(meteorobs) Serious problem about analysis!

[Pavol Habuda]

Hi

>   As you know, I did an analysis of Chinese  Perseids
> observations, but Jin Zhu against me, about the
> analyse method. I've wrote to R.Arlt but he hasn't
> reply until now.
>   As we know, there is a calculator:
> ZHR = (1 + sum n) / sum (Teff/C)
> and my method is, sum all the observers' n, and sum
> all the observers' Teff/C, then recive a ZHR at last.
> But Zhu against it, he thought the correct method
> shold be, calculate every observers' ZHR, and then
> avenge these ZHR. So we have some aboid debates these
> days.



Correct answer is: you are right and you are right too.
But last methot is better for two and more observers,
because it account some more information. If you observe 
during night and see no meteor of shower, what is ZHR?
Zero? Definitely not!


> e.g. Observer A, lm=5, n=2; observer B, lm=6, n=4.
> F=1, Teff=1, h=90, r=2.
> 
> My method: ZHR=6.6+-2.5.
> Zhu's method: [ZHR(A)+ZHR(B)]/2, ZHRave=7.8.
> Because ZHR(A)=8.5+-4.9 ZHR(B)=7.1+-3.2, Zhu thoughts
> the final ZHR should not be smaller than the smaller
> one.


Of course could! I know paradox form statistics, where
there is hospital A and B curing some diseases alpha, beta. 
A cures alpha better than B and A cures beta better than B too.
But for some conditions, if patient does`n know whether it
is alpha or beta, goes to B, where is higher probability
to cure. Of course, you must accept that alpha and beta doesn`t 
have the same probabilities to occur.



> And Zhu thought the calculator should be:
> ZHR = sum n / sum (Teff/C). He said: if an observer
> observe 1 meteors in 1 hours, and h=90 F=1 r=2 lm=6.5,
> the ZHR should be 1. But I thought that's wrong
> because the ZHR should be => 1, not =1.


Zhu is right! But... he must accept, that he count ZHR for
THE ONLY observer! If two observers observe, you know that
they observe together ... 
  If you count ZHR for a few number of observations, it is 
better use average value, or weighted average value (weight 
is statistical error... the higher error the less contribute to
result). If you count ZHR for 100 periods, you have at least 
10 meteors and there 1 meteor doesn`t play a role.



Arlt used (1 + sum N) because of statistics. If you have 
Poissonian distribution (and meteors are randomly distributed 
like this distribution), there is the same probability you
observe 1 or 0 meteors if you don`t know real distribution.
And there is 1 in this case... more in this article:

Title:	
	The Analysis of a Weak Meteor Shower: The June Bootids in 2000
Authors:	
	Arlt, R.
Journal:	
	WGN, Journal of the International Meteor Organization, vol. 28, no. 4, p. 98-108
Publication Date:	
	08/2000
Origin:	
	JIMO
Bibliographic Code:	
	2000JIMO...28...98A



> So we have debate every day but neither of us give in.
> So could the experts give us a correct answer to
> these? Thanks!
> 


Try to read the article.

have a nice day
Pavol Habuda


		
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