float rectHalfHeight = rectangleHeight/2.0f;
    float rectHalfWidth2 = rectangleWidth2/2.0f;
    float rectHalfHeight2 = rectangleHeight2/2.0f;

    float deltax = Math.abs((rectangleX + rectHalfWidth) - (rectangleX2 + rectHalfWidth2));
    float deltay = Math.abs((rectangleY + rectHalfHeight) - (rectangleY2 + rectHalfHeight2));

    float lengthHypotenuseSqure = deltax*deltax + deltay*deltay;

    do{
        // check that distance between the centerse is more than the distance between the circumcircle
        if(lengthHypotenuseSqure > ((rectHalfWidth+rectHalfWidth2)*(rectHalfWidth+rectHalfWidth2) + (rectHalfHeight+rectHalfHeight2)*(rectHalfHeight+rectHalfHeight2))){
            //System.out.println("distance between the centerse is more than the distance between the circumcircle");
            break;
        }

        // check that distance between the centerse is less than the distance between the inscribed circle
        float rectMinHalfSide = Math.min(rectHalfWidth, rectHalfHeight);
        float rectMinHalfSide2 = Math.min(rectHalfWidth2, rectHalfHeight2);
        if(lengthHypotenuseSqure < ((rectMinHalfSide+rectMinHalfSide2)*(rectMinHalfSide+rectMinHalfSide2))){
            //System.out.println("distance between the centerse is less than the distance between the inscribed circle");
            result=true;
            break;
        }

        // check that the squares relate to angles
        if((deltax > (rectHalfWidth+rectHalfWidth2)*0.9) && (deltay > (rectHalfHeight+rectHalfHeight2)*0.9)){
            //System.out.println("squares relate to angles");
            result=true;
        }
    }while(false);

    return result;
  } 
}

I created class for work with shapes
hope you enjoy

public class Geomethry {
  public static boolean intersectionCircleAndRectangle(int circleX, int circleY, int circleR, int rectangleX, int rectangleY, int rectangleWidth, int rectangleHeight){
    boolean result = false;

    float rectHalfWidth = rectangleWidth/2.0f;
    float rectHalfHeight = rectangleHeight/2.0f;

    float rectCenterX = rectangleX + rectHalfWidth;
    float rectCenterY = rectangleY + rectHalfHeight;

    float deltax = Math.abs(rectCenterX - circleX);
    float deltay = Math.abs(rectCenterY - circleY);

    float lengthHypotenuseSqure = deltax*deltax + deltay*deltay;

    do{
        // check that distance between the centerse is more than the distance between the circumcircle of rectangle and circle
        if(lengthHypotenuseSqure > ((rectHalfWidth+circleR)*(rectHalfWidth+circleR) + (rectHalfHeight+circleR)*(rectHalfHeight+circleR))){
            //System.out.println("distance between the centerse is more than the distance between the circumcircle of rectangle and circle");
            break;
        }

        // check that distance between the centerse is less than the distance between the inscribed circle
        float rectMinHalfSide = Math.min(rectHalfWidth, rectHalfHeight);
        if(lengthHypotenuseSqure < ((rectMinHalfSide+circleR)*(rectMinHalfSide+circleR))){
            //System.out.println("distance between the centerse is less than the distance between the inscribed circle");
            result=true;
            break;
        }

        // check that the squares relate to angles
        if((deltax > (rectHalfWidth+circleR)*0.9) && (deltay > (rectHalfHeight+circleR)*0.9)){
            //System.out.println("squares relate to angles");
            result=true;
        }
    }while(false);

    return result;
}

public static boolean intersectionRectangleAndRectangle(int rectangleX, int rectangleY, int rectangleWidth, int rectangleHeight, int rectangleX2, int rectangleY2, int rectangleWidth2, int rectangleHeight2){
    boolean result = false;

