blocks|key|490046|text|你需要一个2维旋转矩阵http://en.wikipedia.org/wiki/Rotation_matrix|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|490047|你的新观点将是|490048|+newX+=+centerX+%2B+(+cosX+*+(point2X-centerX)+%2B+sinX+*+(point2Y+-centerY))
+newY+=+centerY+%2B+(+-sinX+*+(point2X-centerX)+%2B+cosX+*+(point2Y+-centerY))|code-block|syntax|javascript|490049|因为你是顺时针旋转，而不是逆时针旋转。|490050|entityMap|0|LINK|mutability|MUTABLE|url|http://en.wikipedia.org/wiki/Rotation_matrix^0|B|18|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|6|7|Y|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|Z|8|@]|9|@]|C|$I|J]]|$1|K|3|L|5|6|7|10|8|@]|9|@]|C|$]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]]]

You need a 2-d rotation matrix <a href="http://en.wikipedia.org/wiki/Rotation_matrix" rel="noreferrer">http://en.wikipedia.org/wiki/Rotation_matrix</a>

Your new point will be 

<pre><code> newX = centerX + ( cosX * (point2X-centerX) + sinX * (point2Y -centerY))
 newY = centerY + ( -sinX * (point2X-centerX) + cosX * (point2Y -centerY))
</code></pre>

because you are rotating clockwise rather than anticlockwise

blocks|key|490085|text|假设您正在对Java进行签名，请尝试以下代码--|type|unstyled|depth|inlineStyleRanges|entityRanges|data|490086|++++Point2D+result+=+new+Point2D.Double();
++++AffineTransform+rotation+=+new+AffineTransform();
++++double+angleInRadians+=+(angle+*+Math.PI+/+180);
++++rotation.rotate(angleInRadians,+pivot.getX(),+pivot.getY());
++++rotation.transform(point,+result);
++++return+result;|code-block|syntax|javascript|490087|其中，pivot是要旋转的点。|490088|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Assuming you are usign the Java Graphics2D API, try this code -

<pre><code> Point2D result = new Point2D.Double();
 AffineTransform rotation = new AffineTransform();
 double angleInRadians = (angle * Math.PI / 180);
 rotation.rotate(angleInRadians, pivot.getX(), pivot.getY());
 rotation.transform(point, result);
 return result;
</code></pre>

where pivot is the point you are rotating around.

blocks|key|273041|text|这是一种在2D中围绕任何其他点旋转任何点的方法。请注意，在3D中，这可以用作围绕z轴的旋转，因为它不会改变。在3D中绕x轴和y轴旋转也很容易实现。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|273042|代码在JavaScript中。开头的注释行是函数的测试集。它们也是一个用法的例子。|273043|//A+=+new+Array(0,0)
//S+=+new+Array(-1,0)
//fi+=+90
//alert("rotujBod:+"+%2B+rotatePoint(A,+S,+fi))

function+rotatePoint(A,+S,+fi)+{
/**+IN+points+A+-+rotated+point,+S+-+centre,+fi+-+angle+of+rotation+(rad)
*++++points+in+format++[Ax,+Ay,+Az],+angle+fi+(float)
*+++++++OUT+point+B
*/
++++r+=+Math.sqrt((A[0]+-+S[0])*(A[0]+-+S[0])+%2B+(A[1]+-+S[1])*(A[1]+-+S[1]))
++++originOfRotation+=+new+Array(S[0]+%2B+r,+S[1])
++++if+(A[1]+<+S[1])+{
++++++++A2+=+new+Array(A[0],+-1*A[1])
++++++++originalAngle+=+-1*sizeOfAngle(originOfRotation,+S,+A2)
++++}+else+{
++++originalAngle+=+sizeOfAngle(originOfRotation,+S,+A)
++++}
++++x+=+S[0]+%2B+r*Math.cos(fi+%2B+originalAngle)
++++y+=+S[1]+%2B+r*Math.sin(fi+%2B+originalAngle)
++++B+=+new+Array(x,+y)
++++return(B)
}

function+sizeOfAngle(A,+S,+B)+{
++++ux+=+A[0]+-+S[0]
++++uy+=+A[1]+-+S[1]
++++vx+=+B[0]+-+S[0]
++++vy+=+B[1]+-+S[1]
++++if((Math.sqrt(ux*ux+%2B+uy*uy)*Math.sqrt(vx*vx+%2B+vy*vy))+==+0)+{return+0}
++++return+Math.acos((ux*vx+%2B+uy*vy)/(Math.sqrt(ux*ux+%2B+uy*uy)*Math.sqrt(vx*vx+%2B+vy*vy)))
}|code-block|syntax|javascript|273044|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Here is a way to rotate any point about any other point in 2D. Note that in 3D this can be used as rotation about z axis, z-coordinate of a point being ingored since it doesn't change. Rotation about x-axis and y-axis in 3D can be also easily implemented.

The code is in JavaScript. The commented lines at the beginning are a test set for the function. They also serve as an example of usage.

