is my script having 51 equally spaced points?

clc;clear all;
l=25;E=200e9;I=350e-6;w=6e3;
x1=linspace(0,l/2,25)
for n = 1 : length(x1)
y1(n)=(-(w*x1(n))/(384*E*I))*(16*(x1(n)^3)-24*l*(x1(n)^2)+9*(l^3))
end
x2=linspace(l/2,l,26)
for n = 1 : length(x2)
y2(n)=(-(w*x2(n))/(384*E*I))*(8*(x2(n)^3)-24*l*(x2(n)^2) ...
+17*x2(n)*(l^2)-(l^3) )
end
x = [x1, x2];
y = [y1, y2];
plot(x,y, '.')
xlabel('x-axis,length(m)')
ylabel('y-axis,deflection(m)')

3 Kommentare

Roger Stafford
Roger Stafford am 29 Nov. 2014
I would say, most definitely not equally-spaced! The two functions you are plotting are quartic polynomials in which the derivatives are certainly changing over the interval plotted. Even the x values are not equally-spaced, since there are 25 values in the first half interval and 26 in the second half.
ernest
ernest am 29 Nov. 2014
what do u suggest the values i should enter for x to be equally spaced with 51 points. using linspace of course
x = linspace(0,L,51); % I used uppercase L instead of lowercase l
x1 = x(1:25);
x2 = x(26:51);
...

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Antworten (1)

Image Analyst
Image Analyst am 29 Nov. 2014

0 Stimmen

If you want them equally spaced along the curve, you'll have to use interparc():
If you just want uniform spacing along the x, then just use linspace on x once.

5 Kommentare

ernest
ernest am 29 Nov. 2014
Bearbeitet: Image Analyst am 29 Nov. 2014
What do you suggest the values I should put for x to have 51 equally spaced points?
Using linspace() of course.
ernest
ernest am 29 Nov. 2014
using linspace ofcourse
Image Analyst
Image Analyst am 29 Nov. 2014
Bearbeitet: Image Analyst am 29 Nov. 2014
Replace the x2 assignment with this:
deltaX = x1(2) - x1(1)
x2= l/2 : deltaX : 26;
Now x2 will have the same spacing as x1 and your combined [x1,x2] array has 51 elements. Another way using linspace() is to create the whole thing and extract the portions you want for x1 and x2:
x = linspace(0, 26, 51);
x1 = x(1:25);
x2 = x(26:end);
Don't forget to get rid of the x2 assignment in between the for loops if you do it this way!
ernest
ernest am 30 Nov. 2014
thanks man. appreciate the help
Image Analyst
Image Analyst am 30 Nov. 2014
You're welcome. You can also "Thank" people by Accepting and "Voting" for their answers (to give them reputation points).

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