目前在做等值线等值面相干的性能，用户可拖拽控制点批改等值线，再用等值线生成等值面。因为初始的等值线点数据太多，不利于用户操作，所以先应用道格拉斯-普克算法(Douglas–Peucker)进行等值线抽稀，再将抽稀后的控制点应用贝塞尔曲线算法进行平滑。
对于贝塞尔曲线算法的平滑过程，有人做了很具体的示意图，举荐大家看下贝塞尔曲线算法之JS获取点
能够理解到贝赛尔曲线算法平滑失去的曲线是通过起始点的，同时二阶算法须要三个点，三阶算法须要四个点，四阶算法须要五个点，以此类推。
个别的来说，三阶贝塞尔曲线就曾经够用了，而且成果还不错，所以我抉择了三次贝塞尔曲线平滑算法来进行控制点的平滑解决。
贝塞尔曲线平滑后的等值线是根本不通过控制点的，思考到用户操作逻辑，以及点线关系（我的控制点是等值线抽稀失去的，所以等值线是通过控制点的），所以采纳三次贝塞尔曲线过点平滑算法来进行控制点的平滑解决。
过点平滑的原理就是以相邻两个控制点为起始点，而后往起始点两头插入其余过程点（不是在起始点直线上抉择点），这样平滑失去的曲线是通过起始点的，而曲线如何平滑是由插入的点来管制的，三次贝塞尔曲线须要四个点，那就须要在起始点两头插入两个点。
大抵思路就是，先算出相邻原始点的中点，在把相邻中点连成的线段平移到对应的原始点，以平移后的中点作为控制点，相邻原始点为起始点画贝塞尔曲线，这样就保障了连接处的润滑。而贝塞尔曲线自身是润滑的，所以就把这些原始点用润滑曲线连起来了。具体代码及示意图如下：
代码：
function createCurve(originPoint, option){ //控制点膨胀系数 ，经调试0.6较好 let scale = option.tension || 0.6; //平滑插值插入的最大点数 let maxpoints = option.pointsPerSeg let originCount = originPoint.length let curvePoint = [] let midpoints = [] //生成中点 for(let i = 0 ;i < originCount - 1 ; i++){ midpoints.push([ (originPoint[i][0] + originPoint[i + 1][0])/2.0, (originPoint[i][1] + originPoint[i + 1][1])/2.0 ]) } //平移中点 let extrapoints = [] for(let i = 1 ;i < originCount - 1 ; i++){ let backi = i - 1; let midinmid = [ (midpoints[i][0] + midpoints[backi][0])/2.0, (midpoints[i][1] + midpoints[backi][1])/2.0 ] let offsetx = originPoint[i][0] - midinmid[0]; let offsety = originPoint[i][1] - midinmid[1]; let extraindex = 2 * i; extrapoints[extraindex] = [ midpoints[backi][0] + offsetx, midpoints[backi][1] + offsety ] //朝 originPoint[i]方向膨胀 let addx = (extrapoints[extraindex][0] - originPoint[i][0]) * scale; let addy = (extrapoints[extraindex][1] - originPoint[i][1]) * scale; extrapoints[extraindex] = [ originPoint[i][0] + addx, originPoint[i][1] + addy ] let extranexti = extraindex + 1; extrapoints[extranexti] = [ midpoints[i][0] + offsetx, midpoints[i][1] + offsety ] //朝 originPoint[i]方向膨胀 addx = (extrapoints[extranexti][0] - originPoint[i][0]) * scale; addy = (extrapoints[extranexti][1] - originPoint[i][1]) * scale; extrapoints[extranexti] = [ originPoint[i][0] + addx, originPoint[i][1] + addy ] } let controlPoint = [] //生成4控制点，产生贝塞尔曲线 for(let i = 1 ;i < originCount - 2 ; i++){ controlPoint[0] = originPoint[i]; let extraindex = 2 * i; controlPoint[1] = extrapoints[extraindex + 1]; let extranexti = extraindex + 2; controlPoint[2] = extrapoints[extranexti]; let nexti = i + 1; controlPoint[3] = originPoint[nexti]; for(let n = maxpoints; n >= 0; n--){ //存入曲线点 curvePoint.push( bezier3func(n / maxpoints, controlPoint) ); } } return curvePoint} //三次贝塞尔曲线 function bezier3func(uu, controlP){ let partX0 = controlP[0][0] * uu * uu * uu; let partX1 = 3 * controlP[1][0] * uu * uu * (1 - uu); let