Finder scope field of view?
GEOFFREY ORTON
Does anybody know or can you tell me the formula to calculate the
field of view of a 5X24 finder scope?
I used to use a Telrad & calculator to get position of Polaris
vis-a-vis the NCP, but a Telrad's too big for my C5+
I'm guessing the view is about 30' to 1 deg...any confirmation?
John Ongtooguk
GEOFFREY ORTON (ge...@aztec.asu.edu) wrote:
: Does anybody know or can you tell me the formula to calculate the
: field of view of a 5X24 finder scope?
In every one that I've looked thru the field of view is
effectively zero. If they actually used an achromatic
objective and an eyepiece with at least a 40 deg apparent
field the field of view would be 40/5 = 8 deg. The
6x30 finders tend to be a little better as the ones that
I've seen use an achromatic objective and what appears
to be a Kellner eyepiece, so assuming an apparent field
of 40 deg the field would be 40/6 = 7 deg. In any case
I'd avoid 5x24 finders as most appear to be heavily
stopped down single lenses.
John Ongtooguk (jo...@vcd.hp.com)
Michael Packer
actual field of view = (Aparent FOV)/EP Power
You are not going to be able to calc. it given above bcause you are not
given the aparent FOV. However you can measure it by the following and
thus be more assured:
Find two stars that just fit in the FOV (and lay on diameter of FOV) and
determine their anglular distance on sky atlas. if you can't do this,
figure out which stars they were and somebody in the sci.astros will tell
you. Or look at say a house during the day measure how many feet of the
house you see at your distance away. The formula for this is
Tan(Actual FOV/2) = 1/2*(ft of house seen)/(dist cent. of house in ft)
it will probably be around 3 degrees +or- 1.
Michael Packer
In article <5hrshb$n...@news.asu.edu>, ge...@aztec.asu.edu (GEOFFREY ORTON)
wrote:
> Does anybody know or can you tell me the formula to calculate the
> field of view of a 5X24 finder scope?
> I used to use a Telrad & calculator to get position of Polaris
> vis-a-vis the NCP, but a Telrad's too big for my C5+
> I'm guessing the view is about 30' to 1 deg...any confirmation?
>
In article <5hrshb$n...@news.asu.edu>, ge...@aztec.asu.edu (GEOFFREY ORTON)
wrote:
> Does anybody know or can you tell me the formula to calculate the
> field of view of a 5X24 finder scope?
C or G Jordan
> In article <5hrshb$n...@news.asu.edu>, ge...@aztec.asu.edu (GEOFFREY ORTON)
> wrote:
>
> > Does anybody know or can you tell me the formula to calculate the
> > field of view of a 5X24 finder scope?
> > I used to use a Telrad & calculator to get position of Polaris
> > vis-a-vis the NCP, but a Telrad's too big for my C5+
> > I'm guessing the view is about 30' to 1 deg...any confirmation?
> >
Time a star tracking across the FOV and calculat the FOV from 15 degrees of movement per
hour. Quick and dirty but it should work.
-Guy
Mike Skinner
But won't the tracking/drift rate across the FOV be dependant on where the
scope is pointed -- ranging from no tracking/drift at Polaris to maximum
tracking/drift along the celestial equator ? It's true that the angular
drift rate at all points in the sky is 15 degrees/hour, but the physical
distance across the sky (and across a finder's FOV) spanned by that 15 degrees
varies with the declination. Therefore, the drift measurement should be done
along the celestial equator, rather than at any other points in the sky.
Is this true?
Rick Hughes
Mike Skinner wrote:
You must multiply by the cosine of the declination.
Cos(equator) = 1
Cos(pole) = 0
FOV = Time (in seconds) X 15 X Cos(declination)
--
Rick Hughes - email/finger: rick dot hughes at pobox dot com
PGP Key 6073EDCD - 74 4A A2 B8 C8 D2 CB DB D5 F4 66 E2 DF 88 10 36
finger me or http://pobox.com/~rick.hughes/ for my public key
Rick Hughes
Left out one important detail:
FOV (in seconds of arc) = Time (in seconds) X 15 X Cos(declination)
Steve Pattie
ski...@aurfb3aur.alcatel.com (Mike Skinner) wrote:
>|> Time a star tracking across the FOV and calculat the FOV from 15 degrees of
>movement per
>|> hour. Quick and dirty but it should work.
>|>
>|> -Guy
>But won't the tracking/drift rate across the FOV be dependant on where the
>scope is pointed -- ranging from no tracking/drift at Polaris to maximum
>tracking/drift along the celestial equator ? It's true that the angular
>drift rate at all points in the sky is 15 degrees/hour, but the physical
>distance across the sky (and across a finder's FOV) spanned by that 15 degrees
>varies with the declination. Therefore, the drift measurement should be done
>along the celestial equator, rather than at any other points in the sky.
>Is this true?
Yes. You must do it at the celestial equator.
Wouldn't it be easier to point it at the north pole (if you don't
have an equatorial mount) and use a star atlas to determine your
FOV. If the mounting is trackable, then point it anywhere and work
out the FOV along the lines of declination by star chart or by
declination slow motion and setting circle.
Steve Pattie
Chris Marriott
In article <Steve.Pattie...@dsto.defence.gov.au>, Steve Pattie
<Steve....@dsto.defence.gov.au> writes
>>But won't the tracking/drift rate across the FOV be dependant on where the
>>scope is pointed -- ranging from no tracking/drift at Polaris to maximum
>>tracking/drift along the celestial equator ? It's true that the angular
>>drift rate at all points in the sky is 15 degrees/hour, but the physical
>>distance across the sky (and across a finder's FOV) spanned by that 15 degrees
>>varies with the declination. Therefore, the drift measurement should be done
>>along the celestial equator, rather than at any other points in the sky.
>>Is this true?
>
>Yes. You must do it at the celestial equator.
It's not necessary to use a star on the equator. Simply divide the
measured time by the cosine of the star's declination.
Chris
----------------------------------------------------------------
Chris Marriott, SkyMap Software, U.K. e-mail: ch...@skymap.com
Creators of fine astronomy software for Windows.
For full details, visit our web site at http://www.skymap.com