Index exceeds the number of array elements

조회 수: 3 (최근 30일)
N/A
N/A 2023년 1월 11일
댓글: N/A 2023년 1월 13일
Hello All, I'm encountring the following error when I run my code: Index exceeds the number of array elements. My code is simple and shown below:
format longg
syms p1
nm = 3e15;
s01=1;
v1=4.82e20;
tao00=1e-13;
p00=3;
q00=21;
q11=10.6;
R=1.9872036e-3;
T=295;
V1=16.7;
K1=3.27e19;
t1=linspace(1,1e6,10);
taop1=3.6;
S1=4.7;
A1=4.74e3;
Nm = A1*nm;
W1=(Nm/(K1*V1));
y1=q00;
a1= ((s01*tao00*v1)/(nm))*exp((y1)/(R*T));
Z1=(1/(1+a1*p00));
Y1 = (p00/(1+(a1*p00)));
eq1 = ((t1./taop1)==(log(p00./p1))+((a1.*W1).*(Z1-(1./(1+a1.*p1))+log(((1+a1.*p1)./p1).*Y1))));
for i=t1
pp1=0;
pp1(i) = vpa(vpasolve(eq1(i),p1),11);
end
Index exceeds the number of array elements. Index must not exceed 10.

Error in indexing (line 1079)
R_tilde = builtin('subsref',L_tilde,Idx);
pp1;
% loglog(t1,pp1,'o')
% ylim([1e-11 1e-2])
% xlim([10 1e6])
Thank you all for the help in advance.
  댓글 수: 2
N/A
N/A 2023년 1월 12일
When I replace the lincpace with the regular t1=1:1:10e6; i get the following error:
Unable to perform assignment because the left and right sides have a different number of elements.
Error in sym/privsubsasgn (line 1128)
L_tilde2 = builtin('subsasgn',L_tilde,struct('type','()','subs',{varargin}),R_tilde);
Error in sym/subsasgn (line 965)
C = privsubsasgn(L,R,inds{:});
Error in Langmuir_Q21 (line 44)
pp1(i) = vpa(vpasolve(eq1(i),p1),6);
Walter Roberson
Walter Roberson 2023년 1월 12일
Assign the results of the vpasolve to a variable. Test to see if the variable is empty or has more one value.

댓글을 달려면 로그인하십시오.

채택된 답변

Walter Roberson
Walter Roberson 2023년 1월 12일
A number of the solutions involve complex values.
If you start t1 at 1 then the first few output values are enough greater that the rest are not visible on the graph.
When you start t1 at 1 then you get negative components for the real value of some of the results, so you cannot plot on the log scale. There are negative components for most of the imaginary parts so you cannot use log scale for that either.
format longg
syms p1
Q = @(v) sym(v);
nm = Q(3)*Q(10)^15;
s01 = Q(1);
v1 = Q(482)*Q(10)^18;
tao00 = Q(1)*Q(10)^(-13);
p00 = Q(3);
q00 = Q(21);
q11 = Q(106)/Q(10);
R = Q(19872036)*Q(10)^(-10);
T = Q(295);
V1 = Q(167)/Q(10);
K1 = Q(327)*Q(10)^17;
taop1 = Q(36)/Q(10);
S1 = Q(47)/Q(10);
A1 = Q(4740);
Nm = A1*nm;
W1=(Nm/(K1*V1));
y1=q00;
a1= ((s01*tao00*v1)/(nm))*exp((y1)/(R*T));
Z1=(1/(1+a1*p00));
Y1 = (p00/(1+(a1*p00)));
t1 = linspace(1,50,50);
t2 = linspace(50,200,50);
t3 = linspace(200,1e6,500);
syms T1
eq1 = ((T1./taop1)==(log(p00./p1))+((a1.*W1).*(Z1-(1./(1+a1.*p1))+log(((1+a1.*p1)./p1).*Y1))));
pp1 = arrayfun(@(E)vpasolve(E,p1, 1e-5), subs(eq1, T1, t1));
pp2 = arrayfun(@(E)vpasolve(E, p1, 1e-5), subs(eq1, T1, t2));
pp3 = arrayfun(@(E)vpasolve(E, p1, 1e-5), subs(eq1, T1, t3));
subplot(2,3,1);
plot(t1, real(pp1));
title('p1 real');
subplot(2,3,2);
plot(t2, real(pp2));
title('p2 real');
subplot(2,3,3);
plot(t3, real(pp3));
title('p3 real');
subplot(2,3,4);
plot(t1, imag(pp1))
title('p1 imag')
subplot(2,3,5);
plot(t2, imag(pp2))
title('p2 imag');
subplot(2,3,6);
plot(t3, imag(pp3))
title('p3 imag');
  댓글 수: 1
N/A
N/A 2023년 1월 13일
Thank you so much! I understand now whats going on here! I really apprecitae your help.

