error: x(2): out of bound 1
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Lakshmikruthiga Ponnusamy
il 13 Mar 2019
Modificato: KALYAN ACHARJYA
il 13 Mar 2019
I have code of the following. I did everything I could do, but couldnt overcome this error.
When I execute the below function I get the error
function [disp_dot]=stage2(x,t)%Stage_2 calculation for displacement and pressure
m=23.5; %Stage_2:From bottom choked till bottom is exhausted; top is exhausted
ODB=0.122;
IDB=0.04625;
g=9.8;
ODT=0.104;
IDT=0.042;
Patm=100000;
dt=0.0001; %time increment value
L=0.11; %This is the length where adiabatic condition starts in top chamber; the length of the leg of the air distributor
AT = pi*(ODT^2-IDT^2)/4;
AB = pi*(ODB^2-IDB^2)/4;
PT=Patm;
ODSBD=0.1285;
IDSBD=0.122;
ASBD=pi*(ODSBD^2-IDSBD^2)/4;
ASB=0.00451183;
AST=0.001267892;
CFM=840; %air inflow
CFMm=CFM*(1/2118.88);
v=CFMm/AB;
v=5;
PST=Patm;
PSB=Patm;
PT=Patm;
LB=0.0259; %This is the length where bottom chamber becomes adiabatic during upward displacement of the piston
PBinitial=((PT*AT)+(m*g)+(PST*AST)-(PSB*ASB))/AB;
KB=PBinitial*(AB*LB)^1.4;
disp_dot(1) = x(2);
disp_dot(2) = (1/m)*((KB*AB/(0.000058402+((ASBD+AB)*x(1))^1.4)-(PT*AT)-(m*g)+(PSB*ASB)-(PST*AST)); %Constant value is from bottom channel 4 in excel
end
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KALYAN ACHARJYA
il 13 Mar 2019
Modificato: KALYAN ACHARJYA
il 13 Mar 2019
Issue 1:
When you defined the function as
[disp_dot]=stage2(x,t)
There is not role of t function input data t in the function.
Issue 2:
In the following one parenthesis bracket required at last.
disp_dot(2)=(1/m)*((KB*AB/(0.000058402+((ASBD+AB)*x(1))^1.4)-(PT*AT)-(m*g)+(PSB*ASB)-(PST*AST)));
Now I have defined the function as
function [disp_dot]=stage2(x)%Stage_2 calculation for displacement and pressure
m=23.5; %Stage_2:From bottom choked till bottom is exhausted; top is exhausted
ODB=0.122;
IDB=0.04625;
g=9.8;
ODT=0.104;
IDT=0.042;
Patm=100000;
dt=0.0001; %time increment value
L=0.11; %This is the length where adiabatic condition starts in top chamber; the length of the leg of the air distributor
AT=pi*(ODT^2-IDT^2)/4;
AB=pi*(ODB^2-IDB^2)/4;
PT=Patm;
ODSBD=0.1285;
IDSBD=0.122;
ASBD=pi*(ODSBD^2-IDSBD^2)/4;
ASB=0.00451183;
AST=0.001267892;
CFM=840; %air inflow
CFMm=CFM*(1/2118.88);
v=CFMm/AB;
v=5;
PST=Patm;
PSB=Patm;
PT=Patm;
LB=0.0259; %This is the length where bottom chamber becomes adiabatic during upward displacement of the piston
PBinitial=((PT*AT)+(m*g)+(PST*AST)-(PSB*ASB))/AB;
KB=PBinitial*(AB*LB)^1.4;
disp_dot(1)=x(2);
disp_dot(2)=(1/m)*((KB*AB/(0.000058402+((ASBD+AB)*x(1))^1.4)-(PT*AT)-(m*g)+(PSB*ASB)-(PST*AST))); %Constant value is from bottom channel 4 in excel
end
When I ran the code here is the output
>> y1=[1,2]
y1 =
1 2
>> y=stage2(y1)
y =
2.0000 -26.1197
Please note in the fuction code x define as 2 length vector, which require x(1) and x(2), that why I have passed the y=[valuue1 value 2] to function
Hope it helps!
2 Commenti
KALYAN ACHARJYA
il 13 Mar 2019
Modificato: KALYAN ACHARJYA
il 13 Mar 2019
For the 2nd question, please explain the issue in more generalise and simpler way? Or you can ask the question as new question (new thread), so that others can easily see the question and get the fast answer.
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