ios,objective-c,uibezierpath,bezier-curve

You need to set the stroke width before you actually stroke the path: UIGraphicsBeginImageContextWithOptions(CGSizeMake(progressView.frame.size.width, progressView.frame.size.height), NO, 0.0); [[UIColor colorWithRed:246.0/255.0 green:80.0/255.0 blue:36.0/255.0 alpha:1.0] setStroke]; UIBezierPath *path = [UIBezierPath bezierPathWithArcCenter:CGPointMake(progressView.frame.size.width/2 ,progressView.frame.size.height/2) radius:progressView.frame.size.width/2 - 5 startAngle:DEGREES_TO_RADIANS(0) endAngle:DEGREES_TO_RADIANS(angle)...

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 curve. You may also be interested in this more specific example for your application: http://www.codeproject.com/Articles/223159/Midpoint-Algorithm-Divide-and-Conquer-Method-for-D If you really want to get into it, then I suggest this Primer: http://pomax.github.io/bezierinfo/...

Your approach should work. You made a little oversight: Within slide(), you don't update last in your loop. Try: for (++current; current != points.end(); ++current) { result.push_back(last * time + *current * (1.0 - time)); last = *current; // <-- } Note that a different interpretation of bezier curves can...

svg,graphics,trigonometry,bezier-curve,spline

After few tries/errors, I found that the correct ratio is K=0.37. "M" + x1 + "," + y1 + "C" + (x1 + K * (x2 - x1)) + "," + y1 + "," + (x2 - K * (x2 - x1)) + "," + y2 + "," + x2...

r,plot,bezier,bezier-curve,spline

It may not the be the best approach, bit grid certainly isn't inactive. It's included as a default package with the R installation. It's the underlying graphics engine for plotting libraries like lattice and ggplot. You shouldn't need to install it, you should just be able to load it. Here's...

geometry,2d,bezier-curve,spline

Imagine cubic curve between points B and E. If it is defined as cardinal spline with tension parameter s, then tangent vectors in these points are Tb = s * (E - A) Te = s * (F - B) If curve is defined as Bezier one, then tangent vectors...

Yes it is possible and even rather simple to both move (Translate) and enlarge (Scale) the Graphics results by using Graphics.TranslateTransform and Matrix and Graphics.MultiplyTransform: using System.Drawing.Drawing2D; //.. int deltaX = 100; int deltaY = 100; g.TranslateTransform(deltaX, deltaY); float factor = 2.5f; Matrix m = new Matrix(); m.Scale(factor, factor); g.MultiplyTransform(m);...

javascript,css-transitions,bezier,bezier-curve,cubic

If you want only a section of a cubic curve, with t from 0 to 1, there are "simple" formulae to determine what the new coordinates need to be. I say simple because it's pretty straight forward to implement, but if you also want to know why the implementation actually...

c++,opengl,bezier,bezier-curve

Let's look at a single face: if we want a rounded rectangle for that, we need to create a face with sixteen vertices: eight mesh points, and eight control points, arranged on-off-off-on for each corner. Let's assume we want rounded corners of radius d. That's silly and it should be...

javascript,math,svg,bezier-curve

What you want is a vertical slope at the end points. Since the inner control points determine the slope, you have to set them directly above and below the control points. There are two degrees of freedom in how much above and below. For instance, put them at the thirds...

opengl-es,width,line,andengine,bezier-curve

OpenGL ES implementations do not have to support drawing of wide lines. You can query the range of available line widths with: float[] range = new float[2]; GLES20.glGetFloatv(GLES20.GL_ALIASED_LINE_WIDTH_RANGE, range, 0); // range[0] is the minimum supported line width. // range[1] is the maximum supported line width. This gives you the...

javascript,css-transitions,bezier,bezier-curve,cubic

first half is ease-in which is cubic-bezier(.42,0,1,1) and graphically is http://cubic-bezier.com/#.42,0,1,1 Please verify this assumption. (curves original end points are 0,0, and 1,1 in a css timing function) The first half of the bezier curve [0,0, .42,0, .58,1, 1,1] should not be [0,0 .42,0, 1,1, 1,1] The end points...

javascript,jquery,canvas,svg,bezier-curve

It's just a bug in your code. You are not updating lastPoint in your SVG version. http://jsfiddle.net/67haj4nt/4/ And if you update the SVG version to match the canvas version, you get identical curves. http://jsfiddle.net/67haj4nt/5/...

Beizer Path // set up properties for path @property (nonatomic, assign) CGPath startPath; @property (nonatomic, assign) CGPath endPath; @property (nonatomic, strong) CAShapeLayer *pathLayer; // create the startPath UIBezierPath *path = [UIBezierPath //create your path using the paint code code self.startPath = path.CGPath; // create the end path path = [UIBezierPath...

android,opengl-es,andengine,mesh,bezier-curve

Ok I found a problem but I dont really understand why it was a problem. So in: float getBezierX(int triangleIndex, float P1X, float PcX, float P2X){ float t = triangleIndex / (triangleCount * 4); float u = 1 - t; float tt = t * t; float uu = u...

javascript,html5,html5-canvas,bezier,bezier-curve

My guess is, if you want such shape for bezier curves you should not use center of rectangles but first retrieve which side(edge) of rectangle are facing (connected), in your example above, it's Box1' right edge connected to Box2 left edge. After that, use center of edge as start/end points...