Problems with while loop

Hi, I have a problem, I want the subtraction d1 to be less than 1 * 10 ^ -15 after several iterations, but the program stays busy.

[James Tursa]

Please post your code as regular text and highlight it with the CODE button. We can't run pictures.

[Tamia Eli]

Sorry, it is this code:

clear
format LONGG
    a=6373878;
    f=1/297;
k0=0.9996;
e2=2*f-f^2;
e=sqrt(e2);
c0=1+((3/4)*(e^2))+((45/64)*(e^4))+((175/256)*(e^6))+((11025/16384)*(e^8))+((43659/65536)*(e^10));
c2=((3/4)*(e^2))+((15/16)*(e^4))+((525/512)*(e^6))+((2205/2048)*(e^8))+((72765/65536)*(e^10));
c4=((15/64)*(e^4))+((105/256)*(e^6))+((2205/4096)*(e^8))+((10395/16384)*(e^10));
c6=((35/512)*(e^6))+((315/2048)*(e^8))+((31185/131072)*(e^10));
c8=((315/16384)*(e^8))+((3465/65536)*(e^10));
c10=((693/131072)*(e^10));

x=483250.07981339
y=2303647.10551245
zone=39;
hemis='s';
            if(strcmp(hemis,'n')||strcmp(hemis,'N'))
                siglat=1;
            else
                if(strcmp(hemis,'s')||strcmp(hemis,'S'))
                    siglat=-1;
                else
                    fprintf('Wrong');
                end
            end
if siglat==-1
    y=y-10000000;
end
x=x-500000;
tt=1;
phi1=(y/k0)/(a*(1-e2)*c0);
while tt>1e-15
B1=a*(1-e2)*(c0*phi1-c2/2*sin(2*phi1)+c4/4*sin(4*phi1)-c6/6*sin(6*phi1)+c8/8*sin(8*phi1)-c10/10*sin(10*phi1));
d1=(y/k0)-B1; 
d1r=d1/(a*(1-e2)*c0);
phi1=phi1+d1r;
tt=abs(d1); 
end
phip=phi1 %phi'

Answer 1

[Bruno Luong]

"Hi, I have a problem, I want the subtraction d1 to be less than 1 * 10 ^ -15 after several iterations, but the program stays busy."

Well you cannot demand floating point error to be that small.

Double IEEE has about 15 digits relative precision. You compare B1 to (y/k0) which is -7699432.66755457. The most you can demand is error is about

>> tol = eps(y/k0)

tol =

      9.31322574615479e-10

So if you replace the break condition by

tol = eps(y/k0);
while tt>tol
   ...
end

your while loop will stop.

[Tamia Eli]

Thank you, I could understand your answer.