Precision of amean and aave

[brian vant-hull]

Not talking about actual vectors, but n dimensional data vectors...the math will come out correctly if you can imagine abstract angles between n vectors and leave 3D space (and orthogonal area vectors) behind.  But the problem seemed to be elsewhere anyway.

Lemmy <kieucq at YAHOO.CO.UK> 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
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