Parse error at '<': usage might be invalid MATLAB syntax

Question

Arupratan Gupta 2021-8-26

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1441519-parse-error-at-usage-might-be-invalid-matlab-syntax

编辑： Walter Roberson 2021-8-27

% the fluid flow through any of the ipe or chimney can be calculated using
% the realtion of the Moody chart and the Colebrook equation
D = 0:0.1:5;
epsilon = 100:100.5:500;
Re = 2300:2300.5:2900;
f = 0:0.1:10;
i = 0:0.1:1000;
for(i<1100)
    1/((f)^2) = -2.0 log[((epsilon/D)/3.7) + ((2.51/Re*((f)^2)))];
plot(f,D);
title('The Moody chart');

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Ive J 2021-8-26

编辑：Walter Roberson 2021-8-27

what about learning MATLAB before using :) ?

mathworks.com/learn/tutorials/matlab-onramp.html

DGM 2021-8-26

编辑：DGM 2021-8-27

在 MATLAB Online 中打开

For starters:

% these all have different lengths
% some (e.g. Re) appear to be mistakes
D = 0:0.1:5; % 51 elements
epsilon = 100:100.5:500; % 4 elements
Re = 2300:2300.5:2900; % 1 element
f = 0:0.1:10; % 101 elements
i = 0:0.1:1000; % 10001 elements
% this isn't how a for-loop works
for (i<1100) % this expression is assigned to nothing
    % since nothing in the loop depends on the variable that doesn't exist
    % what is the purpose of the loop?
    
    % this is an invalid assignment with several internal problems
    % the RHS is not a target for assignment.
    % log() is the natural log. colebrook eqn needs log10()
    % use parentheses, not square brackets for function scoping
    % use elementwise operators ./ .^ when dealing with non-scalars
    % since none of the vector lengths match, none of this will work anyway
    1/((f)^2) = -2.0 log[((epsilon/D)/3.7) + ((2.51/Re*((f)^2)))];
% the loop structure is never closed