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.
Related
I have many images like the following (only white and black):
My final problem is to find well matching ellipses. Unfortunately the real used images are not always that nice like this. They could be deformed a bit, which makes ellipse matching probably harder.
My idea is to find "break points". I markes them in the following picture:
Maybe these points could help to make a matching for the ellipses. The end result should be something like this:
Has someone an idea what algorithm may be used to find these break points? Or even better to make good ellipse matching?
Thank you very much
Sample the circumference points
Just scan your image and select All Black pixels with any White neighbor. You can do this by recoloring the remaining black pixels to any unused color (Blue).
After whole image is done you can recolor the inside back from unused color (Blue) to white.
form a list of ordered circumference points per cluster/ellipse
Just scan your image and find first black pixel. Then use A* to order the circumference points and store the path in some array or list pnt[] and handle it as circular array.
Find the "break points"
They can be detect by peak in the angle between neighbors of found points. something like
float a0=atan2(pnt[i].y-pnt[i-1].y,pnt[i].x-pnt[i-1].x);
float a1=atan2(pnt[i+1].y-pnt[i].y,pnt[i+1].x-pnt[i].x);
float da=fabs(a0-a1); if (da>M_PI) da=2.0*M_PI-da;
if (da>treshold) pnt[i] is break point;
or use the fact that on break point the slope angle delta change sign:
float a1=atan2(pnt[i-1].y-pnt[i-2].y,pnt[i-1].x-pnt[i-2].x);
float a1=atan2(pnt[i ].y-pnt[i-1].y,pnt[i ].x-pnt[i-1].x);
float a2=atan2(pnt[i+1].y-pnt[i ].y,pnt[i+1].x-pnt[i ].x);
float da0=a1-a0; if (da0>M_PI) da0=2.0*M_PI-da0; if (da0<-M_PI) da0=2.0*M_PI+da0;
float da1=a2-a1; if (da1>M_PI) da1=2.0*M_PI-da1; if (da1<-M_PI) da1=2.0*M_PI+da1;
if (da0*da1<0.0) pnt[i] is break point;
fit ellipses
so if no break points found you can fit the entire pnt[] as single ellipse. For example Find bounding box. It's center is center of ellipse and its size gives you semi-axises.
If break points found then first find the bounding box of whole pnt[] to obtain limits for semi-axises and center position area search. Then divide the pnt[] to parts between break points. Handle each part as separate part of ellipse and fit.
After all the pnt[] parts are fitted check if some ellipses are not the same for example if they are overlapped by another ellipse the they would be divided... So merge the identical ones (or average to enhance precision). Then recolor all pnt[i] points to white, clear the pnt[] list and loop #2 until no more black pixel is found.
how to fit ellipse from selection of points?
algebraically
use ellipse equation with "evenly" dispersed known points to form system of equations to compute ellipse parameters (x0,y0,rx,ry,angle).
geometrically
for example if you detect slope 0,90,180 or 270 degrees then you are at semi-axis intersection with circumference. So if you got two such points (one for each semi-axis) that is all you need for fitting (if it is axis-aligned ellipse).
for non-axis-aligned ellipses you need to have big enough portion of the circumference available. You can exploit the fact that center of bounding box is also the center of ellipse. So if you got the whole ellipse you know also the center. The semi-axises intersections with circumference can be detected with biggest and smallest tangent change. If you got center and two points its all you need. In case you got only partial center (only x, or y coordinate) you can combine with more axis points (find 3 or 4)... or approximate the missing info.
Also the half H,V lines axis is intersecting ellipse center so it can be used to detect it if not whole ellipse in the pnt[] list.
approximation search
You can loop through "all" possible combination of ellipse parameters within limits found in #4 and select the one that is closest to your points. That would be insanely slow of coarse so use binary search like approach something like mine approx class. Also see
Curve fitting with y points on repeated x positions (Galaxy Spiral arms)
on how it is used for similar fit to yours.
hybrid
You can combine geometrical and approximation approach. First compute what you can by geometrical approach. And then compute the rest with approximation search. you can also increase precision of the found values.
In rare case when two ellipses are merged without break point the fitted ellipse will not match your points. So if such case detected you have to subdivide the used points into groups until their fits matches ...
