Precision of amean and aave

[Heiner Körnich]

Hi,

thanks for reply. In fact, I had a mistake in my description, since I
was not comparing amean and aave. Here the difference is indeed the
latitudinal weighting which is only included in aave.

But I thought about the difference of aave, asum and asumg. Since asumg
is without weighting, I assumed mistakenly asum to include latitudinal
weighting. BUT asum has NO latitudinal weighting, thus asum is
equivalent to amean except for the factor of the whole area. I think the
GrADS online documentation could be clearer on asum and asumg.

Cheers,
Heiner



Lemmy wrote:
> Brian,
>
> I would not agree with your explantion simply because you assumed
> implicitly that U and T (in your agrument) are 3D vectors inside the
> operators aave and amean. For model data, this is not applicable
> because model data is stored on grids, and the area-average operator
> is applied only on 2D data. Therefore, U and T lie on the tangent
> surface at each grid point and is thus orthogonal with normal vector A
> and cos(u) and cos(t) should be zero always (surface A as I
> understand in the grads user guide is defined to be the lat-lon
> surface). The differences between amean and aave are explained on
> grads website. It is due solely to the fact that the latter uses
> lattitudes to weight the average (a square area defined by lat-lon
> will shrink to zero at pole) whereas the former does not. So they can
> not be the same, obviously.
>
> Cheers..
>
> */brian vant-hull <brianvanthull at YAHOO.COM>/* wrote:
>
>     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
>
>     A*U/A*T = |U||T|
>
>     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
>
>     |A||U|cos(u)/|A||T|cos(t) = |U|/|T|
>
>     This is only true if the angle between the weighting vector and
>     both data vectors are the same.  Rarely happens.
>
>     */Heiner Körnich <heiner at MISU.SU.SE>/* wrote:
>
>         Hi,
>
>         I am wondering about the precision of the functions amean and
>         aave. If I
>         take the data from the GrADS example (model.ctl and
>         model.dat), I would
>         assume that the following expressions should be equal.
>
>         ga-> d amean(u,g)/amean(T,g)
>         Result value = 0.0550876
>         ga-> d aave(u,g)/aave(T,g)
>         Result value = 0.0609393
>
>         I thought, the difference between aave and amean is only the
>         final weighting
>         of the sum with the area.
>
>         Do I overlook something or what is wrong here?
>
>         Cheers,
>         Heiner
>
>
>
>
>     Brian Vant-Hull
>     301-646-1149
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--
Heiner Körnich
Dept. of Meteorology            Tel:  +46 8 164333
Stockholms University, SE-106 91 Stockholm, Sweden
Email: heiner at misu.su.se   www.misu.su.se/~heiner/