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Jan,

The average error might be the mean of the absolute value of the errors
while the standard deviation would be the square root of the mean of
the squared errors.  Multiplication by 2.5 to get the 95% range would
be for the standard deviation, not the average error.

I wonder why measuring both east and west would yield such a sharp
reduction in the error.  Is it simply a matter of increasing the number
of observations, or are refraction errors being compensated?

Fred

On Dec 24, 2003, at 7:57 AM, Jan Kalivoda wrote:

> Fred,
>
> Maybe you are right.
>
> I have the copy of a German paper from 1889 that deals with errors of
> common astronavigational methods. For lunar distances it gives
> following average errors, gained from observations at sea. Sets of
> observations consisted from five shots in every case, the error was
> obtained by comparing the results with chronometer times interpolated
> for every day at the end of the voyage. (Which is much more accurate
> than working with the assumed rate during the voyage, of course.) In
> these days, vernier sextants were used.
>
> The average errors were:
>
> 31" for one distance of the Moon and a star or planet
> 22" for one distance of the Moon and the Sun
> 9" for the average result from two distances measured eastward and
> westward from the Moon
>
> You should multiplicate these numbers by (approximately) 2.5 to obtain
> the error of 95% of observations and by 3.5 for 99% of observations,
> isn' it?
>
> It should be noted that after cca 1880, the errors of lunar tables
> were negligible compared with the errors of measurements, which wasn't
> the case before. In cca 1830, the errors would have been two times
> greater, owing to the unreliability of the contemporary theory of the
> lunar motion.
>
>
> Jan Kalivoda
>
>
> ----- Original Message -----
> From: "Fred Hebard" 
> To: 
> Sent: Wednesday, December 24, 2003 2:42 AM
> Subject: Re: Lunars: Jupiter's BIG.
>
>
> Frank,
>
> Good point.
>
> In answer to various claims about the accuracy of lunars, it would seem
> to me that the error in lunars should approach the precision of the
> sextant, given enough measurements of decent quality and decent
> reduction procedures.  That would be 0.1 to 0.2' of arc, or 12-24
> seconds.
>
> Fred
>
> On Dec 23, 2003, at 5:40 PM, Frank Reed wrote:
>
>> Fred you wrote:
>> "I believe that to _rate_ a chronometer one needs at least three
>> lunars spread over at least three days. "
>>
>> Just one lunar will do. When you leave port, you know your
>> chronometer's error (assuming it's a port with a well-established
>> longitude). Let's suppose it's sixty seconds slow as you depart. After
>> five months at sea, you get some measure of your longitude. This could
>> be from speaking another ship, from visiting a port, OR from shooting
>> a lunar. Suppose your chronometer now appears to be 4 minutes fast.
>> That means it's gaining 1 minute per month. That's the rate.
>>
>> On the other hand, there is some benefit in doing lunars for a few
>> days in a row. Primarily, this would assure you that you didn't get an
>> accidental close match from your first trial. You could also get the
>> same result by having several navigators shoot lunars each working up
>> an independent longitude. This was certainly the case on at least some
>> American ships at mid-century.
>>
>> Frank E. Reed
>> [X] Mystic, Connecticut
>> [ ] Chicago, Illinois
>
>
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