    float rectHalfWidth = rectangleWidth/2.0f;
    float rectHalfHeight = rectangleHeight/2.0f;
    float rectHalfWidth2 = rectangleWidth2/2.0f;
    float rectHalfHeight2 = rectangleHeight2/2.0f;

    float deltax = Math.abs((rectangleX + rectHalfWidth) - (rectangleX2 + rectHalfWidth2));
    float deltay = Math.abs((rectangleY + rectHalfHeight) - (rectangleY2 + rectHalfHeight2));

    float lengthHypotenuseSqure = deltax*deltax + deltay*deltay;

    do{
        // check that distance between the centerse is more than the distance between the circumcircle
        if(lengthHypotenuseSqure > ((rectHalfWidth+rectHalfWidth2)*(rectHalfWidth+rectHalfWidth2) + (rectHalfHeight+rectHalfHeight2)*(rectHalfHeight+rectHalfHeight2))){
            //System.out.println("distance between the centerse is more than the distance between the circumcircle");
            break;
        }

        // check that distance between the centerse is less than the distance between the inscribed circle
        float rectMinHalfSide = Math.min(rectHalfWidth, rectHalfHeight);
        float rectMinHalfSide2 = Math.min(rectHalfWidth2, rectHalfHeight2);
        if(lengthHypotenuseSqure < ((rectMinHalfSide+rectMinHalfSide2)*(rectMinHalfSide+rectMinHalfSide2))){
            //System.out.println("distance between the centerse is less than the distance between the inscribed circle");
            result=true;
            break;
        }

        // check that the squares relate to angles
        if((deltax > (rectHalfWidth+rectHalfWidth2)*0.9) && (deltay > (rectHalfHeight+rectHalfHeight2)*0.9)){
            //System.out.println("squares relate to angles");
            result=true;
        }
    }while(false);

    return result;
  } 
}

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愁以何悠 2024-07-17 15:11:25

这是修改后的代码 100% 工作：

public static bool IsIntersected(PointF circle, float radius, RectangleF rectangle)
{
    var rectangleCenter = new PointF((rectangle.X +  rectangle.Width / 2),
                                     (rectangle.Y + rectangle.Height / 2));

    var w = rectangle.Width  / 2;
    var h = rectangle.Height / 2;

    var dx = Math.Abs(circle.X - rectangleCenter.X);
    var dy = Math.Abs(circle.Y - rectangleCenter.Y);

    if (dx > (radius + w) || dy > (radius + h)) return false;

    var circleDistance = new PointF
                             {
                                 X = Math.Abs(circle.X - rectangle.X - w),
                                 Y = Math.Abs(circle.Y - rectangle.Y - h)
                             };

    if (circleDistance.X <= (w))
    {
        return true;
    }

    if (circleDistance.Y <= (h))
    {
        return true;
    }

    var cornerDistanceSq = Math.Pow(circleDistance.X - w, 2) + 
                                    Math.Pow(circleDistance.Y - h, 2);

    return (cornerDistanceSq <= (Math.Pow(radius, 2)));
}

Bassam Alugili

Here is the modfied code 100% working:

public static bool IsIntersected(PointF circle, float radius, RectangleF rectangle)
{
    var rectangleCenter = new PointF((rectangle.X +  rectangle.Width / 2),
                                     (rectangle.Y + rectangle.Height / 2));

    var w = rectangle.Width  / 2;
    var h = rectangle.Height / 2;

    var dx = Math.Abs(circle.X - rectangleCenter.X);
    var dy = Math.Abs(circle.Y - rectangleCenter.Y);

    if (dx > (radius + w) || dy > (radius + h)) return false;

    var circleDistance = new PointF
                             {
                                 X = Math.Abs(circle.X - rectangle.X - w),
                                 Y = Math.Abs(circle.Y - rectangle.Y - h)
                             };

    if (circleDistance.X <= (w))
    {
        return true;
    }

    if (circleDistance.Y <= (h))
    {
        return true;
    }

    var cornerDistanceSq = Math.Pow(circleDistance.X - w, 2) + 
                                    Math.Pow(circleDistance.Y - h, 2);

    return (cornerDistanceSq <= (Math.Pow(radius, 2)));
}

Bassam Alugili

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断爱 2024-07-17 15:11:25

这是对此的快速单行测试：

if (length(max(abs(center - rect_mid) - rect_halves, 0)) <= radius ) {
  // They intersect.
}

