Yes, that's it !, the problem posed in a more logic way.
I will save this answer so as not to ask it for a third time ...
Didn't know about the sidereal day, shall investigate this issue further.
Thanks, Chuck !
Fernando.
On Sun, May 8, 2011 at 11:35 PM, Chuck Norcutt
<chucknorcutt@xxxxxxxxxxxxxxxx> wrote:
> I assume the question you're trying to ask is: How do you calculate the
> exposure time to make a star trail of a given length in degrees?
>
> Since the earth rotates 360 degrees in 24 hours a star will leave a
> trail of 360/24 = 15 degrees per hour or 15/60 = 1/4 degree per minute.
>
> To be perfectly precise the 24 hour day only pertains to the sun. When
> talking about tracking the stars across the sky one uses what's called
> the "sidereal" day which at 23 hours, 56 minutes and a few seconds is a
> bit shorter than the solar day. But for your purposes the length of the
> normal solar day will be fine. You're not talking about building a
> stellar tracking device. :-)
>
> Chuck Norcutt
>
>
> On 5/8/2011 9:10 PM, Fernando Gonzalez Gentile wrote:
>> Sorry, forgot the technical data: full moon as only light source,
>> Zuiko 28mm ƒ/2,8 @ ƒ/5,6, tripod mounted plain Olympus OM 2, +2/3
>> exposure compensation, KM.
>>
>> Another question comes again to my mind (asked this before, some 3
>> years ago ... and forgot the answer): how does one calculate the
>> exposure, from the angle of the arc of the star trails?
>> I'm not _that_ smart in Maths ... ;-)
>>
>> Fernando.
>>
>> On Sun, May 8, 2011 at 9:42 PM, Fernando Gonzalez Gentile
>> <fgonzalezgentile@xxxxxxxxx> wrote:
>>
>>> BTW - what do you List, think of this KM, exposed on April 1990?
> --
--
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