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Main
Date: 25 Jul 2007 16:14:37
From: actuary@mchsi.com
Subject: Celestron 8 question
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To: All
I have a vintage 1980 Cestron 8. I'm not sure as to whether my
question is accurately stated but here goes. What is the field of
view at the focal plane of the telescope (Prior to magnification by
eyepiece)?
Larry
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Date: 25 Jul 2007 17:31:37
From: actuary@mchsi.com
Subject: Re: Celestron 8 question
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On Jul 25, 7:26 pm, wsnel...@hotmail.com wrote:
> On Jul 25, 8:10 pm, "actu...@mchsi.com" <actu...@mchsi.com> wrote:
>
>
>
>
>
> > On Jul 25, 7:02 pm, wsnel...@hotmail.com wrote:
>
> > > On Jul 25, 7:41 pm, "actu...@mchsi.com" <actu...@mchsi.com> wrote:
>
> > > > On Jul 25, 6:37 pm, wsnel...@hotmail.com wrote:
>
> > > > > On Jul 25, 7:14 pm, "actu...@mchsi.com" <actu...@mchsi.com> wrote:
>
> > > > > > To: All
>
> > > > > > I have a vintage 1980 Cestron 8. I'm not sure as to whether my
> > > > > > question is accurately stated but here goes. What is the field of
> > > > > > view at the focal plane of the telescope (Prior to magnification by
> > > > > > eyepiece)?
>
> > > > > > Larry
>
> > > > > Assuming an 80-inch focal length, approximately 80 divided by 114.6
> > > > > or about 0.7 degrees per inch, more or less.
>
> > > > Thanks but could explain why "divided by 114.6".
>
> > > 360 degrees divided by Pi (3.14159) If you used 35mm film, the long
> > > dimension of the frame would cover somewhat less than 1 degree. The
> > > maximum field possible would depend roughly on how large the
> > > perforation in the primary is.- Hide quoted text -
>
> > > - Show quoted text -
>
> > Thanks. The reason that I asked was that I found an equation similar
> > to the one you provided excpet that it used "division by 57", about
> > 1/2 of 114.6.
>
> You can also try:
>
> Width of field (degrees) = 57.3 x Size of Film / Focal Length of Lens
>
> and
>
> Size of Object on film = Width of Object x Focal Length / 57.3- Hide quoted text -
>
> - Show quoted text -
Thanks. The formula for the width if in degrees provides the answer
that I'm looking for.
Larry
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Date: 25 Jul 2007 17:26:54
From:
Subject: Re: Celestron 8 question
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On Jul 25, 8:10 pm, "actu...@mchsi.com" <actu...@mchsi.com > wrote:
> On Jul 25, 7:02 pm, wsnel...@hotmail.com wrote:
>
>
>
> > On Jul 25, 7:41 pm, "actu...@mchsi.com" <actu...@mchsi.com> wrote:
>
> > > On Jul 25, 6:37 pm, wsnel...@hotmail.com wrote:
>
> > > > On Jul 25, 7:14 pm, "actu...@mchsi.com" <actu...@mchsi.com> wrote:
>
> > > > > To: All
>
> > > > > I have a vintage 1980 Cestron 8. I'm not sure as to whether my
> > > > > question is accurately stated but here goes. What is the field of
> > > > > view at the focal plane of the telescope (Prior to magnification by
> > > > > eyepiece)?
>
> > > > > Larry
>
> > > > Assuming an 80-inch focal length, approximately 80 divided by 114.6
> > > > or about 0.7 degrees per inch, more or less.
>
> > > Thanks but could explain why "divided by 114.6".
>
> > 360 degrees divided by Pi (3.14159) If you used 35mm film, the long
> > dimension of the frame would cover somewhat less than 1 degree. The
> > maximum field possible would depend roughly on how large the
> > perforation in the primary is.- Hide quoted text -
>
> > - Show quoted text -
>
> Thanks. The reason that I asked was that I found an equation similar
> to the one you provided excpet that it used "division by 57", about
> 1/2 of 114.6.