这是轴对齐的情况，其中 rect_halves 是从矩形中间指向角的正向量。 length() 中的表达式是从center 到矩形中最近点的增量向量。这在任何维度都有效。

Here's a fast one-line test for this:

if (length(max(abs(center - rect_mid) - rect_halves, 0)) <= radius ) {
  // They intersect.
}

This is the axis-aligned case where rect_halves is a positive vector pointing from the rectangle middle to a corner. The expression inside length() is a delta vector from center to a closest point in the rectangle. This works in any dimension.

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べ繥欢鉨o。 2024-07-17 15:11:25

首先检查矩形和与圆相切的正方形是否重叠（简单）。如果它们不重叠，则它们不会碰撞。
检查圆的中心是否在矩形内（简单）。如果在里面，它们就会发生碰撞。
计算从矩形边到圆心的最小平方距离（有点难）。如果小于半径平方，则它们会发生碰撞，否则不会发生碰撞。

它很高效，因为：

首先，它使用廉价算法检查最常见的场景，当确定它们不会发生冲突时，它就会结束。
然后，它使用廉价算法检查下一个最常见的场景（不计算平方根，使用平方值），当确定它们发生冲突时，它就会结束。
然后它执行更昂贵的算法来检查与矩形边框的碰撞。

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厌倦 2024-07-17 15:11:25

对我有用（仅当矩形角度为 180 时才有效）

function intersects(circle, rect) {
  let left = rect.x + rect.width > circle.x - circle.radius;
  let right = rect.x < circle.x + circle.radius;
  let top = rect.y < circle.y + circle.radius;
  let bottom = rect.y + rect.height > circle.y - circle.radius;
  return left && right && bottom && top;
}

worked for me (only work when angle of rectangle is 180)

function intersects(circle, rect) {
  let left = rect.x + rect.width > circle.x - circle.radius;
  let right = rect.x < circle.x + circle.radius;
  let top = rect.y < circle.y + circle.radius;
  let bottom = rect.y + rect.height > circle.y - circle.radius;
  return left && right && bottom && top;
}

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紙鸢 2024-07-17 15:11:25

对于那些必须使用 SQL 计算地理坐标中的圆/矩形碰撞的人，
这是我在 e.James 建议算法的 oracle 11 中的实现。

在输入中，它需要圆坐标、以公里为单位的圆半径和矩形的两个顶点坐标：

CREATE OR REPLACE FUNCTION "DETECT_CIRC_RECT_COLLISION"
(
    circleCenterLat     IN NUMBER,      -- circle Center Latitude
    circleCenterLon     IN NUMBER,      -- circle Center Longitude
    circleRadius        IN NUMBER,      -- circle Radius in KM
    rectSWLat           IN NUMBER,      -- rectangle South West Latitude
    rectSWLon           IN NUMBER,      -- rectangle South West Longitude
    rectNELat           IN NUMBER,      -- rectangle North Est Latitude
    rectNELon           IN NUMBER       -- rectangle North Est Longitude
)
RETURN NUMBER
AS
    -- converts km to degrees (use 69 if miles)
    kmToDegreeConst     NUMBER := 111.045;

    -- Remaining rectangle vertices 
    rectNWLat   NUMBER;
    rectNWLon   NUMBER;
    rectSELat   NUMBER;
    rectSELon   NUMBER;

    rectHeight  NUMBER;
    rectWIdth   NUMBER;

    circleDistanceLat   NUMBER;
    circleDistanceLon   NUMBER;
    cornerDistanceSQ    NUMBER;

