plz help me , why this code is not working ?

m = .02
p = .4
t = .12
x = [0 : .01 : .75]
for i = 1 : length(x)
z(1, i) = (.5 * nacathick(t, x(i)) + nacacamber(m, p, x(i)))
z(2, i) = (-.5 * nacathick(t, x(i)) + nacacamber(m, p, x(i)))
end

3 comentarios

Youssef  Khmou
Youssef Khmou el 2 de Mayo de 2013
hi,
what is the function camber ?
Walter Roberson
Walter Roberson el 2 de Mayo de 2013
And also thick().
It would help if you indicated the difference between what you observe and what you would like.
Image Analyst
Image Analyst el 2 de Mayo de 2013

Respuestas (2)

Jan
Jan el 2 de Mayo de 2013
Editada: Jan el 2 de Mayo de 2013

0 votos

Beside other not useful things, "clear all" removes all breakpoint in the code. Therefore it impedes debugging. But debugging is the best way to solve problems!
So I suggest to omit the "clear all" (and most likely neither clc nor close all are useful also), set a breakpoint in the first and step through the code line by line to inspect, what's going on.
Btw., who suggests such cruel clearing headers like "clc; close all; clear all"? It appears such frequently in the code of beginners that I assume they should be kept away from the debugger for any evil purposes. "clc" hides even valuable warning messages in the command window - how strange.
It would be very nice and useful, if the knowledge, that eval and numbered names of variables like "A1", "A2", ... should be avoided, would be such wide-spread as the "clear all" killer.
shambhu kumar
shambhu kumar el 2 de Mayo de 2013

0 votos

sir, i just want to execute the code,

1 comentario

Jan
Jan el 2 de Mayo de 2013
If you want to get assitence, it would be useful to explain "this code is not working" with any details. Do you get an error message or does the result differ from your expectations?
Did you follow Image Analyst's valuable suggestion to read "How to get a fast answer"?
Youssef and Walter asked for the thick() and camber() functions. As far as I can see, you have renamed them now to nacathick() and nacacamber(), but this does not change anything substantially.

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Preguntada:

el 2 de Mayo de 2013

Cerrada:

el 20 de Ag. de 2021

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