Subject:
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Re: Math/Optics Problem
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Newsgroups:
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lugnet.off-topic.geek
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Date:
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Tue, 14 Aug 2001 21:26:47 GMT
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Viewed:
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112 times
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It's amazing how much physics and math I have forgotten, so let me simply
note that the focus of the camera isn't going to change the proportions of
the objects. The focal length will. The perspective will change as your
lens measurement changes (in 35mm photogrphy sizes, a 20mm lens would be a
wide angle lens with a pronounced shrinkage of background objects, and a
300mm telephoto lens would have a considerably less shrinkage effect).
Using a "zoom" lens allows you to change the focal length of the lens
dynamically and thus the proportions of the closer and further objects.
Bruce
(extremely rusty, so feel free to correct me)
In lugnet.off-topic.geek, David Eaton writes:
> In lugnet.off-topic.geek, Shiri Dori writes:
> > > #1 Will the fisheye constant be consistant for the same camera?
> >
> > Depends. Does it "focus"? The "fisheye" constant you're referring to
> > is, in technical terms, called the focus.
>
> Really? I woulda thought it had more to do with the zoom rather than the
> focus. But then again, I'm really not sure how the lenses in cameras work. I
> guess my assumption (based on nothing at all I think) is:
>
> {A}--------B--------(C)--D--(E)--F--[G]
>
> Wherein {A} is the object, (C) and (E) are the lenses of the camera, and [G]
> is the film/image projection surface. B, D and F are merely the distances
> inbetween.
>
> I had kind of assumed that to affect zoom level, F was modified-- resulting
> in a different fisheye "constant". Meanwhile, D was modified to affect
> "focus"-- lining up the image "crisply" matching the focal points of the
> lenses. Although maybe I have that backwards? Heck, maybe I'm just
> completely and thoroughly wrong about the layout... I dunno.
>
> > It is constant for any individual
> > lens and depends on the curvature radii of the two surfaces, as well as the
> > coefficient of the material you're using (n = speed of light in the
> > first media/speed in 2nd). The equation is:
> >
> > 1/f = (1 - n)*(1/R1 - 1/R2) with R1 and R2 as the radii.
>
> So... pardon my asking, but is f the fisheye constant in the above equasion?
>
> > > #2 If not, will it at least be constant for the same zoom level?
> >
> > No, just for the same "focus". If you "focus" the camera it will be
> > constant.
>
> Hmm... so... wait... when re-focused with a new zoom level, the fisheye
> constant will be the same? Or that if the focus is not altered but the zoom
> is (I.E. the new picture is out of focus, but zoomed in/out) that the
> fisheye constant will be the same?
>
> > > #3 are there any other "missing pieces" to the formula?
> >
> > hmm, well, you've got the magnification of the images. M = -q/p = H'/H ,
> > where q=distance of the image to the lens and p=distance of the object
> > to the lens, H is the length of the object, H' = length of the image. The
> > neg. sign is not relevant here, since the image you get on the camera *is*
> > inverted.
> > ...
> > q is the same for both - it is the distance between the lens of the camera
> > to the film.
>
> This makes muchos sense here-- I guess I was simply associating it with zoom
> level instead of focus? I guess the reason I was hung up on it being NOT the
> focus is that I can envision an out of focus vs. in-focus version having the
> same actual proportions in the image-- they're just harder to determine... I
> dunno...
>
> > A cute little pic for ya:
> > http://www.geocities.com/shiri_writing/optics.jpg
>
> Alright, I nearly started laughing aloud when suddenly saying to myself
> "pee-bob" and "pee-house". Call me infantile.
>
> > > #5 can I use that same formula to find out how far away from the
> > > *camera* the objects are?
> >
> > That *is* the formula above.
>
> Yeah, I was mostly unsure... I thought perhaps there was a formula for
> figuring out relative distance which might not involve distance from the
> camera. So despite the fact that I was thinking that the solution was BASED
> on the difference between camera distances, I wasn't sure enough to think
> that the two might not be unrelated. Gosh, I sure fill up the negatives in
> that sentence, eh? Anyway, yeah, I just wasn't sure in the event that I was
> completely off base.
>
> DaveE
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Message has 1 Reply:
 | | Re: Math/Optics Problem
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| (...) Ahhh-- so perhaps I've been getting stuck on photography terminology. I had assumed the "focus" in Shiri's response to relate to the fuzziness of the resulting image, not the "focal length" which affects the zoom level. That makes just about (...) (23 years ago, 15-Aug-01, to lugnet.off-topic.geek)
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Message is in Reply To:
| | Re: Math/Optics Problem
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| (...) Really? I woulda thought it had more to do with the zoom rather than the focus. But then again, I'm really not sure how the lenses in cameras work. I guess my assumption (based on nothing at all I think) is: {A}--------B--------...E)--F--[G] (...) (23 years ago, 14-Aug-01, to lugnet.off-topic.geek)
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