Effacer les filtres
Effacer les filtres

I am looking to solve for the three theta's using sys in an array. How should my code look to make this happen. Heres what I have:

3 vues (au cours des 30 derniers jours)
syms theta1 theta2 theta3
eq1 = theta1 == 90-asin((L1-hblock_sub-h_ram-L3*sin(theta2)-L4*sin(theta3))/L5);
eq2 = theta2 == acos(-(L6-L4*cos(theta3)-L5*cos(90-theta1))/L3);
eq3 = theta3 == acos((L6+L3*cos(theta2)-L5*cos(90-theta1))/L4);
sol = solve([eq1 eq2 eq3],[theta1 theta2 theta3]);
hblock = subs(sol.theta1,theta2,theta3);hblock_sub = double(hblock_sub);
  1 commentaire
Torsten
Torsten le 10 Mar 2023
Modifié(e) : Torsten le 10 Mar 2023
Replace 90 by pi/2.
I don't think this will solve your problem, but it's an error in your equations.
Even better:
Use sin(pi/2-x) = cos(x) and cos(pi/2-x) = sin(x).

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Réponses (1)

Shushant
Shushant le 13 Mar 2023
From my understanding of your issue, you are trying to solve a set of equations but instead of declaring "theta1", "theta2" and "theta3" separately you want to declare them as part of an array. To accomplish this, you can look through this documentation here. I have tried to implement the same using some random values below.
syms theta [1 3]
L1 = 1;
L2 = 2;
L3 = 3;
L4 = 4;
L5 = 5;
L6 = 6;
hblock_sub = 1;
h_ram = 1;
eq1 = theta1 == 90-asin((L1-hblock_sub-h_ram-L3*sin(theta2)-L4*sin(theta3))/L5);
eq2 = theta2 == acos(-(L6-L4*cos(theta3)-L5*cos(90-theta1))/L3);
eq3 = theta3 == acos((L6+L3*cos(theta2)-L5*cos(90-theta1))/L4);
sol = solve([eq1 eq2 eq3],[theta1 theta2 theta3])
Warning: Unable to solve symbolically. Returning a numeric solution using vpasolve.
sol = struct with fields:
theta1: 91.09127337359374229308322484689 + 0.014498593097431429385469218902618i theta2: 1.4767379109122526840455134003504 + 0.022557641373024566759457348260003i theta3: 0.11249516403464522396375288111294 + 0.0068154950806834305011949243139886i
hblock = subs(sol.theta1,theta2,theta3)
hblock = 
hblock_sub = double(hblock_sub)
hblock_sub = 1

Produits

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