There is a difference between the length of the solar and sidereal days
due to the orbital motion of the earth. Since the sun is at the center
of our orbit the position of the sun with respect to the center of the
earth appears as a constant. We see the sun move only because of the
rotation of the earth and not because of the earth's orbital motion.
But, when we're trying to track the stars, we see the stars move across
the sky from two causes. The major movement is the rotation of the
earth but there is a second smaller movement due to the movement of the
earth in its orbit. Because the earth is shifting its position in orbit
you'll see the same star field you saw last night reach the same
position again almost 4 minutes sooner than the 24 hours of the solar day.
The difference is inconsequential to your attempt to make star trails.
However, it's not inconsequential if you're trying to track the stars
with a guided high magnification telescope since the magnification of
the scope amplifies the movement. If the "clockwork" driving the scope
is turning the scope at the solar rate you'll soon find that the star
you're tracking is slowly drifting out of the field of view.
Chuck Norcutt
On 5/8/2011 11:53 PM, Fernando Gonzalez Gentile wrote:
> 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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