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

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How can I find what is e when sm1=8.1813e-13?

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Star Strider
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'
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.
.

Più risposte (2)

KSSV
KSSV il 12 Ago 2022
Read about interp1

Torsten
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
Gloria
Gloria il 12 Ago 2022
I asked the question wrong.
I have graph for (sm1,e) and W1 and sm1 depend on e;
I want to know what is e value for sm1=8.1813e-13
Torsten
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)
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

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