John Ferman wrote:
> Calculating the distance between two points on a sphere is based on
> one of the solutions of the spherical right triangle. One side of the
> triangle is the difference of latitudes - call that side the
> horizontal base and name it "a." The vertical side, "b", is the
> difference of longitudes. The angle between the "a" and "b" sides is
> 90º and the side opposite is "c," the hypotenuse. The relation, cos(a)
> = cos(a)cos(b). As the angles "a" and "b" are known the angle "c" is
> known. Knowing the earth's radius enables the length of the arc
> between the two points, S = Rc with c expressed in radians. In general
> one second of angular arc is about one nautical mile, it is written.
That should be: "one MINUTE of angular arc is about one nautical mile," but
it's just an approximation since the Earth isn't quite spherical.
Different values for the length of a nautical mile have been used at
various times and in various country's navies, as the size and shape of the
Earth wasn't well known in early times. It has since been fixed as the
"International nautical mile" at 1852 meters or 6076.10333 feet.
If you're going to do the s = Rc calculation based on a spherical Earth,
you should use the MEAN radius of the Earth for better accuracy. For
distance in kilometers, use 6371 for the mean radius of the Earth; for
statue miles, use 3959; for nautical miles, use 3440.
> On Dec 16, 2007, at 4:39 AM, GarminGPSMAP60C_60CS@yahoogroups.com wrote:
>>2a. Distance between 2 points
>> Posted by: "jmconley52" jmconley52@... jmconley52
>> Date: Sat Dec 15, 2007 7:51 am ((PST))
>>On my old Garmin unit I could easily have the GPS calculate the
>>distance between 2 saved waypoints, regardless of the location of the
>>GPS receiver. For instance, the unit would calculate the distance
>>between point A and B, while the receiver was at point C.
>>My new GPSmap 60CSx doesn't seem to offer that option, or should I say
>>I haven't found it yet? Can someone help me?
>>And a second question. I'm in Iraq, and find that the unit calculates
>>a distance of 6174 miles from here to my home. Does the receiver
>>account for the earth's curvature over long distances, or is the
>>measurement given in a straight line from point to point?