partX2 = 3 * controlP[2][0] * uu * (1 - uu) * (1 - uu); let partX3 = controlP[3][0] * (1 - uu) * (1 - uu) * (1 - uu); let partX = partX0 + partX1 + partX2 + partX3; let partY0 = controlP[0][1] * uu * uu * uu; let partY1 = 3 * controlP[1][1] * uu * uu * (1 - uu); let partY2 = 3 * controlP[2][1] * uu * (1 - uu) * (1 - uu); let partY3 = controlP[3][1] * (1 - uu) * (1 - uu) * (1 - uu); let partY = partY0 + partY1 + partY2 + partY3; return [partX, partY]}c++版的能够看穿过已知点画平滑曲线（3次贝塞尔曲线）
然而事件到这里并没有完结，还有坑须要填，间接应用该算法进行平滑，在理论利用中发现控制点间隔近的话，平滑的曲线会有尖角或穿插景象。
比方这样的
还有这样的
这是因为插入的两个新控制点和起始控制点地位非凡，平滑后产生尖角或穿插，如下所示：
优化思路是判断四个点的关系，获取直线控制点1-新控制点1与直线控制点2-新控制点2之间的交点（如果有的话），如果交点在线段控制点1-点1中时，用交点替换点1，线段控制点2-点2同理。或者以控制点1、交点、控制点2三点，而后应用二次贝赛尔曲线算法进行平滑。新的成果示意图如下所示：
优化前与优化后的成果比照如下：右边为优化前的平滑成果，左边是优化后的平滑成果：
优化后的代码：
function createCurve(originPoint, option){ //控制点膨胀系数 ，经调试0.6较好 let scale = option.tension || 0.6; //平滑插值插入的最大点数 let maxpoints = option.pointsPerSeg let originCount = originPoint.length let curvePoint = [] let midpoints = [] //生成中点 for(let i = 0 ;i < originCount - 1 ; i++){ midpoints.push([ (originPoint[i][0] + originPoint[i + 1][0])/2.0, (originPoint[i][1] + originPoint[i + 1][1])/2.0 ]) } //平移中点 let extrapoints = [] for(let i = 1 ;i < originCount - 1 ; i++){ let backi = i - 1; let midinmid = [ (midpoints[i][0] + midpoints[backi][0])/2.0, (midpoints[i][1] + midpoints[backi][1])/2.0 ] let offsetx = originPoint[i][0] - midinmid[0]; let offsety = originPoint[i][1] - midinmid[1]; let extraindex = 2 * i; extrapoints[extraindex] = [ midpoints[backi][0] + offsetx, midpoints[backi][1] + offsety ] //朝 originPoint[i]方向膨胀 let addx = (extrapoints[extraindex][0] - originPoint[i][0]) * scale; let addy = (extrapoints[extraindex][1] - originPoint[i][1]) * scale; extrapoints[extraindex] = [ originPoint[i][0] + addx, originPoint[i][1] + addy ] let extranexti = extraindex + 1; extrapoints[extranexti] = [ midpoints[i][0] + offsetx, midpoints[i][1] + offsety ] //朝 originPoint[i]方向膨胀 addx = (extrapoints[extranexti][0] - originPoint[i][0]) * scale; addy = (extrapoints[extranexti][1] - originPoint[i][1]) * scale; extrapoints[extranexti] = [ originPoint[i][0] + addx, originPoint[i][1] + addy ] } let controlPoint = [] //生成4控制点，产生贝塞尔曲线 for(let i = 1 ;i < originCount - 2 ; i++){ controlPoint[0] = originPoint[i]; let extraindex = 2 * i; controlPoint[1] = extrapoints[extraindex + 1]; let extranexti = extraindex + 2; controlPoint[2] = extrapoints[extranexti]; let nexti = i + 1; controlPoint[3] = originPoint[nexti]; let fn = bezier3func; let cp = intersects(controlPoint.slice(0, 2), controlPoint.slice(-2)) if(cp && isContains(controlPoint[0], controlPoint[1], cp)){ controlPoint[1] = cp } if(cp && isContains(controlPoint[2], controlPoint[3], cp)){ controlPoint[2] = cp } if(controlPoint[1][0] == controlPoint[2][0] && controlPoint[1][1] == controlPoint[2][1]){ fn = bezier2func controlPoint.splice(1, 1) } for(var n = maxpoints; n >= 0; n--){ //存入曲线点 curvePoint.push( fn(n / maxpoints, controlPoint) ); } } return curvePoint} //三次贝塞尔曲线 function bezier3func(uu, controlP){ let partX0 = controlP[0][0] * uu * uu * uu; let partX1 = 3 * controlP[1][0] * uu * uu * (1 - uu); let partX2 = 3 * controlP[2][0] * uu * (1 - uu) * (1 - uu); let partX3 = controlP[3][0] * (1 - uu) * (1 - uu) * (1 - uu); let partX = partX0 + partX1 + partX2 + partX3; let partY0 = controlP[0][1] * uu * uu * uu; let partY1 = 3 * controlP[1][1] * uu * uu * (1 - uu); let partY2 = 3 * controlP[2][1] * uu * (1 - uu) * (1 - uu); let partY3 = controlP[3][1] * (1 - uu) * (1 - uu) * (1 - uu); let partY = partY0 + partY1 + partY2 + partY3; return [partX, partY]} //二次贝塞尔曲线 function bezier2func(uu, controlP){ let partX0 = controlP[0][0] * uu * uu; let partX1 = 2 * controlP[1][0] * uu * (1 - uu); let partX2 = controlP[2][0] * (1 - uu) * (1 - uu); let partX = partX0 + partX1 + partX2; let partY0 = controlP[0][1] * uu * uu; let partY1 = 2 * controlP[1][1] * uu * (1 - uu); let partY2 = controlP[2][1] * (1 - uu) * (1 - uu); let partY = partY0 + partY1 + partY2; return [partX, partY]} /** * Find a point that intersects LineStrings with two coordinates each * 找到一个点，该点与每个线串有两个坐标相交 */function intersects(coords1, coords2) { if (coords1.length !== 2) { throw new Error("<intersects> line1 must only contain 2 coordinates"); } if (coords2.length !== 2) { throw new Error("<intersects> line2 must only contain 2 coordinates"); } const x1 = coords1[0][0]; const y1 = coords1[0][1]; const x2 = coords1[1][0]; const y2 = coords1[1][1]; const x3 = coords2[0][0]; const y3 = coords2[0][1]; const x4 = coords2[1][0]; const y4 = coords2[1][1]; //斜率穿插相乘 k1 = (y4 - y3) / (x4 - x3) ... k2 = (y2 - y1) / (x2 - x1) //k1 k2 同乘 (x4 - x3) * (x2 - x1) 失去denom const denom = ((y4 - y3) * (x2 - x1)) - ((x4 - x3) * (y2 - y1)); const numeA = ((x4 - x3) * (y1 - y3)) - ((y4 - y3) * (x1 - x3)); const numeB = ((x2 - x1) * (y1 - y3)) - ((y2 - y1) * (x1 - x3)); if (denom === 0) { //斜率一样，平行线 return null; } const uA = numeA / denom; const uB = numeB / denom; const x = x1 + (uA * (x2 - x1)); const y = y1 + (uA * (y2 - y1)); return [x, y];}function isContains(sp, ep, p) { return ( (p[0] > ep[0] && p[0] < sp[0]) || (p[0] > sp[0] && p[0] < ep[0]) ) && ( (p[1] > ep[1] && p[1] < sp[1]) || (p[1] > sp[1] && p[1] < ep[1]) )}