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From: Bill (no email)
Date: Wed Oct 20 2004 - 15:13:12 EDT
>> In the worst case scenario, what would be the difference between the
>> averaged Ho and an Ho from from an observation taken at the average time
>> used?
> The following table shows the maximum difference near meridian transit between
> the arithmetic mean of two sun altitudes separated by n minutes, and the
actual altitude
> at half time:
>
> Alt 4m 5m 6m
> 60 deg 0'14" 0'21" 0'31"
> 75 deg 0'30" 0'46" 1'06"
> 80 deg 0'45" 1'10" 1'40"
> 85 deg 1'30" 2'20" 3'21"
Found Herbert's theoretical figures a wee-bit high (on averaging error due
to treating a non-linear function as a linear function) to pass my
common-sense test.
But in the words my former Indiana Congressman Earl Landgrebe on Richard
Nixon's problems, "My mind is made up. Don't confuse me with the facts :-)
Forced to consider the facts by Herbert, computed Hc at LAN for the Sun at
various latitudes on the dates of equinox and solstices, and took their
2-minute differences. NOTE: I did not go as far as picking the exact
longitude for the GMT they occurred, but rather used 000 longitude.
I still find his numbers too high.
Following are the differences between Hc at 2 minutes before LAN and Hc at
the exact meridian passage.
N Lat 22/06/04 22/09/04 21/12/04
00 .3' 25.8' .3'
10 .5' .7' .3'
20 1.9' .3' .2'
23.5 23.2' .2' .2'
30 .9' .2' .2'
40 .3' .1' .1'
50 .2' .0' .1'
60 .1' .0' .1'
70 .1' .0' ----
80 .0' .0' ----
90 .0' .1' ----
Example average for N 20 with HC approx.86.5d, difference of 1.9':
22 June, 2004, N 20, Hc dif 1.9'
11:59:50 86d 31.7'
12:00:50 86d 33.1'
12:01:50 86d 33.6'
12:02:50 86d 33.1'
12:03:50 86d 31.7'
Average:
Averaged 12:01:50 Hc 86d 33.6'
Averaged 12:01:50 Hc 86d 32.6'
Shift 1.0'
Not confining the exercise to meridian passage, we could expect altitude
differences up to 30' in 2 minutes or 1d in 4 minutes.
22 September, 2004 Lat 00, Lon 000, 17:09-17:13 GMT
Time Hc
17:09:00 010d 52.6'
17:10:00 010d 37.6'
17:11:00 010d 22.6'
17:12:00 010d 07.6'
17:13:00 009d 56.6'
4-mintute difference: 59.8'
Actual 17:11:00 Hc 010d 22.6'
Average 17:11:00 Hc 010d 23.4'
Shift from averaging 000d 00.8'
What did this beginner learn from the exercise?
* Alex and Herbert are correct,the error is more pronounced for high
altitude bodies, and they are unacceptable targets for averaging except for
all but last-ditch efforts, especially at meridian passage; even if
double-second-differences and problems with the operator holding the sextant
vertical are discounted.
* Shy away from averaging a body with a declination within 10d-20d of the
observer's latitude. (If it would have trouble casting a significant shadow
of the mast on the deck, leave it alone. This is cel nav SOP.
* Herbert could have conjured up a more extreme LAN "worst-case scenario"
had he chosen. O declination on the equator at an equinox, or N 23d 26.4'd
at the summer solstice, but did not. His answer was, however, not within
the confines of the stated hypothetical question--5 sights averaged--Ha
within altitude 30d-70d.
Having attempted to establish a benchmark, I leave it to the statisticians
to elucidate me.
Bill
>
> I mention the worst case scenario just for fun: If the sun goes through the
> zenith....
>
> The following table shows the maximum difference near meridian transit between
> the arithmetic mean of two sun altitudes separated by n minutes, and the
actual altitude
> at half time:
>
> Alt 4m 5m 6m
> 60 deg 0'14" 0'21" 0'31"
> 75 deg 0'30" 0'46" 1'06"
> 80 deg 0'45" 1'10" 1'40"
> 85 deg 1'30" 2'20" 3'21"
>
> You see that in most 'normal' situations there is not much of an error, but it
> becomes worse RAPIDLY with increasing altitude as well as with increasing time
> span.
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