From: Ronald Wong (ronwong@inreach.com)
Date: Tue Jan 29 2002 - 15:52:12 PST
Message-Id: <l03102802b87c242b5190@[209.209.18.141]> Date: Tue, 29 Jan 2002 15:52:12 -0800 From: Ronald Wong <ronwong@inreach.com> Subject: Re: South Pole Longitude & Compass Question...
Among other things, Raleigh asked:
>>...
>>How wide is a longitude degree near the pole?
>>...
To which Al Sefl replied:
>...
>Your degree question is one that would require spherical trigonometry and I
>am not up to that tonight. Let's see if we can figure something out that
>will be in the ballpark... Grabbing a calculator to find the sin of 89 degrees
>for a rounded figure of 0.01745... That would be 1.16 mile. I don't know
>if this is reasonable or correct.
The sin of 90 degrees is one, so the sin of 89 degrees should be pretty
close to 1.00000 - NOT 0.01745... So either Al hit the wrong button or he
has the wrong trig function in mind. I suspect it's the latter.
In celestial navigation, the earth is considered to be a sphere and, for
most navigational purposes, a minute of latitude is a nautical mile
(approx. 6076 ft.), Navigators have a simple, graphical method for
producing small area mercator charts for use in the immediate area that
they are sailing in. The charts produced aren't true Mercator projections
but, for practical purposes, they are just fine.
In this method, the length of the longitude is proportional to the cosine
of the latitude (i.e. longitude length = latitude length X cosine of
latitude). So at 89° latitude, 1 degree of longitude = (1 degree of
latitude) X (cosine of 89 degrees) in length. This makes it approximately
0.01745 degrees of latitude in length. Since a degree of latitude has 60
minutes and each minute is a nautical mile, multiplying 0.01745 by 60 gives
you the length of 1 degree of longitude at 89 degrees of latitude in
nautical miles. This works out to be 1.047.. nm which is equal to 1.205
statute miles.
Solving the problem by using spherical trigonometry leads to an answer of
1.044.. nm or 1.201 statute miles. The graphical method of map making
differs from this by only 0.3%. So we seem to have very good agreement.
The problem with using spherical trigonometry is that the answer is an arc
along a great circle when in fact the one degree of longitude is along a
parallel of latitude - which is not a great circle. Since the great circle
distance is the shortest distance between two points on a sphere, the
actual degree of longitude must be GREATER.
Here's a third method that is probably closer to the truth: If you believe
that the earth spins about it's axis, then one of the consequences is that
it is flattened at the poles. If we are one degree from the pole - a
distance of 60 nm - then we can think of the earth as being flat. The
problem then reduces itself to a problem in plane geometry where the
parallel of latitude is simply a circle centered on the pole and the task
is simply one of finding the length of a segment of arc subtending one
degree of a circle with a radius of 60 nm (giving an answer for a latitude
that is 1 degree [60 nm] from the pole). It's just the old S = R X (theta)
where S = "length of longitude" (in nm), R = 60 nm, and theta = one degree
(in radians). This yields a value of 1.047..nm (Pay attention mariners: Now
you know why "An error of 1 degree will set you off 1 nautical mile in a
run of 60 nautical miles") which is 1.205 statute miles - very good
agreement with the map maker.
If you want the length of 1 degree of longitude at points closer to the
pole, just reduce the length of the radius. Everything else remains the
same.
If we assume that Al really meant the cosine when he invoked the sine, then
Al Sefl's "back of the envelope" solution, 1.16 miles, is right on the
money. His results are only good to 2 significant figures given the numbers
he chose to use. So 1.16 miles = 1.2 miles which, to two significant
figures, equals 1.205 miles.
So now we have, to two significant figures, four correct solutions to this
problem.
Actually, the earth's equator is closer to 25 000 miles than the 24 000
miles that Al used in his solution. Divide 25 000 miles by 360 degrees and
multiply by the cosine of 89 degrees (that's all you have to do using his
technique. I don't know why he went through the trouble of introducing time
into the solution) and you get 1.21 miles. This result looks even better
then his earlier one but again the answer is actually only 1.2 miles given
the significant figures involved. Still a winner, but no better than Al's
original solution which goes to show that there is nothing wrong in using
24 000 miles - an easier number to remember (the earth goes around once
every 24 hrs and the earth is xx 000 miles around) - for a "back of the
envelope" solution to a problem.
Looks like everyone's a winner.
I better quit while we are ahead.
ron
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