以连续的方式计算圆坐标坐标、方式
我需要有一个给定圆了一个又一个的所有点的坐标，这样我就可以使一个对象去的圈子跳跃，从一个点到下一个。我试过中点画圆算法，但我找到的版本是为了绘制和坐标不在顺序。它们同时产生8象限和在其顶部相对方向。如果至少它们是在同一方向，我可以使一个单独的阵列为每个象限和它们附加到彼此在末端。这是JavaScript改编code我现在有：
函数calcCircle（centerCoordinates，半径）{
VAR coordinatesArray =新的Array（）;
//转换坐标
VAR X0 = centerCoordinates.left;
VAR Y0 = centerCoordinates.top;
//定义变量
变种F = 1 - 半径;
变种ddFx = 1;
VAR ddFy = -radius＆LT;＆LT; 1;
变种X = 0;
变种Y =半径;
coordinatesArray.push（新坐标（X0，Y0 +半径））;
coordinatesArray.push（新坐标（X0，Y0 - 半径））;
coordinatesArray.push（新坐标（X0 +半径，Y0））;
coordinatesArray.push（新坐标（X0 - 半径，Y0））;
//主回路
而（X＆LT; Y）{
如果（f是氟烃基; = 0）{
y--;
ddFy + = 2;
F + = ddFy;
}
X ++;
ddFx + = 2;
F + = ddFx;
coordinatesArray.push（新坐标（X0 + X，Y 0 + Y））;
coordinatesArray.push（新坐标（X0 - X，Y 0 + Y））;
coordinatesArray.push（新坐标（X0 + X，Y0 - Y））;
coordinatesArray.push（新坐标（X0 - X，Y 0 - Y））;
coordinatesArray.push（新坐标（X0 + Y，Y 0 + X））;
coordinatesArray.push（新坐标（X0 - Y，Y0 + X））;
coordinatesArray.push（新坐标（X0 + Y，Y0 - X））;
coordinatesArray.push（新坐标（X0 - Y，Y0 - X））;
}
//返回结果
返回coordinatesArray;
}
我$ P PFER一些快速的算法不三角$，但任何帮助是AP preciated！
修改
这是最终的解决方案。谢谢大家！
函数calcCircle（centerCoordinates，半径）{
VAR coordinatesArray =新的Array（）;
VAR octantArrays =
{OCT1：新的Array（），OCT2：新的Array（），OCT3：新的Array（），OCT4：新的Array（）
oct5：新的Array（），OCT6：新的Array（），oct7：新的Array（），oct8：新的Array（）};
//转换坐标
VAR XP = centerCoordinates.left;
VAR YP = centerCoordinates.top;
//定义添加坐标阵列
VAR setCrd =
功能（targetArray，XC，YC）{
targetArray.push（新坐标（YC，XC））;
};
//定义变量
VAR XOFF = 0;
VAR YOFF =半径;
VAR余额= -radius;
//主回路
而（XOFF＆LT; = YOFF）{
//象限7 - 反转
setCrd（octantArrays.oct7，XP + XOFF，YP + YOFF）;
//象限6 - 直
setCrd（octantArrays.oct6，XP - XOFF，YP + YOFF）;
//象限3 - 反向
setCrd（octantArrays.oct3，XP - XOFF，YP - YOFF）;
//象限2 - 直
setCrd（octantArrays.oct2，XP + XOFF，YP - YOFF）;
//避免重复
如果（XOFF！= YOFF）{
//象限8 - 直
setCrd（octantArrays.oct8，XP + YOFF，YP + XOFF）;
//象限5 - 反向
setCrd（octantArrays.oct5，XP - YOFF，YP + XOFF）;
//象限4 - 直
setCrd（octantArrays.oct4，XP - YOFF，YP - XOFF）;
//象限1 - 反转
setCrd（octantArrays.oct1，XP + YOFF，YP - XOFF）;
}
//一些奇怪的东西
余额+ = XOFF ++ + XOFF;
如果（余额＆GT; = 0）{
平衡 - = --yoff + YOFF;
}
}
//反转逆时针八分阵列
octantArrays.oct7.reverse（）;
octantArrays.oct3.reverse（）;
octantArrays.oct5.reverse（）;
octantArrays.oct1.reverse（）;
//删除逆时针八分数组最后一个元素（避免重复）
octantArrays.oct7.pop（）;
octantArrays.oct3.pop（）;
octantArrays.oct5.pop（）;
octantArrays.oct1.pop（）;
//将所有阵列一起
coordinatesArray =
octantArrays.oct4.concat（octantArrays.oct3）.concat（octantArrays.oct2）.concat（octantArrays.oct1）。
concat(octantArrays.oct8).concat(octantArrays.oct7).concat(octantArrays.oct6).concat(octantArrays.oct5);
//返回结果
返回coordinatesArray;
}
解决方案
使用可以试试下面的办法：用你给的算法，但把你的坐标，以八个不同coordinateArrays。事后必须扭转其中一半（那些与（X0 + X，Y0-y）时，（X0-X，Y 0 + Y），（X0 + Y，Y0 + x）时，（X0-γ，Y0-x）的）事后追加所有阵列以正确的顺序。请注意，您添加的第一个四点到正确的阵列。
I need to have all point coordinates for a given circle one after another, so I can make an object go in circles by hopping from one point to the next. I tried the Midpoint circle algorithm, but the version I found is meant to draw and the coordinates are not sequential. They are produced simultaneously for 8 quadrants and in opposing directions on top of that. If at least they were in the same direction, I could make a separate array for every quadrant and append them to one another at the end. This is the JavaScript adapted code I have now:
function calcCircle(centerCoordinates, radius) {
var coordinatesArray = new Array();
// Translate coordinates
var x0 = centerCoordinates.left;
