NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: Lunars - Even Easier
From: Dan Allen
Date: 2008 Jul 6, 22:44 -0600
From: Dan Allen
Date: 2008 Jul 6, 22:44 -0600
Frank, I am impressed. Great advances simplify things and this is quite a simplification.
-- Dan
--~--~---------~--~----~------------~-------~--~----~
Navigation List archive: www.fer3.com/arc
To post, email NavList@fer3.com
To , email NavList-@fer3.com
-~----------~----~----~----~------~----~------~--~---
-- Dan
On Wed, Jul 2, 2008 at 5:24 PM, <frankreed@historicalatlas.net> wrote:
The LD was 85. The Sun's zenith distance was 45. So if we force the Moon's
George, you wrote:
"I don't see the basis on which he decided to replace an observed zenith
distance for the Moon of 38 degrees by one of 40 degrees. Where did that
value of 40 come from?"
zenith distance to be 40, then the sight is perfectly aligned through the
zenith because that's the only case where the sum of the zenith distances
would be equal to the angle between the objects. And when that happens, the
triangle is degenerate so we don't need any trig to clear the sight which is
of course the goal of all this. Sounds crazy, right? It works because the
altitude of the Moon really doesn't matter much at all under some
circumstances, so we can introduce an "error" with very little downside
which converts the problem into a simple case.
Let's do a realistic example. Let's take the lunar observation we've all
been talking about (Jeremy's observation on June 10) and move the observer
to another location. Instead of being at 15º 14'N DR latitude, we move him
to 25º 14'N. But all of the other setup conditions remain the same. We keep
the DR longitude, temperature/pressure, date and time of observation exactly
the same. That way we don't have to look up a lot of new almanac data. You
can see that the Moon and Sun from that location would no longer
be as nicely aligned. In fact, at that time, their difference in azimuth
would amount to 155º --a good distance away from being aligned in opposite
azimuths, and clearly out of line even to a casual observer.
From the shifted DR, our observer takes these sights at 06:23:00 GMT:
Sun LL 35º 38'
Moon UL 56º 14'
LD Near 85º 40.3'
If you clear this lunar observation, you will find that it is exactly
correct for that time and location. I've set it up that way. Run it through
the lunar distance calculator at www.HistoricalAtlas.com/lunars, and you
will get error=0.0'.
Now let's see if we can adjust this observation and turn it into a simple
lunar with no trig required. We need the observed altitudes of the objects
centers above the true horizon, and we need the observed center-to-center
lunar distance (this is the normal "pre-clearing" step):
Sun LL: 35º38' -10'+16' = 35º 44'
Moon UL: 56º14' -10'-16' = 55º 48'
LD Near: 85º40.3' +15.8' +15.7' = 86º 11.8'
And now we add these up. The total is 177º 43.8'. It doesn't total 180º
because the objects are not aligned in opposite azimuths. And HERE is where
we apply the trick. If we raise the Moon's observed altitude by 2º 16.2'
then, of course, the total WOULD add up to 180º, and as far as the math is
concerned, this means they're now in opposite azimuths. So let's do that...
We work the same lunar observation again, but this time with a Moon UL
altitude of 58º 30.2'. If you do it by any of the standard spherical
triangle approaches, you will find that this modified observation has an
azimuth difference of very nearly 180º. And when we clear this modified
observation, the results are almost exactly the same. The error this time
around is 0.1'. But the important point is that we don't need to use any
spherical trig to solve a degenerate triangle. It reduces to a very simple
case of addition and/or subtraction.
There is one calculation we need to do. We need to make sure that it's
legitimate to shift the Moon's altitude by more than two degrees (legitimate
in the sense that the error introduced is within tolerable limits --the
exact limits of what is "tolerable" depends on the end-user). So we
calculate (6')*tan(LD)/cos(Moon_alt). In this case, this gives 173', nearly
three degrees, so modifying the Moon's altitude should not introduce an
error larger than a tenth of a minute of arc, and sure enough, that's what
we have already found.
Imagine if they had known about this 225 years ago. Back then, a somewhat
larger error in clearing might have been counted as "tolerable". A really
large number of lunar observations could have been reduced to simple cases
of addition or subtraction. The calculational work would have taken five
minutes at most...
Oh well. Can't change history!
-FER
--~--~---------~--~----~------------~-------~--~----~
Navigation List archive: www.fer3.com/arc
To post, email NavList@fer3.com
To , email NavList-@fer3.com
-~----------~----~----~----~------~----~------~--~---