BEGIN
    -- Initialization of remaining rectangle vertices  
    rectNWLat := rectNELat;
    rectNWLon := rectSWLon;
    rectSELat := rectSWLat;
    rectSELon := rectNELon;

    -- Rectangle sides length calculation
    rectHeight := calc_distance(rectSWLat, rectSWLon, rectNWLat, rectNWLon);
    rectWidth := calc_distance(rectSWLat, rectSWLon, rectSELat, rectSELon);

    circleDistanceLat := abs( (circleCenterLat * kmToDegreeConst) - ((rectSWLat * kmToDegreeConst) + (rectHeight/2)) );
    circleDistanceLon := abs( (circleCenterLon * kmToDegreeConst) - ((rectSWLon * kmToDegreeConst) + (rectWidth/2)) );

    IF circleDistanceLon > ((rectWidth/2) + circleRadius) THEN
        RETURN -1;   --  -1 => NO Collision ; 0 => Collision Detected
    END IF;

    IF circleDistanceLat > ((rectHeight/2) + circleRadius) THEN
        RETURN -1;   --  -1 => NO Collision ; 0 => Collision Detected
    END IF;

    IF circleDistanceLon <= (rectWidth/2) THEN
        RETURN 0;   --  -1 => NO Collision ; 0 => Collision Detected
    END IF;

    IF circleDistanceLat <= (rectHeight/2) THEN
        RETURN 0;   --  -1 => NO Collision ; 0 => Collision Detected
    END IF;


    cornerDistanceSQ := POWER(circleDistanceLon - (rectWidth/2), 2) + POWER(circleDistanceLat - (rectHeight/2), 2);

    IF cornerDistanceSQ <=  POWER(circleRadius, 2) THEN
        RETURN 0;  --  -1 => NO Collision ; 0 => Collision Detected
    ELSE
        RETURN -1;  --  -1 => NO Collision ; 0 => Collision Detected
    END IF;

    RETURN -1;  --  -1 => NO Collision ; 0 => Collision Detected
END;

For those have to calculate Circle/Rectangle collision in Geographic Coordinates with SQL,
this is my implementation in oracle 11 of e.James suggested algorithm.

In input it requires circle coordinates, circle radius in km and two vertices coordinates of the rectangle:

CREATE OR REPLACE FUNCTION "DETECT_CIRC_RECT_COLLISION"
(
    circleCenterLat     IN NUMBER,      -- circle Center Latitude
    circleCenterLon     IN NUMBER,      -- circle Center Longitude
    circleRadius        IN NUMBER,      -- circle Radius in KM
    rectSWLat           IN NUMBER,      -- rectangle South West Latitude
    rectSWLon           IN NUMBER,      -- rectangle South West Longitude
    rectNELat           IN NUMBER,      -- rectangle North Est Latitude
    rectNELon           IN NUMBER       -- rectangle North Est Longitude
)
RETURN NUMBER
AS
    -- converts km to degrees (use 69 if miles)
    kmToDegreeConst     NUMBER := 111.045;

    -- Remaining rectangle vertices 
    rectNWLat   NUMBER;
    rectNWLon   NUMBER;
    rectSELat   NUMBER;
    rectSELon   NUMBER;

    rectHeight  NUMBER;
    rectWIdth   NUMBER;

    circleDistanceLat   NUMBER;
    circleDistanceLon   NUMBER;
    cornerDistanceSQ    NUMBER;

BEGIN
    -- Initialization of remaining rectangle vertices  
    rectNWLat := rectNELat;
    rectNWLon := rectSWLon;
    rectSELat := rectSWLat;
    rectSELon := rectNELon;

    -- Rectangle sides length calculation
    rectHeight := calc_distance(rectSWLat, rectSWLon, rectNWLat, rectNWLon);
    rectWidth := calc_distance(rectSWLat, rectSWLon, rectSELat, rectSELon);

    circleDistanceLat := abs( (circleCenterLat * kmToDegreeConst) - ((rectSWLat * kmToDegreeConst) + (rectHeight/2)) );
    circleDistanceLon := abs( (circleCenterLon * kmToDegreeConst) - ((rectSWLon * kmToDegreeConst) + (rectWidth/2)) );

