Since ‘sm1’ is a function of ‘W1’ this is not as straightforward as it might at first appear.
v1=0.034; %viskozite yağ için 0.034 Pas 50C
v2=0.000594; %viskozite tuzlu su için 0.000481 Pas 50C
N=525/60; %hız (rad(s)
DS=0.022; %stern tube çap
R=0.011; %stern tube yarı çap
D=0.02; %şaftın çapı
L1=R; L2=R*2; L3=R*3; %uzunluk (m)
C=1*10^-3; %
%sm=sommerfield number
e=0.1:0.1:0.9;
%%%%%%%%%%%%%%%%
%%%% short journal bearing %%%%
%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%% yağ için %%%%%%%%%%%%%%%%
W1 = v1.*N.*R.*L1.*(L1./C).^2.*(e./4).*((16.*e.^2+pi.^2.*(1-e.^2)).^0.5./(1-e.^2).^2);
sm1 = v1.*N.*L1.*R./(4.*W1.*(L1./C).^2);
sm1q = 8.1813e-13;
W1_fcn = @(e) v1.*N.*R.*L1.*(L1./C).^2.*(e./4).*((16.*e.^2+pi.^2.*(1-e.^2)).^0.5./(1-e.^2).^2);
sm1_fcn = @(e) v1.*N.*L1.*R./(4.*W1_fcn(e).*(L1./C).^2); % Create Function
[e_v,fval] = fsolve(@(x)norm(sm1_fcn(x)-sm1q), rand) % Using 'fsolve'
% [e_v,fval] = fsolve(@(x)sm1_fcn(x)-sm1q, rand) % Using 'fsolve'
e_q = interp1(sm1, e, sm1q, 'linear','extrap') % Using 'interp1'
figure
plot(sm1, e)
xlabel('sm1')
ylabel('e')
Check1 = sm1_fcn(e_v) % 'fsolve'
Check2 = sm1_fcn(e_q) % 'interp1'
There are two different values depending on whether the function is solved numerically or extrapolated, neither of which provides the desired result.
EDIT — (12 Jul 2022 at 13:28)
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