I have an array of different length measurements to the walls of an arbitrarily shaped box from a point. They are taken during a 360 degree rotation and I also have a degree measurement.
Distance(1:k); % distance to wall of arbitrarily shaped box during a rotation
Degree(1:k); % degrees rotated from first measurement
Time(1:k); % time passed since first measurement
How can I used distance and Time/Distance to plot a shape that would look like the shape of the box? I tried the convhull function, wondering if there are better options.
Would this do the trick?
Rads=Degrees*(2*pi/360);
X=Distance.*cos(Rads);
Y=Distance.*sin(Rads);
plot(X,Y);
Related
I have a vector of 1000 random numbers biased towards the bounds of 540 and 600. I need to plot this data as a wriggly circular path between two concentric circles of radii 540 and 600 respectively. How do I do this?
Presently, I'm able to plot the concentric circles of given radii, but if I try to plot the given random data which is between the bounds 540 and 600, it is plotted along the width between the two concentric circles. I want it to be plotted as a noisy circular curve between the concentric circles.
I hope I'm able to explain my point
If anyone can tell me, how do I do that. Thanks
Here is the link to my previous post, wherein I had to generate random numbers biased towards the two well defined bounds
Generating random numbers in matlab biased towards the boundaries
Now I need to plot the same data, as explained above.
This is the image I get:
What you actually want is a circular plot, so the first idea that comes to my mind is to use polar coordinate.
First we get your sample of around 1000 random datas biased to the bounds of your [540 600] interval following that.
Note that this algorithm will not always get you exactly 1000 values as the out of bound values are removed.
So we'll do something like this :
%Get the size of the sample
P=length(X);
% Generate P Angles uniformly distributed between 0 and 2*pi*(P-1)/P
% Note generating an angle equal to 0 and an angle equal to 2*pi would
% mean that we have 2 points with the same angle, and that s something
% you generally want to avoid
Angles=(0:P-1)*(2*pi/P);
% Plot your points with the markers (polar coordinates) in '-+',
% - means that you want a line to be drawn between the points and
% + means that you want a + marker on every point
plot(X.*cos(Angles),X.*sin(Angles),'-+');
%tell matlab to wait so you can add the circles
hold on
% add circle of 540 radius
plot(540*cos(Angles),540*sin(Angles),'-r');
%add circle of 600 radius
plot(600*cos(Angles),600*sin(Angles),'-r');
For the final result of :
Update : About getting the angles randomly distibuted
We'll have to define our Angles diffrently. We still need a vector of length P, but randomly distributed. This can be achieved by taking P random numbers between 0 and 1, and then multiply this vector by 2*pi in order to get randomly distributed values between 0 and 2*pi.
Last step is to order this vector in order to keep the ordering of your data.
Temp=rand(1,P)*2*pi;
Angles=sort(Temp);
For a final result of :
Another lead woult be to take a sample of 1000 gaussian random numbers and use them modulo 2*pi :
Angles=sort(mod(randn(1,P)*2*pi,2*pi));
I think this is very easy question for some of you , but I do not have any experience in Matlab !
I am tracking a point in all the frames in video. now I want to plot this points to show the trajectory
I have (x,y) for the points in all frames
how can I plot this ?
thanks
If you have coordinates x and y, and a frame number t, then plot3 or scatter3 can be useful. The latter allows you to color the points according to the frame number, so that if you rotate it that you only see the xy-plane, you can still distinguish the advancement of time.
Thus:
scatter3(x,y,t,[],t)
should do the trick.
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.
If I explain why, this might make more sense
I have a logical matrix (103x3488) output of a photo of a measuring staff having been run through edge detect (1=edge, 0=noedge). Aim- to calculate the distance in pixels between the graduations on the staff. Problem, staff sags in the middle.
Idea: User inputs co-ordinates (using ginput or something) of each end of staff and the midpoint of the sag, then if the edges between these points can be extracted into arrays I can easily find the locations of the edges.
Any way of extracting an array from a matrix in this manner?
Also open to other ideas, only been using matlab for a month, so most functions are unknown to me.
edit:
Link to image
It shows a small area of the matrix, so in this example 1 and 2 are the points I want to sample between, and I'd want to return the points that occur along the red line.
Cheers
Try this
dat=imread('83zlP.png');
figure(1)
pcolor(double(dat))
shading flat
axis equal
% get the line ends
gi=floor(ginput(2))
x=gi(:,1);
y=gi(:,2);
xl=min(x):max(x); % line pixel x coords
yl=floor(interp1(x,y,xl)); % line pixel y coords
pdat=nan(length(xl),1);
for i=1:length(xl)
pdat(i)=dat(yl(i),xl(i));
end
figure(2)
plot(1:length(xl),pdat)
peaks=find(pdat>40); % threshhold for peak detection
bigpeak=peaks(diff(peaks)>10); % threshold for selecting only edge of peak
hold all
plot(xl(bigpeak),pdat(bigpeak),'x')
meanspacex=mean(diff(xl(bigpeak)));
meanspacey=mean(diff(yl(bigpeak)));
meanspace=sqrt(meanspacex^2+meanspacey^2);
The matrix pdat gives the pixels along the line you have selected. The meanspace is edge spacing in pixel units. The thresholds might need fiddling with, depending on the image.
After seeing the image, I'm not sure where the "sagging" you're referring to is taking place. The image is rotated, but you can fix that using imrotate. The degree to which it needs to be rotated should be easy enough; just input the coordinates A and B and use the inverse tangent to find the angle offset from 0 degrees.
Regarding the points, once it's aligned straight, all you need to do is specify a row in the image matrix (it would be a 1 x 3448 vector) and use find to get non-zero vector indexes. As the rotate function may have interpolated the pixels somewhat, you may get more than one index per "line", but they'll be identifiable as being consecutive numbers, and you can just average them to get an approximate value.
I know Matlab has a function called cylinder to create the points for a cylinder when number of points along the circumference, and the radius length. What if I don't want a unit cylinder, and also don't want it to center at the default axis (for example along z-axis)? What would be the easiest approach to create such a cylinder? Thanks in advance.
The previous answer is fine, but you can get matlab to do more of the work for you (because the results of cylinder separate x,y,z components you need to work a little to do the matrix multiplication for the rotation). To have the center of base of the cylinder at [x0 y0 z0], scaled by [xf yf xf] (use xf=yf unless you want an elliptic cylinder), use:
[x y z] = cylinder;
h=mesh(x*xf+x0,y*yf+y0,z*zf+z0)
If you also want to rotate it so it isn't aligned along the z-axis, use rotate. For example, to rotate about the x-axis by 90 degrees, so it's aligned along the y-axis, use:
rotate(h,[1 0 0],90)
Multiply the points by your favourite combination of a scaling matrix, a translation matrix, and a rotation matrix.