difference result with same equation

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vaya putra
vaya putra 2018-7-19
回答： vaya putra 2018-7-19
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 个评论
Aquatris
Aquatris 2018-7-19
编辑：Aquatris 2018-7-19
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.

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回答（2 个）

vaya putra
vaya putra 2018-7-19
thank you very much, i understand
vaya putra
vaya putra 2018-7-19
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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