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Determining the Distance Between Two Points on the Earth's Surface



 
 
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  #11  
Old July 5th 07, 10:54 PM posted to sci.astro.amateur
[email protected]
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Posts: 169
Default Determining the Distance Between Two Points on the Earth's Surface

On Jul 4, 3:45 pm, laura halliday wrote:
On Jul 4, 2:23 pm, " wrote:

One of the tech notes for a "gcdist" program I wrote and posted to
sci.math about 20 or so years ago provides insight to the calc; only
online ref I found is he


http://groups.google.co.jp/group/sci...891302914fd84?
dmode=source&hl=ja


Maybe I'm behind the times, but for two places that
close together I'd look in an atlas, use the distance
scale, and have an answer in under 10 seconds.

Even if they weren't on the same map, I could still
read off the coordinates and punch them in to a
calculator.

Am I missing something?


Accuracy. and speed. For all intents and purposes, my gcdist
calculates the distance in less than the time required for the
cursor to zip to the left margin to begin displaying the answer.

:-)

A lot depends on one's needs. I often need both bearing and
distance between 2 points (even) in the same town; one example
of how I used the answers appears in the first 7 lines displayed
here (of ClearSkyClocks flanking me within 10 miles):

http://thadlabs.com/ASTRO/view-bracket.htm

  #12  
Old July 7th 07, 03:53 PM posted to sci.astro.amateur
W. Watson
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Posts: 203
Default Determining the Distance Between Two Points on the Earth's Surface

Not at all, but I can find my bookmarks for such matters much faster--until
they expire. :-)

laura halliday wrote:
On Jul 4, 2:23 pm, " wrote:

One of the tech notes for a "gcdist" program I wrote and posted to
sci.math about 20 or so years ago provides insight to the calc; only
online ref I found is he

http://groups.google.co.jp/group/sci...891302914fd84?
dmode=source&hl=ja


Maybe I'm behind the times, but for two places that
close together I'd look in an atlas, use the distance
scale, and have an answer in under 10 seconds.

Even if they weren't on the same map, I could still
read off the coordinates and punch them in to a
calculator.

Am I missing something?

Laura Halliday VE7LDH "Que les nuages soient notre
Grid: CN89mg pied a terre..."
ICBM: 49 16.05 N 122 56.92 W - Hospital/Shafte


--
Wayne Watson (Nevada City, CA)

Web Page: speckledwithStars.net
  #13  
Old July 7th 07, 07:41 PM posted to sci.astro.amateur
SkySea
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Posts: 131
Default Determining the Distance Between Two Points on the Earth's Surface

If you have a spreadsheet, create a formula to find the distance. For
the cell formula below, it refers to four columns to the left:
longitude1 (in Column A...), latitude1, longitude2, latitude2 (...in
Column D):

=(60*DEGREES(ACOS(SIN(RADIANS(B30))*SIN(RADIANS(D3 0))+COS(RADIANS(B30))*COS(RADIANS(D30))*COS(RADIAN S(A30-C30)))))
The result is given in nautical miles. 1M = 1.852km = 1.150782mi
It's wordy because Excel (Quattro for DOS too, and probably for
Windoze) uses radians.

=============
- Dale Gombert (SkySea at aol.com)
122.38W, 47.58N, W. Seattle, WA
  #14  
Old July 12th 07, 01:40 AM posted to sci.astro.amateur
Stuart Levy
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Posts: 30
Default Determining the Distance Between Two Points on the Earth's Surface

laura halliday wrote:
On Jul 4, 2:23 pm, " wrote:

One of the tech notes for a "gcdist" program I wrote and posted to
sci.math about 20 or so years ago provides insight to the calc; only
online ref I found is he

http://groups.google.co.jp/group/sci...e=source&hl=ja


Maybe I'm behind the times, but for two places that
close together I'd look in an atlas, use the distance
scale, and have an answer in under 10 seconds.

Even if they weren't on the same map, I could still
read off the coordinates and punch them in to a
calculator.

Am I missing something?

Laura Halliday VE7LDH "Que les nuages soient notre
Grid: CN89mg pied a terre..."
ICBM: 49 16.05 N 122 56.92 W - Hospital/Shafte


Oh cool. The haversine version is a really handy formula --
adapting from Thad P Floryan's posting above,

hav Distance = hav (Lon1 - Lon2) * cos Lat1 * cos Lat2 + hav (Lat1 - Lat2)

(it even looks like the Pythagorean formula, c^2 = a^2 + b^2,
with the longitude leg weighted by the cosines of both latitudes!)

where hav A = sin^2 (A/2), so
Distance = 2 * asin( sqrt( sin((Lon1-Lon2)/2)^2 * cos(Lat1) * cos(Lat2) + hav((Lat1-Lat2)/2)^2 ) )

This looks much better for computation than the "acos" version mentioned
in the orig. posting, since for small angles (short distances),
the arccos is very insensitive to its input, so errors will be huge.

Not so for the haversine. Its accuracy should be OK except when
the arcsine (asin) is near 90 degrees, so Distance is near 180 deg,
i.e. it only fails for nearly antipodal points.

Stuart Levy in sunny Champaign, IL
 




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