Threejs - How to offset all points on a 2d geometry by distance

问题

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.

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.

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.

see image for example

Thanks in advance.

回答1:

You can read that forum thread.

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;
  }

Feel free to modify it :)

来源：https://stackoverflow.com/questions/50957349/threejs-how-to-offset-all-points-on-a-2d-geometry-by-distance