bezier

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-t)^3 * 100 + 3*(1-t)^2 * t * 600 + 3*(1-t) * t^2 * 800 + t^3 * 800 then rearrange it: 800t^3 + 3*(1-t)

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 Check this in Apple documentation. applyTransform: Transforms all points in the path using the specified affine transform matrix. - (void)applyTransform:(CGAffineTransform)transform I haven't tried this. But here is how to rotate a NSBezierPath from the link rotating

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. polfosol ఠ_ఠ 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 ; i < 1 ; i += 0.01 ) { // The Green Lines xa = getPt( x1 , x2 , i ); ya = getPt( y1 , y2 , i );

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 original curve? The ideal result would be able to able to take a percentage slice of the bezier from the

I have 2 questions about bezier curves, and using them to approximate portions of circles. Given the unit circle arc (1,0)->(cos(a),sin(a)) where 0 < a < pi/2, will it result in a good approximation of this arc to find the bezier curve's control points p1, p2 by solving the equations imposed by the requirements B(1/3) = (cos(a/3), sin(a/3)) and B(2/3) = (cos(2a/3), sin(2a/3)). (In other words, requiring that the bezier curve go through two evenly spaced points in the arc). If we have an affine transformation A which turns the circle arc in an ellipse arc will the transformed control points Ap0

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").getAttribute("cy"); let point2X = document.getElementById("sl").getAttribute("cx"); let point2Y = document

What's the right tool for the job if I want to write a Python script that produces vector graphics in PDF format ? In particular, I need to draw filled polygons with rounded corners (i.e., plane figures that are composed of straight lines and circular arcs ). It seems that matplotlib makes it fairly easy to draw rectangles with rounded corners and general polygons with sharp corners. However, to draw polygons with rounded corners, it seems that I have to first compute a Bézier curve that approximates the shape. Is there anything more straightforward available? Or is there another library that