NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: sight reduction with GPS receiver
From: Frank Reed CT
Date: 2005 Mar 21, 20:06 EST
From: Frank Reed CT
Date: 2005 Mar 21, 20:06 EST
Herbert you wrote:
"Reflecting the economic reality of modern s/w production for the
end
user market, the small importance of the issue is reflected in the want
of documentation rather than quality of implementation. The manuals that
come with my various Garmins just speak of "distance", never ever
specifying what kind. Simple tests with models 45, 12, 12XL reveal that
they all display distance along the geodesic line on the chosen
reference ellipsoid For consistency, I would expect no less from a unit
promising 15m positioning accuracy."
user market, the small importance of the issue is reflected in the want
of documentation rather than quality of implementation. The manuals that
come with my various Garmins just speak of "distance", never ever
specifying what kind. Simple tests with models 45, 12, 12XL reveal that
they all display distance along the geodesic line on the chosen
reference ellipsoid For consistency, I would expect no less from a unit
promising 15m positioning accuracy."
It truly amazes me that the people making these devices would bother with
this issue. But you're right, it's the "economic reality" that drives the
outcome.
And:
"To achieve the 4 digit accuracy on a Garmin display, an approximative
formula is sufficient. Even a full numerical integration would not be a
big deal for an infrequent computation such as this. I see no reason not
to do it right."
"To achieve the 4 digit accuracy on a Garmin display, an approximative
formula is sufficient. Even a full numerical integration would not be a
big deal for an infrequent computation such as this. I see no reason not
to do it right."
And yet, there is no "right" distance. Great circle distances are only
relevant for purposes of comparison. Corrected ellipsoidal differences wouldn't
seem to offer any practical advantage, but, since the devices have spare memory
and spare computing power, why not "do it right"? Thinking on it more, the only
unambiguous "right" distance is the straight-line distance (through the
Earth).
And:
"Paul Hirose obviously has a unit that displays GCD. I would like to hear
which model it is and whether it does this automatically or has to be
coerced into doing it. I have a faint memory of my first Magellan (as
big as a shoe box) being able to display GCD."
"Paul Hirose obviously has a unit that displays GCD. I would like to hear
which model it is and whether it does this automatically or has to be
coerced into doing it. I have a faint memory of my first Magellan (as
big as a shoe box) being able to display GCD."
I would bet that it depends on the age of the receiver (just
speculating).
Maybe you can help me with something here. Years ago, I worked this out as
puzzle. I posted some results on the maps-l list back in 1997 (ouch... time
flies). A copy follows below. I've never had any confirmation that my
approximate expressions are accurate (or how accurate they are). Here's a
copy:
>>>>>
Step 0) Use e = 1/298 for the flattening of the Earth.
Step 1) Convert the latitudes to co-latitudes and all angles to
radians as usual:
phi1' = pi*(90-lat1)/180
phi2' = pi*(90-lat2)/180
theta = pi*(lon2-lon1)/180
radians as usual:
phi1' = pi*(90-lat1)/180
phi2' = pi*(90-lat2)/180
theta = pi*(lon2-lon1)/180
Step 2) Calculate adjusted colatitudes:
phi1 = phi1'+(3/4)*e*sin(2*phi1')
phi2 = phi2'+(3/4)*e*sin(2*phi2')
phi1 = phi1'+(3/4)*e*sin(2*phi1')
phi2 = phi2'+(3/4)*e*sin(2*phi2')
Step 3) Get the standard great circle angular distance:
a' = arccos[cos(phi1)*cos(phi2)+sin(phi1)*sin(phi2)*cos(theta)]
a' = arccos[cos(phi1)*cos(phi2)+sin(phi1)*sin(phi2)*cos(theta)]
Step 4) Calculate the geodesic distance on the ellipsoid:
a = a'*[1+(1/2)*e*(sin(phi1)*sin(phi2)*sin(theta)/sin(a'))^2]
a = a'*[1+(1/2)*e*(sin(phi1)*sin(phi2)*sin(theta)/sin(a'))^2]
Step 5) Convert to miles
D = 3955.5*a.
<<<<<
D = 3955.5*a.
<<<<<
What do you think?
-FER
42.0N 87.7W, or 41.4N 72.1W.
www.HistoricalAtlas.com/lunars
42.0N 87.7W, or 41.4N 72.1W.
www.HistoricalAtlas.com/lunars