NavList:

   

Reply

Bill,

Did you ever take a second-year algebra course where you were asked to graph
parabolas and ellipses and such? They used to call this 'analytic geometry'
but I don't think that term is used much anymore. Anyway, there was a
general problem that you would have worked where you take the equation for a
line:
 y = a*x + b
and you add it to the equation for a basic parabola:
 y = k*x^2.
Now, of course, when you just add them up, you get
 y = b + a*x + k*x^2,
but it's more interesting to re-cast this equation in such a way that the
linear term (a*x) goes away. It's called 'completing the square' (does that
ring a bell? do you remember having to do problems where you completed the
square?). What you find is a parabola that is offset from the origin in both
x and y. Maybe I'll this unfinished in case you want to dust off those
algebra skills...

I have a funny story about this. When I was a 15-year-old sophomore in high
school, just six months after my first course in celestial navigation, I was
on my school's 'math team' (a suprisingly un-nerdy group, except for me, of
course). We went to a meet at some neighboring school, and one of the
problems involved watching a stone tossed up into the air. The stone of
course followed a simple parabola (like most math versions of physics
problems, little details like air resistance could be avoided). Meanwhile,
the observer in the problem was watching the tossed stone from an elevator
descending at constant speed. Can you see the resemblance to the
corresponding navigation problem? The tossed stone moves like the Sun around
noon, following a nearly parabolic arc. The elevator moving down is like a
vessel moving towards the Sun's GP. Anyway, the assigned puzzle was to
determine whether the maximum height of the rock would reach some specified
level as seen by the observer on the descending elevator. I think we had
fifteen minutes to work it out. After I had successfully solved the problem,
I over-heard some competitors at a nearby table. They were complaining that
there was no way to do that problem without using calculus (it can ALSO be
solved by differentiation to find the maximum of the curve), but I knew you
could do it by completing the square with no calculus required --and they
should have known that, too. I wanted to stand up and say, "No. You're
wrong. You just complete the square." but I had just enough competitive
instinct at that time to keep my mouth shut. We won that day. :-) All thanks
to adding a line onto a parabola!

Back to the problem of the Sun around noon as seen from a moving vessel.
Although the motion of the vessel does indeed shift the parabola forward or
backward in time relative to actual local noon, I would emphasize that it is
EASY to correct for this. All you have to do is subtract (or add) a few
minutes of arc to each raw sight (I've outlined this several times so I
won't repeat it right now). There are other ways to do it, too. For example,
you could apply the results of that bit of analytic geometry and then use an
equation to determine the offset from maximum altitude back to true Local
Apparent Noon. This is a SOLVABLE problem. There's no fundamental difficulty
determining longitude around noon, although it's important to remember that
the accuracy of the longitude could be about five times worse than a single
latitude sight at noon. That is, if you have a one minute of arc standard
deviation in your sights normally, then expect a about a five nautical mile
standard deviation in the longitude found this way. Sometimes, it's
considerably better.

 -FER



--~--~---------~--~----~------------~-------~--~----~
Navigation List archive: www.fer3.com/arc
To post, email NavList@fer3.com
To , email NavList-@fer3.com
-~----------~----~----~----~------~----~------~--~---

   

Reply