From: Raleigh McLemore (raleighmclemore@yahoo.com)
Date: Thu Jan 31 2002 - 14:53:11 PST
Message-ID: <20020131225311.38891.qmail@web13005.mail.yahoo.com> Date: Thu, 31 Jan 2002 14:53:11 -0800 (PST) From: Raleigh McLemore <raleighmclemore@yahoo.com> Subject: Re: South Pole Longitude & Compass Question...
Thanks to everyone for the through answers. I suppose,
working with elementary school students, that I will
attempt to interpret this so my students can make
sens,e of it. I took some string and wrapped it around
different latitudes, cut it and then stretched it out.
After measuring it the students see that the distance
around the latitude decreases as you move towards 90.
I just thought of this, I can draw marks at 25% of the
length of each string and have my students measure the
distance and try to find the ratio of change.
Anyway, I appreciate you taking time to explain this
to me. By coincidence I'm taking a "GPS/GIS" class at
the local community college and my student's question
made me think of how to explain GIS to students.
I have to remind myself that the mathematics of
latitude and longitude are human constructs. The
younger students, when confronted with precise lines
draw on a globe, are sometimes wondering it buildings
all taper in towards the poles because the "lines say
so".
Thanks again,
With firm handshake,
Raleigh
--- Ronald Wong <ronwong@inreach.com> wrote:
> 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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