I want to orient my robot based on the distance from the obstacle, but I do not know how to measure the distance from robot center to the obstacle surrounding.
I think it might depend on how your robot moves. For instance, if your robot navigation algorithm guides the robot to move one direction at a time, (x then y, or y then x, along axes in XoY space), the Manhattan distance would be more appropriate, as shown in the picture below. The Yellow and Blue distances are the same, while the Green distance is the Euclidean distance that will be described below.
However, most modern robots may move in any direction rather than one at a time. But from point A to point B the straight line is the obvious best solution to save a great effort. Then, the Euclidean distance should be chosen, as shown below.
Related
I have image of robot with yellow markers as shown
The yellow points shown are the markers. There are two cameras used to view placed at an offset of 90 degrees. The robot bends in between the cameras. The crude schematic of the setup can be referred.
https://i.stack.imgur.com/aVyDq.png
Using the two cameras I am able to get its 3d co-ordinates of the yellow markers. But, I need to find the 3d-co-oridnates of the central point of the robot as shown.
I need to find the 3d position of the red marker points which is inside the cylindrical robot. Firstly, is it even feasible? If yes, what is the method I can use to achieve this?
As a bonus, is there any literature where they find the 3d location of such internal points which I can refer to (I searched, but could not find anything similar to my ask).
I am welcome to a theoretical solution as well(as long as it assures to find the central point within a reasonable error), which I can later translate to code.
If you know the actual dimensions, or at least, shape (e.g. perfect circle) of the white bands, then yes, it is feasible and possible.
You need to do the following steps, which are quite non trivial to do, and I won't do them here:
Optional but extremely suggested: calibrate your camera, and
undistort it.
find the equation of the projection of a 3D circle into a 2D camera, for any given rotation. You can simplify this by assuming the white line will be completely horizontal. You want some function that takes the parameters that make a circle and a rotation.
Find all white bands in the image, segment them, and make them horizontal (rotate them)
Fit points in the corrected white circle to the equation in (1). That should give you the parameters of the circle in 3d (radious, angle), if you wrote the equation right.
Now that you have an analytic equation of the actual circle (equation from 1 with parameters from 3), you can map any point from this circle (e.g. its center) to the image location. Remember to uncorrect for the rotations in step 2.
This requires understanding of curve fitting, some geometric analytical maths, and decent code skills. Not trivial, but this will provide a solution that is highly accurate.
For an inaccurate solution:
Find end points of white circles
Make line connecting endpoints
Chose center as mid point of this line.
This will be inaccurate because: choosing end points will have more error than fitting an equation with all points, ignores cone shape of view of the camera, ignores geometry.
But it may be good enough for what you want.
I have been able to extract the midpoint by fitting an ellipse to the arc visible to the camera. The centroid of the ellipse is the required midpoint.
There will be wrong ellipses as well, which can be ignored. The steps to extract the ellipse were:
Extract the markers
Binarise and skeletonise
Fit ellipse to the arc (found a matlab function for this)
Get the centroid of the ellipse
hsv_img=rgb2hsv(im);
bin=new_hsv_img(:,:,3)>marker_th; %was chosen 0.35
%skeletonise
skel=bwskel(bin);
%use regionprops to get the pixelID list
stats=regionprops(skel,'all');
for i=1:numel(stats)
el = fit_ellipse(stats(i).PixelList(:,1),stats(i).PixelList(:,2));
ellipse_draw(el.a, el.b, -el.phi, el.X0_in, el.Y0_in, 'g');
The link for fit_ellipse function
Link for ellipse_draw function
I am making game in Unity engine, where car is moving along the bezier curve by percentage of bezier curve legth.
On this image you can see curve with 8 stop points (yellow spheres). Between each stop point is 20% gap of total distance.
On the image above everything is working correctly, but when I move handles, that the handles have different length problem occurs.
As you can see on image above, distances between stop points are not equal. It is because of my algorithm, because I am finding point of segment by multiplying segment length by interpolation (t). In short problem is that: if t=0.5 it is not in the 50% percent of the segment. As you can see on first image, stop points are in half of segment, but in the second image it is not in half of segment. This problem will be fixed, if there is some mathematical formula, how to find distance middle point.
As you can see on the image above, there are two mid points. T param mid point can be found by setting t to 0.5 (it is what i am doing now), but it is not half of the distance.
How can I find distance mid point (for cubic bezier curve, that have handles in different distance)?
You have correctly observed that the parameter value t=0.5 is generally not the point in the middle of the length. That is a good start but the difficulty lies in the mathematics beneath.
Denoting the components of your parametric curve by x(t) and y(t), the length of the curve
between t=0 (the beginning) and a chosen parameter value t = u is equal to
What you are trying to do is to find u such that l(u) is one half of l(1). This is sometimes possible but often difficult or impossible. So what can you do?
One possibility is to approximate the point you want. A straightforward way is to approximate your Bezier curve by a piecewise linear curve (simply by choosing many parameter values 0 = t_0 < t_1 < ... < t_n = 1 and connecting the values in these parameters by line segments). Now it is easy to compute the entire length (Pythagoras Theorem is your friend) as well as the middle point (walk along the piecewise linear curve the prescribed length). The more points you sample, the more precise you will be and the more time your computation will take, so there is a trade-off. Of course, you can use a more complicated numerical scheme for the approximation but that is beyond the scope of the answer.
The second possibility is to restrict yourself to a subclass of Bezier curves that will let you do what you want. These are called Pythagorean-Hodograph (shortly PH) curves. They have the extremely useful property that there exists a polynomial sigma(t) such that
This means that you can compute the integral above and search for the correct value of u. However, everything comes at a price and the price here is that you will have less freedom, where to put the control points (for me as a mathematician, a cubic Bézier curve has four control points; computer graphics people often speak of "handles" so you might have to translate into your terminology). For the cubic case you can find the conditions on slide 15 of this seminar talk by Vito Vitrih.