This is what I have in mind with this:
You probably need something like this:
https://en.wikipedia.org/wiki/Circle_Hough_Transform
Your edge points are simply black pixels with at least one white 4-neighbor.
Unfortunately, though, you say that your ellipses may be “tilted”. Generic ellipses are described by quadratic equations like
x² + Ay² + Bxy + Cx + Dy + E = 0
with B² < 4A (⇒ A > 0). This means that, compared to the circle problem, you don't have 3 dimensions but 5. This causes the Hough transform to be considerably harder. Luckily, your example suggests that you don't need a high resolution.
See also: algorithm for detecting a circle in an image
EDIT
The above idea for an algorithm was too optimistic, at least if applied in a straightforward way. The good news is that it seems that two smart guys (Yonghong Xie and Qiang Ji) have already done the homework for us:
https://www.ecse.rpi.edu/~cvrl/Publication/pdf/Xie2002.pdf
I'm not sure I would create my own algorithm. Why not leverage the work other teams have done to figure out all that curve fitting of bitmaps?
INKSCAPE (App Link)
Inkscape is an open source tool which specializes in vector graphics editing with some ability to work with raster (bitmap) parts too.
Here is a link to a starting point for Inkscape's API:
http://wiki.inkscape.org/wiki/index.php/Script_extensions
It looks like you can script within Inkscape, or access Inkscape via external scripts.
You also may be able to do something with zero scripting, from the inkscape command line interface:
http://wiki.inkscape.org/wiki/index.php/Frequently_asked_questions#Can_Inkscape_be_used_from_the_command_line.3F
COREL DRAW (App Link)
Corel Draw is recognized as the premier industry solution for vector graphics, and has some great tools for converting rasterized images into vector images.
Here's a link to their API:
https://community.coreldraw.com/sdk/api
Here's a link to Corel Draw batch image processing (non-script solution):
http://howto.corel.com/en/c/Automating_tasks_and_batch-processing_images_in_Corel_PHOTO-PAINT
I have a turtle-graphics-based algorithm for generating a space-filling Hilbert curve in two dimensions. It is recursive and goes like this:
Wa want to draw a curve of order n, in direction x (where x ∈ {L, R}), and let y be the direction opposite to x. We do as follows:
turn in the direction y
draw a Hilbert curve of order n-1, direction y
move one step forward
turn in the direction x
draw a Hilbert curve of order n-1, direction x
move one step forward
draw a Hilbert curve of order n-1, direction x
turn in the direction x
move one step forward
draw a Hilbert curve of order n-1, direction y
I understand this and was able to implement a working solution. However, I'm now trying to "upgrade" this to 3D, and here's where I basically hit a wall; in 3D, when we reach a vertex, we can turn not in two, but four directions (going straight or backing up is obviously not an option, hence four and not six). Intuitively, I think I should store the plane on which the turtle is "walking" and its general direction in the world, represented by an enum with six values:
Up
Down
Left
Right
In (from the camera's perspective, it goes "inside" the world)
Out (same as above, outside)
The turtle, like in 2D, has a state containing the information outlined above, and when it reaches as vertex (which can be thought of as a "crossing") has to make a decision where to go next, based on that state. Whereas in two dimensions it is rather simple, in three, I'm stumped.
Is my approach correct? (i.e., is this what I should store in the turtle's state?)
If it is, how can I use that information to make a decision where to go next?
Because there are many variants of 3D space filling Hilbert curves, I should specify that this is what I'm using as reference and to aid my imagination:
I'm aware that a similar question has already been asked, but the accepted answer links to a website there this problem is solved using a different approach (i.e., not turtle graphics).
Your 2d algorithm can be summarized as “LRFL” or “RLFR” (with “F” being “forward”). Each letter means “turn that direction, draw a (n-1)-curve in that direction, and take a step forward”. (This assumes the x in step 8 should be a y.)
In 3d, you can summarize the algorithm as the 7 turns you would need to go along your reference. This will depend on how you visualize the turtle starting. If it starts at the empty circle, facing the filled circle, and being right-side-up (with its back facing up), then your reference would be “DLLUULL”.
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.