    IF circleDistanceLon > ((rectWidth/2) + circleRadius) THEN
        RETURN -1;   --  -1 => NO Collision ; 0 => Collision Detected
    END IF;

    IF circleDistanceLat > ((rectHeight/2) + circleRadius) THEN
        RETURN -1;   --  -1 => NO Collision ; 0 => Collision Detected
    END IF;

    IF circleDistanceLon <= (rectWidth/2) THEN
        RETURN 0;   --  -1 => NO Collision ; 0 => Collision Detected
    END IF;

    IF circleDistanceLat <= (rectHeight/2) THEN
        RETURN 0;   --  -1 => NO Collision ; 0 => Collision Detected
    END IF;


    cornerDistanceSQ := POWER(circleDistanceLon - (rectWidth/2), 2) + POWER(circleDistanceLat - (rectHeight/2), 2);

    IF cornerDistanceSQ <=  POWER(circleRadius, 2) THEN
        RETURN 0;  --  -1 => NO Collision ; 0 => Collision Detected
    ELSE
        RETURN -1;  --  -1 => NO Collision ; 0 => Collision Detected
    END IF;

    RETURN -1;  --  -1 => NO Collision ; 0 => Collision Detected
END;

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汐鸠 2024-07-17 15:11:25

有效，一周前才发现这一点，现在才开始测试。

double theta = Math.atan2(cir.getX()-sqr.getX()*1.0,
                          cir.getY()-sqr.getY()*1.0); //radians of the angle
double dBox; //distance from box to edge of box in direction of the circle

if((theta >  Math.PI/4 && theta <  3*Math.PI / 4) ||
   (theta < -Math.PI/4 && theta > -3*Math.PI / 4)) {
    dBox = sqr.getS() / (2*Math.sin(theta));
} else {
    dBox = sqr.getS() / (2*Math.cos(theta));
}
boolean touching = (Math.abs(dBox) >=
                    Math.sqrt(Math.pow(sqr.getX()-cir.getX(), 2) +
                              Math.pow(sqr.getY()-cir.getY(), 2)));

Works, just figured this out a week ago, and just now got to testing it.

double theta = Math.atan2(cir.getX()-sqr.getX()*1.0,
                          cir.getY()-sqr.getY()*1.0); //radians of the angle
double dBox; //distance from box to edge of box in direction of the circle

if((theta >  Math.PI/4 && theta <  3*Math.PI / 4) ||
   (theta < -Math.PI/4 && theta > -3*Math.PI / 4)) {
    dBox = sqr.getS() / (2*Math.sin(theta));
} else {
    dBox = sqr.getS() / (2*Math.cos(theta));
}
boolean touching = (Math.abs(dBox) >=
                    Math.sqrt(Math.pow(sqr.getX()-cir.getX(), 2) +
                              Math.pow(sqr.getY()-cir.getY(), 2)));

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尘曦 2024-07-17 15:11:25

def colision(rect, circle):
dx = rect.x - circle.x
dy = rect.y - circle.y
distance = (dy**2 + dx**2)**0.5
angle_to = (rect.angle + math.atan2(dx, dy)/3.1415*180.0) % 360
if((angle_to>135 and angle_to<225) or (angle_to>0 and angle_to<45) or (angle_to>315 and angle_to<360)):
    if distance <= circle.rad/2.+((rect.height/2.0)*(1.+0.5*abs(math.sin(angle_to*math.pi/180.)))):
        return True
else:
    if distance <= circle.rad/2.+((rect.width/2.0)*(1.+0.5*abs(math.cos(angle_to*math.pi/180.)))):
        return True
return False

