I'm trying to plot some annotations to go along with my step function graphs. I currently have these graphs, and I've been trying to figure out how to draw horizontal arrows that point towards vertical lines. I will also need labeled, vertical lines that are pointing towards horizontal lines.
I have attached an image that shows (in red) what I mean. I've tried the annotation() function, but it's truly a pain to get the arrows where I want them to be. If anyone wouldn't mind explaining how to use that function, or alternative methods for what I'm trying to achieve, that would be amazing!
EDIT: Is there a way to edit the Quiver arrowhead size?
Using Quiver in a 2D Subplot
Not quite sure if this is any better or simpler but I used the quiver() function to plot the lines shown below. The quiver() function takes in a few inputs in this case. In the full script below I used twice the amount of quiver() calls to plot overlapping arrows to create a double headed arrow.
Function Call:
quiver(Start_Point(1),Start_Point(2),X_Displacement,Y_Displacement,0);
• Start_Point → equal to [x y] (x-coordinate y-coordinate)
• Start_Point(1) → The x-coordinate of the arrow's start
• Start_Point(2) → The y-coordinate of the arrow's start
• X_Displacement → The horizontal distance from the start of the array
• Y_Displacement → The vertical distance from the start of the array
Setting the Maximum Size of the Arrow Head:
The maximum size of the arrow head can be set by using the 'MaxHeadSize' property.
clf;
Start_Point(1) = 0;
Start_Point(2) = 0;
X_Displacement = 0; Y_Displacement = 10;
Magnitude = sqrt(X_Displacement.^2 + Y_Displacement.^2);
quiver(Start_Point(1),Start_Point(2),X_Displacement,Y_Displacement,0,'Color','r','MaxHeadSize',1/Magnitude);
hold on
Start_Point(1) = 0;
Start_Point(2) = 0;
X_Displacement = 100; Y_Displacement = 0;
Magnitude = sqrt(X_Displacement.^2 + Y_Displacement.^2);
quiver(Start_Point(1),Start_Point(2),X_Displacement,Y_Displacement,0,'Color','r','MaxHeadSize',1/Magnitude);
Related
Consider the simple code below that generates a straight downward sloping line in MATLAB.
clear, clc, close all
t = 0:0.1:1;
y = -t+1;
plot(t,y)
ax = gca
This is a line with slope -1, so the (acute) angle between the horizontal axis and the line is 45 degrees. Except it isn't when you measure with a protractor on your monitor.
Without changing the range of values displayed on the x and y axes or the height of the figure window, how could I ensure I would measure 45 degrees from the horizontal axis to the line if I held a protractor up to the screen?
My current approach is to change the width of the figure window. In the limit as the figure window is infinitely thin, the line x is a vertical line. Conversely, if the figure window is stretched to the edges of a monitor, it flattens out. Somewhere in the middle, the line has the angle I want. I just can't find a good way to mathematically find this point and instantiate it in code.
Edit: A slightly more generic solution for any acute angle. (I didn't test obtuse angles.)
clear, clc, close all
ang_deg = 70;
slope = tand(ang_deg);
t = 0:0.1:1;
y = -slope*t+1;
f = figure;
f.Position(3) = f.Position(3)*1.5;
plot(t,y)
% For a given height, change the width
ax = gca;
ax.Units = 'pixels';
ax.Position(3) = ax.Position(4)/slope;
You can achieve that with
axis equal
which, according to the documentation,
uses the same length for the data units along each axis.
You may want to also use
axis tight
which
fits the axes box tightly around the data by setting the axis limits equal to the range of the data
Follow up your commands with a declaration that you'll be working in pixels, and then set the width to the height:
ax.Units = 'pixels';
ax.Position(3) = ax.Position(4);
I have a Matlab script that creates a Model Block for each element i found in a text file.
The problem is that all Models are created on each other in the window. So i'm trying to make a loop like:
for each element in text file
I add a Model block
I place right to the previous one
end
So it can look like this:
As you can see on the left, all models are on each other and I would like to place them like the one on the right.
I tried this:
m = mdlrefCountBlocks(diagrammeName)+500;
add_block('simulink/Ports & Subsystems/Model',[diagrammeName '/' component_NameValue]);
set_param(sprintf('%s/%s',diagrammeName,component_NameValue), 'ModelFile',component_NameValue);
size_blk = get_param(sprintf('%s/%s',diagrammeName,component_NameValue),'Position');
X = size_blk(1,1);
Y = size_blk(1,2);
Width = size_blk(1,3);
Height = size_blk(1,4);
set_param(sprintf('%s/%s',diagrammeName,component_NameValue),'Position',[X+m Y X+Width Y+Height]);
Inside the loop but it returns an error Invalid definition of rectangle. Width and height should be positive.
Thanks for helping!
The position property of a block does actually not contain its width and height, but the positions of the corners on the canvas (see Common Block Properties):
vector of coordinates, in pixels: [left top right bottom]
The origin is the upper-left corner of the Simulink Editor canvas before any canvas resizing. Supported coordinates are between -1073740824 and 1073740823, inclusive. Positive values are to the right of and down from the origin. Negative values are to the left of and up from the origin.
So change your code to e.g.:
size_blk = get_param(sprintf('%s/%s',diagrammeName,component_NameValue),'Position');
set_param(sprintf('%s/%s',diagrammeName,component_NameValue),'Position', size_blk + [m 0 0 0]);
I have this 3D image generated from the simple code below.
