On Thu, 2005-03-31 at 23:03 +0200, Oddgeir Kvien wrote:
> Still looking at the calculations and especially the radius of the
> earth.
> By using the formula's in gpsdrive I end up with a radius at 63.5 deg
> north of 6386km, which is larger than the radius at equator which
> are 6378km.
>
> Is this really correct ?
>
>
> GPSDrive uses the following formula:
>
> /*
> the radius of curvature of an ellipsoidal Earth in the plane of the
> meridian is given by
>
> R' = a * (1 - e^2) / (1 - e^2 * (sin(lat))^2)^(3/2)
>
> where a is the equatorial radius,
> b is the polar radius, and
> e is the eccentricity of the ellipsoid = sqrt(1 - b^2/a^2)
>
> a = 6378 km (3963 mi) Equatorial radius (surface to center distance)
> b = 6356.752 km (3950 mi) Polar radius (surface to center distance)
> e = 0.081082 Eccentricity
> */
>
I believe I found an error in the GPSDrive formula's ?
According to http://earth-info.nga.mil/GandG/tr8350.2-a/Chapter%203.pdf
e should be 0.081819, not 0.081082
If you use a = 6378.136km and b = 6356.751km you get e = 0.08182
I will have a close look at this, probably during the weekend.....
--
Oddgeir Kvien <oddgeir-KpobboTMYq3QoR8q8O4SbQ@xxxxxxxxxxxxxxxx>
--
This message was part of the gpsdrive mailinglist
unsubscribing can be done by sending a mail containing a body of:
-quote--
unsubscribe gpsdrive
-unquote--
to majordomo-ahljRfkWE0EwtzFM8z8+AfP6llvjuJOh@xxxxxxxxxxxxxxxx
|
