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

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

Answer 1

[Star Strider]

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.

.

Answer 2

[KSSV]

Read about interp1

Answer 3

[Torsten]

sm1 = v1.*N.*L1.*R./(4.*W1.*(L1./C).^2)

gives

W1 = v1.*N.*L1.*R./(4.*sm1.*(L1./C).^2)

[Gloria]

No opposite of this W1 gives SM1

[Torsten]

You can solve the equation

sm1 = v1.*N.*L1.*R./(4.*W1.*(L1./C).^2)

for W1.

Multiply the equation by W1 and then divide by sm1. This gives

sm1.* W1 = v1.*N.*L1.*R./(4.*(L1./C).^2)
W1 = v1.*N.*L1.*R./(4.*sm1.*(L1./C).^2)

[Gloria]

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]

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