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Convert latitude and longitude data to WGS83 coordinates?

Dear Community,

I have to convert decimal latitude and longitude coordinates to the WGS84 coordinate system. I tried to do it with the GeoPosition function, but I failed - I guess I misunderstand something in the description. Could you help me out and give a small example how I do it correctly? E.g. latitude = 49.74645, longitude = 8.54444, what is the corresponding coordinate pair in WGS84 coordinates? This point should be somewhere in Germany.

Tx for the kind help in advance,
Best regards. Andras

POSTED BY: Andras Gilicz
4 Replies

This simplifies it indeed. Thank you very much

Andras

POSTED BY: Andras Gilicz

Note that the default datum in WL is ITRF00, which is nearly identical to WGS84. Hence perhaps you just need this conversion:

In[22]:= GeoPositionXYZ[GeoPosition[{49.74645, 8.54444}]]
Out[22]= GeoPositionXYZ[{4.08359*10^6, 613536., 4.84461*10^6}]

which performs the whole computation in the ITRF00 datum.

Jose.

Dear Jose,

Ts for the quick answer, GeoPositionXYZ[{4.0835910^6, 613547., 4.8446210^6}, "WGS84Original"] will do it for me.

best regards Andras

POSTED BY: Andras Gilicz

Hi,

One can convert {lat, lon} coordinates from one datum to another (say WGS84) with GeoPosition, but we need to know the original datum as well. Do you know what's the original datum? For example this is a change from WGS72 to WGS84 (in its original form, because there are several versions):

In[20]:= GeoPosition[GeoPosition[{49.74645, 8.54444}, "WGS72"], "WGS84Original"]
Out[20]= GeoPosition[{49.7465, 8.54459}, "WGS84Original"]

Or do you want, perhaps, to convert {lat, lon} coordinates to a triple {x, y, z} in an Earth-Centered frame? That would be done with GeoPositionXYZ:

In[21]:= GeoPositionXYZ[GeoPosition[{49.74645, 8.54444}, "WGS72"], "WGS84Original"]
Out[21]= GeoPositionXYZ[{4.08359*10^6, 613547., 4.84462*10^6}, "WGS84Original"]

Jose.

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