[Proj] Projection conversion problem
Andre Joost
andre+joost at nurfuerspam.de
Thu Jun 30 04:24:12 PDT 2016
Am 30.06.2016 um 10:20 schrieb Damian Rakowski:
> That looks very interesting, thank you for your reply. One more
> question: 6371000 - this is an Earth radius0.0093202511 - how did you
> got this value? I just want to understand :) Best Regards, Damian
>
In a first round, I substituted the ellipsoid by the radius:
echo 2.2945 48.858222 | cs2cs -v +proj=longlat +R=6371000 +no_defs +to
+proj=eqc +lat_ts=0 +lat_0=0 +lon_0=0 +x_0=0 +y_0=0 +R=6371000 +units=m
+no_defs
255136.76 5432786.41
then divided the result in meters by the expected result. Since it is
the same for both coordinates, it seems to be right.
From John P. Snyders "Map Projections - A Working Manual" p.91, you can
take the exact formulas for the sphere:
x = R (lambda-lambda0) cos phi1
y = R phi
with angles in radians. This is linear to the angles, and you just have
to compare them with your scale factor.
Feel free to test it with other coordinate pairs around the globe.
HTH,
André Joost
> Am 28.06.2016 um 19:40 schrieb Hermann Peifer:
>
>>
>> With the given sample coordinates, this will give:
>>
>> $ echo 2.2945 48.858222 | cs2cs -v +proj=longlat +datum=WGS84
>> +no_defs +to +proj=eqc +lat_ts=0 +lat_0=0 +lon_0=0 +x_0=0 +y_0=0
>> +datum=WGS84 +units=m +no_defs # ---- From Coordinate System ----
>> #Lat/long (Geodetic alias) # # +proj=longlat +datum=WGS84 +no_defs
>> +ellps=WGS84 +towgs84=0,0,0 # ---- To Coordinate System ----
>> #Equidistant Cylindrical (Plate Caree) # Cyl, Sph # lat_ts=[,
>> lat_0=0] # +proj=eqc +lat_ts=0 +lat_0=0 +lon_0=0 +x_0=0 +y_0=0
>> +datum=WGS84 +units=m # +no_defs +ellps=WGS84 +towgs84=0,0,0
>> 255422.57 5438872.39 0.00
>>
>> The expected result was: 27374451 582901293
>>
>
> I suggest to take a sphere, and scale with the units:
>
> +proj=eqc +lat_ts=0 +lat_0=0 +lon_0=0 +x_0=0 +y_0=0 +R=6371000
> +to_meter=0.0093202511 +no_defs
>
> which results to 27374451.23 582901292.36 0.00
>
> HTH, André Joost
>
>
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