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Best fit line in log-log scale

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Yen Tien Yap
Yen Tien Yap 2021 年 8 月 24 日
コメント済み: Simon Chan 2021 年 8 月 24 日
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')

採用された回答

Simon Chan
Simon Chan 2021 年 8 月 24 日
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 件のコメント
Simon Chan
Simon Chan 2021 年 8 月 24 日
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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