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Re: [OM] Star trail times [was New Kodak Portra 400]

Subject: Re: [OM] Star trail times [was New Kodak Portra 400]
From: Fernando Gonzalez Gentile <fgonzalezgentile@xxxxxxxxx>
Date: Mon, 9 May 2011 08:55:33 -0300
Great! thanks once again Moose.

Now I know of a practical method to do the measurements !

Fernando.

On Mon, May 9, 2011 at 1:54 AM, Moose <olymoose@xxxxxxxxx> wrote:
> On 5/8/2011 8:53 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 !
>
> I almost posted what Chuck did earlier, but was too lazy at the time to take 
> the next step.
>
> 1. The horizontal AOV of a 28 mm lens is 65.5° Allow ±5% for sample variation.
>
> 2. Then you need to measure the length of the track. Without the film at 
> hand, I measured in PS. I had to assume that
> the presented image was full frame and that the actual camera and scanner 
> masks were, in fact, 24x36 mm, for the above
> angle to be correct.
>
> 3. I measured in pixels, as units don't matter for the calculation. The image 
> height I got is 1013. The bright trail
> upper left-center, interpolated into the foliage came out to be 88 and the 
> dimmer, shorter ones on the left, about 53
> pixels.
>
> 4.  Zo.
>     88/1013 = 8.7%
>     8.7% of 65.5° = 5.7°
>     5.7° times 4 min/° ~= 23 minutes for the long trails.
>     About 14 minutes for the short ones.
>
> A. Calculating Moose
>
>> 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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