You have a mathematical misperception. I can't think of any better way to describe this then by vector mathematics, so go back to first year physics. Think of weighting as a vector dot product. Let A be the area vector, U and T be the other data vectors. Then the weighted averages are A*U/n and A*T/n while the unweighted averages are |U|/n and |T|/n. Canceling out n, you are claiming that<br><br>A*U/A*T = |U||T|<br><br>but A*U = |A||U|cos(u) and A*T = |A||T|cos(t) where u and t are the angles between the vectors A and U or T. So you are claiming that<br><br>|A||U|cos(u)/|A||T|cos(t) = |U|/|T|<br><br>This is only true if the angle between the weighting vector and both data vectors are the same. Rarely happens.<br><br><b><i>Heiner Körnich <heiner@MISU.SU.SE></i></b> wrote:<blockquote class="replbq" style="border-left: 2px solid rgb(16, 16, 255); margin-left: 5px; padding-left: 5px;"> Hi,<br><br>I am wondering about
the precision of the functions amean and aave. If I<br>take the data from the GrADS example (model.ctl and model.dat), I would<br>assume that the following expressions should be equal.<br><br>ga-> d amean(u,g)/amean(T,g)<br>Result value = 0.0550876<br>ga-> d aave(u,g)/aave(T,g)<br>Result value = 0.0609393<br><br>I thought, the difference between aave and amean is only the final weighting<br>of the sum with the area.<br><br>Do I overlook something or what is wrong here?<br><br>Cheers,<br>Heiner<br></blockquote><br><BR><BR>Brian Vant-Hull<br>301-646-1149<p> 
<hr size=1>Want to start your own business? Learn how on <a href="http://us.rd.yahoo.com/evt=41244/*http://smallbusiness.yahoo.com/r-index">Yahoo! Small Business.</a>