i ask about if statement and loop

for H2=256.5:1:269.3
    for Hs=250:1:256
        for p3=675:1:900
            for pm=670:1:674
                for vm=0.0008299:0.0008581
                    ms=(As*(2*0.95*(H2-Hs))^0.5)/((2*0.95*(H2-Hs))^0.5);
                    Vs=(2*0.95*(H2-Hs))^0.5;
                    mm=mp+ms;
                    Vm=mp/Am;
                    muo=(ms/mp);
                    Vbs=((Vp+muo*Vs)/((muo+1)*(1+(0.03*80)/(2*20))))+(((p3+pm)*Am)/((1+(0.03*80))*mm));
                    Hbs=((Hp+((vp^2)/2)+muo*(Hs+((Vs^2)/2)))/(1+muo))-((Vm^2)/2);
                    mmm=(Vm*Am)/vm;
                    if mmm==mm
                        Vs=~0;
                        return
                    else
                        break
                    end
                end
            end
        end
    end
end

[the cyclist]

We can't run your code without defining the constants. If I just arbitrarily set all the undefined constants to 1, your code runs to completion.

So, what is your question?

[Walter Roberson]

                for vm=0.0008299:0.0008581

default increment is 1 so that line will only include the lower bound.

[dpb]

for H2=256.5:1:269.3
  for Hs=250:1:256
    ms=(As*(2*0.95*(H2-Hs))^0.5)/((2*0.95*(H2-Hs))^0.5);
    Vs=(2*0.95*(H2-Hs))^0.5;
    mm=mp+ms;
    Vm=mp/Am;
    muo=(ms/mp);
    for p3=675:1:900
      for pm=670:1:674
        for vm=0.0008299:0.0008581
          Vbs=((Vp+muo*Vs)/((muo+1)*(1+(0.03*80)/(2*20))))+(((p3+pm)*Am)/((1+(0.03*80))*mm));
          Hbs=((Hp+((vp^2)/2)+muo*(Hs+((Vs^2)/2)))/(1+muo))-((Vm^2)/2);
          mmm=(Vm*Am)/vm;
          if mmm==mm
            ...

moving what is invariant to the inner loop variables out for efficiency.

Also NB: that

ms=(As*(2*0.95*(H2-Hs))^0.5)/((2*0.95*(H2-Hs))^0.5);
Vs=(2*0.95*(H2-Hs))^0.5;

can be written more simply/efficiently as

Vs=sqrt(2*0.95*(H2-Hs));
ms=As*Vs/Vs;

which shows that ms is invariant and ==> As as the other factors cancel identically.

A sample case illustrates numerically...

>> As=pi; H2=256;Hs=250;
>> ms=(As*(2*0.95*(H2-Hs))^0.5)/((2*0.95*(H2-Hs))^0.5)
ms =
    3.1416
>> 

As originally coded, the answer for the example is identically pi, the value set for As and is constant.

One would presume this is probably not what was intended...

[Fangjun Jiang]

laziest question yet hard working answers. What is the question?