how can i substitute number with symbol ?

7 vues (au cours des 30 derniers jours)
shahin hashemi
shahin hashemi le 16 Fév 2018
Modifié(e) : shahin hashemi le 16 Fév 2018
dear all
i have code like below :
clc
clear all
N=4;
L=60e-3;
M=3.338e-3;
jx=5.743e-14;
E=2.1e11;
x = sym('x', [1 3*N], 'real');
for j=2:N+1
for i=j-1
f1(j)=(1-cos(x(3*i-2)))/x(3*i-2);
g1(j)=-(1-cos(x(3*i-1)))/x(3*i-1);
f2(j)=sin(x(3*i-2))/x(3*i-2);
g2(j)=sin(x(3*i-1))/x(3*i-1);
pl(:,j)=L*[f1(j) f2(j)*g1(j) f2(j)*g2(j)]';
Rl(:,:,j)=[cos(x(3*i))*cos(x(3*i-2)) -sin(x(3*i))*cos(x(3*i-2)) sin(x(3*i-2));cos(x(3*i))*sin(x(3*i-2))*sin(x(3*i-1))+sin(x(3*i))*cos(x(3*i-1)) -sin(x(3*i))*sin(x(3*i-2))*sin(x(3*i-1))+cos(x(3*i))*cos(x(3*i-1)) -cos(x(3*i-2))*sin(x(3*i-1));-cos(x(3*i))*sin(x(3*i-2))*cos(x(3*i-1))+sin(x(3*i))*sin(x(3*i-1)) sin(x(3*i))*sin(x(3*i-2))*cos(x(3*i-1))+cos(x(3*i))*sin(x(3*i-1)) cos(x(3*i-2))*cos(x(3*i-1))];
p(:,2)=pl(:,2);
end
end
R(:,:,2)=Rl(:,:,2);
R(:,:,1)=[1 0 0;0 1 0;0 0 1];
for i=3:N+1
R(:,:,i)=R(:,:,i-1)*Rl(:,:,i);
p(:,i)=p(:,i-1)+R(:,:,i-1)*pl(:,i);
end
for j=2:N+1
for i=1:N
for r=1:N
d1Pbeta(:,j,i)=diff(p(:,j),x(3*i-2));
d2Pbetabeta(:,j,i,r)=dot([-M*9.81 0 0],diff(d1Pbeta(:,j,i),x(3*r-2)));
end
end
end
for i=1:N
for r=1:N
for j=2:N+1
sigm(i,r)=(E*jx/L)+sum(d2Pbetabeta(:,j,i,r)');
end
end
end
here i want to replace x with number
x=[0.0495837022782378;3.96268247549320e-09;-4.78044641555791e-10;0.0854433755724521;2.89430795211353e-09;-1.43061297211777e-10;0.110831068380515;4.80530147316531e-09;-5.69420213726607e-10;0.125761304817654;1.09464269988965e-08;-4.73181277688372e-10]
but it doesnt work
and after that calculate eigenvalue of sigm
e=eig(sigm)

Réponse acceptée

Birdman
Birdman le 16 Fév 2018
  1 commentaire
shahin hashemi
shahin hashemi le 16 Fév 2018
Modifié(e) : shahin hashemi le 16 Fév 2018
thank you for your attention bird man but when i use this command like below :
subs(sigm,(x(11)),0.1)
it replace x(11) with 0.1 and answer cos(0.1) and doesn't calculate it i mean cos(0.1)=0.9 i think it is still in symbol mod
any solutions ?

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