Polyline Offsetting using Clipper - graphics

I am using Clipper Library in which it is quite easy to do Polygon Offsetting, BUT MY PROBLEM IS TO DO POLYLINE OFFSETTING preferably using Clipper.
While searching on web I found a function called OffsetPaths, but the function offsets in two directions, that is it takes only positive argument as offset distance.
I can understand that unlike in closed Polygons where it is easy to specify direction, enlarging being positive and deflating as negative, Polylines do not have the same luxury. But my requirement is such.

Related

Fitting a transition + circle + transition curve to a set of measured points

I am dealing with a reverse-engineering problem regarding road geometry and estimation of design conditions.
Suppose you have a set of points obtained from the measurement of positions of a road. This road has straight sections as well as curve sections. Straight sections are, of course, represented by lines, and curves are represented by circles of unknown center and radius. There are, as well, transition sections, which may be clothoids / Euler spirals or any other usual track transition curve. A representation of the track may look like this:
We know in advance that the road / track was designed taking this transition + circle + transition principle into account for every curve, yet we only have the measurement points, and the goal is to find the parameters describing every curve on the track, this is, the transition parameters as well as the circle's center and radius.
I have written some code using a nonlinear optimization algorithm, where a user can select start and end points and fit a circle that to the arc section between them, as it shows in next figure:
However, I don't find a suitable way to take the transition into account. After giving it some thought I came to think that this s because, given a set of discrete points -with their measurement error- representing a full curve, it is not entirely clear where to consider it "begins" and where it "ends" and, moreover, it is less clear where to consider the transition, the proper circle and the exit transition "begin" and "end".
Is there any work on this subject which I may have missed? is there a proper way to fit the whole transition + curve + transition structure into the set of points?
As far as I know, there's no method to fit a sequence clothoid1-circle-clothoid2 into a given set of points.
Basic facts are that two points define a straight, and three points define a unique circle.
The clothoid is far more complex, because you need: The parameter A, the final radius Rf, an initial point px,py, the radius Ri at that point, and the tangent T (angle with X-axis) at that point.
These are 5 data you may use to find the solution.
Due to clothoid coords are calculated by expanded Fresnel integrals (see https://math.stackexchange.com/a/3359006/688039 a little explanation), and then apply a translation & rotation, there's no an easy way to fit this spiral into a set of given points.
When I've had to deal with your issue, what I've done is:
Calculate the radius for triplets of consecutive points: p1p2p3, p2p3p4, p3p4p5, etc
Observe the sequence of radius. Similar values mean a circle, increasing/decreasing values mean a clothoid; Big values would mean a straight.
For each basic element (line, circle) find the most probably characteristics (angles, vertices, radius) by hand or by some regression method. Many times the common sense is the best.
For a spiral you may start with aproximated values, taken from the adjacent elements. These values may very well be the initial angle and point, and the initial and final radius. Then you need to iterate, playing with Fresnel and 'space change' until you find a "good" parameter A. Then repeat with small differences in the other values, those you took from adjacents.
Make the changes you consider as good. For example, many values (A, radius) use to be integers, without decimals, just because it was easier for the designer to type.
If you can make a small applet to do these steps then it's enough. Using a typical roads software helps, but doesn't avoid you the iteration process.
If the points are dense compared to the effective radii of curvature, estimate the local curvature by least square fitting of a circle on a small number of points, taking into account that the curvature is most of the time zero.
You will obtain a plot with constant values and ramps that connect them. You can use an estimate of the slope at the inflection points to figure out the transition points.

Calculating average curvature of a polygonal chain with non-uniform segment length