var y0 = centerCoordinates.top;
// Define variables
var f = 1 - radius;
var ddFx = 1;
var ddFy = -radius << 1;
var x = 0;
var y = radius;
coordinatesArray.push(new Coordinates(x0, y0 + radius));
coordinatesArray.push(new Coordinates(x0, y0 - radius));
coordinatesArray.push(new Coordinates(x0 + radius, y0));
coordinatesArray.push(new Coordinates(x0 - radius, y0));
// Main loop
while (x < y) {
if (f >= 0) {
y--;
ddFy += 2;
f += ddFy;
}
x++;
ddFx += 2;
f += ddFx;
coordinatesArray.push(new Coordinates(x0 + x, y0 + y));
coordinatesArray.push(new Coordinates(x0 - x, y0 + y));
coordinatesArray.push(new Coordinates(x0 + x, y0 - y));
coordinatesArray.push(new Coordinates(x0 - x, y0 - y));
coordinatesArray.push(new Coordinates(x0 + y, y0 + x));
coordinatesArray.push(new Coordinates(x0 - y, y0 + x));
coordinatesArray.push(new Coordinates(x0 + y, y0 - x));
coordinatesArray.push(new Coordinates(x0 - y, y0 - x));
}
// Return the result
return coordinatesArray;
}
I prefer some fast algorithm without trigonometry, but any help is appreciated!
EDIT
This is the final solution. Thanks everybody!
function calcCircle(centerCoordinates, radius) {
var coordinatesArray = new Array();
var octantArrays =
{oct1: new Array(), oct2: new Array(), oct3: new Array(), oct4: new Array(),
oct5: new Array(), oct6: new Array(), oct7: new Array(), oct8: new Array()};
// Translate coordinates
var xp = centerCoordinates.left;
var yp = centerCoordinates.top;
// Define add coordinates to array
var setCrd =
function (targetArray, xC, yC) {
targetArray.push(new Coordinates(yC, xC));
};
// Define variables
var xoff = 0;
var yoff = radius;
var balance = -radius;
// Main loop
while (xoff <= yoff) {
// Quadrant 7 - Reverse
setCrd(octantArrays.oct7, xp + xoff, yp + yoff);
// Quadrant 6 - Straight
setCrd(octantArrays.oct6, xp - xoff, yp + yoff);
// Quadrant 3 - Reverse
setCrd(octantArrays.oct3, xp - xoff, yp - yoff);
// Quadrant 2 - Straight
setCrd(octantArrays.oct2, xp + xoff, yp - yoff);
// Avoid duplicates
if (xoff != yoff) {
// Quadrant 8 - Straight
setCrd(octantArrays.oct8, xp + yoff, yp + xoff);
// Quadrant 5 - Reverse
setCrd(octantArrays.oct5, xp - yoff, yp + xoff);
// Quadrant 4 - Straight
setCrd(octantArrays.oct4, xp - yoff, yp - xoff);
// Quadrant 1 - Reverse
setCrd(octantArrays.oct1, xp + yoff, yp - xoff);
}
// Some weird stuff
balance += xoff++ + xoff;
if (balance >= 0) {
balance -= --yoff + yoff;
}
}
// Reverse counter clockwise octant arrays
octantArrays.oct7.reverse();
octantArrays.oct3.reverse();
octantArrays.oct5.reverse();
octantArrays.oct1.reverse();
// Remove counter clockwise octant arrays last element (avoid duplicates)
octantArrays.oct7.pop();
octantArrays.oct3.pop();
octantArrays.oct5.pop();
octantArrays.oct1.pop();
// Append all arrays together
coordinatesArray =
octantArrays.oct4.concat(octantArrays.oct3).concat(octantArrays.oct2).concat(octantArrays.oct1).
concat(octantArrays.oct8).concat(octantArrays.oct7).concat(octantArrays.oct6).concat(octantArrays.oct5);
// Return the result
return coordinatesArray;
}
解决方案
Use can try the following approach: use the algorithm you gave but push your coordinates to eight different coordinateArrays. Afterwards you have to reverse half of them (those with (x0+x,y0-y), (x0-x,y0+y), (x0+y,y0+x), (x0-y,y0-x)) and afterwards append all arrays in the correct order. Take care that you add the first four points to the correct arrays.
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