EAON circular # 10

 

 

 

 

How  good  is  a  prediction  ?

================================

 

 

 

 

Remark. Nature does not play dice : when there is an occultation, the path of the shadow is always perfectly defined, there is nothing random in it.

 

         HOWEVER, the calculator of an occultation does not know EXACTLY neither the position of the star nor the orbit of the occulting body. Uncertainties can reach a tenth of an arcsec, or worse when the star is an unknown double (for instance).

         So, the calculator has to estimate the uncertainty or probable error of his prediction : it is the "1 sigma (1s) uncertainty interval", usually given in fraction of a path width. For instance : * 1 sigma uncertainty interval [path widths]: +/- 0.63

         This 1s is sometimes erroneous, but there is no way to know it before any new observation. So it must be our best guess, although it is nothing but an estimate of errors KNOWN by the calculator to exist. Unknown errors are not taken into account.

 

         How good is a prediction, or rather how bad can it be ? If someone attempt to observe from the predicted central line, what are the odds that he will see nothing (sorry, it was bad) ? If another observer is at a place somewhat far from the predicted line, what are the chances for him to see a chord (luckily, the pred was bad enough) ?

 

         The following table tries to give some answers, for various s values (given in asteroid diameters) ; and for stations on the predicted central line, or halfway to the edges of the shadow, on these edges, and outside the predicted path at distances 0.5s, 1s, 2s, and 3s .

         The probabilities to observe some chord is given in percents (1% = 1 chance out of 100). The probability seems to decrease as the uncertainty and/or the distance increase.

 

            NOTE : of course there is nothing random ; the real path is somewhere, and if you are there you'll see, if you are elsewhere you won't see. The probability gives an estimate of your mean ratio of success when you make many attempts in these conditions. [Taking for granted that there are NO hidden errors in the prediction ; the assumed occulting diameter may be very false - e.g. Diotima, smaller, and Rhodope, 2 times larger than believed ; large shifts (in the range of 3s) have also been noticed, e.g. Dido / HIP59732 on 2005.03.10 and Bronislawa / TYC799-1487 on 2005.03.11, both for unknown causes].

 

 


 

 

 

The diameter (= shadow path width) of the minor planet is always taken here as the unit for lengths. In the example here under, s is about 1.4 pathwidth (1.4 ). The 1s line is at a distance = 1s from the edge of the shadow, etc...

 

 

 

 

 

 

              Probability (in %) to see a chord from the ...

s in diam.

 

central line

    %

halfway to

%

edge

%

0.5 s  line

%

1 s  line

%

2 s  line

%

3 s  line

%

0.1

 

100

99

50

31

16

2.3

0.1

0.2

 

 93

89

50

31

16

2.3

0.1

0.4

 

 79

70

49

31

16

2.3

0.1

0.7

 

 52

50

42

28

15

2.2

0.1

1

 

 38

37

34

24

14

2.1

0.1

2

 

 20

20

19

15

 9

1.7

0.1

4

 

 10

10

 10

  8

  5

1.1

0.1

7

 

   6

  6

   6

  5

  3

0.7

0.1

10

 

   4

  4

   4

  3

  2

0.5

0.0

 

 

 

How to use the above table ?

 

   As an example, let us look at the map for Fides / UCAC2 40827535 on 2006 August 24 (an excerpt from a map by Oliver Kls).

 

 

The shadow path is within the two bold continuous lines : we found there Midstkogen (in Norway).

The 1 s lines are dashed dark blue : we found Stapleton (University of St.Andrews obs.) on the north line, Harper and Jeffreys on the south line.

The 2 s lines are dashed red. Stevens is just inside the south line.

Other stations : Armagh observatory is halfway between the north edge of predicted shadow, and the 1 sigma line : it is around  the 0.5 s line. Mc Cracken is halfway between the south 2 s line and the 3 s line (which is not drawn on the map).

 

                Directly on a screen, with an ordinary graduated scale, one can measure roughly ~37 mm for the path width. From the edge of this shadow path to the 1 s line, one finds ~14mm : this is the width of the 1 s uncertainty. Dividing 15mm / 35mm, one finds that 1 s is thus = 0.4 pathwidth, 0.4  or diameter of Fides. Looking at the table on page 2 of EAON Circular # 10, the first column from the left is headed : "s in diam." [that is to say "sigma in diameters"]. Well, if s = 0.4 , we are on the 3rd line " 0.4 " of the table.

 

                So, an observer like Oernulf Midstkogen on the central line (second column of the table), has 79% of chances of success, or 4 chances out of 5.

 

                Jon Harper is around the 1 s line (3rd column from the right in the table). So, he has a probability of success of 16% : almost 1 chance out of 6 .

 

                Armagh is on the 0.5 s line : the probability of success is 31%, almost 1 chance out of 3.

                [Remark : here 31% is halfway between 49% (prob. on the edge) and 16% (prob. on the 1 s line)].

 

                Stevens is on the 2 s line (2nd column from the right in the table). There the probability of success is no more than 2%, and 2% is 1 chance out of 50.

 

                Now Mc Cracken is halfway between the south 2 s line and 3 s line. From the 2 s line, the probability is 2%, and only 0.1% from the 3 s line (last column). Let us say that the probability of success for Mc Cracken is in between, around 1 chance out of 100.

 

                At last, let us assume that Andrew Elliott is halfway between the south 0.5 s line and the (predicted) south edge of the shadow. His probability of success is thus between 31% and 49%, somewhere around 40%. 2 chances out of five.

 


 

 

                Now, here is another map (somewhat imaginary) for Sicilia 2006.08.12  :

 

 

                Here, one measures roughly ~18 mm for the path width, and ~24mm for the width of the 1 s uncertainty. Dividing 18mm / 27mm, one finds that 1 s is thus = 1.5 pathwidth, 1.5  or diameter of Fides. Looking at the table we fall between the 5th and 6th lines, s = 1 and s = 2.

                Dr. Bredner travelled to the central line. If we had s = 1 , his probability to see a chord would have been 38% . If we had s = 2 , his probability to see a chord would have been 20% only. As we have s = 1.5 , his probability to see a chord is somewhere between 38% and 20%, let us say 29% ?

                Rui Gonalves is on the 1 s line (3rd column from the right). So his probability to see a chord is somewhere between 14% and 9% : some 12%, one chance out of 8 .

                Urania observatory is on the 2 s line (2nd column from the right). So his probability to see a chord is no more than 2% (1 chance out of 50).

                Ewen-Smith is halfway between the 1 s line and the 2 s line. So his probability to see a chord is somewhere between 12% (prob estimated for Rui Gonalves) and 2% (prob estimated for Urania) : about 1chance out of 20 .

                At last, Ricard Casas stopped his travel (north of clouds ? This circular does not deal with probability of clouds !) after crossing the 3 s line, at a third before the 2 s line. A probability between 2% and 0.1%, nearer to 2% ; a little more than 1 chance out of 100 ...

 

 

                However, please remember that SOMETIMES the prediction is wrong of 3 s or even more. Sometimes... The star or the "planet" (or both) can be unknown double. An attempt to observe is always useful.

 

For EAON, Raymond DUSSER

2006 September 1st