bezier | 易学教程

How to draw a continuous curved line from 3 given points at a time

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I am trying to draw a continuous curved line in flash. There are many methods but none of the ones I have found so far quite fit my requirements. First of all, I want to use the flash graphic api's curveTo() method. I DO NOT want to simulate a curve with hundreds of calls to lineTo() per curved line segment . It is my experience and understanding that line segments are processor heavy. Flash's quadratic bezier curve should take less CPU power. Please challenge this assumption if you think I am wrong. I also do not want to use a pre-made method that takes the entire line as an argument (eg mx

Get points from a UIBezierPath

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I drew the above BezierPath by doing: // location is where the user touches screen. // location will be the maximum of the graph CGPoint origin = CGPointMake(xStart, 620.0); CGPoint endpt = CGPointMake(xEnd, 620.0); CGPoint midpt1 = midPointForPoints(origin, location); CGPoint midpt2 = midPointForPoints(location, endpt); UIBezierPath *path = [UIBezierPath bezierPath]; [path moveToPoint:origin]; [path addQuadCurveToPoint:location controlPoint:CGPointMake(midpt1.x, midpt1.y+50)]; [path addQuadCurveToPoint:endpt controlPoint:CGPointMake(midpt2.x, midpt2.y+50)]; [shapeLayer setPath:path.CGPath];

How to Solve Rendering Artifact in Blinn/Loop's Resolution Independent Curve Rendering?

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In implementing Blinn/Loop's algorithm on curve rendering, I realize there is a special case on Loop Curve Type. As described in their paper (subsection 4.4, page 6-7), they said the curve should be divided into two but I'm really confused how to obtain the intersection point. Here's my rendering result: As stated in the paper, this artifact occurs when either td/sd or te/se lie in between value [0, 1]. My source code: ... case CURVE_TYPE_LOOP: td = d2 + sqrt(4.0 * d1 * d3 - 3.0 * d2 *d2); sd = 2.0 * d1; te = d2 - sqrt(4.0 * d1 * d3 - 3.0 * d2 * d2); se = 2.0 * d1; if((td / sd > 0.0 && td/ sd

How can I get a cubic bezier curve closest to given points?

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Given n points: p0, p1, p2, ..., pn; How can I get the point c1, c2 so that the cubic bezier curve defined by p0, c1, c2, pn closest to the given points? I tried least square method. I wrote this after I read the pdf document in http://www.mathworks.com/matlabcentral/fileexchange/15542-cubic-bezier-least-square-fitting . But I can't find a good t(i) function. using System; using System.Collections.Generic; using System.Linq; using System.Windows; namespace BezierFitting { class CubicBezierFittingCalculator { private List<Point> data; public CubicBezierFittingCalculator(List<Point> data) { this

OpenGL Coordinates from Bezier Curves

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问题 Basically, I need to get all the coordinates drawn from a Bezier curve implementation in OpenGL. Specifically, I need the coordinates to move a sphere object (baseball) in my scene along a curved trajectory path. This is what I use to draw my curve: GL2 gl = drawable.getGL().getGL2(); float ctrlpoints[][] = new float[][]{ {0.0f, 0.0f, 60f}, {0.0f, 3.0f, 45.0f}, {0.0f, 2.0f, 15.0f}, {0.0f, 1.0f, 0f}}; FloatBuffer ctrlpointBuf = FloatBuffer.allocate(ctrlpoints[0].length * ctrlpoints.length);

Efficient Line Smoothing and/or Simplification

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I am creating a painting application in Actionscript (although my question is not Actionscript related). The basic idea is to start painting when the mouse is pressed and tracking the mouse movements. What I want to achieve is: reduce mouse "noise" and create more smoother looking lines. Right now, ( 1 ) is problematic because I get thousands of mouse movements within a few seconds. Due to ( 1 ) the line can look jaggy. What current idea: when the user finishes drawing the line, I store all movements in an array and reduce them (median threshold) and then use a spline algorithm to recreate a

Divide bezier curve into two equal halves

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I have the bezier curves between 2 points. I'd like to cut all curves into two equal half. One of my idea is if I can control 't' value I'll draw 2 curves by t = [0,0.5] and t = [0.5,1] but I don't know how. Below is my code. I won't mind any other idea or suggestion <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>D3 test</title> <script src="http://d3js.org/d3.v3.min.js"></script> <script> var

QT下过多点的曲线绘制

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绘制过多点的曲线意义重大。但通过试验，QT的PainterPath不是很如意。当多段曲线围成一个区域时，PainterPath内并不包含该区域的所有面积，只包含曲线和其弦构成的面积。为了解决这一问题，采用如下方法： 1. 生成自己的bezier曲线点集 2. 将多个bezier曲线头尾相联，形成整个polygon的点集 3. 将这个polygon放入一个PainterPath，然后绘制； 4. 这个PainterPath返回留待下次使用。下面是代码： 1. 头文件graphic.h #ifndef GRAPHIC_H #define GRAPHIC_H #include <QPainter> #include <QPoint> #include <QColor> #include <QVector> //step是步长，即t每次的递增量，traceSet返回本曲线的所有生成点 void getBezier3(const QPointF& startPos, const QPointF& controlPos1, const QPointF& controlPos2, const QPointF& endPos, const double step, QVector<QPointF>& traceSet); //画一个多边形的外接曲线，points是多边形的顶点集合

Drawing a Topographical Map

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I've been working on a visualization project for 2-dimensional continuous data. It's the kind of thing you could use to study elevation data or temperature patterns on a 2D map. At its core, it's really a way of flattening 3-dimensions into two-dimensions-plus-color. In my particular field of study, I'm not actually working with geographical elevation data, but it's a good metaphor, so I'll stick with it throughout this post. Anyhow, at this point, I have a "continuous color" renderer that I'm very pleased with: The gradient is the standard color-wheel, where red pixels indicate coordinates

Cubic bezier curves - get Y for given X

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问题 I have a cubic bezier curve where the first and last points are given (namely P0(0,0) and P3(1,1)). The other two points are defined like this: cubic-bezier(0.25, 0.1, 0.25, 1.0) (X1, Y1, X2, Y2, also those values must not be smaller or larger than 0 or 1, respectively) Now what would I have to do to get the Y coordinate for a given X, assuming there's only one? (I know that under certain circumstances there can be multiple values, but let's just put them aside. I'm not doing rocket science