bezier

问题 What is the best way to approximate a cubic Bezier curve? Ideally I would want a function y(x) which would give the exact y value for any given x, but this would involve solving a cubic equation for every x value, which is too slow for my needs, and there may be numerical stability issues as well with this approach. Would this be a good solution? 回答1: Just solve the cubic. If you're talking about Bezier plane curves, where x(t) and y(t) are cubic polynomials, then y(x) might be undefined or

问题 I have bezier curve with about 48 ctrl points and I want to transform it to many cubic bezier curves... Any algorthim , math theory or just a link may help :) ?? 回答1: Math theory: you cannot do this. An nth order Bezier curve cannot be represented as any number of (n-1)th order Bezier curves, because the curvatures cannot be faithfully represented. You can approximate it, but you won't get an identical result. Practice: you can cut up your 48th order curve into sections of simple curve, where

问题 I have this graph: I'm using linearGradient to color the areas between the lines. I'm calculating the cutting point's x, and then calculating the offset for addColorStop(offset, color) of the gradient. When setting the tension property to 0 , (which means straight lines) the graph gets colored as expected. Problem is, that when I set tension to be 0.5 I get the following result: The reason for that is that the tension is calculated by Bezier Curve, which changes the equation I use to

问题 Here is my code for free hand drawing. But when i draw the path, previous path is disappeared. I am not able to figure it out why is it happening so. Can any body help me. Here is my code. - (void)drawRect:(CGRect)rect { for (NSMutableDictionary *dictionary in pathArray) { UIBezierPath *_path = [dict objectForKey:@"Path"]; UIColor *_colors = [dict objectForKey:@"Colors"]; [_colors setStroke]; _path.lineCapStyle = kCGLineCapRound; [_path stroke]; } } -(void)touchesBegan:(NSSet *)touches

问题 I am building a small Silverlight application. In my application I need to draw lines, similar to what's shown in the attached image. I understood that the best way to draw the arched connecting lines (green ones) would be with Bezier curves. My question is, how do I animate the drawing of the lines (getting them to start from a starting (X,Y) coordinate, to the target ones)? Attached Image: 回答1: I've spent a bit of time playing with this and it is possible. The trick is that you don't

问题 Closed. This question is off-topic. It is not currently accepting answers. Want to improve this question? Update the question so it's on-topic for Stack Overflow. Closed 5 years ago . I just watched a very impressive presentation from Siggraph 2012: http://nvidia.fullviewmedia.com/siggraph2012/ondemand/SS106.html My question is, this being a proprietary Nvidia extension, what are the other possibilities to quickly renderer Bezier paths on GPU? Alternatively, is there any hope this will end-up

问题 I have a quadratic curve drawn on html canvas using context.quadraticCurveTo(controlX, controlY, endX, endY); . I have the control-point and the starting and end points, which are not necessarily level with each other horizontally. How can I find the centre point on the curve using these parameters? Actually I want to put a div tag on this center point. Is there any equation solving involved in this process? 回答1: quadraticCurveTo draws a quadratic Bézier curve. The formulas to calculate the

问题 I am relatively new to c# and i am trying to draw the quadratic curve with an X and Y graph to scale with. The i drew curve although appears at the upper left corner of the screen which is very small and barely noticeable. Is there possible way to enlarge my curve line and align it to the middle so it can be shown properly? protected override void OnPaint(PaintEventArgs e) { float a = 1, b = -3, c = -4; double x1, x2, x3, y1, y2, y3, delta; delta = (b * b) - (4 * a * c); x1 = ((b * (-1)) +

问题 Given the points of a Bezier curve (P0, P1, P2, P3) in 2D, I would like to find the y co-ordinate for a given x co-ordinate. The problem is well defined because of the following restrictions: P0 = (0,0), P3 = (1,1) P1 = (t, 1-t) for t between 0, 1 P2 = 1 - P1 (x and y) I have the following function to calculate the answer, having put in all the restrictions above into the Bezier curve formula here CubicBezier.html. I am using Newton-Raphson to work out the parameter of the point I want, and I

问题 We currently have a dynamic image, which holds on it text which is created from user input. This text follows a Bézier curve to define its position and rotation. For various reasons, the text needs to be changed to be a set of images as the font needs to be very specific. We will therefore have one PNG for every allowable character of the alphabet. So if the user enters the word "TEST", the system will pull out the letters T, E, S and T and position them next to each other. This part isn't an