EVAL and INVERTFIT

Robert Janusz wrote on Dec 07, 2009

I have a catalog (vzcat):

1	1

and observation data (vzdat):

1	1.1

and both the config files (vzcfge, vzcfgi):

# for EVAL
catalog
V	2
observation
v	2
transformation
fit	c=0.
FIT:	V = v + c

# for INVERT
catalog
V	2
observation
v	2
transformation
fit	c=0.
FIT:	v = V + c

.
The results of the FITPAR

del vzpare,vzoute
fitpar vzdat vzcat vzcfge vzpare
eval vzdat vzcfge vzpare vzoute catal=vzcat
del vzpari,vzouti
fitpar vzdat vzcat vzcfgi vzpari
invert vzdat vzcfgi vzpari vzouti catal=vzcat

in two cases gives the correct results, for EVAL as well:

1          1.000   INDEF   0.000

but NOT for INVERT:

1          INDEF   INDEF   INDEF

.

When I change the vzcat:

1	1
2	2

the result for EVAL is

1          1.000   INDEF   0.000

and for INVERT is

1          1.000   INDEF   0.000  

, as it should be but the catalog data "2" does not play any role here.

When I cange the catalog and data for:

1	1
2	2

and

1	1.1
2	2

the results for EVALFIT and INVERT are

1          1.050   INDEF   -0.050 
2          1.950   INDEF   0.050  

and

1          1.050   INDEF   -0.050 
2          1.950   INDEF   0.050  

.But for the catalog and data

1	1
2	2
3	3

1	1
2	2
3	3

the results are: for EVAL:

1          1.000   INDEF   0.000  
2          2.000   INDEF   0.000  
3          3.000   INDEF   0.000  

for INVERT

1          1.000   INDEF   0.000  
2          INDEF   INDEF   INDEF  
3          3.000   INDEF   0.000  

.
Why are there these "INDEF" in the INVERT outputs?
rj
PS
When I change the catalog and run again,

1	1
2	2
3	3
4	4

the result for INVERT is correct:

1          1.000   INDEF   0.000  
2          2.000   INDEF   0.000  
3          3.000   INDEF   0.000  

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Last post on Dec 07, 2009