Back-calculating a variable in equation to give a value of zero

3 Ansichten (letzte 30 Tage)
I have a non linear equation with many variables that run through a column array. I am trying to backcalulate the value of one of the variable that would produce a value of zero at each output of the column array. I am using root2d function, which is working fine, but I am only getting the last value. I am not sure how to save the output at each increment of j because the function runs througha for loop and doesn't stop till the last value. I would like the output of my variable rd to be spit out as a column or row array for each increment. Here is what I am doing:
%Calling the function
fun =@root2d;
rd0 = [0.05];
rd = fsolve(fun,rd0)
%loading input and defining variables
function F=root2d(rd)
load 'S71'
depr1=S71.Depth;
qdynr=S71.qdyn;
Velr=S71.velr;
dec2r=S71.dec;
phicv=34;
Q=6;
R=1;
V50=1;
Nkt=12;
c0=300;
c1=0.46;
c2=2.96;
gammap=8; %in kN/m3
k0=0.5;
chmin=0.031;
%for loop inside the function
for j=9:numel (depr1)
pm(j)=gammap*((1+2*k0)/3)*depr1(j);
V(j)=(Velr(j)*0.04375)/chmin;
strainterm(j)=1/(1+(V(j)/V50));
pm (j)
strainterm(j)
V(j)
qdynr(j)
F= qdynr(j)-(((Nkt*0.5*((6*sin(phicv))/(3-sin(phicv)))*exp(Q-1/rd)))+strainterm (j)*((((c0*(pm(j)^c1)*exp(rd*c2))-(Nkt*0.5*((6*sin(phicv))/(3-sin(phicv))))*exp(Q-1/rd)))));
end
end

Akzeptierte Antwort

Paresh yeole
Paresh yeole am 9 Jul. 2020
Bearbeitet: Paresh yeole am 9 Jul. 2020
k = 1; % outside the loop
% inside the loop
F(k) = ... %rest remains unchanged
k = k+1;
  3 Kommentare
Paresh yeole
Paresh yeole am 9 Jul. 2020
The whole point of answering almost the same thing was to put the point that instead of using a cell array, try using an array..
And yes you are correct, preallocating is a good idea.
Reem Jaber
Reem Jaber am 9 Jul. 2020
Well, thanks guys! I have tried doing that before but it didn't work. Once I index F with k, the solution doesn't work anymore. I have attached the data file for your reference.

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (1)

madhan ravi
madhan ravi am 9 Jul. 2020
% outside the loop
F = cell(9:numel(depr1), 1);
k = 0;
% inside the loop
k = k + 1; % the first line
F{k} = ... rest remains unchanged
  2 Kommentare
Reem Jaber
Reem Jaber am 9 Jul. 2020
Thank you! But it doesn't seem to be working, I think the problem is in F{k), it is resulting in an error in the fsolve: "FSOLVE requires all values returned by functions to be of data type double". Also, the size of the cell is resulting in an error too once I define it as you did, it worked when I changed into a scalar.
madhan ravi
madhan ravi am 9 Jul. 2020
I can’t help because I don’t have your data

Melden Sie sich an, um zu kommentieren.

Kategorien

Mehr zu Programming finden Sie in Help Center und File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by