Denote:
the control points,
;
then the Bézier curve is a PH curve if and only if
.
It is up to you to figure out, if you can enforce this condition in your situation or if it is too restrictive for your application.
I am looking for how to calculate the distance along a path in a binary array.
I imported a map as a matrix in matlab. There is a binary image of a river crossing two cities. I only found out how to calculate the distance from the river points to the nearest city but I don't manage to compute the shortest distance along the river.
I made a vector with the indices of all river points but I don't know how to get the distance to the nearest city from that...
Image
So I am looking for the shortest distance through the red line towards one of the light blue points it crosses !
Thnx
If I understand you in the right way it is not very difficult: Just do a dfs or bfs (8-neighbourhood) starting at each river-town and add sqrt(2) if you go diagonal and 1 if you go to a 4-neighbour. At each river pixel you can finally decide by taking the minimum value. You can develop it further stopping at river pixels with already smaller distance to another city...
I really hope I got you in the right way :)
I have a multiple plants in a single binary image. How would I identify each leaf in the image assuming that each leaf is approximately elliptical?
example input: http://i.imgur.com/BwhLVmd.png
I was thinking a good place to start would be finding the tip of each leaf and then getting the center of each plant. Then I could fit the curves starting from the tip and then going to the center. I've been looking online and saw something involving a watershed method, but I do not know where to begin with that idea.
You should be aware that these things are tricky to get working robustly - there will always be a failure case.
This said, I think your idea is not bad.
You could start as follows:
Identify the boundary curve of each plant (i.e. pixels with both foreground and background in their neighbourhood).
Compute the centroid of each plant.
Convert each plant boundary to a polar coordinate system, with the centroid as the origin. This amounts to setting up a coordinate system with the distance of each boundary curve point on the Y axis and the angle on the X axis.
In this representation of the boundary curve, try to identify maxima; these are the tips of the leaves. You will probably need to do some smoothing. Use the parts of the curve before and after the maxima the start fitting your ellipses or some other shape.
Generally, a polar coordinate system is always useful for analysing stuff thats roughly circular.
To fit you ellipses, once you have a rough initial position, I would probably try an EM-style approach.
I would do something like this (I is your binary image)
I=bwmorph(bwmorph(I, 'bridge'), 'clean');
SK=bwmorph(I, 'skel', Inf);
endpts = bwmorph(SK,'endpoints');
props=regionprops(I, 'All');
And then connect every segment from the centroids listed in props.centroid to the elements of endpts that should give you your leaves (petals?).
A bit of filtering is probably necessary, bwmorph is your friend. Have fun!
I have a convex polygon in 3D. For simplicity, let it be a square with vertices, (0,0,0),(1,1,0),(1,1,1),(0,0,1).. I need to arrange these vertices in counter clockwise order. I found a solution here. It is suggested to determine the angle at the center of the polygon and sort them. I am not clear how is that going to work. Does anyone have a solution? I need a solution which is robust and even works when the vertices get very close.
A sample MATLAB code would be much appreciated!
This is actually quite a tedious problem so instead of actually doing it I am just going to explain how I would do it. First find the equation of the plane (you only need to use 3 points for this) and then find your rotation matrix. Then find your vectors in your new rotated space. After that is all said and done find which quadrant your point is in and if n > 1 in a particular quadrant then you must find the angle of each point (theta = arctan(y/x)). Then simply sort each quadrant by their angle (arguably you can just do separation by pi instead of quadrants (sort the points into when the y-component (post-rotation) is greater than zero).
Sorry I don't have time to actually test this but give it a go and feel free to post your code and I can help debug it if you like.
Luckily you have a convex polygon, so you can use the angle trick: find a point in the interior (e.g., find the midpoint of two non-adjacent points), and draw vectors to all the vertices. Choose one vector as a base, calculate the angles to the other vectors and order them. You can calculate the angles using the dot product: A · B = A B cos θ = |A||B| cos θ.
Below are the steps I followed.
The 3D planar polygon can be rotated to 2D plane using the known formulas. Use the one under the section Rotation matrix from axis and angle.
Then as indicated by #Glenn, an internal points needs to be calculated to find the angles. I take that internal point as the mean of the vertex locations.
Using the x-axis as the reference axis, the angle, on a 0 to 2pi scale, for each vertex can be calculated using atan2 function as explained here.
The non-negative angle measured counterclockwise from vector a to vector b, in the range [0,2pi], if a = [x1,y1] and b = [x2,y2], is given by:
angle = mod(atan2(y2-y1,x2-x1),2*pi);
Finally, sort the angles, [~,XI] = sort(angle);.
It's a long time since I used this, so I might be wrong, but I believe the command convhull does what you need - it returns the convex hull of a set of points (which, since you say your points are a convex set, should be the set of points themselves), arranged in counter-clockwise order.
Note that MathWorks recently delivered a new class DelaunayTri which is intended to superseded the functionality of convhull and other older computational geometry stuff. I believe it's more accurate, especially when the points get very close together. However I haven't tried it.
Hope that helps!
So here's another answer if you want to use convhull. Easily project your polygon into an axes plane by setting one coordinate zero. For example, in (0,0,0),(1,1,0),(1,1,1),(0,0,1) set y=0 to get (0,0),(1,0),(1,1),(0,1). Now your problem is 2D.
You might have to do some work to pick the right coordinate if your polygon's plane is orthogonal to some axis, if it is, pick that axis. The criterion is to make sure that your projected points don't end up on a line.