I'm using a polygonal chain to approximate a curve. I want to approximate the average of a function of curvature of all points that lie on the curve. One function of curvature that I need is, for example, the square of curvature.
I can get near that by choosing some points on the chain, calculating the curvature in those points, applying the function on it (for example squaring it), and then averaging the calculated values.
I need both accuracy and speed. I appreciate both — fast, but approximate; as well as accurate, but slow solutions. I'm working in Java, but the answer doesn't need to be written in Java — it doesn't even need to contain any code at all.
Polygonal chain with uniform segment length
If the polygonal chain's segments all have equal length, I can just calculate the curvature in the vertices and then average that. I see two ways to get the curvature in a vertex.
One way is to get the circle that goes through the selected vertex, the vertex before it, and the one after it. The curvature is then 1/radius of the circle.
The other way is to calculate the external angle (in radians) of the two segments connected at the selected vertex and then divide its absolute value by the length of a segment. In the following image, φ marks the external angle:
I am not sure if this method is correct, as I haven't mathematically derived it, but I've noticed through experimentation that it gives similar results to the above method.
Polygonal chain with non-uniform segment length
Unfortunately, though, there's no guarantee that the segments have uniform length.
If I try using the first of the above methods, vertices connected to longer segments give lower curvatures, even if they are visibly sharper. I tried substituting previous and next vertices with a point x units before the selected vertex and a point x units after it. I don't know what to set the x constant to, to get accurate results. All the values I've tried seemed to give inaccurate results.
If I try using the second method, I don't know what length to divide the angle by. If I don't divide by anything at all, I actually get pretty good results for comparing two curves and determining which one is curvier, but I need to be able to determine the actual curvature in a point.
With both of these methods there's also the problem that parts with shorter segments (where points are denser) will affect the average more.
Another possible solution would be to ignore the vertices and instead use an array of points on the chain that are evenly spaced, treat them as a new polygonal chain (connect the points with straight lines), and then calculate curvatures on this new chain instead, using one of the methods I mentioned under the header titled "Polygonal chain with uniform segment length".
Finding such an array of points is not trivial, though, because I have to choose a segment length, and only after producing the points, I can see if the length of the resulting chain is divisible by the chosen segment length.
If you aren't short on space, the last solution you mentioned would be the best, because the "sphere" approximation, as you've perhaps realized, would give awful results in more extreme cases, especially if the curvature is large or changes sign quickly.
There are many ways to do interpolations, the simplest being quadratic and cubic splines. However if you have more pre-processing time, Lagrange polynomials produce very good results: https://en.wikipedia.org/wiki/Lagrange_polynomial.
Side note on your angle division method, consider this diagram:
(From simple geometry the inside angle there is also theta)
For a << l. So the curvature:
So your approximation is in fact correct for small curvatures.
An alternative is to use a local parabola approximation to estimate the curvature. Basically, to estimate the curvature at point P(i), you take P(i-1), P(i) and P(i+1) and construct a parabola from these 3 points. Then, you compute the curvature at P(i) from the parabola. Remember to use chord-length (or centripetal) parametrization when constructing the parabola.

How to get the outline of a stroke? [duplicate]

This question already has answers here:
svg: generate 'outline path'
(2 answers)
Closed 5 years ago.
I want to convert a stroked path to a filled object. (Programmatically, in JavaScript.)
The line is just a simple curved line, a sequence of coordinates. I can render this line as a path, and give it a stroke of a certain thickness... but I'm trying to get a filled shape rather than a stroked line, so that I can do further modifications on it, such as warping it, so the resulting 'stroke' might vary in thickness or have custom bits cut out of it (neither of these things are possible with a real SVG stroke, as far as I can tell).
So I'm trying to manually 'thicken' a line into a solid shape. I can't find any function that does this – I've looked through the docs of D3.js and Raphaël, but no luck. Does anyone know of a library/function that would do this?
Or, even better: if someone could explain to me the geometry theory about how I would do this task manually, by taking the list of line coordinates I have and working out a new path that effectively 'strokes' it, that would be amazing. To put it another way, what does the browser do when you tell it to stroke a path – how does it work out what shape the stroke should be?
There has been a similar question recently:
svg: generate 'outline path'
All in all, this is a non-trivial task. As mentioned in my answer to the linked question, PostScript has a command for generating paths that produce basically the same output as a stroke, called strokepath. If you look at what Ghostscript spits out when you run the code I posted at the linked question, it's pretty ugly. And even Inkscape doesn't really do a good job. I just tried Path => Outline stroke in Inkscape (I think that's what the English captions should say), and what came out didn't really look the same as the stroked path.
The "simplest" case would be if you only have non-self-intersecting polylines, polygons or paths that don't contain curves because in general, you can't draw exact "parallel" Bézier curves to the right and the left of a non-trivial Bézier curve that would delimit the stroked area - it's mathematically non-existent. So you would have to approximate it one way or the other. For straight line segments, the exact solution can be found comparatively easily.
The classic way of rendering vector paths with curves/arcs in them is to approximate everything with a polyline that is sufficiently smooth. De Casteljau's Algorithm is typically used for turning Bézier curves into line segments. (That's also basically what comes out when you use the strokepath command in Ghostscript.) You can then find delimiting parallel line segments, but have to join them correctly, using the appropriate linejoin and miterlimit rules. Of course, don't forget the linecaps.
I thought that self-intersecting paths might be tricky because you might get hollow areas inside the path, i.e. the "crossing area" of a black path might become white. This might not be an issue for open paths when using nonzero winding rule, but I'd be cautious about this. For closed paths, you probably need the two "delimiting" paths to run in opposite orientation. But I'm not sure right now whether this really covers all the potential pitfalls.
Sorry if I cause a lot of confusion with this and maybe am not of much help.
This page has a fairly good tutorial on bezier curves in general with a nice section on offset curves.
http://pomax.github.io/bezierinfo/
A less precise but possibly faster method can be found here.
http://seant23.wordpress.com/2010/11/12/offset-bezier-curves/
There is no mathematical answer, because the curve parallel to a bezier curve is not generally a bezier curve. Most methods have degenerate cases, especially when dealing with a series of curves.
Think of a simple curve as one with no trouble spots. No cusps, no loops, no inflections, and ideally a strictly increasing curvature. Chop up all the starting curves into these simple curves. Find all the offset curves of these simple curves. Put all the offset curves back together dealing with gaps and intersections. Quadratic curves are much more tractable if you have the option to work with them.
I think most browsers do something similar to processingjs, as they have degenerate cases even with quadratic curves. For example, look at the curve 200,300 719,301 500,300 with a thickness of 100 or more.
The standard method is the Tiller-Hanson algorithm (Offsets of Two-Dimensional Profiles, 1984, which irritatingly is not on line for free) which creates a good approximation. The idea is that because the control points of each Bezier curve lie on lines tangent to the start and end of the curve, a parallel curve will have the same property. So we offset the start and the end of the curve, then find new control points using these intersections. However, that gives very bad results for sharp curves, so the first step is to bisect the original curve, which is very easy to do to Bezier curves, until it turns through a sufficiently small angle.
Other refinements are needed to deal with (i) intersections between the parallels, on the inside of each vertex; (ii) inserting an arc of a circle to fill the gap on the outside of each vertex; and (iii) adding end-caps - square, butt or circular.
Tiller-Hanson is difficult to implement, but there's a good open-source implementation in the FreeType library, in ftstroke.c (http://git.savannah.gnu.org/cgit/freetype/freetype2.git/tree/src/base/ftstroke.c).
I'm sorry to say that it can be quite difficult to integrate this code, but I have used it successfully, and it works well.

