How to model 3 Degrees of Freedom at once

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Siddharth Jain 2022-11-9

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1846973-how-to-model-3-degrees-of-freedom-at-once

编辑： Torsten 2022-11-9

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I have been unable to code the follwing equations in matlab-

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Answer 1

Torsten 2022-11-9

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1846973-how-to-model-3-degrees-of-freedom-at-once#answer_1095023

编辑：Torsten 2022-11-9

在 MATLAB Online 中打开

Define a solution vector y = (y1,y2,y3,y4,y5,y6) with

y1 = y_a, y2 = y_a_dot, y3 = z_a, y4 = z_a_dot, y5 = Theta_a, y6 = Theta_a_dot

and write out the differential equations for these 6 variables.

E.g. for y1 and y2:

dy1/dt = y2
dy2/dt = (-F_y(t) - k_ay*y1 - c_ay*y2)/m_a

Then use ODE45 to solve.

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Torsten 2022-11-9

在 MATLAB Online 中打开

You use Fy, Fz and T_a as functions of t. Thus you must define them as such:

Fy = @(t) 1
Fz = @(t) 1
T_a = @(t) 178

instead of

Fy=1;
Fz=1;
T_a = 178;

Torsten 2022-11-9

编辑：Torsten 2022-11-9

在 MATLAB Online 中打开

You defined 6 equations, not 4.

tspan = [0 1];

y0 = zeros(6,1);

[T,Y] = ode45(@reduced,tspan,y0);

plot(T,Y)

function yp = reduced(t,y)

m_a = 0.5;

c_ay=1;

c_az=1;

k_ay= 1;

k_az =1;

Fy=@(t)1;

Fz=@(t)1;

T_a = @(t)178;

I_a = 2000;

R_a=1;

yp = zeros(6,1);

yp(1) = y(2);

yp(2) = (-Fy(t) - k_ay*y(1) - c_ay*y(2))/m_a;

yp(3) = y(4);

yp(4) = (-Fz(t) - k_az*y(3) - c_az*y(4))/m_a;

yp(5) = y(6);

yp(6) = (-T_a(t) - Fy(t)*R_a)/I_a;

end

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