一致地将3D点定为“顺时针”或“逆时针”

Question

我在3D中有以下6点。您可以在下面的代码的pointXYZ变量中看到每个点的XYZ坐标。每个点也有一个id，存储在pointIDs变量中。 pointXYZ[0]对应于pointIDs[0]。

问题：如何以一致的方式从以下位置订购积分：

pointIDs = [463, 478, 464, 1169, 477, 462]首字母

到

pointIDs = [463, 464, 478, 477, 1169, 462]顺时针

或

pointIDs = [463, 464, 462, 1169, 477, 478]逆时针

注意：我了解3D中的cw或ccw取决于从哪一侧看。因此，cw或ccw对我来说无关紧要，应该保持一致，也不要紧要从哪里开始，464只是一个示例。第一个id是总是中心。

现在，在calculate normal和this answer的帮助下，我想到了下面的代码。但是，我很难从那里对其进行排序。非常感谢您的帮助。

function vectorAB(a, b) {
  return [b[0] - a[0], b[1] - a[1], b[2] - a[2]];
}
var pointIDs = [463, 478, 464, 1169, 477, 462];
//console.log(pointIDs);
var pointXYZ = [
  [5.951940485039203, 0, -2.2214890192281618],
  [5.876896688975295, -0.13547705169471963, -2.4092198479124574],
  [5.9742796273835745, 0.22293478422146445, -2.1491607701178843],
  [5.9742796273835745, -0.22293478422146445, -2.1491607701178843],
  [5.876896688975295, 0.13547705169471963, -2.4092198479124574],
  [6.034465603999274, 0, -1.9864123122226367]
];
//console.log(pointXYZ);

var vectors = [];
for (var i = 1; i < pointIDs.length; i++) {
  vectors.push(vectorAB(pointXYZ[i], pointXYZ[0]));
}
//console.log(vectors);

var normal = math.cross(vectors[0], vectors[1]);

console.log("id", pointIDs[1], "and", pointIDs[2],
  math.dot(normal, math.cross(vectors[0], vectors[1])));
console.log("id", pointIDs[2], "and", pointIDs[3],
  math.dot(normal, math.cross(vectors[1], vectors[2])));
console.log("id", pointIDs[3], "and", pointIDs[4],
  math.dot(normal, math.cross(vectors[2], vectors[3])));
console.log("id", pointIDs[4], "and", pointIDs[5],
  math.dot(normal, math.cross(vectors[3], vectors[4])));

console.log("id", pointIDs[5], "and", pointIDs[4],
  math.dot(normal, math.cross(vectors[4], vectors[3])));
console.log("id", pointIDs[4], "and", pointIDs[2],
  math.dot(normal, math.cross(vectors[3], vectors[1])));
console.log("id", pointIDs[4], "and", pointIDs[3],
  math.dot(normal, math.cross(vectors[3], vectors[2])));
 console.log("id", pointIDs[1], "and", pointIDs[5],
  math.dot(normal, math.cross(vectors[0], vectors[4])));

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