I want a function to compute and get the diameter of the circle that circumscribes the object. Is there a built-in function in MATLAB to do this? Otherwise, what can I do?
Try this algorithm:
Compute the average x and average y for every point in the irregular object. This is done by taking the x and y component for every point and add them into the total x and total y and then divide by the number of points. This average x and average y point algorithm gives you a non-weighted center of the object.
Use that center point to compute the distance for every point in the irregular object again. Keeping the largest distance as the radius of the object.
Use the center point and the radius to compute the circumference.
I am submitting proof that the distance between the 2 points that are furthest apart in the object fails with a simple triangle. See image below. Also, the big-O notation for computing the two points that are the furthest apart is x^2. The big-O for this algorithm is 2x. The diameter of the circle in the image would be computed as 20; distance between -10,0 and 10,0. A circle of diameter 20 will not encompass the point # 0,-11. Any movement of the circle would automatically remove at least one of the two points used to compute the diameter of the circle because both points are on tangents.
Suppose M is a mask in BW, just do :
[b_x,b_y] = find(bwperim(M)== 1)
Check this function bwperim
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.
I'm trying to calculate the angle between 2 geographic (Latitude,Longitude) points in MATLAB. The points are:
(-65.226,125.5) and (-65.236,125.433).
I used the MATLAB function, azimuth, as:
azimuth(-65.226,125.5,-65.236,125.433)
I convert the result to radians, and plotting this using quiver, I get the following plot:
I want the red vector to point from the top right dot to the bottom left dot.
The points are at fairly high latitude (~65S), and the separation of the points is low (about 0.1 degrees). Thus, I can't really understand how the curvature of the earth could affect the azimuth prediction that much..
Does anyone have any experience with azimuth in MATLAB, or have a better suggestion to calculating the angle between the coordinate pairs?
Thanks!
Here you can detailed information and formulae on how to find angle between two latitude-longitude points.
I have a certain geographic region defined by the bottom left and top right coordinates. How can I divide this region into areas of 20x20km. I mean in practial the shape of the earth is not flat it's round. The bounding box is just an approximation. It's not even rectangular in actual sense. It's just an assumption. Lets say the bottomleft coordinate is given by x1,y1 and the topright coordinate is given by x2,y2, the length of x1 to x2 at y1 is different than that of the length between x1 to x2 at y2. How can I overcome this issue
Actually, I have to create a spatial meshgrid for this region using matlab's meshgrid function. So that the grids are of area 20x20km.
meshgrid(x1:deltaY:x2,y1:deltaX:y2)
As you can see I can have only one deltaX and one deltaY. I want to choose deltaX and deltaY such that the increments create grid of size 20x20km. However this deltaX and deltaY are supposed to vary based upon the location. Any suggestions?
I mean lets say deltaX=del1. Then distance between points (x1,y1) to (x1,y1+del1) is 20km. BUt when I measure the distance between points (x2,y1) to (x2, y1_del1) the distance is < 20km. The meshgrid function above does creates mesh. But the distances are not consistent. Any ideas how to overcome this issue?
Bear in mind that 20km on the surface of the earth is a REALLY short distance, about .01 radians - so the area you're looking at would be approximated as flat for anything non-scientific. Assuming it is scientific...
To get something other than monotonic steps in meshgrid you should create a function which takes as its input your desired (x,y) and maps it relative to (x_0,y_0) and (x_max,y_max) in your units of choice. Here's an inline function demonstrating the idea of using a function for meshgrid steps
step=inline('log10(x)');
[x,y]=meshgrid(step(1:10),step(1:10));
image(255*x.*y)
colormap(gray(255))
So how do you determine what the function should be? That's hard for us to answer exactly without a little more information about what your data set looks like, how you're interacting with it, and what your accuracy requirements are. If you have access to the actual location at every point, you should vary one dimension at a time (if your data grid is aligned with your latitude grid, for example) and use a curve fit with model selection techniques (akaike/bayes criterion) to find the best function for your data.
I'd like to be able to calculate the surface area of objects seen by the depth camera. Is there an easy way to do this? For example if the kinect is seeing a player I need to calculate how much surface it is covering.
If there is no such functions existing, I can calculate by creating multiple squares with coordinate (x,y) (x+1,y) (x, y+1) (x+1, y+1) and taking into consideration the z value. But I'm not sure how to get the distance in mm or cm between pixels in the x or y axis.
Thanks