How to set custom origin point?

Hello,

I can't manage to find any solution to this problem (do I have to precise I am a complete novice with matlab):

Lets say I have an image named "image" and I show it on screen with axis. By default, my axis will be, from one border to another, zero to end (number of the last column).

But I have a point in "image" that I want to be zero, the origin. I know that point from a mouse input. How do I explain to matlab that I want this point to be the origin of the axis?

Ps: does the following script do something good? (I found it in "How to change the default origin of the axis of rotation in 3d plot matlab") - I don't have matlab today to try that one... But I am not sure if it can even work with my problem.

      origin = [13 58]

If you want some details about the contexte of this problem, here it is:

I am currently working in engineering and I have to do some experiments. In these experiments, I have to take two pictures of a burner: one with the burner cold and a graduated metal rod in it to make the mire (although I don't know if the term exist in english), the other picture is of the burner functionning. The mire help me to transform the image in matlab and know the distances, then I apply the same transformation in the second picture and I can have the flame length simply by mesuring from a point to another.To simplify the operation, I want the origin at the exit of the burner, so that I have directly the distance from the exit when I select a point on the flame. If I am not clear enough, let me know, please.

Thank you for your time and patience

E.L.

Answer 1

[Image Analyst]

Try this:

grayImage = imread('cameraman.tif');
[rows, columns, numberOfColorChannels] = size(grayImage);
origin = [13, 58];  % Assume x, y, NOT row, column
xdata = -origin(1) : columns - origin(1);
ydata = -origin(2) : columns - origin(2);
imshow(grayImage, 'XData', xdata, 'YData', ydata);
axis on;
hp = impixelinfo
grid on;

[Elisa Lafforgue]

Thank you very much, it does work perfectly!

[Phong]

% I'm Nguyen Tien Phong, Hung Yen University of Technical Education

%This is an application program to plot multiple graphs on the same axes

%You guys edit it to be beautiful and suitable for multi-paragraph math problems

[e-mail]

%Good luck

clear all

syms z

y1=3*z^3+2 % Ve tu 0 den 10 Plot this graph from 0 to 10; starting position is 0

y2=3*z^4+2*z^2+6 % Ve tu 10 den 25 (0-10-10-15) Plot this graph from 0 to 15; starting position is 10

L1=10

L2=15

Xtong=linspace(0,L1+L2,20) % Chia thanh 20 doan boi 20 diem de cbi ve cho tat ca

Yt=zeros(1,20)% Tao ra de ve cac diem chia tren z cua tat ca cac doan i

X2=linspace(0,L1,100); % Chia doan 1 thanh 100 diem de ve do thi cho doan 1

A2=subs(y1,z,X2) % Nhan gia tri tai 100 diem de ve do thi cho doan 1

plot(Xtong,Yt,'Linewidth',1) % X tong la tong chieu dai de xac dinh chieu dai truc Z khi ve

hold on

plot(X2,A2,'g','Linewidth',4) % Do thi doan 1

Xve2=linspace(10,25,100); % Chia thanh 100 diem cho doan i=2 tro di

XXX2=linspace(0,L2,100)

B2=subs( y2,z,XXX2) % Nhan 100 gia tri de ve duong do thi

plot(Xve2,B2,'r','Linewidth',4); % Ve do thi

xlabel('Duong truc Z');

title(['Bieu do Nz ', ': kN'])

Answer 2

[Elisa Lafforgue]

Thank you for your answer but with those code we have:

ax.YAxisLocation = 'origin';  % setting y axis location to origin

to set axis location to an automatic origin

Which you can change to put 'left', 'right', 'bottom'... but not a custom coordinates.

My problem is that I want to input some coordinates and make it be the new 'origin' of my axis. How can I do this?

[KSSV]

How about using axis ?

Answer 3

[Mark Saad]

I was trying to do this the other day. I could not find anything, so I just manually shifted all of my data points. Let's say you have

x = [1 2 3 4];
y = [5 6 7 8];

and you want (1,5) to be the origin. Then just do

x = x - 1;
y = y - 5;

Above is an image of the original points on the left, and the points shifted so that (2,6) is the origin on the right.

[Elisa Lafforgue]

This code is very interesting indeed, thank you. I assume you have to precise what to plot such as:

imshow(grayImage, 'XData', xdata, 'YData', ydata);

Did you do something like below?

plot(results, x, y)

[Image Analyst]

I thought you said "Lets say I have an image named "image"" not a plot. Anyway, the concept Mark showed you of subtracting the offset can work with an image too. If you don't like reading the actual shifted (x,y) point directly off the image like I showed you, you can just get the coordinate by whatever means, like impixel() or impixel() or ginput(), and subtract the offset in code to get the shifted/recentered version.

[Mark Saad]

I used the scatter function to plot the points alone.

% Set initial points.
x = [1 2 3 4];
y = [5 6 7 8];

% Plot initial points
subplot(1,2,1);
scatter(x,y,'r');

% Shift the points
x = x - 1;
y = y - 5;

% Plot the shifted points
subplot(1,2,2);
scatter(x,y,'b');

[Elisa Lafforgue]

Thank you Mark Saad, it could be helpful some time.

Yes, Image Analyst, I have an image and not a plot. The answer provided just above by Mark Saad is interesting from my point of vu because I did not know we could such a thing and as I am trying to understand matlab better it is, indeed, very interesting.

Yet your answer was the best because it was exactly what I looked for and it work perfectly with my code.

The reason I picked the other answer instead is simple: I did not know what the "accept the answer" stood for... Actually I thought it was to make the answer seeable or to put an end to the thread as the answer was already given. And I cannot undo this apparently.

Sorry