I have a point and a (curved) line. Now I want to find the distance to the line where the direction from the point to the point on the line stands orthogonal on the line.
Intuitively I think that the shortest connection between the closest point on the line and the point is always orthogonal, but I'm not 100% sure that my geometry intuition is correct. Can you confirm that for finding the orthogonal connection between a point and a point on a line it is enough to check the closed point on the line?
What you wrote can be true, subject to conditions.
Your curve must be either closed, or start/end infinitely far away (like the shape of y=1/x or y=x^2). Otherwise, the closest distance can be to an end point of the curve.
The curve must be smooth. For instance, a triangle is not smooth, the normal is not defined at the 3 vertices and the closest distance can be a distance to the vertex. Another example, cubic Bézier splines may contain a singularity where the normal is not defined, see the top right picture:
Again, closest distance might be a distance to that singularity point.
Also, don’t forget it can be multiple points on the curve with orthogonal connection. You gonna have to find all of them, and use the minimum distance found. Moreover, in some cases “all of them” can be “infinitely many”, if the curve contains a piece of a circular arc.
Related
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.
My app captures the shape of a room by having the user point a camera at floor corners, and then doing a bunch of math, eventually ending up with a polygon.
The assumption is that the walls are straight (not curved). The majority of the corners are formed by walls at right angles to each other, but in some cases might not be.
Depending on how accurately the user points the camera, the (x,y) coordinates I derive for the corner might be beyond the actual corner, or in front of the actual camera, or, less likely, to the left or right. Obviously, in this case, when I connect the dots, I get weird parallelogram or rhomboid shapes. See example.
I am looking for a program or algorithm to normalize or regularize these shapes, provided we know which corners are supposed to be right angles.
My initial attempt involved finding segments which had angles which were "close" to each other, adjust them all to the same angle, and then recalculate the vertices. However, this algorithm proved to be unstable.
My current thinking is to find angles which are most obtuse (as would be caused by a point mistakenly placed beyond the actual corner), or most acute (as would be caused by a point mistakenly placed in front of the actual corner), and find the corner point which would make it a right angle. The problem, however, is that such as adjustment could have side-effects on other corners, such as making them even further away from right angles. I sense I need some kind of algorithm which takes all the information and optimizes/solves it at once--is this a kind of linear programming problem?--but I am stuck.
There is not a unique solution.
For example, take the perpendicular from the middle point of an edge to the two neighboring edges. This will give you two new corners.
Or take the perpendicular from the end point of an edge to other edges.
Or compute the average of angles in the end points of an edge. Use this average and the middle point of the edge to compute new corners.
Or...
To get the most faithful compliance, capture (or calculate) distances from each corner to the other three. Build triangles with those distances. Then use the average of the coordinates you compute for a corner from 2 or 3 triangles.
Resulting angles will not be exactly 90 degrees, but the polygon will represent the room fairly.
Assume you are given the equation of a line (in 2d), and the equations of lines that form a convex polygon (the polygon could be unbounded). How do I determine if the line intersects the polygon?
Furthermore, are there computational geometry libraries where such tasks are pre-defined? I ask because I'm interested not just in the 2D version but n-dimensional geometry.
For the 2D case, I think the problem simplifies a bit.
The line partitions the space into two regions.
If the polygon is present in only one of those regions, then the line does not intersect it.
If the polygon is present in both regions, then the line does intersect it.
So:
Take any perpendicular to the line, making the intersection with the
line the origin.
Project each vertex of the polytope onto the perpendicular.
If those projections occur with both signs, then the polygon
intersects the line.
[Update following elexhobby's comment.]
Forgot to include the handling of the unbounded case.
I meant to add that one could create a "virtual vertex" to represent the open area. What we really need is the "direction" of the open area. We can take this as the mean of the vectors for the bounding edges of the open area.
We then treat the dot product of that direction with the normal and add that to the set of vertex projections.
In geometry, typically see wikipedia a polygon is bounded.
