Precision of amean and aave

[brian vant-hull]

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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