I want to place one polyhedron(Object) into another (Container). Both the polyhedrons are convex and defined by a set of points and triangles. The container has a constant size. The object can be scaled and should be strictly inside the container. I want to compute the position and orientation for the object which makes it the biggest. Approximate and efficient solutions also help.
Any suggestions? Thanks a lot.
Suggestion for a quick & suboptimal solution, based on ellipsoids:
For both vertex sets, center around the gravity center and normalize the coordinates by computing the equivalent ellipsoid of inertia, giving you a more isotropic set.
For the outer set, find the shortest distance between the faces and the origin; for the inner set, find the farthest distance to a vertex. This gives you two spheres, one enclosed, one enclosing.
Now transform the enclosed sphere to the coordinates of the enclosing one, giving an ellipsoid: the longer axis of the ellipsoid tells you how much you can inflate it to fit in the sphere.
This approximation can be poor if the polyhedra are skewed.
You can slightly improve this solution by drawing rays from the center of the inner polyhedron through all vertices, and hitting the outer polyhedron, possibly giving you an extra growth factor.
Another suggestion, if running-time allows it:
For a fixed pose of the outer polyhedron, the pose of the inner polyhedron is defined by 3 translation and 3 rotation parameters (such as Euler angles) around an arbitrary center.
When these parameters are fixed, casting rays from the center through the inner vertices until you hit the outer polyhedron gives you the allowed scaling factor.
Now the problem is recast as the maximization of a function of 6 variables, with local maxima to be expected. This can be addressed by Hooke & Jeeves steps, the up-hill simplex method (Nelder-Mead) and/or simulated annealing.
I do not recommend to start from the solution in my other answer and stay close to it, as you could be trapped in a local maximum.
Related
I am currently creating a feature and patterning it across a flat plane to get the maximum number of features to fit on the plane. I do this frequently enough to warrant building some sort of marcro for this if possible. The issue that I run into is I still have to manually set the spacing between the parts. I want to be able to create a feature and have it determine "best" fit spacing given an area while avoiding overlaps. I have had very little luck finding any resources describing this. Any information or links to potentially helpful resources on this would be much appreciated!
Thank you.
Before, you start the linear pattern bit:
Select the face2 of that feature2, get the outer most loop2 of edges. You can test for that using loop2.IsOuter.
Now:
if the loop has one edge: that means it's a circle and the spacing must superior to the circle's radius
if the loop has more that one edge, that you need to calculate all the distances between the vertices and assume that the largest distance is the safest spacing.
NOTA: If one of the edges is a spline, then you need a different strategy:
You would need to convert the face into a sketch and finds the coordinates of that spline to calculate the highest distances.
Example: The distance between the edges is lower than the distance between summit of the splines. If the linear pattern has the a vertical direction, then spacing has to be superior to the distance between the summit.
When I say distance, I mean the distance projected on the linear pattern direction.
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.
In my meshing application I will have to specify fix points within a domain. The idea is that, the fix points must also be the element points after the domain is being meshed.
Furthermore, the elements around the fix points should be more dense. The general concept is that for the fix points, there should exist a radius r around those points, such that the mesh size inside r is of different sizes than outside of the r. The mesh sizes inside and outside of the r should be specifiable.
Are these two things doable in CGAL 2D Mesh algorithm?
Using your wording, all the input point of the initial constrained Delaunay triangulation will be fix points, because the 2D mesh generator only insert new points in the triangulation: it never removes any point.
As for the density, you can copy, paste, and modify a criteria class, such as CGAL::Delaunay_mesh_size_criteria_2<CDT> so that the local size upper bound is smaller around the fix points.
Now, the difficulty is how to implement that new size policy. Your criteria class could store a const reference to another Delaunay_triangulation_2, that contains only the fixed points you want. Then, for each triangle query, you can call nearest_vertex and then actually check if the distance between the query point is smaller that the radius bound of your circles. For a triangle, you can either verify that for only its barycenter, or for all three points of the triangle. Then, according to the result of that/those query(s), you can modify the size bound, in the code of your copy of CGAL::Delaunay_mesh_size_criteria_2<CDT>.
Yes, no points will be removed from the triangulation by the mesher.
Note however that if you insert points too close to a constraint this will induce a refinement of the constraint while it is not Gabriel.