Intersecting points with a polygon in OpenCV

My inputs
I have a vector<Point2f> that contains the contours of a polygon. I also have a list of points that need to be intersected with this polygon.
The problem
I want to calculate how much of these points intersect with the polygon. I want to repeat this calculation on a number of polygons to see which one contains the highest number of points.
Does OpenCV implement such intersection functionality of its own or will I need to implement an intersection function myself? I'm worried that if I try to implement it myself, the result will be unnecessarily slow. If OpenCV can't do it, are there other free graphics libraries that can perform this task?
pointPolygonTest does exactly what you're looking for, and it's pretty well optimized. The parameter is a Mat which you can make with the constructor that takes your vector of points.
The function determines whether the point is inside a contour, outside, or lies on an edge (or coincides with a vertex). It returns positive (inside), negative (outside) or zero (on an edge) value, correspondingly. When measureDist=false , the return value is +1, -1 and 0, respectively. Otherwise, the return value it is a signed distance between the point and the nearest contour edge.
Your problem seems easily parallelizable, though, i.e. each batch of candidate polygons could run on a different thread, so I'd definitely look into that if you're concerned about performance.

Creating closed spatial polygons

I need to create a (large) set of spatial polygons for test purposes. Is there an algorithm that will create a randomly shaped polygon staying within a bounding envelope? I'm using OGC Simple stuff so a routine to create the well known text is the most useful, Language of choice is C# but it's not that important.
Here you can find two examples of how to generate random convex polygons. They both are in Java, but should be easy to rewrite them to C#:
Generate Polygon example from Sun
from JTS mailing list, post Minimum Area bounding box by Michael Bedward
Another possible approach based on generating set of random points and employ Delaunay tessellation.
Generally, problem of generating proper random polygons is not trivial.
Do they really need to be random, or would some real WKT do? Because if it will, just go to http://koordinates.com/ and download a few layers.
What shape is your bounding envelope ? If it's a rectangle, then generate your random polygon as a list of points within [0,1]x[0,1] and scale to the size of your rectangle.
If the envelope is not a rectangle things get a little more tricky. In this case you might get best performance simply by generating points inside the unit square and rejecting any which lie in the part of the unit square which does not scale to the bounding envelope of your choice.
HTH
Mark
Supplement
If you wanted only convex polygons you'd use one of the convex hull algorithms. Since you don't seem to want only convex polygons your suggestion of a circular sweep would work.
But you might find it simpler to sweep along a line parallel to either the x- or y-axis. Assume the x-axis.
Sort the points into x-order.
Select the leftmost (ie first) point. At the y-coordinate of this point draw an imaginary horizontal line across the unit square. Prepare to create a list of points along the boundary of the polygon above the imaginary line, and another list along the boundary below it.
Select the next point. Add it to the upper or lower boundary list as determined by it's y-coordinate.
Continue until you're out of points.
This will generate convex and non-convex polygons, but the non-convexity will be of a fairly limited form. No inlets or twists and turns.
Another Thought
To avoid edge crossings and to avoid a circular sweep after generating your random points inside the unit square you could:
Generate random points inside the unit circle in polar coordinates, ie (r, theta).
Sort the points in theta order.
Transform to cartesian coordinates.
Scale the unit circle to a bounding ellipse of your choice.
Off the top of my head, that seems to work OK

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