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This is a follow-up to Geoffrey's posting, in which he wrote-

| you can calculate how an altitude will vary
| with time and so you now have a line with the correct slope to fit to your
| data. This is an additional piece of information that simple averaging or
| statistical fitting by method of least squares, say, does not give you.
| Having got your line with the correct slope, you can draw a line parallel
| to this through your data points for a best fit. Yes, no doubt about it,
| this is superior to simple averaging as an extra piece of information is
| being included that was not there before.

We can make a little test to see what this method can do towards 
"eliminating" random scatter in the original observations.

Not having any real data at our disposal, let's take a fictitious star, 
that's rising in altitude at a certain known rate, which we take to be 6 
arc-minutes in each minute of time. Tha rate is chosen quite arbitrarily, 
but is somewhere within the range that's expected.

And let's assume that at a certain arbitrary time T, which we take as our 
time- zero, that star has a certain precise altitude A, which is known (to 
me) but not disclosed (to you). Altogether 9 observations are taken of that 
star altitude, at precisely 1-minute intervals, from T-4 minutes to T+4 
minutes. So, for example, if A happened to be 45�, then the altitudes would 
vary from 44� 36' to 45� 24'. And finally, just to mess things up, I will 
add a computed scatter to that altitude, with a standard deviation of 2 
arc-minutes, and round the result to the nearest tenth.

And now, you are asked to discover the initial value of A, from those 9 
altitudes.

We know that if the altitude had been unchanging, at a meridian passage, 
then the best value would be obtained by averaging the 9, and then on 
average, the scatter in the final result should be 2 arc-minutes divided by 
root 9, or 2/3 arc-minutes.

So the challenge here is to analyse those 9 given altitudes, simulating a 
rising star, in any way you think is best, to get closer than 2/3 of a 
minute to the initial value of A, which I will disclose later. So I will 
provide a set of 9 altitudes below, taken in order from T-4 to T+4, and I 
ask for your best estimate of the altitude at time T.

And to make sure that the result isn't a statistical freak, here are four 
such exercises altogether, with differing initial altitudes, chosen at 
random, but all with the same rate of rise and with similar scatter. And I 
promise that I have done nothing to choose or check or even pre-analyse the 
scattered numbers listed below; they are strictly as they came out of the 
bag.

First:
34� 16.7', 34� 21.0, 34� 25.4, 34� 32.0, 34� 38.6, 34� 46.3, 34� 49.9, 34� 
58.3, 34� 58.8.

Second.
76� 29.7, 76� 39.2, 76� 42.8, 76� 48.2, 76� 57.8, 77� 03.1, 77� 05.9, 77� 
12.8, 77� 19.5.

Third.
42� 47.3, 42� 53.3, 42� 54.9, 43� 05.8, 43� 12.4, 43� 16.9, 43� 24.4, 43� 
29.2, 43� 37.9.

Fourth.

11� 18.0 , 11� 25.9, 11� 30.8, 11� 39.8, 11� 42.7, 11� 44.1, 11� 51.3, 11� 
58.6, 12� 05.3.

George.

contact George Huxtable at george@huxtable.u-net.com
or at +44 1865 820222 (from UK, 01865 820222)
or at 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.



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