I have three vectors in 3D space, one is a light source, one is a ray and one is the point on a circle a ray hits. With this information, how can I work out the vector which points back at the light source from the point the ray hits the circle?
What you really have is two points (light source, circle intersection), and a vector between them, right? The vector is already implied by the two points -- it's the intersection coordinates minus light source coordinations. To reverse it, just negate all the coordinates of the vector!
Related
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 am given a polyhedron which is represented by a list of planes. The volume delimited by these planes is the actual polyhedron.
I have a list of triangles, and I need to remove all the triangles that intersect or is contained in this polyhedron. My idea is to check each plane for an intersection with the triangle. If it does intersect the triangle, then check if the line segment representing the intersection contains a point that lies on the same side of all other planes.
To catch the case where the triangle is fully contained, we can just check if any of the triangle's corners is contained in the polyhedron (by checking that the point lies on the same side of all planes).
I'm not sure if this solution even works for all cases however, or if there is a more elegant solution. I'm also not sure how I can figure out if the line segment of an intersection contains a point on the same side of all the other planes.
I have thought about the separating axis theorem too, but that would require me to convert the polyhedron into some different representation (since the planes are infinite), and I'm not sure how to do that.
Any help would be appreciated!
As #NicoSchertler suggested in a comment, a solution is to take each triangle and clip it on all the planes. If there are no points left (or under 3 points, so it is not a triangle), the triangle intersects the polyhedron. This seems to work well.
I would like to check if two polygons (number of vectors unclear) have the same shape. Without rotations this is easy, but how do I do this with rotated polygons? I need to know the rotation angle, too.
boolean polygonsHaveSameShape(PVector[] polygon1, PVector[] polygon2){
…
}
float getRotationAngle(PVector[] polygon1, PVector[] polygon2){
…
}
With a small number of vertices it might be worth checking the distances between each vertex and the others.
In your square example dist(p1,p2), dist(p1,p3), dist(p1,p4), dist(p2,p3), dist(p2,p4) and dist(p3,p4). These values will exist for each polygon. There will be a point that has the same distance set as p1, and as p2, and so on.
Once you have a vertex in one polygon where all the distances connected to it are the same as those in the second polygon then you can use one of those lines to determine the angle of rotation.
Hope that made sense.
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).
Given a list of points that form a simple 2d polygon oriented in 3d space and a normal for that polygon, what is a good way to determine which points are specific 'corner' points?
For example, which point is at the lower left, or the lower right, or the top most point? The polygon may be oriented in any 3d orientation, so I'm pretty sure I need to do something with the normal, but I'm having trouble getting the math right.
Thanks!
You would need more information in order to make that decision. A set of (co-planar) points and a normal is not enough to give you a concept of "lower left" or "top right" or any such relative identification.
Viewing the polygon from the direction of the normal (so that it appears as a simple 2D shape) is a good start, but that shape could be rotated to any arbitrary angle.
Is there some other information in the 3D world that you can use to obtain a coordinate-system reference?
What are you trying to accomplish by knowing the extreme corners of the shape?
Are you looking for a bounding box?
I'm not sure the normal has anything to do with what you are asking.
To get a Bounding box, keep 4 variables: MinX, MaxX, MinY, MaxY
Then loop through all of your points, checking the X values against MaxX and MinX, and your Y values against MaxY and MinY, updating them as needed.
When looping is complete, your box is defined as MinX,MinY as the upper left, MinX, MaxY as upper right, and so on...
Response to your comment:
If you want your box after a projection, what you need is to get the "transformed" points. Then apply bounding box loop as stated above.
Transformed usually implies 2D screen coordinates after a projection(scene render) but it could also mean the 2D points on any plane that you projected on to.
A possible algorithm would be
Find the normal, which you can do by using the cross product of vectors connecting two pairs of different corners
Create a transformation matrix to rotate the polygon so that it is planer in XY space (i.e. normal alligned along the Z axis)
Calculate the coordinates of the bounding box or whatever other definition of corners you are using (as the polygon is now aligned in 2D space this is a considerably simpler problem)
Apply the inverse of the transformation matrix used in step 2 to transform these coordinates back to 3D space.
I believe that your question requires some additional information - namely the coordinate system with respect to which any point could be considered "topmost", or "leftmost".
Don't forget that whilst the normal tells you which way the polygon is facing, it doesn't on its own tell you which way is "up". It's possible to rotate (or "roll") around the normal vector and still be facing in the same direction.
This is why most 3D rendering systems have a camera which contains not only a "view" vector, but also "up" and "right" vectors. Changes to the latter two achieve the effect of the camera "rolling" around the view vector.
Project it onto a plane and get the bounding box.
I have a silly idea, but at the risk of gaining a negative a point, I'll give it a try:
Get the minimum/maximum value from
each three-dimensional axis of each
point on your 2d polygon. A single pass with a loop/iterator over the list of values for every point will suffice, simply replacing the minimum and maximum values as you go. The end result is a list that has the "lowest" X, Y, Z coordinates and "highest" X, Y, Z coordinates.
Iterate through this list of min/max
values to create each point
("corner") of a "bounding box"
around the object. The result
should be a box that always contains
the object regardless of axis
examined or orientation (no point on
the polygon will ever exceed the
maximum or minimums you collect).
Then get the distance of each "2d
polygon" point to each corner
location on the "bounding box"; the
shorter the distance between points,
the "closer" it is to that "corner".
Far from optimal, certainly crummy, but certainly quick. You could probably post-capture this during the object's rotation, by simply looking for the min/max of each rotated x/y/z value, and retaining a list of those values ahead of time.
If you can assume that there is some constraints regarding the shapes, then you might be able to get away with knowing less information. For example, if your shape was the composition of a small square with a long thin triangle on one side (i.e. a simple symmetrical geometry), then you could compare the distance from each list point to the "center of mass." The largest distance would identify the tip of the cone, the second largest would be the two points farthest from the tip of the cone, etc... If there was some order to the list, like points are entered in counter clockwise order (about the normal), you could identify all the points. This sounds like a bit of computation, so it might be reasonable to try to include some extra info with your shapes, like the "center of mass" and a reference point that is located "up" above the COM (but not along the normal). This will give you an "up" vector that you can cross with the normal to define some body coordinates, for example. Also, the normal can be defined by an ordering of the point list. If you can't assume anything about the shapes (or even if the shapes were symmetrical, for example), then you will need more data. It depends on your constraints.
If you know that the polygon in 3D is "flat" you can use the normal to transform all 3D-points of the vertices to a 2D-representation (of the points with respect to the plan in which the polygon is located) - but this still leaves you with defining the origin of this coordinate-system (but this don't really matter for your problem) and with the orientation of at least one of the axes (if you want orthogonal axes you can still rotate them around your choosen origin) - and this is where the trouble starts.
I would recommend using the Y-axis of your 3D-coordinate system, project this on your plane and use the resulting direction as "up" - but then you are in trouble in case your plan is orthogonal to the Y-axis (now you might want to use the projected Z-Axis as "up").
The math is rather simple (you can use the inner product (a.k.a. scalar product) for projection to your plane and some matrix stuff to convert to the 2D-coordinate system - you can get all of it by googling for raytracer algorithms for polygons.