问题描述
使用Three.js,(尽管我相信这与数学更相关),我拥有一组可以创建2D几何图形的2D点.例如正方形,矩形,五边形或自定义2D形状.基于原始的2D形状,我想创建一种方法,以像附加图像一样的方式均匀地向内或向外偏移点.
Using Three.js, (although I believe this is more math related) I have a set of 2D points that can create a 2D geometry. such as square, rectangle, pentagon, or custom 2D shape. Based of the original 2D shape, I would like to create a method to offset the points inward or outward uniformly in such a way like the attached image.
我不知道是否有一种简单的方法可以向内或向外均匀地偏移/增长/缩小2D形状上的所有点(vector3).如果可以的话,我可以将这些点偏移X距离会很酷吗?有点像是说将2D形状上的点向外或向内偏移X距离.
I don't know if there is a simple way to offset/grow/shrink all the points (vector3) uniformly on the 2D shape inward or outward. And if so, it'll be cool if I can offset the points by X distance? Kinda of like saying offset the points on the 2D shape outward or inward by X distance.
不,我不是指从中心点缩放.虽然缩放可能适用于对称形状,但对于非对称形状则无效.
And no, I'm not referring to scaling from a center point. While scaling may work for symmetrical shapes, it won't work when it comes to non-symmetrical shapes.
例如查看图片
谢谢.
推荐答案
您可以阅读论坛线程.
我已经对ProfiledContourGeometry
进行了一些更改并得到了OffsetContour
,所以我把它留在这里,以防万一,如果有帮助的话:)
I've made some changes with ProfiledContourGeometry
and got OffsetContour
, so I leave it here, just in case, what if it helps :)
function OffsetContour(offset, contour) {
let result = [];
offset = new THREE.BufferAttribute(new Float32Array([offset, 0, 0]), 3);
console.log("offset", offset);
for (let i = 0; i < contour.length; i++) {
let v1 = new THREE.Vector2().subVectors(contour[i - 1 < 0 ? contour.length - 1 : i - 1], contour[i]);
let v2 = new THREE.Vector2().subVectors(contour[i + 1 == contour.length ? 0 : i + 1], contour[i]);
let angle = v2.angle() - v1.angle();
let halfAngle = angle * 0.5;
let hA = halfAngle;
let tA = v2.angle() + Math.PI * 0.5;
let shift = Math.tan(hA - Math.PI * 0.5);
let shiftMatrix = new THREE.Matrix4().set(
1, 0, 0, 0,
-shift, 1, 0, 0,
0, 0, 1, 0,
0, 0, 0, 1
);
let tempAngle = tA;
let rotationMatrix = new THREE.Matrix4().set(
Math.cos(tempAngle), -Math.sin(tempAngle), 0, 0,
Math.sin(tempAngle), Math.cos(tempAngle), 0, 0,
0, 0, 1, 0,
0, 0, 0, 1
);
let translationMatrix = new THREE.Matrix4().set(
1, 0, 0, contour[i].x,
0, 1, 0, contour[i].y,
0, 0, 1, 0,
0, 0, 0, 1,
);
let cloneOffset = offset.clone();
console.log("cloneOffset", cloneOffset);
shiftMatrix.applyToBufferAttribute(cloneOffset);
rotationMatrix.applyToBufferAttribute(cloneOffset);
translationMatrix.applyToBufferAttribute(cloneOffset);
result.push(new THREE.Vector2(cloneOffset.getX(0), cloneOffset.getY(0)));
}
return result;
}
随时对其进行修改:)
这篇关于Threejs-如何按距离偏移2d几何图形上的所有点的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!