Best fit line on Semi log graph

Question

Maiesha Nujhat 2020-10-25

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/625403-best-fit-line-on-semi-log-graph

评论： Maiesha Nujhat 2020-10-25

Hello, I'm new to matlab. i want to plot a graph where x axis is in log scale and y axis is linear. The graph beow though shows straight line but I know the equation is not correct because It didn't pass throgh the scatter points

here is the code. Please help me, how can these line pass through the points and also how can i get the equation?

data=[0.891  35784.525582
      0.91   39142.72
      0.815  17679.26
      0.891  25582
      0.8861 30168.052]
 
p1=data(:,2); %Re
q1=data(:,1); %Cd
scatter(p1,q1)
set(gca,'XScale','log');
pp = polyfit(log(p1), q1, 1)
qq=exp(polyval(pp,log(p1)))
hold on
plot(p1, qq)
hold off

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Alan Stevens 2020-10-25

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/625403-best-fit-line-on-semi-log-graph#answer_523873

在 MATLAB Online 中打开

Like so

data=[0.891  35784.525582
      0.91   39142.72
      0.815  17679.26
      0.891  25582
      0.8861 30168.052];
 
p1=data(:,2); %Re
q1=data(:,1); %Cd
logp1 = log10(p1);
pp = polyfit(logp1, q1, 1);
qq=polyval(pp,logp1);
plot(logp1, qq,'-*',logp1,q1,'o'),grid
xlabel('log10(p1)'),ylabel('q1')
legend('fit','data')

3 个评论
显示 1更早的评论隐藏 1更早的评论

Alan Stevens 2020-10-25

在 MATLAB Online 中打开

The Matlab figures come from fitting Cd to lo10(Re).

The Excel figures come from fitting log10(Cd) to log10(Re).

If we modify the MATLAB program to the following

data=[0.891  35784.525582
      0.91   39142.72
      0.815  17679.26
      0.891  25582
      0.8861 30168.052];
 
p1=data(:,2); %Re
q1=data(:,1); %Cd
logp1 = log10(p1);
logq1 = log10(q1);
pp = polyfit(logp1, logq1, 1);
qq=polyval(pp,logp1);
plot(logp1, qq,'-*',logp1,logq1,'o'),grid
xlabel('log10(Re)'),ylabel('log(Cd)')
legend('fit','data')

we get

pp =
    0.1244   -0.6110

We can manipulate this as follows

%  pp = 0.1244   -0.6110
%  m = pp(1);  c = pp(2)
%  log10(Cd) = m*log10(Re) + c
%  log10(Cd) = log10(Re^m) + c
%  Cd = (10^c)*Re^m
%        10^c = 0.2449
%  Cd = 0.2449*Re^0.1244

So the result depends on the type of fit, Cd vs log10(Re) or log10(Cd) vs log10(Re).

Maiesha Nujhat 2020-10-25