control of a script
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Guido De Angelis
on 8 Jul 2015
Commented: Guido De Angelis
on 8 Jul 2015
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!
0 Comments
Accepted Answer
Walter Roberson
on 8 Jul 2015
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.
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