Related documentation в матлабе ошибка

Код:

Ошибка:

Conversion to logical from sym is not possible.

Error in untitled3 (line 14)

if abs(x1-x0) < eps*(1-M1)/M1 % проверка условия окончания итераций

Пыталась поменять в 13 строке на:

if abs(x1-x0) < double(eps)*(1-double(M1))/double(M1)

Тогда выдает ошибку:

Error using symengine

Unable to convert expression containing symbolic variables into double array. Apply ‘subs’ function

first to substitute values for variables.

Error in sym/double (line 868)

Xstr = mupadmex(‘symobj::double’, S.s, 0);

Error in untitled3 (line 13)

if abs(x1-x0) < double(eps)*(1-double(M1))/double(M1) % проверка условия окончания итераций

Related documentation

error

Описание

msg = 'Error occurred.';
error(msg)

n = 7;
if ~ischar(n)
   error('Error. nInput must be a char, not a %s.',class(n))
end

Error.
Input must be a char, not a double.

if ~ischar(n)
   error('Error. nInput must be a char.')
end

Error. nInput must be a char.

if ~ischar(n)
   error('MyComponent:incorrectType',...
       'Error. nInput must be a char, not a %s.',class(n))
end

Error.
Input must be a char, not a double.

exception = MException.last

exception = 

  MException with properties:

    identifier: 'MyComponent:incorrectType'
       message: 'Error. 
Input must be a char, not a double.'
         cause: {0x1 cell}
         stack: [0x1 struct]

errorStruct.message = 'Data file not found.';
errorStruct.identifier = 'MyFunction:fileNotFound';

errorStruct = 

       message: 'Data file not found.'
    identifier: 'MyFunction:fileNotFound'

function hello(audience)
if nargin < 1
    aac = matlab.lang.correction.AppendArgumentsCorrection('"world"');
    error(aac, 'MATLAB:notEnoughInputs', 'Not enough input arguments.')   
end
fprintf("Hello, %s!n", audience)
end

Error using hello (line 4)
Not enough input arguments.

Did you mean:
>> hello("world")

Представлено до R2006a

function SwitchValueChanged(app, event)
            value = app.Switch.Value;
      
            if (value == 'Off')
                app.SaveOIButton.Enable = 'Off';
                app.OpenOIButton.Enable = 'Off';
            end

            if (value == 'On')
                app.SaveOIButton.Enable = 'On';
                app.OpenOIButton.Enable = 'On';
            end
        end

Arrays have incompatible sizes for this operation.

Error in Mie/SwitchValueChanged (line 82)
            if (value == 'Off')

Related documentation
 
Error while evaluating Switch PrivateValueChangedFcn.

1 ответ

Лучший ответ

function SwitchValueChanged(app, event)
     
    switch app.Switch.Value
        case 'Off'
            app.saveOIButton.Enable = 'off';
            app.openOIButton.Enable = 'off';
        case 'On'
            app.saveOIButton.Enable = 'on';
            app.openOIButton.Enable = 'on';
    end
end

1.- Your codes doesn’t plot V and D: Your code plots D(V) and T(V)

2.- T is completely flat, despite taking part in the inner for loop calculations with T(i)=Tsl*(rho(i)/1.2).^0.75; as it had to be somehow modified.

But in fact it remains constant for all samples of V, constant temperature (°C ?), and for all laps of the outer for loop sweeping variable h within [0:1:8].

The produced T(V) functions are the flat lines.

3.- Then you try building a 3rd function f that you put as if f(V) only but in fact it’s f(V,T) with the right hand side of the function with a numerical expression, without a symbolic expression, the symbolic expression that fzero expects to attempt zero solving.

In MATLAB Zero finding has to be done either symbolically or numerically.

A symbolic zero-finding function like fzero doesn’t work with numerical expressions like the ones you have calculated throughout the 2 loops for h and for alpha .

Examples of function expressions solvable by fzero :

3.1.-

fun = @(x)sin(cosh(x));
x0 = 1;
options = optimset('PlotFcns',{@optimplotx,@optimplotfval});
x = fzero(fun,x0,options)

3.2.-

fun = @sin; % function
x0 = 3; % initial point
x = fzero(fun,x0)

3.3.- put the following 3 lines in a separate file, call this file f.m

function y = f(x)
y = x.^3 - 2*x - 5;

and solve

fun = @f; % function
x0 = 2; % initial point
z = fzero(fun,x0)

3.4.- fzeros can solve parametrically

myfun = @(x,c) cos(c*x);  % parameterized function
c = 2;                    % parameter
fun = @(x) myfun(x,c);    % function of x alone
x = fzero(fun,0.1)

4.- So since you have already done all the numerical calculations and no symbolic expression is supplied, it’s reasonable to solve numerically, not symbolically.

To this purpose there’s a really useful function called intersections.m written by Douglas Schwarz available here

clear all; close all;clc

W = 10000; %[N]
S = 40; %[m^2]
AR = 7;
cd0 = 0.01;
k = 1 / pi / AR;
clalpha = 2*pi;
Tsl=800; 

figure(1);
ax1=gca
hold(ax1,'on');xlabel(ax1,'V');ylabel(ax1,'D');grid(ax1,'on');
title(ax1,'1st graph');
reczeros={}
for h=0:1:8; 
    i=0;
    for alpha = 1:0.25:12
        i=i+1;
        rho(i)=1.2*exp(-h/10.4);
        cl(i) = clalpha * alpha * pi/180;
        V(i) = sqrt(2*W/rho(i)/S/cl(i));
        L(i) = 0.5 * rho(i) * V(i) * V(i) * S * cl(i);
        cd(i) = cd0 + k * cl(i) * cl(i);
        D(i) = 0.5 * rho(i) * V(i) * V(i) * S * cd(i);
        clcd(i) = cl(i)/cd(i);
        p(i) = D(i)*V(i);
        ang(i) = alpha;
        T(i)=Tsl*(rho(i)/1.2).^0.75;
    end
 
    plot(ax1,V,D); hold(ax1,'on'); 
    plot(ax1,V,T);
    
    [x0,y0]=intersections(V,D,V,T,'robust');
    reczeros=[reczeros [x0 y0]];
    for k1=1:1:numel(x0)
        plot(ax1,x0,y0,'r*');hold(ax1,'on')
    end
    
end

Because at each pair D(V) T(V) there may be no roots, 1 root or more than 1 root, it makes sense to use a cell, reczeros, to store whatever roots obtained.

To read obtained roots in let’s say laps 3 and 5:

reczeros{3}
=
55.8850  692.5504
reczeros{5}
=
23.3517  599.5325
55.8657  599.5325

5.- And now the 2nd graph, the function that is defined in a different way as done in the double for loop:

P = 0.5*V.*V.*rho.*S.*cd-T;

figure(2);
ax2=gca
hold(ax2,'on');xlabel(ax2,'V');ylabel(ax2,'P');grid(ax2,'on');
title(ax2,'2nd graph')
plot(ax2,V,P)
plot(ax2,V,T)

[x0,y0]=intersections(V,T,V,P,'robust');
for k1=1:1:numel(x0)
    plot(ax2,x0,y0,'r*');hold(ax2,'on')
end

format short
V0=x0
P0=y0
V0 =
86.9993
P0 =
449.2990