对三次贝塞尔曲线过点平滑中尖角和交叉现象的优化
2020/10/10 5:03:51
本文主要是介绍对三次贝塞尔曲线过点平滑中尖角和交叉现象的优化,对大家解决编程问题具有一定的参考价值,需要的程序猿们随着小编来一起学习吧!
目前在做等值线等值面相关的功能,用户可拖拽控制点修改等值线,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,所以先使用道格拉斯-普克算法(Douglas–Peucker)进行等值线抽稀,再将抽稀后的控制点使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,有人做了很详细的示意图,推荐大家看下贝塞尔曲线算法之JS获取点
可以了解到贝赛尔曲线算法平滑得到的曲线是经过起始点的,同时二阶算法需要三个点,三阶算法需要四个点,四阶算法需要五个点,以此类推。
一般的来说,三阶贝塞尔曲线就已经够用了,而且效果还不错,所以我选择了三次贝塞尔曲线平滑算法来进行控制点的平滑处理。
贝塞尔曲线平滑后的等值线是基本不经过控制点的,考虑到用户操作逻辑,以及点线关系(我的控制点是等值线抽稀得到的,所以等值线是经过控制点的),所以采用三次贝塞尔曲线过点平滑算法来进行控制点的平滑处理。
过点平滑的原理就是以相邻两个控制点为起始点,然后往起始点中间插入其他过程点(不是在起始点直线上选择点),这样平滑得到的曲线是经过起始点的,而曲线如何平滑是由插入的点来控制的,三次贝塞尔曲线需要四个点,那就需要在起始点中间插入两个点。
大致思路就是,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,这样就保证了连接处的光滑。而贝塞尔曲线本身是光滑的,所以就把这些原始点用光滑曲线连起来了。具体代码及示意图如下:
代码:
function createCurve(originPoint, option){ //控制点收缩系数 ,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]方向收缩 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]方向收缩 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++版的可以看穿过已知点画平滑曲线(3次贝塞尔曲线)
然而事情到这里并没有结束,还有坑需要填,直接使用该算法进行平滑,在实际应用中发现控制点距离近的话,平滑的曲线会有尖角或交叉现象。
比如这样的
还有这样的
这是因为插入的两个新控制点和起始控制点位置特殊,平滑后产生尖角或交叉,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,然后使用二次贝赛尔曲线算法进行平滑。新的效果示意图如下所示:
优化前与优化后的效果对比如下:左边为优化前的平滑效果,右边是优化后的平滑效果:
优化后的代码:
function createCurve(originPoint, option){ //控制点收缩系数 ,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]方向收缩 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]方向收缩 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]; //斜率交叉相乘 k1 = (y4 - y3) / (x4 - x3) ... k2 = (y2 - y1) / (x2 - x1) //k1 k2 同乘 (x4 - x3) * (x2 - x1) 得到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]) ) }
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