thnx Tereza<br><br>but I don't understand it. did you mean that my answer [W=0.00912 (m/s)] is not (m/s) and it's in (cm/sec)!? supposing that isn't a 0.912 cm/sec a low speed vertical velocity airflow!? if there's any reference on this subject I'll be glad if you anyone notify me about it. <br>
<br><div class="gmail_quote">On Wed, Nov 16, 2011 at 4:18 AM, <span dir="ltr"><<a href="mailto:tcavazos@cicese.mx">tcavazos@cicese.mx</a>></span> wrote:<br><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex;">
Hi<br>
<br>
The usual unit (magnitude) of vertical velocity is in cm/seg.<br>
So, it is fine.<br>
<br>
Cheers,<br>
Tereza<br>
<div><div class="h5"><br>
> Dear GrADS users<br>
> Hi<br>
><br>
> I'm recently working on a project about the vertical speed of air in<br>
> different elevations. I'm using CFSR data in grb2 format. these data sets<br>
> have omega (Pa/s) instead . I tried to convert them to vertical velocity<br>
> (m/s) using this equation:<br>
> omega= - rho * g * w ( or omega=- (P/RT) *g * w )<br>
> but the results in (m/s ) are in the order of 10^-3. I know it's not a<br>
> Grads problem. but I'm an aerospace student and I need to find out if<br>
> anything is wrong with this equation or the results or not. ( I'm not<br>
> really familiar with meteorology!)<br>
><br>
> for example:<br>
> if omega= 0.10595<br>
> & g=9.807<br>
> & R=287<br>
> & P= 100000 Pa<br>
> & T=294.27<br>
> resulted value is W=0.00912 (m/s)<br>
><br>
><br>
> Thanks in advance,<br>
> Alireza<br>
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><br>
<span class="HOEnZb"><font color="#888888"><br>
<br>
--<br>
Tereza Cavazos<br>
Departamento de Oceanografia Fisica<br>
CICESE<br>
Ensenada, Baja California, MEXICO<br>
<a href="http://usuario.cicese.mx/%7Etcavazos/" target="_blank">http://usuario.cicese.mx/~tcavazos/</a><br>
<br>
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</font></span></blockquote></div><br><br clear="all"><br>-- <br>Alireza Azargoun<br>