I need to find the shortest distance between a 2d circular arc and a finite non-zero length line segment. Would be great to also know the two points, one on the arc and the other on the line segment the make up for the shortest distance. the language doesn't really matter, even pseudo code would be highly appreciated, thank you!
Related
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.
Consider a polygon with two loops i.e. Outer loop& inner loop as shown in the images attached with this question(One can think of an English letter "e" for example). Can someone please explain how exactly the Ray-casting algorithm will work in such cases.? If possible, please put some images/drawings in answer, for better visualization and understanding.
Imagine a point moving from infinity to the target point along a straight line (will also work with a curve).
The point at infinity is outside the shape. Whenever an outline is met, you switch from outside to inside or conversely. This rule defines internal and external points. In the given case, the inside of the rounded rectangle, inner circles excluded.
Algorithmically, you count the intersections of the segments that define the shape with the half-line to the target.
I have a path drawn in Illustrator, and I need to break the path into section of 100 px. I can't figure out the logic. A line consist of 2 points x1,y1 and x2, y2. And this is for a straight line. My line may have angles/curve, so what do I need to do, to figure out the distance between 2 pixels.Here is a graphic illustration of my line and the sections, which I need to select/extract:
From the shape above, I need to break it into section of lines(note these are not straight lines).
Try referencing the Bug Algorithm. It's a very simple intuitive approach to path planning. I've uploaded an example written in LabVIEW here, but I know there are plenty of others available.
The Bug Algorithm generates a continuous line; a lot of data points; however you can keep a running average of the general diretion it's heading in and detect sharp changes in angles as an important node in the path. This allows you to segment paths from possibly thousands of data points into just a handful.
There are two aspects in your question:
how do I break a path at some point,
how do I find points spaced by a certain distance.
To answer the first, the type of primitives that define the path matters. Assuming a sequence of Bezier cubics, you will resort to the de Casteljau's algorithm: it allows you to construct the control points that correspond to a desired section of a given Bezier arc, from the original control points. Then, a section of a path will be obtained as a starting section of an initial Bezier, then (possibly) a sequence of whole Bezier arcs, and finally the ending section of a last Bezier arc.
To answer the second, assuming that you need an accurate answer, you will need to resort to numerical integration of the arc length along the path. Refer to this post: https://math.stackexchange.com/a/1171564/65203.
For a simple approximation, you can flatten the curve (approximate it as a polyline) and compute the accumulated segment lengths (or even count the pixels if your curve renderer gives you access to this information).
This process is not trivial.
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.
This is a programming homework assignment, of which I have no qualms about doing it myself however I'm quite stuck on the geometry of it. I need to be able to determine the exact point of intersection given the center and radius of a circle and two end points of a vertical line segment, and since geometry isn't my forte I was hoping for some help (even pointers in the right direction would be appreciated!)
This probably isn't the best place to ask a question like this but I'm not really sure where else to look for help, my apologies if it's against the rules or something.
edit:
My apologies, what I am really having trouble with is determining what the points of intersection are (and if there is one intersection or two.) I've tried each solution given and they work great for determining if there is an intersection or not but my problem still persists as I mis-worded my question. If anyone can help with that it'd be much appreciated!
Try http://mathworld.wolfram.com/Circle-LineIntersection.html, this covers the geometry aspect of your problem quite well.
If C=(x0,y0) is the center, r the radius, and k the abscissa of the line, you have
y = y0 +/- sqrt(r^2-(k-x0)^2), but no intersection if r < abs(k-x0)
using the centre [x,y] of the circle, find the distance of this particular line from the centre.refer
now if this distance is > radius of the circle => the line won't intersect. otherwise, it will.