>
You can also try:
Width of field (degrees) = 57.3 x Size of Film / Focal Length of Lens
and
Size of Object on film = Width of Object x Focal Length / 57.3
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Date: 25 Jul 2007 17:10:28
From: actuary@mchsi.com
Subject: Re: Celestron 8 question
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On Jul 25, 7:02 pm, wsnel...@hotmail.com wrote:
> On Jul 25, 7:41 pm, "actu...@mchsi.com" <actu...@mchsi.com> wrote:
>
>
>
>
>
> > On Jul 25, 6:37 pm, wsnel...@hotmail.com wrote:
>
> > > On Jul 25, 7:14 pm, "actu...@mchsi.com" <actu...@mchsi.com> wrote:
>
> > > > To: All
>
> > > > I have a vintage 1980 Cestron 8. I'm not sure as to whether my
> > > > question is accurately stated but here goes. What is the field of
> > > > view at the focal plane of the telescope (Prior to magnification by
> > > > eyepiece)?
>
> > > > Larry
>
> > > Assuming an 80-inch focal length, approximately 80 divided by 114.6
> > > or about 0.7 degrees per inch, more or less.
>
> > Thanks but could explain why "divided by 114.6".
>
> 360 degrees divided by Pi (3.14159) If you used 35mm film, the long
> dimension of the frame would cover somewhat less than 1 degree. The
> maximum field possible would depend roughly on how large the
> perforation in the primary is.- Hide quoted text -
>
> - Show quoted text -
Thanks. The reason that I asked was that I found an equation similar
to the one you provided excpet that it used "division by 57", about
1/2 of 114.6.
Larry
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Date: 25 Jul 2007 17:02:29
From:
Subject: Re: Celestron 8 question
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On Jul 25, 7:41 pm, "actu...@mchsi.com" <actu...@mchsi.com > wrote:
> On Jul 25, 6:37 pm, wsnel...@hotmail.com wrote:
>
> > On Jul 25, 7:14 pm, "actu...@mchsi.com" <actu...@mchsi.com> wrote:
>
> > > To: All
>
> > > I have a vintage 1980 Cestron 8. I'm not sure as to whether my
> > > question is accurately stated but here goes. What is the field of
> > > view at the focal plane of the telescope (Prior to magnification by
> > > eyepiece)?
>
> > > Larry
>
> > Assuming an 80-inch focal length, approximately 80 divided by 114.6
> > or about 0.7 degrees per inch, more or less.
>
> Thanks but could explain why "divided by 114.6".
>
360 degrees divided by Pi (3.14159) If you used 35mm film, the long
dimension of the frame would cover somewhat less than 1 degree. The
maximum field possible would depend roughly on how large the
perforation in the primary is.
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Date: 25 Jul 2007 16:41:05
From: actuary@mchsi.com
Subject: Re: Celestron 8 question
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On Jul 25, 6:37 pm, wsnel...@hotmail.com wrote:
> On Jul 25, 7:14 pm, "actu...@mchsi.com" <actu...@mchsi.com> wrote:
>
> > To: All
>
> > I have a vintage 1980 Cestron 8. I'm not sure as to whether my
> > question is accurately stated but here goes. What is the field of
> > view at the focal plane of the telescope (Prior to magnification by
> > eyepiece)?
>
> > Larry
>
> Assuming an 80-inch focal length, approximately 80 divided by 114.6
> or about 0.7 degrees per inch, more or less.
Thanks but could explain why "divided by 114.6".
Larry
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Date: 25 Jul 2007 16:37:22
From:
Subject: Re: Celestron 8 question
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On Jul 25, 7:14 pm, "actu...@mchsi.com" <actu...@mchsi.com > wrote:
> To: All
>
> I have a vintage 1980 Cestron 8. I'm not sure as to whether my
> question is accurately stated but here goes. What is the field of
> view at the focal plane of the telescope (Prior to magnification by
> eyepiece)?
>
> Larry
Assuming an 80-inch focal length, approximately 80 divided by 114.6
or about 0.7 degrees per inch, more or less.
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