boxx = [3;7;7;3;3];
boxy = [3;3;7;7;3];
figure(1)
boxing = plot(boxx,boxy);
xlim([0 11]);
ylim([0 11]);
the code above plots a simple box
how do i remove part of the box such as removing everything above y=6?
to be clearer, to remove the top, y=6 to y=7 part of the line from the left and right lines
ie, removing the blue lines above the green line shown in the picture below
the simple method i know of is to make a new plot by slightly changing the values in the code such as below
boxx = [3;3;6;6];
boxy = [6;3;3;6];
boxing = plot(boxx,boxy);
but since i need it to be dynamic where sometimes i want it at y=5 or y=6.5 and will use the same concept with other shapes such as circulars, triangulars or 2d polygons
is there a different method or code i can use to remove the lines?

6 Kommentare

Rik
Rik am 19 Feb. 2021
You can't remove part of a line object, so you will have to use a similar tactic for your other shapes as well.
If the box edges are always parallel to the axes, it's as easy as,
boxy(boxy>6) = 6;
Adrian Lee
Adrian Lee am 21 Feb. 2021
for this case it will always be parallel, but for some other cases which i will tackle in the future, mostly will not be parallel to axes or each other
Adam Danz
Adam Danz am 21 Feb. 2021
Then you need to rotate the vertices using a rotation matrix before applying the height cuttoff.
Adrian Lee
Adrian Lee am 21 Feb. 2021
ah, i see
hmm,
i would need an equation for identifying the length needed before the rotation
then use the rotation matrix for plotting a new line... or i am mistaken?
Adam Danz
Adam Danz am 21 Feb. 2021
No, that's not quite right.
If the edges of your rectangle are not parallel to the axes you'll need to rotate the rectangle so that it is parallel, then apply the cutoff, and then counter-rotate it back to its original orientation.

Melden Sie sich an, um zu kommentieren.

Antworten (0)

Produkte

Version

R2020a

Tags

Gefragt:

am 19 Feb. 2021

Kommentiert:

am 21 Feb. 2021

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by