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 19:31:55 GMT
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Viewed:
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107 times
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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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| 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 (...) (23 years ago, 14-Aug-01, to lugnet.off-topic.geek)
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Message is in Reply To:
| | Re: Math/Optics Problem
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| (...) Alrighty. Since I have been studying this stuff for the past month and am supposed to pass a test about it, let's see what I can do. (...) Depends. Does it "focus"? The "fisheye" constant you're referring to is, in technical terms, called the (...) (23 years ago, 14-Aug-01, to lugnet.off-topic.geek)
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