I have given two point clouds in an arbitrary base. How can I find the center of rotation (or the best base), while minimizing the translations in a rigid transformation? In order to find the translation and rotation in the given base, I can use the Kabsch' algorithm. I then might have the angles, however, I still don't now the best origin to minimize the translations. Am I missing a point?
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How do I calculate how to fit shape A into shape B? Are there some algorithms I could look up? Neither shape should scale.
This is pretty close except he doesn't specify what routines he's using:
Find maximum area polygon inscribed in larger polygon
The actual use case is fitting a building footprint into an arbitrary area. It's for RTS AI where they need to find a location to place a building.
I have a spherical heightfield, defined by a function f(x, y, z) which returns the distance from the origin of the surface of the heightfield of a line which passes from the origin through (x,y,z).
(In other words, the isosurface for my heightfield is |x,y,z| = f(x,y,z).)
(Also, for the sake of discussion below, I'm going to assume that surface(x,y,z) is the location of the point on the surface directly below (x,y,z).)
When rendering this, I need to calculate the normal for any point on the heightfield. What's the cheapest way of doing this?
To calculate the normal of a point on a rectangular heightfield, the usual trick is to offset (x,y,z) slightly in two directions parallel to the nominal surface, calculate three points on the heightfield to form a triangle, and then use the cross product to calculate the triangle's normal. This is easy as the three points can simply be surface(x,y,z), surface(x+1,y,z) and surface(x,y+1,z) (or similar). But for a spherical heightfield it's a little trickier because the normal can point in any direction. Simply displacing by x and y won't do because if two of my points fall on a radius, then surface() of them will return the same location and I won't get a triangle.
In the past what I've done is to use the vector <x,y,z> as a radius from the sphere's origin; then calculate a vector perpendicular to it; then rotate this vector around <x,y,z> to give me my three points. But this is fiddly and expensive and shouldn't be necessary. There must be a cheaper way. What is it?
Calculate the surface() points and, if they are close enough to cause problems, carry out the more expensive (but accurate) calculation; otherwise, use the cheap/easy calculation.
I am solving a fourth order non-linear partial differential equation in time and space (t, x) on a square domain with periodic or free boundary conditions with MATHEMATICA.
WITHOUT using conformal mapping, what boundary conditions at the edge or corner could I use to make the square domain "seem" like a circular domain for my non-linear partial differential equation which is cartesian?
The options I would NOT like to use are:
Conformal mapping
changing my equation to polar/cylindrical coordinates?
This is something I am pursuing purely out of interest just in case someone screams bloody murder if misconstrued as a homework problem! :P
That question was asked on the time people found out that the world was spherical. They wanted to make rectangular maps of the surface of the world...
It is not possible.
The reason why is not possible is because the sphere has an intrinsic curvature, while the cube/parallelepiped has not. It can be shown that for two elements with different intrinsic curvatures, their surfaces cannot be mapped while either keeping constant infinitesimal distances, either the distance between two points is given by the euclidean distance.
The easiest way to understand this problem is to pick some rectangular piece of paper and try to make a sphere of it without locally stretch it or compress it (you can fold). You can't. On the other hand, you can make a cylinder surface, because the cylinder has also no intrinsic curvature.
In maps, normally people use one of the two options:
approximate the local surface of the sphere by a tangent plane and make a rectangle out of it. (a local map of some region)
make world maps but implement some curved lines everywhere identifying that the measuring distances must be made according to those lines.
This is also the main reason why when traveling from Europe to North America the airplanes seems to make a curve always trying to pass near canada. If we measured the distance from the rectangular map, we see that they should go on a strait line to minimize the distance. However, because we are mapping two different intrinsic curvatures, the real distance must be measured in a different way (and not via a strait line).
For 2D (in fact for nD) the same reasoning applies.
I've been scouring the internet for days, but have been unable to find a good answer (or at least one that made sense to me) to what seems like it should be a common question. How does one scale an arbitrary polygon? In particular, concave polygons. I need an algorithm which can handle concave (definitely) and self-intersecting (if possible) polygons. The obvious and simple algorithm I've been using to handle simple convex polygons is calculating the centroid of the polygon, translating that centroid to the origin, scaling all the vertices, and translating the polygon back to its original location.
This approach does not work for many (or maybe all) concave polygons as the centroid often falls outside the polygon, so the scaling operation also results in a translation and I need to be able to scale the polygon "in place" without the final result being translated.
Is anybody aware of a method for scaling concave polygons? Or maybe a way of finding the "visual center" which can be used as a frame of reference for the scaling operation?
Just to clarify, I'm working in 2D space and I would like to scale my polygons using the "visual center" as the frame of reference. So maybe another way to ask the question would be, how do I find the visual center of a concave and/or self-intersecting polygon?
Thanks!
I'm not sure what your problem is.
You're working in an affine space, and you're looking for an affine transformation to scale your polygon ?
If i'm right, just write the transformation matrix:
scaling matrix
homotethy
And transform your polygon with matrix
You can look up for affine transformation matrix.
hope it helps
EDIT
if you want to keep the same "center", you can just do an homotethy of parameter lambda with center G = barycenter of the polygon:
it verifies :
G won't move since it's the center of the homotethy.
It will still verify the relation below, so it will still be the barycenter. (you just multiply the relation by lambda)
in your case G is easy to determinate: G(x,y) : (average of x values of points, average of y values of points)
and it should do what you need
Perhaps Craig is looking for a "polygon offset" algorithm - where each edge in the polygon is offset by a given value. For example, given a clockwise oriented polygon, offsetting edges towards the left will increase the size of the polygon. If this is what Craig is looking for then this has been asked and answered before here - An algorithm for inflating/deflating (offsetting, buffering) polygons.
If you're looking for a ready made (opensource freeware) solution, I've also created a clipping library (Clipper) written in Delphi, C++ and C# which includes a rather simple polygon offsetting function.
The reason why you can't find a good answer is because you are being imprecise with your requirements. First explicitly define what you mean by "in-place". What is being kept constant?
Once you have figured that out, then translate the constant point to the origin, scale the polygon as usual, and translate back.
I have lot of points (which together form a 3d ellipse) in a given frame (X, Y, Z) and then I have vector (u,v,w). What I want is to orient the ellipse along the vector (u,v,w) . Anyone has useful thoughts on how to go about doing that?
Well I assume you can reverse engineer the ellipse equation by seeing what fits into either 4 or 5 points (I can't remember which -- but it should be easy to figure out from the equations.) Once you have that you can know the two major axes, and center point for the ellipse and the transformation should be straight forward.
Although I support #Paul Hsieh's mathematical approach (and have upvoted it), an alternative brute-force approach which will work for many arbitrary elongated shapes is:
Define the origin as the center of your frame
Find the most distant point from the origin.
Determine rotation that will bring that point into line with your vector.
Apply that rotation to all other points.