def colision(rect, circle):
dx = rect.x - circle.x
dy = rect.y - circle.y
distance = (dy**2 + dx**2)**0.5
angle_to = (rect.angle + math.atan2(dx, dy)/3.1415*180.0) % 360
if((angle_to>135 and angle_to<225) or (angle_to>0 and angle_to<45) or (angle_to>315 and angle_to<360)):
    if distance <= circle.rad/2.+((rect.height/2.0)*(1.+0.5*abs(math.sin(angle_to*math.pi/180.)))):
        return True
else:
    if distance <= circle.rad/2.+((rect.width/2.0)*(1.+0.5*abs(math.cos(angle_to*math.pi/180.)))):
        return True
return False

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土豪 2024-07-17 15:11:25

我在制作这个游戏时开发了这个算法：https://mshwf.github.io/mates/

如果圆与正方形相交，则圆的中心线与正方形的中心线之间的距离应等于(直径+边长)/2。
因此，让我们有一个名为 touching 的变量来保存该距离。问题是：我应该考虑哪条中心线：水平还是垂直？
考虑这个框架：

< img src="https://1.bp.blogspot.com/-kUQjhsDgT2o/XwOb5sln4TI/AAAAAAAAQms/S_K6EW0XRRE8j79h2PI_QN3gKXrs5E-kACK4BGAsYHg/s499/collision%2Balgo-2.jpg" alt="在此处输入图像描述">

每条中心线给出不同的距离，只有一个是无碰撞的正确指示，但使用我们人类的直觉是理解自然算法如何工作的开始。

它们没有接触，这意味着两条中心线之间的距离应该大于接触，这意味着自然算法会选择水平中心线（垂直中心线表明存在碰撞！）。通过观察多个圆圈，您可以看出：如果圆与正方形的垂直延伸线相交，那么我们选择垂直距离（水平中心线之间），如果圆与水平延伸线相交，我们选择水平距离：

另一个示例，圆圈数字4：它与正方形的水平延伸相交，那么我们考虑等于接触的水平距离。

好的，最困难的部分已经揭开了，现在我们知道算法是如何工作的，但是我们如何知道圆与哪个扩展相交呢？
实际上很简单：我们计算最右 x 和最左 x（圆形和正方形）之间的距离，y 轴也是如此，值较大的轴是其延伸部分与圆相交的轴（如果它大于直径+边，则圆位于两个正方形延伸部分之外，如圆#7）。代码如下：

right = Math.max(square.x+square.side, circle.x+circle.rad);
left = Math.min(square.x, circle.x-circle.rad);

bottom = Math.max(square.y+square.side, circle.y+circle.rad);
top = Math.min(square.y, circle.y-circle.rad);

if (right - left > down - top) {
 //compare with horizontal distance
}
else {
 //compare with vertical distance
}

/*These equations assume that the reference point of the square is at its top left corner, and the reference point of the circle is at its center*/

I developed this algorithm while making this game: https://mshwf.github.io/mates/

If the circle touches the square, then the distance between the centerline of the circle and the centerline of the square should equal (diameter+side)/2.
So, let's have a variable named touching that holds that distance. The problem was: which centerline should I consider: the horizontal or the vertical?
Consider this frame:

Each centerline gives different distances, and only one is a correct indication to a no-collision, but using our human intuition is a start to understand how the natural algorithm works.

They are not touching, which means that the distance between the two centerlines should be greater than touching, which means that the natural algorithm picks the horizontal centerlines (the vertical centerlines says there's a collision!). By noticing multiple circles, you can tell: if the circle intersects with the vertical extension of the square, then we pick the vertical distance (between the horizontal centerlines), and if the circle intersects with the horizontal extension, we pick the horizontal distance:

Another example, circle number 4: it intersects with the horizontal extension of the square, then we consider the horizontal distance which is equal to touching.

