CODEWITHSUNDEEP
matlabmatlab-figureaspect-ratio
Unique Worldline
Unique Worldline

Reputation: 201

Consistently generate line at at a specified angle as measured by user in MATLAB

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;

Upvotes: 2

Views: 93

Answers (2)

Luis Mendo
Luis Mendo

Reputation: 112749

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

Upvotes: 4

L. Scott Johnson
L. Scott Johnson

Reputation: 4402

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);

Upvotes: 2

Related Questions