bezier

I'm trying to draw a quarter circle in iOS. In Mac OS, it seems you can use appendBezierPathWithArcWithCenter but this doesn't seem to work in iOS. Does anyone know how to simply draw a quarter circle in iOS? Thanks There are two iOS equivalents, one for UIBezierPath and one for CGPath : UIBezierPath equivalent UIBezierPath *path = [UIBezierPath bezierPath]; [path addArcWithCenter:centerPoint radius:radius startAngle:startAngle endAngle:endAngle clockwise:YES]; CGPath equivalent CGMutablePathRef path = CGPathCreateMutable(); CGPathAddArc(path, NULL, centerPoint.x, centerPoint.y, radius,

What's the best way to draw cubic bezier curves programmatically in AS3? The Graphics class only seems to support quadratic curves. :( I want to be able to do something like: var startPoint:Point = new Point(0, 0); var endPoint:Point = new Point(5, 5); var control1:Point = new Point(5, 0); var control2:Point = new Point(0, 5); var myBezier:Sprite = getBezier(startPoint, control1, control2, endPoint); For a performance target, I'm planning on having ~50 of these on the stage at once. Juan Pablo Califano Note: Flash Player 11 onwards includes a native method to draw cubic curves, cubicCurveTo()

Is there an easy way to approximate the points (if any) where two instances of QuadCurve2D intersect? That is, how could I compute the coordinates of the red dots in this diagram? There is no obvious method in QuadCurve2D to do this. (note: the points are not exact as I have tweaked them manually for the diagram. Also note the "missing" fourth point which does not lie on the curve segment even though it lies on the (infinite) parabola.) These two curve segments were created with the following code: QuadCurve2D curve1 = new QuadCurve2D.Double(-2.00, -2.00, +0.75, +4.75, +2.00, -0.75);

I know: The control points a and d (start and end point of a 2D cubic bezier curve) The slopes a->b, c->d, and b->c (b,c the other control points) Where the halfway point of the Bézier curve is. Now, given this information, what is the formula for the positions of control points b and c ? I know this question is old, but there is no correct or complete answer provided, so I thought I'd chime in with a solution. Note that David's calculations contain several errors and his solution is incomplete even if these errors are corrected. First, define vectors T0 , T1 and T2 using the three slopes: T0

I need to make a cube with smooth corners and smooth edges in C++ with OpenGL. For all I know I have three options: Bezier curves (maybe, is it possible?), a cube with cylinders for edges and spheres for corners or load a .3ds of a cube. Any ideas? You can simulate a cube with smooth lighting by pointing the normals directly out from the center (simulating an 8 cornered sphere). It totally depends on what exactly you are trying to do. Using the above method may be perfectly good enough. If you want to define a cube with curved corners (up close) then you are going to have to subdivide the cube

I'm trying to animate drawing out a circle. In my custom view, I have private final Paint mPaint = new Paint() { { setDither(true); setStyle(Paint.Style.STROKE); setStrokeCap(Paint.Cap.ROUND); setStrokeJoin(Paint.Join.ROUND); setColor(Color.BLUE); setStrokeWidth(30.0f); setAntiAlias(true); } }; ... protected void onDraw(Canvas canvas) { super.onDraw(); if (mOval == null) { mOval = new RectF(getLeft(), getTop(), getRight(), getBottom()); } if (mPath == null) { mPath = new Path(); mPath.moveTo(0, getHeight() / 2); } float sweepAngle = Math.min((float) mElapsedTime / 1000 * 60 * 1, 1) * 360; if

I have a quadratic bezier curve and I want to calculate the slope of the tangent in a given point. For example, let it be the middlepoint of the quadratic bezier curve, therefore t=0.5 (please see the link below for a picture of this). I've calculated the first derivative of the formula for the quadratic bezier curve; however I get 400 as value for the slope, though it should be 0. Maybe I'm using the first derivative in a wrong way? I know I could also calculate the tangents using trigonometric functions; however I'd like to do it using the first derivative, shouldn't this be possible? Thanks

I have an array of control points that represent a Bezier curve. It could be a fifth-order or 100-order Bezier curve, or anything in-between. I am looking for a way to simplify that Bezier curve into multiple cubic Bezier curves. An illustration below shows how tenth-degree curve can be simplified to three-degree curve, but I want to go further and simplify it to several cubic Bezier curves to achieve better approximation. Code example would be very helpful. As mohsenmadi already pointed out: in general this is not a thing you can do without coming up with your own error metric. Another idea