NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: Lunar Distance Puzzle
From: UNK
Date: 2011 Aug 21, 21:15 +0000
From: UNK
Date: 2011 Aug 21, 21:15 +0000
Hi Geoffrey,
I agree largely with Frank's analysis, particularly where he agrees with me!
To see why the problem cannot be solved without refraction:
For the sake of argument let the moon be placed at geocentric RA 0, Dec 0 at a suitable moment. This will also be its topocentric coordinates if I am standing on the equator right under it. A pole, ca 400 km long, is suspended from the center of the moon. I grab it and run along with it always making sure that I keep it pointed towards that same point in the sky where I saw it first, i.e. RA 0 / Dec 0. Basically this means that I always have to keep the pole parallel to itself while running and I have to reduplicate the course of the moon at my end of the pole.
The moon moves through space at 1km/sec. Two minutes later, I will have moved 120 km towards east. I won't have to run all that fast since earth rotation takes care of half of that (0.5 km/sec). My geographical position is now 60 km east of where I started (and a little off the equator to compensate for any change in the moon's Dec. The geocentric RA of the moon has changed to ca. 0 deg 1', while my topocentric RA is still 0. There is no way of distinguishing between the cases at the start and the end of the run or anywhere in-between.
It does not matter for the thought experiment how the topocentric place of the moon is observed. Two distances are plenty. A third one leads to all the problems that arise from over-determination. (For instance, that a local minimum searching algorithm may run straight into the nearest local minimum instead of spitting out the entire (infinite) set of all solutions that are good within given accuracy limits.)
As to the question whether there is a general solution to the original problem with three refracted distances:
Given that refraction in altitude R(h) changes monotonously with h over the interval from 0 to 90, and assuming r(h) <> 0 for h <> 90, maybe there is an (extremely theoretical) solution. But refraction across the celestial sphere is by no means monotonous and the analysis is complex. It remains to be shown that there is a number n (obviously larger than 2) so that ambiguities can always be resolved through n distance observations (of n pairs of celestial bodies).
If I were to think about this, I would start out by looking into what can be learned about sidereal time from the refracted distances of 2, then 3, then 4 stars alone. What would their optimum configuration be? I would not convolute the issue by bringing the moon into the game too early.
Herbert
On 2011-08-21 04:42, Geoffrey Kolbe wrote:
I agree largely with Frank's analysis, particularly where he agrees with me!
To see why the problem cannot be solved without refraction:
For the sake of argument let the moon be placed at geocentric RA 0, Dec 0 at a suitable moment. This will also be its topocentric coordinates if I am standing on the equator right under it. A pole, ca 400 km long, is suspended from the center of the moon. I grab it and run along with it always making sure that I keep it pointed towards that same point in the sky where I saw it first, i.e. RA 0 / Dec 0. Basically this means that I always have to keep the pole parallel to itself while running and I have to reduplicate the course of the moon at my end of the pole.
The moon moves through space at 1km/sec. Two minutes later, I will have moved 120 km towards east. I won't have to run all that fast since earth rotation takes care of half of that (0.5 km/sec). My geographical position is now 60 km east of where I started (and a little off the equator to compensate for any change in the moon's Dec. The geocentric RA of the moon has changed to ca. 0 deg 1', while my topocentric RA is still 0. There is no way of distinguishing between the cases at the start and the end of the run or anywhere in-between.
It does not matter for the thought experiment how the topocentric place of the moon is observed. Two distances are plenty. A third one leads to all the problems that arise from over-determination. (For instance, that a local minimum searching algorithm may run straight into the nearest local minimum instead of spitting out the entire (infinite) set of all solutions that are good within given accuracy limits.)
As to the question whether there is a general solution to the original problem with three refracted distances:
Given that refraction in altitude R(h) changes monotonously with h over the interval from 0 to 90, and assuming r(h) <> 0 for h <> 90, maybe there is an (extremely theoretical) solution. But refraction across the celestial sphere is by no means monotonous and the analysis is complex. It remains to be shown that there is a number n (obviously larger than 2) so that ambiguities can always be resolved through n distance observations (of n pairs of celestial bodies).
If I were to think about this, I would start out by looking into what can be learned about sidereal time from the refracted distances of 2, then 3, then 4 stars alone. What would their optimum configuration be? I would not convolute the issue by bringing the moon into the game too early.
Herbert
On 2011-08-21 04:42, Geoffrey Kolbe wrote:
I have not been following this thread too closely, but I do have a question. In his original challenge, Dave asked,
"Given three lunar distances (only, no altitudes), can one find time and position?"
From the ensuing dialogue about refraction, is the consensus now that the answer is "no", since any refraction model used must have some information about altitudes?
Geoffrey