<pre><code>//A = new Array(0,0)
//S = new Array(-1,0)
//fi = 90
//alert("rotujBod: " + rotatePoint(A, S, fi))

function rotatePoint(A, S, fi) {
/** IN points A - rotated point, S - centre, fi - angle of rotation (rad)
* points in format [Ax, Ay, Az], angle fi (float)
* OUT point B
*/
 r = Math.sqrt((A[0] - S[0])*(A[0] - S[0]) + (A[1] - S[1])*(A[1] - S[1]))
 originOfRotation = new Array(S[0] + r, S[1])
 if (A[1] &lt; S[1]) {
 A2 = new Array(A[0], -1*A[1])
 originalAngle = -1*sizeOfAngle(originOfRotation, S, A2)
 } else {
 originalAngle = sizeOfAngle(originOfRotation, S, A)
 }
 x = S[0] + r*Math.cos(fi + originalAngle)
 y = S[1] + r*Math.sin(fi + originalAngle)
 B = new Array(x, y)
 return(B)
}

function sizeOfAngle(A, S, B) {
 ux = A[0] - S[0]
 uy = A[1] - S[1]
 vx = B[0] - S[0]
 vy = B[1] - S[1]
 if((Math.sqrt(ux*ux + uy*uy)*Math.sqrt(vx*vx + vy*vy)) == 0) {return 0}
 return Math.acos((ux*vx + uy*vy)/(Math.sqrt(ux*ux + uy*uy)*Math.sqrt(vx*vx + vy*vy)))
}
</code></pre>

blocks|key|490225|text|这里是一个关心旋转方向的版本。右(顺时针)为负，左(逆时针)为正。您可以发送一个点或一个2d向量，并在此方法(最后一行)中设置其原语，以避免性能方面的内存分配。您可能需要将vector2和mathutil替换为您使用的库或java的内置point类，您可以使用math.toradians()来代替mathutils。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|490226|/**
+*+rotates+the+point+around+a+center+and+returns+the+new+point
+*+@param+cx+x+coordinate+of+the+center
+*+@param+cy+y+coordinate+of+the+center
+*+@param+angle+in+degrees+(sign+determines+the+direction+%2B+is+counter-clockwise+-+is+clockwise)
+*+@param+px+x+coordinate+of+point+to+rotate+
+*+@param+py+y+coordinate+of+point+to+rotate+
+*+*/

public+static+Vector2+rotate_point(float+cx,float+cy,float+angle,float+px,float+py){
++++float+absangl=Math.abs(angle);
++++float+s+=+MathUtils.sin(absangl+*+MathUtils.degreesToRadians);
++++float+c+=+MathUtils.cos(absangl+*+MathUtils.degreesToRadians);

++++//+translate+point+back+to+origin:
++++px+-=+cx;
++++py+-=+cy;

++++//+rotate+point
++++float+xnew;
++++float+ynew;
++++if+(angle+>+0)+{
++++++++xnew+=+px+*+c+-+py+*+s;
++++++++ynew+=+px+*+s+%2B+py+*+c;
++++}
++++else+{
++++++++xnew+=+px+*+c+%2B+py+*+s;
++++++++ynew+=+-px+*+s+%2B+py+*+c;
++++}

++++//+translate+point+back:
++++px+=+xnew+%2B+cx;
++++py+=+ynew+%2B+cy;
++++return+new+Vector2(px,+py);
}|code-block|syntax|javascript|490227|请注意，这种方式比你在帖子中尝试的方式具有更高的性能。因为您使用sqrt，这是非常昂贵的，并且通过这种方式将度转换为弧度，通过查找表管理，如果您想知道的话。因此它具有非常高的性能。|490228|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Here a version that cares the rotate direction. Right (clockwise) is negative and left (counter clockwise) is positive. You can send a point or a 2d vector and set its primitives in this method (last line) to avoid memory allocation for performance. You may need to replace vector2 and mathutils to libraries you use or to java's built-in point class and you can use math.toradians() instead of mathutils.

<pre><code>/**
 * rotates the point around a center and returns the new point
 * @param cx x coordinate of the center
 * @param cy y coordinate of the center
 * @param angle in degrees (sign determines the direction + is counter-clockwise - is clockwise)
 * @param px x coordinate of point to rotate 
 * @param py y coordinate of point to rotate 
 * */

public static Vector2 rotate_point(float cx,float cy,float angle,float px,float py){
 float absangl=Math.abs(angle);
 float s = MathUtils.sin(absangl * MathUtils.degreesToRadians);
 float c = MathUtils.cos(absangl * MathUtils.degreesToRadians);

 // translate point back to origin:
 px -= cx;
 py -= cy;

 // rotate point
 float xnew;
 float ynew;
 if (angle &gt; 0) {
 xnew = px * c - py * s;
 ynew = px * s + py * c;
 }
 else {
 xnew = px * c + py * s;
 ynew = -px * s + py * c;
 }

 // translate point back:
 px = xnew + cx;
 py = ynew + cy;
 return new Vector2(px, py);
}
</code></pre>

Note that this way has more performance than the way you tried in your post. Because you use sqrt that is very costly and in this way converting from degrees to radians managed with a lookup table, if you wonder. And so it has very high performance.

Imagine you have two points in 2d space and you need to rotate one of these points by X degrees with the other point acting as a center.

<pre><code>float distX = Math.abs( centerX -point2X );
float distY = Math.abs( centerY -point2Y );

float dist = FloatMath.sqrt( distX*distX + distY*distY );
</code></pre>

So far I just got to finding the distance between the two points... any ideas where should I go from that?

<img src="https://i.stack.imgur.com/zS1HB.png" alt="enter image description here">

How to rotate a vertex around a certain point?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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假设你在二维空间中有两个点，你需要将其中一个点旋转X度，另一个点作为中心。float distX = Math.abs( centerX -point2X );float distY = Math.abs( centerY -point2Y );float dist = FloatMath.sqrt( distX*di...

问如何围绕某个点旋转顶点？
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问如何围绕某个点旋转顶点？
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