Only the tilted Cup points

Strange question, but I've had for a while. In my work, there are times where I needed to convert a shape to a series of lines, keeping the curved parts but the break in the line where it's angles. Does anyone know one trick to this other than to manually select them?

kylwell, maybe this script will do.

/**
 * ai.jsx (c)MaratShagiev [email protected] 27.08.2016
 *
 * cutPathByCornerPoints_v1.0
 *
 * Using:
 * * Select a single path item;
 * * Run the script
 *
 * Great destination:
 * * cutting path item into pieces;
 * * the point of cutting path item - corner points;
 * * cut closed and open paths;
 *
 * What is really going on here:
 * * source circuit is used as a model;
 * * on top of the original path created new paths;
 * * the original path is deleted;
 */
//@target illustrator
cutByCornPnts (selection[0]);
function cutByCornPnts (pth) {
  if (_cut (pth)) pth.remove ();
  function _cut (pth) {
    var remPth = false;
    if (pth.closed) {
      for (var i = 0; i < pth.pathPoints.length; i++) {
        var pnt = pth.pathPoints[i];
        if (!__isCornerPoint (pnt)) continue;
        __addPathByTmpl (pth, i);
        remPth = true;
      }
    } else {
      for (var i = 0; i < pth.pathPoints.length; i++) {
        var pnt = pth.pathPoints[i];
        i       = __addPathByTmpl (pth, i);
        remPth  = true;
      }
    }
    return remPth;
    function __addPathByTmpl (pathTmpl, iStartPnt) {
      var doc       = activeDocument,
          j, pntTmpl, pthDupl, pnt,
          endPthAdd = 0,
          pnts      = [pathTmpl.pathPoints [iStartPnt]];
      if (pathTmpl.closed) endPthAdd = 1;
      try {
        for (j = iStartPnt + 1; j < pathTmpl.pathPoints.length + endPthAdd; j++) {
          pntTmpl = pathTmpl.pathPoints[j];
          if (__isCornerPoint (pntTmpl)) {
            pnts.push (pntTmpl);
            break;
          }
          pntTmpl = pathTmpl.pathPoints[j];
          pnts.push (pntTmpl);
        }
      } catch (e) {
        for (j = 0; j < iStartPnt + endPthAdd; j++) {
          pntTmpl = pathTmpl.pathPoints[j];
          if (__isCornerPoint (pntTmpl)) {
            pnts.push (pntTmpl);
            break;
          }
          pntTmpl = pathTmpl.pathPoints[j];
          pnts.push (pntTmpl);
        }
      }
      if (pnts.length > 1) {
        pthDupl = doc.pathItems.add ();
        for (var i = 0; i < pnts.length; i++) {
          var pnt          = pnts[i];
          var p            = pthDupl.pathPoints.add ();
          p.anchor         = [pnt.anchor[0], pnt.anchor[1]];
          p.leftDirection  = [pnt.leftDirection[0], pnt.leftDirection[1]];
          p.rightDirection = [pnt.rightDirection[0], pnt.rightDirection[1]];
          p.pointType      = pnt.pointType;
        }
      }
      iStartPnt = j; // for closed paths
      return --j; // for open paths
    }
    /**
     * Unfortunately the Pen Tool creates all the points as smooth.
     * Even if in fact they are corner.
     * One way to solve this problem -
     * check all the points with this helper.
     *
     * @param {Object} pnt - object of class PathPoint
     * @return {Boolean}
     * */
    function __isCornerPoint (pnt) {
      if (pnt.pointType == PointType.CORNER) return true;
      var x_1, x_2, x_3,
          y_1, y_2, y_3,
          prec    = 100000,
          precLow = 10;
      x_1 = Math.round (pnt.leftDirection [0] * prec) / prec; // x
      y_1 = Math.round (pnt.leftDirection [1] * prec) / prec; // y
      x_2 = Math.round (pnt.rightDirection [0] * prec) / prec;
      y_2 = Math.round (pnt.rightDirection [1] * prec) / prec;
      x_3 = Math.round (pnt.anchor [0] * prec) / prec;
      y_3 = Math.round (pnt.anchor [1] * prec) / prec;
      if ((x_1 == x_2 && x_2 == x_3) && (y_1 == y_2 && y_2 == y_3)) return true;
      if ((x_1 == x_2) && (y_1 == y_2)) return true; // directions is equals
      if ((x_1 == x_3) && (y_1 == y_3)) return true; // left directon equals with anchor
      if ((x_2 == x_3) && (y_2 == y_3)) return true; // right direction equals with anchor
      // The equation of a straight line
      // points 1, 2, 3 are not equal and don't lie on one straight line
      if (
        Math.ceil (((x_3 - x_1) * (y_2 - y_1)) * precLow) / precLow !=
        Math.ceil (((x_2 - x_1) * (y_3 - y_1)) * precLow) / precLow
      ) return true;
      return false;
    }
  }
}

Tags: Illustrator

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Only the tilted Cup points

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