How to vary angles at constant rate in a kinematics problem?

o2 = [0, 0, 0]; % Origin

ain = [26,0,0]; % Initial vector

input_axis = [0,1,0]; % Axis of rotation (z-axis)

theta1 = deg2rad(157.5); % Angle of rotation in radians

% Rotation matrix function

rot_matrix = @(axis, theta) cos(theta) * eye(3) + ...

sin(theta) * [0, -axis(3), axis(2); axis(3), 0, -axis(1); -axis(2), axis(1), 0] + ...

(1 - cos(theta)) * (axis' * axis);

% Compute the rotated vector

a_rotated = rot_matrix(input_axis, theta1) * (ain' - o2') + o2';

% Transpose to row vector for display

a_rotated = a_rotated';

a_final=a_rotated;

bin=[29.5,46,0];

o4=[13.5,46,0];

output_axis=[0,1,0];

theta2= deg2rad(105);

b1_rotated = rot_matrix(output_axis, theta2) * (bin' - o4') + o4';

b1_final=b1_rotated';

u3=[0,0,1];

c1=[125,0,0];

% Compute the rotated vector in terms of phi

syms phi;

rot_matrix_phi = rot_matrix(u3, phi);

c1_afinal_rotated = rot_matrix_phi * (c1 - a_final)'+a_final' ;

c1_afinal_rotated = c1_afinal_rotated'; % Transpose to row vector for display

o4o2=o4-o2;

%disp(o4o2);

coupler=(c1_afinal_rotated-b1_final);

% Define the initial rotated components of the coupler

syms phi;

coupler = subs(coupler, conj(phi), phi);

%disp('Coupler vector without conjugate:');

disp(coupler);

% Parametric substitution

syms t;

cos_phi = (1 - t^2) / (1 + t^2);

sin_phi = 2 * t / (1 + t^2);

% Substitute parametric forms into coupler components

coupler_parametric = subs(coupler, [cos(phi), sin(phi)], [cos_phi, sin_phi]);

% Display the parametric coupler

disp('Parametric form of coupler:');

disp(coupler_parametric);

syms targetvalue % it might be 3.5 ...

normsq = expand(sum(coupler_parametric.^2) - targetvalue^2);

normpoly = simplify(normsq*(t^2+1)^2);

vpa(expand(normpoly),4);

tsolve = solve(normpoly,t,'maxdegree',4,'returnconditions',true);

h=vpa(subs(tsolve.t,targetvalue, 106));

%disp(h);

real_solutions = h(imag(h) == 0);

disp('Real roots:');

disp(real_solutions);

% Convert real values of t to angles using angle = 2 * atan(t)

angles_rad = 2 * atan(real_solutions);

angles_deg = rad2deg(angles_rad);

% Display angles in degrees

disp('Angles in degrees before adjustment:');

disp(angles_deg);

phi=double(angles_rad(1));

p=double(angles_deg(1));

c1_position = double(rot_matrix(u3, phi) * (c1' - a_final') + a_final');

p=(c1_position'-a_final)';

disp(p');

disp(p(2)/norm(c1_position'-a_final));

final_angle_phi=acosd(p(2)/norm(c1_position'-a_final)); % angle with y axis

disp(final_angle_phi) ;

disp(norm(c1_position'-b1_final));

disp(norm(a_rotated-o2));

%% in this code i want to change theta1 at rate of 90deg/sec and theta2 at rate of 60 deg/sec varying from 0 to 360 deg and want to calculate corresponding value of final_angle_phi with the y axis

%please help someone

[Deepak Gupta]

For constant rate, theta1 and theta 2 should have a linear relation with time. i.e.

theta1 = 90*t+c1

and

theta2 = 60*t+c2

where c1,c2 are initial phases.

[Aman]

@Deepak Gupta yes sir,it will be and i am taking c1,c2 equal to 0.

can you please tell how can i write a loop for varying theta1 and theta2 at at given rates to find corresponding final_angle_phi??

[Deepak Gupta]

You should not need to write a for loop. Matlab works on vectors and matrices. So just define time as a vector and calculate corresponding theta values. i.e.

t = 0:0.01:10;
theta1 = 90 * t;
theta2 = 60 * t;

Answer 1

[Aquatris]

For loops are easy to understand but as mentioned in comments, once you understand the logic you should vectorize it. The basic idea is as follow:

t = 0:0.01:4; % 4 seconds since 90*4 = 360, meaning theta1 goes from 0 to 360
theta1_vec = deg2rad(90*t);  % array with theta1 angles at 0 sec upto 4 sec
theta2_vec = deg2rad(60*t);  % array with theta2 angles at 0 sec upto 4 sec

for i = 1:length(t)

   theta1 = theta1_vec(i);
   theta2 = theta2_vec(i);
   
   %%====rest of your code======%%
   
   % store values you want as array
    final_angle_phi(i)=acosd(p(2)/norm(c1_position'-a_final));  
end