Most of the times, I determine contour orientation generating 2D points and computing the closed polygon area. Depending on the area value sign I can understand if the contour is oriented clockwise or not (see How to determine if a list of polygon points are in clockwise order?).
Would it be possible to do the same computations without generating 2D points? I mean, relying only on geometric curve properties?
We are interested in determining the orientation of contours like these ones without sampling them with 2D points.
EDIT: Some interesting solutions can be found here:
https://math.stackexchange.com/questions/423718/general-way-to-find-out-whether-a-curve-is-positively-oriented
Scientific paper: Determining the orientation of closed planar curves, DJ Filip (1990)
How are those geometric curves defined?
Do you have an angle for them? The radius doesn't matter, only the difference between entry-angle and exit-angle of each curve.
In that case, a trivial idea crossing my mind is to just sum up all the angles. If the result is positive, you know you had more curves towards the right meaning it's a clockwise contour. If it was negative, then more curves were leftwards -> anti-clockwise contour. (assuming that positive angels determine a right-curve and vica versa)
After thinking about this for awhile, for polygons that contain arcs I think there are three ways to do this.
One, is to break the arcs into line segments and then use the area formula as described above. The success of this approach seems to be tied to how close the interpolation of the arcs is as this could cause the polygon to intersect itself.
A quicker way than the above would be to do the interpolation of the arcs and then find a vertex in the corner (minimal Y, if tie minimal X) and use the sign of the cross product for that vertex. Positive CCW, negative CW. Again, this is still tied to the accuracy of the interpolation.
I think a better approach would be to find the midpoint of the arc and create two line segments, one from the beginning of the arc to the midpoint and another from the midpoint to the end of the arc and replace the arc with these line segments. Now you have a polygon with only line segments. Then you can add up all the normalized cross products of all the vertices. The sign will tell you the direction. Positive is counter-clockwise, negative is clockwise. In this case it doesn't matter if the polygon self-intersects.
Related
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 working on image segmentation and I thought the convex hull can provide me with a simple solution to my problem. Currently I have polygons with for sides (see image below). Due to image processing issues, the shape does not have clean straight sides and hence when I use the standard convex hull (in Matlab) I may get more than the four main corners to define it.
My goal is to force the convex hull algorithm to find the best 4 vertices that will enclose my polygons (i.e. 4 best enclosing vertices per polygon). Is this possible? An example code will be appreciated.
Thanks
The problem of the minimum area bounding polygon is briefly mentioned in "Geometric applications of a matrix-searching algorithm" (see Applications section). It is not simple and is probably not the way for you.
For an easier (but approximate) answer to your question, you can consider the four cardinal directions and find the farthest points in these, which define a quadrilateral. (Also consider the four intermediate directions, which are more appropriate for an axis-aligned rectangle.)
If you insist having an enclosing quadrilateral, you can translate the four edges to the farthest points in the respective perpendicular directions, and find the pairwise intersections.
If you insist having a rectangle, compute the convex hull and find the minimum area or minimum perimeter bounding rectangle by the Rotating Calipers method. https://geidav.wordpress.com/tag/rotating-calipers/
hi i want to detect fingertips point and valleypoint of hand by using hough transform.Simply the Question is what is the [H,theta,rho]=hough(BW) is good for extract these point.
the image is here:
https://www.dropbox.com/sh/n1lz7b5eedzbui7/AADwy5O1l7sWf5aOx7KWAmhOa?dl=0
tnx
The standard hough transformation is just for detecting straight lines. Not more and not less. The Matlab function hough (please see here) returns the so-called hough space H, a parametric space which is used to find these lines and the parametric representation of each line: rho = x*cos(theta) + y*sin(theta).
You will have to do more than this to detect your desired points. Since your fingers usually won't consist of straight lines, I think you should think of something else anyway, e.g. if you can assume such a perfect curve as the one in your image maybe this is interesting for you.
Another simple technique you might consider is to compare the straight line distance between two points on your hand line to the distance between those two points along the perimeter (geodesic distance). For this you would need an ordered list of points along the perimeter.
Along regions of high curvature, the straight line distance between two points will be smaller than the number of pixels between those two points along the perimeter.
For example, you could check perimeter pixels separate by 10 pixels. That is, you would search through the list and compare the point at index N and the point index N+10. (You'll need to loop back around to the beginning of the list as you approach the end.) If the straight line distance between these two points is nearly 10 pixels as well, then you know those points lie on a straight section of the perimeter. If the straight line distance is much smaller than 10, then you know the perimeter curves in some fashion between those points. Whether you check pixels that are 5, 10, 20, or 30 items apart in the list will depend on the resolution of your image and the curves you're looking for.
This technique is useful because it's simple and quick to implement. Maybe it would work well enough for your needs.
Yet another way: simplify the outline to small line segments, and then you can calculate the line-line angle between adjacent segments. To simplify the curves, implement the Ramer-Douglas-Puecker algorithm. A little experimentation will reveal what parameter settings will work for your application.
https://en.wikipedia.org/wiki/Ramer%E2%80%93Douglas%E2%80%93Peucker_algorithm
Finally, you could look into piecewise curve fitting: a curve would be fitted to small segments of the outline. This can get very complicated, and researchers continue to find ways to decompose complex figures into a limited number of more basic shapes or curves. I recommend trying the simplest technique and then only adding complexity if you need it.
I am doing some work related to eye images.
I did edge detection to it. The edge is like a curve and not continuous. I have to assume it to be continuous and find normals to that curve. How do I find the normals to it using MATLAB?
you can see the image below.
I want to find the normals to the upper curve.
I hope that I was clear enough.
Even though it seems unintuitive, the edge direction at every pixel is a pretty good estimate of the normal. This would be the simplest solution, because it doesn't involve any curve fitting.
In MATLAB, you can find pixel-wise edge directions using the Sobel filter:
[BW,thresh,gv,gh] = edge(I,'sobel');
edgeDir = atan2(gv, gh);
This gives you the edge directions as angles in radians.
You may want to consider curve fitting (MSE based or some other criteria) to the data. I believe a second order will do good for the upper curve, and once you have a model you can can calculate the tangent and normal at each point.
As Zaphod recommended the normal is perpendicular to the edge. You don't need to do curve fitting, you can use back projection to identify the focal point of the curve.
Start at each edge point along the curve and draw a line from curve in the direction of the normal. Draw the line by incrementing the value of each pixel the line passes through. Once you do this for all the edges you would hope to find two pixels with higher values then the rest, one for each of your curves. You should then know by there locations which is the focal point for each curve.
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.