distance computation, sphere or ellipsoid?

When GPS receivers display distance between waypoints, what surface do
they use for the computation? Sphere or ellipsoid? I'm asking because
I have found a way to do celestial navigation spherical trig on a GPS
receiver. This is possible on my Magellan 315 because it uses a
sphere. However, it's a fairly old receiver. Do modern receivers
(consumer grade models) compute leg lengths on the ellipsoid?

To test this, create three waypoints:
P1 at 0 N 0 E
P2 at 40 N 0 E
P3 at 80 N 0 E

Make a route from P1 to P2, then from P2 to P3. If both legs have the
same length, the receiver uses a sphere. If the legs are exactly 2400
nautical miles (they are, on the Magellan 315), the receiver uses 60
miles for the length of a degree.

I would be interested to hear the results of this test with other
receivers.

I realize a GPS receiver is a strange tool for celestial nav sight
reduction! But I'm entertaining myself by thinking up non-obvious ways
to navigate with a receiver even if there are no satellite signals.

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Paul Hirose wrote:

When I checked this previously with my Garmin eMap against some online
sites it agreed with their calculations using the WGS84 ellipsoid.


The eMap showed 2392 nm from P1 to P2 and 2406 nm from P2 to P3.

That's correct to the nearest nm. The unit might be doing an ellipsoid
computation or it might be selecting a radius from a list based on latitude.

Will the eMap output those distances to the nearest meter ?

With 'Geodetic/UTM-Grid Utility' I have P1 to P2 as 4429529 meters and P2 to
P3 as 4455611 meters...


http://www.kbhscape.com/gps.htm

KBH wrote:

Not as far as I know. Setting the units to metric gives the distances
you cite below rounded to the nearest km.

Peter wrote:

John Kabel wrote:


And, as I wrote previously, the Magellan 315 says 2400 miles. Thanks
to both of you for trying the experiment.

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