Ok, the tough part is demystified, now we know how the algorithm will work, but how we know with which extension the circle intersects?
It's easy actually: we calculate the distance between the most right x and the most left x (of both the circle and the square), and the same for the y-axis, the one with greater value is the axis with the extension that intersects with the circle (if it's greater than diameter+side then the circle is outside the two square extensions, like circle #7). The code looks like:

right = Math.max(square.x+square.side, circle.x+circle.rad);
left = Math.min(square.x, circle.x-circle.rad);

bottom = Math.max(square.y+square.side, circle.y+circle.rad);
top = Math.min(square.y, circle.y-circle.rad);

if (right - left > down - top) {
 //compare with horizontal distance
}
else {
 //compare with vertical distance
}

/*These equations assume that the reference point of the square is at its top left corner, and the reference point of the circle is at its center*/

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意中人 2024-07-17 15:11:25

预先检查完全包围矩形的圆是否与圆碰撞。
检查圆内是否有矩形角。
对于每条边，查看是否有一条线与圆相交。将中心点C投影到直线AB上得到点D。如果CD的长度小于半径，则发生碰撞。

    projectionScalar=dot(AC,AB)/(mag(AC)*mag(AB));
    if(projectionScalar>=0 && projectionScalar<=1) {
        D=A+AB*projectionScalar;
        CD=D-C;
        if(mag(CD)<circle.radius){
            // there was a collision
        }
    }

do a pre-check whether a circle fully encapsulating the rectangle collides with the circle.
check for rectangle corners within the circle.
For each edge, see if there is a line intersection with the circle. Project the center point C onto the line AB to get a point D. If the length of CD is less than radius, there was a collision.

    projectionScalar=dot(AC,AB)/(mag(AC)*mag(AB));
    if(projectionScalar>=0 && projectionScalar<=1) {
        D=A+AB*projectionScalar;
        CD=D-C;
        if(mag(CD)<circle.radius){
            // there was a collision
        }
    }

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顾北清歌寒 2024-07-17 15:11:25

有一种非常简单的方法可以做到这一点，您必须在 x 和 y 中夹住一个点，但在正方形内部，而圆心位于垂直轴之一的两个正方形边界点之间，您需要夹住这些点坐标到平行轴，只要确保夹紧的坐标不超出正方形的限制即可。
然后获取圆心与夹紧坐标之间的距离，并检查该距离是否小于圆的半径。

这是我的做法（前 4 个点是方坐标，其余是圆点）：

bool DoesCircleImpactBox(float x, float y, float x1, float y1, float xc, float yc, float radius){
    float ClampedX=0;
    float ClampedY=0;
    
    if(xc>=x and xc<=x1){
    ClampedX=xc;
    }
    
    if(yc>=y and yc<=y1){
    ClampedY=yc;
    }
    
    radius = radius+1;
    
    if(xc<x) ClampedX=x;
    if(xc>x1) ClampedX=x1-1;
    if(yc<y) ClampedY=y;
    if(yc>y1) ClampedY=y1-1;
    
    float XDif=ClampedX-xc;
    XDif=XDif*XDif;
    float YDif=ClampedY-yc;
    YDif=YDif*YDif;
    
    if(XDif+YDif<=radius*radius) return true;
    
    return false;
}

There is an incredibly simple way to do this, you have to clamp a point in x and y, but inside the square, while the center of the circle is between the two square border points in one of the perpendicular axis you need to clamp those coordinates to the parallel axis, just make sure the clamped coordinates do not exeed the limits of the square.
Then just get the distance between the center of the circle and the clamped coordinates and check if the distance is less than the radius of the circle.