As a followup to my previous question about determining camera parameters I have formulated a new problem.
I have two pictures of the same rectangle:
The first is an image without any transformations and shows the rectangle as it is.
The second image shows the rectangle after some 3d transformation (XYZ-rotation, scaling, XY-translation) is applied. This has caused the rectangle to look a trapezoid.
I hope the following picture describes my problem:
alt text http://wilco.menge.nl/application.data/cms/upload/transformation%20matrix.png
How do determine what transformations (more specifically: what transformation matrix) have caused this tranformation?
I know the pixel locations of the corners in both images, hence i also know the distances between the corners.
I'm confused. Is this a 2d or a 3d problem?
The way I understand it, you have a flat rectangle embedded in 3d space, and you're looking at two 2d "pictures" of it - one of the original version and one based on the transformed version. Is this correct?
If this is correct, then there is not enough information to solve the problem. For example, suppose the two pictures look exactly the same. This could be because the translation is the identity, or it could be because the translation moves the rectangle twice as far away from the camera and doubles its size (thus making it look exactly the same).
This is a math problem, not programming ..
you need to define a set of equations (your transformation matrix, my guess is 3 equations) and then solve it for the 4 transformations of the corner-points.
I've only ever described this using German words ... so the above will sound strange ..
Based on the information you have, this is not that easy. I will give you some ideas to play with, however. If you had the 3D coordinates of the corners, you'd have an easier time. Here's the basic idea.
Move a corner to the origin. Thereafter, rotations will take place about the origin.
Determine vectors of the axes. Do this by subtracting the adjacent corners from the origin point. These will be a local x and y axis for your world.
Determine angles using the vectors. You can use the dot and cross products to determine the angle between the local x axis and the global x axis (1, 0, 0).
Rotate by the angle in step 3. This will give you a new x axis which should match the global x axis and a new local y axis. You can then determine another rotation about the x axis which will bring the y axis into alignment with the global y axis.
Without the z coordinates, you can see that this will be difficult, but this is the general process. I hope this helps.
The solution will not be unique, as Alex319 points out.
If the second image is really a trapezoid as you say, then this won't be too hard. It is a trapezoid (not a parallelogram) because of perspective, so it must be an isosceles trapezoid.
Draw the two diagonals. They intersect at the center of the rectangle, so that takes care of the translation.
Rotate the trapezoid until its parallel sides are parallel to two sides of the original rectangle. (Which two? It doesn't matter.)
Draw a third parallel through the center. Scale this to the sides of the rectangle you chose.
Now for the rotation out of the plane. Measure the distance from the center to one of the parallel sides and use the law of sines.
If it's not a trapezoid, just a quadralateral, then it'll be harder, you'll have to use the angles between the diagonals to find the axis of rotation.
I have a point (Lat/Lon) and a heading in degrees (true north) for which this point is traveling along. I have numerous stationary polygons (Points defined in Lat/Lon) which may or may not be convex.
My question is, how do I calculate the closest intersection point, if any, with a polygon. I have seen several confusing posts about Ray Tracing but they seem to all relate to 3D when the Ray and Polygon are not on the same Plane and also the Polygons must be convex.
sounds like you should be able to do a simple 2d line intersection...
However I have worked with Lat/Long before and know that they aren't exactly true to any 2d coordinate system.
I would start with a general "IsPointInPolygon" function, you can find a million of them by googling, and then test it on your poly's to see how well it works. If they are accurate enough, just use that. But it is possible that due to the non-square nature of lat/long coordinates, you may have to do some modifications using Spherical geometry.
In 2D, the calculations are fairly simple...
You could always start by checking to make sure the ray's endpoint is not inside the polygon (since that's the intersection point in that case).
If the endpoint is out of the line, you could do a ray/line segment intersection with each of the boundary features of the polygon, and use the closest found location. That handles convex/concave features, etc.
Compute whether the ray intersects each line segment in the polygon using this technique.
The resulting scaling factor in (my accepted) answer (which I called h) is "How far along the ray is the intersection." You're looking for a value between 0 and 1.
If there are multiple intersection points, that's fine! If you want the "first," use the one with the smallest value of h.
The answer on this page seems to be the most accurate.
Question 1.E GodeGuru