bezier

Algorithm for deriving control points of a bezier curve from points along that curve?

我的梦境 提交于 2019-12-05 03:44:48
I've been looking for, but obviously not finding, an algorithm that will allow me to plug in a list of x,y coordinates that are known to be along a curve so as to get the 4 control points for a cubic bezier curve spit out. To be more precise, I'm looking for an algorithm that will give me the two control points required to shape the curve while inputting a series of discrete points including the two control points which determine the start and end of the curve. Thanks! Edit: Okay, due to math, an old foe, I need to ask for the bezier curve of best fit to a polynomial function. tfinniga So I

Circle approximations using Bezier curves

孤街醉人 提交于 2019-12-05 02:04:03
问题 I have 2 questions about bezier curves, and using them to approximate portions of circles. Given the unit circle arc (1,0)->(cos(a),sin(a)) where 0 < a < pi/2, will it result in a good approximation of this arc to find the bezier curve's control points p1, p2 by solving the equations imposed by the requirements B(1/3) = (cos(a/3), sin(a/3)) and B(2/3) = (cos(2a/3), sin(2a/3)). (In other words, requiring that the bezier curve go through two evenly spaced points in the arc). If we have an

Adjusting the line to fit our head in imageview

末鹿安然 提交于 2019-12-05 01:31:03
I am developing one application same as HairTryOn all things are done. but problem is display in following image. i want to set hair style as per customer face using blue line as per display in image. i used the following code testVw = [[UIView alloc]initWithFrame:CGRectMake(100,100, 100, 100)]; testVw.backgroundColor = [UIColor clearColor]; [self.view addSubview:testVw]; resizeVw = [[UIImageView alloc]initWithFrame:CGRectMake(testVw.frame.size.width-25, testVw.frame.size.height-25, 25, 25)]; resizeVw.backgroundColor = [UIColor clearColor]; resizeVw.userInteractionEnabled = YES; resizeVw.image

Dashed Curves on Html5 Canvas Bezier

跟風遠走 提交于 2019-12-04 20:35:41
问题 For one of my application I would need to draw a dashed curves on the bezier path in Html5 canvas... The dash' length and gaps in between should be variable... It is achivable in JavaFx, see this link... I would like to achieve same effect using Html5 canvas. I know how to draw dashed straight lines, but not curved lines along the bezier... Though I am not an expert, I know the bezier drawing algorithm, problem I see with this algorithm is, it allows you to identify coordinates on the bezier

How to calculate the nearest point of a line and curve? .. or curve and curve?

时光总嘲笑我的痴心妄想 提交于 2019-12-04 17:49:53
问题 Given the points of a line and a quadratic bezier curve, how do you calculate their nearest point? 回答1: I just wanna give you a few hints, in for the case Q.B.Curve // segment : to get a fast enough computation, i think you should first think about using a kind of 'bounding box' for your algorithm. Say P0 is first point of the Q. B. Curve, P2 the second point, P1 the control point, and P3P4 the segment then : Compute distance from P0, P1, P2 to P3P4 if P0 OR P2 is nearest point --> this is

PHP imagick API中文简介

不问归期 提交于 2019-12-04 15:21:47
PHP imagick API中文简介 imagick 类imagick ::adaptiveblurimage 向图像中添加 adaptive 模糊滤镜imagick ::adaptiveresizeimage 自适应调整图像数据依赖关系imagick ::adaptivesharpenimage自适应锐化图像imagick ::adaptivethresholdimage 基于范围的选择为每个像素的亮度imagick ::addimage 图像列表中添加新图像 imagick 对象.imagick ::addnoiseimage 给图像添加随机噪声imagick ::affinetransformimage变换图像imagick ::animateimages 动画图像或图像imagick ::annotateimage annotates 图像的文本imagick ::appendimages 追加一组图像imagick ::averageimages 平均一组图像imagick ::blackthresholdimage 强制所有的像素低于阈值分为黑色imagick ::blurimage 向图像中添加模糊滤镜imagick ::borderimage 四周带有边框的图像imagick ::charcoalimage模拟一个炭笔绘图imagick ::chopimage

Detect&find intersection ray vs. cubic bezier triangle

╄→гoц情女王★ 提交于 2019-12-04 14:53:30
While writing a model editor, besides enabling raytracing I can think about couple of operations where I'd like to find an very good approximation about the intersection point between a ray and a triangular bezier patch. How to do this? I know couple of ways but likely there's better ones. Exact use-cases: I might want to use one bezier triangle patch as a reference surface for drawing detailed shapes with mouse. I might too want to pinpoint a splitting-point from such patch. If there's C source code for it, I might like to see that too. Perhaps even use it instead of rolling my own code. I'd

Rounded Corners Rect in Cocos 2d-x with Bezier

跟風遠走 提交于 2019-12-04 14:39:47
Is it possible to draw a Rect with rounded corners using a DrawNode object? I think that something is possible using Bezier curves, but I have made some tries and I think I can't handle it. Looking at API I've found only these 2 functions: drawQuadBezier (const Vec2 &origin, const Vec2 &control, const Vec2 &destination, unsigned int segments, const Color4F &color) drawCubicBezier (const Vec2 &origin, const Vec2 &control1, const Vec2 &control2, const Vec2 &destination, unsigned int segments, const Color4F &color) [Modified after answer] I have applied the answer in Cocos2dx , maybe somebody

Calculate the arclength, curve length of a cubic bezier curve. Why is not working?

两盒软妹~` 提交于 2019-12-04 13:49:05
I'm calculating the arclength (length of a cubic bezier curve) with this algorithm function getArcLength(path) { var STEPS = 1000; // > precision var t = 1 / STEPS; var aX=0; var aY=0; var bX=0, bY=0; var dX=0, dY=0; var dS = 0; var sumArc = 0; var j = 0; for (var i=0; i<STEPS; j = j + t) { aX = bezierPoint(j, path[0], path[2], path[4], path[6]); aY = bezierPoint(j, path[1], path[3], path[5], path[7]); dX = aX - bX; dY = aY - bY; // deltaS. Pitagora dS = Math.sqrt((dX * dX) + (dY * dY)); sumArc = sumArc + dS; bX = aX; bY = aY; i++; } return sumArc; } But what I get is something like 915. But

How to generate a “thick” bezier curve?

ぐ巨炮叔叔 提交于 2019-12-04 12:59:20
I'm looking for a way to generate a polygon programatically by "thickening" a Bezier curve. Something like this: My initial idea was to find the normals in the line, and generate the polygon from them: But the problem is that the normals can cross each other in steep curves, like this: Are there any formulas or algorithms that generate a polygon from a bezier curve? I couldn't find any information on the internet, but perhaps I'm searching using the wrong words... Yves Daoust If you want a constant thickness, this is called an offset curve and your idea of using normals is correct. This indeed