What you are describing is usually called a polytope or a polyhedron see wikipedia
There are a few geometry libraries available, two that come to mind are boost (polygon) and CGAL. Generally, there is a distinct split between computational methods that deal with 2d,3d, and N-d - for obvious reasons.
For your problem, I would use a somewhat Binary Space Partitioning Tree approach. I would take the first line of your "poly" and trim the query line against it, creating a ray. The ray would start at the intersection of the two lines, and proceed in direction of the interior of the half-space generated by the first line of the "poly". Now I would repeat the procedure with the ray and the second line of the "poly". (this could generate a segment instead of ray) If at some point the ray (or now segment) origin lies on the outer side of a poly line currently considered and does not intersect it, then the answer is no - the line does not intersect your "poly". Otherwise it intersects. Take special care with various parallel edge cases. Fairly straight forward and works for multi-dimensional cases.
I am not fully sure, but I guess you can address this by use of duality. First normalize your line equations as a.x+b.y=1, and consider the set of points (a,b).
These must form a convex polygon, and my guess is that the new line may not correspond to a point inside the polygon. This is readily checked by verifying that the new point is on the same side of all the edges. (If you don't know the order of the lines, first construct the convex hull.)
Let's start from finite polygons.
To intersect polygon a line must intersect one of its edges. Intersection between line and an edge is possible only if two points lie on different sides from the line.
That can be easily checked with sign(cross_product(Ep-Lp,Ld)) for two points of the edge. Ep - edge point, Lp - some point on the line, Ld - direction vector of the line, cross_product(A,B)=Ax*By-Ay*Bx.
To deal with infinite polygons we may introduce "infinite points". If we have a half infinite edge with point E1 and direction Ed, its "second point" is something like E1+infinity*Ed, where infinity is "big enough number".
For "infinite points" the check will be slightly different:
cross_product(Ep-Lp,Ld)=
=cross_product(E1+infinity*Ed-Lp,Ld)=
=cross_product(E1-Lp+infinity*Ed,Ld)=
=cross_product(E1-Lp,Ld)+cross_product(infinity*Ed,Ld)=
=cross_product(E1-Lp,Ld)+infinity*cross_product(Ed,Ld)
If cross_product(Ed,Ld) is zero (the line is parallel to the edge), the sign will be determined by the first component. Otherwise the second component will dominate and determine the sign.
Let's assume I have a polygon and I have computed all of its self-intersections. How do I determine whether a specific edge is inside or outside according to the nonzero fill rule? By "outside edge" I mean an edge which lies between a filled region and a non-filled region.
Example:
On the left is an example polygon, filled according to the nonzero fill rule. On the right is the same polygon with its outside edges highlighted in red. I'm looking for an algorithm that, given the edges of the polygon and their intersections with each other, can mark each of the edges as either outside or inside.
Preferably, the solution should generalize to paths that are composed of e.g. Bezier curves.
[EDIT] two more examples to consider:
I've noticed that the "outside edge" that is enclosed within the shape must cross an even number of intersections before they get to the outside. The "non-outside edges" that are enclosed must cross an odd number of intersections.
You might try an algorithm like this
isOutside = true
edge = find first outside edge*
edge.IsOutside = isOutside
while (not got back to start) {
edge = next
if (gone over intersection)
isOutside = !isOutside
edge.IsOutside = isOutside
}
For example:
*I think that you can always find an outside edge by trying each line in turn: try extending it infinitely - if it does not cross another line then it should be on the outside. This seems intuitively true but I wonder if there are some pathological cases where you cannot find a start line using this rule. Using this method of finding the first line will not work with curves.
I think, you problem can be solved in two steps.
A triangulation of a source polygon with algorithm that supports self-intersecting polygons. Good start is Seidel algorithm. The section 5.2 of the linked PDF document describes self-intersecting polygons.
A merge triangles into the single polygon with algorithm that supports holes, i.e. Weiler-Atherton algorithm. This algorithm can be used for both the clipping and the merging, so you need it's "merging" case. Maybe you can simplify the algorithm, cause triangles form first step are not intersecting.