댓글을 달려면 로그인하십시오.

추가 답변 (1개)

Torsten
Torsten 2023년 1월 12일
편집: Torsten 2023년 1월 12일
I suggest you insert values for p1 and solve for t1. This will be easier than the other way round.
format longg
syms p1 t1
nm = 3e15;
s01=1;
v1=4.82e20;
tao00=1e-13;
p00=3;
q00=21;
q11=10.6;
R=1.9872036e-3;
T=295;
V1=16.7;
K1=3.27e19;
T1=linspace(1,10,10);
taop1=3.6;
S1=4.7;
A1=4.74e3;
Nm = A1*nm;
W1=(Nm/(K1*V1));
y1=q00;
a1= ((s01*tao00*v1)/(nm))*exp((y1)/(R*T));
Z1=(1/(1+a1*p00));
Y1 = (p00/(1+(a1*p00)));
eq1 = ((t1./taop1)==(log(p00./p1))+((a1.*W1).*(Z1-(1./(1+a1.*p1))+log(((1+a1.*p1)./p1).*Y1))));
for i=1:numel(T1)
pp1(i) = double(vpasolve(subs(eq1,t1,T1(i)),p1,11));
end
pp1.'
ans = 10×1
2.27239538495313 1.72126026208779 1.30379462550031 0.987578963504762 0.748056626521194 0.566626808824952 0.429200048722703 0.325104070308059 0.246254996746847 0.186529573242241
  댓글 수: 5
Torsten
Torsten 2023년 1월 13일
편집: Torsten 2023년 1월 13일
I didn't use
pp1(i) = vpa(solve(eq1(i),p1,'IgnoreAnalyticConstraints',1),6);
anywhere in the code I provided.
And the decrease towards 0 has nearly ended at t=10. There is no need to solve up to 1e6.
But if you want: T1 = linspace(1,1e6,100) will also work:
format longg
syms p1 t1
nm = 3e15;
s01=1;
v1=4.82e20;
tao00=1e-13;
p00=3;
q00=21;
q11=10.6;
R=1.9872036e-3;
T=295;
V1=16.7;
K1=3.27e19;
T1=linspace(1,1e6,100);
taop1=3.6;
S1=4.7;
A1=4.74e3;
Nm = A1*nm;
W1=(Nm/(K1*V1));
y1=q00;
a1= ((s01*tao00*v1)/(nm))*exp((y1)/(R*T));
Z1=(1/(1+a1*p00));
Y1 = (p00/(1+(a1*p00)));
eq1 = ((t1./taop1)==(log(p00./p1))+((a1.*W1).*(Z1-(1./(1+a1.*p1))+log(((1+a1.*p1)./p1).*Y1))));
for i=1:numel(T1)
pp1_help = double(vpasolve(subs(eq1,t1,T1(i)),p1,11));
pp1_help = pp1_help(real(pp1_help)>0 & abs(imag(pp1_help))<1e-8);
if ~isempty(pp1_help)
pp1(i) = real(pp1_help(1));
else
pp1(i) = 0.0;
end
end
plot(T1,pp1)
N/A
N/A 2023년 1월 13일
Thank you very much! This is extremely helful. I appreicte your time and efforts.

댓글을 달려면 로그인하십시오.

카테고리

Help CenterFile Exchange에서 Programming에 대해 자세히 알아보기

제품


릴리스

R2021a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by