How can I calculate the length of the longest side of a rectangle and the slope of an imaginary rectangle parallel to the longest side of the first?

I need a suggestion for this easy but confusing dilemma.

I have a rectangle formed by 4 points (x2,y1) (x1,y2) (x3,y4) (x4,y3). The maximum and minimum are respectively (x4,y4) and (x1,y1).

I need to create a script to obtain the length of the longest side of the rectangle and the slope of an imaginary rectangle parallel to the longest side of the rectangle.

The problem is not so easy taking into account that the rectangle could be in many different positions. I have an idea about the slope, using Pythagoras and the maximum and minimum values, but...the length...I'm stuck on that.

[Sean de Wolski]

What have you tried thus far? Show us with code (by editing your question).

[Jules Ray]

i have posted my script below... please lets take a look

Answer 1

[the cyclist]

I don't understand the part about the slope, but there is an easy way to find the longest side. The distance between each pair of points is

 >> dij = sqrt((xi-xj)^2+(yi-yj)^2);

Calculate this for all pairs. You should get three unique length values, corresponding to:

• short side
• long side
• diagonal

The middling value will be the long side of the rectangle.

(You might need to be careful about floating point inexactitudes.)

[Sean de Wolski]

Hence hypot() instead of sqrt((x1-x0).^2+(y1-y0).^2);

[Jules Ray]

Dear cyclist... i like your suggestion and i would like to apply to my data, but i dont know if i should make a loop to obtain these three values, and others questions appears.... like, how to select the middle option ufff... (i'm really sorry i'm new in matlab but i like it). Please check this script, that's what i got till now.

%rectangle propietries for preparing swath profile extraction

p=load('poligono.txt');

%p(:,1)=flipud(p(:,1));

%p(:,2)=flipud(p(:,2));

X=p(:,1);

Y=p(:,2);

%Define each rectangle vertice

A=p(1,:);

B=p(2,:);

C=p(3,:);

D=p(4,:);

minx=min(X);

miny=min(Y);

maxx=max(X);

maxy=max(Y);

%% SLOPE

%defines slope of an imaginary central line

m=((maxy-miny)/(maxx-minx)); %check, posible BUG

an=90-atan(m)*180/pi; %change the slope to degrees

theta=an*pi/180; %theta is the required input "STEP SUCESFULLY COMPLETED"

%% LENGHT

%i will use the pairs of coordinates but i'm not shure how to do it i use :

%but dosen't works

G=p(:,miny); % i nedd that matlab identify the corresponding x coordinate associated %with miny (I indicate that with :)

K=p(maxx,:); % i need that matlab reagrupes the maxx value with its corresponding y %coordinate pair

ddg=sqrt((K(1,1)-(G(1,1)))^2+(K(1,2)-(P(1,2)))^2); %this is my requiered value, the final length of the rectangle

[Sean de Wolski]

-use atand to calculate the angle in degrees

-use the second output of max/min to identify which vertex corresponds to max/min.

-use hypot() to calculate ddg

[Jules Ray]

thanks about your suggestions.... but what's about the length do you have any idea? i want to apply the method proposed by the cyclist, using the major, middle and minor length.... sounds good....

Answer 2

[Image Analyst]

Sorry - I gave you a more extensive answer than the Cyclist but for some reason you (or someone) deleted your first posting of this. I gave you the slope, and told you how to find the various lengths. I'm not inclined to repeat my prior waste of time.

[Jules Ray]

i dont know what's happen but thanx anyway Image Analyst.....

this is my final result, that i want to share with everybody, works percfekt:

p=load('poligono.txt');

X=p(:,1);

Y=p(:,2);

%Define coordinates for each vertice

x1=p(1,1);

x2=p(2,1);

x3=p(3,1);

x4=p(4,1);

y1=p(1,2);

y2=p(2,2);

y3=p(3,2);

y4=p(4,2);

%minimum and maximum for the entire dataset

minx=min(X);

miny=min(Y);

maxx=max(X);

maxy=max(Y);

% SLOPE

%defines slope of an imaginary central line

m=((maxy-miny)/(maxx-minx)); %check, posible BUG

an=90-atan(m)*180/pi; %change the slope to degrees

theta=an*pi/180; %theta is the required input "STEP SUCESFULLY COMPLETED"

nim=numel(p);

% LENGHT

%Alternative 1, calculate all diagonals, the middle one should be the

%lenght

A=sqrt((x2-x1)^2+(y2-y1)^2);

B=sqrt((x3-x2)^2+(y3-y2)^2);

C=sqrt((x3-x4)^2+(y3-y4)^2);

D=sqrt((x4-x2)^2+(y4-y2)^2);

E=sqrt((x3-x1)^2+(y3-y1)^2);

F=sqrt((x4-x1)^2+(y4-y1)^2);

G=sqrt((x3-x1)^2+(y3-y1)^2);

V=[A B C D E F G];

W=sort(V);

S=W'; %max to min ordering of obtained results and transposing

H=round(S); %eliminate decimal places

minV=min(H); % find min value in H

H(H==minV)=0; % set all min values to zero

maxV=max(H); % find max value in H

H(H==maxV)=0; % set all max values to zero

%intermediate=H(H~=0); %matrix with non-zero values, max and min values eliminated

ind=find(H~=0, 1, 'first'); %finds the address of the first non-zero value

ddg=H(ind); %lenght of the rectangle "Requested answer"

clear X Y A B C D E F G V W x1 x2 x3 x4 y1 y2 y3 y4 ind minx miny maxx maxy maxV minV p m an S H nim