Azzera filtri
Azzera filtri

difference result with same equation

1 visualizzazione (ultimi 30 giorni)
vaya putra
vaya putra il 19 Lug 2018
Risposto: vaya putra il 19 Lug 2018
my commands
bhp=100*ones(10,1);
bht=[500;500;500;500;500;300;300;300;300;300];
p= @(bhp,bhT) bhp + g*dz.*rho(bhp,bhT);
tried to validation p(bhp,bhT).*bhp + g*dz.*rho(bhp,bhT)
result ans =
1.0e+09 *
-2.2263
-2.2263
-2.2263
-2.2263
-2.2263
-4.7130
-4.7130
-4.7130
-4.7130
-4.7130
how can be i got difference result if i assume a=g*dz.*rho(bhp,bhT); a =
1.0e+07 *
-2.2042
-2.2042
-2.2042
-2.2042
-2.2042
-4.6664
-4.6664
-4.6664
-4.6664
-4.6664
b=bhp;
c=a+b
result
c =
1.0e+07 *
-2.2042
-2.2042
-2.2042
-2.2042
-2.2042
-4.6664
-4.6664
-4.6664
-4.6664
-4.6664
anyone can help me?
  1 Commento
Aquatris
Aquatris il 19 Lug 2018
Modificato: Aquatris il 19 Lug 2018
because at the top results you are evaluating
p(bhp,bhT).*bhp + g*dz.*rho(bhp,bhT)
and at the bottom
bhp + g*dz.*rho(bhp,bhT)
This can further be proved by looking at the results and seeing the relation
100*(bottom results)+a = (top results)
Also your code is missing g and rho, maybe other variables aswell for me to run it on my Matlab to further help.

Accedi per commentare.

Risposte (2)

vaya putra
vaya putra il 19 Lug 2018
thank you very much, i understand

vaya putra
vaya putra il 19 Lug 2018
actually i want calculate this equation,
that why i define by my self bhp and bhT, ho to prove aot calculate bhp,bhT if i get value (bhp and bhT)
p = @(bhp,bhT) bhp + g*dz.*rho(bhp,bhT);

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