How to solve 3 linear equations symbollically?

1 次查看（过去 30 天）
Aman
Aman 2024-6-5
编辑： Torsten 2024-6-5
Ran in:
b1_position_minus_o4= [ 27.4622 , -4.7904 , 0];
syms omega_output_x omega_output_y omega_output_z real
omega_output = [omega_output_x; omega_output_y; omega_output_z];
% Velocity of point b1
v_b1 = cross(omega_output, (b1_position_minus_o4));
disp(v_b1);
v_a1= [-80.0000 -7.1960 0];
velocity_eq = v_a1 - v_b1
velocity_eq = 
double(solve(velocity_eq(1),omega_output_z))
ans = -16.7001
double(solve(velocity_eq(2),omega_output_z))
ans = -0.2620
omega_output_sol = solve(velocity_eq == 0, [omega_output_x, omega_output_y, omega_output_z])
omega_output_sol = struct with fields:
omega_output_x: [0x1 sym] omega_output_y: [0x1 sym] omega_output_z: [0x1 sym]
% Convert symbolic solution to numeric values
omega_output_numeric = double([omega_output_sol.omega_output_x, omega_output_sol.omega_output_y, omega_output_sol.omega_output_z]);
%disp('Output angular velocity (omega_output):');
disp(omega_output_numeric);
%I want numerical values of omega_output_x, omega_output_y, omega_output_z
%Please help someone

请先登录，再回答此问题。

采纳的回答

Torsten
Torsten 2024-6-5
移动：Torsten 2024-6-5
The first two equations in "velocity_eq" only have "omega_output_z" as variable and contradict each other (see above).

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Symbolic Math Toolbox 的更多信息

标签

Community Treasure Hunt

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

Start Hunting!

Translated by