Powtorzenie czworobok maxima teoria mechanizmów PDF

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Rozwiązanie zadania z czworoboku z teorii mechanizmow w programie maxima...

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powtorzenie_czworobok.wxmx

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(%i1)

eq1:l2·exp(%i·theta_2(t))+l3·exp(%i·theta_3(t))=l1+l4·exp(%i·theta_4(t));

(eq1)

l3 %e %i

(%i2)

d_eq:diff(eq1, t);

(d_eq)

%i l3 %e %i

(%i3)

d_eq:subst([diff(theta_3(t), t)=omega_3, diff(theta_2(t), t)=omega_2, diff (theta_4(t), t)=omega_4], d_eq);

(d_eq)

%i l3 !3 %e %i

Ò3 ( t ) +

l2 %e %i

Ò2 ( t )

Ò3 ( t ) ⎛ d

⎞ ⎞ ⎞ %i Ò2 ( t ) ⎜⎛ d %i Ò4 ( t ) ⎛⎜ d ⎟ ⎟ ⎜ ⎟ ⎜⎝ d t Ò 2 ( t ) ⎟⎠ = %i l4 %e ⎜⎝ d t Ò 4 ( t ) ⎟⎠ ⎜⎝ d t Ò 3 ( t ) ⎟⎠ + %i l2 %e

Ò3 ( t )

+ %i l2 !2 %e %i Ò2 ( t ) = %i l4 !4 %e %i Ò4 ( t )

d_eq:subst([l2=1, l3=1, l4=sqrt(3), theta_2(t)=%pi/2, theta_3(t)=%pi/6, theta_4(t)=2·%pi/3, omega_2=2.6], d_eq); p p p ⎛ 1⎞ 3 ⎞⎟ 3 %i %i !3 − 2.6 = 3 %i ⎜⎜ − ⎟⎟ !4 ⎟ ⎝ 2 2 ⎠ 2⎠

(%i4) (d_eq)

⎛ %i ⎜ ⎜⎝ 2 +

(%i5)

d_eq_real:realpart(d_eq);

(d_eq_real) −

(%i6) (d_eq_imag)

(%i7)

!3 2

3 !4

− 2.6 = −

2

d_eq_imag:imagpart(d_eq); p p 3 !3 2

=−

3 !4 2

solve([d_eq_real, d_eq_imag], [omega_3, omega_4]);

rat: replaced -2.6 by -13/5 = -2.6 (%o7)

[ [ !3 = −

13 10

, !4 =

13 10

]]

(%i8)

float(%);

(%o8)

[ [ !3 = − 1.3 , !4 = 1.3 ] ]

➔

= l4 %e %i Ò4 ( t ) + l1

;...