Best fit line in log-log scale

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Yen Tien Yap
Yen Tien Yap el 24 de Ag. de 2021
Comentada: Simon Chan el 24 de Ag. de 2021
Hi, I want to create a straight best fit line in the first portion of the graph and I don't want a curve best fit line, what should I do? Thank you.
rho=1000; %[kg/m3]
D=12.6*10^-3; %[m]
L=1.5; %[m]
miu=0.001; %[Pas]
g=9.81;
A=pi*D^2/4;
Q0=[1600,1500,1400,1300,1200,1100,1000,900,800,700,600,500,400,300,240,220,...
200,180,160,140,120,100,80,70,70,60,50,40,30,20,10];%[L/hr]
Q=Q0/(1000*3600);
%Wet-wet digital gauge
P_dpg=[20.1,17.5,15.7,13.1,11.6,9.3,8,6.5,5.3,4.1,3,2.1,1.3,0.8]; %[kPa]
%Inverted manometer
h=[6.9,5.9,5,4.1,3.2,2.6,1.8,1.3,0.7,0.5]; %[cm]
h_m=h/100;
P_mtr=rho*g*h_m;
%Capsuhelic gauge
P_cpg=[33,31,28,23,17,12,8];
P=[P_dpg.*1000,P_mtr,P_cpg];
V=Q/A;
Re=rho*V*D/miu;
f=P*D./(2*rho*L*V.^2);
X=Re(25:31);
Y=f(25:31);
p=polyfit(X,Y,1);
y=polyval(p,X);
figure(1)
loglog(Re,f,'x','LineWidth',1)
hold on
loglog(X,y,'--')
grid on
xlim([10^2 10^5])
ylim([0.001 0.1])
xlabel('Reynolds number Re')
ylabel('Friction factor f')
title('f vs Re')

Respuesta aceptada

Simon Chan
Simon Chan el 24 de Ag. de 2021
Like this?
p=polyfit(log(X),log(Y),1);
y=polyval(p,log(X));
figure(1)
loglog(Re,f,'x','LineWidth',1)
hold on
loglog(X,exp(y),'--')
grid on
  2 comentarios
Yen Tien Yap
Yen Tien Yap el 24 de Ag. de 2021
Yes. It is. Thank you so much. Could you briefly explain why would you code it like that?
Simon Chan
Simon Chan el 24 de Ag. de 2021
Simply because you are using loglog scale, so you need the equation:
(log y) = m(log x) + c for function polyfit to fit into a straight line.
This is same for polyval where variable y (but not log(y)) is calculated from (log x) so you need to convert it using exponential in the figure.

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