How to determine an y value at a specific x value?

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Gloria on 12 Aug 2022
Edited: Gloria on 12 Aug 2022
How can I find what is e when sm1=8.1813e-13?

Star Strider on 12 Aug 2022
Edited: Star Strider on 12 Aug 2022
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
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'
Equation solved at initial point. fsolve completed because the vector of function values at the initial point is near zero as measured by the value of the function tolerance, and the problem appears regular as measured by the gradient.
e_v = 0.9969
fval = 6.4533e-10
% [e_v,fval] = fsolve(@(x)sm1_fcn(x)-sm1q, rand) % Using 'fsolve'
e_q = interp1(sm1, e, sm1q, 'linear','extrap') % Using 'interp1'
e_q = 0.9314
figure
plot(sm1, e)
xlabel('sm1')
ylabel('e')
Check1 = sm1_fcn(e_v) % 'fsolve'
Check1 = 6.4614e-10
Check2 = sm1_fcn(e_q) % 'interp1'
Check2 = 3.3070e-07
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)
Changed code to be functions of ‘e’ as stated in this Comment.
.

KSSV on 12 Aug 2022

Torsten on 12 Aug 2022
Edited: Torsten on 12 Aug 2022
sm1 = v1.*N.*L1.*R./(4.*W1.*(L1./C).^2)
gives
W1 = v1.*N.*L1.*R./(4.*sm1.*(L1./C).^2)
Torsten on 12 Aug 2022
Edited: Torsten on 12 Aug 2022
v1=0.034; %viskozite yağ için 0.034 Pas 50C
v2=0.000594; %viskozite tuzlu su için 0.000481 Pas 50C
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; %
sm1 = 8.1813e-13;
W1 = @(e) v1.*N.*R.*L1.*(L1./C).^2.*(e./4).*((16.*e.^2+pi.^2.*(1-e.^2)).^0.5./(1-e.^2).^2);
fun = @(e) sm1 - v1.*N.*L1.*R./(4.*W1(e).*(L1./C).^2);
options = optimset('Display','none','TolX',1e-16,'TolFun',1e-16);
format long
e0 = 0.9;
e1 = fsolve(fun,e0,options)
e1 =
0.993503468416158
fun(e1)
ans =
-2.889044323179053e-09
e0 = 1.1;
e2 = fsolve(fun,e0,options)
e2 =
1.006043289756047
fun(e2)
ans =
-2.487888648942313e-09