bezier

How to draw a bezier curve with variable thickness on an HTML canvas?

由 十年热恋 提交于 2019-12-11 03:35:02
问题 I want to make a visualization with bezier curves connecting boxes. More important edges should be thicker. Every box has one output but many inputs. Therefore, I want to keep the thickness of the incomit edges constant (to save space) and alter only the thickness of the outgoing edges (of which there is only one per box). This is why I want to draw bezier cureves with different thickness at each end. They shall be rendered on an HTML canvas element. I know context.bezierCurveTo() but that

html5 draw gaussian function using bezierCurveTo

由 流过昼夜 提交于 2019-12-11 01:08:22
问题 I have been trying to draw gaussin-like function using bezierCurveTo find the code below <canvas id="thisCan" width="0px" height="0px" style="border:1px solid #d3d3d3;"> Your browser does not support the HTML5 canvas tag. </canvas> <script> (function() { var // Obtain a reference to the canvas element // using its id. htmlCanvas = document.getElementById('thisCan'), // Obtain a graphics context on the // canvas element for drawing. ctx = htmlCanvas.getContext('2d'); var width = 0; var height

joining two bezier curves

由 纵然是瞬间 提交于 2019-12-11 01:04:43
问题 i have two bezier curves placed at a distance apart in space. curve 1 has control points A0, A1,A2, A3. A0 and A3 lie on curve and are its end points Curve 2 has control points C0,C1, C2, C3 . C0 and C3 lie on curve.and are its end points i want to join the two curves A and C with an intermediate bezier curve B. the intermediate Curve B has control points A3 and C0 which lie on the curve and are its end points. the intermediate control points B1 and B2 are unknown to me. also the joining

Verify if a point is part of quadratic Bezier curve in Java

由 不问归期 提交于 2019-12-10 23:37:29
问题 I want to verify if a Point is part of a quadratic Bezier curve defined by points p0, p1 and p2.. This is my function to obtain a Point in the curve with a certain t: public static final Point quadratic (Point p0, Point p1, Point p2, double t) { double x = Math.pow(1-t, 2) * p0.x + 2 * (1-t) * t * p1.x + Math.pow(t, 2) * p2.x; double y = Math.pow(1-t, 2) * p0.y + 2 * (1-t) * t * p1.y + Math.pow(t, 2) * p2.y; return new Point((int)x, (int)y); } Considering that the point B(t) in a quadratic

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

由 房东的猫 提交于 2019-12-10 03:36:29
问题 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

Calculate middle point of Bezier Curve

由 血红的双手。 提交于 2019-12-09 16:53:23
问题 I have a function to draw Bezier Curve through three points. I have already 2 points (start and end) - A and B. How do I calculate middle point between those two points as middle point would be always a little higher or lower than linear function of those two points. Example : Any formulas, ideas would be great! 回答1: I think this is what you're looking for: http://blog.sklambert.com/finding-the-control-points-of-a-bezier-curve/ It goes into detail on calculating the various points on a Bezier

Finding Y given X on a Cubic Bezier Curve?

由 给你一囗甜甜゛ 提交于 2019-12-09 14:48:48
问题 I need a method that allows me to find the Y-coordinate on a Cubic Bezier Curve, given an x-coordinate. I've come across lots of places telling me to treat it as a cubic function then attempt to find the roots, which I understand. HOWEVER the equation for a Cubic Bezier curve is (for x-coords): X(t) = (1-t)^3 * X0 + 3*(1-t)^2 * t * X1 + 3*(1-t) * t^2 * X2 + t^3 * X3 What confuses me is the addition of the (1-t) values. For instance, if I fill in the X values with some random numbers: 400 = (1

BezierPath Rotation in a UIView

由 丶灬走出姿态 提交于 2019-12-09 13:53:16
问题 I am drawing a BezierPath on Touch event. Now I have to rotate that Bezier Path on the same location using Gesture Method. But problem is, after rotation its position become change. Its look like the following image.. How can I fix this? The Upper image is the original image. Share your ideas with me.. Thanks in advance 回答1: Check this in Apple documentation. applyTransform: Transforms all points in the path using the specified affine transform matrix. - (void)applyTransform:

Finding a point on a Bézier curve when given the distance from the start point?

由 五迷三道 提交于 2019-12-09 10:38:08
问题 I created a 4 point Bézier curve, and a distance. Starting at the start point, how do I find the x,y coordinates of a point which is that distance away from the start point? I've looked at the other examples, and from what I can tell, they approximate the values by dividing the curve into several thousand points, then finding the nearest point. This will not work for me. For what I'm doing, I'd like to be accurate to only two decimal places. Below is a simple form of what I have to create my

PaintCode - move object on the path

由 别来无恙 提交于 2019-12-09 07:41:30
I would like draw a curved line and attach an object to it. Is it possible to create fraction (from 0.0 to 1.0) which makes move my object on the path? When fraction is 0 then object is on the beginning, when 0.5 is on half way and finally when is on 1.0 it is at the end. Of course i want a curved path, not a straight line :) Is it possible to do in PaintCode? If you need it only as a progress bar , it is possible in PaintCode. The trick is to use dashed stroke with very large Gap and then just change the Dash . Then just attach a Variable and you are done. Edit: Regarding the discussion under
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