error: x(2): out of bound 1

Question

Lakshmikruthiga Ponnusamy 2019-3-13

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/449831-error-x-2-out-of-bound-1

编辑： KALYAN ACHARJYA 2019-3-13

采纳的回答： KALYAN ACHARJYA

在 MATLAB Online 中打开

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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Answer 1

KALYAN ACHARJYA 2019-3-13

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/449831-error-x-2-out-of-bound-1#answer_365167

编辑：KALYAN ACHARJYA 2019-3-13

在 MATLAB Online 中打开

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!

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Lakshmikruthiga Ponnusamy 2019-3-13

在 MATLAB Online 中打开

example.png

Thank you so much sir. It worked. I also have other question. Can you help on this?

I want the graph to be colored in such a way that stage one is in blue and stage 2 is in red in displ vs time graph, similarly for velocity vs time and pressure vs time graph. I have posted image as example. I want the graph to be like that in image. How do i write the code?

displ=zeros(100,2);
tim=zeros(100,1);
pressure=ones(100,1);
velocity=ones(100,1);
for i=2:100
v=5;
dt=1*10^-4;
function [displ_f,pressure]=stage1(displ_i)
    dt=1*10^-4;
    v=5;
    area=5;
   displ_f=displ_i+v*dt; 
   pressure=area*displ_f;
   velocity=10*area*displ_f;
end
function [displ_f]=stage2(displ_i)
    dt=1*10^-4;
    v=5;
    area=5;
    displ_f=displ_i+v*dt;
    pressure=area*displ_f;
    velocity=10*area*displ_f;
end
if (displ(i-1,1)<0.020 && displ(i-1,2)>=0)
        disp("IN stage 1");
        [displ(i,1)]=stage1(displ(i-1,1));
        displ(i,2)=v;
        tim(i,1)=tim(i-1)+dt;
    elseif (displ(i-1,1)>=0.020 && displ(i-1,1)<=0.0256 && displ(i-1,2)>=0)
        disp("IN stage 2");
        [displ(i,1)]=stage2(displ(i-1,1));
        displ(i,2)=displ(i-1,2);
        tim(i,1)=tim(i-1)+dt;
end
end
plot(tim(:,1),displ(:,1));
plot(tim(:,1),pressure);
plot(tim(:,1),velocity);

KALYAN ACHARJYA 2019-3-13

编辑：KALYAN ACHARJYA 2019-3-13

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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