hi! I wrote this script but i'm not sure if the response can be true, what do you think about it?
clc
clear all
dt=0.4;
[t,a]=ode45(@Law,[0 0+dt],[0;0;0;0]);
where
function [ da ] = Law( t,a )
[Ks,cs,I,m,g,L,K0,Kc,Icw,c0,cc,T0,r,alfa,mcw,lc,lcw,ls]=deal(1.0e+07,133.58,7.3777e-04,0.105795,9.78,0.0709444,0.37,1.0e+06,2.7195e-05,9.3e-05,24.22,0.12,0.028284,pi/4,0.11887,0.12,0.010528,0.09);
A=400;
w=1257;
h=0.003;
yl=lc*tan(a(1));
dyl=lc*a(3)/(cos(a(1)))^2;
yc=r*(sin(alfa)-sin(alfa-a(1)));
dyc=r*a(4)*cos(alfa-a(2));
yu=lc*tan(a(1))+h/cos(a(1));
dyu=((lc+h*sin(a(1)))*a(3))/(cos(a(1)))^2;
%value of Fs
if a(1)>=0
Fs=0;
else
Fs=-Ks*ls*a(1)-cs*ls*a(3);
end
%value of R
if yc<yl
R=lc/cos(a(1));
elseif (yl<yc) && (yc<yu)
R=0;
elseif (yl==yc) && (yc==yu)
R=0;
elseif (yl==yc) && (yc<yu)
R=0;
elseif (yl<yc) && (yc==yu)
R=0;
elseif yc>yu
R=lc/cos(a(1))+h*tan(a(1));
end
%value of Fr
if yc<yl
Fr=Kc*(yl-yc)*cos(a(1))+cc*(dyl-dyc)*cos(a(1))-(yl-yc)*a(3)*sin(a(1));
elseif (yl<yc) && (yc<yu)
Fr=0;
elseif (yl==yc) && (yc==yu)
Fr=0;
elseif (yl==yc) && (yc<yu)
Fr=0;
elseif (yl<yc) && (yc==yu)
Fr=0;
elseif yc>yu
Fr=Kc*(yu-yc)*cos(a(1))+cc*(dyu-dyc)*cos(a(1))-(yu-yc)*a(3)*sin(a(1));
end
%law
da=[a(3);a(4);Fs*ls/I-(R*Fr)/(I)-(m*g*A*L*sin(w*t)*cos(a(1)))/I;-K0*a(2)/Icw-c0*a(4)/Icw-T0/Icw+Fr*r*cos(a(1))*cos(alfa-a(2))/Icw+(mcw*g*lcw*(A*sin(w*t)*cos(a(2))+sin(a(2))))/Icw];
end
thanks in advance!

 采纳的回答

Walter Roberson
Walter Roberson 2015-7-8

0 个投票

When you have two values that are computed through different paths and the two values mathematically should represent the same quantities, then often the two values will not compare equal with == . This is due to floating point round off. For example, ((10/3)*3) might not compare exactly equal to 10.

1 个评论

thank you so much for your disponibility, except this problem do you think that the script and the function can be correct?

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