In a given domain, I need to represent mathematical plane in 3D space, so I intend to create a Plane3D class.
I would also have Point3D, Vector3D, Ray3D, and so on. Main use case would be to test/find line-plane intersections, line-plane angles, and other geometric operations.
Question is:
Which would be a good, canonical, computation-friendly way to implement plane definition in a class?
Most obvious candidates would be "point-normal" form, with six numeric parameters (three coordinates for the point, three for the vector) and "general (algebraic) form", with four numeric parameters (one coefficient per coordinate and one constant). Is one of them computationally preferrable?
Also, is there any open source, high level 3D Geometry library already implementing this class, which would be worth taking a look for inspiration?
OBS: Since .NET has System.Windows.Media.Media3D library with some useful classes, most probably I'll implement this in C#, taking advantage of Point3D and Vector3D structs, but I think the question is language-agnostic.
I'd go for what you call algebraic form. A point (x,y,z) is on a plane (a,b,c,d) if a*x+b*y+c*z+d=0.
To intersect that plane with a line spanned by (x1,y1,z1) and (x2,y2,z2), compute s1=a*x1+b*y1+c*z1+d and s2=a*x2+b*y2+c*z2+d. Then your point of intersection is defined by
x=(s1*x2-s2*x1)/(s1-s2)
y=(s1*y2-s2*y1)/(s1-s2)
z=(s1*z2-s2*z1)/(s1-s2)
To compute the angle between a line and a plane, simply compute
sin(α)=(a*x+b*y+c*z)/sqrt((a*a+b*b+c*c)*(x*x+y*y+z*z))
where (a,b,c) represents the normal vector in this representation of the plane, and (x,y,z) is the direction vector of the line, i.e. (x2-x1,y2-y1,z2-z1). The equation is essentially a normalized dot product, and as such is equivalent to the cosine between the two vectors. And since the normal vector is perpendicular to the plane, and sine and cosine differ by 90°, this means that you get the sine of the angle between the line and the plane itself.
Related
I'm reading a paper the presents a quaternion based complimentary filter but I'm confused on how and why they represent the attitude of an object with a single quaternion denoted by "q" in the paper. I thought that a unit quaternion represents a rotation like a rotation matrix and is defined as an operator. For example to rotate a vector v... v_f=q(v_0)q^-1
In the paper they state "If the device has an attitude q at time t, the attitude at the subsequent time-step t + dt, is denoted q′"
How can an object have an attitude q?
They probably misuse the word attitude. What they mean by attitude is the orientation of an object.
The orientation is usually defined in physics as a frame made of three orthogonal vectors.
You can also define the orientation as a rotation matrix but that forces you to choose a coordinate system. Other way is to use a quaternion, which is an algebraic entity like a vector (and thus coordinate-free).
Hope that helps.
Say I want to construct a 3D cubic Bézier curve, and I already have both end-points, and the direction (normal vector) for both control points. How can I choose the distance of both control points to their respective end-points in order to make the curve as 'nicely rounded' as possible?
To formalize 'nicely rounded': I think that means maximizing the smallest angle between any two segments in the curve. For example, having end-points (10, 0, 0) and (0, 10, 0) with respective normal vectors (0, 1, 0) and (1, 0, 0) should result in a 90° circular arc. For the specific case of 2D circular arcs, I've found articles like this one. But I haven't been able to find anything for my more general case.
(Note that these images are just to illustrate the 'roundness' concept. My curves are not guaranteed to be plane-aligned. I may replace the images later to better illustrate that point.)
This is a question of aesthetics, and if the real solution is unknown or too complicated, I would be happy with a reasonable approximation. My current approximation is too simplistic: choosing half the distance between the two end-points for both control point distances. Someone more familiar with the math will probably be able to come up with something better.
(PS: This is for open-source software, and I would be happy to give credit on GitHub.)
Edit: Here are some other images to illustrate a 3D case (jsfiddle):
Edit 2: Here's a screenshot of an unstable version of ApiNATOMY to give you an idea of what I'm trying to do. I'm creating 3D tubes to represent blood-vessels, connecting different parts of an anatomical schematic:
(They won't let me put in a jsfiddle link if I don't include code...)
