# How to place a value in a function

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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
##### 1 CommentShowHide None
Walter Roberson on 31 Dec 2021

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### Accepted Answer

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 CommentsShowHide 4 older comments
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).

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