How to determine an y value at a specific x value?
8 visualizzazioni (ultimi 30 giorni)
Mostra commenti meno recenti
How can I find what is e when sm1=8.1813e-13?
0 Commenti
Risposta accettata
Star Strider
il 12 Ago 2022
Modificato: Star Strider
il 12 Ago 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
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)
.
0 Commenti
Più risposte (2)
Torsten
il 12 Ago 2022
Modificato: Torsten
il 12 Ago 2022
sm1 = v1.*N.*L1.*R./(4.*W1.*(L1./C).^2)
gives
W1 = v1.*N.*L1.*R./(4.*sm1.*(L1./C).^2)
4 Commenti
Torsten
il 12 Ago 2022
Modificato: Torsten
il 12 Ago 2022
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; %
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)
fun(e1)
e0 = 1.1;
e2 = fsolve(fun,e0,options)
fun(e2)
Vedere anche
Categorie
Scopri di più su Predictive Maintenance Toolbox in Help Center e File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!