displaying indexed values correctly

hi, trying to display some values but am not sure how to go about it. I want to do is; find Vb at half of Vb_ev and then find the corresponding pH value at that point. see last 2 lines. any help will be appreciated.

                  format short
        Kw = 1e-14;
        Ka = 1.755388e-5;
        Ca = 0.5;
        Cb = 0.1;
        Va = 100;
        Vb = 0.05:0.05:2000;
        Ma = (Ca * Va) / 1000;
        Mb = (Cb .* Vb) ./ 1000;
        for i = 1:length(Mb)
          M_excess = Ma - Mb(i);
          if abs(M_excess)<eps
              Hplus = Ka * ((Ma_final * 0.999999) ./ Mb_final);
              Vb_ev = Vb(i); 
              pH_ev = pH(end); 
          elseif M_excess > 0
              Ma_final = (M_excess * 1000) ./ (Va + Vb(i));
              Mb_final = (Mb(i) * 1000) ./ (Va + Vb(i));
              Hplus = Ka * (Ma_final ./ Mb_final);
          elseif M_excess < 0
              OH = (M_excess * 1000 * (-1)) ./ (Va + Vb(i));
              Hplus = Kw ./ OH;
          end
          pH(i) = -log10(Hplus);
        end
  %
  %
         Vb_Mid_Pt1 = uint16(Vb_ev)*0.5;
         pH_Mid_Pt1 = pH(Vb_Mid_Pt1)

i tried the above but the pH_Mid_Pt1 gives me the value in the pH array that is equal to the actual numerical value of that from (Vb_ev)*0.5.

[Azzi Abdelmalek]

I think, you can make your question easier, by giving a short example that you can adapt to your problem

[harley]

apologies for the long questions, as a complete novice i am finding it hard to explain even the smallest issue i face, hopefully i can improve with a little practice. Thanks for your the time again.

[Walter Roberson]

http://www.mathworks.com/matlabcentral/answers/6200-tutorial-how-to-ask-a-question-on-answers-and-get-a-fast-answer

[Walter Roberson]

When you say "half" do you mean the point at which it is 1/2 of the maximum value, or do you mean the point that is half way through the vector ?

[harley]

literally at Vb_ev / 2. Thanks for looking Walter.

[Walter Roberson]

Suppose Vb_ev is 0.9 . Then Vb_ev/2 is 0.45. But there is no Vb that is 0.45 . What do you want to do in such a case? And what if Vb_en is 0.1 the very first Vb value ?

[harley]

thanks for showing some interest Walter. All i want it to do at this point is find the value of half Vb_ev, which is 250 for these parameters and then return the corresponding pH value.

[Walter Roberson]

If half Vb_ev is 250 then is Ph(250) what you want returned? If so then why do you consider your current code to be incorrect?

Are you looking for the point in Ph that is closest to Ph(Vb_ev)/2 ?

[harley]

its gives me the pH value in column 250 of the pH array, (using above code), i want it to give me the value of pH when Vb = 250 (half of Vb_ev). thanks again.

Answer 1

[Richard Brown]

Similar to what you asked before, I think what you want is to find the Vb element that is closest to half of Vb_ev, and then pluck out the corresponding pH element. The answer is very similar to what I gave you last time:

[~, idx] = min(abs(Vb - Vb_ev/2));
Vb_Mid_Pt1 = Vb(idx);
pH_Mid_Pt1 = pH(idx);

I think what's throwing you off is the distinction between values and indices. You are wanting to find certain indices in your array that have values near other specific ones. Hence the use of min with abs, which provides the index of the closest point.

[harley]

thanks Richard, yes i am struggling with all distinctions.

[Walter Roberson]

Note: different versions of MATLAB may return different results for the above. If there are multiple elements with the same difference (as would be the case if Vb_ev/2 falls between two Vb values), then some versions will return the index of the first of them, and some will return the index of the last of them.

[Richard Brown]

Indeed. And for this problem it is likely that you'd end up halfway between two sample values. Using interpolants would be the next step in sophistication