(meteorobs) Serious problem about analysis!

[tom6740]

First, deeply sorry to Zhu, I give the wrong result from his method. 
The correct should be: ZHR(A)=5.656854+-4 ZHR(B)=5.656854+-2.828427, 
I use +1 in his method by mistake to get the result.

--- In meteorobs at yahoogroups.com, RainerArlt  <rarlt at a...> wrote:
>
> Let me also add that this way of averaging ensures
> that the observations are automatically weighed by
> their total correction. Largest corrections get the
> least influence.

Is the "total correction" means C? And I wonder if C has "weighted 
avenge value" function, in that formula?
 

> You can either do
> 
> (1+2+4) / (0.354 + 0.707) = 6.6 as you were doing it, or
> 
> ZHR(A) = 2 * 2^(6.5-1.5) = 5.6
> ZHR(B) = 4 * 2^(6.5-0.5) = 5.6
> avg(ZHR) = 5.6
> This value will come close to the first version, if a
> lot of data is involved. Nevertheless, note that it does
> not weigh the observers by corrections (here accidentally
> irrelevant).

Many thanks for the explain, but I'm afraid I don't catch its mean 
very well, I found IMC has never use Zhu's method to do analysis, so 
which method is more credible? Or the two are different understand 
of the ZHR formula? In a word, what will the result to be, if it is 
a data of IMC? 

> The individual ZHR=6.6 as taken from (1+n) has very little
> meaning and is only valid of no other data are avaliable.
> It is then the best you can say about the ZHR. But it is
> not the basis of averaging!