Here is how I did it (First 4 points are the square coordinates, the rest are circle points):

bool DoesCircleImpactBox(float x, float y, float x1, float y1, float xc, float yc, float radius){
    float ClampedX=0;
    float ClampedY=0;
    
    if(xc>=x and xc<=x1){
    ClampedX=xc;
    }
    
    if(yc>=y and yc<=y1){
    ClampedY=yc;
    }
    
    radius = radius+1;
    
    if(xc<x) ClampedX=x;
    if(xc>x1) ClampedX=x1-1;
    if(yc<y) ClampedY=y;
    if(yc>y1) ClampedY=y1-1;
    
    float XDif=ClampedX-xc;
    XDif=XDif*XDif;
    float YDif=ClampedY-yc;
    YDif=YDif*YDif;
    
    if(XDif+YDif<=radius*radius) return true;
    
    return false;
}

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夕嗳→ 2024-07-17 15:11:25

我的方法：

从OBB/矩形上/内的圆计算closest_point
（最近点将位于边缘/角落或内部）
计算从最近点到圆心的 squared_distance
（平方距离避免平方根）
返回 squared_distance <= 圆半径平方

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錯遇了你 2024-07-17 15:11:25

使用下面我的Java版本的圆-方形碰撞/相交检测的答案：

public static boolean circleIntersectsSquare(Point circle, double radius,
        Point squareCenter, double halfSize) {
    double deltaX = Math.abs(circle.x - squareCenter.x);
    double deltaY = Math.abs(circle.y - squareCenter.y);

    if (deltaX >= (halfSize + radius)) return false;
    if (deltaY >= (halfSize + radius)) return false;

    if (deltaX < halfSize) return true;
    if (deltaY < halfSize) return true;

    double cornerDist = Math.pow(deltaX - halfSize, 2) +
        Math.pow(deltaY - halfSize, 2);

    return (cornerDist < radius*radius);
}

原始答案：圆-矩形碰撞检测（相交）

Using the answer below my Java version of circle - square collision/intersection detection:

public static boolean circleIntersectsSquare(Point circle, double radius,
        Point squareCenter, double halfSize) {
    double deltaX = Math.abs(circle.x - squareCenter.x);
    double deltaY = Math.abs(circle.y - squareCenter.y);

    if (deltaX >= (halfSize + radius)) return false;
    if (deltaY >= (halfSize + radius)) return false;

    if (deltaX < halfSize) return true;
    if (deltaY < halfSize) return true;

    double cornerDist = Math.pow(deltaX - halfSize, 2) +
        Math.pow(deltaY - halfSize, 2);

    return (cornerDist < radius*radius);
}

Original answer: Circle-Rectangle collision detection (intersection)

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谈下烟灰 2024-07-17 15:11:25

假设您有矩形的四个边，检查从边缘到圆心的距离，如果小于半径，则形状相交。

if sqrt((rectangleRight.x - circleCenter.x)^2 +
        (rectangleBottom.y - circleCenter.y)^2) < radius
// then they intersect

if sqrt((rectangleRight.x - circleCenter.x)^2 +
        (rectangleTop.y - circleCenter.y)^2) < radius
// then they intersect

if sqrt((rectangleLeft.x - circleCenter.x)^2 +
        (rectangleTop.y - circleCenter.y)^2) < radius
// then they intersect

if sqrt((rectangleLeft.x - circleCenter.x)^2 +
        (rectangleBottom.y - circleCenter.y)^2) < radius
// then they intersect

Assuming you have the four edges of the rectangle check the distance from the edges to the center of the circle, if its less then the radius, then the shapes are intersecting.

if sqrt((rectangleRight.x - circleCenter.x)^2 +
        (rectangleBottom.y - circleCenter.y)^2) < radius
// then they intersect

if sqrt((rectangleRight.x - circleCenter.x)^2 +
        (rectangleTop.y - circleCenter.y)^2) < radius
// then they intersect

if sqrt((rectangleLeft.x - circleCenter.x)^2 +
        (rectangleTop.y - circleCenter.y)^2) < radius
// then they intersect

if sqrt((rectangleLeft.x - circleCenter.x)^2 +
        (rectangleBottom.y - circleCenter.y)^2) < radius
// then they intersect

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圆-矩形碰撞检测（交叉点）

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圆-矩形碰撞检测（交叉点）

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不再见

真是无聊啊

樱娆

浅语花开

烛光

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