[OSM-dev] Request for PHP programmers for TWISST project

andrzej zaborowski balrogg at gmail.com
Fri Oct 1 17:12:56 BST 2010


Hi,

On 30 September 2010 20:27, Nick Austin <nick.w.austin at gmail.com> wrote:
> I'm a user of TWISST, a free service run by volunteers that sends
> tweets notifying the best viewing times for overhead passes of the
> International Space Station. Earlier today they put out a request for
> help via their web site:
> http://twisst.nl/timezones
>
> In summary they need volunteers to help program in LAMP (Mysql 5.1, PHP 5.2).
>
> If your too lazy to follow the link here's the first few paragraphs
> from the web site:
>
> -----
> "To send people correct ISS-alerts, we need to know in which timezone
> they are. We need a script to tell us that."
>
> "To find out their timezone, we send peoples coordinates to the
> Geonames server. Geonames is an awesome service. However, sometimes
> Geonames is swamped with requests and then we don't get timezones (we
> could opt for a more stable paid subscription, but we don have any
> money to invest here). New followers then are left with UTC as their
> default timezone, which leads to confusion."
>
> "Also, sending requests to Geonames all the time causes a lot of data
> traffic. We would like to cut on our diet of data traffic."
>
> "So, we would like to figure out ourselves which timezone to use for
> certain coordinates. From what we gather, this is not impossible. It
> takes a map with the timezones of the world as polygons and a script
> to find out in which polygon a set of coordinates lies.:"
> ---
>
> It occurs to me that because OSM uses GeoNames that this would improve
> the chances of OSM requests succeeding, so everyone benefits.  For
> further details see the link above.
>
> Once again, I am just a user of their service, I have no connection at
> all with TWISST.

Checking if a point is inside a polygons is trivial, but is it what
they really want?  Wouldn't just the longitude tell you the time
better than the timezone?

E.g. offset_from_gmt = ((lon - 180) / 15) hours

Cheers



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