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.
My question is fairly simple. I have two tetrahedra, each with a current position, a linear speed in space, an angular velocity and a center of mass (center of rotation, actually).
Having this data, I am trying to find a (fast) algorithm which would precisely determine (1) whether they would collide at some point in time, and if it is the case, (2) after how much time they collided and (3) the point of collision.
Most people would solve this by doing triangle-triangle collision detection, but this would waste a few CPU cycles on redundant operations such as checking the same edge of one tetrahedron against the same edge of the other tetrahedron upon checking up different triangles. This only means I'll optimize things a bit. Nothing to worry about.
The problem is that I am not aware of any public CCD (continuous collision detection) triangle-triangle algorithm which takes self-rotation in account.
Therefore, I need an algorithm which would be inputted the following data:
vertex data for three triangles
position and center of rotation/mass
linear velocity and angular velocity
And would output the following:
Whether there is a collision
After how much time the collision occurred
In which point in space the collision occurred
Thanks in advance for your help.
The commonly used discrete collision detection would check the triangles of each shape for collision, over successive discrete points in time. While straightforward to compute, it could miss a fast moving object hitting another one, due to the collision happening between discrete points in time tested.
Continuous collision detection would first compute the volumes traced by each triangle over an infinity of time. For a triangle moving at constant speed and without rotation, this volume could look like a triangular prism. CCD would then check for collision between the volumes, and finally trace back if and at what time the triangles actually shared the same space.
When angular velocity is introduced, the volume traced by each triangle no longer looks like a prism. It might look more like the shape of a screw, like a strand of DNA, or some other non-trivial shapes you might get by rotating a triangle around some arbitrary axis while dragging it linearly. Computing the shape of such volume is no easy feat.
One approach might first compute the sphere that contains an entire tetrahedron when it is rotating at the given angular velocity vector, if it was not moving linearly. You can compute a rotation circle for each vertex, and derive the sphere from that. Given a sphere, we can now approximate the extruded CCD volume as a cylinder with the radius of the sphere and progressing along the linear velocity vector. Finding collisions of such cylinders gets us a first approximation for an area to search for collisions in.
A second, complementary approach might attempt to approximate the actual volume traced by each triangle by breaking it down into small, almost-prismatic sub-volumes. It would take the triangle positions at two increments of time, and add surfaces generated by tracing the triangle vertices at those moments. It's an approximation because it connects a straight line rather than an actual curve. For the approximation to avoid gross errors, the duration between each successive moments needs to be short enough such that the triangle only completes a small fraction of a rotation. The duration can be derived from the angular velocity.
The second approach creates many more polygons! You can use the first approach to limit the search volume, and then use the second to get higher precision.
If you're solving this for a game engine, you might find the precision of above sufficient (I would still shudder at the computational cost). If, rather, you're writing a CAD program or working on your thesis, you might find it less than satisfying. In the latter case, you might want to refine the second approach, perhaps by a better geometric description of the volume occupied by a turning, moving triangle -- when limited to a small turn angle.
I have spent quite a lot of time wondering about geometry problems like this one, and it seems like accurate solutions, despite their simple statements, are way too complicated to be practical, even for analogous 2D cases.
But intuitively I see that such solutions do exist when you consider linear translation velocities and linear angular velocities. Don't think you'll find the answer on the web or in any book because what we're talking about here are special, yet complex, cases. An iterative solution is probably what you want anyway -- the rest of the world is satisfied with those, so why shouldn't you be?
If you were trying to collide non-rotating tetrahedra, I'd suggest a taking the Minkowski sum and performing a ray check, but that won't work with rotation.
The best I can come up with is to perform swept-sphere collision using their bounding spheres to give you a range of times to check using bisection or what-have-you.
Here's an outline of a closed-form mathematical approach. Each element of this will be easy to express individually, and the final combination of these would be a closed form expression if one could ever write it out:
1) The equation of motion for each point of the tetrahedra is fairly simple in it's own coordinate system. The motion of the center of mass (CM) will just move smoothly along a straight line and the corner points will rotate around an axis through the CM, assumed to be the z-axis here, so the equation for each corner point (parameterized by time, t) is p = vt + x + r(sin(wt+s)i + cos(wt + s)j ), where v is the vector velocity of the center of mass; r is the radius of the projection onto the x-y plane; i, j, and k are the x, y and z unit vectors; and x and s account for the starting position and phase of rotation at t=0.
