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Old Jan 29, 2003, 9:27 AM   #1
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Default Hyperfocal length

I found the following on an online photo tutorial site:

“If you want to work out the hyperfocal distance for your lens(es) you can use the equation above - although for digital cameras you will need to know the 135 format "equivalent" focal length (check the camera or lens's manual).”

My Olympus C-4000 owner manual states the following:

“Olympus lens, 6.5mm to 19.5mm, f2.8, 8 elements in 6 groups (equivalent to 32 mm to 96mm lens on a 35mm camera).”

Is this the information I need to determine the hyperfocal length?

If so, is the 32mm without zoom and the 96mm with full zoom? Or vice versa?

Thanks – Ted.
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Old Jan 29, 2003, 12:48 PM   #2
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thats right. Your camera has a wide angle end at 32mm and a telephoto at 96mm. Estimating the HD for those will allow you to use it at those settings. Given the masive depth of field in these cameras, if you are a little off you will be fine.
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Old Jan 29, 2003, 7:37 PM   #3
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Default Re: Hyperfocal length

Quote:
Originally Posted by tedamenta
found the following on an online photo tutorial site:
...
Where did you find that tutoria? I'm pretty sure the Hyperfocal Distance is determined by the actual focal length and f/stop, not the equiv focal length.
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Old Jan 29, 2003, 8:07 PM   #4
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BillDrew is correct.

http://www.wrotniak.net/photo/dof/c3000.html

Has precomputed tables of DOF (in meters) for Oly C-3000, C-3020, and C-4000 cameras as a function of aperature, subject distance and focal length.

This site has some interesting tidbits to offer, you might want to check it out.
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Old Jan 29, 2003, 10:29 PM   #5
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Thank you both! I copied this information from:

http://www.panoguide.com/technique/hyperfocal.html
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Old Jan 30, 2003, 12:04 AM   #6
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Default Follow up question

At

http://www.wrotniak.net/photo/dof/index.html

there is a tutorial about depth of field which seems closely related to the tutorial at

http://www.panoguide.com/technique/hyperfocal.html

on hyperfocal distance.

I am having trouble understanting how that equation of

d = s/ [1 ± ac(s-f)/f˛]

relates to the equation of

h = f˛/ac

Any advise you can offer would be greatly appreciated.
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Old Jan 30, 2003, 12:01 PM   #7
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I suggest using Google's advanced (http://www.google.com/advanced_search?hl=en)search with the exact phrase "circle of confusion" and such things as photograpy, photo, calculation, ... in the "with all of the words" box to narrow the search.

Once you do some reading about the "circle of confusion", you will realize that the hyperfocal distance is not as solid a number as it might seem. You will also find more details on how it is calculated.
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Old Jan 30, 2003, 7:36 PM   #8
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Here is a link to a downloadable, free calculater that will give you depth of field, hyperfocal distances and many other as pects of a lens. I find it easy to use and interesting.

http://tangentsoft.net/fcalc/
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