Given a non-self-intersecting polygon as a list of points (p1...pn), and a point (A) outside that polygon:
I want to calculate the total angular field of view covered by the polygon from the point, as well as the direction from the point towards the middle of that field of view (as either a vector or angle from horizontal).
Visually, I want the angle Θ and direction of the green vector:
Diagram
I tried finding the minimum and maximum angles from horizontal to each of the polygon points, but I don't know how to tell which is the start of the range and which is the end. Assuming the smaller angle is the start gives incorrect results from the left of a simple box.
I'm guessing the solution will have something to do with whether the polygon points are in clockwise or counterclockwise order.
Whether the polygon goes clockwise or counterclockwise doesn't matter. What matters is that the extent of any edge, as seen from the point, must be less than π radians. That will tell us whether the edge -- as seen from the point -- goes counterclockwise from A to B, or from B to A.
For example, suppose the bearings (in radians) from the point to the vertices are {0, 2π/5, 4π/5, 6π/5, 8π/5}. If the edges are represented by the (unordered) pairs (A,C), (A,E), (B,D), (B,E), (C,D). Then the edges run:
0->4π/5
2π/5->6π/5
4π/5->6π/5
8π/5->0
8π/5->2π/5
So the range of the polygon is [8π/5, 6π/5].
I'm trying to build a visualization. The idea is to form a sphere out of circles. Similar to this:
In other words I want to pack certain number, lets say N, of circles in a sphere. All circles should have same radius and closest circles should have same distance from their centers.
I know how to draw a circle so the real problem is to find centers of the circles. How to find points on sphere where all neighboring points will be located on the same distance from each other and what is this distance is equal to.
Describing this task as a function it would have following notation
[RadiusOfSphere, NumberOfPoints] -> [CenterOfCircle, MaximumCircleRadius]
As the problem is stated, there may be no possible solution. The obstacle is that not every number of points N may admit a packing of the kind you are specifying.
You can take the radius to be fixed at 1 and scale afterwards in all cases here.
The requirement that the circles have the same radius and have the same distance to nearest neighbors is tantamount to saying that nearest neighbors are tangent to each other. All nearest neighbors, under this assumption, have the same gap on a line segment connecting their centers. Expand the radius by one-half this gap and now all closest circles are tangent. The centers don't move under this transformation. So we can assume the circles are tangent.
Circle packing is not a trivial mathematical problem, and proofs minimality or non-existence have tended to require computer-assisted proofs. I don't even know about that existence for arbitrary N. There might be a result out there, but it didn't come up in a brief search.
If you don't need every N, but simply a series of them so you can get N big enough, every quasiregular polyhedron and their duals (such as the rhombic triacontahedron, which folks on here might know better as a 30-sided die) all admit circle packings. So do their subdivisions (the figure induced by splitting edges into k pieces). So for given number of circles >N, pick one of the above polyhedrons and subdivide it so that the number of faces is large enough.
In G-codes a clockwise arc can be specified by e.g.
G02 X2 Y0 R2
This code should give an arc between the current position and (2,0) with radius=2.
According to several sources (And Math) eg.
There will always be two circles/arcs that satisfies these conditions each with a clockwise arc.
Which one is chosen and is it implementation/manufacturer dependent?
According to CNCCookbook's G-code Tutorial, it depends on the controller. Some controller use the sign of the radius to choose, some never lets you make an arc of more than 90°.
Given the two choices shown, the controller chooses the path based on the sign of the radius. Negative forces the longer arc, positive the shorter. The negative sign forces the controller to seek a viable arc of more than 180 degrees.
Some controllers are touchier still and will not program an arc that crosses a quadrant line. Hence, the largest angle an arc can follow is 90 degrees, and that angle must not cross 0, 90, 180, or 270 degrees. For angles of 90 degrees that cross a quadrant line, they must be broken into two pieces, with the join between the pieces being right on the quadrant line.
This is related to the arc drawn by HTML5 canvas "arcTo" function. I need to calculate the two tangent points of a circle with the radius R and two lines given by three points Q(x0,y0), P(x1,y1) and R(x2,y2).
The sketch explains the problem more. I need to find the tangent points A(xa,ya) and B(xb,yb). Note that the center of the circle is not given. Please help.
This is a question of solving a triangle with 2 known angles and one known side. Label the centre of the circle C, then the side you know is BC (or AC if you want). Angle PBC (CAP) is a right angle. The line CP bisects the angle RPQ.
Not all such triangles have a solution.
I have two 2d circles in 3d space (defined by a center, normal, and radius) and I'm trying to come up with a pair of points that is one of the set of closest pairs of points. I know that there are anywhere from 1 to an infinite number of point pairs, I just need a single matching pair.
Is there a simple way to do that? Precision is not essential. The radius of both circles are the same, non-zero value.
In case the background is helpful, my overall algorithm takes in a NURBS curve in space and extrudes a 2d polygon along the curve, yielding a deformed cylinder. I just sample several points along the curve. The normal of each circle is the NURBS curve tangent, and I'm trying to figure out how to align adjacent samples, so I don't get weird twisting. It seems that the closest points on adjacent samples should be aligned.
Thanks for all the responses here.. this part of the project got a little delayed, which is why I haven't tested all the answers yet. I'll be sure to toss up some images here and mark an answer when I get to work on this again.
What you are really trying to compute is the pair of points that minimizes the distance between points that lie on 2 different circles in 3 dimensions. The method that you should be employing to find the exact solution (as in almost all optimization problems) is to represent the distance as a function of all possible points and to take its derivate with respect to the independent variables and set the resulting expressions to 0. Since you have 2 circles, you will have 2 independent variables (ie. the angle of a point on one circle and one on the other circle). Once you have solved the minimization equations you would have also found the points on the circles that will satisfy your constraint. (Basically you will find the angles on the circles for the pair of points you are looking for.)
