Iteratively solving using a for loop

Hi there,

I am trying to find the wind shear using the equation (V/Vo) = (H/Ho)^(W) Where W is wind shear.

As of now, I can do it but the for loop using the solve function runs for about 15 minutes to yield 4462 values. WSA490m and WSA3860m are data values enclosed in a csv file, with wind speeds at 49.0m and 38.60m.

if true
Vratio1 = ones(4462,1);
Vratio2 = ones(4462,1);
Hratio1 = (49.0/38.60);
Hratio2 = (59.40/38.60);
LHR1 = log(Hratio1);
LHR2 = log(Hratio2);

 WS1 = ones(4462,1); %Wind Shear exponent with 38.60m as reference and 49.0m as the second point
 WS2 = ones(4462,1); %Wind Shear exponent with 38.60m as reference and 59.40m as the second point
if true
for r = (1:4462)
    Vratio1(r) = (WSA490m(r)/WSA3860m(r));
end
end

if true
for r = (1:4462)
      WS1(r) = solve(x == log(Vratio1(r))/LHR1,x);
end
end

Answer 1

[David Goodmanson]

Hi Rnle,

In Matlab you can divide vectors term-by-term with the ./ command to obtain Vratio1. Then, since

W = solve(x = something,x)    is the same as    W = something

you can determing WS1 in the following way.

WSA490m = rand(4462,1);
WSA3860m = rand(4462,1);
Hratio1 = (49.0/38.60);
Hratio2 = (59.40/38.60);
LHR1 = log(Hratio1);
LHR2 = log(Hratio2);

Vratio1 = WSA490m./WSA3860m;
WS1 = (log(Vratio1)/LHR1);

The two data vectors should be brought into the workspace as complete 4462x1 vectors. After that is done, the rest of it all happens in a couple of milliseconds.

[Rnle]

Thank you! Works like a dream

[Star Strider]

@Rnle —

Since David Goodmanson’s Answer solved your problem, it is appropriate to Accept it.