I cannot comment on the accuracy of the model used as I was not involved in that implementation. I can only confirm that NASA and the US Naval observatory were not the sources used. This does not mean the results are wrong, just that it uses different algorithms. There are many different algorithms for such computations and some are more accurate than others, may be slower than others, and are more/less accurate over longer date ranges. I believe most of the code used in AstronomicalData/SunPosition came from Meeus and/or VSOP87 and are listed in the sources used by AstronomicalData:

http://reference.wolfram.com/language/note/AstronomicalDataSourceInformation.html

As for atmospheric refraction, as documented, it is not taken into account. Its possible this could be made available in a future release, but its not currently available. Wikipedia and other sources mention several correction factors that can be used. You should be able to apply one of those of your choosing to modify the results of SunPosition in the mean time. As with all such algorithms, they also vary in accuracy as modeling the dynamic effects of the atmosphere is highly non-trivial and location/temperature/pressure dependent. http://en.wikipedia.org/wiki/Atmospheric_refraction

If you are looking for a way to make AstronomicalData/.SunPosition match NASA/Naval Observatory, I doubt that is possible without a total reimplementation to use whatever algorithm they use and there are no current plans to do that.

Are you sure the output is ok? I've refreshed with RebuildPacletData[]

If http : // ssd.jpl.nasa.gov/horizons.cgi?s_loc = 1 #top is the gold standard Then for the sample point I looked at {43.953074,4.472485} and at 2014-07-22 17:15:00 UTC AstronomicalData["Sun",.... does not agree with SunPosition[GeoPosition and both are different from the nasa calculation e.g. for the altitude for this point respectively I get 19.85892, 20.06311334010335, and 19.7609

Unless I've done some thing stupid I can't see that 0.3 degrees out is much of a success

copied from mathematica 10 notebook

In[26]:= {AstronomicalData["Sun",{"Azimuth",{2014,7,22,18,15,0},{43.953074,4.472485}},TimeZone->1], AstronomicalData["Sun",{"Altitude",{2014,7,22,18,15,0},{43.953074,4.472485}},TimeZone->1]} Out[26]= {279.28408[Degree],19.85892[Degree]} In[27]:= SunPosition[GeoPosition[{43.953074,4.472485}], DateObject[{2014,07,22,19,15, 0}, TimeZone -> IntegerPart[LocalTimeZone[Entity["City",{"Paris","IleDeFrance","France"}], DateObject[{2014,07,22,19,15, 0},TimeZone-> 0]]]] ]//FullForm Out[27]//FullForm= List[Quantity[279.096879879350734008354638938468698465083.6876323489782243,"AngularDegrees"],Quantity[20.063113340103351466195182502734913517392.5442756941572635,"AngularDegrees"]]

Result of values in the jpl.nasa site 279.3737 19.7609

I suppose I'm still not clear why the two different methods in Mathematica (AstronomicalData and SunPostion) don't agree. If I look at the Azimuth for which refraction presumably doesn't make much difference then I get

AstronomicalData 279.28408
SunPosition 279.09687
and NASA 279.3737

which there's roughly 0.1 degrees between AstronomicalData and NASA and 0.2 degrees between AstronomicalData and SunPosition

=> What would you consider to be the gold standard?

I'm interested as I was wondering how accurate one could plausibly make a sundial for for example the next 100 years and thought it would be fairly easy to set up on Mathematica, but am now not sure. the timezone correction is 4 minutes a degree, i.e. 0.1 degrees is 24 seconds 0.1 degrees is something like 3.5 mm at 2 metres

I spoke too soon. It hasn't pushed to the production server just yet. Should hopefully be on production tomorrow. Sorry if I caused additional confusion.

I have found that, in addition to SunPosition returning the wrong result, DaylightQ and SiderealTime also produce the wrong result. SiderealTime code is almost certainly called by the SunPosition code, so this might be a good place to start looking.

Users here get the same results I do. I'm investigating for any issues I can think of that might cause such behavior.

I checked by setting my time zone to Champaign's

Block[{$TimeZone = -5}, 
  SunPosition[GeoPosition[{52.37`, 4.89`}], DateObject[{2013, 3, 1, #, 0, 0}, TimeZone -> 1]] & /@   Range[0, 23]
]

I now get your results.

So, you get the correct results because part of the calculation somehow assumes your time zone, which yields incorrect results for anyone not in your time zone. You are among the few who can't see the incorrect results ;-)

I can't answer the other issues since I do not use these functions. But on this point you raise

It needs an Internet connection

How would propose that Mathematica get the current ephemeris data for solar system planets and the sun without using an updated database somewhere on the internet?

I use JPL Horizon website myself for this information. You might want to see if there is a way to use this site via an API call to obtain this information. It is free.

I converted to Mathematica a VBS macro that you can find in this Excel sheet. It is written by Greg Pelletier and based on a NOAA Javascript program which was used to drive their sun calculation page. I performed a direct and ugly conversion without any optimization whatsoever. Testing it with the method above I found that it is about 30 times faster than V9 AstronomicalData and about 1100 times faster than V10's SunPosition.

Testing the output and taking the NASA website you refer to as the golden standard I get for the sun positions starting at 1 am (using AstronomicalData for V10):

Azimuth:

Elevation:

I declare winner the one with the smallest absolute deviation from the NASA data.

As to the need for ephemeris: The following plot compares the results for Pelletier's and V10's AstronomicalData in the year 3013, thousand years in the future (CircleDot is Mathematica):