bezier

问题 What's the Matlab way to draw a Bezier curve ? Do you have to prgoram it yourself ? I am not looking for a user made routine, but am asking if Matlab offers a standard way to draw them. 回答1: After looking and searching through the documentation, my answer is No: you'd have to go with one of the 3rd party implementations. Likeliest candidate would be the interp family functions, and they implement no Bezier interpolation. 回答2: With the Curve Fitting Toolbox, Matlab supports B-splines, which

问题 So I've recently picked up graphics programming and I wanted to compute a cubic Bézier curve. I found this excellent answer on quadratic Bézier but I don't know how to convert this to a cubic Bézier curve. 回答1: For cubic Bézier curve, as you see in the link you shared, the green lines are obtained from the same procedure as the quadratic one. the differences are: you have two green lines, and then you need to calculate a blue line based on them. So the for loop changes as: for( float i = 0 ;

问题 Take a typical cubic bezier curve drawn in JavaScript (this example I googled...) http://jsfiddle.net/atsanche/K38kM/ Specifically, these two lines: context.moveTo(188, 130); context.bezierCurveTo(170, 10, 350, 10, 388, 170); We have a cubic bezier which starts at 188, 130 , ends at 388, 170 , and has controls points a: 170, 10 and b: 350, 10 My question is would it be possible to mathematically adjust the end point and control points to make another curve which is only a segment of the

问题 I need to draw a symmetrically curved line between the centers of two circles . <svg> <circle class="spot" id="au" cx="1680" cy="700" r="0"></circle> <circle class="spot" id="sl" cx="1425" cy="525" r="0"></circle> <line id="line1" stroke-width="2" stroke="red"/> </svg> This is the code I wrote so far. < line > element should be replaced with a curved path . function drawNow() { let point1X = document.getElementById("au").getAttribute("cx"); let point1Y = document.getElementById("au")

问题 I've managed to implement quadratic and cubic Bezier curves.They are pretty straightforward since we have a formula. Now I want to represent an n-th order Bezier curve using the generalization: Where and I'm using a bitmap library to render the output, so here is my code: // binomialCoef(n, k) = (factorial(n) / (factorial(k) * factorial(n- k))) unsigned int binomialCoef(unsigned int n, const unsigned int k) { unsigned int r = 1; if(k > n) return 0; for(unsigned int d = 1; d <= k; d++) { r *=

问题 I have read similar topics in order to find solution, but with no success. What I'm trying to do is make the tool same as can be found in CorelDraw, named "Pen Tool". I did it by connecting Bezier cubic curves, but still missing one feature, which is dragging curve (not control point) in order to edit its shape. I can successfully determine the "t" parameter on the curve where dragging should begin, but don't know how to recalculate control points of that curve. Here I want to higlight some

问题 So I need to find out where the control points would be for a cubic bezier curve when only knowing points on the curve, the points can lie in 3D. It would be ideal if I could do this for any number of points on the curve. Most of what I have found deals only with 2D, or only for 4 points. 回答1: Let me see if I understand you: you want an interpolating Bezier curve, going through a given set of points P0 P1 ... but drawn as Bezier curves, with a function like bezier4( nstep, Pj, Cj, Dj, Pj+1 )

问题 I' using KineticJS to drawing a logo, both top an bottom lines are bezierCurveTo. I need to draw lines between them and so I need to locate the points inside both curves. What I thought to use was get the X coordinate and get the Y coordinate. Using method bezierCurveTo I can find the position. The problem is bezierCurveTo use the first parameter as percent and my two berzier are not equivalent, so is not a solution for me. Is there any function that given tree points and X returns the Y ?

问题 Is it possible to get float t back given the resulting value and the four points? If so, how? public static Vector3 GetPoint (Vector3 p0, Vector3 p1, Vector3 p2, Vector3 p3, float t) { t = Mathf.Clamp01(t); float oneMinusT = 1f - t; return oneMinusT * oneMinusT * oneMinusT * p0 + 3f * oneMinusT * oneMinusT * t * p1 + 3f * oneMinusT * t * t * p2 + t * t * t * p3; } Code from this Tutorial by Jasper Flick 回答1: It is, and involves implementing root finding for third degree functions. One direct

问题 I have two separate images. CurvedPath image (Attached Below) PersonImage As you can see this path is not having perfect arc for single angle. I have another PersonImage that I want to animate exactly above the center line for this arc image with zoom-in effect. This animation should be start from bottom-left point to top-right point. How to achieve this kind of animation? I have read about QuartzCore and BeizerPath animation, but as I am having less knowledge about those, it will quiet