How to set up the condition if the accuracy for e should be of 10^-14?

I have one for loop and want to calculate until the difference between reaches upto 10^-14. So should i set up the loop for that .

e = sqrt(1-(1-f^2));
e_bar = e/(sqrt(1-e^2)) ; 

I need set such that it should be less.

Can any one help me to set up the condition for iterative way?

thank you

[KSSV]

difference between what and what? Where is iteration?

[vimal kumar chawda]

I dont have any particular number of iteration but keep iterate until we will get no changing vaule of e for first 14 digits of e.

difference of e with in itself.

same for e_bar

a=6378137;    %semi-major axis of reference ellipsoid(units are in meter)
J2 = 108263e-8;   %non-normalised zonal spherical coecient(unit less)
GM =3986005e8;   %gravity constant G times Mass M (m^3/s^2)
omega = 7292115e-11;  %angular velocity (rad/sec)  
f= 1/298 ;  % approx flattening 

%% first and second numeric eccentricity e and e0 with an accuracy of 10^-14
n=100000;

for i=1:n 
    
e = sqrt(1-(1-f^2));
e_bar = e/(sqrt(1-e^2)) ;           %eq 3
i=i+1

end

[KSSV]

What is the use of iterator here? The same calculation gets done for n times....there is no changed in e and e_bar value.

[vimal kumar chawda]

it makes no sense to give an accuracy between different things. 10^-14 should be the difference between e(i+1) and e(i) and e‘(i+1) and e‘(i) between 2 iterations. You should reach that after 10 iterations.

[vimal kumar chawda]

I got above response from my coordinator, so hope you understood . What they want from me?

[KSSV]

No I didn't get.....what exactly is your problem? What you are trying to achieve? It lookslike you are trying for Newton Rphson method....but the problem is not clear ot me.

[vimal kumar chawda]

I want to iterate nearly 10 to 20 times until the accuray reach to 10e-14. so the value for the e should be differ. We have 2 e values.

e and e_bar. The results after iteration will be in attache pic.

Answer 1

[Ajit Kasturi]

Assuming you want to loop until the accuracy e is of the order 10^-14 you can do the same for e_bar also:

a=6378137;    %semi-major axis of reference ellipsoid(units are in meter)
J2 = 108263e-8;   %non-normalised zonal spherical coecient(unit less)
GM =3986005e8;   %gravity constant G times Mass M (m^3/s^2)
omega = 7292115e-11;  %angular velocity (rad/sec)  
f= 1/298 ;  % approx flattening 
%% first and second numeric eccentricity e and e0 with an accuracy of 10^-14
n=100000;
% keep a variable prev_e
prev_e=0;

for i=1:n 
    
e = sqrt(1-(1-f^2));
e_bar = e/(sqrt(1-e^2)) ;           %eq 3

if i==1
    i=i+1;
    continue;
end

i=i+1

if abs(e-prev_e)<=10^-14
    break;
end

prev_e=e

end

[vimal kumar chawda]

It is not working as the value for e should be e=0.0818191910431508

e_new = 0.082094438152352.

I dont know to work on it.