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...

I finally made it work. Simply added anim.restart() at the end of the incr() function and it works perfectly now....

if (wParam & MK_LBUTTON) ensures that none of the code you have posted actually runs unless the left button is being held down. You can press the right button all you want and it will never make it inside the if body unless the left button is also pressed. Then...

I have solved this by embedding the sprite inside a ccnode. The sprite is animated up and down (forever) using CCMoveBy-s relative to it's parent. Then I animate the parent node along the bezier path. I added an ease out action to smooth things out.

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...

javascript,algorithm,curve,bezier,cubic-spline

It's Catmull-Rom fitting: the code tries to find an appropriate tangent through point X, based on the location of points X-1 and X+1, such that the tangent is parallel to the line (X-1)--(X+1), and then fiddles with the control points that yields, to make sure the "incoming" and "outgoing" tangents...

javascript,geometry,computational-geometry,bezier,adobe-illustrator

Check, if all the control points of the curve are inside the ellipse. If so, your curve lies completely within the ellipse too.

Since your Bezier curve is on the (x, z) plane, I don't quite understand why do you want to take a cross product between first derivative and the z axis (0, 0, 1). The 'normal' vector found in this way is bound to become zero when the first derivative goes...

javascript,svg,bezier,approximation

Turns out this is impossible. I mean, mathematically, at least. You can't do circular arcs with plain cubic Bezier curves, you're always going to have an error, so depending on the arc's angle, you're going to probably need more than one Bezier curve. With that said, cubic Bezier curves work...

Try to set samples to higher value, for example set samples 10000 plot "data.dat" u 1 : 2 t "total bez" smooth bezier From gnuplot help: By default, sampling is set to 100 points. A higher sampling rate will produce more accurate plots, but will take longer. This parameter has...

Bezier curves are used to generate a smooth curve between two points. To move the path from left to right, choose a starting point on the left and choosing an ending point on the right. The two control points determine the shape of the path to take from left to...

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...

You can't. The SVG you're showing uses a cubic path, which uses a third order parametric curve, meaning it has the form: fx(t) = x1 * (1-t)³ + x2 * 3 * (1-t)²t + x3 * 3 * (1-t)t² + x4 * t³ fy(t) = y1 * (1-t)³ + y2...

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,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...

When you add a curve from (100,100) to (300,100), the control points should be (150,50) and (250,50), in your code, control points are instead (350,150) and (250,50), specifying the right control points should make it work.

javascript,math,html5-canvas,bezier,quadratic-curve

To match a quadratic Bezier to this parabola formula and assuming origin is 0, you can use place the control point at -y0 or -y1 from one of the end points. Example First, lets rearrange the formula: y2 = 4ax to: x = y2 / 4a so we can plot...

While there does seem to be a problem with the function for which I have opened an issue, I don't think this function is going to do what you are hoping it would do, assuming that you want a smooth outline. Basically there is not enough information being given in...

javascript,canvas,bezier,quadratic

The question as it stands is a bit unclear about requirements. Here is in any case an approach that does not require much calculations, but takes advantage of draw operations to visualize about the same as shown in the codepen. The main steps are: At an off-screen canvas: Define a...