问生成bezier曲线上的移动

EN

lines = []
x=curve.getX(0), y=curve.getY(0), nx, ny
step=..., interval=1/step, t=step
while(t<=1) {
  nx = curve.getX(t)
  ny = curve.getY(t)
  lines.push( new line(x,y,nx,ny)
  x = nx
  y = ny
  t += interval
}

function arcLength(t) {
  // true fact: computing the arc length of a bezier curve is not
  // a thing you want to end up implementing yourself. It's not hard,
  // but getting to a point where you undestand *why* it's not hard is
  // is certainly time consuming, and depending on how much your brain
  // is unwilling to just take maths at face value, definitely hard.
  // Use someone else's implementation, like this one:
  // https://github.com/Pomax/bezierjs/blob/gh-pages/lib/bezier.js#L87
}

Curve.draw = function() { 
  if (!this.curveLUT) {
    // form a LUT however you like. The following demonstration 
    // code uses something similar to the above flattening:
    for(i=0;i<LUT.length;i++) {
      t = i / (LUT.length-1)
      x = curve.getX(t)
      y = curve.getY(t)
      LUT.push({x:x, y:y, dist: arclength(t)})
    }    
  }
  this.curveLUT.foreach(point -> point.draw())
}

speed = ...
currentPos = SomeLUTindex
if (currentPos < LUT.length) {
  currentDist = LUT[currentPos].dist
  nextDist = currentDist + speed
  nextPos = binarySearch(LUT.slice(currentPos), "dist", nextDist)
}

binarySearch(List, property, target) {
  midIdx = (int) List.length/2
  mid = List[midIdx]
  curr = mid[property]
  if(curr === target) return midIdx
  if(curr > target) return binarySearch(List.slice(0,midIdx), property, target)
  if(curr < target) return midIdx + binarySearch(List.slice(midIdx), property, target)
}