How to place a value in a function

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

[Walter Roberson]

See also https://www.mathworks.com/matlabcentral/answers/1617265-how-to-do-a-cycle-in-the-graph-each-time-of-plot#answer_861745

Answer 1

[Stephen23]

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

[Emilia]

Thank you :)

[Emilia]

I tried to place it and it came out logistical and not valuable.

Theta_swept=Theta_Exit-Theta_Entrance
F_x_angle_swept=F_x(theta==Theta_swept)

[Torsten]

theta = 61.2;
theta_down = floor(theta);
theta_up   = ceil(theta);
index_down = theta_down + 1;
index_up   = theta_up + 1;
F_x_theta  = F_x(index_down)*(theta-theta_down) + ...
             F_x(index_up)*(theta_up-theta)
F_y_theta  = F_y(index_down)*(theta-theta_down) + ...
             F_y(index_up)*(theta_up-theta)
F_theta    = F(index_down)*(theta-theta_down) + ...
             F(index_up)*(theta_up-theta)

[Emilia]

I did not do well, gave an answer that

F_x_theta =1.8199e-12

[Torsten]

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).