How to get an y-value (y1(x)) for the sub condition of 0.8y1=0.25x

Hi there,

i need some help. I have two collum vectors y=force and x=settlement. Now i had to write a scribt which gives me the first y-value (y1(x)) where the subcondition: 0.8*y1=0.25x is true. I am not so familar whith loops and so I would need some help.

Thank you!

[madhan ravi]

What did you try for your assignment?

[Alexander Busch]

The actual try is this one. Dont know if it makes sense. Even "zero" is not working at the moment (Undefined function 'zero' for input arguments of type 'double'.)

x=zero(lenght(settlement),1);
for 
    i=1+1:force(end);
    x(i)=settlement(find(0.8*force(i)))/settlement(find(force(i)));
end
    
I=min(abs(x-0.25));    
end

[madhan ravi]

zeros() not zero()

[Alexander Busch]

sry...

now it is this problem:

    
>> x=zeros(length(weg_fit_mm),1);
for 
    i=1+1:kraft_fit_kN(end);
    x(i)=weg_fit_mm(find(0.8*kraft_fit_kN(i)))/weg_fit_mm(find(kraft_fit_kN(i)));
end
    
I=min(abs(x-0.25));    
end

for 
     ↑
Error: Invalid expression. Check for missing or extra characters.
 

Answer 1

[Vimal Rathod]

In the code from the above comments, there is a problem with defining the for loop condition as the error was shown.

The 'i' variable should be defined in the same line as the for. And the value should not have a semicolon(;) at the end.

Here is a sample in the code which you could try.

for i=1:kraft_fit_kN(end)

You could refer to the following documentation for further details;

https://www.mathworks.com/help/matlab/ref/for.html

[Alexander Busch]

Thank You!

It is working now