How to place a value in a function

3 views (last 30 days)
Emilia
Emilia on 31 Dec 2021
Edited: Torsten on 31 Dec 2021
Hello,
I wrote down a code and I want to place F_x(theta=60) to get the answer . How to do it.
Thanks for the helpers and Happy New Year :)
K_t=750;
K_r=250;
b=5;
f_z=0.1;
theta=0:360 ;
for i = 1: length(theta)
t90 = mod(theta(i), 90);
if (t90 >= 60 & t90 <= 90)
F_x(i)=0;
F_y(i)=0;
F(i)=0;
else
st90 = sind(t90);
ct90 = cosd(t90);
h_cut = f_z * st90;
F_r=K_r*b*h_cut;
F_t=K_t*b*h_cut;
F_x(i) = abs(-F_t .* ct90 - F_r .* st90);
F_y(i) = F_t .* st90 - F_r .* ct90;
F(i)=sqrt((F_x(i)).^2+(F_y(i)).^2);
end
end
plot(theta,F_x,'--r',theta,F_y,'--b',theta,F,'k' )
ylim([0 350])
grid on
legend('F_x' ,'F_y','F')
title('The components of the forces as a function of the angle of chip in the Up milling');
xlabel('theta [deg]');
ylabel('Force [N]');
F_x(theta=60) =?? %%%%%%%% Here help

Accepted Answer

Stephen23
Stephen23 on 31 Dec 2021
Edited: Stephen23 on 31 Dec 2021
F_x is an array, not a function (in the MATLAB sense: https://www.mathworks.com/help/matlab/function-basics.html)
To access elements of an array you can use simple logical indexing:
K_t=750;
K_r=250;
b=5;
f_z=0.1;
theta=0:360 ;
for i = 1: length(theta)
t90 = mod(theta(i), 90);
if (t90 >= 60 & t90 <= 90)
F_x(i)=0;
F_y(i)=0;
F(i)=0;
else
st90 = sind(t90);
ct90 = cosd(t90);
h_cut = f_z * st90;
F_r=K_r*b*h_cut;
F_t=K_t*b*h_cut;
F_x(i) = abs(-F_t .* ct90 - F_r .* st90);
F_y(i) = F_t .* st90 - F_r .* ct90;
F(i)=sqrt((F_x(i)).^2+(F_y(i)).^2);
end
end
plot(theta,F_x,'--r',theta,F_y,'--b',theta,F,'k' )
ylim([0 350])
grid on
legend('F_x' ,'F_y','F')
title('The components of the forces as a function of the angle of chip in the Up milling');
xlabel('theta [deg]');
ylabel('Force [N]');
F_x(theta==60)
ans = 0
  5 Comments
Torsten
Torsten on 31 Dec 2021
It's the correct result since you set F_x, F_y and F to zero for 60 <= theta <=89.
And 1.8199e-12 is zero (numerically).

Sign in to comment.

More Answers (0)

Tags

Community Treasure Hunt

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

Start Hunting!

Translated by