% Input Image size
imageSizeY = 200;
imageSizeX = 120;
imageSizeZ = 100;
%# create coordinates
[rowsInImage, columnsInImage, pagesInImage] = meshgrid(1:imageSizeY, 1:imageSizeX, 1:imageSizeZ);
%# get coordinate array of vertices
vertexCoords = [rowsInImage(:), columnsInImage(:), pagesInImage(:)];
centerY = imageSizeY/2;
centerX = imageSizeX/2;
centerZ = imageSizeZ/2;
radius = 28;
%# calculate distance from center of the cube
sphereVoxels = (rowsInImage - centerY).^2 + (columnsInImage - centerX).^2 + (pagesInImage - centerZ).^2 <= radius.^2;
%# Now, display it using an isosurface and a patch
fv = isosurface(sphereVoxels,0);
patch(fv,'FaceColor',[0 0 .7],'EdgeColor',[0 0 1]); title('Binary volume of a sphere');
view(45,45);
axis equal;
grid on;
xlabel('x-axis [pixels]'); ylabel('y-axis [pixels]'); zlabel('z-axis [pixels]')
I have tried plotting the image with isosurface and some other volume visualization tools, but there remains quite a few surprises for me from the plots.
The code has been written to conform to the image coordinate system (eg. see: vertexCoords) which is a left-handed coordinate system I presume. Nonetheless, the image is displayed in the Cartesian (right-handed) coordinate system. I have tried to see this displayed as the figure below, but that’s simply not happening.
I am wondering if the visualization functions have been written to display the image the way they do.
Image coordinate system:
Going forward, there are other aspects of the code I am to write for example if I have an input image sphereVoxels as in above, in addition to visualizing it, I would want to find north, south east, west, top and bottom locations in the image, as well as number and count the coordinates of the vertices, plus more.
I foresee this would likely become confusing for me if I don’t stick to one coordinate system, and considering that the visualization tools predominantly use the right-hand coordinate system, I would want to stick with that from the onset. However, I really do not know how to go about this.
Right-hand coordinate system:
Any suggestions to get through this?
When you call meshgrid, the dimensions x and y axes are switched (contrary to ndgrid). For example, in your case, it means that rowsInImage is a [120x100x200] = [x,y,z] array and not a [100x120x200] = [y,x,z] array even if meshgrid was called with arguments in the y,x,z order. I would change those two lines to be in the classical x,y,z order :
[columnsInImage, rowsInImage, pagesInImage] = meshgrid(1:imageSizeX, 1:imageSizeY, 1:imageSizeZ);
vertexCoords = [columnsInImage(:), rowsInImage(:), pagesInImage(:)];
I am trying to draw the lines or the edges of a cone using plot3 in matlab. Any help please? I do not need the surface. I need the edges only. SO that I can patch something on it. A useful link. But i need the circle at the bottom:
https://patentimages.storage.googleapis.com/US8514658B2/US08514658-20130820-D00021.png
Few horizontal lines are fine. But no tilted line as i need to patch something inside.
cylinder is your friend here...
You just need to pass it a vector of radii* and transpose the output*...
* negative radii tending to zero will flip the order so the apex is on top...
* so it draws rings not lines from the base to the apex
numRings = 10;
numPointsAround = 100;
[x,y,z] = cylinder(linspace(-1,0,nlines),numPointsAround);
plot3(y.',x.',z.','-k')
I think this is what you want. Most of the answer is directly taken from the above answer by #RTL.
numRings = 2;
numPointsAround = 100;
[x,y,z] = cylinder(linspace(-1,0,numRings),numPointsAround);
plot3(y.',x.',z.','-k')
hold on;line([-0.5878;0], [0.809;0],[0;1]);
hold on;line([0.9511;0], [-0.309;0],[0;1]);
axis square
I have a video of moving hose in an experiment and I need to detect certain points in that hose and calculate the amplitude of their movements, I am using the code below and I am able to extract the required point using detectSURFFeatures, the function get many unnecessary points so I am using cuba = ref_pts.selectStrongest(5); to choose only five points, the problem is I can not get a function to put a bounding box about this 5 points and get their pixel values through the video, Kindly advice what functions can be used, thanks :)
clear;
clc;
% Image aquisition from Video and converting into gray scale
vidIn = VideoReader('ItaS.mp4');
%% Load reference image, and compute surf features
ref_img = read(vidIn, 1);
ref_img_gray = rgb2gray(ref_img);
ref_pts = detectSURFFeatures(ref_img_gray);
[ref_features, ref_validPts] = extractFeatures(ref_img_gray, ref_pts);
figure; imshow(ref_img);
hold on; plot(ref_pts.selectStrongest(5));
cuba = ref_pts.selectStrongest(5);
stats1 = round(cuba.Location);
If you want to find the bounding box which covers all the five points you selected: stats1 now contains (x, y) coordinates of the selected 5 points. Find min and max for x and y coordinates. min values of x and y gives you the starting point of the rectangle. Width and height of the bounding box is now the difference of max and min in y and x directions.
If you want to extract the part of the original image inside the bounding box: just copy that part to another variable as you want. Consider the following example.
img2 = img1(y:h, x:w, :)
Here, x and y are the x and y coordinates of the top left corner of the bounding box. w and h are the width and height of the bounding box.