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FOV and focal reducer

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#1 Guest_**DONOTDELETE**_*

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Posted 18 February 2004 - 10:43 AM

On an 8" SCT, how much will the focal reducer (F/6.3) increase the FOV? With the standard EP (25mm Plossl), Celestron claims a FOV of .64*. Is there a calculation for this in general? Thanks.

#2 Dennis

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Posted 18 February 2004 - 11:06 AM

Currently, you have a TFOV of about .615 degrees. With the focal reducer, you have about .97 degrees.

TFOV = AFOV/mag

There are some great definitions and information in the Astronomy FAQ thread in the "Off Topic" category.

Enjoy!

#3 jmoore

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Posted 18 February 2004 - 11:57 AM

the f6.3 means, I think, that you'll get 63% of the scope's unaltered focal ratio. So, if f10, you get f6.3. If your scope was f12, you get 63% of 12 = about f7.6.

From the new focal ratio, you can calculate your new effective focal length. Instead of 2000mm (assuming 8", f10 scope), your focal length is now 1260mm (63% of 2000).

Put in your 25mm Plossl, and mag = 1260/25 = 50x. Like Dennis said, take AFOV/mag = 52 deg/50x. You get 1.04. A little different than Dennis said, but we're close.

1 degree is A LOT more than .64 degrees; you'll see about 2.5x more area of sky.

#4 Guest_**DONOTDELETE**_*

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Posted 18 February 2004 - 02:35 PM

So if I get the Scopetronix MaxView 40 (40mm, AFoV = 44) in combination with the focal reducer, my resulting TFoV would be 1.4*?

1260mm/40mm = 31.5x mag

44*/31.5 = 1.4*

Also, would a 2x barlow effectively give the 25mm EP without the focal reducer a TFoV of 0.32* due to doubling the focal length? Just trying to wrap my head around all this. Thanks guys.

#5 jmoore

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Posted 18 February 2004 - 02:46 PM

you got it.


#6 Dennis

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Posted 18 February 2004 - 03:29 PM

As a newbie, I would like to correct any misunderstanding I may have. Here's my assumptions:
8" = 2032mm (8 x 25.4)
F/6.3 would yield 1280mm
Plossl has 50* AFOV.
Jeff, your calculations are based on 2000mm and a 52* AFOV. Are my numbers wrong? Please understand, I don't mean to split hairs. I just want to learn this info properly. Thanks!

#7 Suk Lee

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Posted 18 February 2004 - 04:18 PM

Guys, you're leaving out the effect of the maximum light cone you can get into an eyepiece/reducer. i.e. the calculations are correct but they assume you've got an unlimited light cone to work with.

Once you use up all the light coming into a 2" eyepiece, you can't get any wider TFOV (True FOV). You'll just get a vignetted image, like looking into a tube.

There's a really good table of formulas at:

http://www.televue.c...page.asp?ID=107

TFOV can be calculated as:

TFOV in degrees = eyepiece / focal length x 57.3

Televue lists the largest 2" field stop as being 46mm (basically the entire eyepiece barrel), so:

TFOV = 46 / (8 x 25.4 x 10) x 57.3 = 1.3 degrees

The Celestron focal is slightly larger than a 2" drawtube, so if you measure the telescope side ID and plug into the above formula you'll get the maximum TFOV through the reducer.

For example, a 50mm Plossl has the same 46mm field stop as a 41mm Panoptic, so they would both give a TFOV of 1.3 degrees, despite the usual TFOV/AFOV calculations.

The TFOV/AFOV calculations can be used with impunity once the field stop is smaller than the maximum light cone.

Suk

#8 Dennis

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Posted 18 February 2004 - 04:40 PM

Thanks for the explaination, SUK. And that's why they call it a field stop. :foreheadslap:

#9 jmoore

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Posted 18 February 2004 - 04:56 PM

Dennis...you got me right. I was just assuming 2000mm and 52-deg EP.

Suk...thanks for clearing all that up. Yes, I knew I was ignoring the issues you mentioned.

jeff

#10 Guest_**DONOTDELETE**_*

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Posted 18 February 2004 - 05:10 PM

Sweet, thanks.


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