I realized this can be determined in a fairly simple way, using a slight modification of the standard routine that computes the winding number. It is conceptually similar to evaluating the winding both immediately to the left and immediately to the right of the target edge. Here is the algorithm for arbitrary curves, not just line segments:
Pick a point on the target segment. Ensure the Y derivative at that point is nonzero.
Subdivide the target segment at the Y roots of its derivative. In the next point, ignore the portion of the segment that contains the point you picked in step 1.
Determine the winding number at the point picked in 1. This can be done by casting a ray in the +X direction and seeing what intersects it, and in what direction. Intersections at points where Y component of derivative is positive are counted as +1. While doing this, ignore the Y-monotonic portion that contains the point you picked in step 1.
If the winding number is 0, we are done - this is definitely an outside edge. If it is nonzero and different than -1, 0 or 1, we are done - this is definitely an inside edge.
Inspect the derivative at the point picked in step 1. If intersection of the ray with that point would be counted as -1 and the winding number obtained in step 3 is +1, this is an outside edge; similarly for +1/-1 case. Otherwise this is an inside edge.
In essence, we are checking whether intersection of the ray with the target segment changes the winding number between zero and non-zero.
I'd suggest what I feel is a simpler implementation of your solution that has worked for me:
1. Pick ANY point on the target segment. (I arbitrarily pick the midpoint.)
2. Construct a ray from that point normal to the segment. (I use a left normal ray for a CW polygon and a right normal ray for a CCW polygon.)
3. Count the intersections of the ray with the polygon, ignoring the target segment itself. Here you can chose a NonZero winding rule [decrement for polygon segments crossing to the left (CCW) and increment for a crossing to the right (CW); where an inside edge yields a zero count] or an EvenOdd rule [count all crossings where an inside edge yields an odd count]. For line segments, crossing direction is determined with a simple left-or-right test for its start and end points. For arcs and curves it can be done with tangents at the intersection, an exercise for the reader.
My purpose for this analysis is to divide a self-intersecting polygon into an equivalent set of not self-intersecting polygons. To that end, it's useful to likewise analyze the ray in the opposite direction and sense if the original polygon would be filled there or not. This results in an inside/outside determination for BOTH sides of the segment, yielding four possible states. I suspect an OUTSIDE-OUTSIDE state might be valid only for a non-closed polygon, but for this analysis it might be desirable to temporarily close it. Segments with the same state can be collected into non-intersecting polygons by tracing their shared intersections. In some cases, such as with a pure fill, you might even decide to eliminate INSIDE-INSIDE polygons as redundant since they fill an already-filled space.
And thanks for your original solution!!
Given an ordered list of points, I want to draw a smooth curve that passes through all of them. Each part of the curve can either be horizontal, vertical, or an arc with given radius r (all arcs will have the same radius). The transitions should be smooth, i.e., the heading at the end of one part should be the same as the heading at the beginning of the next part. There can be any number of arcs or straight line segments between any two consecutive input points.
It's sort of like a train track that should run orthogonally or along sections with fixed curvature.
Is there a good algorithm to construct such a curve? (or, in cases where such a line is not possible, I would like to know that.)
I looked into Bezier curves, but that seems like overkill and I couldn't find a good way to enforce my constraints.
What you are asking for above implies to me that you seek tangent continuity of your curve across points (similar to a spline with tangent continuity at knots). The train track analogy at least conveys this requirement. Given the strict limitations of straight lines, and fixed radius circular arcs I am fairly certain that you will not be able to do this. Why not consider a spline interpolation of your points if you require such smoothness instead?
To see why consider the following image:
Consider replacing the line segment between B and C with a circular arc. You can do it to make the join continuous, but to make it tangent continuous, you would need a great deal of good fortune as there is only one circle that is tangent continuous to the line segment AB that also touches point C. The chances of that circle having tangent at C matching the tangent of line CD is remote. It is possible that your data will line up like this but you cannot rely on it.
If I have misunderstood your question please let me know and I will adjust the answer.