2) Note that each object has it's own coordinate system to easily represent the motion, but to compare them you'll need to rotate each into a common coordinate system, which may as well be the coordinate system of the screen. (Note though that the different coordinate systems are fixed in space and not traveling with the tetrahedra.) So determine the rotation matrices and apply them to each trajectory (i.e. the points and CM of each of the tetrahedra).
3) Now you have an equation for each trajectory all within the same coordinate system and you need to find the times of the intersections. This can be found by testing whether any of the line segments from the points to the CM of a tetrahedron intersects the any of the triangles of another. This also has a closed-form expression, as can be found here.
Layering these steps will make for terribly ugly equations, but it wouldn't be hard to solve them computationally (although with the rotation of the tetrahedra you need to be sure not to get stuck in a local minimum). Another option might be to plug it into something like Mathematica to do the cranking for you. (Not all problems have easy answers.)
Sorry I'm not a math boff and have no idea what the correct terminology is. Hope my poor terms don't hide my meaning too much.
Pick some arbitrary timestep.
Compute the bounds of each shape in two dimensions perpendicular to the axis it is moving on for the timestep.
For a timestep:
If the shaft of those bounds for any two objects intersect, half timestep and start recurse in.
A kind of binary search of increasingly fine precision to discover the point at which a finite intersection occurs.
Your problem can be cast into a linear programming problem and solved exactly.
First, suppose (p0,p1,p2,p3) are the vertexes at time t0, and (q0,q1,q2,q3) are the vertexes at time t1 for the first tetrahedron, then in 4d space-time, they fill the following 4d closed volume
V = { (r,t) | (r,t) = a0 (p0,t0) + … + a3 (p3,t0) + b0 (q0,t1) + … + b3 (q3,t1) }
Here the a0...a3 and b0…b3 parameters are in the interval [0,1] and sum to 1:
a0+a1+a2+a3+b0+b1+b2+b3=1
The second tetrahedron is similarly a convex polygon (add a ‘ to everything above to define V’ the 4d volume for that moving tetrahedron.
Now the intersection of two convex polygon is a convex polygon. The first time this happens would satisfy the following linear programming problem:
If (p0,p1,p2,p3) moves to (q0,q1,q2,q3)
and (p0’,p1’,p2’,p3’) moves to (q0’,q1’,q2’,q3’)
then the first time of intersection happens at points/times (r,t):
Minimize t0*(a0+a1+a2+a3)+t1*(b0+b1+b2+b3) subject to
0 <= ak <=1, 0<=bk <=1, 0 <= ak’ <=1, 0<=bk’ <=1, k=0..4
a0*(p0,t0) + … + a3*(p3,t0) + b0*(q0,t1) + … + b3*(q3,t1)
= a0’*(p0’,t0) + … + a3’*(p3’,t0) + b0’*(q0’,t1) + … + b3’*(q3’,t1)
The last is actually 4 equations, one for each dimension of (r,t).
This is a total of 20 linear constraints of the 16 values ak,bk,ak', and bk'.
If there is a solution, then
(r,t)= a0*(p0,t0) + … + a3*(p3,t0) + b0*(q0,t1) + … + b3*(q3,t1)
Is a point of first intersection. Otherwise they do not intersect.
Thought about this in the past but lost interest... The best way to go about solving it would be to abstract out one object.
Make a coordinate system where the first tetrahedron is the center (barycentric coords or a skewed system with one point as the origin) and abstract out the rotation by making the other tetrahedron rotate around the center. This should give you parametric equations if you make the rotation times time.
Add the movement of the center of mass towards the first and its spin and you have a set of equations for movement relative to the first (distance).
Solve for t where the distance equals zero.
Obviously with this method the more effects you add (like wind resistance) the messier the equations get buts its still probably the simplest (almost every other collision technique uses this method of abstraction). The biggest problem is if you add any effects that have feedback with no analytical solution the whole equation becomes unsolvable.
Note: If you go the route of of a skewed system watch out for pitfalls with distance. You must be in the right octant! This method favors vectors and quaternions though, while the barycentric coords favors matrices. So pick whichever your system uses most effectively.