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Re: number problem [message #61232 is a reply to message #61198] Thu, 10 July 2008 12:17 Go to previous messageGo to previous message
R.G. Stockwell is currently offline  R.G. Stockwell
Messages: 363
Registered: July 1999
Senior Member
<pgrigis@gmail.com> wrote in message
news:c8f5bb5a-7b15-4abd-bf13-1587add65abe@j22g2000hsf.google groups.com...
> R.G. Stockwell wrote:
>> <d.poreh@gmail.com> wrote in message
>> news:43fbf367-1b18-473e-a047-3ce39612f806@x35g2000hsb.google groups.com...
>> .... snipped ...
>>
>>> yes that was the problem!!1
>>> it is works properly.but for lat-lon data as you can see it is not:
>>> 499690.96 3387795.6
>>
>>> i need more details like this
>>> 499690.95879779 3387795.57157002
>>
>> WHOA WHOA WHOA WHOA!!
>>
>> While we are being pleasant and thinking about what we are doing,
>> let's think about what it means when you say you need 8 digits of
>> lat and lon. (hint, think in millimeters)
>>
>>
>> Granted this is somewhat beside the point of how to read data, but if
>> anyone
>> ever reviews a lat or a lon with more than 2 decimal points, they will
>> flag
>> it.
>
> On the other hand, google maps will pinpoint
> the location of my office at
>
> 42.381009N, 71.128014W
>
> whereas that would be a bit off if it only
> had 2 decimals... ;-)
>
> Ciao,
> Paolo

True, 2 decimals places is about 1km (roughly). But 71.128014W
implies a precision of about 10 cm. That is smaller than the window.
Geophysical data - that is large enough to use lat and lon, is quite
often not taken on a resolution of cms.

Incidentally, three decimal places works just fine.
42.381N, 71.128W (100 m resolution)

I used latitude with minutes and seconds in my phd defense, noting the
position of
an instrument. The examiner called me on it. Luckily I had used
extremely detailed plots of the land to determine the lat and lon,
and it did have an accuracy down to 10 meters. :)


Cheers,
bob
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