Help with iteration!!?
信息
此问题已关闭。 请重新打开它进行编辑或回答。
显示 更早的评论
Can anyone help me with iteration and convergence? I need my new Wd to converge with the old Wd in my loop.
ci=174.0453; %Cone index in psi
hr=15.1; %rear tire section height (in)
br=19; %unloaded tire section width rear (in)
dr=80.8; %unloaded overall diameter (in)
SLRR= 37; %static loaded radius of rear tire (in)
Wsar=22575; %the static rear axle load
t=18*(pi/180); %18 degree angle
deltar= ((80.8/2)-27);
dbh=20.5; %drawbar height (in)
bf=16.5; %unloaded tire section width (in)
hf=14.06; %front tire section height (in)
df=62.6; %unloaded overall diameter (in)
SLRF= 28.4; %static loaded radius of rear tire (in)
Wsaf=17455; %the static rear axle load
deltaf= ((62.5/2)-28.4);
s=.01:.001:.2; %slip range 1-20 increments of .1
Wstf= Wsaf/4;
Wstr=Wsar/6;
Wdf=Wstf;
Wdr=Wstr;
for s=.01:.001:.2;
Bnf=((ci*bf*df)/Wdf)*((1+5*(deltaf/hf))/(1+3*(bf/df)));
Bnr=((ci*bf*dr)/Wdr)*((1+5*(deltar/hr))/(1+3*(br/dr)));
NTf=Wdf*(.88*(1-exp(-.1*Bnf))*(1-exp(-7.5*s))-(1/Bnf)-((.5*s)/sqrt(Bnf)));
NTr=Wdr*(.88*(1-exp(-.1*Bnr))*(1-exp(-7.5*s))-(1/Bnr)-((.5*s)/sqrt(Bnr)));
Px=NTf+NTr;
omega= 72*(pi/180);
Py= ((Px/sin(omega))*sin(t));
P=sqrt((Py).^2+(Px).^2);
Wd=Wstf-Py;
Wd1=Wstr-Py;
while (Wdf-Wd)> 0.0001;
(Wdr-Wd1)>0.0001;
Wd1=Wdr;
Wd=Wdf;
end
end
1 个评论
Kuifeng
2016-4-9
can you edit your message with a empty space in front of the code?.
%space ... then the code
回答（1 个）
Roger Stafford
2016-4-9
0 个投票
Your while-loop does not seem to be very useful. On the first pass through it sets Wd equal to Wdf and because the "(Wdf-Wd)> 0.0001" condition will then fail, that is also its last pass through the loop. You need to rethink what you are trying to accomplish here.
2 个评论
Garrett
2016-4-9
Roger Stafford
2016-4-9
You will have to do some explaining about what you are trying to accomplish. We don't know what you mean by the "old W" and the "new W". Give a good explanation in plain English about what you are trying to achieve with your variable Wd and Wd1.
此问题已关闭。
关闭：
2021-8-20
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by 
