cubic-bezier

问题 I have read the thread to make this happen for 4 points but only in 2D space here . I have implemented the answer for 3D but only for 3 control points here I have read this post but don't understand the sudo code or the math Can anyone simplify in java? I don't want to draw the curve as 2 segments of 3 points 回答1: Formula for cubic Bezier curve component (say X): X(t) = P0.X*(1-t)^3 + 3*P1.X*(1-t)^2*t + 3*P2.X*(1-t)*t^2 + P3.X*t^3 where P0 and P3 are end points, and P1 an P2 are control

问题 I have read the thread to make this happen for 4 points but only in 2D space here . I have implemented the answer for 3D but only for 3 control points here I have read this post but don't understand the sudo code or the math Can anyone simplify in java? I don't want to draw the curve as 2 segments of 3 points 回答1: Formula for cubic Bezier curve component (say X): X(t) = P0.X*(1-t)^3 + 3*P1.X*(1-t)^2*t + 3*P2.X*(1-t)*t^2 + P3.X*t^3 where P0 and P3 are end points, and P1 an P2 are control

来源： https://stackoverflow.com/questions/59290235/express-svg-arc-as-series-of-curves

问题 I want draw bezier curve only by fragment shader to connect nodes in my editor. I know all 4 points that define Bezier Curve. And Fragment Shader is called for every pixel, so i can just check: if "t" for gl_Coord.x is between 0 and 1 then set frag_color to Red for example. I want to avoid loops in shader that's inefficient. Best way, i think, is check for points that lay on the curve. But how to do it for Bezier Curves? Is it possible to express "t" variable from cubic bezier equation? x = (

问题 Following this topic, I am trying to generate a 3D curved triangle as a NURBS surface, but I don't understand how to set up my 3D points to do that. Here is the current implementation : var edges = this.getEdges(), // An edge is a line following 4 dots as a bezier curve. dots = self.getDotsFromEdges(edges), // Get all dots in order for building the surface. ctrlPoints = [ // Is generated only once before, but copy-pasted here for this sample code. [ new THREE.Vector4(0, 0, 0, 1), new THREE

问题 I have a div which contains a some slides and menu. #wrap { position: relative; top: 0; transition: top 1.4s cubic-bezier(.49,.22,.52,1.35); } <div id="wrap"></div> Can anyone explain to me what the transition property does here? I am not able to understand the effect it's going to produce on the div. 回答1: Answer for your question that property is for the animation of <div id="wrap"></div> . It means it will animate using top property! And for Effect: It will Bounce like This Will Be the

问题 I've read a few discussions regarding finding Y at X for a cubic Bezier curve, and have also read this article regarding the matter. My case is more constrained than the general one, and I wonder if there's a better solution than the general ones mentioned in the above discussions. My case: The X value of the different control points is increasing. Ie: X3 > X2 > X1 > X0 . Also, as a result of above, X(t) is strictly monotonically increasing as well. Is there any efficient algorithm that takes

问题 I am trying to make a LineChart with a cubic plot. The result is like in the screenshot below: the cubic bezier displays wrong and has "spikes". Can someone help me make it appear correctly? This is my config : LineDataSet lineDataSet = new LineDataSet(entries,nameLabel); lineDataSet.setColor(Constants.colors.get(i)); lineDataSet.setDrawValues(false); lineDataSet.setDrawCircles(false); lineDataSet.setMode(LineDataSet.Mode.CUBIC_BEZIER); Thank you 回答1: Issues like this can have a number of