What you are basically asking is to have curvature over the spline as constant as possible.
A curve with constant curvature is just a circular arc, so it makes sense to try to fit such an arc to your input parameters. In 2D, this is easy: construct the line which goes through your starting point and is orthogonal to the desired direction vector. Do the same for the ending point. Now intersect these two lines: the result is the center of the circle which passes through the two points with the desired direction vectors.
In your example, this intersection point would just be (0,0), and the desired circular arc lies on the unit circle.
So this gives you a circular arc, which you can either use directly or use the approximation algorithm which you have already cited.
This breaks down when the two direction vectors are collinear, so you'd have to fudge it a bit if this ever comes up. If they point at each other, you can simply use a straight line.
In 3D, the same construction gives you two planes passing through the end points. Intersect these, and you get a line; on this line, choose the point which minimizes the sum of squared distances to the two points. This gives you the center of a sphere which touches both end points, and now you can simply work in the plane spanned by these three points and proceed as in 2D.
For the special case where your two end points and the two known normal vector for the control points happen to make the Bezier curve a planar one, then basically you are looking for a cubic Bezier curve that can well approximate a circular arc. For this special case, you can set the distance (denoted as L) between the control point and their respective end point as L = (4/3)*tan(A/4) where A is the angle of the circular arc.
For the general 3D case, perhaps you can apply the same formula as:
compute the angle between the two normal vectors.
use L=(4/3)*tan(A/4) to decide the location of your control points.
if your normals are aligned in a plane
What you're basically doing here is creating an elliptical arc, in 3D, where the "it's in 3D" part is completely irrelevant, since it's just a 2D curve, rotated/translated to sit in your 3D space. So let's just solve the 2D case, and then the RT is entirely up to you.
Creating the "perfect" cubic Bezier between two points on an arc comes with limitations. You basically can't create good looking arcs that span more than a quarter circle. So, with that said: your start and end point normals give you a 2D angle between your normal vectors, which is the same angle as between your start and end tangents (since normals are perpendicular to tangents). So, let's:
align our curve so that the tangent at the start is 0
plug the angle between tangents into the formula given in the section on Circle approximation in the Primer on Bezier curves. This is basically just dumb "implementing the formula for c1x/c1y/c2x/c2y as a function that takes an angle as argument, and spits out four values as c1(x,y) and c2(x,y) coordinats".
There is no step 3, we're done.
After step 2, you have your control points in 2D to create the most circular arc between a start and end point. Now you just need to scale/rotate/translate it in 3D so that it lines up with where you needed your start and end point to begin with.
if your normals are not aligned in a plane
Now we have a problem, although one that we can deal with by treating the dimensions as separate things entirely. Instead of creating a single 2D curve, we're going to create three: one that's the X/Y projection, one that's the X/Z projection, and one that's the Y/Z projection. For all three of these, we're going to abstract the control points in exactly the same way as before, and then we simply take the projective control points (three for each control point), and then go "okay, we now have X, Y, and Z projective coordinates. That means we have (X,Y,Z) coordinates", and done again.
I have measured points on a two dimensional square lattice.
.
How can I fit the data to a square lattice? I guess some methods like curve fitting or least square approximation would work, but I couldn't find any literature for the same problem.
the lattice has discrete points
the lattice is modeled with multiple parameters: translation (x0 and y0), rotation (θ), lattice spacing (d), and it cannot be represented as y=f(x; a,b,...) form.
I want to detect the best rototraslation matrix between two set of points.
The second set of points is the same of the first, but rotated, traslated and affecteb by noise.
I tried to use least squared method by obviously the solution is usually similar to a rotation matrix, but with incompatible structure (for example, where i should get a value that represents the cosine of an angle i could get a value >1).
I've searched for the Constrained Least Squared method but it seems to me that the constrains of a rototraslation matrix cannot be expressed in this form.
In this PDF i've stated the problem more formally:
http://dl.dropbox.com/u/3185608/minquad_en.pdf
Thank you for the help.
