Understood.
Reminds me of Johannes Kepler. I used to be very good in Astronomy,
but I was fifteen then, and followed another trails.
Fernando.
On Mon, May 9, 2011 at 8:13 AM, Chuck Norcutt
<chucknorcutt@xxxxxxxxxxxxxxxx> wrote:
> 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.
--
_________________________________________________________________
Options: http://lists.thomasclausen.net/mailman/listinfo/olympus
Archives: http://lists.thomasclausen.net/mailman/private/olympus/
Themed Olympus Photo Exhibition: http://www.tope.nl/
|