bezier

来源： https://stackoverflow.com/questions/34399897/svg-polygon-rounding

来源： https://stackoverflow.com/questions/60020136/generating-a-path-between-two-sets-of-pixel-coordinates-x-y

问题 I've been stuck on this for a week now i can't seem to solve it. I have an arc which i can convert to a series of bezier curves quite easily when the arc is flat: But i am struggling to work out how to find the bezier curves when the arc is a helix and the end tangents have different slopes. This is as far as i have gotten so far: As you can see each bezier curve has control points that are not on the right plane, and the start and end tangent (the red vectors in the second image) of the full

问题 I've been stuck on this for a week now i can't seem to solve it. I have an arc which i can convert to a series of bezier curves quite easily when the arc is flat: But i am struggling to work out how to find the bezier curves when the arc is a helix and the end tangents have different slopes. This is as far as i have gotten so far: As you can see each bezier curve has control points that are not on the right plane, and the start and end tangent (the red vectors in the second image) of the full

问题 I have a JPG, BMP, or SVG image (see example below) and I need an algorithm to extract the vertices (X, Y) coordinates and the egdes (i.e., a list that indicates which vertices are connected). The Edges can be of the form of a boolean true/false for each vertex pair or simply a list of vertex pairs that are connected. Any ideas welcome. For example, I would like a function (or series of functions) which input the image and output two lists: Vertices : Vertex 1: X = 1, Y = 2 Vertex 2: X = 3, Y

问题 I have a JPG, BMP, or SVG image (see example below) and I need an algorithm to extract the vertices (X, Y) coordinates and the egdes (i.e., a list that indicates which vertices are connected). The Edges can be of the form of a boolean true/false for each vertex pair or simply a list of vertex pairs that are connected. Any ideas welcome. For example, I would like a function (or series of functions) which input the image and output two lists: Vertices : Vertex 1: X = 1, Y = 2 Vertex 2: X = 3, Y

问题 I want to design the highlighted portion of curve of the below screen snapshot using bezier path in swift. 回答1: It’s just a little trigonometry to figure out the arcs around that button in the bottom right hand corner, e.g. func updatePath() { let buttonCenter = CGPoint(x: bounds.maxX - circleRadius, y: bounds.maxY - circleRadius) circleShapeLayer.path = UIBezierPath(arcCenter: buttonCenter, radius: circleRadius, startAngle: 0, endAngle: .pi * 2, clockwise: true).cgPath // red let angle1 =

问题 I'm writing a TTF parser. For a better understanding of the TTF format, I used TTX to extract the ".notdef" glyph data of C:\Windows\calibri.ttf as follow. <TTGlyph name=".notdef" xMin="0" yMin="-397" xMax="978" yMax="1294"> <contour> <pt x="978" y="1294" on="1"/> <pt x="978" y="0" on="1"/> <pt x="44" y="0" on="1"/> <pt x="44" y="1294" on="1"/> </contour> <contour> <pt x="891" y="81" on="1"/> <pt x="891" y="1213" on="1"/> <pt x="129" y="1213" on="1"/> <pt x="129" y="81" on="1"/> </contour>

问题 I want to give my users an easy way to visually trace a route on a map or a picture. The solution must let the users add control points that they can use to put bends into the route. It should work with html5 canvas - I currently use the Konvajs library so a solution that uses this would be good. In the interests of sharing & learning, if you can suggest solutions using other HTML5 canvas libraries that would be good to see too. Note: This is not the original question posed. However it

问题 I want to give my users an easy way to visually trace a route on a map or a picture. The solution must let the users add control points that they can use to put bends into the route. It should work with html5 canvas - I currently use the Konvajs library so a solution that uses this would be good. In the interests of sharing & learning, if you can suggest solutions using other HTML5 canvas libraries that would be good to see too. Note: This is not the original question posed. However it