loops

ok i made my self a simpler code. what i am trying to accomplish is for MatLab to give me the M at which A and z are equal. Here is my code.

A=1.687;
gamma=1.4;
tol=1e4;
for M=1:1:5
    a=(1/M^2);
    b=(2/(gamma+1));
    c=(1+(((gamma-1)/2)*M^2));
    d=(gamma+1)/(gamma-1);
    z(M)=sqrt(a*((b*c)^d));
end
z'
if abs(A-z)<tol
    M=z(M)
end

i cannot seem to get it to work. i got z' and i know M=2 is pretty close to my A value. FOr M=2 my code returns 1.6875. the if statement prints M=25.00, but i want it to print 2 because thats the answer. Thank you

Answer 1

[Moes]

Try this code instead (no loops):

A=1.687;
gamma=1.4;
tol=1e4;
M = 1:5;
a=(1./M.^2);
b=(2/(gamma+1));
c=(1+(((gamma-1)/2).*M.^2));
d=(gamma+1)/(gamma-1);
z=sqrt(a.*((b.*c).^d));
diffrnc = abs(A*ones(size(z)) - z)
diffrnc(find(diffrnc < tol))

[Moes]

If you just want the index and not the actua;l number replace "diffrnc(find(diffrnc < tol))" with "find(diffrnc < tol)".

Answer 2

[Walter Roberson]

z is an array, so A-z is an array and so abs(A-z) is an array. That array is compared to tol, which gives a logical array in return. When you use a logical array in an "if" statement, the "if" is only true if all of the logical values are true.

Replace the "if" with something like,

Mpos = find(abs(A-z)<tol);
M = z(Mpos);

That is a pretty large tol you have. Do you perhaps mean 1e-4 ?

Answer 3

[marvin corado]

well i initially put 1e-4 but it did not give me any answer. so i decreased it and then increased it until i got something. I was just trying to get a result, but i will try the 1e-4 because that made sense to me. thankyou

Answer 4

[marvin corado]

i tried and here is the result. Really what i am after is the M for which z is equal to A. That is really what i am after. i modified it as you suggested.

A=1.687;
gamma=1.4;
tol=1e-4;
for M=1:1:5
    a=(1/M^2);
    b=(2/(gamma+1));
    c=(1+(((gamma-1)/2)*M^2));
    d=(gamma+1)/(gamma-1);
    z(M)=sqrt(a*((b*c)^d));
end
Mpos=find(abs(A-z)<tol);
M=z(Mpos)

here is the result M =

   Empty matrix: 1-by-0

thank you again for helping

Answer 5

[Matt Fig]

Is this what you are looking for?

A=1.687;
gamma=1.4;
tol=1e-4;
IDX = 0:.0001:2;
z = zeros(1,length(IDX));  % Pre-allocate!
cnt = 0;
for M=IDX
    a = (1/M^2);
    b = (2/(gamma+1));
    c = (1+(((gamma-1)/2)*M^2));
    d = (gamma+1)/(gamma-1);
    cnt = cnt + 1;
    z(cnt) = sqrt(a*((b*c)^d));
    if abs(A-z(cnt))<tol
        M % Turns out there is more than one!
        z(cnt)
%         break
    end     
end

.

.

.

Here is another method of finding these two locations:

g = gamma;  % For convenience...
f = @(M) sqrt((1./M.^2).*(((2./(g+1)).*(1+(((g-1)./2).*M.^2))).^((g+1)/(g-1)))) - A;
rt1 = fzero(f,2)
rt2 = fzero(f,.5)