I have found a paper online (at this site) that rigorously goes through with the calculations but the end result is solving an 8th order polynomial equation. You might try to simplify the equations and come up with a less exact solution that satisfies your needs.
There is also an paper that claims to have a much faster algorithm for finding the distance between two circles in 3d; however, I cannot view the contents and, thus, cannot tell if it also gives you the pair of points that satisfy that condition.
UPDATE: Having re-read your question, I see that even though you are asking for a way to find the closest pair of points on two circles in 3 dimensions, I think, you should pay more attention to the properties of the NURBS curve that you are trying to extrude the 2D polygon along. You mention that the orientation of the circle at a given point on the curve is specified by the tangent vector at that point. However, there is more to 3D curves than just the tangent vector; there is the normal (or curvature) vector that points towards the center of curvature of the curve at a given point and then there is the torsion vector that basically specifies the amount of "lift" of the curve from the plane given by the tangent and the normal vectors. All of these define a (what is called) Frenet frame. You can read up more on these at the Wikipedia article.
My suspicion is that you can achieve the effect you desire by joining the points of consecutive circles that each lie along the the normal vector direction of the underlying 3D curve. That way, you will have twisting only when the curve is actually twisting, ie when the torsion vector is non-zero and the normal vector is changing direction as well. In other circumstances, this should satisfy your actual need.
You probably don't need the overkill of finding closest points on consecutive circles.
For what you describe, it is sufficient to select a point on the perimeter of the first circle and find the point on the perimeter of each circle along that is closest to the one selected for the previous circle; this will completely constrain the polygonization, with no twisting, and should be much easier to solve than the general case - simply find the point on the plane containing the second circle that is closest to that selected in the first, and intersect the line passing through that point and the second circle's center with the second circle's perimeter.
However, this might not yield as pleasing a polygonisation for the extruded cylinder as keeping the polygon area constant as possible, and to do that will require some twisting between adjacent circles.
Yikes, unless the circles happen to be on the same plane or parallel planes I think the only way to do it is to find a minimum on the equation of the distance between two points on the circle.
http://www.physicsforums.com/showthread.php?t=123168
That link shows how to get the equation of each circle in 3D space, then minimize for the distance formula between those equations. Not pretty though, hopefully someone will come up with something more clever.
I think with the two closest points you might still get weird twisting... An extreme example: Let's assume both circles have the R=1. If the first circle's centre is O, and it is sitting on X-Y plane, and the second circle's centre is sitting at X=1,Y=0,Z=0.01, and it just slightly tilted in the growing direction of X, the closest points on the two circles will for sure get the "weird twist" you are trying to avoid. Since the closest points would not get you the weird twist in case the second circle is at X=0,Y=0,Z=0.01 and is equally tilted, then at some point the statements "aligned to two closest points on two circles" and "no weird twisting seen" no longer correspond to each other.
Assuming this can happen within the constraint of NURBS, here's another idea. In the start, take the three points on the NURBS curve - two that belong to the centers of your circles, and the third one precisely inbetween. Draw a plane between the three. This plane will cross the two circles at 4 points. Two of these points will be on the same "side" of the line that connects the centers of the circles - they are your alignment points.
For the next alignment points you would take the alignment point of the "previous circle", and draw the plane between the center of the "previous circle", this alignment point, and the center of the "new circle". From this you get the "next alignment point" based on the intersection with the other circle.
Next step - "previous circle" = "new circle", and the "new circle" - your next one according to the NURBS curve.
If the radii from the centers of the circles to the selected alignment points cross, you know you the picture will look a bit ugly - that's the scenario where with the "closest point" algorithm you'd still get the weird twisting.
I think the coordinates of the point on the circle that is intersection with the plane going via its center should be easy to calculate (it's a point on the line made by intersection of the two planes, one of the circle and the target plane; at the distance R from the center).
I don't have the rigorous proof to fully assert or deny the above - but hopefully it helps at all, and I think it should be quick enough to verify, compared to calculating the closet points on the two circles... (If there are any flaws in my logic, the corrections in the comments are very welcome).
The thread here, mentioned in another answer gives the parameterization formula for a 3D circle: P = R cos(t) u + R sin(t) nxu + c, where u is a unit vector from the centre of the circle to any point on the circumference; R is the radius; n is a unit vector perpendicular to the plane and c is the centre of the circle, t goes from 0 to 2pi, and by nxu I mean "n cross u". Parameterize one circle this way, and another similarly with a different parameter, say s. Then each point Pt on the first circle will have coordinates in the variable t, and each point Ps on the second circle will have coordinates in the variable s.
Write the distance function d(s,t) between Ps and Pt in the usual way (or better, the square of the Euclidean distance so you don't have to mess with the square root when you take derivatives). The graph of this function d of two variables is a surface over a 2pi by 2pi square in the s,t plane, and it's minimum is what you're after. You can determine it with the standard calculus methods, e.g. as explained here.
Extend the circles to planes (using the center points and normals). If the planes are parallel, then any points will do. If the planes are not parallel, then they intersect in a line. Construct the plane through the two centers of the circles perpendicular to the line. The two circles intersect this new plane in four points. These four points are the two nearest points and the two farthest points on the circles.
Isn't this just a matter of constructing the line between the two centers of the circles/spheres and finding the intersection of the line and the circles? The solutions that are closest are it (unless the circle intersect, then the answer depends on how you want to interpret that case).