John
Make a simple drawing and you will find that the answer for the sphere is
dl = dlong * R * cos( lat ) (*)
if dlong is in radians then dl is in whatever units are used for R. Radians
are fine in formulae, it is a bit harder to measure them directly (BTW,
does anyone have protractor scaled in radians?). The practical formula
known for centuries in navigation is
dl[NM] = dlong['] * cos( lat ) (**)
That is express dlong in minutes of arc and you will get the answer in
Nautical Miles. 1 Nautical Mile is 1 minute of arc of the meridian, so it
is really useful as opposed to another mile. Advocating SI nearly anywhere
I would hate to use kilometers (and radians or graduses) in navigation but
if you insist on km then 1 NM = 1.852 km. dl expressed in NM is called the
departure. Gordon cited the exact formula for an ellipsoid.
You are close at arriving at Mercator formula for his famous projection.
Slawek
At 11:15 AM 8/20/99 -0700, John Carmichael wrote:
>Hello dialists:
>
>I've had another basic question that has bothered me for a while. In fact,
>it is such an elementary question that I'm almost embarassed to ask it, but
>as they say, no question is a dumb, just the person who asks it! And I know
>none of you are judgemental.
>
>I am trying to come up with a simple formula for determining the length of a
>degree of longitude at a specific latitude. I wish someone of you would
>check my reasoning and my math to see if my conclusions are correct.
>
>I am basing my calculations on the premise (true or false?) that the length
>of a degree of longitude is directly proportional to the latitude because
>the length of a degree of longitude is maximum at the equator (90* or 111.32
>km.) and zero at the poles. (to obtain 111.32 km. as the length of one
>degree longitude at the equator, I divided the circumference of the earth by
>360 degrees: 111.32 km./degree=40,074.16 km./360*)
>
>For example, if 45* latitude (north or south) is one half the distance from
>the pole to the equator, then the length of one degree of longitude at 45*
>latitude would be equal to one half the length of a degree of longitude at
>the equator. This is 111.32 km./ 2 = 55.66 km. Is this correct?
>
>Using similar logic, if I wanted to know the length of one degree of
>longitude at latitude 32* 13' 18", first I need to convert this latitude to
>decimals. This would be: 32.2216*.
>Then I determine the percentage of the distance this latitude is from the
>pole to the equator. This is: (90*-32.2216*)/90*=.64198 or 64.2%. 64.2% of
>111.32 km. = 71.46 km.
>Is this correct?
>
>32* 13' 18" is the latitude for Tucson Arizona as given by Fer in his
>sundial program in his list of world cities and their coordinates. I
>wondered to which part of the city these coordinates refered. Was it City
>Hall, The main plaza, the geographical center or what? (Fer, if you are
>reading this would you let us know?)
>
>If one degree of longitude at this latitude is 71.46 km. then one minute of
>longitude is 71.46*/60'/deg.=1.19 km, and one second of longitude is
>1.198km./60"/min.=.01985km.=19.85 meters. Correct?
>
>The length of a degree of latitude is the same everywhere on earth. It would
>equal the circumference of the earth divided by 360. This is 111.32 km., the
>same length as a degree of longitude at the equator. This makes the length
>of a minute of latitude=111.32 km./60'/deg.=1.855 km., and one second of
>latitude is 1.855 km./60"/min.= .03092 km.=30.92 meters.
>
>If this is so, then Fer's coordinates are accurate to 30 meters of latitude
>and 20 meters of longitude at Tucson's latitude. Am I correct?
>
>This leads me to ask if this degree of precision and accuracy even
>necessary. I doubt that it would even be possible to build a sundial to
>these exacting requirements. When dialists label their sundials with
>latitude and longitudes accurate to the second, is this not presumtious, or
>are they just giving the precise coordinates of the place in which the
>sundial is located?
>
>I think this is an important concept for all dialists to understand,
>especially the beginners. You might have delt with this before I joined the
>list, but I'm sure most of us who are newcomers would appreciate a replay of
>the thread.
>
>Thanks so much,
>John Carmichael
>http://www.azstarnet.com./~pappas
>
>
Slawek Grzechnik
32 57.4'N 117 08.8'W
http://home.san.rr.com/slawek
