I didn't see any response to your mail so here is one :<br><br>2.5° is not 250km !!!!<br>Along an earth circumference (a meridian for example) 1 minute of angle is 1 nautic mile (nm) which is 1852m. you may apply this rule to latitudes (1°=60min=60nm=~111km). For longitudes unless you live on equator you have to multiply by cos(phi) where phi is the longitude.<br>
<br>Cordially<br>Thierry HEDDE<br><br><div class="gmail_quote">2010/12/7 Asnor Muizan Ishak <span dir="ltr"><<a href="mailto:asnorjps@yahoo.com.my">asnorjps@yahoo.com.my</a>></span><br><blockquote class="gmail_quote" style="margin: 0pt 0pt 0pt 0.8ex; border-left: 1px solid rgb(204, 204, 204); padding-left: 1ex;">
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<div>Dear Everyone,</div>
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<div>May I ask you all a question about the degree of resolution for global data for intance ECMWF. This global data is consider as a input for WRF or MM5 as well. According to ECMWF, the data could be downloaded from 2.5 degree up to 0.0 degree. The 2.5 degree is equal to 250km, am I right? But in my case I have used 1.0 degree data and therefore should I set my courses domain with 100km resolution as well due to the 1.0 degree resolution? Is this will make huge error after post processing? At the moment my course domain is 81km and 27km and finest domain are 3km and 1km (I have 2 cases here). Therefore, does anyone can help me about this conflict of information. I tried to find this information via the guide line or procedures but until now I cant find it. Please everyone to help me about this information.</div>
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<div>Many thanks in advance.<br></div></div>
<div style="font-family: arial,helvetica,sans-serif; font-size: 12pt;">Asnor<br></div></div>
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<br></blockquote></div><br><br clear="all"><br>-- <br>Thierry HEDDE<br>Laboratoire de Modélisation des Transferts dans l'Environnement<br>CEA/CADARACHE<br>DEN/DTN/SMTM/LMTE<br>Bât. 307 Pièce 9<br>13108 ST PAUL LEZ DURANCE CEDEX<br>
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