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

4 个评论

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?
for vm=0.0008299:0.0008581
default increment is 1 so that line will only include the lower bound.
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...
laziest question yet hard working answers. What is the question?

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