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

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vimal kumar chawda
vimal kumar chawda 2020-6-19
评论： vimal kumar chawda 2020-6-23
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
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KSSV
KSSV 2020-6-19
编辑：KSSV 2020-6-19
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
vimal kumar chawda 2020-6-23
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.

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回答（1 个）

Ajit Kasturi
Ajit Kasturi 2020-6-19
编辑：Ajit Kasturi 2020-6-19
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
  1 个评论
vimal kumar chawda
vimal kumar chawda 2020-6-23
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.

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