The short answer: What you will need here is "Principal Component Analysis".
Apply this to both sets of points centered at their respective centers of mass. The PCA will effectively give you a rotation matrix for each aligned to the data set principal components. Multiplying the inverse matrix of the original set by the new rotation will give you a matrix that takes the old (centered) set to the new. Inverse translations and translations can similarly be applied to the rotation to create a homogeneous matrix that maps the one set to the other.
The book PRINCE, Simon JD. Computer vision: models, learning, and inference. Cambridge University Press, 2012.
gives, in Appendix "B.4 Reparameterization", some info about how to constrain a matrix to be a rotation matrix.
It seems to me that your problem has also a solution based on SVD: see the Kabsch algorithm also described by Olga Sorkine-Hornung and Michael Rabinovich in
Least-Squares Rigid Motion Using SVD and, more practically, by Nghia Kien Ho in FINDING OPTIMAL ROTATION AND TRANSLATION BETWEEN CORRESPONDING 3D POINTS.
Suppose you have a list of 2D points with an orientation assigned to them. Let the set S be defined as:
S={ (x,y,a) | (x,y) is a 2D point, a is an orientation (an angle) }.
Given an element s of S, we will indicate with s_p the point part and with s_a the angle part. I would like to know if there exist an efficient data structure such that, given a query point q, is able to return all the elements s in S such that
(dist(q_p, s_p) < threshold_1) AND (angle_diff(q_a, s_a) < threshold_2) (1)
where dist(p1,p2), with p1,p2 2D points, is the euclidean distance, and angle_diff(a1,a2), with a1,a2 angles, is the difference between angles (taken to be the smallest one). The data structure should be efficient w.r.t. insertion/deletion of elements and the search as defined above. The number of vectors can grow up to 10.000 and more, but take this with a grain of salt.
Now suppose to change the above requirement: instead of using the condition (1), let's request all the elements of S such that, given a distance function d, we want all elements of S such that d(q,s) < threshold. If i remember well, this last setup is called range-search. I don't know if the first case can be transformed in the second.
For the distance search I believe the accepted best method is a Binary Space Partition tree. This can be stored as a series of bits. Each two bits (for a 2D tree) or three bits (for a 3D tree) subdivides the space one more level, increasing resolution.
Using a BSP, locating a set of objects to compare distances with is pretty easy. Just find the smallest set of squares or cubes which contain the edges of your distance box.
For the angle, I don't know of anything. I suppose that you could store each object in a second list or tree sorted by its angle. Then you would find every object at the proper distance using the BSP, every object at the proper angles using the angle tree, then do a set intersection.
You have effectively described a "three dimensional cyclindrical space", ie. a space that is locally three dimensional but where one dimension is topologically cyclic. In other words, it is locally flat and may be modeled as the boundary of a four-dimensional object C4 in (x, y, z, w) defined by
z^2 + w^2 = 1
where
a = arctan(w/z)
With this model, the space defined by your constraints is a 2-dimensional cylinder wrapped "lengthwise" around a cross section wedge, where the wedge wraps around the 4-d cylindrical space with an angle of 2 * threshold_2. This can be modeled using a "modified k-d tree" approach (modified 3-d tree), where the data structure is not a tree but actually a graph (it has cycles). You can still partition this space into cells with hyperplane separation, but traveling along the curve defined by (z, w) in the positive direction may encounter a point encountered in the negative direction. The tree should be modified to actually lead to these nodes from both directions, so that the edges are bidirectional (in the z-w curve direction - the others are obviously still unidirectional).
These cycles do not change the effectiveness of the data structure in locating nearby points or allowing your constraint search. In fact, for the most part, those algorithms are only slightly modified (the simplest approach being to hold a visited node data structure to prevent cycles in the search - you test the next neighbors about to be searched).
This will work especially well for your criteria, since the region you define is effectively bounded by these axis-defined hyperplane-bounded cells of a k-d tree, and so the search termination will leave a region on average populated around pi / 4 percent of the area.